Martin Gardnerin the Twenty-First Century

Michael Henle and Brian Hopkins, Editors

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Martin Gardner enormously expanded the field of

recreational mathematics with the Mathematical Games

columns he wrote for Scientific American for over 25 years

and the more than 70 books he published. He also had

a long relationship with the Mathematical Association

of America, publishing articles in the MAA journals

right up to his death in 2010. This book collects articles

Gardner wrote for the MAA in the twenty-first century,

together with other articles the MAA published from

1999 to 2012 that spring from and comment on his work.

Martin Gardnerin the Twenty-First Century

Michael Henle and Brian Hopkins, Editors

9 780883 859131

ISBN: 978-0-88386-913-1

v1 vh vi vj vn

Martin G

ardner in the Twenty-First Century

Michael Henle and

Brian Hopkins, Editors

MAA-Gardner cover v7POD_Layout 1 11/19/12 4:10 PM Page 1

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Martin Gardner

in the

Twenty-First Century

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c 2012 by the Mathematical Association of America, Inc.

Library of Congress Catalog Card Number 2012954077

Print edition ISBN: 978-0-88385-913-1

Electronic edition ISBN: 978-1-61444-801-3

Printed in the United States of America

Current Printing (last digit):

10 9 8 7 6 5 4 3 2 1

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Martin Gardner

in the

Twenty-First Century

edited by

Michael Henle

Oberlin College

and

Brian Hopkins

Saint Peter’s University

Published and Distributed by

The Mathematical Association of America

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Preface

Martin Gardner and the MAA share a long history. In 1958, around the time he started his

famous “Mathematical Games” column for Scientific American, he submitted the first of

many problems to The American Mathematical Monthly. In 1982, as his column wound

down, Gardner’s first MAA article was published in The Two Year College Mathematics

Journal. He wrote for MAA journals the rest of his life, particularly The College Mathe-

matics Journal and Math Horizons. Gardner contributed to the latter almost annually from

its founding in 1993 until 2005.

Gardner’s prodigious writing activity continued right until his death in 2010. Articles,

stories, problems, solutions, Quickies, and other kinds of contributions continued to flow.

His last mathematical article to appear in an MAA journal, “L-tromino Tiling of Mutilated

Chessboards,” was the centerpiece of a special puzzle issue of The College Mathematics

Journal in 2009, and it is included here.

Early in 2010, Math Horizons editors Steve Abbott and Bruce Torrence were surprised

to receive a typescript manuscript. Gardner used a typewriter his whole life, never email.

The submission was accompanied by a note, “Is this short story something you can use? I

wrote the math column in Scientific American for 25 years. If my piece is not right for Math

Horizons, there is no need to send it back. All best, Martin.” There was not enough time

for the editors to thank Martin for his submission [6]. Fittingly, this story, “Superstrings

and Thelma,” is the last piece in this collection.

Apart from his own work, Martin Gardner, by enormously expanding the field of recre-

ational mathematics, opened up vast mathematical tracts for exploration by others. This

was quite deliberate. In an interview in The Two-Year College Mathematics Journal [1],

Gardner said, “I’m defining [recreational mathematics] in the very broad sense to include

anything that has a spirit of play about it.” Gardner, of course, had a refined and very

well-developed sense of play, one quality that made his pieces so enjoyable to read. In

almost everything he wrote, Gardner posed problems to challenge his readers, and they

responded. He maintained an extensive correspondence with mathematicians, both profes-

sional and amateur. Their work fueled his own pieces, but then his correspondents turned

around and wrote their own articles.

One consequence was the expansion of recreational mathematics into a major research

area (also helped by the development of the computer and the corresponding expanded in-

terest in discrete mathematics) that is such a feature of the current mathematical landscape.

The CMJ devoted the January 2012 issue to papers on topics that Gardner introduced to the

mathematical public. There were so many articles to include that half of the March 2012

issue continued the theme.

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vi Preface

Another consequence of the flowering of recreational mathematics is this volume. We

have collected MAA journal articles starting from 1999 on topics that Gardner developed.

Some are written by Gardner, but most are by others. The tribute articles from the January

and March 2012 CMJ issues are all here, but they constitute less than half of this collection.

All the MAA journals are represented, Mathematics Magazine and the Monthly, as well as

CMJ and Math Horizons. The limitation to pieces published roughly in the twenty-first

century is a practical one. Even so, some puzzle collections, longer articles, and pieces less

directly linked to Martin Gardner have been omitted.

The 41 pieces collected here are grouped around common themes, such as geometry,

number theory and graph theory, and cards and probability. Flexagons, the topic of Gard-

ner’s first Scientific American column, are seen to be associated with Catalan numbers and

together merit their own section. Geometric tiling and various “magic” number puzzles

are all about “Making Things Fit,” and there are enough other puzzles and games to fill

another section.

Gardner’s interests ranged far beyond mathematics. A fan of magicians and magic

tricks from childhood (“I waste a lot of time on it” [2]), he wrote several books on magic.

He annotated Lewis Carroll’s Alice books and other classics, and produced two novels of

his own. Other topics included philosophy, religion, literature and pseudo-science, leading

to some 70 books.

The last section of this volume highlights some of these other facets of Martin Gard-

ner’s wide-ranging interests. It includes two short stories by Gardner and several other

pieces that demonstrate his support for amateur mathematicians, his love of play (about an

April Fool’s joke he played on his Scientific American readers), and his interest in debunk-

ing false science. Also included is Gardner’s review of a 2004 novel in which an important

character seems to be based on him.

Our hope is that this volume will play a role in perpetuating the memory of Martin

Gardner, modest celebrity, larger-than-life character, self-confessed amateur as a mathe-

matician, popularizer of recreational mathematics in the broadest sense, prolific and bril-

liant writer. Given the lasting impression he made on several generations of mathematics

enthusiasts of all backgrounds, we are confident that the MAA and others will be publish-

ing articles inspired by Gardner’s work for a long time.

Gardner, like the readers of this book, loved mathematics. We close this preface with

Gardner’s own words on his background, the community, and why he enjoyed the field so

much (from [5], [4], and [3], respectively).

I took no math in college. I’m like a person who loves music but can’t hum

a tune or play an instrument. My understanding of math does not go beyond

a minimal understanding of calculus. I hasten to add that I consider this one

reason for the success of my Scientific American column. I had to work hard

to understand whatever I wrote about, and this made it much easier for me to

write in a way that readers could understand.

When I first started the column, I was not in touch with any mathemati-

cians, and gradually mathematicians who were creative in the field found out

about the column and began corresponding with me. So my most interesting

columns were columns based on material that I got from them, so I owe them

a big debt of gratitude.

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Preface vii

[I enjoy mathematics] because it has a strange kind of unearthly beauty.

There is a strong feeling of pleasure, hard to describe, in thinking through an

elegant proof, and even greater pleasure in discovering a proof not previously

known.

Aclnowledgments

The editors are grateful for the encouragement and hard work of the MAA publications

staff: Director of Publications Ivars Peterson, who conceived the book, Production Man-

ager Carol Baxter, who designed it, and especially Electronic Publication Manager Beverly

Joy Ruedi, who set the book in type.

Bibliography

[1] Anthony Barcellos and Martin Gardner, A Conversation with Martin Gardner, Two-Year College

Math. J. 10 (1979) 233–244.

[2] Don Albers and Martin Gardner, On the Way to “Mathematical Games”: Part I of an Interview

with Martin Gardner, College Math. J. 36, (2005) 178–190.

[3] Don Albers and Martin Gardner, “Mathematical Games” and Beyond: Part II of an Interview

with Martin Gardner, College Math. J. 36 (2005) 301–314.

[4] Colm Mulcahy and Martin Gardner, An Interview with Martin Gardner, Card Colm, October

2006, available at www.maa.org/columns/colm/cardcolm200610.html

[5] Michael Henle and Martin Gardner, Interview with Martin Gardner, College Math. J. 40 (2009)

158–161.

[6] Bruce Torrence and Stephen Abbott, To Our Readers, Math Horizons 18 (2010) 2–4.

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Contents

Preface v

I Geometry 1

1 The Asymmetric Propeller 3

Martin GardnerGardner, paying tribute to dentist and geometer Leon Bankoff, discusses some of his unpub-

lished results and concludes with an open question.

2 The Asymmetric Propeller Revisited 7

Gillian Saenz, Christopher Jackson, and Ryan CrumleyThree University of Texas students use dynamic geometry software to confirm Bankoff’s re-

sults and resolve Gardner’s question.

3 Bracing Regular Polygons As We Race into the Future 11

Greg W. FredericksonA problem Gardner published in 1963 continues to spur generalizations and improved solu-

tions around the world.

4 A Platonic Sextet for Strings 19

Karl SchafferThe professor and dance company co-director details string polyhedra constructions for ten

participants.

5 Prince Rupert’s Rectangles 25

Richard P. Jerrard and John E. WetzelA 17th century puzzle that Gardner posed in higher dimensions is here solved in the case of

three-dimensional boxes.

II Number Theory and Graph Theory 35

6 Transcendentals and Early Birds 37

Martin GardnerGardner moves from Liouville to an “innocent but totally useless amusement” that nonetheless

captured the attention of Solomon Golomb.

7 Squaring, Cubing, and Cube Rooting 39

Arthur T. BenjaminThe professor and “mathemagician,” inspired as a high school student by Gardner’s tricks for

mental calculations, extends some of them here.

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x Contents

8 Carryless Arithmetic Mod 10 45

David Applegate, Marc LeBrun, and N. J. A. SloaneInspired by the carryless arithmetic of the game Nim, this trio of authors explores the number

theory of a South Pacific island.

9 Mad Tea Party Cyclic Partitions 53

Robert Bekes, Jean Pedersen, and Bin ShaAnother playful trio analyzes cyclic arrangements that build from integer partitions in a Lewis

Carroll setting.

10 The Continuing Saga of Snarks 65

sarah-marie belcastroA type of graph, given a fanciful name by Gardner from Lewis Carroll, was the subject of a

Branko Grunbaum conjecture for 39 years.

11 The Map-Coloring Game 73

Tomasz Bartnicki, Jaroslaw Grytczuk, H. A. Kierstead, and Xuding ZhuDaltonism and half-dollar coins are used in this exploration of a Steven Brams game theory

approach to the Four Color Theorem.

III Flexagons and Catalan Numbers 85

12 It’s Okay to Be Square If You’re a Flexagon 87

Ethan J. Berkove and Jeffrey P. DumontThis article details the 1939 origin of flexagons at Princeton University and focuses on the

neglected tetraflexagons.

13 The V-flex, Triangle Orientation, and Catalan Numbers in Hexaflexagons 103

Ionut E. Iacob, T. Bruce McLean, and Hua WangThis trio of Georgia Southern University authors examines a once-illegal variety of flex and

makes a connection between “pat classes” and Catalan numbers.

14 From Hexaflexagons to Edge Flexagons to Point Flexagons 109

Les PookAn engineer and author of two books on flexagons considers the more general edge flexagons

and recently discovered point flexagons.

15 Flexagons Lead to a Catalan Number Identity 113

David CallanExamining the descent permutation statistic on flexagon pats leads the author to full binary

trees and a combinatorial proof.

16 Convergence of a Catalan Series 119

Thomas Koshy and Z. GaoCalculus is brought to bear on the infinite sum of Catalan number reciprocals and related

series; � and the golden ratio make appearances.

IV Making Things Fit 125

17 L-Tromino Tiling of Mutilated Chessboards 127

Martin GardnerIn his last MAA mathematics article, Gardner moves from classic chessboard domino tiling

problems to new results.

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Contents xi

18 Polyomino Dissections 135

Tiina Hohn and Andy LiuThe authors introduce a new technique for solving dissection problems, often presented in the

context of quilts, leaving several puzzles for the reader.

19 Squaring the Plane 143

Frederick V. Henle and James M. HenleA father and son team resolve Golomb’s “heterogenous tiling conjecture” and discuss another

dozen open questions.

20 Magic Knight’s Tours 153

John BeasleyThe author surveys results combining a knight’s tour on the chessboard with magic squares,

including a computer-aided solution to a Gardner question.

21 Some New Results on Magic Hexagrams 159

Martin GardnerHere Gardner focuses on three types of puzzles about placing numbers on six-pointed stars,

mentioning a “rare mistake” of the British puzzlist Henry Dudeney.

22 Finding All Solutions to the Magic Hexagram 167

Alexander Karabegov and Jason HollandThe authors relate magic hexagrams to magic edge labelings of cubes, using card shuffling to

enumerate distinct solutions.

23 Triangular Numbers, Gaussian Integers, and KenKen 173

John J. WatkinsMiyamoto’s contemporary puzzle is expanded to complex numbers where a different unique

factorization adds to the challenge.

V Further Puzzles and Games 179

24 Cups and Downs 181

Ian StewartOne of Gardner’s mathematical successors at Scientific American uses graph theory and linear

algebra on two related puzzles.

25 30 Years of Bulgarian Solitaire 187

Brian HopkinsSome recent math history explains this oddly-named puzzle on integer partitions, visualized

with state diagrams and generalized to a new two-player game.

26 Congo Bongo 195

Hsin-Po WangA high school student uses state diagrams and Dennis Shasha’s detectives to open a tricky

treasure chest.

27 Sam Loyd’s Courier Problem with Diophantus, Pythagoras,

and Martin Gardner 201

Owen O’SheaA Classroom Capsule extends Gardner’s solution of related Sam Loyd puzzles to other army

formations.

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xii Contents

28 Retrolife and The Pawns Neighbors 207

Yossi ElranAn inverse version of Conway’s game Life, famously popularized by Gardner, is examined

using chessboards.

29 RATWYT 213

Aviezri FraenkelThe combinatorial game theorist uses the Calkin Wilf tree to devise a rational number version

of Wythoff’s Nim.

VI Cards and Probability 219

30 Modeling Mathematics with Playing Cards 221

Martin GardnerIn addition to probability applications, Gardner uses a deck of cards for a discrete version of a

fluid mixing puzzle and mentions a correction to W. W. Rouse Ball.

31 The Probability an Amazing Card Trick Is Dull 227

Christopher N. SwansonRook polynomials and the principle of inclusion-exclusion help determine the likelihood that

the author’s students were underwhelmed.

32 The Monty Hall Problem, Reconsidered 231

Stephen Lucas, Jason Rosenhouse, and Andrew ScheplerThese authors remind us of Gardner’s early role in this infamous problem that still “arouses

the passions” and examine variations.

33 The Secretary Problem from the Applicant’s Point of View 243

Darren GlassChanging perspective, the author reconsiders a classic strategy in order to help job seekers

choose the best interview slot.

34 Lake Wobegon Dice 249

Jorge Moraleda and David G. StorkLake Wobegon Dice, named after Garrison Keillor’s Minnesota town, have the property that

each is “better than the set average.”

35 Martin Gardner’s Mistake 257

Tanya KhovanovaAnother controversial problem about probability and information is carefully discussed,putting

Gardner in the company of Dudeney and Ball.

VII Other Aspects of Martin Gardner 263

36 Against the Odds 265

Martin GardnerIn this short story, a principal recognizes the potential in a student whose unconventional

thinking irritates his teacher.

37 A Modular Miracle 271

John StillwellGardner used an obscure result of Hermite and the limitations of 1970’s calculators for an

April Fool’s Day prank.

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Contents xiii

38 The Golden Ratio—A Contrary Viewpoint 273

Clement FalboBuilding on a Gardner article in The Skeptical Inquirer, the author argues that � “is not entirely

astonishing.”

39 Review of The Mysterious Mr. Ammann by Marjorie Senechal 285

Philip StraffinThis Media Highlight discusses an example of Gardner’s support of an amateur mathematician

who independently discovered Penrose tiles.

40 Review of PopCo by Scarlett Thomas 287

Martin GardnerThis popular 2004 novel includes a character based on Gardner, so he was a natural choice to

review the book.

41 Superstrings and Thelma 289

Martin GardnerGardner’s last MAA submission, a short story about a physics graduate student and a waitress

who quips, “How are strings?”

Index 293

About the Editors 297

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IGeometry

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1The Asymmetric Propeller

Martin Gardner

The late Leon Bankoff (he died in 1997) was a Beverly Hills, California, dentist who also

was a world expert on plane geometry. (For G. L. Alexanderson’s interview with Bankoff,

see [1].) We became good friends. In 1979 he told me about a series of fascinating discov-

eries he had made about what he called the asymmetric propellor theorem. He intended to

discuss them in an article, but never got around to it. This is a summary of what he told me.

The original propeller theorem goes back to at least the early 1930’s and is of unknown

origin. It concerns three congruent equilateral triangles with corners meeting at a point as

shown shaded in Figure 1.1. The triangles, which resemble the blades of a propeller, need

not form a symmetrical pattern, but may be in any position. They may touch one another or

even overlap. Lines BC , DE , and FA are drawn to form a hexagon inscribed in a circle.

The midpoints of the three lines mark the vertices of an equilateral triangle.

A

B

C

DE

F

Figure 1.1.

A proof of the theorem, using complex numbers, appeared in [2] as the answer to

Problem B-1 in the annual William Lowell Putnam Competition. H. S. M. Coxeter sent the

proof to Bankoff on a Christmas card, asking him if he could provide a Euclidean proof of

the theorem.

Reprinted from The College Mathematics Journal, Vol. 30, No. 1 (Jan. 1999), pp. 18–22. This article receivedthe George Polya Award in 2000.

3

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4 Part I. Geometry

Bankoff had no difficulty finding such a proof. In a paper titled “The Asymmetric

Propeller” [3], Bankoff, Paul Erdos, and Murray Klamkin made the first generalization of

the theorem. They showed that the three equilateral triangles need not be congruent. They

can be of any size, as shown in Figure 1.2, and the theorem still holds. Two proofs are

given, one a simple Euclidean proof, the other with complex numbers. As before, and in

all subsequent extensions, the triangles may touch one another or even overlap.

Figure 1.2. Figure 1.3.

Later, Bankoff made three further generalizations. As far as I know they have not been

published.

Second generalization: The propeller triangles need not meet in a point. They may meet

at the corners of any equilateral triangle, as shown in Figure 1.3.

Third generalization: The propeller triangles need not be equilateral! They need only

be similar triangles of any sizes that meet at a point. The midpoints of the three added lines

will then form a triangle similar to each of the propellers, as shown in Figure 1.4.

Figure 1.4.

Fourth generalization: The similar triangles need not meet at a point! If the propellers

meet at the corners of a fourth triangle of any size, provided it is similar to each propeller,

the midpoints of the added lines will form a triangle similar to each propeller. Vertices of

the interior triangle must touch corresponding corners of the propellers.

Here is how Bankoff proved his final generalization on a sheet that he typed in 1973. It

makes use of Figure 1.5.

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1. The Asymmetric Propeller 5

A

B C

D

E

F

G

H

J

X

Y

Z

P

QR

S

Figure 1.5.

The propellers shown are right triangles, although they can be any type of triangle.

Perhaps the proof that follows can be simplified.

If ABC , AHJ , DBE , and F CG are similar triangles, all labelled in the same sense

and situated so that corresponding angles meet at the vertices of triangle ABC , then

X; Y; Z, the midpoints of DF , GH and JE , are vertices of a triangle similar to the other

four.

Proof. Let †BCA D †GCF D ˛; †DBE D †ABC D ˇ; †JAH D †CAB D ; and

let P; Q; R; S denote the midpoints of the segments DC , AC , AE and CH respectively.

We proceed stepwise to show that triangles PQR, PSZ and finally XYZ are similar to

triangle ABC .

If triangle ABD is pivoted about B so that AB falls along BC and DB along EB , it is

seen by the relation AB=BC D DB=BE and by the equality of angles ABD and CBE that

triangles ABD and CBE are similar and that EC=AD D BC=AB , with †EC; AD D†BC; BA D ˇ.

Since RQ is parallel to and equal to half EC while QP is parallel to and equal to

half AD, we extend the previous relation to read RQ=QP D EC=AD D BC=BA, with

†RQ; QP D †EC; AD D †BC; BA D ˇ. It follows that triangles PQR and ABC are

similar.

In like manner, because of the relationship of AJ and AH to RZ and QS as well as

to RP and QP in both relative length and in direction, we find triangles ZRP and SQP

similar. Then ZP=SP D ZR=QS D AJ=AH D AC=AB , with the angles between the

segments in the numerator and in the denominator all equal to . As a result, triangles

PSZ and ABC are similar.

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6 Part I. Geometry

Continuing as before, we find triangles ZPX and ZSY similar since PX=SY DCF=CB D CA=CG and †PX; SY D †CF; CG D †CA; CB D ˛.

Noting that in the similar triangles ZPX and ZSY we have ZX=ZY D ZP=ZS DCA=CB and †ZX; ZY D †CA; CB D ˛, we conclude that triangles XYZ and ABC

are similar.

And now a question for interested readers to explore. Do the propellers have to be

triangles? It occurred to me that if squares are substituted for triangles, as in Figure 1.6,

that equilateral triangle still shows up.

Figure 1.6.

I have written this piece as a tribute to one of the most remarkable mathematicians I

have been privileged to know.

Bibliography

[1] G. L. Alexanderson, A conversation with Leon Bankoff, College Math. J. 23 (1992) 98–117.

[2] J. H. McKay, The William Lowell Putnam Mathematical Competition, Amer. Math. Monthly 75

(1968) 732–739.

[3] Leon Bankoff, Paul Erdos, and Murray Klamkin, The asymmetric propeller, Math. Mag. 46

(1973) 270–272.

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2The Asymmetric Propeller Revisited

Gillian Saenz, Christopher Jackson, and Ryan Crumley

In [1] Martin Gardner proved an asymmetric propeller theorem that was originally pro-

posed by Leon Bankoff. He showed that by connecting one vertex from each of three

equilateral triangles inscribed in a circle to the center of the circle, one can form a fourth

equilateral triangle. This fourth triangle is formed by connecting the remaining vertices

with segments as seen in Figure 2.1. Then the midpoints of these segments are connected

to form a triangle. This works regardless of the arrangement of the original triangles. We

confirmed Gardner’s findings and proceeded to consider his final question: when working

with squares, does the same phenomenon occur? We show that the answer is no. Here we

give counterexamples.

H

J

I

Figure 2.1.

Using Geometer’s Sketchpad, we attempted to show that the propeller theorem also

worked for squares and possibly other polygons. In [1, Fig. 6] a midpoint of a side of each

of three squares occurred at the vertices of an equilateral triangle. We had the squares meet

at a single point by shrinking the equilateral triangle to a point. If the conjecture is false for

this case, then it will not hold true if the point is expanded to a triangle. By experimenting

with Geometer’s Sketchpad we found that even when using squares inscribed in a circle

Reprinted from The College Mathematics Journal, Vol. 31, No. 5 (Nov. 2000), pp. 347–349.

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8 Part I. Geometry

I

J

K

Figure 2.2.

the propeller theorem was not true, see Figure 2.2. The triangle shown in this figure was

not equilateral not did its angles remain constant when the squares were moved.

Seeing that the square conjecture would not work, we set up the following proof. We

start with three congruent squares inscribed in a circle. The squares meet at the midpoints

of one of their edges. The vertices of the squares are then connected as in Figure 2.3. The

midpoints of these new segments are then connected as before to form a triangle. Using

the distance formula, it is evident that the edges of the triangle are not congruent, because

3 ¤p

34=2.

(–1, 2) (1, 2)

(2, 1)

(2, –1)

(–1, –2) (1, –2)

Figure 2.3.

We also investigated whether the midpoint connection was essential when working with

equilateral triangles. We used the simplest case possible, three congruent equilateral trian-

gles inscribed in a circle meeting at a point. Rather than constructing the midpoints of the

connecting segments, we constructed a point one quarter along each segment. These points

were connected to form a triangle. Figure 2.4 shows that the triangle formed is not an equi-

lateral triangle. The theorem seems to be true only when the triangle formed is constructed

using the midpoints of the connected vertices. Attempts using other polygons were unsuc-

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2. The Asymmetric Propeller Revisited 9

K

L

M

Figure 2.4.

cessful. Based on our results, a propeller theorem for squares and other polygons seems

unlikely.

Bibliography

[1] Martin Gardner, The asymmetric propeller, College Math. J., 30 (Jan. 1999) 18–22; Chapter 1 in

this volume.

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3Bracing Regular Polygons

As We Race into the Future

Greg N. Frederickson

If today’s math-and-science-oriented kids could see the resources available to like-minded

youths of the 1960s and 70s, they might pity their counterparts back then: no laptop com-

puters, no high-speed internet, no online bookstores. Yet such pity would be wasted on that

earlier generation, who could well respond, “Ah, but we had Martin Gardner.”

For a quarter century, Gardner wrote Mathematical Games, a wonderful column in

Scientific American [3]. Each month he treated his readers to a tour of some mathematical

recreation that he made tantalizing for people who, like himself, had no advanced training

in mathematics. He was at the center of an amazing human network, referred to as “Martin

Gardner’s mathematical grapevine” by Doris Schattschneider [7].

Long before our current electronic networks, Gardner gathered juicy mathematical tid-

bits from his far-flung contacts, turned them into mouth-watering morsels, and encouraged

readers to send back their own variations. How many thousands leaped into mathemati-

cal, scientific, and engineering fields because of the enthusiasm nurtured by this singular

man? Let’s reach back in time to sample a problem that Gardner introduced and explore the

magic that he created, magic which would expand as we raced towards our technological

future.

Bracing a square

In his November 1963 column, Gardner dished up a tasty problem devised by Raphael

Robinson, a mathematician at the University of California, Berkeley, who is best known

for determining the minimum number of pieces for the Banach-Tarski paradox. Gardner

wrote:

Imagine that you have before you an unlimited supply of rods all of the same

length. They can be connected only at the ends. A triangle formed by joining

three rods will be rigid but a four-rod square will not: it is easily distorted into

other shapes without bending or breaking a rod or detaching the ends. The

simplest way to brace the square so that it cannot be deformed is to attach

eight more rods to form the rigid octahedron.

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 51–57.

11

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12 Part I. Geometry

Suppose, however, you are confined to the plane. Is there a way to add

rods to the square, joining them only at the ends, so that the square is made

absolutely rigid? All rods must, of course, lie perfectly flat in the plane. They

may not go over or under one another or be bent or broken in any way. The

answer is: Yes, the square can be made rigid. But what is the smallest number

of rods required?

Readers are asked to try to solve the problem and, if they succeed, to save

their solution or send it to this department. Next month we shall publish the

minimum answer Robinson has so far been able to achieve.

The next month, Gardner displayed two solutions by Robinson, each with 31 rods in

addition to the four rods of the square. Robinson used equilateral triangles to form lengths

of 2, 3, andp

13, producing a right triangle whose right angle braces the square. Clever?

Yes! And yet, instead of administering the coup de grace to this intriguing little problem,

the high-powered Berkeley mathematician had merely set in motion a long sequence of

improvements as the problem began to bounce around in Gardner’s network.

Superior solutions from readers

Two months later, in the February 1964 column, Gardner reported that 44 readers had

surprised him with a 25-rod solution, bettering each of Robinson’s solutions by 6 rods!

Moreover, Gardner was stunned that seven readers had designed an even niftier 23-rod

solution!

AB C

D

E

AB C

D

E

Figure 3.1. (a) 23-rod solution for bracing a square; (b) Underlying geometry.

The 23-rod solution appears in Figure 3.1a. Two pairs of equilateral triangles guarantee

that the distances BE and CE are eachp

3. Two trusses, each consisting of three equilateral

triangles, guarantee that the rods form straight lines between B and D, and between C and

D. These straight lines force points A, B , and C to be collinear. That the distances BE and

CE are equal then forces reflection symmetry along the axis containing A and D. Thus

†BAE and †CAE are both right angles, forcing distances AB and AC to both bep

2. We

can then conclude that the distance AD is alsop

2, forcing the rhombus containing A and

D to actually be a square. Figure 3.1b depicts this underlying structure.

Typically, Gardner collected his columns from Scientific American into books that he

published every several years. When doing this, he added new material that had come to

light once the columns had appeared. Including this column in his Sixth Book of Mathe-

matical Games [4], Gardner noted on page 55 about the square bracing problem:

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3. Bracing Regular Polygons As We Race into the Future 13

The problem can be extended to other regular polygons. The hexagon is simply

solved with internal braces (how many?), but the pentagon is a tough one. T. H.

O’Beirne managed to rigidify a regular pentagon with 64 additional rods, but

it is not known if this number is minimal.

O’Beirne’s solutions for various polygons

In 1999 I was selecting material to include in Hinged Dissections [1]. The 23-rod solution

for bracing a square, along with O’Beirne’s 64-rod solution for bracing a regular pentagon,

seemed perfect for a short segment in the book. There was just one hitch—I couldn’t find

O’Beirne’s braced pentagon anywhere. In desperation I contacted Gardner.

As it turned out, neither he nor O’Beirne had ever published the pentagon bracing.

Gardner generously sent me all of O’Beirne’s correspondence on this problem. One aero-

gramme from O’Beirne showed three progressively better solutions for the pentagon, each

added as he excitedly discovered a further improvement. During one remarkable week in

December 1963, O’Beirne had found solutions not only for the pentagon but also for three

other regular polygons as well! Almost certainly prompted by Robinson’s square bracings

in Gardner’s December column, O’Beirne had found ways to brace the pentagon (64 rods),

the octagon (105 rods), the decagon (183 rods), and the dodecagon (45 rods). I included

the pentagon in [1].

Armed with an M.A. in math and physics from the University of Glasgow, Thomas H.

O’Beirne (1915–1982) had worked in turn as a government scientist, a chief mathematician

for an engineering firm, and a lecturer in the university’s computing department. In 1961–

62, he wrote a weekly series for the New Scientist, which was collected into a book, Puzzles

and Paradoxes.

Lying dormant for more than thirty years, O’Beirne’s nifty constructions might well

have been forgotten. But that would have belied the residual strength of Gardner’s network

and the growing strength of modern networks.

An improved octagon bracing

After publishing my book with its discussion of the braced square and braced pentagon,

I continued to think about bracing problems. In 2003, I discovered an improvement to

O’Beirne’s solution for the octagon, using only 51 additional rods, as shown in Figure 3.2a.

The solution incorporates several ideas. First, the rods in the interior of the octagon

form rhombuses, forcing opposite pairs of sides of the octagon to be parallel. For every

pair of opposite sides, not only are those two rods parallel, but also a pair of bracing rods

in the interior.

Second, some rods outside of the octagon force other rods to be parallel to rods on the

boundary of the octagon. Thus a truss consisting of five equilateral triangles forces three

rods to be on the line LM. Also, a truss of three equilateral triangles forces two rods to be

on the line CN.

Third, other rods outside the octagon enforce certain distances between points on the

inside of the octagon. The group of four equilateral triangles between M and C force a

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14 Part I. Geometry

M

E

CI

N

B

H

J

L

K

F

G D

A CH

DG

L

BA

I

EF

JK

Figure 3.2. (a) 51-rod solution for an octagon; (b) Underlying geometry.

distance ofp

3 between these vertices. Since rods ME and CI are parallel, distance EI

must also bep

3.

Similarly, a group of four equilateral triangles between H and N force a distance ofp3. Rods HJ and BN transmit this restriction to points J and B . A final group of four

equilateral triangles forces a distance ofp

3 between H and L. Two sets of parallel rods

force distance BK to also bep

3. Yet it is not so clear that the octagon is then rigid.

We can verify this claim by modeling the braced structure (Figure 3.2b) with equations

that we solve using Mathematica. First, fix the location of an edge, say AB, by xA D 0,

yA D 0, and xB D 1, and yB D 0.

Next, for each of the three rods along the boundary from B to E , such as CD, force its

length to be 1 by .xC � xD/2 C .yC � yD/2 D 1. Then, for each of the six rhombuses

inside the octagon, such as ABCI, force an opposite pair of rods to be parallel by xB �xC DxA � xI and yB � yC D yA � yI . Do the same for the (degenerate) “flat rhombus” LFEF.

And to enforce a distance ofp

3 between each of the three pairs of two vertices, such as

H and L, introduce .xH � xL/2 C .yH � yL/2 D 3.

This gives us 24 equations determining the 24 total coordinates of the twelve points.

However there may be more than one solution to this system of equations, because we

have not specified, for example, whether rod IC is above or below rod AB. So introduce

more constraints to force a specific topology as in Figure 3.2b. These constraints would be

yC > yB , yD > yB , xC > xB , and xD > xB . The resulting set of restrictions leaves us

with a unique solution that then guarantees rigidity.

Theobald’s solution for bracing a pentagon

In 2004, after studying O’Beirne’s braced pentagon (in [1]), Gavin Theobald improved

it, reducing the number of additional rods from 64 to 56, as in Figure 3.3a. Theobald

adapted key ideas of O’Beirne and also introduced new ones. Let’s verify carefully that the

pentagon is braced.

O’Beirne had first set out to produce an angle of 108ı, the interior angle of the regular

pentagon. He had known that a diagonal in a regular pentagon divides the pentagon into

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3. Bracing Regular Polygons As We Race into the Future 15

a trapezoid and an isosceles triangle with this angle. Furthermore, he had recognized that

if the two equal sides of the triangle are of lengthp

3, then the length of the third side isp3.1 C

p5/=2. Trusses consisting of three equilateral triangles will create 4ABC with

sides 1, 2, and 2, as in Figure 3.3a. By the Pythagorean Theorem, such a triangle has height

.p

15/=2. Adding the height of an equilateral triangle (top of Figure 3.3b) gives the desired

distance ofp

3.1 Cp

5/=2.

A

B

C

D

E

G

I

H

J

K

L

M

N

F

A

B

C

B

C

D

I

H

J

22

1

2

2

108 ◦

Figure 3.3. (a) Theobald’s braced pentagon; (b) Underlying geometry.

While O’Beirne had then gone on to generate the angle of 108ı and transfer it elsewhere

in the structure, Theobald chose instead to transfer the distance itself. Theobald noted

that he already had a suitable angle, namely †BCD, so that the desired length would be

between points B and D (See the middle of Figure 3.3b).

Theobald then transferred the desired length by constructing parallelogram BDHI as

follows. He made a truss of four equilateral triangles to force distance BI to bep

7, and

then made a “zigzag truss” with five equilateral triangles to force distance DH to bep

7.

He forced BE to be parallel to DG by creating a sequence of three parallelograms between

them. Then BI must be parallel to DH . Since BDHI is a parallelogram, its opposite sides

must be the same length, and thus distance HI is alsop

3.1 Cp

5/=2.

Theobald positioned the vertices of an isosceles triangle HIJ with HI as the base.

Then two equilateral triangles force distance IJ to bep

3 and four equilateral triangles

force distance HJ to bep

3. This gives the isosceles triangle with †IJH D 108ı. The

two equilateral triangles touching at J supply two of the sides of the regular pentagon.

Once O’Beirne had positioned one angle of 108ı in the regular pentagon, he had po-

sitioned the opposite side by using a sequence of an equilateral triangle, two rhombuses,

and another equilateral triangle. Theobald used a similar idea to position side LM of his

pentagon with just one rhombus.

Theobald’s construction forces the side LM of the pentagon to be at the correct angle

to make the pentagon be regular: First, rods EL and FM form a rhombus, so that rod LM

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16 Part I. Geometry

is parallel to rod BC . Rod BC is at an angle of 30ı with the line BD. Then rod LM is at an

angle of 30ı with line HI . Pentagon edges JK and JN are rotated 30ı counterclockwise

from the lines IJ and HJ , which means that diagonal KN is parallel to LM . Thus LM is

at the correct angle. Even knowing O’Beirne’s approach, Theobald still needed creativity

and persistence to reduce the number of rods by eight.

Gavin Theobald was born in London, England, in 1961. Graduating with a B.Sc. in

Computer Studies at Loughborough University in 1983, he worked as a computer program-

mer on low-level high-performance graphics rendering engines. Theobald is an expert in

geometric dissections [8], an area into which he was drawn after reading one of Gardner’s

books.

Khodulev’s solutions when rods cross

Even more progress was made by Andrei Khodulev, who had learned of the problem from

the Russian translation of Gardner’s book [5]. Noting that one can use fewer rods if they

are allowed to cross, Khodulev discarded the restriction on crossing. Then, for example, a

regular octagon can be rigidified with only 23 additional rods: eight rods to create rhom-

buses within the octagon, and three sets of five rods each to produce three distances ofp

3.

Published in the bulletin of the Russian puzzle club Diogen (Diogenes) [6], Khodulev’s

results are reproduced on Erich Friedman’s webpage [2].

Andrei Khodulev (1953–1999)earned a Ph.D. in computer science in 1984 from Keldysh

Institute of Applied Mathematics, in Moscow, where he was a senior research scientist. His

research areas were in graphics, imaging, approximation theory, and stochastic simulation.

Gardner’s enduring challenge

Thanks to Martin Gardner’s gift for connecting people together, a problem that originated

with Raphael Robinson (1911–1995) spread to Gardner’s wide readership, resulting in im-

provements from dozens of his correspondents. Thomas H. O’Beirne then generalized the

problem and produced solutions for four other polygons, one of which Gardner mentioned

in his subsequent book. That led the current author to include O’Beirne’s solution in a

book, which in turn attracted the attention of Gavin Theobald, who found an improvement

for the pentagon, as I found an improvement for the octagon. Separately, Andrei Khodulev

made progress by allowing rods to cross. So don’t be surprised when further delectable

developments are announced over our networks—just brace yourself!

Acknowledgment I thank Gavin Theobald for permission to include his pentagon brac-

ing, Serhiy Grabarchuk for copies of Sharada, Svitlana Mayboroda for translating Russian,

and the referees for good suggestions.

Bibliography

[1] G. N. Frederickson, Hinged Dissections: Swinging and Twisting, Cambridge Univ. Press, New

York, 2002.

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3. Bracing Regular Polygons As We Race into the Future 17

[2] E. Friedman, Problem of the month (January 2000); available at

www.stetson.edu/˜efriedma/mathmagic/0100.html, 2000.

[3] M. Gardner, Mathematical games, Scientific American, Nov., 1963, p. 144; Dec., 1963,

pp. 144, 146; Feb., 1964, p. 128.

[4] ——, Martin Gardner’s Sixth Book of Mathematical Games from Scientific American, Charles

Scribner’s Sons, New York, 1971.

[5] ——, Mathematical Leisure (Matematiqeskie Dosugi), Mir (Mir), Moscow, 1972.

[6] A. Khodulev, A task about a rigid polygon (zadaqa o �estkom mnogougolbnike),

Sharada (Xarada) 7, no. 61 (1998) 2–8.

[7] D. Schattschneider, In praise of amateurs, in The Mathematical Gardner, D. A. Klarner, ed.,

Prindle, Weber & Schmidt, Boston, 1981, 140–166.

[8] G. Theobald, Geometric dissections; available at

home.btconnect.com/GavinTheobald/Index.html.

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4A Platonic Sextet for Strings

Karl Schaffer

Our dance company, the Dr. Schaffer and Mr. Stern Dance Ensemble, began performing

mathematical dance shows in 1990. Scott Kim joined us in 1993, and in 1995 Scott, another

Bay area dancer Barbara Susco, and I developed a school-age show entitled “Through the

Loop, In Search of the Perfect Square,” which included the performance of many string

figures.

Martin Gardner wrote several columns on polyhedra and one on traditional string fig-

ures, in addition to his many references to magic tricks and puzzles with string [5, Ch. 1],

[6, Ch. 17], [7, Ch. 19], [8, Ch. 10], [6, pp. 74–75]. Traditional string figures, made on

the hands with a simple loop of string, are found worldwide, particularly associated with

ancient cultures [7]. In 1994 our dance company began experimenting with oversize loops

carried on stage by dancers, and manipulated to form a variety of string figures. String

figure constructions and the associated mathematics make a pleasurable classroom activ-

ity that introduces graph theory and polyhedra to geometry, liberal arts, and discrete math

students, or as a hands-on activity for workshops and math festivals.

In the process of creating one of our shows, “Through the Loop,” which includes over

20 string figures, we began altering well-known figures, using large loops held by several

dancers, or using a sequence of finger manipulations different from the customary methods.

Traditional string figures often display shapes or actions reminiscent of the natural world,

as indicated by such titles as: Bear Climbing a Tree, the Yam Thief, and Mosquito. Yet

these figures also involve complex mathematical patterns in both their construction and

final designs. We began experimenting with polyhedral designs which we thought might

be more recognizable by contemporary audiences as being mathematical.

Scott Kim and the author came up with several string constructions of the tetrahedron,

octahedron, and cube [8, 9], though Scott’s are particularly beautiful for their symmetry

and simplicity. All these constructions keep the string loops taut, and are simplified by

involving symmetrical transfers by participants. Several of our mathematical designs, in-

cluding that of the octahedron and its collapsed form as a six pointed star, were used in

one of the dances in the Loop show; other string figures in that show were traditional ones

as well as a series of star polygons [7]. Subsequently the author found several ways to

build all five Platonic solids sequentially, including a method employing identical loops

of three colors which creates interesting symmetric color patterns. This six loop version

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 64-69.

19

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20 Part I. Geometry

was demonstrated for the first time at a Math and the Arts conference at UC Berkeley in

1998 and at the Gathering for Gardner 3 in Atlanta in 1998. It has since has been handed

around at conferences and workshops, in the manner of folk mathematics familiar to fans

of Martin Gardner. It is published here for the first time.

These multiple loop constructions are inspired by contemporary modular origami which

uses identical copies of modular units to construct complex polyhedral forms. Though

some observers insist on calling this string sequence a dance, these more complex string

polyhedra are too stationary in space and require too many careful manipulations by the

hands of participants crowding around the figures to be usable as a part of actual staged

dances.

Euler circuits

An Euler circuit is a sequence of adjacent edges in a graph that includes each edge once.

The name refers to Leonhard Euler’s 1736 solution of the Seven Bridges of Konigsberg

Problem, though such figures also appear in sand drawings found in West Africa and else-

where [1, pp. 30–43]. Traditional string figures, which invariably use simple unknotted

loops of string, accomplish the same thing.

Think of the points of support, where hands or fingers keep the string in tension, as the

vertices, and the strands of string as the edges of a polyhedral graph. For our polyhedra we

use the thumb and first finger of one hand for each vertex, so that a polyhedron with 20

vertices, such as the dodecahedron, requires ten people. As Euler found, a connected graph

can be decomposed into a circuit (or collection of circuits) only if all vertex degrees are

even. Therefore some edges of our polyhedra need to be doubled and made with at least

two strands to even out the vertex degrees. For example, the four vertices of the tetrahedron

are each of degree three, and so we need to double at least two opposite edges in order to

Eulerize its graph and so allow it to be made with a circuits (i.e., loops of string). Four of

the Platonic solids have odd degree vertices, and require edge doublings; the octahedron

has degree four vertices, and so can be made from loops without edge-doubling.

By utilizing the symmetries of the polyhedra, we allow all the loops to be of the same

size, and this simplifies the constructions and facilitates the directions, as multiple partic-

ipants do the same string manipulations. After choosing edges to double, we essentially

decompose the graph of each polyhedron into cycles, but the struggle to insure that the

strings do not end in tangles nor the participants interfere with each other makes the search

for these decompositions more challenging. The most pleasing figures seem to allow the

least slack in the strings and make use of simple and surprising transformations.

Transformations

The overall manner in which we handle the string can be modeled by several standard

graph theory operations. An edge subdivision replaces an edge uv having vertices u and

v with a new vertex w and two new edges uw and vw. This is accomplished with string

when one grasps and pulls on the middle of a string edge. In order to manipulate the string

smoothly, each hand involved holds it by making a loop with the thumb and index finger,

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4. A Platonic Sextet for Strings 21

so that the string can slide easily “through” the vertex. This is why we require a whole

hand for a single vertex.

We also use the reverse operation, smoothing a vertex, releasing a vertex w while si-

multaneously pulling outwards at all other vertices to take up the slack, and replacing two

edges uw and wv by single edge uv. Pulling two or more vertices together and holding

them in one hand is vertex amalgamation. We retain the separate identification of the amal-

gamated vertices, since we sometimes need to reverse the operation and release some of

the vertices from the amalgamation. Vertex separation is the release of a string from an

amalgamated vertex. We may also perform an edge contraction by sliding two vertices

joined by an edge together to amalgamate their endpoint vertices, while removing the edge

between them by pulling outward on the other vertices, as with vertex smoothing. In all

cases the participants need to adjust their hands slightly to make the polyhedra more regu-

lar; letting the strings slide through the loop made by the thumb and forefinger facilitates

this, while grabbing the strings makes it more difficult.

Six-loop platonic solids

Figure 4.1 shows the six-string transformations of the Platonic solids. All six strings should

be same length, and it is helpful to use two loops of each of three colors. Each vertex is

labeled Ri or Li , indicating that it is held by person i with the Right or Left hand. Figure

4.1a shows the opening pattern, seen from above, made with the two loops of the same color

held by opposite participants. It is important to keep the doubled loops from crisscrossing,

as otherwise the later figures will contain extra edges. One of the pairs of loops must sit

inside the other two pairs (that held by R1 and R3), and one pair must be external to both

the others (that held by R5 and R6), so this construction is not as symmetrical as other of

the polyhedral string figures; a more symmetric alternative is to link the strings like the

Borromean rings, but this leads to tangles!

Figure 4.1a shows that first R5 and R6 subdivide and hand their loops off, forming

the square R1R2R3R4 in Figure 4.1b. R5 and R6 then grasp new vertices as shown in

Figure 4.1b, forming squares R4R5R2R6, and R1R5R3R6, and thus the octahedron shown

in Figure 4.1c. The six participants still use only one hand apiece, though actually both

hands are needed to perform the transition from Figure 4.1a to Figure 4.1b. The octahedron

is usually held with one vertex high and the opposite near the floor to make the instructions

easier to follow, though one person must reach high and another low in order to grasp

the vertices. The octahedron thus lines up an axis of four-fold rotational symmetry with

the vertical; we might also sit the octahedron on one of its triangles to accentuate the

3-fold symmetries.

In order to transform the octahedron to the icosahedron, we must double the number of

vertices, and transform the 12 edges of the octahedron to the 30 edges of the icosahedron.

This is accomplished by adding one vertex and 3 edges for each of the octahedron’s 8

vertices as follows (see the arrows in Figure 4.1c for the R1, R2, and R6 vertices.) Note that

all edges in the octahedron are doubled, but only six non-adjacent edges in the icosahedron

are doubled. The two vertices from the horizontal R1R2R3R4 loop held at the R1 vertex

are separated, one sliding upwards, the other downwards along the R1R5R3R6 loop (in the

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22 Part I. Geometry

(e)

L1L7 L3L4

R2

R4

R3R1

L5

L6

R5

R6

L2

L8L9

L10

R7

R8R9

R10

L10

R7R9

(g)

(d)

L1

L2

L3L4

R2

R4

R3R1

L5

L6

R5

R6

(a)

R1 R3

R2

R4 R5

R6 (b)

R1

R3

R2

R4

R5

R6

L8(f)

L7

L8L9

L10

R7

R8R9

R10

R1

R3

R2

R4

R5

R6

(c)

R10

Figure 4.1. Six loop Platonic solids.

plane of the drawing), to form the new R1 and L1 vertices. Each participant will execute a

similar maneuver. The exact same operation as at R1 in Figure 4.1c is performed at the R3

vertex. At R2 and R4 the vertical R4R5R2R6 loop’s amalgamated vertices are separated

horizontally to form new R2; L2; R4, and L4 vertices. At R5 and R6 the R1R6R3R5

loop’s vertices are separated horizontally to form new R5; L5; R6, and L6 vertices. All

this produces the icosahedron in Figure 4.1d. When performing these operations make

sure that the pairs of same colored loops do not cross. Note that the remaining 6 doubled

edges are parallel in pairs, forming a perfect matching of the vertices of the icosahedron,

and hence a minimal Eulerization.

At this point notice that there are 8 triangles, 4 high and 4 low, which have edges of

3 colors, none doubled. To transform the icosahedron, with 12 vertices and 30 edges, to

the dodecahedron, with 20 vertices and 30 edges, we need to add 8 vertices, and some-

how change 20 triangles to 12 pentagons. To accomplish this, each edge of those eight,

3-colored triangles will be subdivided, and the three new vertices on each such face amal-

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4. A Platonic Sextet for Strings 23

gamated as one vertex. We call this operation on all the edges of a single face squeezing

the face. This is illustrated in Figure 4.1d by arrows (just for the R2R3; L5 and R2L3; R6

triangles), resulting in the dodecahedron with all edges doubled (some doubled edges not

shown), in Figure 4.1e. Removing these 8 triangles results in a polyhedron with 20�8 D 12

pentagons. This operation requires 4 new participants, who each will most conveniently

squeeze two triangles above and below each other, with say, right hands high, left hands

low. Two of the same colored loops are shown as part of sets of doubled edges in Figure

4.1e. These sets of edges form what are known as the Petrie polygons of the dodecahedron,

skew polygons no three of whose successive edges are part of the same face.

The step from Figure 4.1e to 4.1f is simple: participants 1 through 6 release their ver-

tices, while the new participants (7 through 10) hang on to theirs. Their 8 vertices (white

circles in Figure 4.1e) are the vertices of a cube in Figure 4.1f, which magically appears

with all its edges doubled. Strings of like colors appear on opposite faces.

The transition to the tetrahedron (Figure 4.1g) is equally simple: two of the remaining

participants opposite each other on the upper face of the cube, say 7 and 9, hold on to

their high vertices with their right hands and release their left hand vertices, while 8 and

10 hold on to their left hand vertices and release their right. In the tetrahedron formed

each pair of string loops of the same color are non-incident and skew perpendicular. To

complete the cycle back to the octahedron, each of these four participants uses their free

hand to squeeze the edges of the three colors of one triangle back to a new vertex, re-

forming the cube; this operation is indicated by the arrows on one of the faces in Figure

4.1g. Participants 1 through 6 now return to form the rhombic dodecahedron by squeezing

one face per participant to a new vertex, taking care to use edges of two colors only per

vertex, and such that the same two colors are used for the vertices on opposite faces of the

cube. Also, one of the pairs of same color loops must be external to the others, and one

pair internal, or the octahedron will become tangled. Participants 1 through 6 now hang on

to their vertices, and 7 through 10 release theirs, forming the octahedron again.

Conclusion

Many, many other polyhedral string figures similar to these are possible. For example,

the author has found fairly direct transformations of the six loop polyhedra shown to the

Archimedean polyhedra, as well as a series of constructions of the Platonic solids using

either one or four loops. See also [8, pp. 85–97], [9], [3, pp. 280–287], and [7]. These con-

structions make excellent classroom investigations of Euler circuits and polyhedra. Start

simply. For example, a five-pointed star is easy to draw but making one with a loop of

string requires 3 people cooperating carefully. Have students find their own constructions

of a tetrahedron and octahedron; there are many. If the students find a nice construction

have them practice several times so the actions are quick and fluid. Directions and lore on

traditional string figures can be found at [7]. Many variations on the figures mentioned in

this article are possible.

Note on making loops for use in these constructions

We use loops which can be stretched to “arm span,” length or about 5 to 6 feet. We either

use cotton string, taping the ends together with cloth or duct tape, or nylon string with the

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24 Part I. Geometry

ends melted together over a candle flame. Use string that is at least 1=8th to 1=4 inch in

diameter. For the Platonic solids, we use 1=2 inch diameter flexible cotton rope, purchased

in 3 colors. One can also dye cotton rope a variety of colors with fabric dye. Make sure the

rope or string is manufactured with a soft weave, rather than using brittle or inflexible string

or rope. The string polyhedra constructed in this paper use multiple loops of the same size.

To store the loops without getting them all tangled, double each one over several times and

tie it in a knot.

Videos

George Csicsery has produced several videos of sequences of string polyhedra, including a

three-polyhedra sequence similar to, but shorter than, the one described here, and involving

only four participants. This video and other videos relevant to string polyhedra and to Mar-

tin Gardner are available on YouTube at http://www.youtube.com/user/g4gman.

Bibliography

[1] M. Ascher, Ethnomathematics, a Multicultural View of Mathematical Ideas, Brooks/Cole Pub-

lishing Company, Pacific Grove CA, 1991.

[2] M. Gardner, The Second Book of Mathematical Diversions, Simon and Schuster, New York,

1961.

[3] ——, The Unexpected Hanging and Other Mathematical Diversions, Simon and Schuster, New

York, 1968.

[4] ——, Martin Gardner’s Sixth Book of Mathematical Diversions from “Scientific American,”

Simon and Schuster, New York, 1971.

[5] ——, Wheels, Life, and other Mathematical Amusement, W.H. Freeman and Company, New

York, 1983.

[6] ——, The Last Recreations Hydras, Eggs, and Other Mathematical Mystifications, Springer-

Verlag, New York, 1997.

[7] K. Schaffer, Star String Polygons, Gathering for Gardner 9, 2010 (to appear).

[8] K. Schaffer, E. Stern, and S. Kim, Math Dance with Dr. Schaffer and Mr. Stern, MoveSpeak-

Spin; available at www.movespeakspin.org, 2001.

[9] ——, math dance activities; available at www.mathdance.org.

[10] M. Sherman, ed., International String Figure Association, string figure references and instruc-

tions; available at www.isfa.org.

[11] I. Stewart, Cows in the Maze and Other Mathematical Explorations, Oxford Univ. Press,

Oxford, 2010.

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5Prince Rupert’s Rectangles

Richard P. Jerrard and John E. Wetzel

Introduction

More than three hundred years ago, according to the contemporaneous John Wallis [11,

pp. 470–471], Prince Rupert� (1619–1682) won a wager that a hole can be cut in one of

two equal cubes large enough to permit the second cube to pass through.

Nearly a century later the Dutch scientist Pieter Nieuwland (1764–1794) showed that

the largest cube that can be so passed through a cube of side one has side 3p

2=4 � 1:061.

An accessible discussion with enlightening anaglyphs appears in Ehrenfeucht [8]. In 1950,

D. J. E. Schreck [10] gave an interesting, historically based survey of Prince Rupert’s

problem and Nieuwland’s extension. Schreck includes a photograph of a model showing

the cube in transit.

Nieuwland’s “passage” problem of finding the largest cube that can pass through a unit

cube is equivalent to finding the largest square that fits in the unit cube, because once the

largest square is located, the hole through the cube having that largest square as its cross

section clearly provides the desired passage. In higher dimensions one might seek the side

of the largest m-dimensional cube that fits in an n-dimensional cube of side one. Not much

is known about this question, which apparently was first raised, at least in the case of a cube

in a hypercube, by Martin Gardner (see Gardner [2, pp. 172–173] or Guy and Nowakowski

[7, pp. 967–68]).

The literature on Prince Rupert’s and Nieuwland’s problems is extensive and we cannot

claim to have examined it all, but nowhere have we found any mention of the analogous

questions for rectangles and “boxes” (i.e., rectangular parallelepipeds):

Question 1. Find the largest rectangle R with given aspect ratio � that fits in the unit cube,

i.e., for given � with 0 � � � 1, find the largest L so that an L � �L rectangle fits in the

unit cube.�

This question is an instance of a general unsolved fitting problem for boxes in higher

dimensions: in d -dimensional Euclidean space Ed , find necessary and sufficient conditions

Reprinted from The American Mathematical Monthly, Vol. 111, No. 1 (Jan. 2004), pp. 22–31.�Count Palatine of the Rhine and Duke of Bavaria, son of Frederick V, the Winter King, Elector Palatine, and

king of Bohemia, and Elizabeth, daughter of James I of England.�It is convenient to regard a line segment as a rectangle with one side 0, and a rectangle as a box with one

edge 0.

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26 Part I. Geometry

on the k C d edges for a k-dimensional box .k � d/ with edges p1, p2; : : : ; pk to fit

in a d -dimensional box with edges s1, s2; : : : ; sd . For k D 1 the matter is trivial for

each d : a line segment of length p fits in the d -dimensional box precisely when p �q

s21 C s2

2 C � � � C s2d

. For k D d D 2 the question of when one rectangle fits in another

was asked in 1956 by L. R. Ford [4], and a necessary and sufficient condition was soon

given by W. B. Carver [2] (see also Wetzel [12]). The problem was posed for k D d D 3

by F. M. Garnett in 1923, and an incomplete answer was supplied in 1925 by Carver [1].

Question 2. Find the largest box of given shape that can be passed through a suitably

perforated unit cube, i.e., for given � and � with 0 � � � � � 1, find the largest D so that

a D � �D � �D box can pass through a suitable hole in the unit cube.

This “passage” question bears the same relation to the “fitting” question 1 as Rupert’s

does to Nieuwland’s: once the largest rectangle with aspect ratio � D �=� in the unit cube

is known, then the hole through the cube having that rectangular cross section provides the

desired passage; and no smaller hole with similar cross section suffices. As an example,

suppose that � D 0:7 and � D 0:2. The largest D � �D � �D box that can pass through

the unit cube has D D 2.p

143 �p

8/=9 � 2:02885 (see Theorem 1 and its corollary),

so that �D � 1:42020 and �D � 0:40577. Figure 5.1a shows the largest rectangle with

aspect ratio � D 2=7 that fits in the unit cube and the cube perforated to accommodate the

box. Figure 5.1b shows the box in transit through the perforated cube.

(a) The perforated unit cube. (b) The box in transit.

Figure 5.1. A box through the unit cube.

The Fitting Lemma

Let C be the unit cube, and let @C denote its boundary, viz., the union of its six closed

faces. Once the aspect ratio � is fixed in Œ0; 1�, there is a largest real number Lmax so that

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4. Prince Rupert’s Rectangles 27

the Lmax � �Lmax rectangle Rmax fits in C. The first thing we need to ask is precisely how

this largest rectangle Rmax is situated in C. The answer is given by the “fitting lemma,”

Lemma 4, whose proof is the objective of this section.

Note first that, because of the central symmetry, any rectangle R that fits in C must also

fit in C with its center at the center of symmetry of the cube.

Lemma 1. If a rectangle R D ABPQ (Figure 5.2) fits in the unit cube C in any way

whatsoever, then it also fits in C with its center at the center of C.

A

B

Q

P

Figure 5.2. A rectangle with aspect ratio 0.4.

Proof. Suppose that the rectangle R D ABPQ fits in C and let A0, B 0, P 0, Q0 be the

points symmetric to A, B , P , Q in the center O of C (Figure 5.3). Then the rectangle

R0 D A0B 0P 0Q0 is congruent to R and forms with R a parallelepiped that fits in C with

center at the center O of C. The medial section of this parallelepiped midway between R

and R0 is a rectangle that is congruent to R and has its center at O .

R

R'

A A'

O

Figure 5.3. Centering.

As a consequence of this lemma, we may always assume that a rectangle under consid-

eration is centrally placed in C. We call such a rectangle centered. A corner of a centered

rectangle lies on an edge or face of C precisely when the opposite corner of the rectangle

lies on the opposite edge or face of C.

Corners at vertices If one corner of a centered rectangle R in C lies at a vertex of C, then

the opposite corner of R lies at the opposite vertex of C, and the two remaining corners of

R are at opposite vertices of C. There are only two possibilities: � D 0 and R is a diagonal

line segment of C regarded as ap

3�0 rectangle, or � D 1=p

2 and R is ap

2�1 diagonal

cross section of C. In the following analysis, we usually suppose that the corners of R are

not at the vertices of C.

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28 Part I. Geometry

Centered maximal rectangles How does a centered maximal rectangle R fit in C? We

show next that its corners must lie on the boundary @C of C.

Lemma 2. The corners of a centered maximal rectangle R lie on the boundary @C of C.

Proof. Our principal tool is the observation that a rectangle that fits in C with all four of

its corners inside C is not maximal. In particular, if a centered rectangle has two adjacent

corners in the interior of C, then it is not maximal. It follows that we need examine only

those centered rectangles having at least one corner on a face (but not at a vertex) of C.

Suppose that one corner A of a centered rectangle R D ABPQ lies inside C and an

adjacent corner B lies on a closed face f of C but not at a vertex (Figure 5.4). If B lies in

the interior of a face f , then an appropriate small rotation of R in its circumcircle (Figure

5.4a) moves B inside C as well; and consequently R is not maximal. If B lies on an edge e

of f (Figure 5.4b), then a suitable small rotation of R about the line ` normal to f through

O leaves A inside C and moves B off e into f , the case just considered; so again R is not

maximal. It follows as claimed that all four corners of a maximal centered rectangle must

lie on faces of C.

Of

A

B`

e

f

O

AB

`

(a) Corner B on an open face. (b) Corner B on an edge.

Figure 5.4. Corner A inside the cube.

But more is true: the corners of a maximal centered rectangle must be on the edges of

C.

Lemma 3. The corners of a centered maximal rectangle R lie on edges of C.

Proof. We show first that at least one of the corners of a centered maximal rectangle R

must lie on an edge of C. If adjacent corners A and B of R both lie on the same (open) face

of C (Figure 5.5a), then the other two corners P and Q lie on the opposite face of C, and

a suitable small rotation about the axis ` through O parallel to AB moves R completely

inside C, contrary to Lemma 2. If adjacent corners A and B of R lie on opposite (open)

faces of C, then the adjacent corners A and Q lie on the same (open) face of C, which as we

have just seen is a contradiction. If adjacent corners A and B of R lie on adjacent (open)

faces of C (Figure 5.5b), a suitable small rotation of R about the diagonal AP moves B

inside C, again contrary to Lemma 2. Consequently, at least one corner of R must lie on an

edge of C; of course, the opposite corner of R then lies on the opposite edge of C.

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4. Prince Rupert’s Rectangles 29

A

B

O`

A

B

P

`

(a) Corners A and B on the same face. (b) Corners A and B on adjacent faces.

Figure 5.5. Corners A and B on open faces.

Suppose next that a corner A of the centered maximal rectangle R lies on an edge e of

C. If a corner B adjacent to A lies on an (open) face f of C adjacent to e (Figure 5.6a),

then a suitable small rotation of R about the line ` through O perpendicular to f moves

both A and B into the (open) face f , contrary to the first part of this proof. If a corner B

adjacent to A lies on an (open) face f that does not meet e, then the opposite corner Q is

adjacent to A and lies on the face f 0 opposite the face f , which is adjacent to e. Again it

follows that R is not maximal. Finally, if a corner A of R lies on an edge e and an adjacent

corner B lies on a face f that meets e only at a vertex of C (Figure 5.6b), the corner P

opposite A lies on the edge opposite e, and a suitable small rotation about the diagonal AP

of R moves both B and Q into the interior of C, once more contrary to Lemma 2.

A

B

e

f O

`

A

P

e

f

B

(a) Corner B on an open face adjacent to e. (b) Corner B on an open face normal to e.

Figure 5.6. Corner A on an edge.

It follows, as claimed, that all four corners of a centered maximal rectangle lie on edges

of C.

Lemma 4 (Fitting Lemma). A centered maximal rectangle R whose corners are not at the

vertices of C fits in C in exactly one of the following two ways:

a. two adjacent corners of R lie on the same edge of C (Figure 5.8); or

b. two adjacent corners of R lie on adjacent edges of C (Figure 5.9).

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30 Part I. Geometry

Figure 5.7. Rectangle is not maximal.

Proof. The four corners of R lie on a sphere centered at the center O of C, and each lies

at the same distance from the nearest vertex of C. If (a) and (b) are both false, then the

corners must also lie on four parallel edges of C, one on each. It follows that there are two

adjacent corners of R (labeled A and B in Figure 5.7) whose edge is parallel to an edge of

C. But then a small rotation about an axis through O parallel to AB would move all four

vertices off the edges of C into the (open) faces, contrary to Lemma 3.

The Maximal Rectangles

According to Lemma 4, there are only two possibilities to examine.

Situation (a). Suppose that two adjacent corners of the maximal centered rectangle R lie

on the edge e of C (Figure 5.8). Write a for the distance from A to the nearest vertex V of

C. Since AB � 1 � AQ, we see that L D AQ Dp

2 and �L D AB D 1 � 2a. Since

0 � a � 1=2, we find that when 0 � � � 1=p

2

L Dp

2: (5.1)

V

A

B

Q

P

O

Figure 5.8. Situation (a). Figure 5.9. Situation (b).

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4. Prince Rupert’s Rectangles 31

Situation (b). Suppose that two adjacent corners A and B of a centered maximal rect-

angle R of aspect ratio � are on the adjacent open edges eA and eB that share a vertex V

(Figure 5.9). Write a for the distance VA and b for the distance VB . Then

AB Dp

a2 C b2; BP Dp

1 C .1 � a/2 C .1 � b/2: (5.2)

Since OA D OB , it is easy to see that b must be either a or 1�a. Further, AB could be the

smaller dimension �L or the larger dimension L of R. Thus situation (b) splits into four

apparent subcases, one of which proves to be impossible.

Case b1:

b D a and AB D �L. Setting a D b and � D 1 in (5.2), we find that a D 3=4 so that

L D 3p

2=4, which is Nieuwland’s result. More generally, setting a D b, eliminating a,

and writing f1.�/ D L, we learn that when 0 � � � 1

L D 3p3 � �2 C

p2�

D f1.�/ (5.3)

and

a D 3p2

�p3 � �2 C

p2�

: (5.4)

Case b2:

b D 1 � a and AB D �L. Substituting in (5.2), eliminating a, requiring that 0 � a � 1,

and writing f2.�/ D L, we see that for � satisfying 1=p

3 � � � 1=p

2

L D 1p

1 � �2D f2.�/: (5.5)

In this case,

a D 1

2˙ 1

2

r

3�2 � 1

1 � �2:

Case b3:

b D a and AB D L. Eliminating a in (5.2) leads to the equation

.1 � �2/L2 � 2p

2L C 3 D 0:

Writing L D f3.�/, we discover that when 1=p

3 � � � 1

L D 3p

2 Cp

3�2 � 1D f3.�/: (5.6)

(The second root is too large when � < 1.) Then

a D 2 �p

2p

3�2 � 1

1 � �2; (5.7)

and to ensure that a � 1 we require additionally that � � 1=p

2. In other words, the

appropriate domain for f3.�/ in (5.6) is described by 1=p

2 � � � 1:

The apparent possibility with b D 1�a and AB D L leads to an equation for L having

no real solutions, so this possibility can not be realized.

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32 Part I. Geometry

L

ff

f1

23

1

2

max

3

2

10 l1 l2

l

3 4

Figure 5.10. The longer side Lmax in terms of the aspect ration �.

Conclusions

All that remains is to collect our results. We find the following answer to Question 1 of the

introduction (see Figure 5.10).

Theorem 1. Let �1 D 1 �p

2=2 � 0:29289 and �2 D 1 � �1 Dp

2=2 � 0:70711. For

each � with 0 � � � 1, the longer side Lmax.�/ of the largest rectangle Rmax with aspect

ratio � that fits in the cube of side 1 is given as follows:

Lmax.�/ D

8

ˆˆˆ<

ˆˆˆˆ:

3p3 � �2 C

p2�

if 0 � � � �1;

p2 if �1 � � � �2;

3p

2 Cp

3�2 � 1if �2 � � � 1:

(5.8)

Proof. When 0 � � � �1 the longer side Lmax.�/ is the larger of the number f1.�/,

where f1 is the function of case b1, andp

2 (from (5.1)). If �1 � � � �2, Lmax.�/ Dp

2,

because f1.�/ �p

2 on the entire interval Œ�1; �2� and f2.�/ �p

2 whenp

3=3 � � ��2. Similarly Lmax.�/ D f3.�/ for � satisfying �2 � � � 1.

Figure 5.11 shows the positions of the maximal rectangles. When 0 � � � �1, there are

twelve maximal rectangles, all centered and located as shown in Figure 5.11a. Their longer

dimension L and the distance a D VA are given by formulas (5.3) and (5.4), respectively.

If �1 � � � �2, then L Dp

2, and there are six centered maximal rectangles and infinitely

many that are not centered, all located on the diagonal planes of C as shown in Figure 5.8.

If �2 � � � 1 there are again twelve maximal rectangles, all centered and located as

shown in Figure 5.11b, and their longer dimension L and the distance a D VA are given

by the formulas (5.6) and (5.7).

Theorem 1 gives a necessary and sufficient condition for a rectangle to fit in a unit

cube: an a � b rectangle with a � b fits into the unit cube C if and only if a � Lmax.b=a/,

where Lmax.�/ is given by (5.8).

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4. Prince Rupert’s Rectangles 33

A

(a) The case 0 � � � �1. (b) The case �2 � � � 1.

Figure 5.11. Maximal rectangles.

The answer to Question 2 of the introduction follows immediately from this theorem.

Corollary. Given � and � with 0 � � � � � 1, a D � �D � �D box can pass through a

suitable hole in a unit cube C if and only if

D � 1

�Lmax

� �

�

�

:

Proof. The largest L � .�=�/L rectangle that fits in C has L D Lmax.�=�/, so the box can

pass through the cube precisely when �D � L.

Final Remarks

Greg Huber, a physicist at the University of Massachusetts, and Terry J. Ligocki of Lawrence

Berkeley National Laboratory have generated data for the diagonal dmax.�/ of Rmax.�/ by

modeling the problem with an iterative numerical procedure known as Boltzmann simu-

lated annealing. Professor Huber emailed us as follows, “The basic idea comes from sta-

tistical mechanics. The size of the embedded rectangle is taken to be an “energy,” and con-

figurations (orientations of the rectangle) are perturbed and accepted stochastically with

respect to a continuously changing “temperature,” according to an annealing schedule.

Asymptotic convergence to a global extremum is guaranteed under a few strong conditions

on the acceptance function and scheduling, but in practice [we] chose heuristic functions.”

In March 2002 he forwarded to us two hundred computer-generated values of dmax.�/ for

� in Œ0; 1�. The graph of the corresponding values of Lmax.�/ D dmax=p

1 C �2 matches

closely the function described in Theorem 1 and sketched in bold in Figure 5.10.

In response to a problem proposed by Mauldon [9], Chapman [3] provided an algebraic

answer for the question of finding the largest square that fits in the unit cube, thereby

supplying a solution to this well-known problem of intuitive geometry that is “not overly

dependent on geometric intuition,” as the Monthly’s 1995 Problems Editors remarked. An

algebraic argument independent of the fitting lemma can probably be given for Theorem 1

along the same lines.

Finally, it is a pleasure to acknowledge the assistance of Jonathan P. Green of the UIUC

Department of Germanic Languages and Literatures with John Wallis’s Latin.

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34 Part I. Geometry

Bibliography

[1] W. B. Carver, Solution to Problem 3036, Amer. Math. Monthly 32 (1925) 47–49.

[2] ——, Solution to Problem E1225, Amer. Math. Monthly 64 (1957) 114–116.

[3] R. J. Chapman, Solution to Problem 10251, Amer. Math. Monthly 102 (1995) 465–467.

[4] L. R. Ford, Problem E1225, Amer. Math. Monthly 63 (1956) 421.

[5] M. Gardner, The Colossal Book of Mathematics, W. W. Norton, New York, 2001.

[6] F. M. Garnett, Problem 3036, Amer. Math. Monthly 30 (1923) 337.

[7] R. K. Guy and R. J. Nowakowski, Monthly unsolved problems, 1969–1997, Amer. Math.

Monthly 104 (1997) 967–973.

[8] A. Ehrenfeucht, The Cube Made Interesting, Pergamon Press, Oxford, 1964.

[9] J. G. Mauldon, Problem 10251, Amer. Math. Monthly 99 (1992) 782.

[10] D. J. E. Schreck, Prince Rupert’s problem and its extension by Pieter Nieuwland, Scripta Math.

16 (1950) 73–80, 261–267.

[11] J. Wallis, De Algebra Tractatus, 1685, in Opera Mathematica, vol. 2, Oxoniae: E Theatro Shel-

doniano, 1693, pp. 470–471.

[12] J. E. Wetzel, Rectangles in rectangles, Math. Mag. 73 (2000) 204–211.

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IINumber Theory

and

Graph Theory

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6Transcendentals and Early Birds

Martin Gardner

It’s hard to believe that it was not until 1844 that transcendental numbers were known to

exist! But first, a few definitions.

A rational number is one that can written as a=b, where a and b are integers. In decimal

form, rational numbers either terminate (1=4 D :25) or they have a pattern of consecutive

digits that repeat endlessly (1=7 D :142857 142857 142857 : : :).

An irrational number is one that cannot be expressed as a=b where a and b are inte-

gers. In decimal form, it never ends, and it has no pattern of consecutive digits that keep

repeating.

An algebraic number is the root of a polynomial equation with rational coefficients. It

can be rational or irrational. Thus 2 is algebraic because it is the root of such equations as

10x D 20. The square root of 2 is algebraic because it is the root of x2 D 2.

Transcendental numbers, such as � and e, are irrational numbers that are not the roots

of algebraic equations with rational coefficients.

In 1844, the French mathematical Joseph Liouville (1809-1882) first proved the ex-

istence of transcendentals by actually constructing an infinite number of them. Liouville

numbers, as they are known, can be called artificial because Liouville did not find them

anywhere in mathematics. He simply made them up out of whole cloth.

The simplest Liouville number is the binary transcendental shown below:

:110001000000000000000001 : : :

The ones are at positions given by consecutive factorials. The first 1 is at position

1Š D 1, the second one is at position 2Š D 2, the third is at position 3Š D 6, the fourth is at

position 4Š D 24, and so on.

It’s easy to construct artificial numbers that are obviously irrational but not transcen-

dental (like the square root of two). It is not so easy to construct such numbers that are

transcendental. Consider

:101001000100001 : : :

Liouville showed that any number of this form, where the ones can be replaced by any

constant digit from 2 through 9, is transcendental.

It was not long until � , e, and scores of other famous irrationals were shown to be

transcendental. In spite of the great recent progress in proving numbers transcendental,

Reprinted from Math Horizons, Vol. 13, No. 2 (Nov. 2005), pp. 5, 34.

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38 Part II. Number Theory and Graph Theory

many deep unsolved questions remain. For example, e� has been proved transcendental,

but no one yet knows if �e is transcendental or even if � C e or �e or ee or �� are

irrational! All are believed to be transcendental, but until some genius proves otherwise, it

is possible that the sum of � and e is a rational number!

Other than Liouville numbers, the most famous of all artificial transcendentals is made

simply by putting a decimal point in front of the counting numbers, like so:

:123456789101112131415 : : :

This is called Mahler’s number, after Kurt Mahler, a mathematician who first proved

that this counting number is transcendental is all bases. You will find a neat proof for

base 10 in Ivan Niven’sNiven, Ivan marvelous little book Irrational Numbers, recently

republished by the Mathematical Association of America.

A number is normal if every digit, every doublet, every triplet, or any specified pattern

of digits appear the number of times you would expect if the number were generated by a

randomizer. So far, � , e, and the irrational roots of all algebraic numbers have passed all

statistical tests for normalcy, but none has yet been proved normal.

In 1934, it was shown ab is transcendental if a is an algebraic number not 0 or 1, and

b is any irrational number. Thus 2� and 10p

2 are transcendental.

Now for a startling coincidence. Note the last five digits of Mahler’s number. They

are � to four decimals! (I’m indebted to Jaime Poniachik, an Argentine puzzle maker, for

telling me about this.)

Every positive integer obviously appears somewhere in Mahler’s number, but many

numbers show up far ahead of where they normally would occur as a counting number. Pi

turns up so early because 3 is the last digit of 13, followed by 14 and 15. Poniachik points

out that � to 5 decimals appears as early as 141593-141594. Consider 666. Of course it

shows up in the counting sequence, but you can spot it early in the sequence 65-66-67.

Much innocent but totally useless amusement can be had by searching for the earliest

appearance of such familiar numbers as integral powers of roots, the last four digits of

your phone number, your house number, and so on. Better still, see if you can formulate

a procedure for determining the earliest appearance of any given integer in the counting

number.

If a number appears ahead of its position as a counting number, let’s call it an early bird.

The famous year 1492, for another example, is an early bird (491-492). Poniachik points

out that 8192, the thirteenth power of 2, is an early bird (18-19-20). Like � , the transcen-

dental number e, to four decimals (2.7182) is also an early bird (2718-2719), though not

as early as � .

What is the smallest early bird prime? When will the next coming year be an early

bird? Mathematician Solomon W. Golomb, in a personal letter, mentioned that his home

zip code 91011 (9-10-11) is a very early bird. The early bird numbers, Golomb wrote, are

a subset of the counting numbers, and therefore countably infinite. He is convinced that

“almost all” integers are early birds. There are no one-digit early birds, but half (45 out of

90) two-digit numbers are early birds, with rapidly increasing members of the species as

the number of their digits increases.

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7Squaring, Cubing, and Cube Rooting

Arthur T. Benjamin

I still recall the thrill and simultaneous disappointment I felt when I first read Mathematical

Carnival [4] by Martin Gardner. I was thrilled because, as my high school teacher had told

me, mathematics was presented there in a playful way that I had never seen before. I was

disappointed because Gardner quoted a formula that I thought I had “invented” a few years

earlier. I have always had a passion for mental calculation, and the formula (7.1) below

appears in Gardner’s chapter on “Lightning Calculators.” It was used by the mathematician

A. C. Aitken to square large numbers mentally.

Squaring

Aitken took advantage of the following algebraic identity.

A2 D .A � d/.A C d/ C d 2: (7.1)

Naturally, this formula works for any value of d , but we should choose d to be the distance

to a number close to A that is easy to multiply.

Examples To square the number 23, we let d D 3 to get

232 D 20 � 26 C 32 D 520 C 9 D 529:

To square 48, let d D 2 to get

482 D 50 � 46 C 22 D 2300 C 4 D 2304:

With just a little practice, it’s possible to square any two-digit number in a matter of

seconds. Once you have mastered those, you can quickly square three-digit numbers by

rounding up and down to the nearest hundred.

Examples

2232 D 200 � 246 C 232 D 49;200 C 529 D 49;729

9522 D 1000 � 904 C 482 D 904;000 C 2;304 D 906;304:

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 58–63.

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40 Part II. Number Theory and Graph Theory

To do mental calculations of this size, one needs to be quick at multiplying 2-digit and

3-digit numbers by 1-digit numbers, generating the answer from left to right. More details

and examples are given in Secrets of Mental Math [1], which Martin encouraged me to

write. As I prepared a DVD version of this book [2] recently, I learned some new methods

for cubing and cube-rooting. Although much of that material did not end up in the DVD, I

thought it would be of interest to readers of the College Mathematics Journal.

Cubing

To cube a two-digit number in your head quickly, it is worth learning how to multiply two

two-digit numbers that are near each other. I call it the “close together method” and it is

based on the formula

.z C a/.z C b/ D z.z C a C b/ C ab; (7.2)

where z is typically a number that ends in zero.

Example For the problem 23 � 26, z D 20, a D 3, b D 6, leads to

23 � 26 D 20 � 29 C 3 � 6 D 580 C 18 D 598:

Notice that on both sides of (7.2), the two-digit numbers being multiplied have the

same sum (23 C 26 D 49 D 20 C 29).

Example Thus, to do a problem like 88 � 86, since the numbers sum to 174 D 90 C 84,

we can begin with the product 90 � 84:

88 � 86 D 90 � 84 C .�2/ � .�4/ D 7560 C 8 D 7568:

To cube a two-digit number, we can exploit the algebra

A3 D .A � d/A.A C d/ C d 2A:

Example Thus to cube 23, we can do

233 D 20 � 23 � 26 C 32 � 23 D 20 � 598 C 9 � 23 D 11;960 C 207 D 12;167;

where we used the close-together method to do 23 � 26.

But there is a faster way. Since we know that the problem will begin by doing 20 � 29,

let’s build that into the algebra. Writing the problem a different way, to cube z C d , we do

.z C d/3 D zŒz.z C 3d/ C 3d 2� C d 3: (7.3)

Example To cube 23, z D 20 and d D 3 leads to the easier calculation

233 D 20 � Œ20 � 29 C 27� C 33 D 20 � 607 C 33 D 12;140 C 27 D 12;167:

Equation (7.3) is an instance of Horner’s method for polynomial evaluation. Notice that

when cubing a two-digit number, d will always be one of the numbers ˙1; ˙2; ˙3; ˙4; ˙5

and so 3d 2 will always be 3; 12; 27; 48 or 75, which reduces the mental effort. Also notice

that in the first two-digit multiplication, the number z C 3d will always end with a digit

that is three times the original last digit.

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7. Squaring, Cubing, and Cube Rooting 41

Example For 883, by tripling the last digit, we know that we will multiply our rounded

number (90) by the number closest to 88 that ends in 4, namely 84. Thus,

883 D 90 � Œ90 � 84 C 12� C .�2/3 D 90 � 7572 � 23 D 681;480 � 8 D 681;472:

As a side note, the last part of the calculation is not as hard as it looks. By splitting 7572 D7500 C 72,

9 � 7572 D .9 � 7500/ C .9 � 72/ D 67;500 C 648 D 68;148:

To complete the calculation, simply multiply this number by 10, and subtract 8 at the same

time.

The calculation of two-digit cubes this way is surprisingly fast. With practice, the

squares and cubes of two-digit numbers below 50 will be so quick that you will be able to

cube three-digit numbers as well.

Example Exploiting the previous calculations:

3233 D 300 � Œ300 � 369 C 3.232/� C 233

D 300 � Œ110;700 C 1587� C 233

D 3 � 112;287 � 100 C 233

D 33;686;100 C 12;167

D 33;698;267:

Another calculating tip. After the second multiplication, you can almost always say

the millions part of the answer (here, 33 million) and hold the hundreds digit on your

fingers. Here, just raise 1 finger to “hold onto” the hundreds digit so you only have to

remember 686 while you calculate the cube of 23. By the way, if you just want a good

cubing approximation, simply compute z.z.z C 3d//.

For example,

3233 � 369 � 300 � 300 D 33;210;000:

This will come in handy in the calculation of cube roots.

Cube rooting

One of the easiest feats of lightning calculation is determining the cube root of a perfect

cube when the cube root is two digits. The following description comes from Martin Gard-

ner’s classic book, Mathematics, Magic, and Mystery [5].

The cube root demonstration begins by asking members of the audience to

select any number from 1 through 100, cube it, then call out the result. The

performer instantly gives the cube root of each number called. To do the trick

it is necessary first to memorize the cubes of the numbers from 1 through 10.

An inspection of this table reveals that each cube ends in a different digit.

The digit corresponds to the cube root in all cases except 2 and 3 and 7 and

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42 Part II. Number Theory and Graph Theory

Table 7.1. Table of cubes.

1 2 3 4 5 6 7 8 9 10

1 8 27 64 125 216 343 512 729 1000

8. In these four cases the final digit of the cube is the difference between the

cube root and 10.

To see how this information is used by the lightning calculator, let us sup-

pose that a spectator calls out the cube 250047. The last number is a 7 which

tells the performer immediately that the last number of the cube root must be

3. The first number of the cube root is determined as follows. Discard the last

three figures of the cube (regardless of the size of the number) and consider

the remaining figures—in this example they are 250. In the above table, 250

lies between the cubes of 6 and 7. The lower of the two figures—in this case 6

—will be the first figure of the cube root. The correct answer, therefore, is 63.

One more example will make this clear. If the number called out is 19,683,

the last digit, 3, indicates that the last digit of the cube root is 7. Discarding

the final three digits leaves 19, which falls between the cubes of 2 and 3. Two

is the lower number, therefore we arrive at a final cube root of 27.

I learned this trick in a book on mental magic [3] as a high school student, and won-

dered how it might be extended to larger problems, where the cube root was a three-digit

number. I didn’t pursue that question seriously at the time, since the cube of a three-digit

number can be as large as nine digits, and most calculators back then only went up to

eight. Now that calculators with greater capacity are quite common, I have learned two

quick ways to do this problem, using casting out nines and casting out elevens.

Casting out nines

Ask someone to cube a three-digit number, and ask how many digits are in the answer.

(It should be seven, eight, or nine digits.) Ask the volunteer to recite (or write down) the

answer, and you can pretty easily determine the cube root in your head. The first digit and

last digit of the cube root are determined exactly as before. The last digit of the cube root

is uniquely determined by the last digit of the cube. (In fact, the last digit of the cube root

is the last digit of the cube of the last digit of the cube!) For the first digit, we look at the

magnitude of the millions.

Example Let’s find the cube root of the perfect cube 377,933,067. The first digit of the

cube root must be 7 (since 377 lies between 73 D 343 and 83 D 512) and the last digit of

the cube root must be 3 (since the cube ends in 7). Hence the answer must be of the form

7‹3. How do we determine the middle digit? The method I use is simply to add the digits of

the cube, mod 9. Here the digits sum to 45, which is a multiple of 9. This can only happen

if the cube root is a multiple of three. In other words, the cube root must be 723 or 753 or

783. Since 377 is much closer to 343 than 512, we would go with a cube root of 723. For

extra verification, we could estimate 7203 � 700 � 700 � 760 D 372;400;000.

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7. Squaring, Cubing, and Cube Rooting 43

This same approach can be used when the cube is not a multiple of nine, since there

are only two other possible outcomes, as seen in Table 7.2. If the cube reduces to 1 (mod

9), then the cube root reduces to 1 (mod 3). If the cube reduces to 8 (mod 9), then the cube

root reduces to 2 (mod 3).

Table 7.2. Table of cubes mod 9.

n 0 1 2 3 4 5 6 7 8

n3 0 1 8 0 1 8 0 1 8

Example If the 3-digit cube results in 19,248,832, then we know that the cube root is

of the form 2‹8. To determine the middle digit, notice that the cube has digit sum 37,

which reduces to 1 mod 9. Hence the original number is congruent to 1 mod 3, so the

cube root must be either 208 or 238 or 268 or 298. Judging from the proximity of 19 to its

surrounding cubes, 8 and 27, the answer 268 seems most likely. Since 2703 � 300�300�210 D 18;900;000, we have more confidence in our answer of 268.

Casting out elevens

With a little extra mental effort and memory, we can determine the middle digit without

any guesswork, because the cubes of the numbers 0 through 10 are all distinct mod 11, as

shown in Table 7.3.

Table 7.3. Table of cubes mod 11.

n 0 1 2 3 4 5 6 7 8 9 10

n3 0 1 8 5 9 4 7 2 6 3 10

Example revisited For the cube root of 19,248,832, we know that the answer will begin

with 2 and end in 8 as in the last example. By alternately adding and subtracting its digits

from right to left, we see that it reduces to 2 � 3 C 8 � 8 C 4 � 2 C 9 � 1 D 9 .mod 11/:

Hence the cube root must reduce to 4 (mod 11). Since the answer is of the form 2?8, then

2 C 8�‹ D 4 (mod 11) results in an answer of 268, as well. Note that if the cube has 7 or 9

digits, then you can alternately add and subtract the digits in their natural order. If the cube

has 8 digits, you can start with 11 then alternately subtract and add the subsequent digits.

Example To find the number with cube 111,980,168, we see immediately that the answer

will be of the form 4‹2. Casting out 11s from left to right, we compute 1 � 1 C 1 � 9 C8 � 0 C 1 � 6 C 8 D 3. Here, the cube root must reduce to 9 (mod 11). Thus 4 C 2�‹ D 9

.mod 11/ tells us that the missing digit is 8. Hence the cube root is 482.

Between the two approaches, I prefer casting out nines over casting out elevens, since

it is easier to add the digits of the cube as they are called out. I recommend that the reader

play with a few examples using both methods, choose one to master, and cast out the other.

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44 Part II. Number Theory and Graph Theory

Acknowledgment The author wishes to thank the referee for many helpful suggestions,

and especially Martin Gardner for being such an important factor in his life.

Bibliography

[1] A. Benjamin and M. Shermer, Secrets of Mental Math: The Mathemagician’s Guide to Lightning

Calculation and Amazing Math Tricks, Random House, New York, 2006.

[2] A. T. Benjamin, The Secrets of Mental Math (DVD Course), The Teaching Company, Chantilly,

VA, 2011.

[3] T. Corinda, 13 Steps to Mentalism, 4th ed., Tannen Magic Publishing, New York, 1968.

[4] M. Gardner, Mathematical Carnival: A New Round-up of Tantalizers and Puzzles from Scientific

American, Random House, New York, 1965.

[5] ——, Mathematics, Magic, and Mystery, Dover, New York, 1956.

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8Carryless Arithmetic Mod 10

David Applegate, Marc LeBrun, and N. J. A. Sloane

Nim

Forms of Nim have been played since antiquity and a complete theory was published as

early as 1902 (see [3]). Martin Gardner described the game in one of his earliest columns

[7] and returned to it many times over the years ([8]–[16]).

Central to the analysis of Nim is Nim-addition. The Nim-sum is calculated by writing

the terms in base 2 and adding the columns mod 2, with no carries. A Nim position is a

winning position if and only if the Nim-sum of the sizes of the heaps is zero [2], [7].

Is there is a generalization of Nim in which the analysis uses the base-b representa-

tions of the sizes of the heaps, for b > 2, in which a position is a win if and only if the

mod-b sums of the columns is identically zero? One such game, Rimb (an abbreviation of

Restricted-Nim) exists, although it is complicated and not well known. It was introduced in

an unpublished paper [6] in 1980 and is hinted at in [5]. Despite his interest in Nim, Martin

Gardner never mentions Rimb , nor does it appear in Winning Ways [2], which extensively

analyzes Nim variants.

In the present paper we focus on b D 10, and consider, not Rim10 itself, but the arith-

metic that arises if calculations, addition and multiplication, are performed mod 10, with

no carries. Along the way we encounter several new and interesting number sequences,

which would have appealed to Martin Gardner, always a fan of integer sequences.

The Carryless Islands

The fabled, carefree residents of the Carryless Islands in the remote South Pacific have

very few possessions, which is just as well, since their arithmetic is ill-suited to accurate

bookkeeping. When they add or multiply numbers, they follow rules similar to ours, except

that there are no carries into other digit positions. Sociologists explain this by noting that

the Carryless Islands were originally penal colonies, and, as penal institutions are generally

known to have excellent dental care, the islanders were, happily, generally free of carries.

We will use C and � for their operations�, and C and � for the standard operations used

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 43–50.�We prefer not to use outlandish symbols such as | and �, since C and � are perfectly reasonable operations,

although to our eyes they have rather strange properties. As Marcia Ascher remarks, writing about mathematics

in indigenous cultures, “in many cases these cultures and their ideas were unknown beyond their own boundaries,or misunderstood when first encountered by outsiders” [1].

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46 Part II. Number Theory and Graph Theory

by the rest of the world. Addition and multiplication of single-digit numbers are performed

by “reduction mod 10.” Carry digits are simply ignored, so 9C4 D 3, 5C5 D 0, 9C4 D 6,

5�4 D 0, and so on. Adding or multiplying larger numbers also follows the familiar

procedures, but again with the proviso that there are no carries. For example, adding 785

and 376 produces 51, and the product of 643 and 59 is 417 (see Figure 8.1).

(a) 7 8 5

C 3 7 6

0 5 1

(b) 6 4 3

� 5 9

4 6 7

0 0 5

0 4 1 7

Figure 8.1. (a) Carryless addition. (b) Carryless multiplication.

What does elementary number theory look like on these islands? Let’s start with the

carryless squares n�n. For n D 0; 1; 2; 3 we get 0, 1, 4, 9, Then for n > 3 we have

4�4 D 6, 5�5 D 5, 6�6 D 6, 7�7 D 9, 8�8 D 4, 9�9 D 1, 10�10 D 100; : : : ; giving

the sequence

0; 1; 4; 9; 6; 5; 6; 9; 4; 1; 100; 121; 144; 169; 186; 105; 126; 149; 164; : : : :

It turns out that this is entry A059729 in the OEIS [17], contributed by Henry Bottomley

on February 20, 2001, although without any reference to earlier work on these numbers.

Bottomley also contributed sequence A059692, giving the carryless multiplication table,

and several other sequences related to carryless products. Likewise the sequence of values

of nCn,

0; 2; 4; 6; 8; 0; 2; 4; 6; 8; 20; 22; 24; 26; 28; 20; 22; 24; 26; 28; 40; 42; : : : ;

is entry A004520, submitted to the OEIS by one of the present authors around 1996, again

without references. (If these numbers are sorted and duplicates removed, we get the carry-

less “evenish” numbers, that is, numbers all of whose digits are even, A014263.) Carryless

arithmetic must surely have been studied before now, but the absence of references in [17]

suggests that it is not mentioned in any of the standard texts on number theory.

The carryless primes

If we require that a prime � is a number whose only factorization is 1 times itself, we are

out of luck, since every carryless number is divisible by 9, and there would be no primes

at all. (For 9�1 D 9, 9�2 D 8, 9�3 D 7; : : : ; 9�9 D 1. So if we construct a number � by

replacing all the 1’s in � by 9’s, all the 2’s by 8’s, : : : then � D 9��, and � would not be

a prime.)

There are primes, when defined in the right way. Since 1�1 D 1, 3�7 D 1 and 9�9 D1, all of 1; 3; 7 and 9 divide 1 and so divide any number. We call 1; 3; 7 and 9 units, the usual

name for integers that divide 1. Units should not be counted as factors when considering if

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8. Carryless Arithmetic Mod 10 47

a number is prime (just as factors of �1 are ignored in ordinary arithmetic: 7 D .�1/�.�7/

doesn’t count as a factorization when considering if 7 is a prime).

So we define a carryless prime to be a non-unit � whose only factorizations are of the

form � D u�� where u is a unit. Computer experiments suggest that the first few primes

are

21; 23; 25; 27; 29; 41; 43; 45; 47; 49; 51; 52; 53; 54; 56; 57; 58; 59; 61; 63; : : : ; (8.1)

but there are surprising omissions in this list, resulting from some strange factorizations:

2 D 2�51, 10 D 56�65, 11 D 51�61. It is hard to be sure at this stage that the above list

is correct, since there exist factorizations where one of the numbers is much larger than the

number being factored, such as 2 D 4�5005505553. One property that makes carryless

arithmetic interesting is the presence of zero-divisors: the product of two numbers can be

zero without either of them being zero: 2�5 D 0, 628�55 D 0. Perhaps 21 is the product of

two really huge numbers? Nonetheless, the list is correct, as we will see (it is now sequence

A169887 in [17]).

Algebra to the rescue

The secret to understanding carryless arithmetic is to introduce a little algebra. Let R10 de-

note the ring of integers mod 10, and R10ŒX� the ring of polynomials in X with coefficients

in R10. Then we can represent carryless numbers by elements of R10ŒX�: 21 corresponds

to 2X C 1, 109 to X2 C 9, and so on. Carryless addition and multiplication are simply

addition and multiplication in R10ŒX�: our first example,

785C376 D 51

corresponds to

.7X2 C 8X C 5/ C .3X2 C 7X C 6/ D 5X C 1;

where the polynomials are added or multiplied in the usual way, and the coefficients then

reduced mod 10. Conversely, any element of R10ŒX� represents a unique carryless number

(just set X D 10 in the polynomial). In fact arithmetic in R10ŒX� is clearly exactly the

same as the arithmetic of carryless numbers. This could be used as a formal definition of

carryless arithmetic mod 10. It also shows that this arithmetic is commutative, associative

and distributive.

Since R10ŒX� is a ring, we can not only add and multiply, we can also subtract, some-

thing the Carryless Islanders never considered. The negatives of the elements of R10 are

�1 D 9, �2 D 8; : : : ; �9 D 1, and similarly for the elements of R10ŒX�. So the nega-

tive of a carryless number is its “10’s complement,” obtained by replacing each nonzero

digit d by 10 � d , for example �702 D 308. To subtract A from B , we add �A to B:

650�702 D 650C308 D 958. This is equivalent to doing elementary school subtraction

where we can “borrow” but don’t have to pay back!

The units in R10ŒX�, that is, the elements that divide 1, are the constants 1; 3; 7; 9, and

the carryless primes that we defined are the irreducible elements in R10ŒX�, that is, non-

units f10.X/ 2 R10ŒX� whose only factorizations are of the form f10.X/ D ug10.X/,

where u is a unit and g10.X/ 2 R10ŒX�. The units can also be written as 1; �1; 3 and �3,

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48 Part II. Number Theory and Graph Theory

which more closely relates them to the units 1 and �1 in ordinary arithmetic (3 and �3 act

in some ways like the imaginary units i and �i , squaring to �1, for example).

The key to further progress is to notice that R10 is the direct sum of the ring R2 of

integers mod 2 and the ring R5 of integers mod 5. Given r10 2 R10, we read it mod 2 and

mod 5 to obtain a pair Œr2; r5� with r2 2 R2, r5 2 R5. The elements 0; 1; : : : ; 9 2 R10

(or equivalently the carryless digits 0; 1; : : : ; 9) and their corresponding pairs Œr2; r5� are

given by the following table. The Chinese Remainder Theorem guarantees that this is a

one-to-one correspondence.

0 1 2 3 4 5 6 7 8 9

[0,0] [1,1] [0,2] [1,3] [0,4] [1,0] [0,1] [1,2] [0,3] [1,4](8.2)

As a check, we note that f1g is the (singleton) set of units in R2, while f1; 2; 3; 4g is the set

of units in R5, so the pairs Œ1; 1�, Œ1; 2�, Œ1; 3� and Œ1; 4� correspondingly produce the units

1, 7, 3 and 9 of R10.

Similarly, polynomials f10.X/ 2 R10ŒX� correspond to pairs of polynomials Œf2.X/;

f5.X/�, obtained by reading f10.X/ respectively mod 2 and mod 5. Conversely, given any

such pair of polynomials Œf2.X/; f5.X/�, there is a unique f10.X/ 2 R10ŒX� that corre-

sponds to them, which can be found using (8.2). We indicate this by writing f10.X/ $Œf2.X/; f5.X/�. If also g10.X/ $ Œg2.X/; g5.X/�, then

f10.X/ C g10.X/ $ Œf2.X/ C g2.X/; f5.X/ C g5.X/�

and

f10.X/g10.X/ $ Œf2.X/g2.X/; f5.X/g5.X/�:

We are now in a position to answer many questions about carryless arithmetic.

The carryless primes, again

What are the irreducible elements f10.X/ 2 R10ŒX�? If f10.X/ $ Œf2.X/; f5.X/� is

irreducible then certainly f2 and f5 must be either units or irreducible, for if f2 D g2h2

then we have the factorization Œf2; f5� D Œg2; f5�Œh2; 1�. Also Œf2; f5� D Œf2; 1�Œ1; f5�, so

one of f2, f5 must be irreducible and the other must be a unit. So the irreducible elements

in R10ŒX� are of the form Œf2.X/; u�, where f2.X/ is an irreducible polynomial mod 2

of degree � 1 and u 2 f1; 2; 3; 4g, together with elements of the form Œ1; f5.X/�, where

f5.X/ is an irreducible polynomial mod 5 of degree � 1.

The irreducible polynomials mod 2 are X , X C 1, X2 C X C 1; : : : ; and the irre-

ducible polynomials mod 5 are uX , uX C v; : : : ; where u; v 2 f1; 2; 3; 4g (see entries

A058943, A058945 in [17]). The first few irreducible elements in R10ŒX� are therefore

ŒX; 1�, ŒX; 2�, ŒX; 3�, ŒX; 4�, ŒX C1; 1�, ŒX C1; 2�; : : : ; and Œ1; X�, Œ1; 2X�, Œ1; 3X�, Œ1; 4X�,

Œ1; X C 1�, Œ1; 2X C 1�; : : : : The corresponding carryless primes, according to (8.2), are

56; 52; 58; 54; 51; 57; : : : ; and 65; 25; 85; 45; 61; 21; : : : : And so we can verify that the list

in (8.1) is correct.

We will call a number with at least two digits in which all digits except the rightmost

are even but the rightmost is odd an e-type number (A143712), and a number with at

least two digits in which all digits except the rightmost are 0 or 5 and the rightmost is

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8. Carryless Arithmetic Mod 10 49

neither 0 nor 5 an f-type number (A144162). Similarly, we call the primes corresponding

to the irreducible elements Œ1; f5.X/� e-type primes, and the primes corresponding to the

irreducible elements Œf2.X/; u� f-type primes.

We also see that our earlier concern about the primality of 21 was groundless. It is

impossible for the length (in decimal digits) of a nonzero carryless product to be less than

the length of both of the factors. This follows from the fact that if `.n/ is the number of

decimal digits in the number n > 0 corresponding to a pair Œf2.X/; f5.X/�, then `.n/ D1 C maxfdeg f2; deg f5g. So if mn > 0, `.mn/ � minf`.m/; `.n/g.

Also, since we know how many irreducible polynomials mod 2 and mod 5 there are of

given degree (see A001037, A001692 in [17]), we can write down a formula for the number

of k-digit carryless primes, something that we cannot do for ordinary primes, namely

4

k � 1

X

d dividesk�1

�

�k � 1

d

�

.2d C 5d /;

for k � 2, where � is the Mobius function (A008683). There are 28 primes with two digits

(the twenty listed in (8.1), together with 65; 67; 69; 81; 83; 85; 87; 89), 44 with three digits,

: : : (A169962). For large k the number is about 4 � 5k�1=.k � 1/, whereas the number of

ordinary primes with exactly k digits is much larger, about 9�10k�1=.k log 10/, so carryless

primes are much rarer than ordinary primes.

Incidentally, the prime ideals in R10ŒX�, as distinct from the irreducible elements, all

have a single generator, which is one of Œ0; 1�; Œ1; 0�; Œ1; 1�; Œf2.X/; 1�; Œ1; f5.X/�, where

f2.X/, f5.X/ are irreducible (cf. [18, Chap. III, Thm. 30]).

The carryless squares, again

Squaring a mod 2 polynomial is easy: f2.X/2 D f2.X2/. So if n corresponds to the

pair Œf2.X/; f5.X/�, n2 corresponds to Œf2.X2/; f5.X/2� D Œf2.X2/; 0� C Œ0; f5.X/2�.

This gives a two-step recipe for producing all carryless squares. First find, using (8.2), the

carryless number m corresponding to Œ0; f5.X/2�, where f5.X/ is any polynomial mod 5.

The effect of adding a nonzero Œf2.X2/; 0� changes some subset of the digits in positions

0; 2; 4; : : : of m by the addition of 5 mod 10.

For example, if f5.X/ D X C 2, f5.X/2 D X2 C 4X C 4, and by (8.2) Œ0; f5.X/2 �

corresponds to the carryless square m D 644. We now add 5 mod 10 to any subset of

the digits in positions 0; 2; 4; 6; : : : of m (considering m extended by prefixing it with

any number of zeros), obtaining infinitely many squares 644; 649; 144; 149; : : : ;

50644; 5050649; : : : :

This also leads to a formula for the number of k-digit carryless squares. For even k the

number is 0, and for odd k it is

1

29 � 10.k�1/=2 C 2.k�3/=2

(zero is excluded from the count). There are five squares of length 1 (namely 1; 4; 5; 6 and

9), 46 of length 3, : : : (see A059729, A169889, A169963). For large odd k there are about

twice as many k-digit carryless squares as ordinary squares.

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50 Part II. Number Theory and Graph Theory

Divisors and factorizations

What about the factorization of numbers into the product of carryless primes? Unfortu-

nately, the existence of zero-divisors complicates matters, and it turns out that there is no

natural way to define, for example, an analog of the usual sum-of-divisors function �.n/.

In our analysis we define several classes of carryless numbers:

U WD f1; 3; 7; 9g, the units,

E WD f0; 2; 4; 6; 8; 20; 22; : : :g, the “evenish” numbers, in which all digits are even

(A014263),

F WD f0; 5; 50; 55; : : :g, the “fiveish” numbers, in which all digits are 0 or 5 (A169964),

Z WD E [ F D f0; 2; 4; 5; 6; 8; 20; 22; : : :g, the zero-divisors (A169884),

N WD f1; 3; 7; 9; 10; 11; 12; 13; : : :g, the positive numbers not in Z (A169968).

Suppose d is a carryless divisor of n, that is, there is a number q such that d�q D n.

What can be said about the possible choices for q? One can show—we omit the straight-

forward proofs—that

� if d 2 N then there is a unique q,

� if d 2 E then d�q0 D n if and only if q0 D q C v for some v 2 F ,

� if d 2 F then d�q0 D n if and only if q0 D q C e for some e 2 E .

The same distinctions are needed to describe factorizations into primes.

� If n 2 N then n has a unique factorization as a carryless product of primes, up to

multiplication by units. For example, we already saw 10 D 56�65. But we also

have 10 D .3�56/�.7�65/ D 58�25 D .9�56/�.9�65/ D 54�45 D 9�52�25,

etc., illustrating the nonuniqueness. Also 11 D 51�61; 101 D 21�29�51, 1234 D23�23�23�51�51�52. It follows that any non-unit in N can be written both as

e�f and e0Cf 0, where e and e0 are e-type numbers and f and f 0 are f-type num-

bers. For example, 12 D 81�52 D 61C51.

� If n 2 E then n has a unique factorization as 2 times a product of e-type primes, up

to multiplication by units (in this case, every f-type prime divides n). For example,

20 D 2�65, 22 D 2�61, 2468 D 2�69�69�69.

� If n 2 F then n has a unique factorization as 5 times a product of f-type primes, up

to multiplication by units (in this case, every e-type prime divides n). For example,

50 D 5�52, 505 D 5�51�51.

Here are the analogous statements about divisors:

� if n 2 N , n has only finitely many divisors. If d divides n and u 2 U , then d�u

divides n. The divisors may be grouped into equivalence classes d�U . Since the sum

of the elements of U is zero, so is the sum of the divisors of n.

� if n 2 E , d divides n, u 2 U and v 2 F , then d�u C v divides n. So n has infinitely

many divisors, belonging to equivalence classes d�U C F .

� if n 2 F , d divides n, u 2 U and e 2 E , then d�u C e divides n. So n has infinitely

many divisors, belonging to equivalence classes d�U C E .

Any attempt to define a sum-of-divisors function must specify how to choose repre-

sentatives from the equivalence classes. There seems to be no natural way to do this. One

possibility would be to choose the smallest decimal number in each class, but this seems

unsatisfactory (since it depends on the ordering of decimal numbers, another concept the

islanders seem not to be familiar with).

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8. Carryless Arithmetic Mod 10 51

Further number theory

In summary, we can help the Carryless Islanders by defining subtraction, prime numbers,

and factorization into primes. But further concepts such as the number of divisors, the sum

of divisors and perfect numbers seem to lie beyond these Islands.

However, many other carryless analogs are well-defined, including including triangu-

lar numbers (A169890), cubes (A169885), partitions (A169973), greatest common divi-

sors and least common multiples, and so on. Some seem exotic, while other familiar se-

quences simply become periodic. For example, the analog of the Fibonacci numbers co-

incides with the sequence of Fibonacci numbers read mod 10, A003893, which becomes

periodic with period 60 (the periodicity of the Fibonacci numbers to any modulus being

a well-studied subject, see sequence A001175). Similarly, the analogue of the powers of

2 (A000689) becomes periodic with period 4. We might also generalize beyond simple

squares, cubes, etc., and investigate the properties of polynomials or power series based on

carryless operations—How do these factor? What are their fixed points?—and so on.

Taking a different tack, carryless mod 10 partitions are enumerated in A169973, which

may be derived as the coefficients of zn in the formal expansion of the analog of the classic

partition generating functionQ1

kD1.1 C zk/, wherein powers of z are multiplied together

by combining their exponents with carryless mod 10 addition instead of the ordinary sum.

Afterword

There’s a great deal yet to be explored in these Carryless Islands! Watch for our next

paper on another carryless arithmetic, in which operations on single digits are defined by

a ˚ b D maxfa; bg, a ˝ b D minfa; bg. We call this “dismal arithmetic.”

When the Handbook of Integer Sequences was published 39 years ago, Martin Gardner

was kind enough to write in his Mathematical Games column of July 1974 that “every

recreational mathematician should buy a copy forthwith.” That book contained 2372 se-

quences: today its successor, the On-Line Encyclopedia of Integer Sequences (or OEIS)

[17], contains nearly 200,000 sequences. We were about to write to Martin about carryless

arithmetic when we heard the sad news of his death. This article, the first of a series on

various kinds of carryless arithmetic, is offered in his honor.

Bibliography

[1] Marcia Ascher, Mathematics Elsewhere: An Exploration of Ideas Across Cultures, Princeton

Univ. Press, Princeton, NJ, 2002.

[2] E. R. Berlekamp, J. H. Conway, and R. K. Guy, Winning Ways for Your Mathematical Plays,

A K Peters, Wellesley MA, 2nd ed., 4 vols, 2004.

[3] C. L. Bouton, Nim, a game with a complete mathematical theory, Ann. Math. 3 (1902) 35–39;

available at dx.doi.org/10.2307/1967631

[4] J. H. Conway, On Numbers and Games, Academic Press, NY, 1976.

[5] T. S. Ferguson, Some chip transfer games, Theoret. Comput. Sci. 191 (1998) 157–171; available

at dx.doi.org/10.1016/S0304-3975(97)00135-7

[6] J. A. Flanigan, NIM, TRIM and RIM, unpublished document, Mathematics Department, Uni-

versity of California at Los Angeles, 1980; available at

citeseer.ist.psu.edu/viewdoc/summary?doi=10.1.1.74.955.

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52 Part II. Number Theory and Graph Theory

[7] Martin Gardner, Nim and Tac Tix, The Scientific American Book of Mathematical Puzzles and

Diversions, Simon & Schuster NY, 1959.

[8] ——, Jam, Hot and Other Games, Mathematical Carnival, Vintage Books NY, 1977.

[9] ——, Nim and Hackenbush, Wheels, Life and Other Mathematical Amusements, W. H. Free-

man NY, 1983.

[10] ——, Sim, Chomp and Race Track, Knotted Doughnuts and Other Mathematical Entertain-

ments, W. H. Freeman NY, 1986.

[11] ——, Dodgem and Other Simple Games, Time Travel and Other Mathematical Bewilderments,

W. H. Freeman NY, 1988.

[12] ——, Wythoff’s Nim, Penrose Tiles to Trapdoor Ciphers : : : and the Return of Dr. Matrix,

W. H. Freeman NY, 1989.

[13] ——, Matches, Mathematical Circus, Mathematical Association of America, Washington DC,

revised ed., 1992.

[14] ——, The Rotating Table and Other Problems, Fractal Music, Hypercards and More : : : ; W. H.

Freeman NY, 1992.

[15] ——, Lavinia Seeks a Rule and Other Problems, The Last Recreations:Hydras, Eggs and Other

Mathematical Mystifications, Springer NY, 1997.

[16] ——, Surreal Numbers, The Colossal Book of Mathematics, W. W. Norton NY, 2001.

[17] The OEIS Foundation Inc., The On-Line Encyclopedia of Integer Sequences (2011); available

at oeis.org.

[18] O. Zariski and P. Samuel, Commutative Algebra, Van Nostrand NY, vol. I, 1958.

The Mad Tea Party Ride

Alice and friends begin a mathematical adventure (see next page).

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9Mad Tea Party Cyclic Partitions

Robert Bekes, Jean Pedersen, and Bin Shao

Alice, the March Hare, the Hatter, and the Dormouse, were standing by the Mad Tea Party

(MTP) ride (see opposite page). “We’ll start with the same number of teacups as people,”

said the Hatter, bossy as usual. “The teacups are arranged in a circle, and each person sits

in his or her own teacup.”

“Won’t that be lonely?” objected Alice.

“Don’t worry,” replied the Hatter, “At the end of this ride, everyone stands up, one of

the teacups is removed, then everyone finds a new place to sit. Every teacup has to be

occupied by at least one person, so on the second ride there will be one teacup with two

people in it. After that ride, another cup will disappear and eventually, when there is just

one teacup, no one will be lonely.”

The rest said in unison, “You must be mad. Everyone will be squished and it will be

quite unbearable.”

“Not my problem,” the Hatter said, and walked away.

“Wait,” said Alice, who was a keen amateur mathematician, “If there are n people, how

many ways are there to fit them into k cups arranged in a circle, assuming, naturally, all

people are alike?”

She quickly wrote out the possibilities for n D 5 and k � 5 in Table 9.1, where

the actual partitions are in the second row and the number of partitions are in the third

row (consistent with the results in [7]). Alice explained to the group that the sum of the

numbers in the third row (which is 7) is the answer to her question. Then she noticed that

product of the numbers in the third row (which is 4) tells us the number of different paths

through the various partitions available at each stage. (This paper obtains both results, but

concentrates on the nature and number of the actual partitions.)

Table 9.1.

k D Number of cups 5 4 3 2 1

The listing of partitions for k cups 1 1 1 1 1 1 1 1 2 1 1 3 or 1 2 2 1 4 or 2 3 5

The number of paritions for k cups 1 1 2 2 1

But finding them became much tougher with more teacups. “This is hard,” the March

Hare said, “Martin Gardner wrote several times about partitions [5]–[7]. The scariest part,

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 25–36.

53

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54 Part II. Number Theory and Graph Theory

for us, is in [7] where Gardner wrote:

Although there are many recursive procedures for counting unordered parti-

tions, using at each step the number of known partitions for all smaller num-

bers, an exact asymptotic formula was not obtained until recent times. The

big breakthrough was made by the British mathematician G. H. Hardy, work-

ing with his Indian friend Srinivasa Ramanujan. Their not quite exact for-

mula was perfected by Hans A. Rademacher in 1937. The Hardy-Ramanujan-

Rademacher formula is a horribly shaggy infinite series that involves (among

other things) pi, square roots, complex roots, and derivatives of hyperbolic

functions!

So, how can we find the answer to an even more difficult problem?”

Now Alice had read the paper-folding results of Hilton and Pedersen ([4, pp. 129–

136], [5, pp. 107–109], and [6]) and thought to apply their ideas. A recursive algorithm

would do. Though it wouldn’t give a closed form, it would be, conceptually, far sim-

pler than the Hardy-Ramanujan-Rademacher formula. “Sometimes it’s easier to solve a

more difficult problem and use that solution to answer the easier problem,” Alice re-

minded the others, “That’s what we should do here. The first thing we need to do is

give our cyclic arrangements a name and decide on an appropriate notation. Let’s call

them MTP partitions. Our problem is, how many MTP partitions are there for any positive

integer n?”

For notation we will write, as an example, .2; 1; 3/ to mean there are 2, 1, and 3 people

in three cups arranged clockwise in a circle beginning with a teacup containing 2 people.

Sometimes, when no confusion occurs, instead of .2; 1; 3/ we will simply write 2 1 3.

The March Hare commented that the partition .1; 3; 2/ must be the same as .1; 2; 3/,

because if you turned the MTP ride upside down they would be the same. “No way.” the

Hatter responded contemptuously, “Disney won’t turn the ride upside down. And if you

can’t do that, then these two partitions are different. People in teacups on the left and right

side of the teacup with 2 people in it are clearly different in these 2 cases.”

Notice that in Table 9.1 the partition 1 1 1 2 would be the same as 1 1 2 1, 1 2 1 1, or

2 1 1 1, because they are cyclic permutations of each other.

The modest Dormouse said, “All I know how to do is subtract odd numbers from odd

numbers, divide even numbers by 2, and add numbers that sum to less than 100.”

“That’s great!” said Alice, ”you have just the mathematical skills we need to tackle this

problem.” And she began to explain:

First, I’ll teach you, with some examples, how to construct coaches, and how they

constitute symbols. We always start with an odd number b and use all the odd numbers

ai , where ai < b2

. You’re going to wonder what on earth these symbols are about—

but I can’t take time to discuss that; you simply must consult [4]–[6] if you want to

know more.

Second, I’ll describe how to construct the coaches in symbols in purely mathematical

language. After more examples I’ll state two theorems about the coaches and symbols, give

a reconstruction algorithm, and observe some persistent patterns among coaches in two

special sequences. All this information gives us a non-recursive algorithm that answers our

main question: How many MTP partitions are there for any positive integer n?

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9. Mad Tea Party Cyclic Partitions 55

Constructing coaches and symbols

We are going to show how to write a coach that involves an initial odd number b and odd

ai , such that ai < b2

. As an example, I’ll begin with b D 35. The possible values for ai

are 1; 3; 5; 7; 9; 11; 13; 15 and 17. We begin with the smallest value, which we call a1 and

write the following

b D 35

ˇˇˇˇ

a1 D 1:

Now subtract 1 from 35 and divide the answer (34) as many times as possible by 2 (in

this case, once), obtaining remainder 17. The number of factors of 2 that were divided out

(1) is written at the bottom of the first column. Then 17 is written at the top of next column.

This gives

b D 35

ˇˇˇˇ

a1 D 1 a2 D 17

k1 D 1:

We repeat this process (subtracting ai from b and dividing by factors of 2 until we get

an odd number) with a2; a3; : : : until the next value of ai is equal to the original a1 D 1 (in

this case), as must happen (see [4] or [5]). In the last step we subtract 3 from 35 and divide

the answer (32) as many times as possible by 2 (5 times), obtaining 1. We write 5 under

the 3 and stop, since 1 is our original number a1 for this coach. This completed verbose

coach is:

b D 35

ˇˇˇˇ

a1 D 1 a2 D 17 a3 D 9 a4 D 13 a5 D 11 a6 D 3

k1 D 1 k2 D 1 k3 D 1 k4 D 1 k5 D 3 k5 D 5

ˇˇˇˇ:

Eliminating the labels ai ; ki ; we have

b D 35

ˇˇˇˇ

1 17 9 13 11 3

1 1 1 1 3 5

ˇˇˇˇ:

In sum, the arithmetic for this coach is

35 � 1 D 34 D 21 � 17; k1 D 1; a2 D 17I35 � 17 D 18 D 21 � 9; k2 D 1; a3 D 9I35 � 9 D 26 D 21 � 13; k3 D 1; a4 D 13I35 � 13 D 22 D 21 � 11; k4 D 1; a5 D 11I35 � 11 D 24 D 23 � 3; k5 D 3; a6 D 3I35 � 3 D 32 D 25 � 1; k6 D 5; a7 D 1 D a1 .StopŠ/:

Now not all the odd numbers ai < 352

have appeared in the top row of this coach. So

we begin again, using the smallest unused ai (so we begin with a1 D 5), and construct the

next coach of the symbol obtaining:

b D 35

ˇˇˇˇ

5 15

1 2

ˇˇˇˇ:

The arithmetic for this coach is:

35 � 5 D 30 D 21 � 15; k1 D 1; a2 D 15I35 � 15 D 20 D 22 � 5; k2 D 2; a3 D 5 D a1 .StopŠ/:

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56 Part II. Number Theory and Graph Theory

But 7 still hasn’t appeared in the top row of a coach, so we begin again, with a1 D 7, and

construct the next coach of the symbol obtaining:

b D 35

ˇˇˇˇ

7

2

ˇˇˇˇ:

The arithmetic for this coach is:

35 � 7 D 28 D 22 � 7; k1 D 2; a2 D 7 D a1 .StopŠ/:

Now that all possible ai s have appeared in the top row of some coach we have all the

possible coaches. We combine them to obtain the complete symbol for b D 35:

b D 35

ˇˇˇˇ

1 17 9 13 11 3

1 1 1 1 3 5

ˇˇˇˇ

5 15

1 2

ˇˇˇˇ

7

2

ˇˇˇˇ:

In order to make our symbols unique we will always require that each coach have the

smallest available ai in the position of a1. In general, each coach of a symbol is written:

b

ˇˇˇˇ

a1 a2 � � � ar

k1 k2 � � � kr

ˇˇˇˇ; (?)

where b and ai are odd, with each ai < b2

, and

b � ai D 2ki aiC1; i D 1; 2; : : : ; r; arC1 D a1: (??)

Note that ki is maximal (that is, all powers of 2 have been factored from b � ai /, since

aiC1 must be odd.

In [4]–[6] it is shown that, given any two odd numbers a and b, with a < b2

, there is

always a completely determined, unique symbol consisting of coaches as determined by

(?) and .??); sometimes a symbol may consist of only one coach.

If for any i , gcd.b; ai / D 1, we say that the coach (?) is reduced, and if there are no

repeats among all the ai ’s we say that the coach (?) is contracted. At this stage, we do not

assume that our coaches are reduced, but we do assume they are contracted.

The Dormouse said, “Aren’t all coaches contracted because of the way we constructed

them?” “Yes,” said Alice, “but later we’ll have good reason to construct non-contracted

coaches.”

Alice produces examples and theorems

If a coach in a symbol is contracted, but not reduced (allowing gcd.b; ai / ¤ 1) we call that

coach and the symbol that contains it hybrid. If all the coaches in a symbol are contracted

and reduced we call it pure. Now let’s look at some examples (including b D 35) of pure

and hybrid symbols. “Look for patterns,” Alice instructed the Dormouse and the March

Hare. “See if you can anticipate the theorems. There is a special reason for singly- or

doubly-underlining some coaches and boxing some entries in Figure 9.1.” Can you see

why?

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9. Mad Tea Party Cyclic Partitions 57

b D 23

ˇˇˇˇ

1 11 3 5 9 7

1 2 2 1 1 4

ˇˇˇˇ

b D 35

ˇˇˇˇ

1 17 9 13 11 3

1 1 1 1 3 5

ˇˇˇˇ

5 15

1 2

ˇˇˇˇ

7

2

ˇˇˇˇ

b D 41

ˇˇˇˇ

1 5 9

3 2 5

ˇˇˇˇ

3 19 11 15 13 7 17

1 1 1 1 2 1 3

ˇˇˇˇ

b D 57

ˇˇˇˇ

1 7 25

3 1 5

ˇˇˇˇ

3 27 15 21 9

1 1 1 2 4

ˇˇˇˇ

5 13 11 23 17

2 2 1 1 3

ˇˇˇˇ

19

1

ˇˇˇˇ

b D 63

ˇˇˇˇ

1 31

1 5

ˇˇˇˇ

3 15

2 4

ˇˇˇˇ

5 29 17 23

1 1 1 3

ˇˇˇˇ

7

3

ˇˇˇˇ

9 27

1 2

ˇˇˇˇ

11 13 25 19

2 1 1 2

ˇˇˇˇ

21

1

ˇˇˇˇ

b D 65

ˇˇˇˇ

1

6

ˇˇˇˇ

3 31 17

1 1 4

ˇˇˇˇ

5 15 25

2 1 3

ˇˇˇˇ

7 29 9

1 2 3

ˇˇˇˇ

11 27 19 23 21

1 1 1 1 2

ˇˇˇˇ

13

2

ˇˇˇˇ

b D 91

ˇˇˇˇ

1 45 23 17 37 27

1 1 2 1 1 6

ˇˇˇˇ

3 11 5 43

3 4 1 4

ˇˇˇˇ

7 21 35

2 1 3

ˇˇˇˇ

ˇˇˇˇ

9 41 25 33 29 31 15 19

1 1 1 1 1 2 2 3

ˇˇˇˇ

13 39

1 2

ˇˇˇˇ

Figure 9.1. Examples of symbols.

Did you notice that:

� Where the symbol has no underlined coaches, b is prime?

� The underlined coaches are not reduced, furthermore there is a common divisor

(other than 1) between all of the numbers in the top row (the ai ) and b?

� Each of the boxed numbers divides into b and all the ai for that coach?

� In the doubly-underlined hybrid coaches the sum of the bottom row is a proper divi-

sor of the sum of the bottom row for the pure coaches?

If we eliminate the underlined coaches in Figure 9.1, those remaining constitute pure

symbols (gcd.b; ai / D 1). For a given b what do you notice about the sum of the ki s in

the coaches of the pure symbols? For a given b what is r (the number of columns) in each

pure coach? These are crucial questions to attempt to answer before going on.

Your observations may help you to understand the following (proved in [4]–[6]):

The Quasi-Order Theorem. Suppose that a coach of a pure symbol has the form (?),

and formula (??) holds. LetPr

iD1 ki D k. Then k is the smallest positive integer such that

2k � ˙1 .mod b/. In fact, 2k � .�1/r .mod b/.

In other words the sum of the numbers, ki , in the bottom row of any coach of a pure

symbol is the smallest number, k, such that b exactly divides 2k � .�1/r .

For the pure symbols in Figure 9.1 the Quasi-Order Theorem results are in Table 9.2.

Notice that the conclusion of the Quasi-Order Theorem can be obtained from any coach

of the pure symbol since k is the same and r is always either even or odd.

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58 Part II. Number Theory and Graph Theory

Table 9.2.

b 23 35 41 57 63 65 91

k 11 12 10 9 6 6 12

r 6 6 3 or 7 3 or 5 2 or 4 1; 3; or 5 4; 6; or 8

2k � .1/r 4047 4096 1025 513 63 65 4095

Here is another pertinent Hilton-Pedersen theorem [4] connected with the Quasi-Order

Theorem.

The Coach Theorem. Let � be the Euler phi function. In a complete, pure symbol, withPr

iD1 ki D k, in .?/, and c the number of coaches, we have �(b/ D 2kc.

Recall that if p is a prime then �.pn/ D .p � 1/pn�1, and if p1, p2 are distinct prime

numbers then �(pn1

1 � pn2

2 / D �.pn1

1 /�.pn2

2 ). The data confirming that �.b/ D 2kc for

the pure symbols in Figure 9.1 is in Table 9.3.

Table 9.3.

b 23 35 41 57 63 65 91

k 11 12 10 9 6 6 12

c 1 1 2 2 3 4 3

�.b/ 22 24 40 36 36 48 72

The Dormouse complained bitterly that Alice had said they knew enough arithmetic to

work this problem and now she was describing unbelievable things—things about divisi-

bility, primes, and other stuff. Alice said that she had once been told by the White Queen

that, “Why, sometimes I’ve believed as many as six impossible things before breakfast.”

Nevertheless Alice told them that if they didn’t want to apply themselves by thinking about

extraordinary things they should take a walk and return later. The Dormouse left and Alice

began to explain to the March Hare the

Reconstruction Algorithm. Given .k1, k2, : : : ; kr/, one can recover b, and a1, a2, : : : ;

ar , to obtain the coach .?/, which may be pure or hybrid.

The proof for pure coaches is in [4]. Here we present an example of a particular, but not

special, case that should enable you to reproduce the process for yourself in other instances.

Suppose you are given k1 D 2; k2 D 1; k3 D 3; and you are asked to produce a coach

of the form

b

ˇˇˇˇ

a1 a2 a3

2 1 3

ˇˇˇˇ:

By .??/ we have the three simultaneous linear equations in a1; a2; a3:

b � a1 D 22 � a2; b � a2 D 21 � a3; b � a3 D 23 � a1:

Alice solved this system of linear equations using Cramer’s rule, evaluating the de-

terminants by a condensation method invented by Charles Dodgson (Lewis Carroll) ([8];

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9. Mad Tea Party Cyclic Partitions 59

[9, pp. 118–119]). She told the March Hare to pay particular attention to the form of the

solution compared with the numbers in the original problem:

a1

bD 26�3 � 26�3�1 C 26�3�1�2

26 � .�1/3D 23 � 22 C 20

26 C 1D 5

65; so a1 D 5; b D 65I

a2

bD 26�2 � 26�2�3 C 26�2�3�1

26 � .�1/3D 24 � 21 C 20

26 C 1D 15

65; so a2 D 15; b D 65I

a3

bD 26�1 � 26�1�2 C 26�1�2�3

26 � .�1/3D 25 � 23 C 20

26 C 1D 25

65; so a3 D 25; b D 65:

The March Hare checked this result by constructing the coach of the hybrid symbol

for b D 65, starting with a1 D 5, getting the following coach, which satisfies the original

data:

b D 65

ˇˇˇˇ

5 15 25

2 1 3

ˇˇˇˇ:

But, as Alice noticed immediately, all the b’s are the same and you really didn’t need to

solve for a2, or a3, because once you had the solution for a1 and b you could then construct

the symbol to get the remaining values for ai .

“To obtain a coach for a pure symbol with the given values of ki ;” Alice explained, “all

you need to do is reduce each of the fractions 565

D 113

; 1565

D 313

, and 2565

D 513

. Then,

from the first reduced fraction you see that you can begin with b D 13 and a1 D 1 and

construct the following coach, which also satisfies the original data:

b D 13

ˇˇˇˇ

1 3 5

2 1 3

ˇˇˇˇ:”

The Dormouse and the Hatter returned at this point. Alice convinced them to construct

systematically some (possibly) hybrid symbols involving b D 2n � 1, and b D 2n C 1.

Persistent coach patterns

The Dormouse set to work, constructing the symbols in Figure 9.2, with only doubly-

underlined markings, while Alice, the Hatter, and the March Hare looked for patterns.

As soon as the last symbol was written down the March Hare said, “For all n � 3 it

looks as if when b D 2n � 1 and a1 D 1, we have a persistent coach pattern

2n � 1

ˇˇˇˇ

1 2n�1 � 1

1 n � 1

ˇˇˇˇ:”

Alice proved it using .??/.

Not to be outdone, the Hatter said, “For n � 3, except for n D 4, every coach with

b D 2n � 1 that begins with a1 D 3 always has the form

2n � 1

ˇˇˇˇ

3 2n�2 � 1

2 n � 2

ˇˇˇˇ:” (~n)

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60 Part II. Number Theory and Graph Theory

22 � 1 D 3

ˇˇˇˇ

1

1

ˇˇˇˇ

23 � 1 D 7

ˇˇˇˇ

1 3

1 2

ˇˇˇˇ

24 � 1 D 15

ˇˇˇˇ

1 7

1 3

ˇˇˇˇ

3

2

ˇˇˇˇ

5

1

ˇˇˇˇ

25 � 1 D 31

ˇˇˇˇ

1 15

1 4

ˇˇˇˇ

3 7

2 3

ˇˇˇˇ

5 13 9 11

1 1 1 2

ˇˇˇˇ

26 � 1 D 63

ˇˇˇˇ

1 31

1 5

ˇˇˇˇ

3 15

2 4

ˇˇˇˇ

5 29 17 23

1 1 1 3

ˇˇˇˇ

7

3

ˇˇˇˇ

9 27

1 2

ˇˇˇˇ

11 13 25 19

2 1 1 2

ˇˇˇˇ

21

1

ˇˇˇˇ

Figure 9.2. The Dormouse’s symbols.

The cautious Dormouse began to check the Hatter’s work. When n D 3, in (~n) he got the

coach

23 � 1 D 7

ˇˇˇˇ

3 1

2 1

ˇˇˇˇ;

which is the same as

7

ˇˇˇˇ

1 3

1 2

ˇˇˇˇ

because it is cyclic. But when he substituted n D 4 in (~n) he got a coach no one had ever

seen before (since they had always stopped when a repeat occurred). It looked like this:

24 � 1 D 15

ˇˇˇˇ

3 3

2 2

ˇˇˇˇ: (~4)

At the same time the Hatter came up with another coach, valid for all n � 5, except for

n D 6 using b D 2n � 1 and a1 D 9 in (??),

2n � 1

ˇˇˇˇ

9 2n�1 � 22 � 1 2n�2 � 2n�3 C 1 2n�1 � 2n�3 C 2n�4 � 1

1 2 1 n � 4

ˇˇˇˇ: (�n)

By substituting n D 6 into (�n), the group obtained:

26 � 1

ˇˇˇˇ

9 27 9 27

1 2 1 2

ˇˇˇˇ: (�6)

Both (~4) and (�6) caused great puzzlement. They could see that these coaches were

not contracted, but they really didn’t want to reject coaches that fit the persistent coach

patterns (~n/ and (�n/.

“Maybe these curious coaches give us a clue that we shouldn’t always stop at the first

repeat in a contracted symbol,” Alice suggested, “Maybe we should go on until the bottom

row adds up to n.”

The group studied (~n), (~4), (�n), and (�6) (as the reader should also), along with

other coaches with a fixed a1 where b is always 2n � 1, or 2n C 1. They discovered that

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9. Mad Tea Party Cyclic Partitions 61

in each sequence of coaches, every time a new number ai enters the top row of a coach

it represents the beginning of yet another persistent coach pattern, where if ai is the first

element in the top row, the ki values all remain the same, except for kr (the last entry in the

bottom row) which will always increase by 1 every time n increases by 1. To see another

example of this look at the form of the coaches where b D 2n C 1, and a1 D 3, for n D 4,

5, and 6. It turns out that if a coach, with fixed a1 where b is always 2n �1 or 2nC1, is such

that the k value for that coach is less than n then the process of duplicating the columns in

that coach until the resultant value of k in that coach equals n, will always make that coach

appear as it would in some persistent coach sequence without being contracted.

Now the Dormouse asked Alice when they were going to get to the main problem.

Alice said “That’s next! We have everything we need to answer the question now.”

The big payoff—A non-recursive algorithm for finding the

number of MTP partitions

Let’s begin with n D 5. Because we know the MTP partitions in this case (Table 9.1) it will

give us a plausibility check. Alice instructed them to look at the following two, (possibly)

hybrid, symbols:

b D 25 � 1 D 31ˇˇˇ1 15

1 4

ˇˇˇ3 7

2 3

ˇˇˇ5 13 9 11

1 1 1 2

ˇˇˇ;

b D 25 C 1 D 33ˇˇˇ1

5

ˇˇˇ

3 15 9

1 1 3

ˇˇˇ5 7 13

2 1 2

ˇˇˇ

11

1

ˇˇˇ:

Because of the form of b in these two cases, we know in advance (by the Quasi-Order

Theorem) that if the symbol is reduced and contracted the value of k for each coach is 5.

In one coach the value of k is less than 5. However, in that deficient case it is possible to

repeat columns until the sum of entries in the bottom row sums to 5. For our purposes, we

only need to know the duplicate entries in the bottom row.

Thus, in bottom of the fourth coach for b D 33 we repeat (1) to get 1 1 1 1 1. Having

done this, you can simply record the numbers that appear in the bottom row of the various

coaches to obtain all the MTP partitions of 5. The partitions contributed by b D 31 are

all the even MTP partitions of 5. The values contributed by b D 33 are all the odd MTP

partitions of 5. In Table 9.4 we record the MTP partitions by reading across the bottom row

of the coaches, repeating the doubly-underlined entries where appropriate.

Table 9.4. MTP partition for n D 5.

Even MTP partitions from the coaches of b D 31 1 4, 2 3, 1 1 1 2

Odd MTP partitions from the coaches of b D 33 5, 1 1 3, 2 1 2, 1 1 1 1 1

There were, in fact, exactly the same partitions that they had found earlier, even though

some were in a different cyclic order. In this case if each partition is written in ascending

order there are no repeats so these yield the 7 unordered partitions for n D 5.

Then the Dormouse asked, “How can we be sure we have all the MTP partitions?”

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62 Part II. Number Theory and Graph Theory

Alice explained, “This is easy! If you think one is missing, just tell me which one it is.

By the reconstruction algorithm, I can construct the coach having that partition. But, we

know such a coach must already exist because in [4]–[6] it was shown that (??) defines a

unique permutation of the ai s and our symbol contains all of the ai s. Finally, it turns out

that, if the coach connected with a set of the ai s involves a repeat of the columns then this

coach must fit one of the persistent coach patterns. This means we must have all the MTP

partitions!” She then summarized their algorithm for finding MTP partitions, for a given n:

1. Construct the (possibly) hybrid symbol for b D 2n � 1 and for b D 2n C 1.

2. In any hybrid coaches for either symbol repeat the bottom row, if necessary, so that

the k value for that coach is n.

3. From the bottom row of the coaches record the MTP partitions.

Example with n D 8

That’s it! The bottom row of the coaches in the symbols for b D 2n � 1, or b D 2n C 1, are

the even or odd, MTP partitions, respectively. And, you can always write each of the MTP

partitions in descending order, strike out the duplicates, and obtain the unordered partitions

of n. This algorithm will work for any n—and a computer can be programmed to produce

the symbols.

The March Hare, the Hatter, and the Dormouse were really into the process now. They

decided to try it for n D 8. Figures 9.3 and 9.4 give the (possibly hybrid) symbols for

b D 28 � 1 D 255 and b D 28 C 1 D 257, as calculated by the Dormouse.

255

ˇˇˇˇ

1 127

1 7

ˇˇˇˇ

3 63

2 6

ˇˇˇˇ

5 125 65 95

1 1 1 5

ˇˇˇˇ

7 31

3 5

ˇˇˇˇ

9 123 33 111

1 2 1 4

ˇˇˇˇ

ˇˇˇˇ

11 61 97 79

2 1 1 4

ˇˇˇˇ

13 121 67 47

1 1 2 4

ˇˇˇˇ

15

4

ˇˇˇˇ

17 119

1 3

ˇˇˇˇ

19 59 49 103

2 2 1 3

ˇˇˇˇ

ˇˇˇˇ

21 117 69 93 81 87

1 1 1 1 1 3

ˇˇˇˇ

23 29 113 71

3 1 1 3

ˇˇˇˇ

25 115 35 55

1 2 2 3

ˇˇˇˇ

ˇˇˇˇ

27 57 99 39

2 1 2 3

ˇˇˇˇ

37 109 73 91 41 107

1 1 1 2 1 2

ˇˇˇˇ

43 53 101 77 89 83

2 1 1 1 1 2

ˇˇˇˇ

ˇˇˇˇ

45 105 75

1 1 2

ˇˇˇˇ

51

2

ˇˇˇˇ

85

1

ˇˇˇˇ

Figure 9.3.

In Table 9.5 they recorded the MTP partitions from the bottom row of the coaches,

repeating the doubly-underlined coaches as appropriate. Ordering the numbers in each of

these 35 MTP partitions and striking out the duplicates they verified that the number of

ordinary, unordered partitions for n D 8 is 22.

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9. Mad Tea Party Cyclic Partitions 63

257

ˇˇˇˇ

1

8

ˇˇˇˇ

3 127 65

1 1 6

ˇˇˇˇ

5 63 97

2 1 5

ˇˇˇˇ

7 125 33

1 2 5

ˇˇˇˇ

9 31 113

3 1 4

ˇˇˇˇ

11 123 67 95 81

1 1 1 1 4

ˇˇˇˇ

ˇˇˇˇ

13 61 49

2 2 4

ˇˇˇˇ

15 121 17

1 3 4

ˇˇˇˇ

19 119 69 47 105

1 1 2 1 3

ˇˇˇˇ

21 59 99 79 89

2 1 1 1 3

ˇˇˇˇ

ˇˇˇˇ

23 117 35 111 73

1 2 1 1 3

ˇˇˇˇ

25 29 57

3 2 3

ˇˇˇˇ

27 115 71 93 41

1 1 1 2 3

ˇˇˇˇ

ˇˇˇˇ

37 55 101 39 109

2 1 2 1 2

ˇˇˇˇ

43 107 75 91 83 87 85

1 1 1 1 1 1 2

ˇˇˇˇ

ˇˇˇˇ

45 53 51 103 77

2 2 1 1 2

ˇˇˇˇ

Figure 9.4.

As we leave our friends the Dormouse went off to buy a small Mickey Mouse hand

calculator so that he could more easily do the MTP calculations for n D 9.

Table 9.5. MTP partitions for n D 8.

Even MTP partitions

from the

coaches of Figure 3

1 7, 2 6, 1 1 1 5, 3 5, 1 2 1 4, 2 2 1 4, 1 1 2 4, 4 4, 1 3 1 3, 2 2 1 3

1 1 1 1 1 3, 3 3 1 3, 1 2 2 3, 2 1 2 3, 1 1 1 2 1 2, 2 1 1 1 1 2,

1 1 2 1 1 2, 2 2 2 2, 1 1 1 1 1 1 1 1

Odd MTP partitions

from the

coaches of Figure 4

8, 1 1 6, 2 1 5, 1 2 5, 3 1 4, 1 1 1 1 4, 2 2 4, 1 3 4, 1 1 2 1 3

2 1 1 1 3, 1 2 1 1 3, 3 2 3, 1 1 1 2 3, 2 1 2 1 2, 1 1 1 1 1 1 2, 2 2 1 1 2

Acknowledgment The authors are grateful to Don Albers, Gerald L. Alexanderson, Vic-

tor Garcia, Jennifer Hooper, Geoffrey Shephard, and Robin Wilson for reading earlier ver-

sions of this manuscript and offering valuable suggestions for improving the text; and we

also thank Veronica Garcia for producing the cartoon.

The authors accept full responsibility for any typos that may have escaped our scrutiny—

but we blame any arithmetical errors on the Dormouse who tended to nod off doing the

longer calculations.

Bibliography

[1] M. Gardner, Fractal Music, Hypercards, and More Mathematical Recreations from Scientific

American Magazine, W. H. Freeman, 1992, 24–38.

[2] ——, Mathematical Carnival, Mathematical Association of America, 1989, 137.

[3] ——, The Last Recreations: Hydras, Eggs, and Other Mathematical Mystifications, Copernicus

Series, Springer, NY, 1997, 37–42.

[4] P. Hilton, D. Holton, and J. Pedersen, Mathematical Reflections—In Room with Many Mirrors,

Second printing, Undergraduate Texts in Mathematics, Springer, NY, 1998.

[5] P. Hilton and J. Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Math-

ematics, Cambridge Univ. Press, 2010.

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64 Part II. Number Theory and Graph Theory

[6] P. Hilton, J. Pedersen, and B. Walden, A property of complete symbols: An ongoing saga con-

necting geometry and number theory, Homage to a Pied Puzzler, A K Peters, 2009, 47–50.

[7] A. Knopfmacher and N. Robbins, Some properties of cyclic compositions, Fibonacci Quart. 48

(2010) 249–255.

[8] A. Rice and E. Torrence, “Shutting up like a Telescope,” Lewis Carroll’s “curious” condensation

method for evaluating determinants, College Math. J. 38 (2007) 85–95.

[9] R. Wilson, Lewis Carroll in Numberland: His Fantastical Mathematical Logical Life, W. W.

Norton, NY, 2010.

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10The Continuing Saga of Snarks

sarah-marie belcastro

Way back in 1852, Francis Guthrie conjectured that every map drawn on the plane can be

colored so that regions sharing a border have different colors—and only four colors are

necessary. This became known as the Four Color Conjecture. In 1880, Tait [16] proved

that the Four Color Conjecture is equivalent to a problem of edge coloring graphs. This is

where our story begins, because the study of snarks grew from exactly this edge-coloring

problem. To avoid confusion, we note that there is no relationship between the English

word ‘snark’ (or its derivatives snarky, snarkiness, etc.) and the mathematical term; in fact,

the origin of the latter is a pivotal point in our story.

Recall that a graph is a collection of nodes, called vertices, and a collection of connec-

tions between these nodes, called edges. You may also remember that a graph is planar if

it can be drawn in the plane with no edge-crossings, at which point the regions of the plane

demarcated by the graph are called faces. You are less likely to know that a bridgeless

graph has no edge whose deletion disconnects the graph, that a 3-regular graph has three

edges meeting at every vertex, and that a proper edge coloring assigns a color to every

edge of a graph such that no two edges meeting at a vertex have the same color. A graph to

which all three of these terms apply is shown in Figure 10.1.

Figure 10.1. A bridgeless, 3-regular, properly 3-edge colored planar graph, but not a snark.

In this language, Tait showed that the Four Color Conjecture holds if and only if every

bridgeless, 3-regular, planar graph can be properly edge colored using only three colors.

That is, instead of coloring regions of maps, one can approach the Four Color Conjecture

by coloring edges of bridgeless, 3-regular, planar graphs. As Martin Gardner pointed out

in his 1976 introduction to this subject [9], it follows that the search for a counterexample

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 82–87.

65

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66 Part II. Number Theory and Graph Theory

to the Four Color Conjecture is really a search for a bridgeless, 3-regular, planar graph that

is not 3-edge colorable.

In Gardner’s language, this was a search for a nontrivial (no bridges or cycles of lengths

4 or less), uncolorable (using only three colors), trivalent (another word for 3-regular)

planar graph. Because “Nontrivial Uncolorable Trivalent” is a mouthful, Gardner sought a

shorter name. He considered ‘NUT’, decided it wasn’t serious enough, and then proposed

‘Snark’ from the Lewis Carroll poem The Hunting of the Snark—which does not seem, at

least to this author, to be a more serious name. (Carroll invented the name Snark, but his

poetic use was not related to the already existing English noun, or, it seems, to the verb

and adjective that came later [15].) The name stuck, and has invigorated all subsequent

discourse on the topic.

Classical snarkiness

We now reveal the full technical definition.

Definition. A snark is a bridgeless, 3-regular graph that

� is not edge colorable using only three colors,

� has smallest cycle length (girth) at least 5, and

� is cyclically 4-edge connected, meaning that at least 4 edges must be removed in

order to separate the graph into two components that each contain a cycle.

Figure 10.2. The Petersen graph, BlanuLsa-1, and BlanuLsa-2.

The three smallest snarks are shown in Figure 10.2. In 1976, Gardner gave several

examples both of individual snarks and of infinite families of snarks [9]. Much of this

column was a translation into lay language of an excellent paper by Isaacs [11], published

the year before and still cited in modern research. As with many of his columns, Gardner’s

writing on snarks was exciting partly because he gave puzzle-like exercises, some of which

led toward proofs!

Martin Gardner’s showcasing (and naming) of snarks is only one part of their story. Ear-

lier, in 1968, Branko Grunbaum [10] posed a generalization of the Four Color Conjecture

that can be phrased in terms of snarks. To understand it, we need a few more definitions.

A planar graph should really be drawn on a sphere to avoid the unbounded face present

in a planar drawing. Focusing on the faces of a graph takes us beyond ordinary graph theory

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10. The Continuing Saga of Snarks 67

into topological graph theory. The topology referred to here is the topology of surfaces such

as the sphere, torus, and Mobius band. (The lovely classification theorem for surfaces is

presented superbly in [8].)

(a) (b)

Figure 10.3. The complete graph K5 is nonplanar (a) but does embed on the torus (b).

In Figure 10.3b we show a small nonplanar graph embedded (drawn without edges

crossing) on the torus, as well as a way of representing the torus that allows us to avoid the

use of dotted lines for unseen parts of edges. The arrows indicate gluing; when an edge of a

graph goes off the boundary of the rectangle, it returns on the matching arrowed boundary

segment. It is quite an enjoyable exercise to embed the Petersen graph on a torus.

We only consider cellular embeddings, in which every face is a topological disk. (In

other words, faces cannot have punctures or doughnut holes.) A graph is polyhedrally

embeddable on a surface if it can be drawn so that no face uses an edge twice and no two

faces share more than one edge. Figure 10.3b depicts a cellular embedding, but because

some edges do not separate distinct faces, the embedding is not polyhedral.

Grunbaum conjectured that every 3-regular graph polyhedrally embeddable on an ori-

entable surface (a sphere, torus, or n-holed torus) is three-edge colorable. Phrased in terms

of snarks, this says that no snark has a polyhedral embedding on any orientable surface.

When Martin Gardner published [9], there had been no progress on Grunbaum’s Conjec-

ture, but a few months later, Appel and Haken turned the Four Color Conjecture into the

Four Color Theorem [2], [3]. They did not use edge coloring in their proof (for an excel-

lent exposition of the saga, see [20]), and so by Tait’s theorem we obtain, as a corollary of

the Four Color Theorem, that every bridgeless 3-regular planar graph is 3-edge colorable;

equivalently, no snark embeds on a sphere.

It’s not just the connection to the Four Color Theorem that makes snarks interesting.

Vizing’s theorem (proved only a few years before Grunbaum made his conjecture) implies

that every 3-regular graph needs at most 4 colors to properly color its edges. Of course, at

least 3 colors are needed, so there is not much wiggle room in the land of edge coloring.

Snarks are simultaneously barely not 3-edge colorable, and as difficult to edge-color as

possible.

Modern snarkiness

For a long time, the main research goal was to find snarks, as very few were known. In 1950

there were only four known snarks, but, during the 1970s, five infinite families of snarks

and about five individual snarks (not in any families) were found [7]; some are pictured

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68 Part II. Number Theory and Graph Theory

Figure 10.4. Six sample snarks from the seventies (drawn by Mathematica).

in Figure 10.4. Twenty years later, the computer age made it possible to generate snarks,

so in the 1990s, most work on snarks took a structural approach. These results are highly

technical; thus, we give only some flavor of the ideas.

First, we consider ways to break down snarks. If one can remove some set of k edges

from a snark so that the remaining graph is a disconnected collection of subgraphs of

snarks, then the original snark is considered reducible, and if this is not possible, the snark

is irreducible. Many snarks are reducible in this sense, and irreducible snarks have the

property that removal of two vertices leaves a 3-edge colorable graph. By definition, a

snark cannot have a Hamilton circuit (a circuit containing all vertices), as otherwise it

would be 3-edge colorable (2-edge color the circuit and give the remaining edges a third

color). However, many snarks almost have Hamilton circuits in the sense that they have

circuits including all but one vertex. And, some snarks have the stronger property that

leaving out any one vertex, the remainder of the graph forms a circuit. All such snarks are

irreducible [6].

Tutte conjectured in 1966 that every snark contains a subgraph that is a Petersen graph

but subdivided with extra vertices [18]. This can be rephrased by saying that every snark

has a Petersen minor. Robertson, Sanders, Seymour, and Thomas announced a proof in

1999, much of which is yet to be published [17, p.14]. Because the Petersen graph occupies

such a revered position in graph theory, the fact that every snark has a Petersen minor is in

some sense the ultimate structural result.

In contrast to the structural approach is the topological approach. From this perspec-

tive, we are interested in knowing on which surface(s) a given snark embeds, and with

what type of embedding. This essentially amounts to investigating Grunbaum’s Conjec-

ture. Aside from the resolution of the Four Color Conjecture in 1976—proved using a

structural approach!—no serious progress was made on Grunbaum’s Conjecture for more

than 30 years after it was posed. The graph theory community was divided on whether the

conjecture was true, with some leading figures taking opposite sides of the issue at a 2003

conference.

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10. The Continuing Saga of Snarks 69

Very little was known about embedding properties of snarks until the 2000s, though

thousands of snarks were tested by computer and proved not to be polyhedrally embed-

dable on any orientable surface [14], [4] (and therefore none provided a counterexample

to Grunbaum’s Conjecture). It was only known that the Petersen graph and one of the two

Blanusa snarks embed on the torus. In 2004, several infinite families of snarks were ex-

hibited that embed on the torus [5], [19] and on the 2-holed torus [5] (none polyhedrally).

Because they are pretty, two seed snarks for these families are shown in Figure 10.5.

1 5

3 4 6

82

7 11 15

1 14 16

1812

179

10

19 23

21 22 24

2620

25

1 5

3 4 6

82

7 11 15

13 14 16

1812

179

10

19

23

21

22

2426

20

25

Figure 10.5. The smallest snarks in two distinct infinite families of toroidal snarks.

That same year, it was proved that for each number of Mobius-band-like twists g �1, there exists a snark with a polyhedral embedding on the corresponding nonorientable

surface [14]. (Whereas orientable surfaces are classified by the number of torus holes they

have, nonorientable surfaces are classified by the number of Mobius-band-like twists they

have.) Both of these results provided evidence in favor of Grunbaum’s Conjecture.

At a conference late in 2006, Grunbaum himself was asked whether he still thought

his eponymous conjecture was true. He demurred, but noted that saying it should be true

caused more people to work on it! Then, in 2007 everything blew up. Counterexamples to

Grunbaum’s Conjecture were found on the 9-holed torus, and then shortly thereafter, on the

5-holed torus [13] using snarks exhibited in [5]. (In fact, infinitely many counterexamples

were found! The powerful superposition technique of [12] did the trick.) At the same time,

Grunbaum’s Conjecture was affirmed for most three-regular graphs embedded on the torus

[1]. While this deflates an inspiring balloon, a smaller (and perhaps more manageable?)

balloon still exists: What is true about Grunbaum’s Conjecture restricted to n-holed tori for

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70 Part II. Number Theory and Graph Theory

1 � n � 4? While perhaps all of mathematical research is about refining the boundaries

between different categories of objects, it somehow seems that this is especially true for

snarks. It often feels that doing research on snarks is like honing the edge of a knife used to

separate two crumbly loaves of bread. Much, much more can (and should!) be said about

snarks. The interested reader is encouraged to consult the references and contact the author

for further information!

Bibliography

[1] M. O. Albertson, H. Alpert, s-m. belcastro, and R. Haas, Grunbaum colorings of toroidal

triangulations, J. Graph Theory 63 (2010), 68–81; available at dx.doi.org/10.1002/

jgt.20406

[2] K. Appel, W. Haken, and J. Koch, Every planar map is four colorable, I: Discharging, Illinois

J. Math. 21 (1977) 429–490.

[3] K. Appel and W. Haken, Every planar map is four-colorable, II: Reducibility, Illinois J. Math.

21 (1977) 491–567.

[4] D. Archdeacon, Problems in Topological Graph Theory: Three-Edge-Coloring Orientable

Triangulations (1995); available at www.emba.uvm.edu/˜archdeac/problems/

grunbaum.htm.

[5] s.-m. belcastro and J. Kaminski, Families of dot-product snarks on orientable surfaces of

low genus, Graphs Combin. 23 (2007), 229–240; available at dx.doi.org/10.1007/

s00373-007-0729-9.

[6] A. Cavicchioli, T. E. Murgolo, B. Ruini, and F. Spaggiari, Special classes of snarks, Acta Appl.

Math. 76 (2003), 57–88; available at dx.doi.org/10.1023/A:1022864000162.

[7] A. G. Chetwynd and R. J. Wilson, Snarks and supersnarks, in The Theory and Applications of

Graphs (Kalamazoo, Mich., 1980), Wiley, New York, 1981, 215–241.

[8] G. K. Francis and J. R. Weeks, Conway’s ZIP proof. Amer. Math. Monthly 106

(1999), 393–399; available at dx.doi.org/10.2307/2589143; also available at

new.math.uiuc.edu/zipproof/.

[9] M. Gardner, Mathematical games, Scientific American, April 1976, 126–130.

[10] B. Grunbaum, Conjecture 6, in Recent Progress in Combinatorics, W.T. Tutte, ed., Academic

Press, New York, 1969, 343.

[11] R. Isaacs, Infinite families of nontrivial trivalent graphs which are not Tait colorable, Amer.

Math. Monthly 82 (1975), 221–239; available at dx.doi.org/10.2307/2319844.

[12] M. Kochol, Snarks without small cycles, J. Combin. Theory Ser. B 67 (1996), 34–47; available

at dx.doi.org/10.1006/jctb.1996.0032.

[13] ——, Polyhedral embeddings of snarks in orientable surfaces, Proc. of the Amer. Math.

Soc., 137 (2009), 1613–1619; available at dx.doi.org/10.1090/S0002-9939-

08-09698-6.

[14] B. Mohar and A. Vodopivec, On polyhedral embeddings of cubic graphs, Combin. Prob. Com-

put. 15 (2006), 877–893; available at dx.doi.org/10.1017/S0963548306007607.

[15] snark, v. and n. The Oxford English Dictionary, 2nd ed., Oxford Univ. Press, 1989.

[16] P. G. Tait, Note on a theorem in geometry of position. Trans. Roy. Soc. Edinburgh 29 (1880),

657–660.

[17] R. Thomas, Recent excluded minor theorems for graphs, Surveys in combinatorics, 1999 (Can-

terbury), London Math. Soc. Lecture Note Ser., 267, Cambridge Univ. Press, Cambridge, 1999,

201–222; available at people.math.gatech.edu/˜thomas/PAP/bcc.pdf.

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10. The Continuing Saga of Snarks 71

[18] W. T. Tutte, On the algebraic theory of graph colorings, J. Combin. Theory Ser. B 1 (1966),

15–50; available at dx.doi.org/10.1016/S0021-9800(66)80004-2.

[19] A. Vodopivec, On embeddings of snarks in the torus, Discrete Math. 308 (2008), 1847–1849;

available at dx.doi.org/10.1016/j.disc.2006.09.051.

[20] R. Wilson, Four Colors Suffice: How the Map Problem Was Solved, Princeton Univ. Press,

2003.

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11The Map-Coloring Game

Tomasz Bartnicki, Jarosław Grytczuk,

H. A. Kierstead, and Xuding Zhu

Introduction

Suppose that Alice wants to color a planar map using four colors in a proper way, that

is, so that any two adjacent regions get different colors. Despite the fact that she knows

for certain that it is eventually possible, she may fail in her first attempts. Indeed, there are

usually many proper partial colorings not extendable to proper colorings of the whole map.

Thus, if she is unlucky, she may accidentally create such a bad partial coloring.

Now suppose that Alice asks Bob to help her in this task. They color the regions of a

map alternately, with Alice going first. Bob agrees to cooperate by respecting the rule of a

proper coloring. However, for some reason he does not want the job to be completed—his

secret aim is to achieve a bad partial coloring. (For instance, he may wish to start the color-

ing procedure over and over again just to stay longer in Alice’s company.) Is it possible for

Alice to complete the coloring somehow, in spite of Bob’s insidious plan? If not, then how

many additional colors are needed to guarantee that the map can be successfully colored,

no matter how clever Bob is?

This map-coloring game was invented about twenty-five years ago by Steven J. Brams

with the hope of finding a game-theoretic proof of the Four Color Theorem, avoiding per-

haps the use of computers. Though this approach has not been successful, at least we are

left with a new, intriguing map-coloring problem: What is the fewest number of colors

allowing a guaranteed win for Alice in the map-coloring game in the plane?

Brams’s game was published by Martin Gardner in his Mathematical Games column

in Scientific American in 1981. Surprisingly, it remained unnoticed by the graph-theoretic

community until ten years later, when it was reinvented by Hans L. Bodlaender [1] in the

wider context of general graphs. In this version Alice and Bob play as before by coloring

properly the vertices of a graph G. The game chromatic number �g.G/ of G is the smallest

number of colors for which Alice has a winning strategy. As every map is representable

by a graph whose edges correspond to adjacent regions of the map, Brams’s question is

equivalent to determining the game chromatic number of planar graphs. Since then the

problem has been analyzed in serious combinatorial journals and gained the attention of

Reprinted from The American Mathematical Monthly, Vol. 114, No. 9 (Nov. 2007), pp. 793–803.

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74 Part II. Number Theory and Graph Theory

several experts. In this article we present what is known and what remains unknown about

this fascinating game.

Four colors do not suffice

One can see quickly that there are maps demanding more colors in the game than normally.

Perhaps the simplest example of this phenomenon is due to Robert High. Consider the map

with six regions corresponding to the faces of a cube. Clearly, the map can be 3-colored,

but Bob can win even if there are four colors available. Suppose that Alice uses color 1

in her first move. Then Bob answers with color 2 on the opposite face. Thus colors 1 and

2 cannot be used any further, and Alice has to play color 3 (see Figure 11.1). Then Bob

answers with color 4 on the opposite face and wins the game.

1

2

Alice

Bob Bob

1

1

Alice

Figure 11.1. First two moves on the cube and the dodecahedron.

Earlier, Lloyd Shapley gave another elegant example based on a regular dodecahedron,

showing that even five colors do not suffice. Here Bob’s strategy is similar, but this time he

replies with the same color on the face opposite to the face that Alice has just colored (see

Figure 11.1). In this way, Alice must use a new color in each of her moves, which leads to

a win for Bob after the fifth round.

When Steven Brams and Martin Gardner started to think that perhaps six colors would

be the maximum number that Bob could force, Robert High found a map for which Alice

needed seven colors to win. Indeed, there is a modification of the dodecahedron map for

which a win by Alice requires eight colors. Unfortunately, it is not possible to verify this

without a tedious case-by-case analysis. A natural suspicion arose that perhaps there was

no finite bound at all, a suspicion that persisted for thirteen years.

Possible strategies for Alice

Let us consider the problems posed for Alice in designing an effective strategy for the map-

coloring game. Whenever Bob colors a vertex v, he poses a threat to each of its uncolored

neighbors. Alice can counter this threat at one of these neighbors w by immediately color-

ing it, but this in turn creates new threats for the neighbors of w. Moreover, the remaining

uncolored neighbors of v are still open to attack, and Bob can strengthen his position on

subsequent moves by coloring their uncolored neighbors. Thus Alice should design a strat-

egy that prioritizes responses to threats to uncolored vertices so that she can keep the total

threat under control. She must also limit the damage done when she colors a vertex.

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11. The Map-Coloring Game 75

The following idea allows Alice to avoid creating new points of vulnerability when she

colors a vertex. Let �.G/ denote the usual chromatic number of a graph G, that is, the

least number of colors in a proper coloring of the vertices of G. Suppose that f W V.G/ !fc1; : : : ; ctg is an optimal coloring, with t D �.G/, and that the game is played on a set of

kt colors of the form fcij W i 2 Œk�; j 2 Œt �g, where we use the standard abbreviation Œn� for

f1; 2; : : : ; ng. We refer to cij as a shade of cj . We say that a color is correct for a vertex v

if it is a shade of f .v/; otherwise it is incorrect. Alice should always try to color a vertex

v with a correct color. If she is successful, then her early moves will never conflict with

her later moves. This is possible as long as she can play in such a way that at any time any

uncolored vertex is adjacent to fewer than k vertices incorrectly colored by Bob.

This leaves Alice with the problem of limiting the number of neighbors colored in-

correctly by Bob that an uncolored vertex can have. In ordinary (noncompetitive) graph-

coloring there is a well-known strategy for managing such threats. The coloring number

col.G/ of a graph G is the smallest integer k such that every subgraph of G has a vertex

with degree less than k. For example, it is a simple consequence of Euler’s formula that the

coloring number of an arbitrary planar graph is at most six. The vertices of a graph with

coloring number at most k can be ordered as v1; : : : ; vn so that each vertex vi has fewer

than k neighbors that precede it in the ordering: we simply construct the ordering from

right to left, always choosing a vertex whose degree among the unchosen vertices is less

than k. It is also clear that the existence of such an ordering implies that col.G/ � k. Now,

if Alice colors the vertices from left to right, she will need at most k colors, since each un-

colored vertex will have fewer than k colored neighbors. In other words, �.G/ � col.G/

for all graphs G.

Suppose that Alice attempts to modify this strategy for the map-coloring game. She

might naturally try to color the left-most (in the ordering) neighbor of the last vertex col-

ored by Bob. This vertex will have fewer than k backward (left) neighbors colored by Bob,

but the number of its forward (right) neighbors colored by Bob is unbounded. This was

the situation with the problem in the early nineties when help came from an unexpected

source.

Ramsey numbers and arrangeability

A simple graph G consists of a vertex set V.G/ and an edge set E.G/ where each edge

is an unordered pair of distinct vertices. A complete graph or clique is a simple graph

in which every pair of vertices forms an edge. A clique on t vertices is denoted by Kt .

According to Ramsey’s famous theorem, any graph G has a Ramsey number r.G/ defined

as the smallest integer t such that for any red-blue coloring of the edges of Kt there is

a monochromatic copy of G in Kt . For a class C of graphs define rC.n/ by rC.n/ Dmaxfr.G/ W G 2 C; jV.G/j D ng. Burr and Erdos conjectured that rC.n/ is linear in n

for any class of graphs satisfying jE.G/j D O.jV.G/j/. This conjecture remains open

in general, although some special cases have been proved. Chvatal, Rodl, Szemeredi, and

Trotter [5] verified the conjecture for the special case of graphs with bounded degree by

using Szemeredi’s regularity lemma. The proof involved constructing an embedding of

G into a monochromatic part of Kn one vertex at a time in an order v1; : : : ; vn. It was

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76 Part II. Number Theory and Graph Theory

crucial to their argument that each vertex have a bounded number of backward and forward

neighbors. If the argument had required only that the backward degree be bounded, they

would have established the whole conjecture. Notice that planar graphs satisfy jE.G/j DO.jV.G/j/ but may have arbitrarily large degree.

Chen and Schelp made the brilliant observation that something less was required: the

number of vertices vh with h < i that are adjacent to vi or to a forward neighbor vj of vi

should be bounded (see Figure 11.2). This idea, in one form or another, has turned out to

be absolutely crucial for the theory of coloring games.

v1 vh vi vnvj

Figure 11.2. The number col2.G/ bounds the number of predecessors of vi that are adjacent to vi

(grey nodes) or to a forward neighbor of vi (black nodes).

Assume that the vertices of a graph G are ordered as v1; v2; : : : ; vn. A vertex vh is

called a loose backward neighbor of vi if h < i and either (a) vh is adjacent to vi or (b)

vh is adjacent to vj and vj is adjacent to vi for some j with j > i (see Figure 11.2). We

define the 2-coloring number col2.G/ of a graph G to be the least integer k such that there

is an ordering of V.G/ for which each vertex has fewer than k loose backward neighbors.

This is an inessential variant of the parameter that Chen and Schelp called arrangeability

in [3], where it is proved that the conjecture of Burr and Erdos is true for the family of

graphs with bounded 2-coloring numbers. They then proved the Burr-Erdos conjecture for

planar graphs by showing that the 2-coloring number of any planar graph is at most 761.

Kierstead and Trotter [14] realized that this was the missing tool for bounding the game

chromatic number �g.G/ of planar graphs G. They proved the following theorem:

Theorem 1. Every graph G satisfies �g.G/ � �.G/.1 C col2.G//.

Proof. Let �.G/ D t and col2.G/ D k. To prove the assertion we describe a strategy for

Alice using at most t.kC1/ colors. Alice should begin by fixing an optimal coloring f of G

using the colors c1; : : : ; ct and an ordering v1; : : : ; vn of G that satisfies col2.G/ D k. She

prioritizes responses to threats according to this ordering. Whenever Bob colors a vertex v,

Alice colors the least-indexed uncolored backward neighbor u of v with a correct color; if

v has no such backward neighbor then Alice correctly colors the least-indexed uncolored

vertex. To be certain that u has a correct color available (i.e., a color not used by any of

its colored neighbors), it suffices to show that u has at most k neighbors that are colored

by Bob. Assume that Bob has colored s loose backward neighbors of u and that Alice has

colored s0 loose backward neighbors of u. In accordance with the stated strategy, each time

Bob colors a forward neighbor z of u, Alice colors the least-indexed uncolored backward

neighbor u0 of z. Since u is an uncolored backward neighbor of z, u0 precedes u, so u0 is

a loose backward neighbor of u. As Alice has colored s0 loose backward neighbors of u,

we conclude that Bob has colored at most s0 forward neighbors of u. Therefore the number

of neighbors of u colored by Bob is at most s C s0 .< k/. Note that this counts only the

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11. The Map-Coloring Game 77

number of colored neighbors of u at the end of Alice’s previous move. After that move,

Bob may have colored another neighbor of u, meaning that u now has at most k neighbors

colored by Bob. Thus u has an available correct color, and Alice can win the game with

t.k C 1/ colors.

This reasoning implies that �g.G/ � 3044 for a planar graph G. Kierstead and Trot-

ter obtained much tighter bounds on the game chromatic number of planar graphs. They

demonstrated that col2.G/ � 10 holds for each planar graph G (which gives �g.G/ � 44)

and provided an example of a planar graph G with col2.G/ D 8. They further improved

their bound on the game chromatic number of planar graphs to �g.G/ � 33 by using a

more carefully designed parameter than 2-coloring number.

Acyclic coloring

A cycle on vertices v1; : : : ; vn is a graph whose edges are v1v2; v2v3; : : : ; vnv1. A forest

is a graph without cycles. A coloring of the vertices of a graph G is acyclic if it is a proper

coloring such that no cycle of G is 2-colored. In other words, a subgraph of G whose vertex

set is a union of any two color classes is a forest. The minimum number of colors needed

is the acyclic chromatic number of G, denoted by a.G/. This notion was introduced by

Grunbaum, who conjectured that a.G/ � 5 holds for any planar graph G. The conjecture

was turned into a theorem by Borodin in 1979 [2].

Just as the chromatic number of a graph G is bounded by its coloring number, its acyclic

chromatic number is bounded by its 2-coloring number. Let v1; : : : ; vn be an ordering of

the vertices of G that realizes its 2-coloring number. Color vertices recursively so that no

vertex is colored the same color as a loose backward neighbor. This yields a proper coloring

using at most col2.G/ colors. We still must check that each cycle C in G receives at least

three colors. The largest-indexed vertex vj in C has two backward neighbors vh and vi

with h < i < j . Since vh is a loose backward neighbor of vi , they have different colors.

Since vj is adjacent to both vi and vh, it has a third color.

Twenty years after Grunbaum introduced acyclic coloring, Dinski and Zhu [6] applied

the acyclic chromatic number to the Brams game for arbitrary graphs. They proved the

following theorem, which implies that thirty colors suffice for Alice to win on any planar

map:

Theorem 2. Every graph G satisfies �g.G/ � a.G/.a.G/ C 1/.

Proof. The argument is similar to the proof of Theorem 1. Suppose that G has an acyclic

coloring using the color set fc1; : : : ; ctg, and let V1; : : : ; Vt be the corresponding color

classes of vertices. When i ¤ j , the subgraph Fij of G consisting of all edges with ends

in Vi [ Vj is a forest. Fix an orientation of each Fij so that every vertex has outdegree at

most one. Since every edge of G appears in exactly one of the forests Fij , this provides

an orientation of G such that every vertex has at most one outneighbor in each color class.

Now a vertex v in Vi is endangered if it is as yet uncolored but there is an inneighbor u

of v colored with a shade of ci (see Figure 11.3). Since Alice always colors with a correct

color, an endangered vertex can be created only by Bob. Clearly, in a single move Bob can

create at most one endangered vertex v. Alice should color this vertex immediately after it

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78 Part II. Number Theory and Graph Theory

Viv

u

Figure 11.3. An endangered vertex v will be colored by Alice in her next move.

is created by Bob. This is always possible, for there are at most t neighbors of v colored

with shades of ci at this moment (at most t � 1 outneighbors and only one inneighbor u

that has just been colored by Bob).

Daltonism may help

Both strategies for Alice considered earlier were local. Alice responded to any threat posed

when Bob colored a vertex by coloring one of its neighbors. As any Go player knows,

global strategies can be more powerful. Our next aim is to develop a global strategy for

Alice.

Curiously, further progress in bounding the game chromatic number was achieved by

making the game more difficult for Alice. Suppose Alice suffers from full daltonism (col-

orblindness): she just cannot distinguish between colors. Because of her disease she pro-

poses to modify the rules of the coloring game so that she can still have some fun playing

it. Rather than coloring vertices, the players create an ordering <g of vertices by taking

turns choosing the next vertex in the ordering. For a fixed positive integer k Alice’s goal

is to have <g demonstrate that col.G/ � k. Bob, wishing to remain in her company, tries

to thwart her. For the purposes of actual play it is convenient for the players to indicate

that they have chosen a vertex by coloring it gray. Bob wins if at some time some uncol-

ored vertex has k gray neighbors, and Alice wins if this never occurs. We denote the least

number k allowing success for Alice by colg.G/ and call it the game coloring number of

G. Clearly, if Alice can win this ordering game for some integer k, she can also win the

coloring game with k colors (even as a completely colorblind person!). In other words,

�g.G/ � colg.G/ is true for every graph G.

Alice is now forced to use a strategy that does not rely on shades of correct colors to

control threats created by her moves. Suppose that she again selects a specific ordering

v1; : : : ; vn (not to be confused with <g ) of the vertex set for the purpose of prioritizing

responses to threats. Suppose also that no vertex has too many backward neighbors in this

order. When Bob chooses a vertex v (colors it gray), Alice faces conflicting requirements.

On the one hand, she should choose a backward neighbor x of v, addressing the threat

to x posed by v. But this in turn produces a threat from x to its backward neighbors.

Alternatively, she could put off choosing x and instead choose a backward neighbor y of

x so that y would already be protected when later she finally did choose x. Or should she,

perhaps, choose a backward neighbor of y? Here we propose a compromise plan: the first

time Alice contemplates choosing x she should move on to considering y, but the second

time she considers choosing x she should actually do so.

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11. The Map-Coloring Game 79

Here are the details of Alice’s activation strategy. Imagine that the vertices of G are

made of light bulbs that shine a warning when activated. Let N C.v/ and N �.v/ denote the

sets of backward and forward neighbors of v, respectively. On her first move Alice activates

vertex v1 and colors it. The basic step of the strategy is to activate a vertex (leaving a warn-

ing that it is in danger) and then jump to its least-indexed uncolored backward neighbor.

Suppose that Bob has just colored a vertex v. Then Alice starts by activating v (provided

it has not been activated hitherto) and jumps to the uncolored vertex x of smallest index in

N C.v/. (See Figure 11.4.) If x is already active, then it is endangered, so Alice stops and

colors it. Otherwise she repeats the activation step for x, that is, she activates x and jumps

to the least-indexed uncolored vertex y in N C.x/. This process goes on until she stops

at some vertex u, either because u is active or because there are no uncolored vertices in

N C.u/. In each case she activates and colors u. If it happens that N C.v/ (see Figure 11.4,

second line) already contains no uncolored vertices, then she picks the uncolored vertex

of smallest index, activates it (provided it has not yet been activated), and colors it. Notice

that after Alice’s move all gray vertices are active, whereas after Bob’s move there can be

only one gray inactive vertex.

u y x v

u y x v

Figure 11.4. Basic step of the activation procedure.

For example, let’s see how this strategy works for trees (see Figure 11.5). Any tree

T can be ordered so that each vertex has at most one backward neighbor. Suppose that

Bob has colored a vertex v, and let x be its unique backward neighbor. Then Alice should

activate all vertices on the unique backward path starting from v and ending at some active

vertex u. Notice that Alice can jump to the same vertex only twice (the first time she

activates it, the second time she colors it). It follows that if x stayed uncolored after her

move, it could have no more than two active neighbors (one backward and one forward).

Hence, after Bob’s move there are at most three colored vertices around x, which gives

u

x

v v v

xx

u u

Figure 11.5. Activation on a tree.

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80 Part II. Number Theory and Graph Theory

colg.G/ � 4. This result was proved by Faigle, Kern, Kierstead, and Trotter in [8]. The

following result reveals the information provided by the activation strategy for general

graphs:

Theorem 3. Every graph G satisfies colg.G/ � 3 col2.G/ � 1.

Proof. Fix an order of the vertices v1; : : : ; vn realizing col2.G/ D k, and assume that

Alice applies the activation strategy with respect to this order. For each vertex vj let S.vj /

be the set of loose backward neighbors of vj . By the definition of col2.G/, the size of

S.vj / is at most k � 1. We count the number of active forward neighbors that vj could

have before it gets colored. From each active vertex in N �.vj / Alice jumps to the least-

indexed uncolored backward neighbor in S.vj / or possibly to vj . If she jumps to vj , then

she immediately colors it or immediately jumps to another vertex in S.vj /. Since she can

jump only twice to a vertex, there can be at most 2.k � 1/ active vertices in N �.vj / after

Alice’s move. Hence, there are at most 3.k � 1/ active vertices around vj after Alice’s

move. Counting Bob’s last move, there are at most 3k � 2 colored vertices around vj when

it gets colored.

v

v

v

1

2

3

v

Figure 11.6. Construction of a 3-tree.

There is a nice class of graphs for which the bound in Theorem 3 is optimal (as long

as we assume that Alice cannot distinguish between colors). Fix a number k and start with

a clique on k vertices v1; : : : ; vk . Then in each subsequent step add a new vertex v and

join it to k vertices of any clique in the graph constructed so far. In this recursive manner

we produce graphs called k-trees (see Figure 11.6). The resulting ordering of the vertices

shows that col2.G/ D k C 1, which gives colg.G/ � 3k C 2 for any k-tree G. It can be

proved that this bound is optimal for each k (> 1). For planar graphs the theorem gives

�g.G/ � 29, which is not bad but is still not optimal.

Six dollar order

We now demonstrate that the estimate �g.G/ � 18 holds for every planar graph G, which

is almost the best upper bound discovered so far. This time we take more care in ordering

the vertices v1; : : : ; vn so as to make the activation strategy more efficient. We define this

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11. The Map-Coloring Game 81

order inductively (in reverse order), starting with any vertex vn of degree at most five.

Suppose that we have already picked vertices vn; vn�1; : : : ; viC1 and are looking for a

candidate for vi . Partition the vertex set V into two parts, C D fvn; vn�1; : : : ; viC1g and

U D V n C , and construct a new graph H as follows:

1. Delete each edge between pairs of vertices in C .

2. Delete each vertex v in C with at most three neighbors in U .

3. For each deleted vertex v in C add edges between its neighbors in U so that these

neighbors form a clique.

It can be verified that the new graph H is planar: the new edges in H can be drawn

near deleted edges incident to deleted vertices of C . By Euler’s formula H satisfies the

inequality jE.H/j � 3jV.H/j � 6.

Now suppose that each edge of H is assigned $2 (in half-dollar coins). These coins are

distributed to the vertices of H as follows: if an edge e links two vertices in U , then e gives

$1 to each of the two vertices; if e links a vertex x in C and a vertex y in U , then e gives

$1.50 to x and $0.50 to y (see Figure 11.7). As the total amount of dollars is equal to 2jEj(< 6jV.H/j), there is a vertex v that receives less than $6. This will be the i th vertex vi in

our order. Notice that vi cannot belong to C , for each remaining vertex in C is incident to

at least four edges of H , hence receives at least $6.

$1.5

$0.5

$1

$1

vi

CU

Z

D

N-(vi)

Figure 11.7. Ordering the set of vertices of G by finding a poor vertex in a graph H .

Now we are ready to prove the result stated at the beginning of this section:

Theorem 4. Every planar graph G satisfies colg.G/ � 18.

Proof. Let v1; : : : ; vn be the ordering of the vertices of G obtained in the manner just de-

scribed. It suffices to show that if Alice applies the activation strategy, then at the end of

each of her turns there will be at most sixteen active neighbors around any uncolored ver-

tex. Let v D vi be any uncolored vertex, let H , C , and U be as indicated in the discussion

preceding Theorem 4 (notice that all of them depend on vi ), let D be the set of those for-

ward neighbors of v (in G) that were deleted while constructing H , and let Z be the set of

those vertices of U that are adjacent to v in H . Since we didn’t delete any edges between

vertices of U , we see that N C.v/ is contained in Z. Also Z includes all vertices, other

than v, to which Alice can jump after activating any vertex in the set D (see Figure 11.7).

Hence, arguing as in the previous proof, there can be at most

3jZj C jN �.v/ n Dj

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82 Part II. Number Theory and Graph Theory

active vertices around v at the end of any move by Alice. Since v received strictly less than

$6 in the ordering procedure,

1 � jZj C 0:5 � jN �.v/ n Dj < 6:

This implies that 3jZj C jN �.v/ n Dj � 16, which completes the proof.

Variants of the game

Currently, the best upper bounds for a planar graph G are �g.G/ � colg.G/ � 17, as

proved in [20]. The proof is based on a modification of the activation strategy. At present,

the best lower bounds for planar graphs are 8 � �g.G/ and 11 � colg.G/. There is

evidence that the upper bound on the game coloring number colg.G/ may be sharp. How-

ever, we cannot even anticipate what the correct answer is for the game chromatic number

�g.G/ of planar graphs.

This situation probably exists because of the rather strange general behavior of the

parameter �g.G/. For instance, �g.G/ is already unbounded for graphs satisfying �.G/ D2 and is also not monotone with respect to subgraphs. Indeed, let U D fu1; : : : ; ung and

V D fv1; : : : ; vng be disjoint sets of vertices, and let Kn;n be the graph on these vertices

obtained by joining each vertex in U with all vertices in V . Let M be a perfect matching in

Kn;n (i.e., a set of n edges no two of which are incident with the same vertex; for instance,

M D fu1v1; : : : ; unvng). Clearly we have �g.Kn;n/ D 3, but �g.Kn;n � M/ D n. Bob’s

strategy is to copy Alice’s moves on the other ends of the missing matching (see Figure

11.8).

Alice

Bob

Figure 11.8. Copy cat strategy allows Bob to win on Kn;n �M if fewer than n colors are available.

More surprisingly, it is not clear what influence increasing the number of colors has

on the coloring game. Suppose that Alice wins with k colors on a graph G. Then it would

seem trivial that she should also win on G with k C 1 colors. But can you prove it? As yet,

nobody has been able to do so!

In view of such pathologies it is natural to look for more predictable variants of the

coloring game. For instance, in a modification considered in [3] Bob is not allowed to use

a new color until it is absolutely necessary, that is, unless all uncolored vertices are sur-

rounded by all previously used colors. Chen, Schelp, and Shreve prove there that ��g.T / �

3 holds for any tree T , where ��g.G/ is the related game chromatic number. However, no

better bounds for arbitrary planar graphs are provided. In this setting it is clear that increas-

ing the number of colors cannot help Bob. However, we are still left with the anomaly that

��g.Kn;n � M/ D n.

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11. The Map-Coloring Game 83

There are many natural variations of the coloring and ordering games. One variation is

to allow Alice more moves so as to limit Bob’s ability to interfere [12], [13], [17]. Other

variants consider oriented coloring [15], [16], [18] and d -relaxed coloring [4], [7]. It is

remarkable that the activation strategy is still effective in these diverse settings.

Acknowledgment We would like to thank Steven Brams for useful materials and com-

ments on the origin of the map-coloring game. The research of the fourth author is sup-

ported in part by the National Science Council under grant NSC95-2115-M-110-013-MY3.

Bibliography

[1] H. L. Bodlaender, On the complexity of some coloring games, Internat. J. Found. Comput. Sci.

2 (1991) 133–147.

[2] O. V. Borodin, On acyclic colorings of planar graphs, Discrete Math. 25 (1979) 211–236.

[3] G. Chen, R. H. Schelp, and W. E. Shreve, A new game chromatic number, European J. Combin.

18 (1997) 1–9.

[4] C. Chou, W. Wang, and X. Zhu, Relaxed game chromatic number of graphs, Discrete Math.

262 (2003) 89–98.

[5] V. Chvatal, V. Rodl, E. Szemeredi, and W. T. Trotter, The Ramsey number of a graph with a

bounded maximum degree, J. Combin. Theory Ser. B 34 (1983) 239–243.

[6] T. Dinski and X. Zhu, Game chromatic number of graphs, Discrete Math. 196 (1999) 109–115.

[7] C. Dunn and H. Kierstead, A simple competitive graph coloring algorithm III, J. Combin. The-

ory Ser. B 92 (2004) 137–150.

[8] U. Faigle, U. Kern, H. A. Kierstead, and W. T. Trotter, On the game chromatic number of some

classes of graphs, Ars Combin. 35 (1993) 143–150.

[9] M. Gardner, Mathematical games, Scientific American (April, 1981) 23.

[10] D. Guan and X. Zhu, The game chromatic number of outerplanar graphs, J. Graph Theory 30

(1999) 67–70.

[11] H. A. Kierstead, A simple competitive graph coloring algorithm, J. Combin. Theory Ser. B 78

(2000) 57–68.

[12] ——, Asymmetric graph coloring games, J. Graph Theory 48 (2005) 169–185.

[13] ——, Weak acyclic coloring and asymmetric graph coloring games, Discrete Math. 306 (2006)

673–677.

[14] H. A. Kierstead and W. T. Trotter, Planar graph coloring with an uncooperative partner, J. Graph

Theory 18 (1994) 569–584.

[15] ——, Competitive colorings of oriented graphs, Electron. J. Combin. 8 (2001) #12, 15 pp.

[16] H. A. Kierstead and Z. Tuza, Marking games and the oriented game chromatic number of partial

k-trees, Graphs Combin. 8 (2003) 121–129.

[17] H. A. Kierstead and D Yang, Very asymmetric marking games, Order 22 (2005) 93–107.

[18] J. Nesetril and E. Sopena, On the oriented game chromatic number, Electron. J. Combin. 8

(2001) #14, 13 pp.

[19] X. Zhu, The game coloring number of planar graphs, J. Combin. Theory Ser. B 75 (1999) 245–

258.

[20] ——, Refined activation strategy for the marking game, J. Combin. Theory Ser. B (2007, in

press).

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IIIFlexagons and

Catalan Numbers

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12It’s Okay to Be Square

If You’re a Flexagon

Ethan J. Berkove and Jeffrey P. Dumont

It has been said that a mathematician can be content with only paper and pencil. In fact,

there are times when one doesn’t even need the pencil. From a simple strip of paper it

is possible to make a surprisingly interesting geometric object, a flexagon. The flexagon

can credit its creation to the difference in size between English-ruled paper and American

binders. The father of the flexagon, Arthur Stone, was an English graduate student studying

at Princeton University in 1939. To accommodate his smaller binder, Stone removed strips

of paper from his notebook sheet. Not being wasteful, he creased these lengths of paper

into strips of equilateral triangles, folded them in a certain way, and taped their ends. Stone

noticed that it was possible to flex the resulting figure so that different faces were brought

into view—and the flexagon was born [1]. Stone and his colleagues, Richard Feynman,

Bryant Tucker, and John Tukey, spent considerable time cataloging flexagons but never

published their work.

Like many geometric objects, flexagons can be appreciated on many levels of mathe-

matical sophistication (the first author remembers folding flexagons in elementary school).

With so adaptable a form, it is not surprising that flexagons have been studied from points

of view that vary from art to algebra. Our interest in flexagons was sparked by a ques-

tion posed in a paper by Hilton, Pedersen, and Walser [9]. They studied one of the hex-

aflexagons, so-named because the finished model has the shape of a hexagon. They calcu-

lated the group of motions for a certain hexaflexagon, then inquired about other members

of the hexaflexagon family. We have determined that the trihexaflexagon is exceptional,

as it is the only member of the hexaflexagon family whose collection of motions forms a

group.

Working to generalize this result, we shifted our attention to tetraflexagons, which are

constructed from strips of squares folded into a 2 � 2 square final form. We discovered

that tetraflexagons are, if anything, more complicated and interesting than their hexagonal

cousins. These results convinced us that tetraflexagons, often only mentioned in passing

in the literature, deserve to be brought into the limelight. In this paper, we will summarize

the results of our investigations. Although most of the material on hexaflexagons is known,

the material on tetraflexagons includes new results and open questions. It is our intent to

Reprinted from Mathematics Magazine, Vol. 77, No. 5 (Dec. 2004), pp. 335-348.

87

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88 Part III. Flexagons and Catalan Numbers

give interested readers enough material to start their own explorations of these fascinating

objects.

The hexaflexagon family

a

b

c

d

e

f

g

h

i

tab

Figure 12.1. Trihexaflexagon net.

It is easy to fold a flexagon, and we highly recommend making one of your own as this

experience will be helpful in following the results in this section (and it’s fun). Construct

a strip of nine equilateral triangles and a tab as in Figure 12.1; this strip is called the net

of the flexagon. Each triangle in the strip, and in general each polygon in a flexagon net, is

called a leaf of the flexagon. You may want to label both sides of each leaf and precrease

all edges in both directions. Hold the leaf marked a in your hand, fold leaf c over leaf b, f

over e, and i over h. Finish the flexagon by gluing or taping the tab onto leaf a. The final

model should look like the flexagon depicted in Figure 12.2. Clockwise from the top, one

can read off the leaves f; d; c; a; i; g. We call .f; d; c; a; i; g/ a face of the flexagon.

ci

a

f

dg

Figure 12.2. The .f; d; c; a; i; g/ face of the trihexaflexagon.

To flex your new creation, bring the three corners at the dashed lines down together

so they meet. The hexagon will form a Y, at which point it will be possible to open the

configuration at the middle. (This is the only possible way to perform a flex-down for

this flexagon. There is also an inverse operation, a flex-up.) The result is a different face

.f; e; c; b; i; h/ of the flexagon. The flex can be repeated to get a third face .g; e; d; b; a; h/,

and one more flex returns the flexagon to .g; f; d; c; a; i/, the original face rotated clock-

wise through an angle of �=3. Since it has three distinct faces, this flexagon is known as

the trihexaflexagon—it is the simplest member of the hexaflexagon family. The three faces

can be seen more easily if they are marked somehow: Wheeler [15] shows how to color the

net so each flex brings out a new color, and Hilton et al. [9] give a way of marking the net

so flexes bring out happy and sad pirate faces.

We would like a way to keep track of all the faces of a flexagon while we flex, which

we can do using a graph: vertices represent the faces of the flexagon, and an edge joins

two vertices if there is a flex that takes one face of the flexagon to the other. On occasion

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12. It’s Okay to Be Square If You’re a Flexagon 89

fdcaig

fecbih gedbah

Figure 12.3. The structure diagram for the trihexaflexagon: vertices denote faces, edges denote flexes

between faces.

we will use a directed graph, where an arrow points towards the face that is the result of a

flex-down. We choose to ignore the orientation of a face in the graph as this has a tendency

to make the graph overly complicated. The completed graph is called a structure diagram.

The cycle in Figure 12.3 is the structure diagram for the trihexaflexagon. It shows the three

distinct faces, as well as their relationship via flexing.

The flex-down we described is a motion of the flexagon, a transformation that takes one

hexagonal face of the flexagon to another hexagonal face. We require that our flexagons

have no faces containing loose flaps that can be unfolded or moved so the hexagonal shape

is lost. This becomes a significant issue as the number of triangles in the net increases.

Indeed, in larger nets it is increasingly likely that a random folding of the net yields a face

containing a loose flap, which in turn causes the entire flexagon to fall apart into a Mobius

band with multiple twists. Therefore, we only consider flexagons that are folded in such a

way that every flex is a motion.

To determine the structure of the set of motions, let f denote a flex-down (hence f �1

denotes a flex-up), and denote by � a flip along the x-axis of the flexagon. Then f 3 D�2 D id and it is easy to confirm that �f � D f �1. The group with this presentation is

well known as S3, the symmetric group on 3 letters, which has 6 elements.

This analysis ignores the fact that f 3 is not strictly the identity flex, but is instead a ro-

tation through �=3 degrees. The complete set of motions of the trihexaflexagon, including

rotations, is analyzed in [9]; the only difference from the argument above is that f 18 D id ,

and the resulting group is D18, the dihedral group with 36 elements.

Motions of the hexahexaflexagon

We perform a similar analysis for the member of the hexaflexagon family with six faces.

This hexahexaflexagon can be constructed from the net of 18 triangular leaves with a tab

as in Figure 12.4. To create the flexagon, label all leaves front and back and precrease all

edges as before. Fold leaf a under the rest of the strip. Then fold the edge between leaves c

ab

cd

ef

gh

ij

kl

mn

op

qr

tab

Figure 12.4. The hexahexaflexagon net.

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90 Part III. Flexagons and Catalan Numbers

and d so that a and d are adjacent. Next fold the edge between e and f so that c and f are

adjacent. Continue rolling the strip in this manner until o and r are adjacent. The finished

roll should look like the initial strip for the trihexaflexagon with leaf b on the far left. Fold

this like the trihexaflexagon, then tape the tab to a to complete the model.

In contrast to the trihexaflexagon, from the initial face of the hexahexaflexagon either

alternating set of corners flexes down. We can distinguish the two flexes by looking at

what they do to the number of leaves in a triangular segment of the hexagon. Following

Oakley and Wisner [11], we call the entire triangular segment a pat. In our newly folded

hexahexaflexagon, pats alternately contain 2 and 4 leaves, or .2; 4/ for short. One of the

flexes, which we call f , preserves thicknesses so is pat-preserving. The other flex, g, is

pat-changing, from .2; 4/ to .1; 5/ and vice versa. The pat thicknesses are invariant under

� , a flip along the x-axis.

Starting with the initial face of the hexahexaflexagon, one finds that f 3 rotates the

hexagon clockwise through an angle of �=3, so f 18 D id . In addition, as �f � D f �1, f

and � generate a copy of D18 in the collection of hexahexaflexagon motions. On the other

hand, g leads to a face where the only possible flex is the pat-preserving f , which leads

to a face where the only possible flex is the pat-changing g. The three-flex combination

gfg rotates the hexagon clockwise through an angle of �=3, and g satisfies �g� D g�1.

Therefore, there is at least one other copy of D18 in the motions of the hexahexaflexagon.

We turn this information into the structure diagram in Figure 12.5. Solid arcs between

faces denote pat-preserving flexes, while dotted arcs denote pat-changing flexes. There is a

primary cycle that can be traversed using pat-preserving flexes, and three subsidiary cycles

that can be entered at faces where two flexes are possible. From the cycle structure we

learn an important fact: the collection of motions of the hexahexaflexagon must have at

least two generators, f and g. However, there are faces where only one of f and g can be

applied, so it is not always possible to apply f or g twice in a row. In fact, g3 does not even

make sense. We now have an answer to the question posed in [9]: the hexahexaflexagon’s

motions do not form a group!

This conclusion surprised us, as the collection of motions for every other geometric

object we know of has a group structure. Furthermore, all hexaflexagons except the trihex-

aflexagon share this characteristic, as they contain pat-changing flexes. Pat-changing flexes

f

g

f

Figure 12.5. Structure diagrams for the hexahexaflexagon: dotted edges are pat-changing flexes,

solid edges are pat-preserving flexes.

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12. It’s Okay to Be Square If You’re a Flexagon 91

occur at faces with two possible flexes, and the only hexaflexagon structure diagram with-

out intersecting cycles is the trihexaflexagon’s. Structure diagrams are necessarily finite,

and results in Wheeler [15] imply that cycles cannot form a closed link. As a result, in any

hexaflexagon there are only so many times the pat-changing flex can be applied. Thus, for

every hexaflexagon with a pat-changing flex g there is a k such that gk is undefined, which

implies that the collection of motions for that hexaflexagon cannot form a group.

The astute reader might have noted that the structure diagram in Figure 12.5 seems to

have nine faces, not six, as the hexahexaflexagon’s name suggests. This discrepancy can

be explained by carefully studying the hexahexaflexagon’s faces. Upon closer inspection,

three identical sets of triangles occur in two separate faces, but in different orders. If a face

depends on both the triangles and their order, then indeed the hexahexaflexagon has nine

faces. If order is disregarded, however, there are only six faces. The latter is the standard

accounting, hence the name (although Oakley and Wisner [11] distinguish between the six

physical faces and the nine mathematical faces).

We can construct other members of the hexaflexagon family by increasing the number

of triangles in the initial net. For example, starting with a straight strip of 36 leaves, the

strip can be rolled, then rerolled to yield the net in Figure 12.1, then folded as in the

trihexaflexagon case to yield the dodecahexaflexagon. One can fold other members of the

family by starting with nets that are not straight. The article by Wheeler [15] and Pook’s

book [14] contain some nice directions for folding the tetrahexa- and pentahexa- cases.

There is also a HexaFind program [3], which generates all nets for hexaflexagons with any

given number of faces.

We have only discussed a bit of what is known about hexaflexagons. In [11], Oakley

and Wisner introduce the concept of the pat, then use it to count the total number of hex-

aflexagons with a particular number of faces. Madachy [10], O’Reilly [12], and Wheeler

[15] describe connections between hexaflexagons and their structure diagrams. McIntosh

[7], Madachy [10], and Pook [14] provide lengthy bibliographies to other work on hex-

aflexagons. Gilpin [8], Hilton et al. [9], and Pedersen [5] study the motions of the trihex-

aflexagon. Furthermore, the group of motions of the trihexaflexagon is identified in [8]

and [9].

The distant cousins: tetraflexagons

We were introduced to tetraflexagons in a Martin Gardner Mathematical Games column

in Scientific American [6]. We immediately noticed some differences; the direct analog

of a strip of triangles, a straight strip of squares, makes for a poor flexagon; when you

fold square over square you end up with a roll that does not flex at all (and has a trivial

structure diagram). Therefore, we allow the nets to have right-angled turns. These turns

occur at what we call corner squares, those attached to their neighbors on adjacent edges

rather than opposite ones. Because so much is known about hexaflexagons, we felt that the

tetraflexagon family would be readily analyzed. Our actual experience mirrored Stone’s,

as reported by Gardner [6]: “Stone and his friends spent considerable time folding and an-

alyzing these four-sided forms, but did not succeed in developing a comprehensive theory

that would cover all their discordant variations.” However, we have established some nota-

tion and proved some results; our investigations have convinced us that tetraflexagons are

as fascinating as, and more subtle than, their hexaflexagon relatives.

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92 Part III. Flexagons and Catalan Numbers

a b c

d

e

f g h i

j

k

l

Figure 12.6. A “4 � 4 ring” net.

Here is one way to fold a tetraflexagon using a 4 � 4 net with a cut adjacent to a corner

square. Label a net as in Figure 12.6 and lay it flat. Mark the cut edges of a and l so you

can tape them together after you have folded the tetraflexagon.

1. Fold leaf c over leaf d . Leaves a and b will remain to the left of c, but will flip over.

2. Fold c and d over e; the back of leaf c will touch the front of leaf e, and again, a

and b are left free.

3. Fold f over g; move the flap with a and b so it points to the right.

4. Fold f and g over h so f and h are touching. Then slip the flap with a and b under j .

5. Fold j over i (without moving a and b), and flip the partially folded tetraflexagon

upside down. Position the tetraflexagon so k and l are at the top of the model and a

and b are upside down and facing to the left.

6. Fold i and j over k so i and k are touching.

7. Slide the flap with a and b under l , then fold a over l . Finally, tape a and l together

on the right, along the edges that were originally cut.

The finished tetraflexagon will have 90-degree rotational symmetry, and should look like

Figure 12.8a.

Now that you have a tetraflexagon in front of you, let’s introduce some notation (see

also Figure 12.7). The tetraflexagon has four pats, which conveniently look like the four

quadrants in Cartesian coordinates. Therefore, we will refer to pats by the quadrant they

are in: pat I, pat IV, etc. If you look carefully at your tetraflexagon, you’ll notice that

adjacent pats are connected by a layer of paper, called a bridge. The two leaves, one in

pocket

looseflap

foldable edge

freesides

connectedsides

Figure 12.7. Tetraflexagon notation.

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12. It’s Okay to Be Square If You’re a Flexagon 93

each quadrant, that make up this layer are bridge leaves. Leaves b and c are bridge leaves,

as are k and l . Bridges will figure prominently in our analysis.

Define a pocket as a strict subset of leaves within a pat attached to the rest of the

tetraflexagon on two adjacent sides. Without the strictness condition, an entire pat is a

pocket, connected to the rest of the flexagon by the two bridges; this is an extreme case we

wish to avoid. On the other hand, a pocket may consist of a single leaf, in which case the

pocket is a bridge leaf. As an example, squares f and g form a pocket in the tetraflexagon

you just folded and this is the only pocket in its pat. The attached sides of the pocket are

connected, whereas the other two sides are free. When a free side of a pocket lies along

the outside edge of the tetraflexagon (like the outside edge of h and i ), we call the entire

tetraflexagon edge a foldable edge. The foldable edge will be folded in half during the flex.

There are two conditions that a tetraflexagon must satisfy in order to flex: there must be

a pocket with a component bridge leaf, and the bridge in the pocket must be at right angles

to the foldable edge. (Since the bridge goes from a pocket to an adjacent pat it cannot

cross the pocket’s free side.) To perform a flex, orient the tetraflexagon so the foldable

edge is forward and the pocket lies on top of the tetraflexagon, as in Figure 12.8a. Fold the

tetraflexagon in half perpendicular to the foldable edge so that the pocket remains on the

outside. Put your thumbs into the pocket and its kitty-corner companion (Figure 12.8b),

then pull outwards in the direction of the arrows. The pocket layers will separate from the

rest of the pat, rotating outwards 180 degrees but staying in the same quadrant. The rest

of the pat layers will maintain their orientation but move to the top of the adjacent pat.

Simultaneously, the layers in the adjacent pats will rotate outwards 180 degrees. When you

flatten the tetraflexagon, you will see a new face as in Figures 12.8c and 12.8d.

a b c d

Figure 12.8. Flexing a tetraflexagon.

We call the flex with the pockets up and the foldable edge forward a book-down flex.

It is only one of four possibilities, depending on the position of the pocket and foldable

edge. We therefore distinguish between four types of flexes: flexes up or down along the

y-axis (book-up and book-down) and flexes up or down along the x-axis (laptop-up and

laptop-down). From Figure 12.8, we see that an up flex is the inverse of a down flex and

vice versa. As before, the top layer of leaf of the tetraflexagon, like .a; d; g; j /, is called a

face, and a flex is a motion if it takes one tetraflexagon face to another.

Before we go further into our exploration, we introduce some assumptions about the

tetraflexagons we consider to aid in our analysis.

1. Net assumption. All tetraflexagon nets have exactly four corner squares. The two

vertical strips are each composed of m leaves, and the two horizontal strips of n

leaves. All nets are folded into four-pat (2 � 2) tetraflexagons.

2. Winding assumption. There is only one bridge between adjacent pats of the tetraflex-

agon. In other words, as the net is folded, the strip of paper enters and exits each

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94 Part III. Flexagons and Catalan Numbers

quadrant exactly once. This assumption is easily justified; we will see in the Flexing

Lemma below that subsequent flexes maintain all four bridges as single layers.

3. Rigidity assumption. No tetraflexagon face contains a loose flap. A loose flap is a col-

lection of two or more leaves connected to the rest of the tetraflexagon on only one

side (see Figure 12.7). Since a loose flap determines one of two possible tetraflex-

agon faces depending on its position, movement of a loose flap can be considered

a motion. The rigidity assumption ensures that the only tetraflexagon motions are

flexes.

4. Symmetry assumption. Folded tetraflexagons have 180-degree rotational symmetry

(in contrast to the hexaflexagons’ 120-degree symmetry).

Figure 12.9. Rings, bolts, and snakes.

The net assumption implies that nets are one of three shapes, depending on parity. All

nets are shown in Figure 12.9: when m and n are both even, the only possible net is a ring;

when both m and n are odd, the only possible net is a lightning bolt; and when exactly one

of m and n are odd, the only possible net is a snake. We remark that analogs of the other

three assumptions hold for most hexaflexagons.

Basic analysis: folding 101 and structure diagrams with dead ends

Flexing for tetraflexagons is a bit more complicated than for hexaflexagons. We carefully

describe what happens during a flex in the following

Flexing Lemma. The following hold for tetraflexagons under our assumptions:

1. Corner squares remain in their pats during flexes.

2. A pocket that has a component bridge leaf is either a single layer (that is, it is the

bridge leaf), or consists of every leaf in the pat except for either the top or bottom

one. In the latter case, the leaf not belonging to the pocket is also a bridge leaf.

3. A flex maintains all four bridges as single layers.

Proof. To simplify the arguments, we will assume for this proof that the pocket is always

on top of pat IV with the foldable edge forward, as in Figure 12.8a. The other cases will

follow by rotation and mirror images. Moreover, by the symmetry assumption we only

need focus on pats III and IV.

We start with the first claim. When we perform the flex shown in Figure 12.8, all leaves

in pat III remain in pat III, which means that the only corner square that can change pats

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12. It’s Okay to Be Square If You’re a Flexagon 95

lies in pat IV. However, if the book-down flex moves this corner square to pat III, then the

flex is also forced to move the strip of leaves connecting the corner squares in pats I and

IV. This is impossible as the corner square in pat I is fixed by the symmetry assumption.

For the second claim, assume that the pocket with a bridge leaf consists of more than

one layer. We claim that the corner square in pat IV must be a leaf of the pocket. By

way of contradiction, assume the corner square is below the pocket. Notice that the corner

squares in pats I and IV are connected by a vertical strip. As this strip is woven from the

corner square in pat IV to the corner square in pat I, it must at some point become the

unique bridge to pat I (by the winding assumption). One of the bridge leaves is a leaf of

the pocket, so the vertical strip of squares must also cross from the bottom of pat IV to the

pocket. Because the vertical strip is woven back and forth, it can only cross to the pocket

on the front or back edge of pat IV. The vertical strip cannot cross the gap at the front edge

of pat IV because that edge is free by assumption. The vertical strip cannot cross the gap

at the back edge of the pocket either, as that would lock the pocket to the rest of pat IV,

making a flex impossible. We conclude that the corner square must be a leaf of the pocket.

An analogous argument shows that the vertical strip cannot cross from the corner

square to the bottom layer either. By the rigidity assumption, this bottom layer must there-

fore consist of a single leaf. Now if the bridge leaf to pat III were any layer but the bottom,

it would force the left side of the pocket in pat IV to be connected, rather than free. This is

impossible as all pockets have two free sides.

For the final claim, the winding assumption implies that, when first folded, the tetra-

flexagon has single layer bridges between pats. Consider the effect of a book-down flex.

This flex flips the bridge leaves that lie in the “pages” of the book upside down, so those

bridges remain single layers. For the two bridges that cross the spine of the book, focus

on pats III and IV, and assume first that the pocket is a single layer; then the book-down

flex leaves a single layer in pat IV, which must be the bridge leaf. Otherwise, the pocket

has more than one leaf and the bottom leaf of pat IV is a single layer. In this case, the

completed flex moves the single layer from pat IV to pat III, again maintaining the bridge

as a single layer.

The Flexing Lemma gives us insight about what happens to a tetraflexagon during a

single flex. What about the larger picture? What can we say about the set of all the motions

of the tetraflexagon we folded? Again, we answer these questions with a structure diagram,

but now use the orientation of the graph’s edges to distinguish among the four possible

types of flexes. Denote a book flex by a horizontal line segment in the structure diagram,

a laptop flex by a vertical one, and let arrows point to the faces that are the result of a flex

down. With a little flexing and experimentation, we find that the structure diagram for the

tetraflexagon we folded is an L shape, with five vertices (corresponding to faces), and four

edges, two in each part of the L (corresponding to flexes). On all edges, the arrows point

away from the corner of the L. This structure diagram, which appears in the bottom left of

Figure 12.11, has a feature we have not seen before: there are vertices incident to a single

edge. We call the faces associated to these vertices dead ends, because once these faces are

reached via a flex, the only motion that can be applied is the flex’s inverse.

For the hexaflexagon cases, a flex of finite order appeared in the structure diagram as a

cycle. The structure diagram we just constructed has no cycles, hence no elements of finite

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96 Part III. Flexagons and Catalan Numbers

order. In addition, any flex can be applied at most twice, at which point some other flex

must be applied. In other words, there is really no group structure at all in the motions of

the tetraflexagon we constructed. An immediate question is whether there is a tetraflexagon

whose collection of motions forms a group.

There is, and it can be folded from the 3 � 3 lightning bolt net shown in Figure 12.10.

Start with a copy of this net—you may want to specially mark the shaded sides of a and

h as shown in the diagram, as those are the edges that get taped together at the end. Fold

leaf b over c, d (and a, b, and c) over e, and g over f . Make sure h lies on top of a, then

tape a and h together on the right, along their shaded sides. Your finished tetraflexagon

should have 180-degree rotational symmetry, as per the symmetry assumption. From this

initial face, the only possible down flex is a laptop-down. If this is followed with book-

down, laptop-down, and book-down flexes (again, the only possible down flexes) you get

back to the initial face. The resulting structure diagram is a square cycle, corresponding to

the symmetry group Z=4Z, the cyclic group with 4 elements. We believe that this is the

only tetraflexagon whose collection of motions forms a group (although a rigorous proof

eludes us).

a

b

c d e

f

g h

Figure 12.10. A 3 � 3 bolt net.

You may have noticed that although we call the net in Figure 12.10 a 3 � 3 lightning

bolt, there are only two leaves, g and h, in one row. Actually, that row does contain a third

leaf, a, which becomes part of the row after we finish taping. In general, when we refer to

a net as m by n, we include the corner squares in the count. For the cases of the bolt and

snake nets, this means that either one row or one column will be a leaf short when the net

is first constructed.

By reducing larger nets to the 3 � 3 case, one can construct many tetraflexagons whose

structure diagrams contain cycles. For example, starting with the net in Figure 12.6, folding

b over c and i over h turns the 4 � 4 ring net into a 4 � 3 snake net. Another pair of folds

results in a 3�3 lightning bolt net, which can then be folded as above. More generally, one

can turn an m�n net into either an m�1�n net or an m�n�1 net by making appropriate

folds. Here, appropriate means that the pair of folds results in a net that satisfies the sym-

metry assumption. Also, a given sequence of edge choices must result in a tetraflexagon

that satisfies our four assumptions. Once the net is a 3 � 3 lightning bolt, it is folded to

guarantee at least one cycle in the structure diagram. This process is reminiscent of how a

hexaflexagon is folded from a strip containing 9.2n/ triangles by rolling it n times to form

the trihexaflexagon net with 9 triangles. A difference, though, is that there are many ways

of choosing edges in pairs to reduce an m�n net to the 3�3 net, so a generic tetraflexagon

net can yield a large number of tetraflexagons with distinct structure diagrams.

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12. It’s Okay to Be Square If You’re a Flexagon 97

Figure 12.11. Tetraflexagon structure diagrams folded from a 4 � 4 ring.

As an example, through exhaustive folding, both figurative and literal, we have identi-

fied in Figure 12.11 most (all?) of the possible structure diagrams that result from a 4 � 4

ring. Some of the diagrams are very basic, but notice that three of them contain cycles, all

but one contain dead ends, and two contain both. This variety in structure diagrams from

the same net demonstrates how difficult it is to classify even simple tetraflexagons.

Despite their “discordant variations,” the tetraflexagons we investigated have some fea-

tures in common. In a number of the examples we folded, we came across faces where all

four flexes were possible. We call such faces crossroads. For the reader who wants to see a

crossroad in action, start with the net in Figure 12.6, labeled front and back with each letter.

Fold b under c, c over d , e (and a–d ) under f , g (and the rest of the net) over f , i over

h, j over i , and l over k. Lift a over l , then tape a and l together on the right (outside)

edge. The final tetraflexagon is shown in Figure 12.12d. (What is the resulting structure

diagram?) Crossroads are among the most interesting features in a structure diagram, and

we wondered how many crossroads a structure diagram could contain. We noticed that

after we flexed a crossroad, we never saw another crossroad, which led us to prove the

Crossroad Theorem. Crossroad faces are never adjacent in tetraflexagons built from ring,

lightning bolt, or snake nets.

Proof. Let’s analyze the shape of a tetraflexagon at a crossroad face. In order to perform all

four types of flexes, every edge of the tetraflexagon must be a foldable edge and there must

be two pockets per edge to allow both types of flexes. It is easy to show that the pockets

cannot be on the same side of the foldable edge, as in Figure 12.12a. If the pockets were

on the same side, the remaining leaves of the two pats would be the bridge between the

pats, by the second claim of the Flexing Lemma. By the symmetry assumption, the same

would be true for the other two pats of the tetraflexagon, and the entire tetraflexagon would

simply fall apart.

We conclude that one pat along the foldable edge must have a pocket on the top of

the pat to allow a flex down, while the adjacent pat along the edge must have a pocket

on the bottom of the pat to allow a flex up. There are essentially two cases to consider,

shown in Figures 12.12b and 12.12c (Figure 12.12d is Figure 12.12c’s mirror image). We

can show that Figure 12.12b is impossible by a careful analysis of pat IV. Since one of the

two pockets in pat IV contains all the layers but one, assume, without loss of generality,

that it is the pocket associated to the foldable edge on the right. Then the bridge to pat III

crosses the edge marked by the arrow, and the component bridge leaf in pat IV is one of

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98 Part III. Flexagons and Catalan Numbers

a b c d

Figure 12.12. Two impossible configurations and two crossroads.

the leaves in the pocket by the second claim of the flexing lemma. However, this implies

that the edge marked by the arrow is a connected edge, which makes a book-down flex

impossible. Therefore, the only possible crossroad faces have configurations like Figures

12.12c and its mirror image 12.12d.

We next determine when a flex from a crossroad yields another crossroad. By symmetry

arguments, it is sufficient to consider Figure 12.12c. We claim that is impossible to perform

a book-down flex from Figure 12.12c and get the crossroad in Figure 12.12d. This time,

consider pat III. Note that by the second claim of the Flexing Lemma one of the two pockets

in pat III must consist of every layer in the pat except one; assume (by symmetry again)

that it is the bottom pocket. After a book-down flex is applied, the original layers in pat III

are flipped upside down, and are covered by a single layer from pat IV. This single layer

must be part of the new bridge to pat IV. Therefore, it is not possible for the right edge of

pat III to be free, as in Figure 12.12d.

Thus, if two crossroads are connected by a flex, they must be identical. We rule out

this possibility too. Consider the enlargement of pat IV of Figure 12.12c, shown in Figure

12.13. In order for a book-down flex to yield the same configuration, pat IV must contain

a pocket as part of the bridge. But the bridge must be a single layer, by the second claim of

the Flexing Lemma.

Figure 12.13. Closeup of pat IV.

The Crossroad Theorem tells us that cycles cannot be too dense in a structure diagram.

For example, we have a corollary to the theorem, named after the symbol that appears on

all Purina pet foods.

Purina Corollary. Under our assumptions, there is no “Purina Tetraflexagon” with the

structure diagram shown in Figure 12.14.

Surgery: adding new parts to an old flexagon

So far, we have a very general idea of what can and cannot be part of a structure diagram.

On the other hand, it would be desirable to add a given component to a structure diagram,

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12. It’s Okay to Be Square If You’re a Flexagon 99

Figure 12.14. Purina structure diagram.

allowing us to tailor-make tetraflexagons with interesting properties. There is a technique

we learned from Harold McIntosh’s online notes [7] that allows us to do exactly that at

an appropriate spot in a structure diagram. We call this technique surgery. One performs

surgery on a tetraflexagon by symmetrically grafting two strips of squares to the model to

add the new feature.

An instance of surgery is shown in Figure 12.15. Start with a tetraflexagon at a face

where one of the pats is a single leaf, that is, a corner square. Place this tetraflexagon so

the single leaves are in pats II and IV with the foldable edge forward, as in the top left

picture in Figure 12.15. By considering mirror images, if necessary, we may assume that

the pocket in pat III is on the bottom of the pat. Tape a 2 � 1 strip to the right edge at pat

IV and make a cut in the tetraflexagon along the positive x-axis as in the top right picture.

Fold the strip on top of pat IV, then fold the strip up and in half as in the bottom left picture.

Finally, tape the top edge of the strip (in pat IV) to the cut edge in pat I as in the bottom

right picture. Perform the same procedure symmetrically on pat II; when you are done,

the front and back foldable edges will have pockets on both top and bottom. The resulting

tetraflexagon’s structure diagram will be the same as before, but with a cycle tacked on at

the appropriate face.

Following this technique, it is straightforward, at least in theory, to build a tetraflexagon

whose structure diagram consists of an arbitrarily long row of four-cycles connected one

to the other at opposite corners. (In light of the Crossroad Theorem, in some sense this is

Figure 12.15. How to add a cycle to a tetraflexagon.

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100 Part III. Flexagons and Catalan Numbers

Figure 12.16. A challenging structure diagram.

as crowded as collections of cycles can become.) In practice, the tetraflexagons quickly

become too thick to flex easily as the size of the net increases. We challenge the reader to

use surgery to construct a tetraflexagon with the structure diagram shown in Figure 12.16.

One can also use surgery to add dead ends to a structure diagram. Start with a tetraflex-

agon in the same position as before and make the same cut along the positive x-axis, but

this time add a 1 � n strip horizontally to the right edge of pat IV. Roll the strip counter-

clockwise until it lies on top of pat IV, then tape the top edge of the middle square of the

roll to the cut edge of pat I. Repeat this symmetrically on pat II. This procedure will add n

horizontal segments to the structure diagram at the appropriate place.

To finish this section we mention that besides appearing in Gardner articles [2] and [6],

“square” flexagons are the topic of a short note by Chapman [2] where he uses primary

and secondary colors to distinguish tetraflexagon faces. This note also contains directions

for constructing tetraflexagons whose structure diagrams are a cycle and two linked cycles.

For more information on surgery and tetraflexagons, the reader should consult McIntosh’s

notes [7] or Pook’s book [14].

Where do we go from here?

At this point, we know something about tetraflexagons—how they flex, how and where

cycles can occur, and why some structure diagrams are impossible—but there are many

questions we haven’t answered. What are the possible tetraflexagon structure diagrams?

How many tetraflexagons, up to rotation and mirror image, can be made from an m � n

net with four corners? We can get a rough upper bound by recalling that the corner squares

in a tetraflexagon stay fixed in their quadrant. Flex a tetraflexagon so that one pat consists

of a single leaf and its adjacent pats contain n C m � 1 leaves. With the exception of the

bottom (or top) leaf, these might be in any order, so there are at most .nC m� 2/Š possible

tetraflexagons that can be folded from a given m � n net. Many of these tetraflexagons will

not satisfy our four assumptions; is there a better upper bound? Finally, there is the question

that provided our initial motivation to study flexagons: Are there any other tetraflexagons

besides the one built from the net in Figure 12.10 whose collection of motions forms a

group?

We developed most of our tetraflexagon results before looking through McIntosh’s

material [7], and you can imagine our surprise when we saw directions for the Purina

Tetraflexagon in his notes! McIntosh introduced this creature in a discussion on surgery,

and we were no less surprised when we constructed it and confirmed that it worked. The

apparent discrepancy between beautiful theory and ugly counterexample was explained

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12. It’s Okay to Be Square If You’re a Flexagon 101

once we took a scissors to the flexagon and opened it up; in the net, every one of the

24 leaves was a corner square! Clearly, in order to do surgery while maintaining the net

assumption, one must be careful that the transplanted leaves do not add any more corner

squares.

More importantly, this example shows that there are many interesting tetraflexagons

that do not satisfy the net assumption. In fact, perhaps our favorite flexagon violates both

the net and symmetry assumptions, as it is folded from regular pentagons. Of course, this

flexagon cannot be folded flat, but we were impressed that such an object could exist at all.

This flexagon, and many others, are described in McIntosh’s notes [7] as well as Pook’s

book [14]. We highly recommend these references, as well as an impressive report by

Conrad and Hartline [1] (also found in [7]) for the reader interested in deepening his or her

background about flexagons and their kin.

We have found flexagons to be interesting mathematical objects at many levels. Since

they are easily folded, they are a good addition to an introductory mathematics class, giving

students an opportunity to look for patterns and explore a topic at their own pace. In a

more advanced setting, such as a course in geometry or algebra, flexagons can be used

to introduce notions of symmetry and transformation. They are also an excellent topic of

study for a senior thesis, as most of the flexagon materials are written at an elementary level

and are appropriate for a student’s introduction to the reading of mathematical articles.

And, of course, there are many open questions that are easily asked but difficult to answer.

Despite their long history, flexagons still give an exciting twist to an otherwise boring strip

of paper, and are well worth a little study.

Acknowledgments We would like to thank Harold McIntosh for informative correspon-

dence, and Liz McMahon, Kyra Berkove, and the anonymous referees, whose comments

greatly improved the exposition of this paper.

Bibliography

[1] Anthony Conrad and Daniel Hartline, Flexagons, RIAS Technical Report 62-11, Baltimore,

MD, 1962.

[2] P. B. Chapman, Square flexagons, Math. Gaz. 45 (1961) 192–194.

[3] Antonio De Queioroz, The HexaFind Program,

www.coe.ufrj.br/˜acmq/hexaflexagons.

[4] Martin Gardner, Hexaflexagons and Other Mathematical Diversions, Univ. of Chicago Press,

Chicago, 1988.

[5] ——, The 2nd Scientific American Book of Mathematical Puzzles and Diversions, Simon and

Schuster, New York, 1961.

[6] ——, Mathematical games, tetraflexagons and tetraflexigation, Scientific American, May 1958,

122–126.

[7] Harold McIntosh, Flexagons,

delta.cs.cinvestav.mx/˜mcintosh/oldweb/pflexagon.html.

[8] Michael Gilpin, Symmetries of the trihexaflexagon, Math. Mag. 49 (1976) 189–192.

[9] Peter Hilton, Jean Pedersen, and Hans Walser, The faces of the trihexaflexagon, Math. Mag. 70

(1997) 243–251.

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102 Part III. Flexagons and Catalan Numbers

[10] Joseph Madachy, Mathematics on Vacation, Charles Scribner, New York, 1966.

[11] C. O. Oakley and R. J. Wisner, Flexagons, Amer. Math. Monthly 64 (1957) 143–154.

[12] Thomas O’Reilly, Classifying and counting hexaflexagrams, J. Recreat. Math. 8 (1975) 182–

187.

[13] Jean Pedersen, Sneaking up on a group, Two Year College Math. J. 3 (1972) 9–12.

[14] Les Pook, Flexagons Inside Out, Cambridge Univ. Press, Cambridge, 2003.

[15] Roger Wheeler, The flexagon family, Math. Gaz. 42 (1958) 1–6.

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13The V-flex, Triangle Orientation, and

Catalan Numbers in Hexaflexagons

Ionut E. Iacob, T. Bruce McLean, and Hua Wang

Using only the pinch flex, the flex described by Martin Gardner in [4], the six triangles of

each hexagonal face of a hexaflexagon stay together. If the faces are colored, the face facing

up is always monochrome. To scramble the triangles and mix the colors, you need other

flexes. In this paper, we describe the V-flex. With the V-flex, faces become multi-colored

when flexed. It takes persistence to master, but the V-flex is worth it. A hexahexaflexagon

has only 9 mathematical faces under the pinch flex; with the V-flex it has 3,420.

We conjecture that many people performed a V-flex accidentally as they read Martin

Gardner’s 1956 article and had no expert help. In 1963, Bruce McLean discovered the

V-flex and, in 1978, provided a graph of the faces. He presented it to Arthur Stone at an

MAA meeting that year and was told “I never allowed my students to do that!” In that

paper, McLean approached the V-flex systematically as a new flex. The view that the V-

flex is somehow wrong, however, persists to the present. In 2011, a referee of this paper

wrote, “I’ve taken the majority view, that it was an illegal move and that my only purpose

in understanding it was to fix ‘broken’ flexagons.” The V-flex is now understood as a well-

defined operation (see [8] for an overview) that adds considerable complexity to an already

interesting object. After the meeting with McLean, Stone expressed pleasure that order still

prevailed under the V-flex. We hope the reader will be pleased too.

V-flexing

Begin by constructing a hexahexaflexagon (see Figure 12.4 in the previous chapter). For

clarity, we prefer to number all triangles in the strip of paper from left to right on the front

and back of the paper as in our Figure 13.1, and add arrows pointing to each triangle edge

not shared with another triangle. The tab is helpful for construction, but not required.

Fold the triangle UP 2 under the 1 and continue to spiral in the same direction to obtain

a shorter strip of 9 triangles as in Figure 13.1c. Next, fold triangle UP 5 under the 4, UP

11 under the 10, and UP 17 under the 16. Finish by taping the tab to UP 1. The result

is the hexahexaflexagon shown in Figure 13.2a. Note that each triangular region of the

hexahexaflexagon, called a pat, has a thickness, i.e., the number of triangles. This number

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 6–10.

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104 Part III. Flexagons and Catalan Numbers

Face UP Face DOWN1 2 3 17 18 1 2 17 18

tab

(a) (b)

1

4

5 17

16

(c)

Figure 13.1. Constructing a hexahexaflexagon.

is the degree of the pat. The strip would continue straight and a hexagon could not be

constructed if the degree of any pat was divisible by three, hence pat degrees are always

congruent to 1 or 2 (mod 3).

Pat 2

Pat 1

Pat 6

Pat 5

Pat 4

4u

1u16u

13u10u

7

6

4

116

14

13

3

UP

(a) (b)

Figure 13.2. Starting the V-flex.

For the V-flex, separate Pat 2 with your left thumb into two halves (see Figure 13.2a).

It may help to fold the whole flexagon in half, so that 7, 10 and 13 are on one side. You

may also need your right thumb to assist in separating Pat 5. The flexagon should begin to

open (Figure 13.2b) and eventually look like Figure 13.3a (as seen from the top). Note that

Figure 13.3a is three-dimensional: an inverted pyramid with a square base and tetrahedra

Pat 2

Pat 1

Pat 6

Pat 5Pat 3

Pat 4

76

3

14

13

16

6

18

15

14

113

(a) (b)

Figure 13.3. Completing the V-flex.

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13. The V-flex, Triangle Orientation, and Catalan Numbers in Hexaflexagons 105

on two opposite faces. Pinch 13 with your right thumb on top and pull it towards you while

pushing 6 away from you. Rotating 13 under 11 completes the flex (see Figure 13.3b). (An

online video may be helpful; see [9].)

Note that a V-flex is not possible if either Pat 2, 5, or 6 contains only 1 triangle. To

return to the original, flip the flexagon on the horizontal axis, rotate 60 degrees clockwise,

and V-flex again. Enjoy!

Triangle orientation

Many flexagon lovers wrote Martin Gardner [5] to point out that numbers alone are not

enough to describe a flexagon configuration completely. A triangle can come up face up

or face down and with different orientations, that is, the arrow mentioned in the previous

section may point in different directions within the face.

We define the state of a triangle as the following three pieces of information: (i) face

position: up (U) or down (D); (ii) triangle number: even (E) or odd (O); and (iii) triangle

orientation (the arrow’s direction): outside the hexagon (O), towards the next pat in the

face counterclockwise (N), or towards the previous pat in the face (P). There are thus 12

distinct states. For instance, the state UOO means that the triangle is facing up, is odd-

numbered, and the arrow points outwards in the hexaflexagon; the state DOP means a face

down triangle, odd-numbered, and with the arrow pointing clockwise.

Simple observations lead to the fact that from one triangle to the next, in increasing

numerical order, parity changes (evidently), but facing up or down changes only if both

triangles are in the same pat. We say that a transition from one triangle to the next one

is a folding transition (FT), if both triangles are in the same pat; or a pat transition (PT),

if they are in consecutive pats. Then, it is easy to observe that, for instance, from a DON

triangle we can only have a folding transition (FT) to the next triangle in the sequence,

which therefore must be UEO (U because it is in the same pat; E because it comes after

an odd numbered triangle; and O as one can determine by looking at the original strip

of paper).

We summarize these observations with the labeled, directed graph in Figure 13.4,

where each edge is either a FT (folding transition) or a PT (pat transition). The diagram

shows clearly which states allow a FT or PT, or both, and that the number of triangles

folded in consecutive pats forms a sequence consisting of numbers congruent alternately

to 1 and 2 (mod 3).

Given any snapshot of triangle numbers in a pat, we can easily determine the complete

state (with orientation) of every triangle. For instance, consider the pat consisting of trian-

gles 2, 1, 4, 3 (from top to bottom) in a hexahexaflexagon. Triangle 1 must be at a tip of

a PT edge (hence in either DOO or UON states) and must begin a path of three FT edges

leading to a state that allows the next PT (since the pat contains 4 triangles). This puts

triangle 1 in DOO state. Then, as we move to triangle 2, we must follow a FT edge (it is

the same pat), which places triangle 2 in UEP state. We continue with a FT to 3 in DON,

and another FT to 4 in UEO.

Catalan numbers in flexagons

The enumeration of pat classes (under a natural equivalence relation according to how the

degree splits, see [1]) enjoys a beautiful connection to the well-known Catalan numbers,

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106 Part III. Flexagons and Catalan Numbers

UEO

DEO

UOO

UOP

DEN

UON

DEP

DOP

UEN

DON

UEP

DOO

FT FT FT FT

FT FT

FT FT

FT FT

PT PT

FT FT

PT PT

Figure 13.4. Triangle state diagram.

which we proceed to describe. A pat of degree n; n > 1, splits naturally where a “thumb-

hole” is formed, the one place in the physical pat where the thumb can be inserted without

encountering a pocket [4]. This divides the pat into two parts whose degrees sum to n.

Thus each pat class can be represented as a partition of n into an ordered sum of two pos-

itive integers representing the degrees of its two principle parts. Like n, these integers are

equivalent either to 1 or to 2 (mod 3). Then each part in this partition is in turn a sum of

two such parts and so on, until every part equals 1. For example, a pat class of degree 10

may be represented as follows:

10 D 2 C 8

D .1 C 1/ C .1 C 7/

D .1 C 1/ C .1 C .5 C 2//

D .1 C 1/ C .1 C ..1 C 4/ C .1 C 1///

D .1 C 1/ C .1 C ..1 C .2 C 2// C .1 C 1///

D .1 C 1/ C .1 C ..1 C ..1 C 1/ C .1 C 1/// C .1 C 1///:

From a combinatorial point of view, such a representation assigns pairs of parenthesis

to the sum of n 1’s keeping the sum within each pair of parenthesis either 1 or 2 modulo 3.

The number of pat classes is the number of ways to complete this process. The well known

Catalan numbers (which Martin Gardner also popularized [3]) count, among other objects

(including planar trees, lattice paths, triangulations of a polygon), exactly this quantity

without the (mod 3) constraint. The number of pat classes can therefore be analyzed using

the same combinatorial tools used for the Catalan numbers (see [2]). In the recent paper of

Anderson et al. [1], a pair of recursive relations for the number of pat classes with a given

degree is given, namely

bn DnX

iD0

ai an�i and an Dn�1X

iD0

bibn�i�1;

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13. The V-flex, Triangle Orientation, and Catalan Numbers in Hexaflexagons 107

where an is the number of pat classes of degree 3n C 1 and bn is the number of pat classes

of degree 3n C 2. The reader may have already noticed that when partitioning a number

of the form 3n C 1, the two parts must be of the form 3n C 2, and vice versa. This is the

reason for this bi-variable recursion.

The enumeration of various objects related to the flexagons that Martin Gardner intro-

duced to a wide audience in 1952 is now an interesting area of research. Both sequences

fang and fbng are in the On-Line Encyclopedia [11]. Among other interesting concepts

associated with an are the number of 4-ary (or quartic) trees with 3n C 1 leaves. It turns

out that

an D 1

3n C 1

4n

n

!

and bn D 2

3n C 2

4n C 1

n

!

:

(For more on the combinatorics of flexagons, see [1] and [4].)

Bibliography

[1] T. Anderson, T. B. McLean, H. Pajoohesh, C. Smith, The combinatorics of all regular

flexagons, European J. Combin. 31 (2010) 72–80; available at dx.doi.org/10.1016/

j.ejc.2009.01.005.

[2] R. Brualdi, Introductory Combinatorics, 5th ed., Pearson, 2008.

[3] M. Gardner, Catalan Numbers, Time Travel and Other Mathematical Bewilderments, W. H.

Freeman, 1987.

[4] ——, Hexaflexagons, College Math. J., 43 (2011) 2–5; available at dx.doi.org/

10.4169/college.math.j.43.1.002.

[5] ——, Hexaflexagons and Other Mathematical Diversions, Cambridge Univ. Press, 2008.

[6] I. E. Iacob, T. B. McLean, and H. Wang, About general order regular flexagons (submitted).

[7] T. B. McLean, V-Flexing the hexahexaflexagon, Amer. Math. Monthly 86 (1979) 457–466;

available at dx.doi.org/10.2307/2320415.

[8] ——, Flexagons; available at

math.georgiasouthern.edu/˜bmclean/flex/index.html.

[9] R. Moseley, Flexagon.net; available at

www.flexagon.net/index.php?option=com_content&task=view&id=22.

[10] C. O. Oakley and R. J. Wisner, Flexagons, Amer. Math. Monthly 64 (1957) 143–154; available

at dx.doi.org/10.2307/2310544.

[11] N. Sloane, The On-Line Encyclopedia of Integer Sequences; available at oeis.org/.

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14From Hexaflexagons to Edge

Flexagons to Point Flexagons

Les Pook

Edge flexagons

Hexaflexagons, the subject of Martin Gardner’s column [5], were the first family of flexagons

to be discovered. They are, however, only one example of an edge flexagon, a folded band

of hinged, usually identical, convex polygons (called leaves). The strip of polygons used to

construct a flexagon is called a net. In a main position an edge flexagon has the appearance

of a ring of hinged polygons (see Figure 14.1). The rings are not necessarily flat and what

look like single polygons are folded piles of one or more leaves, called pats. By defini-

tion, edge flexagons can be flexed to other main positions to display other pairs of faces,

usually in a cyclic order. When flexing, leaf-bending is sometimes allowed, but in main

positions the leaves are flat. The preceding chapter in this part concerns hexaflexagons

(Figure 14.1a). Here we describe some non-hexagonal forms and say something about

their classification.

(a) Trihexaflexagon. (b) Square flexagon. (c) Pentagonal flexagon.

Figure 14.1. Main positions of some edge flexagons.

Workable paper models of flexagons are easy to make; however, the mathematics is

complex and just how a flexagon works is not immediately obvious from casual examina-

tion of a paper model. Even square flexagons (Figure 14.1b), the second family of flexagons

to be discovered and the subject of Chapter 12, are far from simple. Martin Gardner wrote

[6], “Stone and his friends spent considerable time folding and analyzing these four-sided

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 11–14.

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110 Part III. Flexagons and Catalan Numbers

forms, but did not succeed in developing a comprehensive theory that would cover all their

discordant variations,” a significant statement as Stone’s friends included John Tukey and

Richard Feynman.

Classification of edge flexagons

This is a difficult subject, as Martin Gardner’s remark indicates. Certain edge flexagons,

however, can be classified on the basis of their characteristic polygon, which describes

how the hinge edges are linked by the leaves and is related to the vertex figures used to

characterize polyhedra [1]. In particular, by describing how the leaves are linked, the char-

acteristic polygon contains most of the information needed to construct the flexagon. For

those edge flexagons that have a characteristic polygon, classification amounts to classifi-

cation of all possible types of characteristic polygons.

We give two examples (in Figures 14.2 and 14.3) of edge flexagons and their char-

acteristic polygons. To make paper models of any of the flexagons described here, use

heavy paper stock and enlarge the nets to about twice the printed size. Transfer the num-

bers in parentheses from the upper side to the lower side and delete the numbers from the

upper side.

In Figure 14.2a, we have the net for a square edge flexagon. To assemble this model,

fold together pairs of leaves numbered 3 and 4, and join the ends (dashed lines). As assem-

bled the flexagon is in main position (Figure 14.1b). This is flat and consists of alternate

pats of single leaves and fan folded piles of three leaves. The flexagon can be flexed around

a 4-cycle by folding it in two (to an intermediate position) and then unfolding into a new

main position, and so on. The operation of this flexagon is similar to the traditional Chinese

falling block toy sometimes called Jacob’s Ladder [8].

3(2)

1(2)

1(4)

3(4)

3(2)

1(2)

1(4)

3(4)

a a

aa

a ab b

b b

b b

c c

c

c c

c

cd

a

bc

d

d

d d

d d

(a) Net.

(b) Characteristic

polygon.

Figure 14.2. A square edge flexagon.

The characteristic polygon is obtained by joining the midpoints of the hinge edges on

each leaf in the order they are linked by the leaves. This results in a twice-traced square

(Figure 14.2b). Incidentally, joining end-to-end two copies of the net shown in Figure 14.2a

results in the net for another square edge flexagon, the octopus flexagon. Main positions are

not flat, but it can still be flexed around a 4-cycle using a pinch flex similar to the flex used

for the trihexaflexagon. Other flexes are possible, and its overall behaviour is bewilderingly

complex [6].

There are two regular pentagons: the usual convex one and the pentagram. This gives

rise to two series of pentagonal flexagons, one of whose net is in Figure 14.3a. To assemble,

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14. From Hexaflexagons to Edge Flexagons to Point Flexagons 111

3(2)

1(2)

5(4)

3(4) 5(1)

aa

a

a

bb

b

b c

cc

c

dd

d

dd

ea

b

c

d

e

ee

(a) Half the net.

(b) Characteristic

polygon.

Figure 14.3. A pentagonal edge flexagon.

first make two copies of Figure 14.3a, turn one over and join them end-to-end. Next, fold

together pairs of leaves numbered 2, 3, 4 and 5, and join the ends. Assembled, the flexagon

is in an intermediate position which looks like a pair of pentagons with a common edge.

From here, it can be opened into a type of main position (Figure 14.1c), and then closed

into another main position in one continuous aesthetically satisfying movement, and so

on around a 5-cycle. It can also be opened into another type of main position, but flexing

round a 5-cycle is then complicated [5]. This is a difficult flexagon to handle.

Point flexagons

Point flexagons are the duals of edge flexagons. They are a recent discovery, first described

by Scott Sherman in 2007 [5]. Nets are derived by truncating the leaves of the net of an

edge flexagon at the vertices, so that vertices become edges and edges become vertices

(see Figure 14.4). Edge hinges, therefore, become point hinges, usually modelled by short

paper strips. By definition, a point hinge has only two degrees of freedom. The triangles in

Figure 14.4 may rotate relative to each other about the y-axis, and the z-axis, but not be

twisted relative to the x-axis.

Edge hinge Point hinge

x

y

Figure 14.4. Truncating the edge-hinged triangles (dashed) creates point-hinged triangles (solid).

Figure 14.5a gives a net for a point flexagon with three faces. To assemble, crease the

lines between the leaves to form hinges. Fold pairs of leaves numbered 2 and 3 together and

glue the two part paper strips together at A-A. Assembled, the visible leaves, above and

below, will both be numbered 1. The flexagon can be flexed around a 3-cycle by turning

over the top leaf so that it becomes the bottom leaf, and so on, rather like the way cards

are flipped in a Rolodex rotary file. This is analogous to the 3-cycle of the trihexaflexagon

described by Martin Gardner [5]. The flex is not completely smooth since the hinge folds

must be reversed and they are not parallel to the axis of rotation.

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112 Part III. Flexagons and Catalan Numbers

1(2)

3(2)

3(1)

A A

(a) Net. (b) Assembled.

Figure 14.5. A triangle point flexagon (dual to the trihexaflexagon).

Conclusion

Martin Gardner opened a Pandora’s box when he introduced flexagons to a wide audience.

Fifty years later, they are a mature topic with an extensive literature. They continue to fas-

cinate with new results and new flexagons still being discovered and still to be discovered.

Acknowledgment Thanks to the anonymous reviewers for helpful comments. The fig-

ures are copyrighted by the author.

Bibliography

[1] H. S. M. Coxeter, Regular Polytopes, 2nd ed., The Macmillan Company, New York, 1963.

[2] M. Gardner. Hexaflexagons, College Math. J. 43 (2012) 2–5; available at

dx.doi.org/10.4169/college.math.j.43.1.002.

[3] ——, Mathematical games: tetraflexagons and tetraflexigation, Scientific American, May 1958,

122–126; available at dx.doi.org/10.1038/scientificamerican0558-122.

[4] E. Iacob, T. B. McLean, and H. Wang, The V-flex, College Math. J. 43 (2012) 6–10; Chapter 13

in this volume and available at dx.doi.org/10.4169/college.math.j.43.1.006.

[5] L. P. Pook. Serious Fun with Flexagons. A Compendium and Guide. Springer, Dordrecht, 2009.

[6] ——, Flexagons; available at www.pook.org.uk; accessed 11 April 2011.

[7] S. Sherman. Point Flexagons; available at loki3.com/flex/point-flexagon.html,

Accessed 17 April 2007.

[8] Wikipedia, Jacob’s Ladder, Wikipedia: the free encyclopedia; available at

en.wikipedia.org/wiki/Jacob%27s_ladder_%28toy%29.

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15Flexagons Lead to a

Catalan Number Identity

David Callan

Introduction

Flexagons are remarkable objects obtained by suitably folding and glueing a strip of paper.

Several websites give detailed instructions on how to construct various kinds. After con-

struction, the flexagon lies flat with a number of faces, delineated by creases in the paper,

visible on top and bottom. The name derives from the fact that they can be pinched, or

flexed, in various ways into a 3-dimensional shape and then flattened again to reveal differ-

ent faces. First discovered by Arthur Stone at Princeton in 1939, they were popularized by

Martin Gardner [2] in his 1956 debut column in Scientific American and continue to attract

mathematical interest [1].

At about the same time as Gardner’s column appeared, Oakley and Wisner [4] consid-

ered the variety now called hexahexaflexagons—hexagonal shape when flat, with 6 trian-

gular faces visible. They showed that these flexagons can be constructed recursively using

a critical fold that splits a flexagon into two smaller ones. In particular, they are equinu-

merous with Dyck paths, counted by the Catalan numbers Cn D 1nC1

�2nn

�

[5]. A Dyck

path of size n is a lattice path of n upsteps .1; 1/ and n downsteps .1; �1/ that starts at the

origin and never goes below ground level. The first downstep that returns the path to the

x-axis splits a Dyck path into two smaller ones, in complete analogy to the decomposition

of flexagons into smaller ones. Oakley and Wisner used this decomposition to show that

hexahexaflexagons can be encoded by permutations called pats. Pats themselves are de-

fined recursively, with permutations represented as lists. A singleton permutation is a pat,

and a permutation p of length n � 2 is a pat if and only if (i) there is a unique split point

that divides p into subpermutations p1 and p2 such that all entries in p1 are greater than

all entries in p2, and (ii) the reverse of each of p1 and p2 is a pat. The only pat on [[2]] is

21; the pats on [[3]] are 231 and 312, and on [[4]] are 2431, 3241, 3412, 4132, 4213. The

number of pats on ŒŒn C 1�� is Cn.

The purpose of this note is to find the distribution of the statistic “number of descents”

on pats (a descent in a permutation p is a pair of adjacent entries .pi ; piC1/ with pi >

piC1). To do so, we establish a recurrence that leads to a quartic equation for the generating

Reprinted from The American Mathematical Monthly, Vol. 119, No. 5 (May 2012), pp. 415–419.

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114 Part III. Flexagons and Catalan Numbers

function. Lagrange inversion then yields a simple closed form for the number of pats on

ŒŒn�� with k descents, thereby giving a combinatorial interpretation of the identity

nX

kD0

1

2n � 2k C 1

2n � 2k C 1

k

!

2k

n � k

!

D Cn: (15.1)

The last section exhibits a bijection from pats to full binary trees showing that descents on

pats are distributed as even-level vertices on binary trees and concludes with a generaliza-

tion of (15.1).

Lagrange inversion and the distribution of descents on pats

From its definition, a pat p of length n determines two pats p1; p2 of lengths i and n � i

respectively for some i 2 Œ1; n � 1�. The connecting entries from p1 to p2 contribute a

descent to p and the remaining descents of p correspond to the ascents in p1 and p2.

Hence we obtain the recurrence

u.n; k/ Dn�1X

iD1

i�1X

j D0

u.i; j /u.n � i; n � j � k � 1/ for n � 2 and k � 0

for the number u.n; k/ of pats on Œn� with k descents, with u.1; 0/ D 1. The recur-

rence leads directly to a functional equation for the generating function F.x; y/ WDP

n�1; k�0 u.n; k/xnyk ,

F.x; y/ D x C 1

yF

�

xy;1

y

�2

: (15.2)

Iterating (15.2) once yields the algebraic equation

F D x C y�

x C F 2�2

; (15.3)

where F.x; y/ is now abbreviated to F .

The Lagrange inversion formula is a powerful technique for solving functional equa-

tions such as (15.3). In its basic form, it says that for a function f .x/ analytic about the

origin and nonzero at the origin, the coefficient of xn in the (compositional) inverse f ı of

f is given by Œxn�f ı.x/ D 1nŒxn�1�g.x/n where g is defined by f .x/ D x=g.x/. The

standard analytic proof (see [6, Chapter 5.1], for example) is based on the special property

of the case n D �1 in the contour integral of zn around the origin; it is the only value

of n that yields a nonzero integral. This is a little mysterious, but in the context of for-

mal power series, there is a transparent combinatorial proof involving ordered (plane) trees

with weighted edges (see [5, Theorem 5.4.2], for example). To apply the Lagrange inver-

sion formula, one typically formulates an equation of the form y D x=g.x/ and wishes to

solve for x in terms of y. With f .x/ WD x=g.x/, we then have x.y/ D f ı.y/ and so the

formula applies (with a little care not to get x and y mixed up).

Direct application of the formula to solve (15.3) is cumbersome but a trick shown

to me by Ira Gessel (see [3]) rapidly solves it. Introduce a new variable z and consider

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15. Flexagons Lead to a Catalan Number Identity 115

F D F.x; y; z/ defined by

F D z�

x C y�

x C F 2�2�

: (15.4)

Equation (15.4) has the form z D F=�.F /where

�.F / D x C y�

x C F 2�2

;

and so Lagrange inversion says that

h

ykz2j C1i

F.x; y; z/ D 1

2j C 1

h

ykF 2ji

�.F /2j C1

D 1

2j C 1

2j C 1

k

!

2k

j

!

xj CkC1: (15.5)

The coefficient of xnyk in F.x; y; 1/ is obtained by setting j D n � k � 1 in (15.5),

yielding

u.n; k/ D Œxnyk �F.x; y/

D Œxnyk �F.x; y; 1/

D 1

2n � 2k � 1

2n � 2k � 1

k

!

2k

n � k � 1

!

;

and hence identity (15.1) after replacing n by n C 1. Here are the first few values of

u.n; k/.Table 15.1. Values of u.n; k/, the distribution of descents on pats

nnk 0 1 2 3 4 5 6

1 1 0 0 0 0 0 0

2 0 1 0 0 0 0

3 0 2 0 0 0 0 0

4 0 1 4 0 0 0 0

5 0 0 12 2 0 0 0

6 0 0 12 30 0 0 0

7 0 0 4 100 28 0 0

8 0 0 0 140 280 9 0

9 0 0 0 90 980 360 0

10 0 0 0 22 1680 2940 220

Pats as trees

A pat on ŒŒn C 1�� can be represented, using its successive split points, by a vertex-labeled

full binary tree on 2n edges as illustrated in Figure 15.1.

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116 Part III. Flexagons and Catalan Numbers

4

7 5 6

2 1

3

479856213

479856

79856

798

98

56

213

21

Figure 15.1. Full binary tree for pat 4 7 9 8 5 6 2 1 3, interior vertices at even level are enlarged.

In fact, the labels are unnecessary; they can be uniquely recovered from the underlying

tree. So we have a bijection from pats on ŒŒn C 1�� to full binary trees on 2n edges. Under

this bijection, a descent in the pat shows up as an interior vertex at even level in the tree

(where rain water would collect between the two leaves corresponding to the descent).

Pruning the leaf edges in a full binary tree to obtain a binary tree, we obtain the follow-

ing.

Theorem. Descents on pats are distributed as even-level vertices in binary trees.

Similar considerations for ternary and higher order trees yield a generalization of (15.1),

nX

kD0

1

rn � rk C 1

rn � rk C 1

k

!

rk

n � k

!

D 1

rn C 1

rn C 1

n

!

for r � 2.

Acknowledgments I thank Ira Gessel for suggesting the Lagrange inversion technique

used in the second section.

Bibliography

[1] T. Anderson, T. B. McLean, H. Pajoohesh, C. Smith, The combinatorics of all regular

flexagons, European J. Combin. 31 (2010) 72–80; available at dx.doi.org/10.1016/

j.ejc.2009.01.005.

[2] M. Gardner, Hexaflexagons and Other Mathematical Diversions: The First Scientific American

Book of Puzzles and Games, Univ. of Chicago Press, 1988.

[3] I. Gessel, A combinatorial proof of the multivariable Lagrange inversion formula, J. Combin.

Theory Ser. A 45 (1987) 178–195; available at

dx.doi.org/10.1016/0097-3165(87)90013-6.

[4] C. O. Oakley, R. J. Wisner, Flexagons, Amer. Math. Monthly 64 (1957) 143–154; available at

dx.doi.org/10.2307/2310544.

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15. Flexagons Lead to a Catalan Number Identity 117

[5] R. P. Stanley, Enumerative Combinatorics, Cambridge Studies in Advanced Mathematics, Vol.

62, Cambridge Univ. Press, Cambridge, 1999.

[6] H. S. Wilf, Generating Functionology, Academic Press, New York, 1990; also available at

www.math.upenn.edu/˜wilf/DownldGF.html

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16Convergence of a Catalan Series

Thomas Koshy and Zhenguang Gao

The well known Catalan numbers Cn are named after Belgian mathematician Eugene

Charles Catalan (1814–1894), who found them in his investigation of well-formed se-

quences of left and right parentheses. As Martin Gardner (1914–2010) wrote in Scien-

tific American [2], they have the propensity to “pop up in numerous and quite unexpected

places.” They occur, for example, in the study of triangulations of convex polygons, planted

trivalent binary trees, and the moves of a rook on a chessboard [1, 2, 3, 4, 6].

The Catalan numbers Cn are often defined by the explicit formula Cn D 1nC1

�2nn

�

,

where n � 0 [1, 4, 6]. Since .nC1/ j�

2nn

�

, it follows that every Catalan number is a positive

integer. The first five Catalan numbers are 1, 1, 2, 5, and 14. Catalan numbers can also be

defined by the recurrence relation CnC1 D 4nC2nC2

Cn, where C0 D 1. So limn!1

CnC1

CnD 4.

Here we study the convergence of the seriesP1

nD01

Cnand evaluate the sum. Since

limn!1

CnC1

CnD 4, the ratio test implies that the series

P1nD0

xn

Cnconverges for jxj < 4. Con-

sequently, the seriesP1

nD01

Cnconverges. We evaluate this infinite sum using generating

functions, plus fundamental tools from the differential and integral calculus.

Sum of the series1P

nD0

1Cn

To this end, let f .x/ be the generating function of the reciprocals of Catalan numbers,

f .x/ DP1

nD0xn

Cn. We compute the sum in three steps. First, we find an ordinary differen-

tial equation satisfied by f .x/, then, after solving the differential equation in the interval

.0; 4/, we compute f .1/.

We first rewrite f .x/ as

f .x/ D 1 C1X

nD1

xn

Cn

:

So

f 0.x/ D1X

nD1

nxn�1

Cn

D1X

nD0

n C 1

CnC1

xn:

Reprinted from The College Mathematics Journal, Vol. 43, No. 2 (March 2012), pp. 141–146.

119

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120 Part III. Flexagons and Catalan Numbers

Since nC2Cn

D 4nC2CnC1

, by the recurrence relation, this yields

1X

nD0

n C 2

Cn

xn D1X

nD0

4n C 2

CnC1

xn;

1X

nD0

n

Cn

xn C 2

1X

nD0

xn

Cn

D1X

nD0

4.n C 1/

CnC1

xn � 2

1X

nD0

xn

CnC1

;

xf 0.x/ C 2f .x/ D 4f 0.x/ � 2

xŒf .x/ � 1�;

x.x � 4/f 0.x/ C .2x C 2/f .x/ D 2: (16.1)

This is a first-order differential equation for f .x/ with the initial conditions f .0/ D 1 Df 0.0/.

To facilitate solving (16.1) for x 6D 0, we introduce the function g.x/ Dˇˇ4�x

x

ˇˇ3=2

.

Then g0.x/g.x/

D �6x.4�x/

. This implies that

Œx.x � 4/g.x/�0 D .2x C 2/g.x/: (16.2)

Multiplying (16.1) by g.x/, we get

x.x � 4/f 0.x/g.x/ C .2x C 2/f .x/g.x/ D 2g.x/:

Using (16.2), this can be rewritten as

Œx.x � 4/f .x/g.x/�0 D 2g.x/:

But, again using (16.2),

fx.x � 4/Œf .x/ � 1�g.x/g0 D Œx.x � 4/f .x/g.x/�0 � Œx.x � 4/g.x/�0

D 2g.x/ � .2x C 2/g.x/

D �2xg.x/;

consequently,

x.x � 4/Œf .x/ � 1�g.x/ D �2

Z

xg.x/ dx C C1

f .x/ D 1 C 2R

xg.x/ dx � C1

x.4 � x/g.x/;

where C1 is a constant.

Suppose 0 < x < 4. Then

Z

xg.x/ dx DZ

x

�4 � x

x

�3=2

dx

DZ

.4 � x/3=2

x1=2dx:

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16. Convergence of a Catalan Series 121

Letting x D u2, this implies that

Z

xg.x/ dx D 2

Z

.4 � u2/3=2 du

D1

2u.4 � u2/3=2 C 3u.4 � u2/1=2 C 12 arcsin

u

2C C2

D 1

2

px.4 � x/3=2 C 3

px.4 � x/1=2 C 12 arcsin

px

2C C2;

where C2 is another constant. Therefore, we have

f .x/ D 1 Cp

x.4 � x/3=2 C 6p

x.4 � x/1=2 C 24 arcsinp

x

2C 2C2 � C1

x.4 � x/�

4�xx

�3=2

D 1 Cx.4 � x/3=2 C 6x.4 � x/1=2 C 24

px arcsin

px

2C C

px

.4 � x/5=2;

where C D 2C2 � C1. Since f .0/ D 1 D f 0.0/, C D 0. Thus

f .x/ D 1 Cx.4 � x/3=2 C 6x.4 � x/1=2 C 24

px arcsin

px

2

.4 � x/5=2D

1X

nD0

xn

Cn

:

When x D 1, this yields

1X

nD0

1

Cn

D 1 C 33=2 C 6 � 31=2 C 24 arcsin 1=2

35=2

D 2 C 4p

3

27�: (16.3)

Thus, the seriesP1

nD01

Cnconverges to the limit 2 C 4

p3

27� , which is approximately

2.80613305077. Note that alreadyP22

nD01

Cn� 2:80613305077; so the series converges

to the limit remarkably fast.

Additional consequences of the differential equation

Letting x D 1 in (16.1), we get

3f 0.1/ D 4f .1/ � 2

D 4

2 C 4p

3

27�

!

� 2

f 0.1/ D 2 C 16p

3

81�:

Since f 0.x/ DP1

nD0nC1

CnC1xn, it follows that

1X

nD0

n C 1

CnC1

D 2 C 16p

3

81�:

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122 Part III. Flexagons and Catalan Numbers

Since the differential equation is infinitely differentiable, it follows from (16.1) that

x.x � 4/f 00.x/ C .4x � 2/f 0.x/ C 2f .x/ D 0: (16.4)

This yields 3f 00.1/ D 2f 0.1/ C 2f .1/, so f 00.1/ D 83

C 56p

3243

� . Since

f 00.x/ D1X

nD0

.n C 1/.n C 2/

CnC2

xn;

this implies that1X

nD0

.n C 1/.n C 2/

CnC2

D 8

3C 56

p3

243�:

Differentiating (16.4) with respect to x, it follows similarly that f 000.1/ D 2f 0.1/, so

1X

nD0

.n C 1/.n C 2/.n C 3/

CnC3

D 4 C 32p

3

81�:

Clearly, this technique can be employed to evaluate further sums of the form

1X

nD0

.n C 1/.n C 2/ � � � .n C k/

CnCk

;

where k � 1.

Sum of the series1P

nD0

.�1/n

Cn

We now turn to to solving (16.1) for �4 < x < 0. We have

Z

xg.x/ dx DZ

x

ˇˇˇˇ

4 � x

x

ˇˇˇˇ

3=2

dx

D �Z

.4 � x/3=2

px

dx:

Letting x D �u2, this gives

Z

xg.x/ dx D �Z

.u2 C 4/3=2

u.�2u/ du

D 2

Z

.u2 C 4/3=2 du

D 2

"

u.u2 C 4/3=2

4C 3

2up

u2 C 4 C 6 lnˇˇˇu C

p

u2 C 4ˇˇˇ

#

C C3

D 1

2u.u2 C 4/3=2 C 3u

p

u2 C 4 C 12 lnˇˇˇu C

p

u2 C 4ˇˇˇC C3

D 1

2

p

jxj.4 � x/3=2 C 3p

jxj.4 � x/ C 12 ln�p

jxj Cp

j4 � xj�

C C3;

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16. Convergence of a Catalan Series 123

where C3 is a constant.

Consequently, as before, we have

f .x/ D 1 C

p

jxj.4 � x/3=2 C 6p

jxj.4 � x/ C 24 ln�p

jxjCp

j4�xjC4

�

x.4 � x/ˇˇ4�x

x

ˇˇ3=2

D 1 �jxj.4 � x/3=2 C 6

p

jxj.4 � x/ C 24p

jxj ln�p

�xCp

4�x

C4

�

.4 � x/5=2:

where C4 is a nonzero constant. Since f .0/ D 1 D f 0.0/, C4 D 2. Thus

f .x/ D 1 �jxj.4 � x/3=2 C 6

p

jxj.4 � x/ C 24p

jxj ln�p

�xCp

4�x

2

�

.4 � x/5=2:

Since f .�x/ generates the alternating seriesP1

nD0.�x/n

Cnfor 0 < x < 4, setting x D 1

we get

1X

nD0

.�1/n

Cn

D 1 � 53=2 C 6 � 51=2 C 24 ln�

55=2

D 14

25� 24

p5

125ln �; (16.5)

where � is the well-known golden ratio 1Cp

52

.

Formulas (16.3) and (16.5) can be employed to compute the sums of the subseriesP1

nD01

C2nand

P1nD0

1C2nC1

:

1X

nD0

1

C2n

D 32

25C 2

p3

27� � 12

p5

125ln �;

1X

nD0

1

C2nC1

D 18

25C 2

p3

27� C 12

p5

125ln �:

Acknowledgment The authors would like to thank the editor and reviewers for their

unwavering enthusiasm, and suggestions for the improvement of the exposition.

Bibliography

[1] D. I. A. Cohen, Basic Techniques of Combinatorial Theory, Wiley, New York, 1978.

[2] M. Gardner, Catalan Numbers: An Integer Sequence that Materializes in Unexpected Places,

Scientific American, 234 (1976), 120–125; available at

10.1038/scientificamerican0676-120.

[3] H. W. Gould, Bell and Catalan Numbers: Research Bibliography of Two Special Number Se-

quences, Revised Edition, Combinatorial Research Institute, Morgantown, West Virginia, 1978.

[4] T. Koshy, Catalan Numbers with Applications, Oxford Univ. Press, New York, 2009.

[5] ——, Fibonacci and Lucas Numbers with Applications, Wiley, New York, 2001.

[6] R. P. Stanley, Enumerative Combinatorics, Vol. 1, Wadsworth & Brooks/Cole, Monterey, Cali-

fornia, 1986.

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IVMaking Things Fit

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17L-Tromino Tiling of Mutilated

Chessboards

Martin Gardner

Suppose a standard chessboard is ‘mutilated’ by the removal of two diagonally opposite

corner cells. Can the remaining 62 squares be tiled with 31 dominos? The answer is ‘no’

because the removed squares are the same color. Say the color is white. The remaining

62 squares will have an excess of two black cells. Each domino covers one black and one

white cell. After 30 are placed, two black cells will remain uncovered. They cannot be

adjacent, therefore they can’t be covered by a domino. This famous puzzle, solved by a

simple parity check, is a simple example of a tiling problem on a mutilated chess board.

Less well known is the following related problem. Assume the chessboard is mutilated

by having two cells removed of opposite color from anywhere on the board. Can the re-

maining 62 squares always be tiled by dominoes? The answer is yes, and there is a lovely

proof by Ralph Gomory [2].

Figure 17.1. Gomory’s proof.

Imagine heavy lines drawn on the chessboard as shown in Figure 17.1. They outline

a closed path along which the squares are like beads of alternating color on a necklace.

If any two cells of opposite color are taken from the path, it will cut the path into two

open-ended segments, or one segment if the removed cells are adjacent. Each segment

will consist of an even number of cells of alternating colors, therefore it can be tiled with

Reprinted from The College Mathematics Journal, Vol. 40, No. 3 (May 2009), pp. 162–168.

127

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128 Part IV. Making Things Fit

dominoes. Gomory’s clever proof is easily generalized to all square boards with an even

number of cells.

If, instead of dominos, we tile with L-trominos, also called bent, or V, or right trominos,

then all square boards with a number of cells divisible by 3 can be tiled except for the 3 � 3

board. We will not be concerned with such ‘whole’ boards, but only with mutilated boards

with a number of cells that is a multiple of 3 after a single cell has been removed from any

spot on the board. We will call such boards deficient. In other words, a board of side n is

deficient if n2 � 1 is a multiple of 3, i.e., n is not a multiple of 3. The sides of such boards

form the sequence

2; 4; 5; 7; 8; 10; 11; 13; 14; : : : .�/

We will call these numbers the orders of a board and, from now on, the word tromino will

mean an L-tromino exclusively.

Our basic question is this: What deficient boards with sides in the sequence .�/ can be

tiled without gaps or overlaps with L-trominos after a cell has been taken from anywhere

on the board? We will take up these boards roughly in numerical order, culminating with a

statement of the complete solution.

Powers of 2

Consider the order-2 board first. It obviously is tilable with any cell missing (see Figure

17.2, left). Figure 17.2, right, shows how the order-4 can be tiled. The 2 � 2 square takes

care of a missing cell in each of its four corners. The rest of the board is tiled by taking

advantage of what Solomon Golomb named a rep-tile—a tile that can form an enlarged

replica of itself. The top left 2 � 2 square rotates to put its missing cell in four places, and

the entire order-4 square rotates to carry the missing cell to any of its sixteen places.

Figure 17.2. Orders 2 and 4.

In 1953 Golomb, the “father” of polyominoes (he named them and was the first to study

them in depth) discovered a beautiful proof by induction that all boards with sides in the

doubling sequence 2; 4; 8; 16; : : : could be tiled with trominoes when any cell is missing.

The proof was first published in [3]. It is repeated on pages 27–28 of [4]. Numerous math-

ematicians have since included the proof in their books, often without credit to Golomb.

Roger Nelsen, in [6], gives Golomb’s proof with a wordless single diagram.

Golomb’s famous proof starts with the 2�2 case shown on the left of Figure 17.3. This

square is placed in the corner of the order-4 as shown at the center of Figure 17.3. The

4 � 4 then goes in the corner of an order-8 (shown on the right) and a tromino placed at the

corner of the shaded order-4. We know the dark square can be tiled with any cell missing,

and we know the three unshaded quadrants can be tiled with trominoes because each has a

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17. L-Tromino Tiling of Mutilated Chessboards 129

Figure 17.3. Golomb’s induction proof.

missing corner cell. By rotating the board, a missing cell at any spot in the shaded quadrant

can be brought to any spot on the order-8 board.

Orders 5 and 7

The order-5 board is next, as 5 is the next unsolved number in the sequence .�/. It has a

neat symmetrical tiling when the center cell is gone, as shown in Figure 17.4, left. I have

tiled this board with four 2 � 3 tiles. Each is tilable with two trominoes in two different

ways. Using 2 � 3 tiles is a valuable device for solving tromino problems.

When the missing cell is the one shown black in Figure 17.4, center, the cell above it

must be covered by a tromino on either side. In each case, shown here with a tromino above

and on the right, this produces two cells (numbered 1 and 2) that cannot be covered with

a tile. Indeed, the order-5 square can be tiled only when the missing cell is one of the nine

shown in black in Figure 17.4 right. As a pleasant exercise, see if you can tile the board

when the missing cell is at a corner.

1

2

Figure 17.4. The order-5 square.

The order 7 board is more difficult to analyze. I was unable to find a single diagram

that would prove this board tilable, but Golomb sent me his unpublished way of proving

tilability with the aid of three diagrams.

His proof goes like this. Figure 17.5 shows three tilings of the order-7 board. In each

tiling, the 2 � 2 square obviously can be tiled with a tromino so that the missing cell is at

any of the four corners. By rotating the three patterns, the missing cell can be placed at any

spot on the board.

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130 Part IV. Making Things Fit

Figure 17.5. Golomb’s proof that order-7 is tilable.

Somewhat more difficult is to find tilings that maximize the number of 2 � 3 tiles. As a

challenge, can the reader find a tiling of the 7�7 board using six 2�3 tiles and 4 trominoes

(see Figure 17.6)? The solution is unique except for a single reflection (see page 134).

?

Figure 17.6. A challenge to the reader.

Note in Figure 17.5, that in each pattern the number of free trominoes—trominoes not

in any 2 � 3 tile—is always even. This is no coincidence. It led me to the following trivial

little law. When a board’s order is even, the number of free trominoes in a tiling pattern is

odd, and vice versa. When the board’s number is odd, the number of free trominoes must

be even.

The parity proof is simple. If a board’s order is even, after a cell is removed there will be

.n2 � 1/=3 trominoes in any tiling, an odd number. Each 2 � 3 tile contains two trominoes,

so the total number of trominoes in 2 � 3 tiles will be even. Subtracting this number from

the odd total of trominoes and you get an odd number of trominoes not in any 2 � 3 tiles.

Suppose the board’s order is odd. After a cell is removed there will remain an even

number of cells. Subtracting the even number of trominoes in the 2 � 3 tiles leaves an even

number of trominoes not in a 2 � 3 tile.

Beyond 7

Golomb’s induction proof can be applied to an infinity of other doubling sequences. In

particular, now that we have tiled the 7 � 7 board, we can tile boards of size n � n where n

is of the form 2k7. For example, consider the order-14 board. Divide it into quadrants with

a shaded order-7 board in the top left corner, and attach a tromino to its lower right corner

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17. L-Tromino Tiling of Mutilated Chessboards 131

as before. Because the 7-board is tilable, the proof for order-14 follows, and of course leads

by induction to proofs for orders 28; 56; 112; : : : :

A similar proof for the order-10 board can’t be obtained by placing an order-5 in the

corner because order-5 is not tilable, but we can handle it in a slightly different way. Put

in the top left corner an order-8 which we know is tilable. The remaining area forms a

path of width 2 along the bottom and right sides of the large square (see Figure 17.7). By

rotations and reflections, each missing cell in the order-8 can be transferred to any cell on

the board. This leads to proofs for orders 20, 40, 80, and so on. A similar proof for order-11

has an order-7 square in the corner, and a path of width 4 along bottom and side. It leads by

induction to solutions for orders 22; 44; 88; : : : : Clearly this technique provides an infinity

of doubling sequences for tilable boards. Simply, put in the top left corner of any board a

tilable board with a side equal to or smaller than the larger board. If you can tile the path it

leaves at the bottom and side, then the board is tilable.

8x8

Figure 17.7. Proof that order-10 is tilable.

Boards with sides that are primes are usually the hardest to tile. Order 17 is solved by

a corner square of side 13 and a path of width 4. Order 19 is solved by a corner square of

order 14, in turn based on order-7, and a path of width 5. (See Figure 17.8.)

The complete result

By working with these patterns I came close, but not close enough, to finding an induction

proof that all deficient squares are tilable except for order-5. A proof was finally obtained

by I. Ping Chu and Richard Johnsonbaugh [1].

Chu and Johnsonbaugh not only took care of all deficient squares, but also all defi-

cient rectangles! Their induction proof is too technical to repeat here. To summarize, they

showed tilability for all m by n rectangles (including squares when m D n) which have a

number of cells that is a multiple of 3 after a cell is removed. Such boards are tilable if and

only if all of the following are true:

1. m is equal to or greater than 2.

2. n is equal to or greater than m.

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132 Part IV. Making Things Fit

14

Figure 17.8. Order-19 is tilable.

3. if m is 2, n must be 2,

4. m is not 5.

A 4 � 7 rectangle is the smallest deficient rectangle, not a square, that is tilable with

L-trominoes. As another exercise, see how long it takes you to tile it when the missing cell

is at a corner, and there are two 2 � 3 tiles.

Christopher Jensen, in an unpublished paper, showed that if two cells are taken from

a corner of any board, as shown in Figure 17.9, the board obviously cannot be tiled with

trominoes. However, if none of these five cases is allowed, a 3m � 1 by 3n C 1 board, with

any two cells missing can be tiled if an only if n D 1 or m and n are each equal to or

greater than 3.

A final word

Kate Jones, who founded and runs Kadon Enterprises, a firm that makes and sells hand-

some mechanical puzzles, games, and other recreational math items, has on the market a

game called Vee-21 [5]. The Vee is for V-trominoes, and 21 for the 21 tromino tiles in the

set. The trominoes are brightly colored, and there is an order-8 board on which to place

them. The basic task is to put a monomino (order-1 tile) at any spot on the board, then

cover the remaining 63 cells with the trominoes, thus solving an order-8 board. A 40-page

brochure comes with the set. It contains a short article on “The Deficient Checkerboard”

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17. L-Tromino Tiling of Mutilated Chessboards 133

Figure 17.9. Impossible tiling patterns when two cells are missing at a corner.

by Norton Starr, and pictures of rectangular fields that offer other challenges.

Our final tiling (see Figure 17.10) is a beautiful, symmetric tiling of the standard chess-

board.

Figure 17.10. An order-8 tiling with no 2 � 3 tiles and 5 rep-tiles.

Bibliography

[1] I. P. Chu and R. Johnsonbaugh, Tiling deficient boards with trominoes, Math. Mag. 59 (1986)

34–40.

[2] M. Gardner, The Eight Queens and Other Chessboard Diversions, in The Unexpected Hanging

and Other Mathematical Diversions, Univ. of Chicago Press, 1991.

[3] S. W. Golomb, Checker boards and polyominoes, Amer. Math. Monthly 61 (1954) 675–682.

[4] ——, Polyominoes, Scribner, New York, 1965.

[5] K. Jones, Vee-21; available at www.gamepuzzles.com/polycub2.htm#V21.

[6] R. Nelsen, Proofs Without Words II: More Exercises in Visual Thinking, Mathematical Associa-

tion of America, Washington, 2000.

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134 Part IV. Making Things Fit

Solution to the puzzle on page 130

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18Polyomino Dissections

Tiina Hohn and Andy Liu

The damaged patchwork quilt

Those of us fortunate to have been born before Martin Gardner’s retirement can relate

to the excitement of eagerly awaiting his monthly column Mathematical Games in the

magazine Scientific American. Particularly popular were those containing a selection of

short puzzles. Here is one from the very last column of this kind, for April, 1981.

The patchwork quilt in Figure 18.1 was originally made up of 108 unit squares. Part of

the quilt’s center became worn, making it necessary to remove 8 squares as indicated. Cut

the quilt along the lines into just two parts that can be sewn together to make a 10�10 quilt.

Figure 18.1.

This beautiful problem is rather old, and had appeared earlier in [1] and [10, p. 64–65].

Gardner’s version was later anthologized in [7, pp. 249–250] and [8, p. 168]. We give the

readers a chance to solve this problem on their own now, before we return to it later.

The father of the field of geometric dissections is generally acknowledged to be Harry

Lindgren who wrote the fundamental treatise [12]. The torch has now been passed on

to Greg Frederickson, with three outstanding books so far, [2]–[4]. He was motivated by

Lindgren’s book, which might not have been written without the influence of Gardner’s

November 1961 column, later anthologized in [5, pp. 43–51].

Martin Gardner devoted another column in July 1977 to geometric dissections, titled

“Cutting Shapes into N Congruent Parts.” This is later anthologized in [6, pages 165–

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (January 2012), pp. 88–94.

135

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136 Part IV. Making Things Fit

181]. About half of the problems share a common feature with the Damaged Patchwork

Quilt Problem, in that the figures to be dissected, as well as the pieces resulting from the

dissections, are polyominoes, plane figures consisting of unit squares joined edge to edge.

They have a long history. Solomon Golomb was the first to popularize them, and Gardner

wrote several columns about them.

One of Gardner’s passions was to introduce puzzles into the classroom. From a the-

oretical point of view, polyomino dissections are an excellent topic. They require little

background, and provide training in geometric visualization. The most important reason of

all is that they are fun.

However, there are practical difficulties. How we should we present such problems, say

the Damaged Patchwork Quilt Problem? Providing lots of scissors and pre-cut rectangle

with holes leaves a lot to be desired. Nobody would look forward to the preparation, and

mixing kids and scissors is never a good idea. Besides, there will eventually be a ton of

litter on the floor.

Practical objections aside, there is a more important shortcoming of this model. It does

not help anyone find a solution to the problem! Cutting at random will not work. In fact,

one has to have the solution already in mind before any meaningful cutting can be done.

Preliminary work

How do we solve the Damaged Patchwork Quilt problem? Since it appears quite difficult,

we “downsize” it. This also yields the benefit of providing subsidiary problems at lower

levels. We generalize the 9 � 12 rectangle with a 1 � 8 hole in the middle to a .2n � 1/ �.2n C 2/ rectangle with a 1 � .2n � 2/ hole, and the 10 � 10 square to an 2n � 2n square.

The original problem is the case n D 5.

For n D 1, we have a 1 � 4 rectangle with a 1 � 0 hole in the middle and a 2 � 2

square. This is going too far down. The case n D 2 turns out to be quite easy. Since the

top row of the rectangle has length 6 while the target square has side 4, we must have a

cut as indicated in Figure 18.2. Another cut in the symmetric position yields the desired

two parts.

-

Figure 18.2.

In the case n D 3, we also have to make a cut along the top edge of the 5 � 8 rectangle

at the point 6 units from a corner. However, if we cut through to the hole as we did in the

case n D 2, it will not work. After a bit of fiddling around, we may arrive at the solution

in Figure 18.3.

From these preliminary investigations, it is not unreasonable to assume that the two

parts in the solution are congruent to each other and rotationally symmetric. However, we

are certainly not confident at this point of solving the original problem.

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18. Polyomino Dissections 137

-

Figure 18.3.

Eureka!

After much deep thought, we finally came up with a fantastic model. Not only does it solve

the practical problems mentioned before, it actually leads us step by step to a solution. We

illustrate with the case n D 4. (In [8], there is an erroneous statement that there are no

two-piece solutions to this case. That statement is removed in [7].)

We draw two playing boards, one the initial 7 � 10 rectangle with a 1 � 6 hole in the

middle, and the other the target 8 � 8 square, and provide ourselves with a large supply of

bingo chips of two different colors, say yellow and blue. The objective is to fill both boards

with chips in such a way that those of the same color form the same shape on both boards.

This solves the problem.

We begin with a yellow chip at the top left corner of the square board and a blue chip

at the bottom right corner. This signifies that these two squares do not belong to the same

part, a most reasonable assumption. We define these two corners as the principal corners

of the respective parts. We mark them in the same orientation since we anticipate that the

two parts are in rotational symmetry. See Figure 18.4a.

-?

� 6

e

u

-?

� 6e

e

e

e

e

e

b

e

e

e

e

e

e

e

u

u

u

u

u

u

u

r

u

u

u

u

u

u

(a) (b)

b

b

b

b

b

b

b

b

b

b

b

b

b

b e e e e e e

r

r

r

r

r

r

r

r

r

r

r

r

r

r

u u u u u ub

b

b

b

b

b

b

b

b

b

b

b

b

b b b b b b b

e e e e

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r r r r r r

u u u u

(c) (d)

Figure 18.4.

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138 Part IV. Making Things Fit

We now move over to the punctured rectangular board and mark the principal corners.

Besides playing the chips at the corners, we observe that in the square board, the yellow

chips cannot extend beyond the eighth column from the left while the blue chips cannot

extend beyond the eight column from the right. This allows us to place the additional chips

on the rectangular board. See Figure 18.4b.

Moving back to the square board, we can fill the top row with yellow chips and the

bottom row with blue ones, because here, the yellow chips cannot extend beyond the sev-

enth row from the top and the blue chips cannot extend beyond the seventh row from the

bottom. See Figure 18.4c. Moving over to the rectangular board, we notice that the two

blue chips on the seventh row from the bottom cannot extend at all to the left. This means

that we must have six yellow chips on the second row from the top, going from left to

right. Similarly, there must be six blue chips on the second row from the bottom, going

from right to left. See Figure 18.4d. Continuing with the same strategy, we can complete

the solution of the problem as in Figure 18.5.

b b b b b b b b

b b b b b b

b b

b b

b b

b b

b b

e e

r r r r r r r r

r r r r r r

r r

r r

r r

r r

r r

u u

b b b b b b b b

b b b b b b

b b

b b

b b

b b b b

b b

e e

r r r r r r r r

r r r r r r

r r

r r

r r

r rr r

r r

u u

(a) (b)

e e e e

u u u u

(c) (d)

Figure 18.5.

Onward

This approach may be used to solve related problems. In the above problem, the two parts

we end up with are congruent to each other. Of course, they also have to fit together to form

a square. Since the punctured rectangle has rotational symmetry, there are many trivial di-

visions into two congruent parts. However, if a figure does not have any obvious symmetry,

the task may be quite challenging. Here is an example from [11, p. 37].

There are two natural choices for the pair of principal corners. Since the figure does

not have rotational symmetry, we choose opposite orientations for the corners. In the first

attempt as shown in Figure 18.7a, the boundary squares are easily filled out. We observe

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18. Polyomino Dissections 139

Figure 18.6. Dissect into two congruent pieces.

that in the third row from the bottom, we can have at most two blue chips going from

right to left. This means that in the third column from the left, we must have at least four

blue chips going from bottom to top. This in turn means that in the second row from the

bottom, we must have at least four yellow chips going from left to right. However, this is

impossible as the blue chips just placed are in the way. The other choice leads easily to the

solution shown in Figure 18.7b.

-?

�6

b

b

b

b

b

b b b b b b

r r r r r r

r

r

r

r

r

u

u

u

�?

-6

(a) (b)

Figure 18.7.

Many polyomino dissection problems have appeared in recent years, no doubt influ-

enced by Gardner’s columns. (Serhiy Grabarchuk dedicated his book [9] to Martin Gard-

ner.) We offer the readers a choice selection here.

1. Solve the “Damaged Patchwork Quilt” problem.

2. Dissect each of the following figures into two congruent pieces. The first is taken

from [13] and the last two from [6].

3. The patchwork quilt in the diagram below was originally made up of 35 unit squares.

Part of the quilt’s became worn, making it necessary to remove 5 of the squares as

indicated, Cut the quilt along the lines into just two parts that can be sewn together

to make a 5 � 6 quilt. This is taken from [13].

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140 Part IV. Making Things Fit

4. Dissect each of the following figures into two congruent pieces. The first two are

taken from [6] and the last four from [11]. The last one is a composition of Solomon

Golomb.

5. Dissect each of the following figures into three congruent pieces. The first one is

taken from [13] and the second from [9].

6. Dissect each of the following figures into two parts that can be reassembled to form

an 8 � 8 square. The first one is taken from [13] and the second one from [9].

A final word

Any trick used twice becomes a technique. The technique discovered by us and described

here is good enough at solving certain polyomino dissection problems that it can be pre-

sented to school children. On October 21, 2010, which would have been Martin Gardner’s

96th birthday, The Damaged Patchwork Quilt was one of Gardner’s problems presented

at a Math Fair by students of Stratford Junior High School at the Telus World of Science

Edmonton. We hope this was something that would please Martin.

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18. Polyomino Dissections 141

Bibliography

[1] H. Dudeney, Puzzles And Curious Problems, Thomas Nelson & Sons, 1931.

[2] G. Frederickson, Dissections: Plane & Fancy, Cambridge Univ. Press, Cambridge, 1997.

[3] ——, Hinged Dissections: Swinging & Twisting, Cambridge Univ. Press, Cambridge, 2002.

[4] ——, Piano-Hinged Dissections: Time To Fold!, A K Peters, Natick MA, 2006.

[5] M. Gardner, The Unexpected Hanging, Univ. of Chicago Press, Chicago, 1991.

[6] ——, Penrose Tiles To Trapdoor Ciphers, Mathematical Association of America, Washington,

1997.

[7] ——, The Last Recreations, Springer-Verlag, New York, 1997.

[8] ——, The Colossal Book Of Short Puzzles and Problems, ed. Dana Richards, W. W. Norton &

Co., New York, 2006.

[9] S. Grabarchuk, The New Puzzle Classics, Sterling Publishing Co., New York, 2005.

[10] B. Kordemsky, The Moscow Puzzles, Dover Publications Inc., New York, 1987.

[11] R. Kurchan, Mesmerizing Math Puzzles, Sterling Publishing Company, New York, 2001.

[12] H. Lindgren, Geometric Dissections. Dover Publications, New York, 1972.

[13] N. Yoshigahara, Geometric dissections, Puzzle World 1 (1992) 48–51.

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19Squaring the Plane

Frederick V. Henle and James M. Henle

Introduction

This research was inspired by two lovely pieces of mathematics. The first is the discovery

by William T. Tutte, A. H. Stone, R. L. Brooks, and C. A. B. Smith of squares with integral

sides that can be tiled by smaller squares with integral sides, no two alike. Tutte tells the

story in “Squaring the Square,” a beautifully written article that conveys vividly the excite-

ment of mathematical research [9]. It became widely-read in 1958 when it was reprinted

in Martin Gardner’s Mathematical Games column in Scientific American. It undoubtedly

played a role in inspiring many to become mathematicians.

The second piece is the well-known tiling of the plane by squares whose sides are

the Fibonacci numbers (Figure 19.2).� The tiling is elegant, but in light of Tutte’s work,

possesses a minor flaw—it contains two squares of the same side. The flaw is easily reme-

died. We can use the squared square in Figure 19.1, together with a square of side 175

(Figure 19.3). This works, but some of the elegance is lost.

64

33 3543

31 29

8

51

56

3830

81

18 20

55

16 149

5539

Figure 19.1. 175� 175 perfect square.

Reprinted from The American Mathematical Monthly, Vol. 115, No. 1 (Jan. 2008), pp. 3–12.�Coincidentally, this tiling appears in the same volume of Gardner’s columns as “Squaring the Square” [1].

143

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144 Part IV. Making Things Fit

1 1

23

5

8

13

21

34

55

Figure 19.2. Fibonacci tiling.

175

350525

875

1400

2275

3675

5950

9625

Figure 19.3. Fibonacci rectangle seeded with Tutte’s

perfect square.

In looking for a natural tiling that doesn’t repeat squares, one quickly discovers that

the first five squares fit together neatly (Figure 19.4) but it becomes progressively more

difficult to add consecutive squares without overlapping or leaving gaps. This suggests the

following question: Can the squares with whole-number sides, one of each size, be fitted

together to tile the plane? The answer is that they can. In succeeding sections we will prove

this, discuss the history of the problem, and pose a number of questions.

1

23

45

Figure 19.4. The first five squares.

Squaring the plane

We are going to show that the plane admits a tiling that uses exactly one square of side-

length n (n D 1; 2; : : :). Our approach focuses on rectangles and ells. By “ell,” we mean

any six-sided figure whose sides are parallel to the coordinate axes.

Figure 19.5. An ell.

The following extends Tutte’s use of “perfect” for rectangles and squares.

Definition 1. A figure is perfect if it is composed entirely of squares of different sides.

All the remaining figures in this paper will be perfect. The key to our result is Lemma 9,

which states that given any perfect ell, it is possible to add squares to it to form a perfect

rectangle.

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19. Squaring the Plane 145

5

2

Figure 19.6. A perfect ell.

5

7 9

16

28

33

25

36

Figure 19.7. The ell of Figure 19.6 squared up.

When we add squares to a perfect figure, keeping it perfect, we’ll say we are “puffing

it up.” When we puff an ell up to form a perfect rectangle, we’ll say we are “squaring up

the ell.” We can then “square the plane” as follows:

1. Start with any perfect ell and square it up.

2. Create a new ell by attaching to the rectangle the smallest square not yet used.

3. Square this ell up, making sure that new squares are added in all four directions.

4. Repeat steps 2 and 3 ad infinitum.

Definition 2. An ell in standard position is an ell oriented so that the single reflex angle

is at the upper right. We refer to the sides by the uppercase letters in Figure 19.8 and their

lengths by the corresponding lower case letters.

a

b

c

d

e

f

A

B

C

DE

F„ ƒ‚ …

8

ˆˆˆ<

ˆˆˆ:

)

9

>=

>;

‚ …„ ƒ‚ …„ ƒ

Figure 19.8. An ell in standard position.

When we add squares to a figure, we always do so in such a way that an added square’s

side-length matches that of at least one side of the figure against which it is placed. That

means that the only squares we add to an ell in standard position are squares adjacent to

sides A, B , C , D, E , or F of the appropriate lengths. We call these operations A, B, C, D,

E, and F. So, for example, applying B then A to an ell produces a larger ell (Figure 19.9).

We denote sequences of operations by sequences of letters with the understanding that they

are applied from left to right (so the combination of operations in Figure 19.9 would be

written BA).

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146 Part IV. Making Things Fit

B

A

Figure 19.9. An application of BA to the ell of Figure 19.8.

Note that in any ell in standard position, the length of side A is the sum of the lengths

of sides C and E ,

a D c C e;

and the length of side F is the sum of the lengths of sides B and D,

f D b C d:

Thus, to describe the dimensions of such an ell, we need describe only the lengths of sides

B , C , D, and E .

Definition 3. A 4-tuple hb; c; d; ei describes the dimensions of an ell if the sides B , C , D,

and E of the ell are b, c, d , and e, respectively.

Definition 4. A side of an ell is composite if it is not the side of a single square in the ell.

In squaring up an ell, we make particular use of the operations B, F, and ED. For that

reason, the following definition is most useful:

Definition 5. A perfect ell is regular if each of the moves B, F, and ED either results in a

perfect ell in standard position or results in a perfect rectangle.

Figure 19.10 shows the smallest regular ell.

31

4

7

10

17

Figure 19.10. The smallest regular ell has dimensions h27; 12; 1; 5i.

If an ell with dimensions hb; c; d; ei is regular, then there can be no squares with sides

b, f , e, or d Ce in the ell. In particular, sides B , F , and E must be composite. In addition,

it must be that c � d C e (or else move ED would create an ell that is not in standard

position). These properties—no squares of sides b, f , e, or d C e, and c � d C e—are

both necessary and sufficient for regularity.

Lemma 5. Every perfect ell P in standard position can be puffed up to form a regular ell

without increasing the length of side D.

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19. Squaring the Plane 147

Proof. Let the dimensions of P be hb; c; d; ei. In every perfect ell in standard position,

either A or F is the longest side, and whichever side is longest is necessarily composite.

We may assume that a � f , since if not, then side A is the longest side and composite, so

A can be performed, and in the new ell the length of side A is less than the length of side

F (Figure 19.11).

A

Figure 19.11. A makes a � f .

Since a � f , side F is the longest side and is composite, so F can be performed.

This allows us to assume further that c > d . If not, we can perform FABA, and in the

new ell, the length of side C is greater than the length of side D (Figure 19.12). Note that

FABA does indeed puff up the ell—the first square, added by F, is larger than any in the

ell, and each new square is larger than the one before. Note also that the length of side D

is unaffected.

F

A

B

A

Figure 19.12. The operation FABA.

With our assumptions now (f � a, c > d ), we perform FABA (Figure 19.13). Let

hb0; c0; d; e0i be the dimensions of the ell formed after FABA (d , since FABA doesn’t

affect side D). It should be clear that performing either F or B to this new ell would result

in a perfect ell in standard position.

To check move ED, note first that move E adds a square of side e0 D b C d C e. This

is larger than b C d but smaller than b C c C d C e, hence a square not previously used.

Applying D after E adds a square of side d C e0 D b C 2d C e. This is larger than the

square just added, but again less than b C c C d C e, since c > d . The result then, is a

perfect ell. To show that the ell is either a rectangle or in standard position, we need only

the fact that c0 � d C e0. But c0 D 2b C 2c C d C e and e0 D b C d C e, and using c > d ,

we have c0 D b C 2c C e0 > d C e0.

Key to proving our main lemma (Lemma 9) is an analysis of c � e modulo d . For a

regular ell c � d C e. Consequently, we can express c as kd C e C i where k � 1 and

0 � i < d .

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148 Part IV. Making Things Fit

c + e

b

b + db + c + d + e

b 0 = 5b +3c +3d +3e

c 0 = 2b + 2c + d + e

d

e0 = b + d + e

8

ˆ<

ˆ:

n‚ …„ ƒ

„ ƒ‚ …

9

>=

>;

9

>>>>>>>>>>=

>>>>>>>>>>;

‚ …„ ƒ

Figure 19.13. New dimensions after FABA.

Lemma 6. If a regular ell has dimensions hb; c; d; ei such that c D kd CeC i , with k � 2

and 0 � i < d , then the sequence of moves BFA produces a regular ell with dimensions

hb0; c0; d; e0i, where c0 D .k � 1/d C e0 C i .

Proof. Performing BFA (Figure 19.14) produces an ell with dimensions

hb0; c0; d; e0i D h3b C c C d C e; b C c; d; b C d C ei:

Computing, we obtain

c0 D b C c D b C kd C e C i D .k � 1/d C b C d C e C i D .k � 1/d C e0 C i:

To see that the new ell is regular, first note that a move of B or F would result in a perfect

ell in standard position.

c + ec

e

B

b

d

F

b + d

A2b + c + d + e

(

8

ˆˆˆˆˆˆˆˆ

<

ˆˆˆˆˆˆˆˆ

:

‚ …„ ƒ

„ ƒ‚ …

o

Figure 19.14. The operation BFA.

To see that ED would add only new squares, observe that BFA added squares of sides

f D bCd , b, and a0 D 2bCcCd Ce. Move ED would add squares of sides e0 D bCd Ce

and d C e0 D b C 2d C e. On the other hand, we have

2b C c C d C e > b C 2d C e > b C d C e > b C d > b;

the first inequality following from the fact that c > d .

All that remains to show that the ell is regular is that c0 � d C e0. This follows from

c0 D .k � 1/d C e0 C i , where k � 2.

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19. Squaring the Plane 149

Lemma 7. If a regular ell with dimensions hb; c; d; ei is such that c D kd C e for some

positive integer k, then the ell can be squared up.

Proof. We can apply Lemma 6 until we have a regular ell whose dimensions

hb0; c0; d; e0i satisfy c0 D d C e0. This ell can then be squared up with ED.

Lemma 8. Every regular ell can be squared up.

Proof. Given a regular ell with dimensions hb; c; d; ei, we have c D kd C e C i with

k � 1 and 0 � i < d . We do induction on d to show that the ell can be squared up.

If d D 1, then i D 0 and we are done by Lemma 7. Now assume that the result

holds for any ell with dimensions hb�; c�; d �; e�i such that d � < d . We apply Lemma 6

iteratively until we have an ell with dimensions hb0; c0; d; e0i with c0 D d C e0 C i . Then

ED produces an ell with dimensions hb0; i; d C e0; d C 2e0i (Figure 19.15). By flipping

this over we obtain in standard position an ell with dimensions h2e0 C d; e0 C d; i; b0i(Figure 19.16).

E

D

i

Figure 19.15. The operation ED.

E

D

i

Figure 19.16. A flip after ED.

By Lemma 5 we can puff this up to a regular ell without increasing the length of side

D (which is i ). Since 0 � i < d , by the inductive hypothesis this flipped ell can be

squared up. Clearly, the unflipped ell can also be squared up, completing the proof of the

lemma.

Our main lemma follows immediately:

Lemma 9. Every perfect ell can be squared up.

Proof. Given an ell, we puff it up to a regular ell (Lemma 5), then square it up (Lemma 8).

In other words, every perfect ell is a part of a perfect rectangle. We have now completed

the spadework necessary to answer the tiling question raised in the introduction:

Theorem. It is possible to tile the plane with nonoverlapping squares using exactly one

square of each integral side.

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150 Part IV. Making Things Fit

Proof. It is easy to see how to carry out the plan described earlier. We start with any perfect

ell and grow it by adding squares as described in the algorithm whose steps we presented

in Section 2.

Note that we don’t move squares around; once added to the figure they are fixed. In par-

ticular, we never actually flip the figure as in the proof of Lemma 9—the flipping described

there is simply to demonstrate that we can square up a particular ell. Note also that we are

guaranteed to fill the plane since we are careful to grow the figure in all four directions in

each cycle of the algorithm.

Finally, for any n the square of side n will be added to the figure by or before the nth

time we perform step 2 of the algorithm. Since we perform step 2 infinitely many times,

we are guaranteed to incorporate in the tiling a square of each integral side.

Reflections

In preparing this paper, we learned some of the history of the problem. The question was

first posed by Solomon Golomb in a 1975 article in the Journal of Recreational Mathemat-

ics. He called it the “heterogeneous tiling conjecture” and challenged readers to prove or

disprove it.

Martin Gardner wrote about the question four years later in his column in Scientific

American, later anthologized in [2]. He described one approach to a solution, a fairly or-

derly tiling of roughly three-quarters of the plane with squares and reported that Verner

Hoggatt Jr., then editor of The Fibonacci Quarterly, had shown that no square in the tiling

appeared more than once.

Grunbaum and Shepard wrote about the problem in their 1987 book Tilings and Pat-

terns [4]. They described there a second way in which a squared square S can generate a

tiling of the plane (in addition to the method indicated in the present paper): Take a second

copy of S and expand it to a square S1 such that the smallest square in S1 is the size of the

original square S and fit S into that square. Take another copy of S and expand it to S2

so that its smallest square is the size of S1, and so on. Grunbaum and Shepard record the

observation of Carl Pomerance that in every tiling of the plane by unique squares known

at that time, the sides of the squares grow exponentially.

In 1997, Karl Scherer [6] succeeded in tiling the plane using multiple copies of squares

of all integral sides. The number t.n/ of squares of side n is finite but not bounded. He

describes his tiling as “size-alternating,” in that no two squares of the same side share any

portion of an edge (though they may share a corner).

Questions

Tiling is an enormous field. This theorem might be said to reside in the subfield that con-

cerns infinite tilings of the plane that use exactly one specimen each of a well-defined

collection of similar figures. Much work has been done here and many questions remain:

Efficiency The algorithm presented in this paper is extravagant in that the ratio of the

largest square used so far to the smallest square not yet used rapidly diverges. The proce-

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19. Squaring the Plane 151

dure for squaring up, for example, when applied to the smallest possible ell, a 2 � 2 square

next to a 1 � 1 square, ends in a rectangle with dimensions 1106481365205154721693 �2659648557852203795117. The smallest square not used at this point is 4 � 4.

The squaring-up procedure can certainly be improved. By way of illustration, take

the ell in Figure 19.6. This can be squared up to a 69 � 61 rectangle with the sequence

of operations FEFEABC (Figure 19.7). Our procedure, however, doesn’t square the ell up

until a rectangle is reached with dimensions approximately .5:0�1014272/�.5:8�1014272/.

Can our squaring-up procedure be improved in some well-defined way? Does an algo-

rithm exist for tiling the plane that methodically expands a connected island of squares in

such a way that the ratio of the largest square used to the smallest not yet used is bounded

by a polynomial?

Simple tilings A perfect figure is simple if it contains no perfect subrectangle. Our tiling

is far from simple. Is there a simple tiling of the plane using one specimen of each integral

side?

The half-plane and quarter plane Can the half-plane be squared? Can the quarter-plane

be squared? We are especially interested in this question because if it is possible to tile a

quarter plane four times using, altogether, every integral square just once, then it’s possible

to tile the plane using all the integral squares plus one square of any given side (say � ).

¼

Figure 19.17. � � � .

Rational squares The algorithm of our theorem can easily be used to tile the plane using

one square of every rational side. As with integral squares, we don’t know if a similar

procedure works for the half- or quarter-plane.

Positive and negative squares We can tile the plane with squares whose sides are natural

numbers and with squares whose sides are rationals. What about squares whose sides are

(positive and negative) integers? We interpret the effect of placing a small negative square

on top of a large positive one as removing a part of the large square. Once again, our

algorithm works easily for this. Just as placing a positive square next to a rectangle creates

an ell, so does placing a negative square on a corner of a rectangle. With care, no point of

the plane will be touching more than three squares (one negative, two positive).

Odd squares Can the plane be squared with all the odd squares? This seems unlikely to

us. Can the plane be squared with some of the odd squares? In general, what well-defined

subsets of the natural numbers can be used to tile the plane?

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152 Part IV. Making Things Fit

Coloring Neither our tiling nor the Fibonacci tiling can be 3-colored. Is there a 3-colorable

tiling of the plane using exactly one square of each integral side? Is there a simple algo-

rithm for 4-coloring the tiling described in this paper?

Space Can space be cubed?

Triangles Scherer proved that the plane cannot be tiled with equilateral triangles of dif-

ferent sizes if one triangle is smallest [7]. He has found a way of tiling the plane, however,

with different sizes of iscoceles right triangles and with enlargements of certain other tri-

angles [8]. Pomerance proved that it is possible to tile the plane with one rational triangle

of each congruence class such that any two neighboring triangles share either an entire

side or just a vertex [5]. Left open is the question: Can the plane be tiled with all rational

equilateral triangles so that no triangle has an infinite number of neighbors?

Acknowledgments We would like to thank Joe O’Rourke for his encouragement, sug-

gestions, and enthusiasm. We are also grateful for the improvements to the paper suggested

by the anonymous referees.

Bibliography

[1] M. Gardner, The Second Scientific American Book of Mathematical Puzzles & Diversions: A

New Selection, Simon and Schuster, New York, 1961.

[2] ——, Fractal Music, Hypercards and More . . . , W. H. Freeman, New York, 1992.

[3] S. W. Golomb, The heterogeneous tiling conjecture, J. Recreat. Math. 8 (1975) 138–139.

[4] B. Grunbaum and G. C. Shephard, Tilings and Patterns, W. H. Freeman, New York, 1987.

[5] C. Pomerance, On a tiling problem of R. R. Eggleton, Discrete Math. 18 (1977) 63–70.

[6] K. Scherer, New Mosaics, privately printed, 1997.

[7] ——, The impossibility of a tessellation of the plane into equilateral triangles whose sidelengths

are mutually different, one of them being minimal, Elem. Math. 38 (1984) 1–4.

[8] ——, Nutts and Other Crackers, privately printed, 1994.

[9] W. Tutte, Squaring the square, Canad. J. Math. 2 (1950) 197–209.

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20Magic Knight’s Tours

John D. Beasley

In Mathematical Magic Show [4], Martin Gardner looked at the classic problem of the

knight’s tour on a chess board, paying particular attention to tours in which the numbers in

each row and column added to 260. At that time, the question of whether the numbers in

each long diagonal could also be made to add to 260 was still open. Thanks to advances

in computer power since Martin wrote, this question has now been decided, and perhaps

readers will be interested in the complete statement that it is now possible to make.

A brief history of magic knight’s tours

In knight’s tour literature, the term magic is used for all tours in which the rows and

columns add to the same number. Any diagonal or other properties are a bonus. Figure

20.1a is an example. This is a closed tour (the starting and finishing squares are a knight’s

move apart), and it has two-fold rotational symmetry (numbers which are diametrically

opposite differ by 32, so if we draw the lines joining each number to the next, and 64 to 1,

we get a pattern which repeats itself on rotation through 180 degrees). No knight’s tour on

an 8�8 board can be laterally or diagonally symmetric, nor can it have more than two-fold

rotational symmetry, so Figure 20.1a has as much symmetry as we can hope for. We may

notice that the numbers along a row or column are alternately odd and even.

2 11 58 51 30 39 54 15 55 30 17 42 27 36 15 38

59 50 3 12 53 14 31 38 18 43 54 29 16 39 26 35

10 1 52 57 40 29 16 55 31 56 41 20 33 28 37 14

49 60 9 4 13 56 37 32 44 19 32 53 40 13 34 25

64 5 24 45 36 41 28 17 57 2 45 8 21 64 51 12

23 48 61 8 25 20 33 42 46 5 60 1 52 9 24 63

6 63 46 21 44 35 18 27 3 58 7 48 61 22 11 50

47 22 7 62 19 26 43 34 6 47 4 59 10 49 62 23

(a) Wenzelides, 1849 (b) Jaenisch, 1859

Figure 20.1. Two magic knight’s tours, with starting and finishing squares in bold.

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (January 2012), pp. 72–75.

153

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154 Part IV. Making Things Fit

The first magic knight’s tour to appear in print was published by William Beverley in

1848, but this was an open tour (there was no knight’s move from the final square back

to the first square). Figure 20.1a, which was found by Karl Wenzelides and appeared in

Schachzeitung in 1849, was the first closed magic knight’s tour [12, 18]. Several more

closed magic tours appeared during the next few years, including Figure 20.1b, by C. F.

Jaenisch, also rotationally symmetric, which appeared in The Chess Monthly in 1859 [5]

and was described by its discoverer as “La solution la plus parfaite du probleme du cav-

alier” ([5] is presented bilingually, but the French was the author’s original). Although in

neither of these tours do the long diagonals add to 260, the two long diagonals together do

add to 520, and Figure 20.1b has a further property at which we look in a moment.

By 1940, 126 magic tours had been discovered, 59 of them closed. They were listed in

1951 in an unpublished monograph The Magic Knight’s Tours, a Mathematical Recreation

by H. J. R. Murray [9, 16], but there was no convenient list in print until George Jelliss gave

one in Chessics in 1986 [7, 8]. He accompanied it with a note that with modern computer

methods it should be possible to ascertain whether all such tours had been discovered, and

T. W. Marlow immediately showed that they hadn’t by discovering five more, which he

published in 1987 and 1988 [14, 15]. T. S. Roberts found two more in January 2003 [10].

Still, none had proved to be diagonally magic, and a distributed computing exercise was

set up by Hugues Mackay, Jean-Charles Meyrignac, and Guenter Stertenbrink later in 2003

to put the matter finally to rest. This exercise rediscovered all the magic tours previously

known and added seven new ones, three open and four closed, giving a grand total of 140

(63 closed), none diagonally magic [11, 17]. These conclusions were subsequently verified

by Yann Denef using a program written independently [17].

The complete catalogue is on at least two web sites [11, 17] and in Variant Chess [1, 2].

Each tour can be presented in any of eight orientations and can be numbered from either

end, so these 140 tours give rise to 2240 different arithmetical matrices.

Job done?

Not quite. Only the 8 � 8 case was settled. We can ignore boards of odd side, because the

row sums of a tour on such a board will be alternately odd and even, and so cannot all be

the same. We can also ignore boards of side 4n C 2, since George Jelliss has produced

an elegant and ingenious proof implying that no magic tour on such a board exists [11].

Proofs in this field tend to be either elementary or non-existent, but this is an exception.

Theorem (G. P. Jelliss). A magic knight’s tour on a board of side 4n C 2 is impossible.

Lemma. In a magic knight’s tour, the number of entries of the form 4k C 2 or 4k C 3 in

any row or column, counted together, must be even.

Proof of the lemma. Let h be half the length of the side. The row sum is now

4h3 C h. If we express the numbers in the row in binary, we see that h of them, being

odd, have a non-zero units digit, and adding these digits gives h. Remove these units digits

from the entries, and the numbers represented by the remaining digits must sum to 4h3.

This is a multiple of four and the units digits in the contributing numbers have been re-

moved, so the number of non-zero twos digits in them must be even. But each entry of the

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20. Magic Knight’s Tours 155

form 4k C 2 or 4k C 3 contributes one such digit and no entry of the form 4k or 4k C 1

contributes any, so the result follows. A similar argument applies to the columns.

Proof of the theorem. Label the squares of the chessboard thus:

A B A B A B . . .

C D C D C D . . .

A B A B A B . . .

C D C D C D . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We reflect the tour, if necessary, so that the number in the top left corner is odd; then all

the A and D squares will contain odd numbers, and all the B and C squares even numbers.

Now consider the distribution of the entries of the form 4k C 3. There are 4n2 C 4n C 1

of these, which is an odd number, and they must all be in A or D squares, so either the A

squares contain an odd number and the D squares an even number, or vice versa. Similarly,

either the B squares contain an odd number of entries of the form 4k C 2 and the C squares

an even number, or vice versa.

Now suppose first that the A squares contain an odd number of 4k C 3 entries and the

B squares an odd number of 4k C2 entries; then the C squares will contain an even number

of 4k C 2 entries, and the total number of 4k C 3 and 4k C 2 entries on A and C squares,

counted together, will be odd. But these squares appear only in alternate columns, and by

the lemma each individual column must contain an even number of such entries. This is a

contradiction, so the situation is impossible. A similar argument applies whichever of the

A or D squares contain the odd number of 4k C 3 entries and whichever of the B or C

squares the odd number of 4k C 2 entries.

Thus there can be no magic knight’s tour on a board of side 4n C 2.

So only boards of side 4n need be considered. T. H. Willcocks had discovered an open

magic tour on a 12 � 12 board in which one diagonal added to the magic constant, and in

April 2003 Awani Kumar found four such tours with both diagonals magic [10, 17]. As far

as I know, the existence of closed magic tours on this board with both diagonals also magic

remains an open question, but two such tours on a 16 � 16 board were discovered by H. E.

de Vasa and published in 1962 [3, 11], and others have been discovered since [13, 17].

Return to the 8 � 8 board

In the field of ordinary magic squares, where the numbers are not constrained to form a

knight’s tour, particular attention is paid to pan-diagonal squares, whose numbers add to

the magic constant not only along rows, columns, and principal diagonals, but also along

broken diagonals (in algebraic chess notation, diagonals such as a4-e8/f1-h3). Even while

the possibility of a magic knight’s tour with each long diagonal also adding to 260 was

open, it was realised that no pan-diagonal tour could exist, because the numbers on a diag-

onal are either all odd or all even, and the 32 odd numbers from 1 to 64 don’t add to 4�260;

they add to 4 � 256, and the 32 even numbers to 4 � 264. So while preparing the catalogue

of magic tours for Variant Chess, I looked to see if any had a property which I called

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156 Part IV. Making Things Fit

quasi-magic, where the odd diagonals, principal and broken, each add to 256, and the even

diagonals add to 264. Given the inherent disparity between odd and even diagonals, this

seemed to be as close to a truly pan-diagonal magic knight’s tour as we could hope for.

It turned out that none had this property in full, but Figure 20.1b is half-way there. The

principal odd diagonal adds to 256, as do each of the three parallel odd broken diagonals,

and the principal even diagonal and the three parallel even broken diagonals each add

to 264. The same is true of two further tours which Jaenisch published in 1862 [6], but

those tours lack the rotational symmetry of Figure 20.1b. Because of this combination of

properties, therefore, Figure 20.1b is indeed the “most perfect” solution to the problem of

the knight’s tour on the 8 � 8 board.

Now comes a curious question: Was Jaenisch aware of this property of the broken

diagonals in Figure 20.1b, which to my mind adds greatly to the elegance of this tour? I can

find no mention in any of the sources I have examined from the nineteenth to the twenty-

first century. The property will have become obvious as soon as it occurred to somebody

to look for it, and I wrote in 2008 that I could not believe it had remained unspotted until

then. Even so, nobody has yet drawn my attention to an earlier report of its existence.

Perhaps a reader will be able to enlighten me.

Acknowledgment I have been merely the reporter, and in addition to the workers named

above I am grateful to the Bodleian Library for access to The Magic Knight’s Tours, a

Mathematical Recreation and to other Murray papers, to Cambridge University Library

for access to Traite des Applications de l’Analyse Mathematique au Jeu des Echecs, and

to George Jelliss for photocopies of the relevant pages from Schachzeitung and The Chess

Monthly (made for him by Marian Stere and Ken Whyld). I am also grateful to George

Jelliss for constructive comments on my original draft.

Bibliography

[1] J. D. Beasley, Another look at 8 � 8 magic knight’s tours, Variant Chess 57 (2008) 50–53.

[2] ——, Credit where credit is due, Variant Chess 58 (2008) 72.

[3] G. D’Hooghe, Les Secrets du Cavalier, Brepols, Bruxelles, 1962.

[4] M. Gardner, Knights of the square table, in Mathematical Magic Show, George Allen and Un-

win, London, 1977, 188–202, 283.

[5] C. F. Jaenisch, De la solution la plus parfaite du probleme du cavalier, The Chess Monthly

(1859) 110–115, 146–151, 176–179.

[6] ——, Traite des Applications de l’Analyse Mathematique au Jeu des Echecs, vol. 2, St Peters-

burg, 1862.

[7] G. P. Jelliss, Catalogue of 8 � 8 magic knight’s tours, Chessics 26 (1986) 122–128.

[8] ——, Notes on Chessics 26, Chessics 29/30 (1987) 163.

[9] ——, H. J. R. Murray’s history of magic knight’s tours, The Games and Puzzles Journal 14

(1996) 238–244, 15 (1997) 266–267.

[10] ——, Recent advances in magic knight’s tours, Variant Chess 43 (2003) 40–41.

[11] ——, Knight’s tour notes (2001–2005); available at www.mayhematics.com.

[12] D. E. Knuth, letter to G. P. Jelliss, quoted in The Games and Puzzles Journal 14 (1996) 243.

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20. Magic Knight’s Tours 157

[13] A. Kumar, Studies in magic tours of knight on 16�16 board, J. Recreat. Math. 34 (2005–2006)

275–285.

[14] T. W. Marlow, Magic knight tours, The Games and Puzzles Journal 1 (1987) 11.

[15] ——, Magic knight tours, The Problemist 12 (1988) 379.

[16] H. J. R. Murray, The Magic Knight’s Tours, a Mathematical Recreation, manuscript in the

Bodleian Library, shelfmark MS Eng d.2370.

[17] G. Stertenbrink, Computing magic knight’s tours; available at magictour.free.fr.

[18] K. Wenzelides (as “...l .........s in P.......”), Bemerkungen uber den Rosselsprung, Schachzeitung

4 (1849) 41–97.

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21Some New Results on

Magic Hexagrams

Martin Gardner

Combinatorial problems involving magic squares, stars, and other geometrical structures

often can be solved by brute force computer programs that simply explore all possible

permutations of numbers. When the number of permutations is too large for a feasible run-

ning time, an algorithm can frequently be reduced to manageable time by finding ingenious

shortcuts. Such planning makes computer solving less trivial and much more interesting.

A superb example of such planning was described in a little-known short article in

the Mathematical Gazette (vol. 75, June 1991, pp. 140–142). The authors, Brian Bolt,

Roger Eggleton, and Joe Gilks, posed for the first time a problem based on the pattern

shown in Figure 21.1. It is the traditional hexagram or Star of David with its inner hexagon

divided into six equilateral triangles. Can numbers 1 through 12 be placed inside the twelve

triangles so that each of the six rows, indicated by the arrow, have the same sum?

The authors point out that there are 12Š ways of arranging the numbers. When the

pattern’s six rotations, and six reflections, are excluded, the number reduces to 11Š or

39,916,800. Can this number be further reduced?

Figure 21.1.

Reprinted from The College Mathematics Journal, Vol. 31, No. 4 (September 2000), pp. 274–280.

159

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160 Part IV. Making Things Fit

It can. Each of the six rows must have the same sum, and this “magic constant” must

be either odd or even. It is not hard to discover that, by neglecting rotations, reflections,

and complements, there are just three ways that odd and even numbers from 1 to 12 can be

distributed on the hexagram (see Figure 21.2).

A B C

0

0 0

0

0

0 1 1 1

11

1

1 1 1 1

11

0

00

0

0

0

0

00 0

0

0

1

11 1

1

1

Figure 21.2.

A complement, I should explain, is obtained by subtracting each number of a magic

square or star from the pattern’s largest number plus 1. For example, consider the Chinese

lo shun or 3 � 3 magic square in Figure 21.3. If each number is replaced by the remainder

when it is taken from 10, we obtain the complement (Figure 21.4).

8 1 6

3 5 7

4 9 2

Figure 21.3.

2 9 4

7 5 3

6 1 8

Figure 21.4.

Note that the complement in this case is the same as before except for a rotation and

reflection. In the case of magic hexagrams, complements are always different, but are con-

sidered trivial variations. The complements of the three patterns shown in Figure 21.2 are

obtained by replacing each 1 with 0 and each 0 with 1. Each 1 stands for an odd number,

each 0 for an even number. By neglecting complements, as well as rotations and reflec-

tions, the three odd-even patterns provide a way of establishing upper and lower bounds

for the star’s magic constant.

In pattern A all the even numbers go in outside triangles. They add to 42, but because

each number appears in two rows, we double 42 to get 84 as the contribution the even

numbers make to the total of the sums of all six rows.

The odd numbers in pattern A go on interior triangles. They add to 36, but because

each number is in three rows we triple 36 to get 108 as the contribution the odd numbers

make to the total of all six rows.

The sum of all the rows sums is 84 C 108 D 192. There are six rows, so to find the

magic constant for this pattern we divide 192 by 6 to get 32. This is the magic constant

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21. Some New Results on Magic Hexagrams 161

for pattern A on the assumption that the numbers can be arranged to make the hexagram

magic. As it turns out, pattern A has no solution.

We turn now to patterns B and C. Each has four odd numbers and two even numbers

on the interior triangles. To make the magic constant as low as possible, we place on these

inside triangles, where each number appears in three rows, the four lowest odd numbers

.1; 3; 5; 7/ and the two lowest even numbers .2; 4/. The sum of the inside numbers is 22.

Tripling it gives 66.

On the outside triangles go odd numbers 9; 11 and even numbers 6; 8; 10; 12. They add

to 56. Each number appears in two rows so we double 56 to get 112. Adding 66 to 112

gives 178.

We know that this total must be a multiple of 6 because six rows must have the same

sum. So we raise 178 by the smallest amount to arrive at a multiple of 6, namely 180.

Dividing 180 by 6 yields 30. This is the lower bound for the magic constants of patterns B

and C.

To obtain the upper bound we put the largest two even numbers .10; 12/ and the four

largest odd numbers .5; 7; 9; 11/ on the interior triangles. They add to 54 and three times

54 is 162. The outside numbers .1; 2; 3; 4; 6; 8/ add to 24 and twice 24 is 48. The sum of

162 and 48 is 210. Dividing 210 by 6 gives 35 as the upper bound for the magic constants

of patterns B and C.

We now have shown that the magic constant, if such a magic hexagram does indeed

exist, must be 30; 31; 32; 33; 34, or 35. Knowing this, and knowing how odd and even

numbers must be distributed to make the hexagrams magic, allowed the three authors to

reduce their computer program to a manageable length. To the authors’ vast surprise, their

computer search produced just one solution! It is shown on the left of Figure 21.5. Its com-

plement (each number taken from 13) is on the right. “When we discovered the beautiful

solution,” the authors write, “we rushed from our baths into the street shouting ‘Eureka!’”

The two patterns make excellent puzzles. Place numbers 1 through 12 on the hexagram so

each row adds to 33, or so each row adds to 32. In both cases the solution is unique. Note

that in the pattern on the left each pair of opposite corners add to 12, and on the right, they

total 14.

1

2 4

10

11

83

57

69

12

510

86

79

12

111

4

3

2

Figure 21.5.

A much older magic hexagram problem, now thoroughly explored, consists of placing

numbers 1 through 12 on the vertices of the traditional Star of David shown in Figure 21.6.

Because each number appears in two rows, the sum of all the row sums is twice the sum of

numbers 1 through 12, or 78�2 D 156. There are six rows, so 156=6 D 26, the hexagram’s

magic constant.

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162 Part IV. Making Things Fit

Figure 21.6.

The British puzzlist Henry Ernest Dudeney, in Modern Puzzles (1926) claimed there

are 37 fundamental solutions, or 74 if complements are included. This was one of Du-

deney’s rare mistakes. He missed three basic patterns. Von J. Christian Thiel, in the German

periodical Archimedes (vol. 5, September 1963, pp. 65–72) displays 15 basic solutions. By

applying well known transformations in which numbers are interchanged a certain way,

Thiel raises the total of patterns, not counting rotations and reflections, to 40. These, to-

gether with their complements, make 80 solutions.

Dudeney, in Modern Puzzles, and later in A Puzzle-Mine, identifies six solutions which

have the additional feature that the points of the star also add to 26. They are shown in

Figure 21.7. Each has a complement in which the interior numbers add to 26. Note that on

each hexagram each large triangle has a sum of 13, so together they add to 26, and the sum

of any small triangle is the same as the sum of the small triangle opposite.

8

10

9

5

5

10

6 9 8 3

5 7

2 11 12 1

4

7 10 4 5

6 11

2 12 9 3

1

3 11 10 2

49

1 6 12 7

84

3 12 10 1

8 5

2 6 11 7

4 7 9 6

58

1 11 12 2

3 8

2 7 11 6

10 3

1 9 12 4

Figure 21.7.

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21. Some New Results on Magic Hexagrams 163

29 137

149 157 163 191

193 179

173 167 139 181

15137

73 43 41 47

71 67

53 31 61 59

Figure 21.8.

Akira Hirayama and Gakuho Abe, in Researches in Magic Squares (Osaka, 1983)—the

text is in Japanese—give a slightly different way of arriving at 80 fundamental solutions.

The book also includes Abe’s amazing constructions of two magic hexagons made with

consecutive primes! They are shown in Figure 21.8.

There is now a sizeable literature on stars with more than six points. The five-pointed

star, or pentagram, is easily shown to have no solution for numbers 1 through 10. You’ll

find one such proof in the chapter on magic stars in my Mathematical Carnival. In the

same chapter I explain how the numbers in a solution for the traditional hexagram can be

transferred to the twelve edges of the cube so that four numbers around each face total

26. Because the cube’s “dual” is the octahedron (faces and vertices exchanged), the same

numbers can be placed on the octahedron’s edges so that the four edges around each vertex

add to 26. Figure 21.9 shows how a hexagram solution is transferred to a magic cube and

to a magic octahedron.

10

3

1 11 12 2

8 5

4 7 9 6

7

64 9

5108

1 112

3

1211

121

23

8

105

9

47 6

Figure 21.9.

A third type of hexagram is formed by adding three diagonals to the traditional structure

as shown in Figure 21.10. The question arises: Can numbers 1 through 19 be placed on

the vertices of this pattern so eveiy line of five has the same sum? Harold Reiter, at the

University of North Carolina, Charlotte, posed this as an unsolved problem in his article

“A Magic Pentagram,” in Mathematics Teacher (March 1983, pp. 174–177).

By trial and error I discovered the solution shown; it has 1; 2; 3; 4; 5; 6 on the out-

side points, and a magic constant of 46. Its complement (numbers taken from 20) puts

14; 15; 16; 17; 18; 19 on the outside points, and raises the magic constant to 54.

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164 Part IV. Making Things Fit

19

1

24 12 9

10 15

18 13 8

11 16

75 17 14 3

6

Figure 21.10.

Finding all solutions to this problem by a brute force computer algorithm is out of

the question because it would require examining 19Š=12 possible permutations, Reiter and

his associate David Ritchie, in “A Complete Solution To the Magic Hexagram Problem,”

in The College Mathematics Journal (vol. 21, September 1989, pp. 307–316), describe the

clever shortcuts they used to reduce their algorithm to an examination of a mere 18,264,704

patterns. In less than five minutes their Pascal program found 2,190 basic solutions, not

counting rotations, reflections, and complements, or 4,380 solutions if complements are

included. The magic constants range from 46 through 54.

The hexagram pattern in Reiter’s problems is an interesting one. It solves an old tree-

plant puzzle of placing 19 trees to form six straight rows with four trees in each row. I used

this pattern for a checker-like game I named Solomon. It can be purchased from Kadon

Enterprises, 1227 Lorene Drive, Pasadena, Maryland 21122.

Frank Bernhart wrote to point out that there are many symmetric patterns with vertices

and lines that are combinatorially equivalent to Reiter’s hexagram, and therefore have the

same set of solutions. Three are shown in Figure 21.11.

Figure 21.11.

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21. Some New Results on Magic Hexagrams 165

The traditional hexagram obviously contains eight different triangles. The modified

hexagram of the kind shown in Figure 21.1 has 20 triangles. (Many old puzzle books

give this as a problem.) How many different triangles can you find in Reiter’s hexagram?

Counting them all is not so easy. I leave this as a question to be answered in the next issue

(see next page).

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166 Part IV. Making Things Fit

In The College Mathematics Journal (Vol. 31, No. 5, page 146), the number of triangles

in Reiter’s hexagram was given as 50. However, the current editors have found 56 distinct

triangles.

19

1

24 12 9

10 15

18 13 8

11 16

75 17 14 3

6

The triangles are grouped by number of regions below with the count and a representative

for each of seven types of congruent triangles.

1 region: 12 triangles congruent to 1; 9; 12,

2 regions: 6 triangles congruent to 1; 12; 19,

2 regions: 12 triangles congruent to 1; 10; 13,

4 regions: 6 triangles congruent to 1; 5; 13,

5 regions: 6 triangles congruent to 1; 6; 18,

6 regions: 12 triangles congruent to 1; 5; 17, and

12 regions: 2 triangles congruent to 1; 3; 5.

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22Finding All Solutions

to the Magic Hexagram

Alexander Karabegov and Jason Holland

Problem 394 in Henry Dudeney’s 536 Puzzles and Curious problems [1] poses this puzzle:

Put the numbers from 1 to 12 in the circles on the left in Figure 22.1 so that the sum of the

four numbers on each line is 26. One solution is shown on the right.

6

2 5 9 10

G

I H F E

12 7DJ

1

3 11 48CB

A

LK

Figure 22.1. The hexagram along with one solution.

The solution pictured in Figure 22.1 is listed as solution number 16 in Table 22.1 at the

end. Table 22.1 contains 20 solutions to the puzzle. The letters in Figure 22.1 on the left

are used as positions for the solutions in Table 22.1.

There are many puzzles referred to as magic stars (see, for example, [2, Ch. 5]). While

such puzzles are usually fun, they often require the “brute force” method (also called the

method of exhaustion) for their solution. In this brief article, we present a systematic ap-

proach to finding all solutions of Dudeney’s puzzle. The key to this approach is something

known as an out-shuffle in cards.

It is known that if two solutions are considered to be the same when one can be obtained

from the other by rotation or reflection, then there are 80 different solutions to this puzzle.

(Curiously, Dudeney erred in his count, finding only 74, see [1, p. 350]). Martin Gardner

[2, chapter 21 in this collection] called this puzzle the Magic Hexagram and pointed out

Reprinted from The College Mathematics Journal, Vol. 39, No. 2 (Mar. 2008), pp. 102–106.

167

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168 Part IV. Making Things Fit

Table 22.1. The twenty non-equivalent solutions to the Magic Hexahedron.

A B C D E F G H I J K L

1 1 7 2 8 10 4 12 3 9 5 6 11

2 1 10 2 8 7 4 12 6 9 5 3 11

3 1 5 7 9 11 4 6 8 3 10 2 12

4 1 8 4 6 11 7 9 5 3 10 2 12

5 1 5 7 11 9 6 2 8 3 12 4 10

6 1 10 8 4 11 5 9 3 7 12 2 6

7 1 7 3 12 6 2 9 8 10 4 5 11

8 1 5 3 12 8 2 9 6 10 4 7 11

9 1 4 2 10 11 5 9 3 7 6 8 12

10 1 9 3 11 5 4 8 10 7 6 2 12

11 1 9 2 6 10 7 11 4 5 8 3 12

12 1 6 3 11 8 2 10 7 9 4 5 12

13 1 4 7 11 10 2 6 9 5 8 3 12

14 1 7 2 10 8 3 11 6 9 4 5 12

15 1 9 2 10 6 11 3 4 5 12 7 8

16 1 8 4 7 10 9 6 5 2 12 3 11

17 1 2 5 12 11 3 6 4 8 7 9 10

18 1 9 5 12 4 3 6 11 8 7 2 10

19 1 7 3 8 10 9 6 5 2 11 4 12

20 1 6 5 10 9 7 4 8 2 11 3 12

that there are indeed 80 different solutions. He also observed that it is equivalent to placing

the numbers from 1 to 12 on the edges of a cube so that the sum of the four labels on every

face is 26. This identification is illustrated in Figure 22.2.

There seems little choice but to call Gardner’s version the Magic Hexahedron (unless

it be called the Magic Cube). The two versions are not entirely equivalent however—the

6

2 5 9 10

12 7

1

3 11 48

9

6

5

4

7

10

8

11

121

2

3

Figure 22.2. The identification between the hexagram and the cube.

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22. Finding All Solutions to the Magic Hexagram 169

6

25 910

12 7

1

311 4 8

9

6

5

4

7

10

8

11

121

2

3

Figure 22.3. Solution generated from Figure 22.2 by a reflection.

cube has more symmetries than the hexagram (or, equivalently, the hexagon). While the

symmetry group of a hexagon has order 12, that of the cube has order 48. For each solution

to the Magic Hexahedron, there are four Magic Hexagram solutions. We use Figure 22.2

to illustrate the latter point. If one reflects the cube across the vertical plane through the

midpoints of the edges labeled 7, 6, 12, and 1, one gets the solution shown on the left in

Figure 22.3.

The corresponding Magic Hexagram is shown on the right in Figure 22.3. Clearly, this

is different from the solution of the Magic Hexagram in Figure 22.2, that is, no symmetry

of one of these hexagrams can produce the other. (For an excellent discussion of the full

symmetry group of the cube, see [3, pp. 112–113].) We now turn to determining that there

are twenty different Magic Hexahedron solutions.

To facilitate our enumeration, we consider two ways of classifying a face in a solution.

First, a face is parity-balanced if it has two odd labels and two even ones. Note that the

only non-parity balanced faces have all labels odd or all even. A solution is then called

parity-balanced if all of its faces have this property. Observe that in a solution that is not

parity-balanced, one face must have all labels odd, one must have all labels even, and the

other four labels must alternate in parity around the cube. Since the only possibilities for a

face that is not parity-balanced are f1; 5; 9; 11g, f3; 5; 7; 11g, f2; 4; 8; 12g, and f2; 6; 8; 10g,

it is straightforward (and not difficult) to find the ten solutions that are not parity-balanced.

These are the first ten solutions in Table 22.1.

We now turn to our second way of classifying a solution. We call a face magnitude-

balanced if two of its labels are 6 or less and the other two are 7 or more, and a solution

is magnitude-balanced if all of its faces are. Note that of the solutions 1–10 in Table 22.1,

only the first two do not have this property.

Given a magnitude-balanced face, its four labels can be written as i , j , k C6, and l C6,

where each of i , j , k, and l is at most 6. Observe that since i C j C k C l D 14, the set

f2i � 1; 2j � 1; 2k; 2lg then is a parity-balanced set with sum 26. Hence, the mapping M

with

M.n/ D(

2n � 1 if 1 � n � 6,

2.n � 6/ if 7 � n � 12,

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170 Part IV. Making Things Fit

transforms a magnitude-balanced solution to one that is parity-balanced. If we apply the

mapping M to the eight solutions 3–10 in Table 1 (perhaps twice), we get the ten solutions

11–20 in Table 22.1. This process is illustrated in Table 22.2 where arrows to the right

indicate that the mapping M has been applied.

Table 22.2. Solution numbers from Table 22.1 along with what happens when M is applied.

3 ! 11 ! 12

4 ! 13 ! 14

5 ! 15

6 ! 16

7 ! 17

8 ! 18

9 ! 19

10 ! 20

All that remains is to show that there are no other solutions to the Magic Hexahedron

puzzle. For this, we use the inverse P of the mapping M :

P.n/ D

8

<

:

12.n C 1/ if n is odd,

12n C 6 if n is even.

This is a permutation of f1; 2; : : : ; 12g that takes a parity-balanced solution to one that is

magnitude-balanced. In a deck of 12 cards, this permutation is known as an out-shuffle [4].

We now show that every parity-balanced solution eventually leads through repeated

application of P to a non-parity solution. Suppose not, and let S be such a solution. Note

that our out-shuffle P leaves 1 and 12 fixed and permutes the other ten numbers in the

cycle .2; 7; 4; 8; 10; 11; 6; 9; 5; 3/. One of the faces in the solution S contains 12 and not

1. Sooner or later, if we apply P repeatedly, the labels on this face include 12, 11, and

two other numbers, neither of which can be 1. Since their sum is 26, this is impossible.

Hence, every parity-balanced solution eventually leads to a non-parity-balanced solution.

Therefore, every parity-balanced solution arises from a non-parity-balanced one by apply-

ing the mapping M , either once or twice. Thus, Table 22.1 gives all solutions to the Magic

Hexahedron puzzle.

Theorem. The Magic Hexahedron puzzle has precisely twenty non-equivalent solutions.

Corollary. The Magic Hexagram puzzle has precisely eighty non-equivalent solutions.

Acknowledgments The authors wish to thank John Barton who wrote a program that

verified all 960 solutions to the Magic Hexagram and thus gave us plenty of data to work

with. We also wish to thank Amy Niemeyer who pointed out the term out-shuffle, Stephanie

Wong who brought this wonderful puzzle to our attention, and Lowell Beineke for suggest-

ing the name Magic Hexahedron.

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22. Finding All Solutions to the Magic Hexagram 171

Bibliography

[1] H. E. Dudeney, 536 Puzzles & Curious Problems, Charles Scribner’s Sons, 1967.

[2] M. Gardner, Mathematical Carnival, Mathematical Association of America, 1989.

[3] F. Goodman, Algebra Abstract and Concrete, Prentice Hall, 1998.

[4] E. Weisstein, Out-Shuffle Mathworld webpage,

mathworld.wolfram.com/Out-Shuffle.html.

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23Triangular Numbers, Gaussian

Integers, and KenKen

John J. Watkins

One of the first of Martin Gardner’s Mathematical Games columns I ever read was

“Euler’s Spoilers,” in November 1959 [1], and after all these years I think it is still my

favorite (maybe this is just because I love the way he rhymed ‘Euler’ and ‘spoiler’). It dealt

with what we now call Latin squares, n � n arrays using n symbols such that each symbol

appears exactly once in each row and column.

Latin squares are extremely useful in the design of statistical experiments, but they

are perhaps better known now for their appearance in recreational puzzles such as sudoku

where each row and each column of a 9 � 9 array contains each of the integers 1; 2; : : : ; 9

exactly once. In 2004, a Japanese mathematics teacher, Tetsuya Miyamoto, invented a

sudoku-like puzzle called KenKen (loosely translated, this means “wisdom squared”).

KenKen quickly became so popular that it is now a standard feature in newspapers around

the world, and there are two very good web sites ([2], [3]) that offer new puzzles daily

online.

As in sudoku, the goal in KenKen is to fill an n � n grid with the numbers 1 through

n so as not to repeat a number in any row or column. Despite the fact that it uses num-

bers, sudoku is not an arithmetical puzzle; any nine symbols could be used in place of the

integers 1–9. Solving a KenKen puzzle, on the other hand, depends heavily upon several

important ideas about numbers.

Triangular numbers

In Figure 23.1, we see a typical KenKen where each heavily outlined ‘cage’ indicates

what must happen inside that particular cage. For example, inside the cage labeled ‘24�’

the product of the three numbers must be 24; hence, the three numbers in this cage are

forced to be 4, 3, and 2. The cage labeled ‘5�’ must contain the three numbers 5, 1, and

1 (because 5 is prime); moreover, we can immediately place these three numbers in this

cage in their correct positions since the two 1s must appear in different rows and different

columns. Since the only ways to partition the integer 10 into a sum of three distinct integers

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 37–42.

173

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174 Part IV. Making Things Fit

24× 5×

12×

10+

10+

2÷4 –

1–

1–

2

3

(a)

Figure 23.1. A typical 5 � 5 KenKen.

1; 2; 3; 4, and 5 are as 10 D 5 C 4 C 1 or as 10 D 5 C 3 C 2, we see that we have two

choices for the numbers to put in each of the cages labeled ‘10C’.

It almost goes without saying that prime factorization and partitions of integers are

important in solving KenKen puzzles. What is quite surprising is that another important

notion in number theory, the ancient Greek concept of triangular numbers, can also often

be used to solve a KenKen because in any solution to a KenKen the sum of the numbers

in any row or column of the grid is 1 C 2 C � � � C n; that is, the sum is the nth triangular

number. In Figure 23.1, the sum of the numbers in any row or column must be 15 (that is,

the triangular number 15 D 1 C 2 C 3 C 4 C 5). For example, since in the bottom row the

sum in the first cage is 10, the sum in the second cage must be 5. Hence, that cage contains

2 and 3. But, there is already a 3 in the fourth column, so we can immediately place the 2

and 3 in their correct positions in the bottom row.

At this point we can use a similar argument for the left-hand column to conclude that

the sum of the two numbers in the upper and lower left-hand corners must also be 5. Thus,

these numbers are either 2 and 3, or 1 and 4. But, they can’t be 2 and 3 because we have

already placed both 2 and 3 in the bottom row. Therefore, we know these two numbers

must be 1 and 4, and we even know that the 4 goes in the upper corner since the cage

labeled ‘24�’ contains 4, 3, and 2. It is now a straightforward matter to finish solving this

puzzle. (Solutions for the ordinary KenKen puzzles in this article are on p. 178.)

In Figure 23.2, we see a 4 � 4 KenKen puzzle that normally would require a lot of trial

and error to solve. For example, it is clear that the ‘3�’ cage contains a 1 and 4 and the

cage below contains a 2 and 3, but a good deal of experimenting would need to take place

to determine the exact placement of these four numbers. Fortunately, we can use triangular

numbers to avoid a tedious trial and error process.

It is clear that the two numbers a and b in the ‘2�’ cage must be 1 and 2 in some order.

Since the sum of the numbers in the two bottom rows is an even number .20 D 10 C 10/,

we immediately conclude that b is also even (because 9 is odd and the sum of the two

numbers in the ‘1�’ cage must also be odd). Therefore, b D 2, and it is now easy to solve

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23. Triangular Numbers, Gaussian Integers, and KenKen 175

2÷

12×3– 1–

1–

4

9+

a

b

(b)

Figure 23.2. Using triangular numbers to find b.

this puzzle without any trial and error (the first column can now be completed and the

‘12�’ cage cannot contain a 1). Note that we could have added all the known quantities in

the bottom two rows and then subtracted the resulting sum, 18, from 20 to find b, but that

takes more work, and often, as in this case, a simple parity argument provides the needed

information.

In Figure 23.3, we have a rather difficult 6 � 6 KenKen that can also be solved with

the aid of triangular numbers. We focus first on the ‘4�’ cage in the bottom two rows. The

question is whether the 4 factors as 4 � 1 � 1 or as 2 � 2 � 1. Note that, in either case, since the

120 in the ‘120�’ cage in the two bottom rows must factor as 6 � 5 � 4, we can conclude that

the ‘2�’ cage in the bottom row does not contain both 2 and 4 (because in the first case

there are already two 4s in the bottom two rows, and in the second case there are already

12×

120×

120×4×

11+

12+

10+

2

3

1–

3–

3÷

3÷2÷

2÷

7+

a

b

c

d

(c)

Figure 23.3. Using triangular numbers to find a; b and c.

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176 Part IV. Making Things Fit

two 2s in the bottom two rows). Hence, this ‘2�’ cage contains either 1 and 2, or 3 and 6,

and either way the sum of the two numbers in this cage is odd. Now, 3 is also odd, as is the

sum of the numbers in the ‘120�’ cage (6 C 5 C 4 D 15); however, the sum in the ‘10C’

cage is even, as is the entire sum of the bottom two rows (being twice the triangular number

21). Therefore, we immediately see that a C b C c must be odd. Hence, the numbers in

the ‘4�’ cage are 2, 2, and 1. With this information we can quickly fill in the bottom two

rows (using the fact that the ‘11C’ cage in the top row must contain a 5 and 6), and easily

complete the rest of the puzzle without ever having to resort to trial and error.

More tricks

The puzzle in Figure 23.3 can also be used to illustrate a very sneaky technique I learned

from Barry Cipra in Atlanta at the eighth Gathering for Gardner conference. He used his

idea to solve sudoku puzzles, but the same idea can also be helpful for KenKen. So, let’s

start from scratch on the puzzle in Figure 23.3, and try to determine the value of d based

on an argument involving the assumption that there is a unique solution to this problem.

It is usually implicit, though rarely stated, in puzzles such as KenKen and sudoku, that

there is a unique solution to the puzzle, and this fact can sometimes be invoked as an aid

in finding that solution. Again, we consider the situation in the ‘120�’ cage at the bottom

of the puzzle. This time we look at the ‘11C’ cage in the top row directly above this cage.

Since the ‘11C’ cage must contain 5 and 6, if we assume that there is a unique solution to

this puzzle then there cannot also be a 5 and 6 in the two squares in the bottom row in the

‘120�’ cage (otherwise, simply switching the 5s and 6s in the top and bottom rows yields

two distinct solutions). And, of course, 5 and 6 can’t both appear in the 4th column in the

‘120�’ cage since either a 5 or a 6 appears in that same column in the ‘11C’ cage in the

top row. The only remaining option in the ‘120�’ cage then is if d D 4.

Factorials, the multiplicative analog of triangular numbers, can also be used in solving

KenKen puzzles. For example, in a 6 � 6 puzzle, if there is a straight ‘120�’ cage with

four squares in a row, then the product of the two extra squares in this row is 6 (because6Š

120D 6). So, if this ‘120�’ cage is next to an angled ‘12�’ cage with two squares in the

same row and one square in the row above, then the single square in the row above must

contain a 2.

Gaussian puzzles

One of the most enjoyable features of KenKen puzzles, especially larger ones, is the way

in which the various possibilities that arise from the prime factorization of an integer come

into play when dealing with irregularly shaped cages. Another number system with unique

factorization into primes is the Gaussian integers, all complex numbers of the form a C bi

where a and b are integers. What is fun about the Gaussian integers is that some familiar

numbers such as 2 and 5 are no longer prime since they can be factored: 2 D .1 C i/.1 � i/

and 5 D .1 C 2i/.1 � 2i/.

We conclude this tribute to Martin Gardner with several KenKen puzzles for the reader

to solve featuring the Gaussian integers (solutions are on p. 186). Here are a few hints

to get you started. Consider the ‘4�’ cage in puzzle (d). Since 4 D 2 � 2 and we are not

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23. Triangular Numbers, Gaussian Integers, and KenKen 177

4×

2×2×

3+

3+

i –

1+ i – i – i –

i –

1– i –

8×

4×

(d) (e)

Figure 23.4. Solve using the four numbers 1, 1C i , 1 � i , and 2.

allowed to repeat a 2 in a column, there is only one useful way to factor 4 as a product of

three distinct numbers: 4 D 2.1 C i/.1 � i/. Thus, we know which three numbers belong

in this cage (though not their order). This tells us immediately that a 1 goes in the lower

left-hand corner (and then a 2 next to it in the ‘2�’ cage). There are two ‘3C’ cages in

this puzzle, one with three squares and one with two squares; it is easy to see that the only

possibility is for the cage with three squares to contain 1; 1 C i , and 1 � i , and for the cage

with two squares to contain 1 and 2.

What about the ‘i�’ cage in this puzzle? Clearly 2 cannot appear here since 2 minus

any of the other three numbers leaves a 1 behind. Also, this cage cannot contain both 1 C i

and 1� i since .1C i/� .1� i/ D 2i . So, the only options for this cage are for it to contain

1 and 1 C i (since .1 C i/ � 1 D i ), or for it to contain 1 and 1 � i (since 1 � .1 � i/ D i ).

5× 2×

5×4×2+2 i ×

2–2 i ×

1+2 i –

1+2 i–

i –

3 i –

4+ 3+

3+

i ÷

(f)

Figure 23.5. Solve using the six numbers 1, 1C i , 1 � i , 1C 2i , 1 � 2i , and 2.

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178 Part IV. Making Things Fit

Similarly, the ‘1 C i�’ cage must contain a 2 (since, otherwise, the 1s completely cancel

during subtraction). Thus, the only possibility is for this cage to contain 2 and 1 � i (since

2 � .1 � i/ D 1 C i ). The Gaussian integers, as lattice points in the complex plane, are

also vectors. For example, 1 C i can be thought of as the vector from the origin .0; 0/ to

.1; 1/. Geometrically, the difference ˛ � ˇ between any two complex numbers ˛ and ˇ is

just the vector from point ˇ to point ˛. Thus, in puzzle (d), it is easy to spot which pairs

of numbers can go in the ‘i�’ cage because the i vector can only be drawn from .1; 0/ to

.1; 1/, or from .1; �1/ to .1; 0/; similarly, we can determine the pair of numbers that must

go in the ‘1 C i�’ cage merely by observing that we can only draw a 1 C i vector from

.1; �1/ to .2; 0/.

Bibliography

[1] M. Gardner, Euler’s spoilers: the discovery of an order-10 Graeco-Latin square, New Mathemat-

ical Diversions, Mathematical Association of America, Washington DC, 1995, 162–172.

[2] The official KENKEN web site with puzzles posted daily; available at www.kenken.com.

[3] The New York Times puzzle web site with puzzles posted daily; available at

www.nytimes.com/kenken.

Solutions to the ordinary KenKen puzzles (a)–(c) on pages

174–175

(a)4 3 2 1 5

3 2 4 5 1

2 5 1 3 4

5 1 3 4 2

1 4 5 2 3

(b)1 3 4 2

4 1 2 3

3 2 1 4

2 4 3 1

(c)3 4 1 5 6 2

5 1 4 3 2 6

6 5 2 1 3 4

4 6 5 2 1 3

1 2 3 6 4 5

2 3 6 4 5 1

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VFurther Puzzles and Games

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24Cups and Downs

Ian Stewart

Martin Gardner was fond of mathematically-based magic tricks using simple apparatus

—in particular, pennies. Chapter 2 of his Mathematical Carnival [2, pages 12–26] is en-

tirely about penny puzzles. About five years later, Gardner and Karl Fulves invented a

delightfully simple trick with three pennies (see Demaine [1]). Its explanation involves an

important mathematical idea called a state diagram, useful in many combinatorial prob-

lems.

A related trick, usually described using cups instead of coins, has the same state dia-

gram with ‘heads’ and ‘tails’ replaced by ‘up’ and ‘down.’ This article explores the rela-

tions between the two tricks, develops the deeper mathematics that they have in common,

and discusses a generalization of the cups trick. (For other developments from Gardner’s

penny puzzles, see [1].)

Gardner’s three-penny trick

The trick is performed by a blindfolded magician. A volunteer places three pennies in a

row, and chooses at will whether each coin shows heads or tails. However, both heads and

tails must appear, otherwise the trick ends before it begins. The magician announces that

even though she cannot see the coins, she will give instructions to turn coins over so that

all three coins show the same face, heads or tails.

The instructions are:

1. Flip the left-hand coin.

2. Flip the middle coin.

3. Flip the left-hand coin.

After steps 1 and 2 the magician asks whether all three coins show the same face, and

if the answer is ‘yes’, the trick stops, otherwise the magician requests the third flip. Figure

24.1 shows an example. Although it is plausible that enough flips will eventually get all

coins the same way up, it is a little surprising that at most three flips are needed.

In this case it is straightforward to list all possible starting configurations, and check

that three moves always suffice. However, this is not a very satisfactory explanation. All be-

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 15–19.

181

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182 Part V. Further Puzzles and Games

(a) Typical start. . . (b) Flip the left. . .

(c) Flip the middle. . . (d) Flip the left.

Figure 24.1.

comes crystal-clear, if we draw the state diagram, which shows all possible configurations,

together with the transitions that occur when the instructions are obeyed.

The diagram itself naturally forms a cube, Figure 24.2a, in which each coin corresponds

to an axis in 3-dimensional space, and H, T (respectively) are the coordinates 0, 1. The

effects of moves 1 and 2 are shown in the figure, and it can easily be checked that starting

anywhere except HHH and TTT, a sequence of at most three moves always leads to HHH

or TTT.

⋂⋃⋂

⋂⋂⋂

⋂⋂⋃

⋂⋃⋃

⋃⋃⋂

⋃⋂⋂

⋃⋂⋃

⋃⋃⋃

THT

TTT

TTH

THH

HHT

HTT

HTH

HHH

flip

middle

flip

middle

flip

middle

flip

middle

flip left

flip left

flip left

flip left

Figure 24.2. (a) State diagram for the coin trick. (b) State diagram for the cup trick.

Three cups

The same state diagram occurs in the analysis of an old trick, which starts by placing three

cups upright on a table or a bar[[[

;

then turning the middle one over:[\[

:

You announce that you will turn all three of them upside down in exactly three moves,

where a move inverts exactly two cups. Like this:

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24. Cups and Downs 183

[\[

\[[

[[\

\\\

:

It can be done in one move, but asking for three is a classic piece of misdirection, setting

up the second part of the trick. Which is to casually turn the middle cup upright:

\[\

;

and invite someone to repeat the trick. But now, whatever they do, they can’t get all three

cups upside down.

What they haven’t spotted is that you’ve altered the initial position, and the parity of the

upright cups has changed from even to odd. The original start position has even parity, and

so does the required end position. But the new start position has odd parity. Since inverting

two cups preserves parity, the task becomes impossible.

There are several different ways to think about this puzzle mathematically. One is to

draw its state diagram, which shows all possible states of the three cups and uses connecting

lines to show what happens if you turn cups upside down one at a time. Figure 24.2b shows

this. Of course, you have to invert them two at a time, so you redraw the diagram to show

which states are connected by two 1-cup moves, obtaining Figure 24.3. This falls into two

disconnected pieces: one contains the first starting state and the required finishing one, and

the other contains the second starting state. Obviously you can’t change components by

going along the lines.

⋃⋂⋂ ⋂⋃⋂

⋂⋂⋃

⋃⋃⋃

⋂⋃⋃ ⋃⋂⋃

⋃⋃⋂

⋂⋂⋂

Figure 24.3. State diagram for 2-cup moves. (Each edge represents two or three distinct combinations

of 1-cup moves.)

The problem doesn’t depend on the order in which the cups are arranged, just how

many are up (and hence how many are down). So you can ‘factor out the symmetry’ by

considering only the number of ‘up’ cups, getting Figure 24.4a for 1-cup moves and Figure

24.4b for 2-cup moves. Now the obstacle is even more obvious.

3 2 1 0 3 1 2 0

Figure 24.4. State diagrams identifying symmetrically related states. (a) 1-cup moves. (b) 2-cup

moves.

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184 Part V. Further Puzzles and Games

You can also represent parity by colouring odd numbers grey and even ones white. The

result, for Figure 24.4a, is Figure 24.5. Now any two-step move preserves colour (that is,

parity) so again there can’t be a solution.

3 2 1 0

Figure 24.5. Parity diagram for 1-cup moves.

Matrix solution and generalization

If you prefer algebra to graph theory, you can use the transition matrix. For 1-cup moves

this is a 4 � 4 matrix A whose rows and columns correspond to the four states in Figure

24.4, with entry aij being 1 if there is a move from state j to state i , and 0 if not. Clearly

A D

2

664

0 1 0 0

1 0 1 0

0 1 0 1

0 0 1 0

3

775

:

The transition matrix for 2-cup moves is just the square of A, and

A2 D

2

664

1 0 1 0

0 2 0 1

1 0 2 0

0 1 0 1

3

775

:

The entry 2 shows not just that there is a connection, but that there are two ways to realise

it, but this is information we don’t really need. So let’s introduce a new symbol, 1, subject

to the rules 1 C 1 D 1; 1 � 1 D 1. Then we can use 1 to represent the existence of

a connection, and 0 when there isn’t one. Now

A2 D

2

664

1 0 1 0

0 1 0 11 0 1 0

0 1 0 1

3

775

:

Sequences of 2-cup moves now correspond to powers of A2. But you can use the properties

of 1 to show that .A2/2 D A2, so such sequences can’t achieve anything new. Since the

start and end states are not connected, the problem is impossible.

A few years ago Man-Keung Siu, a mathematician at Hong Kong University, wondered

what happens if you allow more than three cups to start with, and each move inverts a

fixed number of cups. In particular, when can you start with all cups up and end with all

cups down, and what is the shortest sequence of moves that achieves this? Together, we

discovered that this problem has a surprisingly complicated answer. If there are n cups,

and any m of them can be inverted at each move, then this minimal number is shown in

Table 24.1. Here dxe is the ceiling function, the smallest integer � x.

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24. Cups and Downs 185

Table 24.1.

n even, m even n odd, m odd n odd, m even n even, m odd

2m � n dn=me 2d.n�m/=2me C 1 no solution 2dn=2me2m > n 1 if m D n; 3 if m < n 1 if m D n; 3 if m < n no solution 2dn=2.n�m/e

The proof uses the transition matrix and is found in [3]. Exploring some of the simpler

cases is highly recommended: you will quickly appreciate that the general pattern is not at

all obvious.

Bibliography

[1] Erik D. Demaine. Puzzles and tricks from Martin Gardner inspire math and science,

American Scientist 98 (2010) 452–456; available at www.americanscientist.org/

issues/pub/2010/6/recreational-computing.

[2] M. Gardner. Penny puzzles, in Mathematical Carnival, Knopf, New York, 1975.

[3] I. Stewart and M.-K. Siu, How to invert n cups m at a time? Math. Today (Southend-on-Sea) 46

(2010) 34–38.

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186 Part V. Further Puzzles and Games

Solutions to the Gaussian integer KenKen puzzles (d)–(f)

on page 177

(d)1 C i 1 2 1 � i

2 1 � i 1 C i 1

1 � i 1 C i 1 2

1 2 1 � i 1 C i

(e)1 C i 2 1 � i 1

1 1 C i 2 1 � i

2 1 � i 1 1 C i

1 � i 1 1 C i 2

(f)2 1 � 2i 1 C 2i 1 1 � i 1 C i

1 C 2i 1 � i 1 1 C i 2 1 � 2i

1 � 2i 1 1 C i 1 � i 1 C 2i 2

1 � i 1 C 2i 1 � 2i 2 1 C i 1

1 1 C i 2 1 C 2i 1 � 2i 1 � i

1 C i 2 1 � i 1 � 2i 1 1 C 2i

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2530 Years of Bulgarian Solitaire

Brian Hopkins

“Oh, you’re a mathematician! Let me show you something interesting.”

But I’m trying to work on my talk, I thought. As the train sped along, the man sitting

across from me looked eager. OK, let’s get this over with.

“Here are fifteen playing cards. Arrange them into piles; as many piles as you like, each

with as many cards as you like.”

I made five piles with heights 3, 1, 4, 1, 6. The � reference went unnoticed.

“Now take one card from each pile to make a new pile.”

The operation left me with piles of 3 � 1 D 2 cards, 4 � 1 D 3 cards, 6 � 1 D 5 cards,

two empty piles from 1 � 1 D 0, and a new pile of 5 cards. I realized that the order of the

piles does not matter, so for consistency I put them in non-increasing order, 5; 5; 3; 2.

“Now do it again and again. I know what will happen!” He looked away.

Did he already think through the iterations? How long will this go? I was curious now.

Here is the sequence of pile sizes:

.6; 4; 3; 1; 1/ ! .5; 5; 3; 2/ ! .4; 4; 4; 2; 1/ ! .5; 3; 3; 3; 1/

! .5; 4; 2; 2; 2/ ! .5; 4; 3; 1; 1; 1/ ! .6; 4; 3; 2/

Oh no, that’s almost where I started. When will this end? But then, suddenly, it did end:

.6; 4; 3; 2/ ! .5; 4; 3; 2; 1/ ! .5; 4; 3; 2; 1/ again.

“Hmm,” I said.

“You ended with one pile of 5 cards, one of 4, one of 3, one of 2, and one of 1, didn’t

you?” He looked at the cards. “Yes! That’s always what happens. Try again!”

I started with a single pile of 15 cards. It took more moves, but I did indeed end up with

the 5; 4; 3; 2; 1 pattern. Then I started with three piles of 5 cards. It took even more moves,

but ended at the same fixed point.

Three examples is not a proof, but the claim was now reasonable. Why would it al-

ways go there? How long could it take to reach this fixed point? What happens with other

numbers of cards?

“This is interesting,” I admitted.

Reprinted from The College Mathematics Journal, Vol. 43, No. 2 (Mar. 2012), pp. 135–140.

187

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188 Part V. Further Puzzles and Games

This puzzle was popularized by Martin Gardner in 1983 [9] with the unusual name

Bulgarian solitaire. In this article, we examine the earlier history of the puzzle including

its name, summarize subsequent research, and consider a new two-player variation.

Before Gardner

Around 1980, Konstantin Oskolkov of the Steklov Mathematical Institute in Moscow trav-

eled by train to give a talk in Leningrad (now Saint Petersburg). A man on the train told

him of the problem, although the details of the dialog above are fictional. Oskolkov shared

this with his colleagues at the institute; reportedly when one number theorist heard about it

“his face took a Satanic expression, he ran to his office, closed the door and did not come

out until he solved the problem.”

The puzzle reached Victor Gutenmacher, editor of the journal Kvant. The problem ap-

peared there in 1980, in the context of stacks of books. Andrei Toom, a research scientist

at Moscow State University, published a solution in 1981 [19]. The material was also in-

cluded in a 1981 book on mathematics olympiads [20] whose authors include Gutenmacher

and Toom.

Later in 1980, Anatolii Alexeevich Karatsuba traveled from Steklov to the Institute of

Mathematics of the Bulgarian Academy of Sciences in Sofia. After a lecture on approxi-

mation theory, he shared Oskolkov’s story and the puzzle. Again, it captured the attention

of many mathematicians, and Milko Petkov published it in the “competition problems”

section of a high school mathematics journal in 1980, in the context of heaps of balls. No

student solved the puzzle, so the solution of Petkov’s institute colleague Borislav Bojanov

was published in 1981 [4].

Gert Almkvist of Lund University was visiting Sofia while Karatsuba was there. When

Almkvist returned to Sweden, he shared the puzzle with colleagues, including Henrik

Eriksson of the KTH Royal Institute of Technology, who published a solution also in 1981

[7] with the name “Bulgarisk patiens,” in the context of piles of cards.

Jørgen Brandt, finishing his masters at Aarhus University, somehow learned of the

puzzle as well (without the name). “The problem appeared so pure, that I did not think

much about where it came from,” he explained recently. His thorough solution, in the

setting of integer partitions, was submitted to Proceedings of the American Mathematical

Society in 1981 and published in 1982 [5].

Meanwhile, Eriksson traveled from Stockholm to California, where he called the puzzle

Bulgarian solitaire. He recently explained, “The silly name is my invention, silly because

it is neither Bulgarian nor a solitaire.” Donald Knuth started a fall 1982 programming

and problem-solving seminar with Bulgarian solitaire [12]. Ron Graham passed it on to

Gardner, who used Bulgarian solitaire as the culminating example of “tasks you cannot

help finishing no matter how hard you try to block finishing them” [9]. The silly name

has stuck.

Gardner and beyond

A partition of n is a collection of positive integers � D .�1; : : : ; �t/ whose sum is n; as

the order does not matter, we index the parts in non-increasing order, i.e., �1 � �2 �

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25. 30 Years of Bulgarian Solitaire 189

� � � � �t . Bulgarian solitaire can be considered as an operation on partitions, B.�/ D.t; �1 � 1; : : : ; �t � 1/ where there may be zeros to remove and the parts may need to be

reordered. This can also be visualized through a graphic representation of partitions. The

Ferrers diagram of a partition has a column of �1 dots followed by a column of �2 dots, etc.

Figure 25.1 shows the Bulgarian solitaire operation on .6; 4; 3; 1; 1/, choosing to remove

the bottom dot of each column, that is, the bottom row.

�

� � �

� � � �

� � � ! � � �

� � � � � � �

� � � � � � � � �

6 4 3 1 1 5 5 3 2

Figure 25.1. Example of the Bulgarian solitaire operation.

The number of partitions of n grows very quickly as n increases, but it is helpful to

look at the effect of B on all partitions of n for small n. Figure 25.2 shows all partitions of

6 under the operation.

2211 - 411 - 33 -

16

23

-

-

6

313

-

-

51

6

42

�

-

214

321�

Figure 25.2. Bulgarian solitaire on all partitions of 6; exponents denote repetition.

Like the 15 card puzzle of the introduction, Bulgarian solitaire on partitions of 6 leads

to a fixed point, the partition �3 D .3; 2; 1/. Both 6 and 15 are triangular numbers, i.e., can

be written as Tk D 1 C � � �C k D k.k C 1/=2. The effect of Bulgarian solitaire can be very

different on other values of n.

Exercise. Work out Bulgarian solitaire on the 22 partitions of 8. (For the solution, see [13,

Figure 4].)

The three 1981 solutions in European journals [4, 7, 19] use very similar ideas to show

that, for n D Tk , there is indeed just the single fixed point �k D .k; k � 1; : : : ; 2; 1/. Each

author argues that a ball / book / card cannot move to a longer diagonal in the configuration,

and will eventually move into the shortest diagonal possible, filling any gaps and leaving a

triangular shape. Brandt [5] and Toom [19] showed that for other values of n, the “almost

triangular” partitions form cycles. For example, the partitions .4; 2; 2/ and .3; 3; 1; 1/ form

a 2-cycle in one connected component of the partitions of 8 from the exercise; see Figure

25.3. From this characterization, Brandt used Polya enumeration theory to determine the

number of connected components. For instance, the partitions are in a single component

exactly when n is within one of a triangular number. See also [11], which revisits and

generalizes Toom’s solution, and [1], which elaborates on Brandt’s work.

The partitions of Figure 25.3 illustrate another relationship. Reading the row lengths of

.4; 2; 2/ rather than the column heights gives .3; 3; 1; 1/. Considered another way, reflect-

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190 Part V. Further Puzzles and Games

�

� � �

� � � ! � �

� � � � � � �

4 2 2 3 3 1 1

Figure 25.3. A two-cycle among the partitions of 8.

ing the Ferrers diagram for .3; 3; 1; 1/ across the diagonal line y D x gives .4; 2; 2/. These

partitions are said to be conjugates.

Gardner’s article [9] focuses on the fixed point result for triangular numbers, with

a discussion of using 45 cards, and the analog of Figure 25.2 for partitions of 10. He

mentions two other conjectures, which were both proven within a few years. First, for

n D Tk , data suggest that the greatest possible distance to the fixed point is linear in

n, specifically k.k � 1/. Igusa [15] shows that the partition k D .k � 1; k � 1; k � 2;

k � 3; : : : ; 3; 2; 1; 1/ is at distance k.k � 1/ from �k and that this distance is maximal; see

also [3] and [8]. Second, look carefully at the sequence of partitions from

3 D .2; 2; 1; 1/ to �3 D .3; 2; 1/ in Figure 25.2: until the fixed point, the sequence con-

sists of nested conjugate pairs, shown in Figure 25.4. Bentz [3] proves that this occurs

along the “main trunk” of the diagram for all n D Tk , which distinguishes k among the

partitions at maximal length from the fixed point (for k � 4, there are several partitions at

that distance). See also [2].

2211 - 411 - 33 - 23 - 313 - 42 - 321�

Figure 25.4. Conjugate pairs on the “main trunk” of Figure 25.2.

Gardner also mentioned “Eden” partitions that have no predecessor under the opera-

tion, such as 2 D .2; 2; 1; 1/, .2; 1; 1; 1; 1/, and .1; 1; 1; 1; 1; 1/ in Figure 25.2. Recently,

Hopkins and Sellers [14] determined a formula for the number of these partitions for gen-

eral n that resembles a recurrence result of Euler.

There remain many open questions about Bulgarian solitaire, several of which are sur-

veyed in Hopkins and Jones [13]. Griggs and Ho [10] conjectured maximal lengths to a

cycle partition for n ¤ Tk . How many partitions are at given length from a cycle partition

or fixed point? What is the relation between B and conjugation outside the “main trunk”?

When there are multiple components, is there an easy way to tell if two partitions are in

the same component? What can be said about component sizes? There are many questions

about asymptotic and random behavior; some are addressed by Popov [18].

Bulgarian solitaire continues to arise in the teaching literature, such as Nicholson [16]

and Doree [6]. The operation has been rediscovered at least once, see [17]. Several variants

have been considered in the literature, but there is not space here to discuss them.

A new game

Figure 25.1 shows the effect of changing the bottom row of the Ferrers diagram of

.6; 4; 3; 1; 1/ to a column. Suppose you could change any row to a column. Figure 25.5

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25. 30 Years of Bulgarian Solitaire 191

shows the effect of changing the length 2 row to a column. Under this operation, .6; 4; 3; 1; 1/

could be followed by one of three partitions: .5; 5; 3; 2/ by selecting the bottom row,

.5; 3; 3; 2; 1; 1/ selecting the length 2 or a length 3 row, or .5; 4; 3; 1; 1; 1/ selecting a length

1 row.

�

� �

� � �

� � � ! � � �

� � � � � � �

� � � � � � � � � � �

6 4 3 1 1 5 3 3 2 1 1

Figure 25.5. Changing the length 2 row of .6; 4; 3; 1; 1/ to a column gives .5; 3; 3; 2; 1; 1/.

For a 2-player game, we start with the single-part partition .n/. In turn, each player

chooses a single row to change into a column. The loser is the first player who creates a

partition that has occurred before in play.

The first player can only chose a row of length 1, giving the partition .n � 1; 1/. The

second player then can choose a row of length 1 or 2, making the partition .n � 2; 1; 1/ or

.n � 2; 2/, respectively (for sufficiently large n). All possible moves for partitions of 6 are

shown in Figure 25.6.

16 - 6 - 51

��� ?

411

��� ?

42

??33

?

313

����

?

321�

??23

����

214

- �

2211�B

BBB

BB

BBM

Figure 25.6. All possible moves in the two-player game on partitions of 6. Double arrows indicate

where selecting either of two different row lengths produces the same partition.

Here are some open questions about this game.

� As a finite impartial two-player game of perfect information with no draws, either

the first or second player has a winning strategy. Which player? What is the strategy?

Do the answers depend on n?

� Notice that there is a Hamiltonian cycle in the directed graph of Figure 25.6. That

is, there is a sequence of play where every partition arises before there is repeti-

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192 Part V. Further Puzzles and Games

tion. I conjecture that this occurs for all n. If so, is there a nice way to describe the

corresponding sequence of moves?

Conclusion

A mathematical recreation that reportedly arose from a conversation in a train has gone

on to inspire several lines of research and student activities. This is due largely to Martin

Gardner’s attention to it in his mathematics column from Scientific American. Bulgarian

solitaire is just one example of the impact his work has had on all of us who enjoy mathe-

matics.

Acknowledgment The history of Bulgarian solitaire comes from personal communica-

tion with Borislav Bojanov, Jørgen Brandt, Vesselin Drensky, Victor Gutenmacher, Pencho

Petrushev, Andrei Toom, and especially Henrik Eriksson. Recollections of events thirty

years ago can be imperfect; hopefully the account provided here approaches what occurred.

Aleksandar Nikolov kindly provided translations from Bulgarian. Suggestions from two

anonymous referees improved the article. I would like to thank Eriksson, Suzanne Doree,

and Mizan Khan for their interest and help.

Bibliography

[1] E. Akin and M. Davis, Bulgarian solitaire, Amer. Math. Monthly 92 (1985) 237–250; available

at dx.doi.org/10.2307/2323643.

[2] T. Bending, Bulgarian Solitaire, Eureka 50 (1990) 12–20.

[3] H.-J. Bentz, Proof of the Bulgarian solitaire conjectures, Ars Combin. 23 (1987) 151–170.

[4] B. Bojanov, Problem solution 4, Obuchenieto po matematika (Mathematics Education) 24 (5)

(1981) 59–60.

[5] J. Brandt, Cycles of partitions, Proc. Amer. Math. Soc. 85 (1982) 483–486; available at

dx.doi.org/10.1090/S0002-9939-1982-0656129-5.

[6] S. Doree, Bulgarian Solitaire, Resources for Teaching Discrete Mathematics, B. Hopkins, ed.,

Mathematical Association of America, Washington DC, 2009, 83–92.

[7] H. Eriksson, Bulgarisk patiens, Elementa 64 (4) (1981) 186–188.

[8] G. Etienne, Tableaux de Young et solitaire bulgare. J. Combin. Theory Ser. A 58 (1991) 181–

197; available at dx.doi.org/10.1016/0097-3165(91)90059-P.

[9] M. Gardner, Mathematical Games: Tasks you cannot help finishing no matter how hard

you try to block finishing them, Scientific American 249 (1983) 12–21; available at

dx.doi.org/10.1038/scientificamerican0883-12. Also available as Bulgarian

solitaire & other seemingly endless tasks, The Last Recreations, Springer-Verlag NY, 2007, 27–

43.

[10] J. Griggs and C.-C. Ho, The cycling of partitions and compositions under repeated shifts, Adv. in

Appl. Math. 21 (1998) 205–227; available at dx.doi.org/10.1006/aama.1998.0597.

[11] T. Hart, G. Khan, and M. Khan, Revisiting Toom’s proof of Bulgarian solitaire, Ann. Sci. Math.

Qu., to appear.

[12] J. Hobby and D. Knuth, A Programming and Problem-Solving Seminar, Stanford University

Department of Computer Science report, 1983.

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25. 30 Years of Bulgarian Solitaire 193

[13] B. Hopkins and M. Jones, Shift-Induced Dynamical Systems on Partitions and Compositions,

Electr. J. Comb. 13 (2006) R80 19 pp.

[14] B. Hopkins and J. Sellers, Exact Enumeration of Garden of Eden Partitions, Integers 7 (2)

(2007) A19.

[15] K. Igusa, Solution of the Bulgarian solitaire conjecture, Math. Mag. 58 (1985) 259–271.

[16] A. Nicholson, Bulgarian Solitaire, The Mathematics Teacher 83 (1996) 84–86.

[17] H. D. Phan and E. Thierry, Dynamics of the Picking transformation on integer partitions,

Discrete Math. Theor. Comput. Sci. AB(DMCS) (2003) 43–56.

[18] S. Popov, Random Bulgarian Solitaire, Random Struct. Alg. 27 (2005) 310–330; available at

dx.doi.org/10.1002/rsa.20076.

[19] A. Toom, Problem solution M655, Kvant (Quantum) 1981 (7) 28–30.

[20] N. Vasilyev, V. Gutenmacher, J. Rabbot, and A. Toom, Zaochnye matematicheskie olimpiady

(Mathematical olympiads by correspondence), Nauka, Moscow, 1981.

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26Congo Bongo

Hsin-Po Wang

An expedition into Congo uncovered a treasure chest in the shape of a regular octagon. At

each corner was a bongo drum. A scroll attached to the chest, written in French, explained

that there was a genie inside each bongo drum. A genie is either standing upright or doing

a handstand. One may strike a number of bongo drums at the same time. When a bongo

drum is struck, the genie inside will change its posture from right side up to upside down,

or vice versa. The treasure chest will open if and only if all genies are right side up, or

all are upside down. However, each time some bongo drums are hit the treasure chest will

spin rapidly on its vertical axis. As the bongo drums are all identical in appearance, after

the rotation it is impossible to tell which of them had just been hit.

Unfortunately, the scroll did not record the exact procedure by which the treasure chest

might be opened. However, it mentioned that such a procedure had been documented.

This document was considered so valuable that it was put inside the treasure chest for

safekeeping. It was a lot safer than was originally thought.

The sponsor of the expedition was definitely not pleased with the current state of affairs,

so she hired a team of mathematicians to try to open it. The task was seemingly hopeless,

and many gave up, until Dr. Jacob Ecco arrived with his sidekick, Professor Justin Scarlet.

“My dear professor,” said Dr. Ecco, “we must approach this challenge systematically.

What is the most basic principle in problem solving?”

“Downsizing,” replied the professor, who was well versed with the methods of Dr.

Ecco. “Suppose there is only one bongo drum. Then the treasure chest will open automat-

ically.”

“What if there are two bongo drums?” asked Dr. Ecco. “Well, if the treasure chest is

not already open, hitting either of the bongo drums will do it.”

“Excellent! As usual, my dear friend, you have provided me with the inspiration that

leads to the solution.”

“How?” cried Professor Scarlet. “I have made but the most trivial observation.”

“Nevertheless, a solution to the case with two bongo drums leads to a solution to the

case with four bongo drums, which in turn leads to a solution to the case with eight bongo

drums,” Dr. Ecco proclaimed.

“I can smell that you are going to use induction, but l still do not see the connection

from one case to the next,” replied Professor Scarlet.

Reprinted from Math Horizons, Vol. 18, No. 1 (Sept. 2010), pp. 18–21.

195

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196 Part V. Further Puzzles and Games

“Be patient. Obviously, we will be treating pairs of bongo drums as a single bongo

drum. How can we identify pairs of bongo drums? Adjacency is unsatisfactory because

after the treasure chest spins around, it is impossible to tell whether a bongo drum is to be

paired with its neighbor to the left or its neighbor to the right.”

“The obvious solution is to identify opposite pairs of bongo drums,” pronounced Pro-

fessor Scarlet.

“What insight!” said Dr. Ecco with a slight smile. “This is fine if the genies in each

opposite pair of bongo drums are both right side up or both upside down. But there is no

guarantee that they start off like that,” his companion pointed out.

“True, but there is a way to make them end up that way. Let 0 or 1 indicate whether

a genie is right side up or upside down, and consider a four-sided chest. If we are really

lucky, the initial state may already be (0,0,0,0) or (1,1,1,1).”

“You may be good enough to be that lucky, but that never works with me,” complained

Professor Scarlet.

“I am not counting on luck. I am just singling out the most favorable scenario. We may

regard (0,0,0,0) and (1,1,1,1) as the same state. Call it state 0. It is also referred to as an

absorbing state, in that once we enter it we do not leave (for the problem is then solved).

How many other states are there?” asked Dr. Ecco.

Professor Scarlet paused. “Well, we have three other states, namely, (0,1,1,1), (0,0,1,1),

and (0,0,0,1).”

“You are correct as far as the number of other states is concerned, but wrong about their

composition,” Dr. Ecco said. “By symmetry, we may consider (0,1,1,1) and (0,0,0,1) as the

same state. We call it state 3. Your (0,0,1,1) is state 2, which is not the same as (0,1,0,1).

This I call state 1.”

“Yes!” exclaimed Professor Scarlet excitedly. “In state 1, the genies in each opposite

pair of bongo drums are both right side up or both upside down. In the case with two bongo

drums, we hit one of them. So here we hit an opposite pair of bongo drums, and the treasure

chest will open.”

“We will call hitting an opposite pair of bongo drums operation A, and as you keenly

observed, performing this operation from state 1 will unlock the chest. What happens when

I perform operation A if we are in state 2 or state 3?”

“Let me see,” Professor Scarlet said. “Ah, we will remain in the same state as before.”

“How can I move from state 2 to state 0 or 1?” Dr. Ecco asked.

“Well, we may hit an adjacent pair of bongo drums. I suppose this will be operation B.

If we hit both zeros, or both ones, we will be in state 0 immediately. If we hit one 0 and

one 1, we will be in state 1.”

“Indeed, you can check that state 3 is left unchanged by operation B, just as it was

unchanged by operation A,” Dr. Ecco agreed.

“The rest is easy now. We perform operation C by striking just one bongo drum. This

will change state 3 into state 0, state 1, or state 2,” Professor Scarlet said triumphantly.

“Not so fast,” Dr. Ecco cauntioned. “After operation B, we may be in state 1, and if we

rush into operation C now, state 1 will become state 3, and we will be going in circles.”

“I was hasty,” admitted Professor Scarlet. “We must perform operation A once more

to clear state 1 before taking care of state 3. After operation C, we will clear the lower

states with the sequence ABA again. So the overall procedure for the four-sided chest is

ABACABA.”

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26. Congo Bongo 197

“Excellent. Here, l have drawn a state transition diagram for you,” Dr. Ecco said.

State 0 State 1 State 2 State 3

0

00

0

1

11

1

1

1

1

0

0

0 0

10

0

1

1

1 1

0

0

C

A

A/B

B A

Here is the same drawing, with an emphasis on the structure of the transition opera-

tions:

BB

B

C

C

C AA

A

State 0

State 1

State 2

State 3

“That explains everything clearly!” said the professor. “Now I am confident we are onto

something that will crack the eight-sided treasure chest. Let’s see. We wish to treat each

pair of opposing bongo drums as a single entity. As before, the simplest states are those in

which the genies in each opposite pair of bongo drums are both right side up or both upside

down. Here is my drawing of this part of the state transition diagram. It is really the same

as yours, except that operation A means hitting every other pair of opposing bongo drums;

operation B means hitting any two adjacent pairs of opposing bongo drums; operation C

means hitting any pair of opposing bongo drums. By performing the sequence ABACABA

when starting from any such simple state, the treasure chest will open.”

“Excellent, my dear professor,” exclaimed Dr. Ecco. “These states together form an

expanded absorbing state in the overall diagram below. It is the box marked 4, where the

box marked m contains all states with m matching opposite pairs, for 0 D m D 4.”

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198 Part V. Further Puzzles and Games

0 0

0 01

11

11

1

1

1

00

0

A

BA

A/B

0

00 0

0

000

1 111

11 1

1

C

0

00 0

1

111 0

000

1

1 110

“This diagram looks suspiciously like the ones before, except that states 4 and 0 are no

longer equivalent. Why do you have a state 2 and a state 20? And what are the operations

D, E, F, and G?”

D

E

E

F

F

GG

F

E

G

GD

D

D

F

E 1 or 3

4

0

2

2¢

“The states with 2 matching pairs are classified according to whether those matching

pairs are alternating or adjacent,” Dr. Ecco said. “The former are grouped under state 2

while the latter are grouped under state 20. Operation D hits any four adjacent bongo drums.

Operation E hits any two bongo drums separated by one other drum. Operation F hits any

two adjacent bongo drums. Operation G hits any one bongo drum.”

“Let me see. If we denote the sequence ABACABA by X, then the sequence that will

open the treasure chest is XDXEXDXFXDXEXDXGXDXEXDXFXDXEXDX,” summed

up Professor Scarlet.

“We keep repeating X so that whenever we enter state 4, we will not return to another

state. Whatever the initial state of the treasure chest may be, it will open by the end of this

sequence.”

The sponsor of the expedition was delighted with their analysis. She hired a team of

Congo natives to strike the bongo drums in the prescribed fashion. Lo and behold, the

treasure chest opened right before her eyes. Unable to contain her excitement, she reached

inside, but found only another scroll. It was written in Kituba, but the most prominent line

read something like XDXEXDXFXDXEXDXGXDXEXDXFXDXEXDX.

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26. Congo Bongo 199

Later, back in his New York apartment, Dr. Ecco said to Professor Scarlet, “That was

hilarious, wasn’t it? By the way, have you thought of the obvious question?”

“Yes, l have,” replied the professor. “For what number of bongo drums can such a

treasure chest be opened? From our analysis, the general procedure when the number is

a power of two is clear. We treat each opposing pair as a single entity, thereby reducing

to the preceding case. Then we progressively move all states into the expanded absorbing

state. But what happens if the number of drums is not a power of two?”

“Then unlocking the chest is not always possible,” Dr. Ecco replied. “For in this case

the number of bongo drums has an odd prime factor p. Suppose the pagan god Hemba is

having fun with us. He will choose p evenly spaced bongo drums and make sure that the

genies inside are not all right side up and not all upside down. Now ignore all other bongo

drums except these p. The two types of bongo drums are not equal in number since p is

odd. In order for us to succeed, we must strike precisely the bongo drums of one type. But

Hemba will spin the treasure chest so that the bongo drums we plan to hit include at least

one from the opposite group. This way, he can keep us from opening the treasure chest

forever.”

See problem 249 in the Playground below for more fun with these treasure chests.

Further Reading

This problem was posed in the Senior A-Level paper in the 2009 Fall Round of the Inter-

national Mathematics Tournament of the Towns. It is related to a problem posed in Martin

Gardner’s famous “Mathematical Games” column in Scientific American (February 1979),

which is reproduced in his anthology Fractal Music, Hypercards and More Mathematical

Recreations (W.H. Freeman, 1992). Gardner’s version of the problem is also treated in

three other sources:

Ted Lewis and Steve Willard, The rotating table, Math. Mag., 53 (1980): 174–179.

William Laaser and Lyle Ramshaw, Probing the rotating table, in The Mathematical

Gardner, edited by David Klarner, (Wadsworth, 1981) 288–307.

Albert Stanger, Variations on the rotating table problem, J. Recreat. Math., 19 (1987):

307–308 and 20 (1988): 312–314.

Jacob Ecco and Justin Scarlet are fictitional characters created by Dennis Shasha of the

Courant Institute, New York. He is a leading creator of mathematical puzzles. Shasha has

written a number of puzzle books featuring these clever detectives. A good place to start is

his book The Puzzling Adventures of Doctor Ecco (Dover, 1998).

Playground Problem 249

The end of the article explains why no sequence of drum strikes is guaranteed to open

an n-drum chest unless n is a power of 2. However, for other n it may be possible to

have a pretty good chance of opening the chest during a certain sequence of drum strikes,

assuming that the spinning between strikes is truly random. This leads us to Problem 249,

Congo Bongo not so Wrongo: What is the smallest number of drum strikes needed on a

three-drum treasure chest to give you at least a 90 percent chance of opening it? (As a

separate further challenge, which may require computer assistance, try to answer the same

question for a five-drum chest.)

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200 Part V. Further Puzzles and Games

Problem 249 Solution

The Math Club from the University of Minnesota Duluth submitted a correct solution that

6 drum strikes are required; there were two incorrect submissions. First, the club members

realized that there are only two possible states for the chest: either all three of the genies

have the same orientation (state A, which causes the chest to open) or just two of the

genies have the same orientation (state B). Then, they realized that striking either one or

two drums will cause a chest in state B to move to state A with probability 1/3 and stay in

state B with probability 2/3.

Assuming that the chest is initially in state B (otherwise it would open immediately!),

the probability that the chest opens in k or fewer strikes is

1=3 C .2=3/.1=3/ C .2=3/2.1=3/ C � � � C .2=3/k�1.1=3/ D 1 � .2=3/k

which first becomes greater than 0.9 when k D 6.

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27Sam Loyd’s Courier Problem

with Diophantus, Pythagoras,

and Martin Gardner

Owen O’Shea

In his classic collection Cyclopedia of Puzzles, published in 1914, Sam Loyd has two

versions of the Courier Problem ([2, p. 315]):

For the reason that many communications are being received relating to a very

ancient problem, the authorship of which has been incorrectly accredited to

me, occasion is taken to present the original version which has led to consid-

erable discussion. It has been reproduced, in many forms, generally accompa-

nied by an absurd statement regarding the impossibility of solving it, which

produced letters of inquiry as well as correct answers from some, who, under

the misapprehension of having mastered a hitherto unsolved problem, desire

to have the same published.

It is a simple and pretty problem which yields readily to ordinary methods,

and can be solved by experimental analysis upon the plan generally adopted

by puzzlists. The trouble is that the terms of the problem are seldom given

correctly and are not generally understood, for which reason,. . . , we will first

look at the ancient version which appears in the oldest mathematical works:

A courier starting from the rear of a moving army, fifty miles long, dashes

forward and delivers a dispatch to the front and returns to his position in the

rear, during the exact time it required the entire army to advance just fifty

miles.

How far did the courier have to travel in delivering the dispatch, and returning

to his previous position in the rear of the army?

A better puzzle is created by the following extension of the theme given as

problem No. 2:

If a square army, fifty miles long by fifty miles wide, advances fifty miles

while a courier makes the complete circuit of the army and returns to the

starting point in the rear, how far does the courier have to travel?

Reprinted from The College Mathematics Journal, Vol. 39, No. 5 (Nov. 2008), pp. 387–391.

201

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202 Part V. Further Puzzles and Games

The problem appeared in print long before Loyd published it. The first known ap-

pearance was in A Companion to the Gentleman’s Diary of 1798, and a year later it was

published in The Gentleman’s Mathematics Companion. It also appeared much later in

William R. Ransom’s One Hundred Mathematical Curiosities in 1955 [3]. Loyd describes

the puzzle as an ancient problem and gives the answers to both parts. However, he does not

indicate how to tackle the problems.

In his book More Mathematical Puzzles of Sam Loyd [1, p. 103], Martin Gardner dis-

cusses both problems and their solutions. The answer to the first question is that the courier

travels 1 Cp

2 times the length of the army, or about 120.71 miles for a fifty-mile-long

army. Here is Gardner’s solution: For convenience, assume that the length of the army is

1 unit, and also assume that the time it takes the army to march its length is 1. Thus its

speed is also 1. Let x denote the total distance traveled by the courier, so that is also his

speed. On his forward trip, the courier’s speed relative to the army is thus x � 1, and on the

return trip it is x C 1. Since each trip is a distance of 1 relative to the moving army and the

courier’s total journey is completed in unit time, it follows that

1

x � 1C 1

x C 1D 1;

or equivalently, x2 � 2x � 1 D 0. Consequently, x D 1 Cp

2. This means that in Loyd’s

problem the courier travels 50�

1 Cp

2�

miles, or, as asserted earlier, about 120.71 miles.

Gardner’s solution to the second question is similar. As before, let the length of the

army and the time it takes to travel its length be 1 unit, so its speed is also 1. Additionally,

we let x be the total distance traveled by the courier and also his speed. On the forward

trip, the courier’s speed relative to the army is x � 1, on the return trip, x C 1, and on each

of the diagonal portions,p

x2 � 1. Since each trip has length 1 relative to the army and the

total journey is completed in unit time, it follows that

1

x � 1C 1

x C 1C 2p

x2 � 1D 1:

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27. Sam Loyd’s Courier Problem with Diophantus, Pythagoras, and Martin Gardner 203

This time a fourth-degree equation results: x4 �4x3 �2x2C4x C5 D 0. The only solution

that fits the conditions of the problem is approximately 4.18. Thus, in traversing the entire

army, the courier now travels approximately 4.18 times as far as the army, or about 209

miles.

In the spirit of classroom exercises, it would be nice to have versions of the courier

problems with integer solutions; that is, Diophantine problems. In exploring this possi-

bility, we discovered that certain Pythagorean triples, such as .3; 4; 5/, .5; 12; 13/, and

.7; 24; 25/, that is, those in which the two largest numbers differ by 1, can be used to

generate such problems.

Army in single file

Assume that an army 60 miles long is marching at a constant rate. A courier, also going

at a constant rate, rides from the rear of the army up to the front, delivers a message, and

returns, arriving at the rear just as the army has gone 45 miles. How far did the courier go?

Following Gardner, we again let the speed of the army (but not its length) be 1, and we

let x be the speed of the courier. As before, the speed of the courier relative to the army

is x � 1 on the outward journey and x C 1 on the return journey, so the total time that

he travels is 60x�1

C 60xC1

. This must equal 45, the time it takes the army to go the given

distance. Gardner’s equation is therefore

60

x � 1C 60

x C 1D 45:

This reduces to 3x2 � 8x � 3 D 0, for which the only realistic solution is x D 3. Thus, the

courier travels three times as fast as the army, and therefore rides 135 miles. We thus have

a Diophantine version of the courier problem.

Consider now a general single-file problem, in which the army is ` miles long and

travels d miles during the time that the courier makes the trip. Gardner’s equation becomes

`

x � 1C `

x C 1D d:

Setting ˛ D `d

, we find that this reduces to

x2 � 2˛x � 1 D 0;

which has x D ˛ Cp

˛2 C 1 as its only positive solution.

This is where Pythagorean triples come to the fore. If .a; b; b C 1/ is such a triple and

˛ D ba

, then x D a is a solution to Gardner’s equation. Our example used the triple .3; 4; 5/

with distances multiplied by 15. Since in these triples b D 12.a2 � 1/ and c D 1

2.a2 C 1/,

we can pose a single-file courier problem with an integer solution for any odd positive

integer a and any constant k:

Problem 1 (single file army) An army 12.a2 � 1/k miles long travels a distance of ak

miles during the time that the courier makes his journey. How far does the courier travel?

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204 Part V. Further Puzzles and Games

Army in square formation

We can do something similar for the second of Loyd’s problems. Here is an example (also

based on the .3; 4; 5/ triple):

An army in the formation of a square 36 miles on a side is advancing at a constant

rate. A courier, also moving at a steady rate, travels all the way around the army just as it

advances 256 miles. How far does the courier go?

Again we take the speed of the army to be 1 and that of the courier x, so Gardner’s

equation becomes36

x � 1C 36

x C 1C 2 � 36p

x2 � 1D 256:

In polynomial form, this is

81x4 � 576x3 � 162x3 C 576x C 1105 D 0;

but all we care about is that x D 54

is a root (this is easy to check in the original equation,

not so easy to discover).

The same Pythagorean triples .a; b; c/ as before (with a an odd integer, b D 12.a2 � 1/,

and c D 12.a2 C 1/) all generate square-army courier problems with Diophantine solu-

tions:

Problem 2 (square army) An army b2 miles on a side advances 2a .b C c/ miles during

the time that a courier completes his circuit. How far does the courier go?

Gardner’s equation for this army can be written as

2b2

�x

x2 � 1C 1p

x2 � 1

�

D 2a .b C c/ ;

which has x D ca

as a solution. That is, the courier travels ca

times as fast as the army and

hence goes 2c.b C c/ miles.

Note that the roles of a and b can be interchanged here; that is, the army could be b2

on a side and travel 2b.a C c/ miles while the courier completes his circuit.

Army in rectangular formation

Of course, armies do not always march in either single-file or square formation; sometimes

they are proper rectangles. Suppose that an army is ` miles long and w miles wide and that

it goes a distance d in the time that the courier transverses it, going at a speed x times that

of the army. Gardner’s equation for this situation is

`

x � 1C `

x C 1C 2wp

x2 � 1D d:

For example, assume that the army is 40 miles long and 30 miles wide, and that it

travels 120 miles while the courier completes his circuit. Then the courier rides 5=3 as fast

as the army travels, and hence goes 200 miles.

Thus, Pythagorean triples of the given type can be used in this equation to generate Dio-

phantine solutions for a rectangular army in two ways, with .a; b; c/ again a Pythagorean

triple with c � b D 1.

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27. Sam Loyd’s Courier Problem with Diophantus, Pythagoras, and Martin Gardner 205

Problem 3 (rectangular army) An army’s length is a multiple of 12b2 and its width a

multiple of 12b. It advances 2a .b C c/ miles during the time that courier completes his

circuit. How far does the courier travel?

Acknowledgment The author thanks the referees and the editors for helpful suggestions.

Bibliography

[1] Martin Gardner, More Mathematical Puzzles of Sam Loyd, Dover, 2007.

[2] Sam Loyd, Cyclopedia of Puzzles, Lamb Publishing, 1914. www.mathpuzzle.com/loyd/

[3] William R. Ransom, One Hundred Mathematical Curiosities, J. Weston Walch, 1955.

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28Retrolife and The Pawns Neighbors

Yossi Elran

Martin Gardner created no less than a revolution in the popularization of mathematics

through his many books, and, in particular, his Scientific American columns. Some of his

most important columns, which spurred not only popular interest but also a wealth of in-

novative research and even practical applications, were his columns describing John Con-

way’s “Game of Life” [5, 6]. Gardner himself said [7],

Probably my most famous column was the one in which I introduced Con-

way’s game of Life. Conway had no idea, when he showed it to me, that it

was going to take off the way it did. He came out on a visit, and he asked

me if I had a Go board. I did have one, and we played Life on the Go board.

He had about 50 other things to talk about besides that. I thought that Life

was wonderful—a fascinating computer game. When I did the first column on

Life, it really took off. There was even an article in Time magazine about it.

“It really took off” is an understatement.

The Game of Life

The “Game of Life” was invented in 1970 by British mathematician John Conway. It is

best described as the archetype of cellular automata. The player (there is only one) places

an initial distribution of checkers on an infinite checkerboard; one checker per square. The

squares are called cells and the initial distribution is called a population. This terminology

reflects one of the original objects of the game—to simulate biological evolution. Cells are

live if they contain a checker and dead if empty. The population evolves in time in discrete

time steps known as generations, according to a simple set of rules, applied simultaneously.

The rules determine the fate of each cell according to the number of live cells in its eight

neighboring cells. If a cell is empty and exactly three of its eight neighboring cells are alive,

the cell itself will be born—a checker is placed there. If a cell is alive and two or three of its

eight neighboring cells are also alive, nothing changes—the cell survives. A live cell dies

(i.e., the checker is removed from the board) in two situations. If a cell is alive and four or

more of its eight neighboring cells are also alive, the cell dies (from overcrowding). A live

cell also dies from isolation, if it has less than two live neighbors. Figure 28.1 shows three

consecutive generations of Life.

Reprinted from The College Mathematics Journal, Vol. 43, No. 2 (Mar. 2012), pp. 147–151.

207

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208 Part V. Further Puzzles and Games

Figure 28.1. Three generations of the Game of Life.

Although these rules are simple and straightforward, iterating Life for many genera-

tions can result in a surprising variety of intriguing outcomes. A population can die out.

Another possibility is that the population evolves into a stable, constant form which doesn’t

change anymore. A 2-by-2 block of live cells is an example of such a form. Populations

may also cycle, regenerating themselves after a certain time. Such populations come in

forms known as oscillators and gliders. Three isolated live cells in a row are an example

of a simple oscillator.

Populations can also grow indefinitely. In 1970, a team from the Artificial Intelligence

Project at M.I.T. won a 50 dollar prize from Conway for proving that an initial finite popu-

lation can grow without limit. Such populations are generated by shapes variously known

as glider guns, agars and puffer-trains. Martin Gardner wrote: “You will find the popula-

tion constantly undergoing unusual, sometimes beautiful and always unexpected change.”

Stephen Silver’s lexicon [10] lists over 700 terms applied to various forms of Life.

Much work on Life was done as a result of Martin Gardner’s original publication in

1970, in particular by Bill Gosper from the original M.I.T. group, Robert Wainwright

(who managed a newsletter called “Lifeline”) and other “Life-enthusiasts” (as Gardner

calls them). The three articles that Martin Gardner wrote in Scientific American and there-

after reprinted in Wheels, Life and Other Mathematical Amusements [6] have spurred so

much research that it is impossible to summarize it all without doing injustice to some peo-

ple. Life is considered the prototypical computer simulation. Among other things, it is a

universal Turing machine [3]. Paul Rendell proved the universality of Life by showing a di-

rect simulation of counter machines [9], and Dean Hickerson created an initial population

that, when evolved, generates the Prime Numbers [8].

Variations on Life [2] have also been developed. Some interesting games, where only

the rules are changed, are Highlife, which has the same rules as Life and in addition six

live neighbors also cause a birth, and Seeds where a birth is caused when a cell has exactly

2 live neighbors and in all other cases the cell dies. The two-dimensional Moore neigh-

borhood grid can also be changed, resulting in games played on triangular and hexagonal

boards. Andrew Adamatzky recently commemorated Life’s 40th birthday by publishing a

unique collection of works that provide a comprehensive and up-to-date study of Life, with

contributions from many renowned mathematicians and computer scientists [1].

Retrolife

An important aspect of Life is that applying the rules to any given generation results in

a unique outcome. The reverse, however, is not true. In fact, there are an infinite number

of possible population distributions (hereafter referred to as states) in the preceding gen-

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28. Retrolife and The Pawns Neighbors 209

eration. Finding a particular state from the variety of possible states is the essence of the

Retrolife puzzles [4].

In its original form, the task in Retrolife is to find predecessors of a given initial state

that comply with some extra constraints. Many constraints can be chosen, each generating

a slightly different puzzle. One might, for instance, demand that all the cells in the initial

state survive, that is, cells live in the puzzle (the initial state in Life) are also live in the

solution (the previous generation). Another constraint is to demand the opposite, that all

the live cells in the initial state are born. Figure 28.2 shows three Retrolife puzzles and

their solution using this last constraint.

Figure 28.2. Three Retrolife puzzles. Top row: puzzles. Bottom row: possible solutions.

To explain the rules of Retrolife, one first needs to explain Life and then suggest the

reverse problem. Needless to say, this is rather cumbersome. Instead, we therefore suggest

the following rephrasing of Retrolife.

The Pawns Neighbors

A player is given an infinite chessboard, any number of pawns, and some tokens. Each

square on the board has eight neighboring squares (in the directions: north, south, west, east

and on the diagonals). The pawns are placed in some pattern on the board. The challenge:

surround each pawn (and no other square on the board) with exactly three tokens so that

no token itself has either two or three tokens as neighbors. A token may have less than

two tokens as neighbors, or more than three. Note also that no empty square in the board

is allowed to be surrounded with three tokens; however it may be surrounded by more or

less. Figure 28.3 shows one setup and a possible solution.

The Pawns Neighbors is really Retrolife in disguise, however it can be described in a

straightforward manner and solved without any prior knowledge of Life or Retrolife.

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210 Part V. Further Puzzles and Games

Figure 28.3. A Pawns Neighbors puzzle and a solution.

Variations on The Pawns Neighbors

The Pawns Neighbors is isomorphic to Retrolife with the constraint that all the cells in the

initial state are born. We can change the rules to create an isomorphic Retrolife problem

with the constraint that all the cells in the initial state survive. In this case, we have to

surround each pawn with tokens so that the pawns have either two or three neighbors,

either pawns or tokens, but no token has two or three neighbors (again, neighbors can be

either pawns or tokens). No empty square on the board is allowed to have three neighbors

(pawns or tokens). Many other variations are possible.

The computational complexity of Pawn Neighbor games is an open question. We leave

the reader the following Pawns Neighbors problem to solve. Surround each pawn (and

no other square on the board) with exactly three tokens. No token can have either 2 or 3

neighbors. The solution is on p. 211.

Figure 28.4. A Pawns Neighbors puzzle.

Martin Gardner—in memoriam

Martin Gardner was particularly fond of Life and related problems. Shortly before he died,

he wrote to me enthusiastically about Retrolife and offered to add a piece about them in

one of his upcoming books. It’s a pity I can’t share with him this article. I’m sure he would

have enjoyed it. I miss Martin a lot, and for all the many books he wrote, I still yearn for

more.

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28. Retrolife and The Pawns Neighbors 211

Bibliography

[1] A. Adamatzky (ed.), Game of Life Cellular Automata, Springer-Verlag, London, 2002.

[2] B. Bays, The Game of Life in Non-square Environments, in Game of Life Cellular Automata,

A. Adamatzky (ed.), Springer-Verlag, London, 2002, 319–330.

[3] E. R. Berlekamp, J. H. Conway and R. K. Guy III, Winning Ways (for your Mathematical

Plays), Volume 2, A. K. Peters, Wellesley, MA, 2003.

[4] Y. Elran, Retrolife, in Homage to a Pied Puzzler, E. Demaine, M. Demaine and T. Rodgers

(ed.), A. K. Peters, Wellesley, MA, 2008, 129–136.

[5] M. Gardner, Scientific American 223 (1970) 120–123.

[6] ——, Wheels, Life and Other Mathematical Amusements, W. H. Freeman, New York, 1983.

[7] ——, “Mathematical Games” and Beyond: Part II of an Interview of Martin Gardner, College

Math J. 36 (2005) 301–314.

[8] C. Hickerson, Dean Hickerson’s Game of Life Page; available at radicaleye.com/DRH/

life.html

[9] P. Rendell, Turing Universality of the Game of Life, in Collision Based Computing, A.

Adamatzky (ed.), Springer-Verlag, London, 2002, 513–540.

[10] S. Silver, The Life Lexicon; available at www.argentum.freeserve.co.uk/lex.htm

Solution to the problem on p. 210.

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29RATWYT

Aviezri S. Fraenkel

In memory of Martin Gardner, who was and remains enchantingly influential and inspiring.

WYTHOFF

In 1907, the Dutch mathematician, Willem Abraham Wythoff [13] invented this game, later

vividly explained by Martin Gardner in [7].

WYTHOFF is played on a pair of nonnegative integers, .M; N /. A move consists of

either (i) subtracting any positive integer from precisely one of M or N such that the result

remains nonnegative, or (ii) subtracting the same positive integer from both M and N such

that the results remain nonnegative. The first player unable to move loses.

Given the position .3; 3/, say, the next player wins in a single move: .3; 3/ ! .0; 0/.

The position .3; 3/ is called an N -position, because the Next player wins. If M D N D 0,

the next player loses, and the previous player, the one who moved to .0; 0/, wins. Thus

.0; 0/ is a P -position, because the Previous player wins.

If M > 0, it is easy to see that .0; M/ and .M; M/ are N -positions, since the next

player can win in one move. On the other hand, .1; 2/ is a P -position because all its fol-

lowers—positions reached in one move—are N -positions. The first few P -positions are

listed in Table 29.1. Note that every N -position has at least one P -follower, but all fol-

lowers of a P -position are N -positions. From an N -position, in order to win, a player

must move to a P -position. Further, the P - and N -positions partition the set of all game

positions: every game position is either a P -position or an N -position but never both.

Table 29.1. The first few P -positions .An; Bn/ for Wythoff’s game.

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

An 0 1 3 4 6 8 9 11 12 14 16 17 19 21 22 24 25 27 29 30 32 33 35 37 38 40 42 43

Bn 0 2 5 7 10 13 15 18 20 23 26 28 31 34 36 39 41 44 47 49 52 54 57 60 62 65 68 70

The sequences, An and Bn in Table 29.1, each strictly increasing, have remarkable

properties. Note that Bn D An C n for all n. But how is An computed? A study of the

table reveals that for n � 1, An is the smallest positive integer not yet appearing in the

sequences. Thus, the next entries in Table 29.1 are A28 D 45, B28 D 73. It follows that the

Reprinted from The College Mathematics Journal, Vol. 43, No. 2 (Mar. 2012), pp. 160–164.

213

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214 Part V. Further Puzzles and Games

sequences are complementary: every positive integer appears precisely once in these two,

recursively defined sequences.

Can a take-away game, such as WYTHOFF, be played on the rational numbers, rather

than the integers? We present such a game here.

A rational take-away game

Given a rational number p=q in lowest terms, a step is defined by

p

q!

p � q

q;

if p=q � 1, otherwisep

q! p

q � p:

The game RATWYT is played on a pair of reduced rational numbers .p1=q1, p2=q2/.

A move consists of either (i) doing any positive number of steps to precisely one of the

rationals, or (ii) doing the same number of steps to both. The first player unable to play

(because both numerators are 0) loses.

For example, from .3=5; 2=3/, suppose that Alice moves to .3=2; 2=3/. Then Bob can

do three steps to both rationals: .3=2; 2=3/ ! .1=2; 2=1/ ! .1=1; 1=1/ ! .0=1; 0=1/, and

thereby win. Could Alice have made a better initial move? Can Alice win? Suppose she

does two steps to each of the initial rationals: .3=5; 2=3/ ! .3=2; 2=1/ ! .1=2; 1=1/. If

Bob then moves to .1=1; 1=1/, Alice can move to .0=1; 0=1/ D .0; 0/, winning. We leave

it to the reader to verify that in the three remaining possible moves for Bob, Alice can also

win in one move. Thus .3=5; 2=3/ is an N -position in RATWYT.

Is there a nice winning strategy for RATWYT? If so, what is it?

Continuing with our example, we expand 3=5 into a continued fraction:

3=5 D 0 C 1

1C 1

1C 12

;

or 3=5 D Œ0; 1; 1; 2�, to use the short notation for continued fractions. Similarly, 2=3 DŒ0; 1; 2�.

The sum of the partial quotients: 0 C 1 C 1 C 2 D 4 is called the integer induced by

the corresponding rational. We claim that playing RATWYT on .3=5; 2=3/ is equivalent to

playing WYTHOFF on their induced integers, .3; 4/!

Supposing this established, then, because .3; 4/ is not in Table 1, it is not a

P -position, rather an N -position, in WYTHOFF. Hence .3=5; 2=3/ is an N -position in

RATWYT. So Alice can, in fact, win. Let us re-examine her initial move: .3=5; 2=3/ !.3=2; 2=3/. Now 3=2 D Œ1; 2� � 3 (where � denotes the inducing process),

so .3=2; 2=3/ � .3; 3/. Alas, .3; 3/ is another N -position in WYTHOFF, so Alice missed

her opportunity to move to a P -position! Had she made the two-step move .3=5; 2=3/ !.1=2; 1=1/ D .Œ0; 2�; Œ1�/ � .1; 2/, she could have won, since .1; 2/ is a P -position in

WYTHOFF.

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29. RATWYT 215

The CW-tree

To substantiate our claim, that playing RATWYT on .p1=q1; p2=q2/ is equivalent to play-

ing WYTHOFF on their induced integers, we resort to a method of counting the rationals,

described in the short, elegant, and influential paper of Calkin and Wilf [4].

The Calkin Wilf tree, CW-tree for short, is a binary tree whose nodes are all the non-

negative rational numbers without repetition! The root is the fraction 0=1. The rest of the

tree is described inductively by the rule that each vertex i=j has at most two children: a

right child .i C j /=j ; and, if i > 0, a left child i=.i C j /. The first five levels (levels 0 � 4)

are drawn in Figure 29.1. Note that at each step in RATWYT we move towards the root of

the tree by one level, so we are guaranteed to eventually reach 1=1 on level 1, and then the

root on level 0.

0/1

1/1

1/2

1/3

1/4 4/3

3/2

3/5 5/2

2/1

2/3

2/5 5/3

3/1

3/4 4/1

Figure 29.1. The first five levels of the CW-tree of reduced fractions.

Here are some key properties of the CW-tree.

1. The numerator and denominator at each vertex are relatively prime: proved by in-

duction on the level of the tree.

2. The induced integer of every rational on level k of the tree is k. Indeed, an element

r has the right child r C 1. Incrementing by 1 means incrementing the first term

of the continued fraction, hence incrementing the induced integer. The left child is

1=.1 C 1=r/. Since taking the reciprocal of a continued fraction amounts to either

prefixing or removing a leading 0, neither of which changes the induced integer [8],

this also increments the induced integer by one. (See Figure 29.2.)

3. Every rational number p=q in reduced form occurs precisely once in the CW-tree,

namely on level k, where k is the integer induced by p=q. There is a unique path

from p=q to the root whose length is precisely k—the same as for every element

on level k of the tree, in particular of the rational k=1 D k. Hence playing on the

rational p=q is equivalent to playing on its induced integer k, as we set out to show.

More games

In [13], Wythoff also discovered that the sequences An, Bn can be described explicitly:

An D bn�c, Bn D bn�2c, where bxc is the floor function; and � D .1 Cp

5/=2 is the

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216 Part V. Further Puzzles and Games

[0]

[1]

[0,2]

[0,3]

[0,4] [1,3]

[1,2]

[0,1,1,2] [2,2]

[2]

[0,1,2]

[0,2,2] [1,1,2]

[3]

[0,1,3] [4]

Figure 29.2. The first five levels of the CW-tree of continued fractions.

golden ratio. This observation leads to a winning strategy for Wythoff that is polynomial

time in the succinct input size log.xy/ of any given game position .x; y/.

The preceding section explained how playing RATWYT on the rationals is equivalent

to playing on their induced integers. The same holds for any take-away game. Playing on

the integers means restricting travel along the right edge of the CW-tree. Playing on the

integer k is equivalent to playing on any of its 2k�1 rational siblings on level k of the tree

.k � 1/. Here are a few sample take-away games, played on reduced rationals. Keep in

mind that the number of steps performed must always preserve non-negativity, and that the

first player unable to move loses.

Game I. Let t be a fixed positive integer. A position is a pair .p1=q1, p2=q2/ of rationals.

There are two types of moves: (i) do any positive number of steps on precisely one of the

rationals, or (ii) do k > 0 steps on one of the rationals and ` > 0 on the other, such that

jk � `j < t .

This game, if played on the integers induced by p1=q1, p2=q2, is a generalization of

WYTHOFF (the case t D 1 is WYTHOFF). The winning strategies given in [5] apply di-

rectly to every rational number with the same induced integers. P -positions of generalized

WYTHOFF partition the integers, so Game I partitions the rationals. Are there any mean-

ingful partitions of the rationals that transcend tree level?

Game II. Let t be fixed positive integer. A position is a single rational, p=q. A move

consists of performing up to t steps. The P -positions are all the rationals on nonnegative

integer multiples of level .t C 1/.

Game III. A position consists of m rationals .p1=q1; : : : ; pm=qm/. A move consists of

selecting any one of the rationals and performing a positive number of steps. This game

amounts to playing Nim on the integers induced by those rationals.

Though playing on the integer k is equivalent to playing on any of its 2k�1 rational

siblings on level k of the tree .k � 1/, there are games on the rationals with no obvious

integer parallels. For example, one may restrict every move to a single tree direction: either

move up right, or move up left, both for impartial (Wythoff-like) and partizan (chess-like)

games.

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29. RATWYT 217

Our game steps amount to single steps in the “slow Euclidean Algorithm.” The game

EUCLID corresponds to playing along the Euclidean Algorithm at a pace chosen by the

player. There is a large bibliography on EUCLID including [3], [11], [6]. For related games,

see [2] and [10].

Like the CW-tree, the older Stern-Brocot tree [12], [1] also contains all the reduced

rationals. The former has the advantage that the transition from any level to its neighbor

is simple; the latter that it is a search tree. Both trees are intimately related to the famous

Stern diatomic sequence available at http://oeis.org/A002487. A construction of

the rationals using recurrence is given in [9].

Final remark In a different context, an anonymous referee recently asked whether there

are any “bridges” between combinatorial game theory (CGT) and classical game theory. In

the latter, of course, the notion of “rationality” plays a prominent role (in the form of the

rational players). Now we see that rationality also plays a role in CGT.

Bibliography

[1] A. Brocot, Calcul des rouages par approximation, nouvelle methode, Revue Chronometrique 6

(1860) 186–194.

[2] G. Cairns and N. B. Ho, Min, a combinatorial game having a connection with prime

numbers, Integers 10 G03 (2010) 765–770; available at dx.doi.org/10.1515/

INTEG.2010.103.

[3] G. Cairns, N. B. Ho, and T. Lengyel, The Sprague-Grundy function of the real game

Euclid, Discrete Math. 311 (2011) 457–462; available at dx.doi.org/10.1016/

j.disc.2010.12.011.

[4] N. Calkin and H. S. Wilf, Recounting the rationals, Amer. Math. Monthly 107 (2000) 360–363;

available at dx.doi.org/10.2307/2589182.

[5] A. S. Fraenkel, How to beat your Wythoff games’ opponent on three fronts, Amer. Math.

Monthly 89 (1982) 353–361; available at dx.doi.org/10.2307/2321643.

[6] A. S. Fraenkel, Euclid and Wythoff games, Discrete Math. 304 (2005) 65–68; available at

dx.doi.org/10.1016/j.disc.2005.06.022.

[7] M. Gardner, Wythoff’s Nim, Ch. 8, pp. 103–118 in: Penrose Tiles to Trapdoor Ciphers, Revised

Edition, Mathematical Association of America, 1997.

[8] J. Gibbons, D. Lester and R. Bird, Functional pearl: Enumerating the rationals, J.

Funct. Programming 16 (3) (2006) 281–291; available at dx.doi.org/10.1017/

S0956796806005880.

[9] S. P. Glasby, Enumerating the rationals from left to right, Amer. Math. Monthly 118 (2011)

830–835; available at dx.doi.org/10.4169/amer.math.monthly.118.09.830.

[10] S. Hofmann, G. Schuster, and J. Stending, Euclid, Calkin & Wilf—playing with rationals, Elem.

Math V. 63 (2008) 109–117; available at dx.doi.org/10.4171/EM/95.

[11] G. Nivasch, The Sprague-Grundy function of the game Euclid, Discrete Math. 306 (2006)

2798–2800; available at dx.doi.org/10.1016/j.disc.2006.04.020.

[12] M. A. Stern, Uber eine zahlentheoretische Funktion, J. Reine Angew. Math. 55 (1858) 193–220;

available at dx.doi.org/10.1515/ crll.1858.55.193.

[13] W. A. Wythoff, A modification of the game of Nim, Nieuw Arch. Wisk. 7 (1907) 199–202.

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VICards and Probability

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30Modeling Mathematics

with Playing Cards

Martin Gardner

Because playing cards have values 1 through 13 (jacks 11, queens 12, kings 13), come

in two colors, four suits, and have fronts and backs, they provide wonderfully convenient

models for hundreds of unusual mathematical problems involving number theory and com-

binatorics. What follows is a choice selection of little known examples.

One of the most surprising of card theorems is known as the Gilbreath principle after

magician Norman Gilbreath who first discovered it. Arrange a deck so the colors alternate.

Cut it so the bottom cards of each half are different colors, and then riffle shuffle the halves

together. Take cards from the top in pairs. Amazingly, every pair will consist of a red and

black card!

Here is a simple proof by induction that this must happen. Assume that the first card

to fail on the table during the shuffle is black. If the next card to fail is the card directly

above it in the same half, that card will be red. This places on the table a red-black pair. If

the next card after the first one comes from the other half, it too will be red and will put a

red-black pair on the table. In either case, after two cards have dropped, the bottom cards

of each half will be of different colors, so the situation is exactly the same as before, and

the same argument applied for the rest of the cards. No matter how careful or careless the

shuffle, it will pile red-black pairs on the table.

Gilbreath’s principle generalizes. Arrange the deck so the suits are in an order, say

spades, hearts, clubs, diamonds, that repeats throughout the pack. Deal as many cards as

you like to form a pile. This of course reverses the order of the suits. When the pile is about

the same size as the remaining portion of the deck, riffle shuffle the two portions together.

If you now take cards in quadruplets from the top of the shuffled pack, you will find that

each set of four contains all four suits. The ultimate generalization is to shuffle together

two decks, one with its cards in the reverse order of the other deck. After the shuffle, divide

the 104-card pack exactly in half. Each half will be a complete deck of 52 different cards!

What mathematician David Gale has called the “non-messing-up theorem” is another

whimsical result. From a shuffled deck, deal the cards face up to form a rectangle of any

proportion. In each row, rearrange the cards so their values do not decrease from left to

right. In other words, each card has a value higher than the one on its left, or two cards

Reprinted from The College Mathematics Journal, Vol. 31, No. 3 (May 2000), pp. 173–177.

221

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222 Part VI. Cards and Probability

of the same value are side by side. After ordering the rows, do the same thing with the

columns. This of course drastically alters the order of cards in the rows. After rearranging

the columns, you may be amazed to find that the rows are still ordered!

The theorem is at least a hundred years old. You will find it proved as the answer to a

problem in the American Mathematical Monthly (70:2, 1963, 212-13), and in a monograph

by Gale and Richard Karp, published in 1971 by the operations research center of the

engineering school of the University of California, Berkeley. Donald Knuth discusses the

theorem in the third volume of The Art of Computer Programming in connection with a

method of sorting called “shellsort.” In my The Last Recreations, Chapter 11, I describe a

clever card trick based on the theorem.

Is it possible to arrange a deck so that if you spell the name of each card by moving a

card from top to bottom for each letter, then turning over the card at the end of the spell

and discarding it, it will always be the card you spelled? For example, can you so arrange

the cards that you can first spell all the spades, taking them in order from ace through

king, then do the same thing with the hearts, clubs, and diamonds? You might imagine it

would take a long time to find out how to arrange the deck, assuming it is possible to do

so, in a way that permits the spelling of all 52 cards. Actually, finding the order is absurdly

easy. First arrange the deck from top down in the order that is the reverse of your spelling

sequence. Take the King of Diamonds from the top of the deck, then take the queen, place

it on top of the king, and spell “Queen of Diamonds” by moving a card at each letter from

bottom to top. In brief, you are reversing the spelling procedure. Continue in this way until

the new deck is formed. You are now all set to spell every card in the predetermined order.

Of course you can do the same thing with smaller packets, such as the thirteen spades, or

with cards bearing pictures, say of animals whose names you spell.

Remember the old brainteaser about two glasses, one filled with water, the other with

wine? You take a drop of water, put it into the wine, stir, then take a drop of the mixture,

move it back to the water, and stir. Is there now more or less wine in the water than water

in the wine? The answer is that the two quantities are exactly equal. The simplest proof is

to realize that, after the transfers, the amounts of liquid in each glass remain the same. So

the quantity missing from the water is replaced by wine, and amount of wine missing from

the other glass is replaced by the same amount of water.

This is easily modeled with cards. Divide the deck into two halves, one of all red cards,

the other of all black. Randomly remover n red cards, insert them anywhere in the black

half, and shuffle. Now randomly remove n cards from the half you just shuffled, put them

back among the reds, and shuffle. Inspection will show that the number of black cards in

the red half exactly equals the number of red cards in the black half. It doesn’t matter in

the least if the red and black portions are not equal at the start.

Closely related to this demonstration is the following trick. Cut a deck exactly in half,

turn over either half and shuffle the two parts together. Cut the mixed-up deck in half

again, and turn over either half. You’ll find that the number of face-down cards in either

half exactly equals the number of face-down cards in the other half. The same is true, of

course, for the face-up cards. Do you see why this is the case? The trick is baffling to

spectators if they don’t know that the deck is initially divided exactly in half, and if you

secretly turn over one half as you spread its cards on the table.

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30. Modeling Mathematics with Playing Cards 223

Playing cards provide a wealth of counterintuitive probability questions. The notorious

Monty Hall problem can be modeled with cards. There are three cards face down on the

table and you are told that one card only is an ace. Put a finger on a card. Clearly the

chance you have selected the ace is 1=3. A friend now secretly peeks at all three cards and

turns face up a card that is not the ace. Two cards remain face down, one of which you

know is the ace. What now is the probability your finger is on the ace? Many persons think

the probability has risen from 1=3 to 1=2. A little reflection should convince you that it

remains 1=3 because your friend can always turn a non-ace. Now switch your finger from

the card it is on to the other card. The probability you have now chosen the ace jumps form

1=3 to 2=3. This is obvious from the fact that the card you first selected has the probability

of 1=3 being the ace. Because the ace must be one of the two face-down cards, the two

probabilities must add to 1 or certainty.

A similar seeming paradox also involves three face-down cards dealt from a shuffled

deck. A friend looks at their faces and turns over two that are the same color. What’s the

probability that the remaining face-down card is the same color as the two face-up cards?

You might think it is 1=2. Actually it is 1=4. Here’s the proof. The probability that three

randomly selected cards are the same color is two out of eight equal possibilities, or 1=4.

Subtract 1=4 from 1 (the card must be red or black) and you get 3=4 for the probability

that the face-down card differs in color from the two face-up cards. This is the basis for

an ancient sucker bet. If you are the operator, you can offer even odds that the card is of

opposite color from the two face-up cards, and win the bet three out of four times.

Here’s a neat problem involving a parity check. Take three red cards from the deck.

Push one of them back into the pack and take out three black cards. Push one of them back

into the deck and remove three reds. Continue in this manner. At each step you randomly

select a card of either color, return it to the deck and remove three cards of opposite color.

Continue as long as you like. When you decide to stop you will be holding a mixture of reds

and blacks. Is it possible that the number of black cards you hold will equal the number of

reds? Unless you think of a parity check it might take a while to prove that the answer is

no. After each step you will always have in your hand an odd number of cards, therefore

the two colors can never be equal.

Magicians have discovered the following curiosity. Place cards with values ace through

nine face down in a row in counting order, ace at the left. Remove a card from either end of

the row. Take another card from either end. Finally, take a third card from either end. Add

the values of the three cards, then divide by six to obtain a random number n. Count the

cards in the row from left to right, and turn over the nth card. It will always be the four!

I leave it to readers to figure out why this works and perhaps to generalize it to longer

rows of numbers. For example, use twelve cards with values 2; 3; 4; 5; 6; 7; 8; 9; 10; J; Q; K

to make the row. Take a card three times from either end, divide their sum by 9, and call

the result n. The nth card from the left will always be the five.

A classic card task, going back more than two centuries, is to arrange all the aces,

kings, queens, and jacks—sixteen cards in all—in a square array so that no two cards of

the same value, as well as no two cards of the same suit, are in the same row, column,

or diagonal. Counting the number of different solutions is not trivial. W. W. Rouse Ball,

in his classic Mathematical Recreations and Essays, said there are 72 fundamental solu-

tions, not counting rotations and reflections. This is a mistake that persisted through the

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224 Part VI. Cards and Probability

book’s eleventh edition, but was dropped from later editions revised by H. S. M. Coxeter.

Dame Kathleen Ollerenshaw, a noted British mathematician who was once Lord Mayor of

Manchester, found there are twice as many fundamental solutions, 144, making the number

of solutions including rotations and reflections 8 � 144 D 1;152. She recently described a

simple procedure for generating all 1;152 patterns in an article written for the blind. (Dame

Ollerenshaw, now 87, is slowly losing her vision, and energetically learning how to read

Braille.)

This is her procedure. Number the sixteen positions of the square from 1 through 16,

left to right, top down. Place an arbitrary card, say the Ace of Spades, in position 1, the top

left corner. A second ace, say the Ace of Hearts, goes in the second row. It can’t go in the

same column or diagonal as the Ace of Spades, so it must go in either space 7 or 8. Place

it arbitrarily in space 7. Two aces remain to go in rows 3 and 4. Put the Ace of Diamonds

in the third row. It can go only in space 12. The Ace of Clubs is now forced into space 14

of the bottom row. Had the second ace gone in space 8, the last two aces would be forced

into spaces 10 and 15.

Consider the other three spades. They can’t go in the top row or leftmost column, or

in a main diagonal. This forces them into spaces 4, 10, and 15. Arbitrarily place the King

of Spades in 4, the Queen of Spades in 10, and the Jack of Spades in 15. The pattern now

looks like this:

A¨

Kª

Qª

Jª

Aª

A©

A§

The remaining nine cards are forced into spaces that complete the following pattern:

A¨

Kª

Qª

Jª

Aª

A©

A§

Q¨

J¨

K¨ Q©

J©

Q§

K§

K© J§

Multiply the number of choices at each step, 16 � 3 � 2 � 2 � 3 � 2, and you get the total

of 1,152 patterns.

For more examples of mathematical theorems, problems, and tricks with playing cards,

see my Dover paperback Mathematics, Magic, and Mystery, and Karl Fulves’s Self-Working

Card Tricks, also a Dover soft-cover, and the following chapters in my collections of Scien-

tific American columns: The Scientific American Book of Mathematical Puzzles and Diver-

sions, Chapter 10; Mathematical Carnival, Chapter 10 and 15; Mathematical Magic Show,

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30. Modeling Mathematics with Playing Cards 225

Chapter 7; Wheels, Life, and Other Mathematical Amusements, Chapter 19; Penrose Tiles

to Trapdoor Ciphers, Chapter 19, and The Last Recreations, Chapter 2.

Now for two puzzles that can be modeled with cards. Solutions appear on page 226.

1. Arrange nine cards as shown in Figure 30.1. Assume the aces have a value of 1. Each

row, each column, and one diagonal has a sum of 6. The task is to alter the positions of

three cards so that the matrix is fully magic for all rows, columns, and diagonals.

A

A

A

2

2

3

3

3

2

Figure 30.1.

A 2 3

4 5 6

7 8 9

Figure 30.2.

2. Nine cards arranged and shown in Figure 30.2 have the property of minimizing the

sum of all absolute differences between each pair of cells that are adjacent vertically and

horizontally. Assume that the matrix is toroidal; that is, it wraps around in both directions.

The sum of the differences is 48. This was proved minimal by Friend Kirstead, Jr., in the

Journal of Recreational Mathematics (18, 1985–86, 301). The challenge is to take nine

cards of distinct values (court cards may be used) and form a toroidal square that will

maximize the sum of all absolute differences.

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226 Part VI. Cards and Probability

Solutions to the playing card puzzles above

1. Move the bottom two of three cards to the top, or move the leftmost column to the

right. Either change produces the desired magic square in Figure 30.3a.

A

Y

J

K

2

Z

3

Q

X

2

2

2

3

3

A

A

A

3

(a) (b)

Figure 30.3.

2. Figure 30.3b shows how nine cards of distinct values can be placed in a toroidal

square so as to maximize the sum of absolute differences of adjacent values in rows

and columns. Any value 4 through 10 can go in cells, X, Y, and Z to make a total

of 120. This was proved maximal by Brian Maxwell, of Middlesex, England, in the

Journal of Recreational Mathematics, vol. 18 (1985–86), p. 300.

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31The Probability an Amazing

Card Trick Is Dull

Christopher N. Swanson

The Ashland University student chapter of the MAA holds biweekly meetings, typically

consisting of a short business meeting to plan activities that the group is sponsoring fol-

lowed by a social time when the group plays mathematical games and munches on brown-

ies. As a new feature in Fall 2002, I told the students that I would perform a new math-

ematical card trick at each of these meetings. My source for most of these card tricks is

the delightful book [2]. One of the finest tricks described there is due to an amateur New

York magician named Henry Christ. A spectator shuffles a deck of cards several times, and

then the magician deals nine cards face down on a table. The spectator selects one of these

cards, looks at it, and stacks the nine dealt cards on the table with the selected card on

top. The magician places this stack on the bottom of the deck, and hands the deck to the

spectator. The spectator is told to deal the cards out face up in a pile, counting backwards

out loud from ten while dealing the cards. If at some point, the denomination of the card

matches the number the spectator says, then the spectator repeats this procedure with a sec-

ond pile, again counting backwards from ten. If 1 is reached without any cards matching,

the spectator places the next card face down on top of the pile, and repeats this procedure

with the second pile. The spectator does this until four piles are created in this manner,

and then adds the numbers appearing on the cards that are face up. The spectator counts

through the remaining cards of the deck to find the card in the position represented by this

sum, and this card is the one that was originally selected.

The reason why this card trick works is that the final card will always be the 44th card

in the deck. This can be seen by noting that any unmatched pile will contain 11 cards that

are set aside. If a pile has a match with a value i showing, then the match occurred after

counting down 11� i cards. This card with value i will contribute i to the sum determining

how many cards in the final deck to count down, and thus, a total number of 11�i Ci D 11

cards will be set aside for this pile.

After practicing this card trick a few times, I was ready to present it to the students

at our meeting. Unfortunately, when I performed the trick at the meeting, a match did

not occur in any of the four piles. This is unfortunate for two reasons. First, nothing is

done with the pile of cards left in the spectator’s hands, and the magician must tell the

Reprinted from The College Mathematics Journal, Vol. 36, No. 3 (May 2005), pp. 209–212.

227

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228 Part VI. Cards and Probability

spectator to simply turn over the card appearing face down on the final pile. Second, and

more serious from the perspective of the magician, the trick seems quite simplistic when

no matches occur because it is clear that you set aside 44 cards. Thus, a magician may ask,

“What is the probability that no matches occur and my trick is dull?” This question falls

within a class of problems known as counting permutations with restricted positions. For a

more general discussion of this class of problems, see [1]. For a more advanced treatment

of the topic, see [7] or [8], and for additional applications, see [3], [4], [5], [6].

The probability of no matches

Our first step in studying this card trick problem is to note that after the nine cards are

returned to the bottom of the deck, the order of the cards in the deck is one of the possible

52! permutations. Hence, for our purposes, we just need to focus on the latter part of the

card trick. Let D be the number of ways to shuffle the deck of cards such that there is not

a 10 in the 1st, 11th, 21st, or 31st positions, there is not a 9 in the 2nd, 12th, 22nd, or 32nd

positions, etc. Thus, each card from ace and ten has four positions where it cannot occur.

We find D by using the principle of inclusion-exclusion: Given a collection of finite sets

A1; A2; : : : ; An, the number of elements in the union of these sets A1 [ A2 [ � � � [ An is

jA1 [ A2 [ � � � [ Anj DX

1�i�n

jAi j �X

1�i<j �n

jAi \ Aj j C � � �

C .�1/nC1 jA1 \ A2 \ � � � \ Anj:

Essentially this means to find the number of elements that appear in at least one of these

sets, count the number of elements in each set, then subtract off the number of elements

that appear in two sets, then add the number of elements that appear in three sets, and

continue to alternate these additions and subtractions up to the number of elements that

appear in all of the sets. To apply this principle to our problem, we number the forty cards

with restricted positions and let Ai be the set of permutations that have card numbered i

in one of its restricted positions. Then the number of permutations without any cards in

restricted positions is

D D 52Š � jA1 [ A2 [ � � � [ A40j

D 52Š �

0

@X

1�i�40

jAi j �X

1�i<j �40

jAi \ Aj j C � � � � jA1 \ A2 \ � � � \ A40j

1

A

D 52Š C40X

kD1

.�1/kX

1�i1<i2<���<ik�40

jAi1 \ Ai2 \ � � � \ Aik j:

The inner sum of this formula overcounts the number of ways to place at least k cards in

restricted positions. Let rk (sometimes called a rook number due to an interpretation con-

cerning placing non-attacking rooks on a chessboard) be the number of ways of placing k

cards in restricted positions, ignoring the placement of any other cards. Then the remaining

cards can be placed in .52 � k/Š ways, and we see thatX

1�i1<i2<���<ik�40

jAi1 \ Ai2 \ � � � \ Aik j D rk.52 � k/Š:

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31. The Probability an Amazing Card Trick is Dull 229

Taking r0 D 1 (as is standard), we see that our formula becomes

D D40X

kD0

.�1/krk.52 � k/Š; (31.1)

for which we need the numbers rk . To calculate rk , we find the generating function (called

the rook polynomial) for this sequence, R.x/ D r0 C r1x C r2x2 C � � � C r40x40. (Note

that ri is simply the coefficient of xi .) We do this by first finding the generating polynomial

P.x/ for the number of ways for having i cards for a particular denomination in a restricted

position, and then observing that R.x/ D .P.x//10 .

For ease of description, we focus on placing 10’s in restricted positions, but clearly the

same holds for each denomination. There are�

41

�

� 4 D 16 ways to place one 10 since there

are�

41

�

ways to choose the 10 and four positions for it. Similarly, for two 10’s, there are�

42

�

ways to choose the cards, and then 4 � 3 ways to place them in restricted positions (with the

suits in alphabetical order say, four choices for the “first” and three for the “second”) for a

total of�42

�

� 4 � 3, or 72. In the same way, we find there are 96 ways to place three 10’s and

24 ways to place all four 10’s in restricted positions. Therefore, with there being just one

way to have no 10’s in restricted positions, P.x/ D 1 C 16x C 72x2 C 96x3 C 24x4.

Now let’s consider the cards of two denominations, say the 10’s and the 9’s. We can

find the number of ways of having various combinations of these eight cards in restricted

positions (without regard to any others) by squaring P.x/ W 1 C 32x C 400x2 C 2496x3 C8304x4C14592x5C12672x6C4608x7C576x8. For example, consider the term 2496x3.

This says there are 2496 ways of having three 9’s and 10’s in restricted positions, arising

from no 9’s and three 10’s (1 � 96 ways), one 9 and two 10’s (16 � 72 ways), two 9’s and

one 10 (72 � 16 ways), and three 9’s and no 10’s (96 � 1 ways). The multiplicative rule is

justified since the restricted positions for different denominations are disjoint. From this

line of reasoning it follows that the coefficient of xk in .P.x//10 is the number of ways of

having k number cards in restricted positions, ignoring the placement of the other cards.

Thus, R.x/ D .1 C 16x C 72x2 C 96x3 C 24x4/10.

We used MAPLE to expand R.x/ to find the numbers rk , and then used equation (31.1)

to get

D D 3286792937574955359428826782591912851876783001405845125895657881600

� 3:29 � 1066:

Therefore, the probability that no card is in a forbidden position is D52Š

� 0:0407. We

see that although it is unlikely that the magician would not get any cards to match while

performing the magic trick, it will happen often enough that the magician better know how

to adjust the trick. (An anonymous referee noted that we can also produce the approximate

value of 0:0407 using double-precision arithmetic.)

Exercises

1. The game Euchre uses only 6 cards .9; 10; Jack; Queen; King; Ace/ from each suit.

Suppose you are at a party and you wish to demonstrate the card trick using such

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230 Part VI. Cards and Probability

a deck. You plan to have the spectator create 4 piles while counting backwards

Jack-10-9. How many cards should you first deal? What is the probability that no

match occurs in the four piles?

2. Consider the general version of Henry Christ’s trick. You have a deck of cards con-

sisting of k suits having f face values and in the trick the spectator is to create p

piles while counting backwards from m to 1. What are the possible values for m?

How many cards should you first deal? Find an expression for the probability that

no match occurs in the p piles.

Acknowledgment The author would like to thank the editor and the anonymous referees

for improving the clarity of this article. He would also like to thank the students of Ashland

University’s student chapter of the MAA for putting up with his bad humor and card tricks

during meetings.

Bibliography

[1] R. A. Brualdi, Introductory Combinatorics, 3rd ed., Prentice-Hall, 1999, Ch. 6.

[2] M. Gardner, Mathematics, Magic and Mystery, Dover Publications, 1956.

[3] A. W. Joseph, A problem in derangements, J. Inst. Actuaries Students’ Soc. 6 (1946) 14–22.

[4] F. F. Knudsen and I. Skau, On the asymptotic solution of a card-matching problem, Math. Mag.,

69 (1996) 190–197.

[5] B. H. Margolius, The dinner-diner matching problem, Math. Mag., 76 (2003) 107–118.

[6] S. G. Penrice, Derangements, permanents, and Christmas presents, Amer. Math. Monthly 98

(1991) 617–620.

[7] J. Riordan, An Introduction to Combinatorial Analysis, Princeton Univ. Press, 1978.

[8] R. P. Stanley, Enumerative Combinatorics: Volume I, Cambridge Univ. Press, 1998.

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32The Monty Hall Problem,

Reconsidered

Stephen Lucas, Jason Rosenhouse, and Andrew Schepler

In its classical form, the Monty Hall Problem (MHP) is the following:

Version 1. (Classic Monty) You are a player on a game show and are shown three iden-

tical doors. Behind one is a car, behind the other two are goats. Monty Hall, the host of the

show, asks you to choose one of the doors. You do so, but you do not open your chosen

door. Monty, who knows where the car is, now opens one of the doors. He chooses his door

in accordance with the following rules:

1. Monty always opens a door that conceals a goat.

2. Monty never opens the door you initially chose.

3. If Monty can open more than one door without violating rules one and two, then he

chooses his door randomly.

After Monty opens his door, he gives you the choice of sticking with your original choice

or switching to the other unopened door. What should you do to maximize your chances of

winning the car?

In the entire annals of mathematics, you would be hard-pressed to find a problem that

arouses the passions like the MHP. It has a history going back at least to 1959, when Mar-

tin Gardner introduced a version of it in Scientific American [4, 5]. When statistician Fred

Moseteller included it in his 1965 anthology of probability problems [9], he remarked that

it attracted far more mail than any other problem. In his 1968 book Mathematical Ideas in

Biology [18], biologist John Maynard Smith wrote, “This should be called the Serbelloni

problem since it nearly wrecked a conference on theoretical biology at the villa Serbel-

loni in the summer of 1966.” In its modern game show format the problem made its first

appearance in a 1975 issue of the academic journal The American Statistician [16]. Math-

ematician Steve Selvin presented it as an interesting classroom exercise on conditional

probability. Though he presented the correct solution, (that there is a big advantage to be

gained from switching), he found himself strongly challenged by subsequent letters to the

editor [17].

Reprinted from Mathematics Magazine, Vol. 82, No. 5 (Dec. 2009), pp. 332–342.

231

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232 Part VI. Cards and Probability

The problem really came into its own when Parade magazine columnist Marilyn vos

Savant responded to a reader’s question regarding it. There followed several rounds of

angry correspondence, in which readers challenged vos Savant’s solution. The challengers

later had to eat crow when it was shown by a Monte Carlo simulation that vos Savant was

correct, but not before the fracas reached the front page of the New York Times [20]. The

whole story is recounted in the books by Rosenhouse and vos Savant [13, 15].

In the end, the situation has been best summed up by cognitive scientist Massimo

Palmatelli-Palmarini who wrote that, “. . . no other statistical puzzle comes so close to fool-

ing all the people all the time. . . The phenomenon is particularly interesting precisely be-

cause of its specificity, its reproducibility, and its immunity to higher education” [10].

Why all the confusion?

The trouble, you see, is that most people argue like this: “Once Monty opens his door only

two doors remain in play. Since these doors are equally likely to be correct, it does not

matter whether you switch or stick.” We will refer to this as the fifty-fifty argument.

This intuition is supported by a well-known human proclivity. A negative consequence

incurred by inaction hurts less than the same negative consequence incurred through some

definite action. In the context of the MHP, people feel worse when they switch and lose

than they do after losing by sticking passively with their initial choice.

There is a large literature in the psychology and cognitive science journals documenting

and explaining the difficulty people have with the MHP. Burns and Wieth [3] summarized

the findings of numerous such studies by writing,

These previous articles reported 13 studies using standard versions of the

MHD, and switch rates ranged from 9% to 23% with a mean of 14.5%. This

consistency is remarkable given that these studies range across large differ-

ences in the wording of the problem, different methods of presentation, and

different languages and cultures.

(Note that MHD stands for “Monty Hall Dilemma.”)

Gilovich, Medvec, and Chen [6] studied people’s reactions to losing by switching ver-

sus their reactions to losing by sticking. They used boxes instead of doors, and crafted an

experimental situation in which players would lose regardless of their decision to switch

or stick. Their findings?

Because action tends to depart from the norm more than inaction, the indi-

vidual is likely to feel more personally responsible for an unfortunate action.

Thus, subjects who switched boxes in our experiment were more likely to

experience a sense of “I brought this on myself,” or “This need not have hap-

pened,” than subjects who decided to keep their initial box.

It would seem the defenders of sticking can point both to a plausible mathematical

argument and to certain fine points of human psychology. How can the forces for switching

fight back?

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32. The Monty Hall Problem, Reconsidered 233

Focus on Monty, not the doors

There are a variety of elementary methods for solving the MHP. Working out a tree diagram

for the problem, as in Figure 32.1, establishes that switching wins with probability 23

,

while sticking wins with probability 13

. Consequently, we double our chances of winning

by switching.

MontyopensDoor 2

MontyopensDoor 3

MontyopensDoor 3

MontyopensDoor 2

1

3

1

3

1

21 1

1

2

CarbehindDoor 1

CarbehindDoor 2

CarbehindDoor 3

Switching loses Switching loses Switching wins Switching wins

Probability =16

Probability =16

Probability =13

Probability =13

1

3

Figure 32.1. Probability tree for the classical MHP, when the player initially chooses door one.

Monte Carlo simulations are also effective for establishing the correct answer. The

Monty Hall scenario is readily simulated on a computer. The large advantage to be gained

from switching quickly becomes apparent by playing the game multiple times.

Such methods, however, do little to clarify why the fifty-fifty argument is incorrect.

Practical results obtained from a simulation can show you that something is wrong with

your intuition, but they will not make the correct answer seem natural. The trouble lies in

the difficulty people have in recognizing what is and is not important in reasoning about

conditional probability.

The mantra about focus goes a long way towards pointing people in the right direction.

When Monty opens door X , there is a tendency to think, “I have learned that door X

conceals a goat, but I have learned nothing of relevance about the other two doors.” This

is what we mean by “focusing on the doors.” The proper approach involves focusing on

Monty, specifically on the precise manner in which he chooses his door to open. We should

think, “Monty, who makes his decisions according to strict rules, chose to open door X .

Why this door as opposed to one of the others?”

Let us assume the player initially chose door one and Monty then opened door two.

According to the rules, we can be certain that one of the following two scenarios has

played out:

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234 Part VI. Cards and Probability

1. The car is behind door one. Monty chose door two at random from among doors two

and three.

2. The car is behind door three. Since the player initially chose door one, Monty was

now forced to open door two.

The second of these scenarios is more likely than the first. Since the car is behind the

first door one-third of the time, and since Monty then opens door two in one-half of those

cases, we see that scenario one occurs one-sixth of the time. Scenario two, on the other

hand, happens whenever the car is behind door three (and the player has chosen door one).

That happens one-third of the time. Scenario two is twice as likely as scenario one.

Thus, we should think, “I have just witnessed an event that is twice as likely to occur

when the car is behind door three than it is when the car is behind door one. Consequently,

the car is more likely to be behind door three, and I am more likely to win the car by

switching.”

An exotic selection procedure

The general principle here is that anything affecting Monty’s decision-making process is

relevant to updating our probabilities after Monty opens his door. To further illuminate this

point, let us consider an altered version of the problem:

Version 2. (High-Numbered Monty) As before, we have three identical doors conceal-

ing one car and two goats. The player chooses a door that remains unopened. Monty now

opens a door he knows to conceal a goat. This time, however, we stipulate that Monty al-

ways opens the highest-numbered door available to him (keeping in mind that Monty will

never open the door the player chose). Will the player gain any advantage by switching

doors?

For reasons of concreteness, we will assume once more that the player initially chooses

door one.

Any time door one conceals a goat, Monty has no choice regarding which door to open.

He can not open door one (since the player chose that door), and he can not open the door

that conceals the car. This leaves only one door available to him.

The interesting cases occur when door one conceals the car. Unlike Classic Monty,

who now chooses randomly, High-Numbered Monty will always open door three when he

can. It follows that if we see him open door two instead we know for certain that the car is

behind door three.

And if High-Numbered Monty opens door three? Since Monty is certain to open door

three whenever the car is behind door one or door two, we now have no basis for deciding

between them. It really is a fifty-fifty decision in this case.

Take this as a cautionary tale. Whether we are playing Classic Monty or High-Numbered

Monty, it is certain that Monty will open a goat-concealing door. In the former case the

probability that our initial choice concealed the car did not change while in the latter case

it did. This shows that any proposed solution to the MHP failing to pay close attention to

Monty’s selection procedure is incomplete.

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32. The Monty Hall Problem, Reconsidered 235

Monty meets Bayes

The main point thus far is that the probability that door X conceals the car, given that

Monty has shown us the goat behind door Y , depends on a detailed consideration of

Monty’s selection procedure. More precisely, it depends on the probability that Monty

will open door Y under the assumption that door X conceals the car. The precise manner

in which these probabilities are related is given by Bayes’ theorem.

We denote by Ci the event that the car is behind door i , and by Mj the event that

Monty opens door j to reveal a goat. Also assume the player initially chooses door one,

and Monty then opens door two. The probability that the player’s door conceals the car,

given Monty’s display, can be found using Bayes’ theorem:

P.C1jM2/ DP.C1/P.M2jC1/

P.M2/:

Expanding the bottom of this fraction via the law of total probability leads to

P.C1jM2/ D P.C1/P.M2jC1/

P.C1/P.M2 jC1/ C P.C2/P.M2jC2/ C P.C3/P.M2 jC3/:

In both of our versions of the MHP we have P.M2jC2/ D 0, since it is given that Monty

will never open the door concealing the car. Also, since we are given that the doors are

identical, we have

P.C1/ D P.C2/ D P.C3/ D 1

3:

Making these substitutions leads to

P.C1 jM2/ D P.M2jC1/

P.M2jC1/ C P.M2jC3/:

In both versions of the game we have P.M2jC3/ D 1. That is, when the player chooses

door one and the car is behind door three, Monty is certain to open door two.

In Classic Monty we have P.M2jC1/ D 12

, since Monty chooses at random when

the car is behind the door initially chosen by the player. In High-Numbered Monty we

have P.M2jC1/ D 0, since Monty is required by his rules to open door three. Plugging

everything into Bayes’ Theorem shows that for Classic Monty we now have

P.C1jM2/ D12

12

C 1D

1

3[Classic],

while for High-Numbered Monty we have

P.C1jM2/ D12

0 C 1D 1

2[High-Numbered].

These are precisely the answers we obtained in the previous section.

Let us go one more round:

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236 Part VI. Cards and Probability

Version 3. (Random Monty) As always, assume that the player has initially chosen door

one and Monty subsequently opened door two to reveal a goat. This time, however, you

know that Monty chose his door randomly, subject only to the restriction that he not open

the door the player chose. Should we switch?

The novelty here lies in the nonzero probability of Monty opening the door concealing

the car. Intuitively we would reason as follows: Since Monty opened door two after I se-

lected door one, since door two concealed a goat, and since I know Monty chose randomly

between doors two and three, I conclude that one of two scenarios has played out:

1. The car is behind door one, Monty chose door two randomly.

2. The car is behind door three, Monty chose door two randomly.

Since the car is equally likely to be behind doors one and three, these scenarios are equally

likely to occur. The conclusion is that the remaining doors are equiprobable, and therefore

there is no advantage to switching.

Our intuition is confirmed via Bayes’ Theorem. We know that Monty will not open

door one, and we know that door two conceals a goat. We now have

P.C1/ D P.C2/ D P.C3/ D 1

3;

P.M2jC1/ D P.M2jC3/ D 1

2;

P.M2jC2/ D 0:

Bayes’ Theorem now says

P.C1jM2/ D13

�12

�

13

�12

�

C 13.0/ C 1

3

�12

� D 1

2:

The tree diagram in Figure 32.2 might be helpful for visualizing the situation.

Two-player Monty

Three-door versions of the MHP can become remarkably complex. The following version

comes from a paper by philosopher Peter Baumann [1, 2]. For the remainder of the paper

we will refer simply to the probability of door X, thereby avoiding the more cumbersome

expression, “The probability that door X conceals the car.”

Version 4. (Two-Player Monty) We begin with three identical doors concealing two

goats and one car. There are two players in the game. Each player chooses one of the

doors, but does not open it. Each player knows there is another person in the game, but

neither knows which door the other player selected. Monty now opens a door according to

the following procedure.

1. If both players selected the same door, then everything proceeds as in the classical

game. Monty opens a goat-concealing door, choosing randomly if he has a choice.

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32. The Monty Hall Problem, Reconsidered 237

MontyOpensDoor 2

MontyOpensDoor 3

MontyOpensDoor 2

MontyOpensDoor 3

MontyOpensDoor 2

MontyOpensDoor 3

Switching loses

Probability =16

Switching loses

Probability =16

1

3

1

3

Switching wins

Probability =16

Game ends

Switching wins

Probability =16

Game ends

1

2

1

2

1

2

1

2

1

2

1

2

CarbehindDoor 1

CarbehindDoor 2

CarbehindDoor 3

1

3

Figure 32.2. Probability tree for Random Monty, when the player initially chooses door one.

2. If the players selected different doors, then Monty opens the one remaining door,

regardless of what is behind it.

We assume that both players select their initial doors randomly. If you are one of the players

and you have just seen Monty open a goat-concealing door, should you switch?

This will be a fine test of our new-found intuition. For concreteness, suppose that Player

A initially chose door one, and Monty has now opened the goat-concealing door two. What

do Monty’s actions tell us about Player B’s choice? Initially we consider it equally likely

that Player B chose door one, door two or door three. After seeing Monty open door two

we reason that one of three scenarios has played out:

1. Player B chose door three. In this case Monty was forced to open door two, which

conceals a goat with probability 23

.

2. Player B chose door one and door one conceals the car. In this case Monty opens

door two with probability 12

. Since door one conceals the car with probability 13

, this

scenario occurs with probability 16

.

3. Player B chose door one and door three conceals the car (which happens with prob-

ability 13

). In this case Monty is again forced to open door two.

Combining items two and three above shows that Player B chose door one with probability13

C 16

D 12

. Item one shows that Player B chose door three with probability 23

. We conclude

that the event in which Monty opens the goat-concealing door two after Player A chooses

door one is 43

more likely to occur when Player B has chosen door three than when he has

chosen door one.

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238 Part VI. Cards and Probability

It is a consequence of Bayes’ theorem that the probabilities we now assign to “Player B

chose door three,” and “Player B chose door one” must preserve this 4 W 3 ratio. (A proof

of this assertion can be found in the paper by Rosenthal [14].) Consequently, we assign

probabilities of 47

and 37

respectively.

To continue the analysis, note that from Player A’s perspective there are now four

possibilities. Player B could have chosen door one or door three, and the car could be

behind either of those doors. Let us denote these possibilities via ordered pairs of the form

(Player B’s Door, Location of the Car).

Thus, the four remaining possibilities are

.3; 1/; .3; 3/; .1; 1/; .1; 3/:

Consider the first two pairs. If Player B chose door three, then Monty was forced to

open door two. Consequently, we learn nothing regarding the probability of doors one and

three. Since these two scenarios collectively have a probability of 47

, and since they are

equally likely, we now assign the following probabilities:

P.3; 1/ D P.3; 3/ D 2

7:

The remaining two pairs, however, are not equiprobable. Suppose that Player B chose

door one, just as Player A did. If the car is behind door one, then Monty chose door two

randomly, which happens with probability 12

. If the car is behind door three, then Monty

was forced to choose door two. It follows that it is twice as likely that the car is behind door

three than that it is behind door two. Since these scenarios have a collective probability of37

, we assign the following probabilities:

P.1; 1/ D 1

7and P.1; 3/ D 2

7:

Of the four scenarios, the two in which Player A wins by switching are .1; 3/ and .3; 3/.

Since both have probability 27

, this gives a total probability of winning by switching of 47

.

That is our solution.

The really amusing part is that both players will go through this analysis, and both will

decide to switch doors. In those scenarios in which the players chose different doors, this

implies that someone is definitely making the wrong decision. Such are the cruelties of

probability.

Two-player Monty has also been discussed by Baumann [1], Levy [7], Rosenhouse

[13], and Sprenger [19].

Many doors

Ready for the final exam?

Version 5. (Progressive Monty) This time there are n identical doors, concealing one

car and n � 1 goats. The player chooses a door, but does not open it. Monty now opens a

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32. The Monty Hall Problem, Reconsidered 239

goat-concealing door, choosing randomly from among his options. The player is now given

the choice of sticking or switching. The player makes his choice, but again does not open

his chosen door. Monty opens another goat-concealing door. The player is again given the

opportunity to stick or switch. This continues until Monty has opened n � 2 doors. The

player makes his final selection, and wins whatever is behind his door. What strategy will

maximize his chances of winning the car?

To help us get our bearings, let us try a case study. Suppose we begin with five doors.

At any stage of the game we represent the probabilities of the doors, based on all available

knowledge, via an ordered 5-tuple, which we call the probability vector. As the game

begins, we have probability vector�

1

5;

1

5;

1

5;

1

5;

1

5

�

:

As always, let us assume the player chooses door one and Monty now opens door two.

Our past experience suggests that the probability of our door does not change, and this

is confirmed by Bayes’ theorem. In the following calculation, the notation Ci denotes the

event where the car is not behind door i .

We now compute

P.C1jM2/ DP.C1/P.M2 jC1/

P.C1/P.M2 jC1/ C P.C1 and C2/P.M2jC1 and C2/

D15

�14

�

15

�14

�

C 35

�13

� D 1

5:

Since the other doors are identical and since their probabilities must sum to 45

, we now

have probability vector �1

5; 0;

4

15;

4

15;

4

15

�

:

What if we now switch to door three and then see Monty open door five? Known

probabilities are now

P.C3/ D P.C4/ D P.C5/ D 4

15;

P.M5jC1/ D P.M5jC4/ D 1

2;

P.C1/ D 1

5and P.M5jC3/ D 1

3:

If we use the law of total probability to write

P.M5/ D P.C1/P.M5jC1/ C P.C3/P.M5 jC3/ C P.C4/P.M5jC4/ D29

90;

and plug the results into Bayes’ Theorem, the result is the probability vector

�9

29; 0;

8

29;

12

29; 0

�

:

The probabilities of all the remaining doors went up.

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240 Part VI. Cards and Probability

What if Monty had opened door one after we switched to door three? The reader can

supply the details that lead to the vector�

0; 0;1

4;

3

8;

3

8

�

:

Notice that the probability of door three went down, from 415

to 14

. Our chosen door actually

seems less likely as the result of Monty’s actions. A surprising result!

Things get messy indeed in this version. Plainly we need some guidelines to aid our

intuition.

The first principle is simple. Any time Monty chooses not to open a door different from

your present choice, the probability of that door goes up. In our case study, Monty opened

door two after we chose door one. The event, “Monty does not open door three,” is more

likely to happen when the car is behind door three than when it is elsewhere. Consequently,

we will revise upward our probability of door three.

The second principle is that if the doors different from your present choice are equiprob-

able, then the probability of your choice does not change when Monty opens a door. In our

case study, after Monty opened door two, we reason that the event, “Monty does not open

door one,” has probability one regardless of the location of the car. Consequently, we learn

nothing from the occurrence of that event. The calculation in our case study confirms this

intuition.

Why, though, does it matter that the other doors are equiprobable? The answer is that

Monty’s failure to open a door is not the only source of information to which we have

access. The probability of the event, “Monty opens door X ,” depends in part on the prob-

ability of the event, “Door X conceals the car.” Specifically, the more likely a door is to

conceal the car, the less likely Monty is to open that door. Once more returning to our

case study, we switched to door three at a moment when doors three through five were

equiprobable and collectively very likely to conceal the car. By opening door five, Monty

eliminated one element of this collection. This revelation does nothing to shake our confi-

dence that the car is more likely to be found among doors three through five than it is to be

found among any collection of three doors that includes door one. Consequently, we will

revise upward the probability of our chosen door.

Why did the probability of door three go down when Monty opened door one? This one

is harder to explain, but our calculation suggests the proper way to think about it. If all four

doors had been equiprobable at the moment we switched to door three, then we would sim-

ply be playing Classic Monty on four doors. In that case, our chosen door would retain its 14

probability after Monty opens a door. The probability vector we computed for our present

situation is identical to what we would have obtained were we playing four-door Classic

Monty. The implication is that by eliminating door one, Monty has essentially erased the

prior history of the game. We are now faced with three doors that were equiprobable at the

moment we chose among them, and these doors were among an ensemble of four doors,

just as in four-door Classic Monty. That door one had a different probability from the other

doors does not distinguish our situation in a relevant way.

This observation leads to our final clue. If we select a door at a moment when precisely

k doors remain, the probability of that door can never be smaller than 1k

. Even if we have

been careless in extracting the maximum amount of information from Monty’s actions, we

still know the door was chosen from among k possibilities.

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32. The Monty Hall Problem, Reconsidered 241

As a test of our principles, let us go another round with our case study. We left off with

the player having chosen door three and with probability vector,

�

0; 0;1

4;

3

8;

3

8

�

:

Imagine that we now switch to door four.

If Monty now opens door three then only doors four and five remain in play. We would

reason that these two doors were equiprobable at the moment we switched to door four,

but that door four was selected from among three possibilities. We are, in effect, playing

Classic Monty, and we would expect our updated probability vector to be

�

0; 0; 0;1

3;

2

3

�

:

The calculation is

P.C4jM3/ D P.C4/P.M3 jC4/

P.C4/P.M3 jC4/ C P.C5/P.M3jC5/

D38

�12

�

38

�12

�

C 38.1/

D 1

3:

And if Monty opens door five instead? Our intuition tells us that both doors should see

their probabilities go up: door three, because it might have been opened but was not; door

five, because it was part of an equiprobable ensemble that has decreased in size. Bayes’

Theorem confirms our intuitions. We compute

P.C3 jM5/ DP.C3/P.M5 jC3/

P.C3/P.M5 jC3/ C P.C4/P.M5jC4/D

14.1/

14.1/ C 3

8

�12

� D4

7;

and obtain probability vector�

0; 0;4

7;

3

7; 0

�

:

Remarkably, our arguments to this point are already enough to justify the correct so-

lution to Progressive Monty. Consider the strategy in which we switch at the last minute

(SLM). That is, we will stick with our initial choice until only two doors remain, and then

we will switch. Our initial choice has probability 1n

. Since the other doors are equiprobable,

this probability will not change so long as we keep it as our choice. At the moment when

only two doors remain, the other door will have probability n�1n

. That is the probability

that we win with SLM.

We also know that there will never be a moment in the game when a door has a proba-

bility smaller than 1n

. Thus, at the moment when only two doors remain it is impossible to

produce a door with probability greater than n�1n

. This shows that SLM is optimal.

Very nice. A full, rigorous proof that SLM is, in fact, uniquely optimal can be found

in the book by Rosenhouse [13]. You might also wonder what can be said about other

strategies. For example, what if we are playing with fifty doors and we are absolutely

determined to switch exactly seven times during the game? What is our best strategy? A

consideration of such questions can be found in the paper by Lucas and Rosenhouse [8].

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242 Part VI. Cards and Probability

Progressive Monty receives further attention in one paper by Paradis, Viader, and Bibiloni

[11] and another by Rao and Rao [12]. For a variation on the problem, see the paper by

Zorzi in the December 2009 Mathematics Magazine.

It would seem that a bit of clear thinking can steer us through even the densest of

Monty-inspired forests. Once our intuition has been tuned to what is important, it is not so

difficult to ferret out the correct answer.

Bibliography

[1] Peter Baumann, Three doors, two players, and single-case probabilities, American Philosophi-

cal Quarterly 42(1) (January, 2005) 71–79.

[2] ——, Single-case probabilities and the case of Monty Hall: Levy’s view, Synthese 162(2) (May

2008) 265–273. dx.doi.org/10.1007/s11229-007-9185-6

[3] Bruce Burns and Mareike Wieth, The collider principle in causal reasoning: why the Monty

Hall Problem is so hard, Journal of Experimental Psychology, General 103(3) (2004) 436–449.

[4] Martin Gardner, Problems involving questions of probability and ambiguity, Scientific Ameri-

can 201(4) (April, 1959) 174–182.

[5] ——, How three modern mathematicians disproved a celebrated conjecture of Leonhard Euler,

Scientific American 201(5) (May, 1959) 188.

[6] T. Gilovich, V. H. Medvec, and S. Chen, Commission, omission, and dissonance reduction:

coping with regret in the Monty Hall Problem, Personality and Social Psychology Bulletin

21(2) (February 1995) 182–190. dx.doi.org/10.1177/0146167295212008

[7] Ken Levy, Baumann on the Monty Hall Problem and single-case probabilities, Synthese 158(1)

(September 2007) 139–151. dx.doi.org/10.1007/s11229-006-9065-5

[8] Stephen K. Lucas and Jason Rosenhouse, Optimal Strategies for the Progressive Monty Hall

Problem, Math. Gaz. (to appear).

[9] Fred Mosteller, Fifty Challenging Problems in Probability, With Solutions, Addison-Wesley

Inc., Reading, 1965.

[10] Massimo Piattelli-Palmarini, Probability blindness, neither rational nor capricious, Bostonia

(March/April 1991) 28–35.

[11] J. Paradis, P. Viader, and L. Bibiloni, A mathematical excursion: from the three-door

problem to a Cantor-type set, Amer. Math. Monthly 106 (1999) 241–251. dx.doi.org/

10.2307/2589679

[12] V. V. Bapeswara Rao and M. Bhaskara Rao, A three-door game show and some of its variants,

Math. Sci. 17(2) (1992) 89–94.

[13] Jason Rosenhouse, The Monty Hall Problem, Oxford University Press, New York, 2009.

[14] Jeffrey Rosenthal, Monty Hall, Monty fall, Monty crawl, Math Horizons (October 2008) 5–7.

[15] Marilyn vos Savant, The Power of Logical Thinking, St. Martin’s Press, New York, 1996.

[16] Steve Selvin, A problem in probability, (Letter to the Editor), Amer. Statist. 29(1) (1975) 67.

[17] ——, On the Monty Hall Problem, (Letter to the Editor), Amer. Statist. 9(3) (1975) 134.

[18] John Maynard Smith, Mathematical Ideas in Biology, Cambridge University Press, London,

1968.

[19] Jan Sprenger, Probability, rational single-case decisions, and the Monty Hall Problem, Synthese

(to appear).

[20] John Tierney, Behind Monty Hall’s doors, puzzle, debate, and answer? The New York Times,

July 21, 1991, 1A.

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33The Secretary Problem from the

Applicant’s Point of View

Darren Glass

Searching for a job is always stressful and, with unemployment rates at their highest levels

in years, never more so than now. Applicants can and should use every advantage at their

disposal to obtain a job which is rewarding, financially and otherwise. While this author

believes a math major gives applicants many advantages as they search for their dream job,

one often overlooked is the ability to strategize and schedule their interviews to maximize

the chance of landing that job.

The secretary problem helps an employer find the optimal candidate for a job out of a

large pool of applicants. The set up is as follows: only one person can be hired, and, for

any pair of applicants, the employer has a strict preference for one of them that they can

discern after seeing both. However, after each interview the employer must either accept

or reject the candidate. If the candidate is the final person, the interviewer simply must

accept them, as rejected candidates cannot be recalled. In the classical formulation of the

problem, the goal is to select the best applicant overall.

What strategy can the employer use to maximize the probability of hiring the best

overall applicant? It is clear that, other than the final candidate, you only hire an applicant

if they are the best applicant you have seen to that point. Otherwise you are certainly not

hiring the best person. It follows that the best strategy is to reject an initial number of

candidates and then hire the first candidate who is better than everyone seen so far. With

a little work, one can show that if you know there will be a total of n candidates, where

n is large, then you should initially reject k � ne

candidates. For a proof, see [3]. A more

recent article discussing this problem and its implications for students’ dating lives can be

found in [6].

While there is some uncertainty regarding the origin of this problem, its first published

appearance was in Martin Gardner’s Mathematical Games column in February of 1960

(reprinted in [2]). A detailed history of the secretary problem and other classical stopping

problems is in [1].

The secretary problem is an outstanding example of how a seed planted by one of

Gardner’s columns has grown and flourished. Over the intervening half-century, people

have studied many variants. For example, how does the answer change if the interviewer

is allowed to recall the last few applicants [7]? Or what if there is a full committee of

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 76–81.

243

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244 Part VI. Cards and Probability

interviewers rather than a single interviewer [4]? Or how should one deal with the costs

of interviewing and pressures to hire an early applicant to reduce costs [5]? As of this

writing, there are over 130 papers in MathSciNet referring to the Secretary Problem and

its extensions.

Almost all published variants look at the problem from the employer’s perspective.

At a recent senior thesis presentation on one such variant at Gettysburg College, another

senior, clearly concerned about his own job prospects, asked “What does this mean for

applicants? If I know that an employer is going to behave optimally, when should I schedule

my interview?” This is the question we examine here.

When rank is unknown

First we consider the situation of an applicant who has no idea how strong they are com-

pared to the rest of the pool. Throughout this note, we assume that there are n applicants

for a single job, and that the employer’s strategy is initially to reject the first k and then

hire the first subsequent candidate who is better than all candidates already seen, if such

a candidate exists, or the final candidate otherwise. We call this the optimal strategy. We

assume that all nŠ possible ordered rankings of the n candidates are equally likely and ex-

amine the probability that each position in the interview order is the one where the chosen

applicant is found.

Let us define the random variable X to be the interview position of the applicant who

is finally chosen. Using the optimal strategy, the employer rejects the first k applicants, and

therefore P.X D i/ D 0 if 1 � i � k. In order for the .k C 1/st candidate to be chosen,

they must be better than all the candidates in the rejected group. This occurs exactly when

the .k C 1/st candidate is the best among the first k C 1 candidates, which occurs with

probability 1kC1

. More generally, for k C 1 � i � n � 1, the i th candidate will be chosen

if and only if:

� The i th candidate is the best of the first i candidates.

� The best of the first i � 1 candidates is in the initial rejected group of k candidates.

The second condition ensures that we get to the i th candidate without choosing someone

else; the first condition ensures that we then choose the i th candidate. The probability of

these two conditions simultaneously holding is 1i

� ki�1

.

Finally, there are two separate situations in which the last candidate is selected: either

the best candidate overall was in the initial group of rejected candidates, in which case

the final candidate is chosen as a last resort, or the best overall candidate is the last one

interviewed and the second best candidate was in the initial group of rejectees. The first

scenario occurs with probability kn

; the second with probability 1n

� kn�1

. Adding these two

cases together, P.X D n/ D kn

C kn.n�1/

D kn�1

. In summary, we have the following

result.

Theorem 1. The probability that the i th candidate is chosen is

P.i/ D

8

ˆ<

ˆ:

0 if i � k,

ki.i�1/

if k < i < n,

kn�1

if i D n.

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33. The Secretary Problem from the Applicant’s Point of View 245

The goal of the applicant is to obtain interview slot i which maximizes P.i/. Note thatk

i.i�1/is a decreasing function in i . Therefore the candidate should target the .k C 1/st slot

or the final slot, depending on whether 1kC1

or kn�1

is larger. An applicant will prefer the

final slot if kn�1

> 1kC1

. This is the case when

k >�1 C

p4n � 3

2; (33.1)

agreeing with the intuition that rejecting a larger number of candidates makes it more

likely that the best candidate is in the rejected group and that the employer accepts the final

candidate.

If the employer uses the optimal strategy with k D ne

, then the applicant must check

whether (33.1) holds, which happens when k2 C .1 � e/k C 1 > 0. A simple calculation

shows this is always the case, and therefore the candidate should choose to be interviewed

in the final slot if they have no information about their relative standing with the other

candidates. Of course, ne

is never an integer, so in reality the employer chooses either

k D bnec or dn

ee. In the latter case, the candidate will still wish to choose the final interview

slot, because if ne

> �1Cp

4n�32

then certainly dnee is as well. If the employer rounds down

then one can check that the candidate will be better off choosing last as long as k > 2:5. In

particular, being interviewed in the final position is optimal if n � 10. One can manually

check that, if the interviewer is going to reject k D bnec applicants, one will be no worse

off being interviewed last except when n D 4; 5, or 8. Thus we have proved

Theorem 2. If the number of applicants to a position is at least nine and an employer

uses the optimal strategy then the probability that they hire the final person interviewed is

higher than any other person.

When rank is known

On the whole, in the absence of information about their standing, a candidate should choose

to be interviewed last. This is also true if you know you are the worst candidate, as you

will never be better than all the candidates in the rejected group, so you can only be chosen

if you are the last resort. On the other hand, if you know you are the strongest candidate,

then you prefer the .k C 1/st slot over all others. You are then guaranteed to be chosen:

you are certainly better than anyone in the rejected group, and in any later slot there is a

chance that someone else is picked before you are interviewed. In general, then, it seems

that stronger applicants want to be interviewed early and weaker applicants later. In this

section, we prove that this is so by considering the point of view of a candidate who knows

they are the j th best in a pool of n candidates.

Assume first that our candidate interviews last. As in the previous section, if j ¤ 1

then the only way that they can be chosen is if the best candidate is in the rejected group,

which happens with probability kn�1

, since there are n � 1 slots remaining for the best

applicant to be in, all equally likely. On the other hand, if j D 1 then the only way to be

chosen from the last slot is if the second best candidate is rejected, which also happens

with probability kn�1

. In other words, the probability that the candidate in the final slot is

chosen is kn�1

independent of how strong they are.

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246 Part VI. Cards and Probability

As before, if a candidate is interviewed in slot i , k < i < n, they will be chosen if

and only if they are the best candidate seen so far and the second best candidate to that

point is rejected. Assume that i > 1, which will is the case as long as k > 0, i.e., at

least one candidate is rejected. The probability that the j th best candidate overall is the

best candidate seen so far, when they are interviewed in the i th slot, is the probability that

all i � 1 candidates seen previously come from the n � j candidates who are worse than

them. This happens with probability�

n�ji�1

�

=�

n�1i�1

�

. Moreover, the probability that the best

candidate seen before the i th slot was in the rejected group is ki�1

. Putting this together,

we get the following result:

Theorem 3. Assume the j th best candidate is interviewed in the i th slot. The probability

of getting chosen is:

Pj .i/ D

8

ˆ<

ˆ:

0 if i � k,

.n�ji�1/

.n�1i�1/

ki�1

if k < i < n,

kn�1

if i D n.

Again, Pj .i/ is a decreasing function in the range k < i < n. Explicitly, we note that

if i and i C 1 are in this range, then

Pj .i/

Pj .i C 1/D�

n�ji�1

�

�n�1i�1

�k

i � 1

�n�1

i

�

�n�j

i

�i

k

D .n � i/Š .n � i � j /Š

.n � i � 1/Š .n � i � j C 1/Š

i

i � 1

D n � i

n � i � j C 1

i

i � 1> 1;

and therefore Pj .i/ > Pj .i C 1/, as desired. This means that for any fixed value of j ,

an applicant should either choose to be interviewed in the .k C 1/st slot or in the nth

slot. In order to see which of these options gives the higher probability, we must compare

Pj .k C 1/ D�n�j

k

�

=�n�1

k

�

with Pj .n/ D kn�1

. After some algebraic manipulation, we find

that Pj .k C 1/ > Pj .n/ if and only if�

n�jk

�

>�

n�2k�1

�

. We wish to consider this inequality

separately for different values of j .

If j D 1, then we compare�

n�1k

�

to�

n�2k�1

�

. Since�

n�1k

�

D�

n�2k

�

C�

n�2k�1

�

, the best

candidate will prefer to be interviewed in the .k C 1/st slot, as we argued earlier.

If j D 2, then we compare�

n�2k

�

and�

n�2k�1

�

. The former is larger exactly when k <n�1

2. If the employer chooses k � n=e, then this holds for large values of k, implying that

the second best candidate should also choose the .k C 1/st slot.

If j D 3, then we compare�

n�3k

�

and�

n�2k�1

�

. Looking at the quotient of these terms and

expanding algebraically, we see that the former is larger exactly if

k2 C .5 � 3n/k C .n2 � 3n C 2/ > 0;

which occurs if k < 3n�5�p

5n2�18nC172

. For large n, this is true if k < 3�p

52

n � :38n;

which is (barely) guaranteed if the employer chooses k D n=e � :367n. More specifically,

one can work out that the third-best candidate should choose the .k C 1/st slot for all cases

except for 33 specific values of n, the largest of which is n D 98.

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33. The Secretary Problem from the Applicant’s Point of View 247

For j � 4, comparing�n�j

k

�

and�

n�2k�1

�

is difficult. In particular, one can show that the

former term is greater exactly when

�n � k � 1

k

� j �2Y

`D1

n � k � ` � 1

n � ` � 1> 1:

Assuming that n D ek and letting n be large, we see that this holds if and only if.e�1/j �1

ej �2 � 1, which is false for j � 4. In particular, for sufficiently large n one prefers to

be interviewed last if ones ranking is fourth or worse. We summarize the preceding results:

Theorem 4. If you are the j th best applicant then you want to be interviewed in the

.k C 1/st slot for j � 3, and in the final slot if j � 4 and n is sufficiently large.

We also note that for any n and j � 3 and k > n�j2

we will have that�

n�2k�1

�

>�n�jk�1

�

��

n�jk

�

. If the employer sets k D dnee then k will be greater than n�j

2if j

n> e�2

e� :26. In

particular, this says that regardless of n, if a candidate suspects they are not in roughly the

top quarter of candidates, then they would prefer to be interviewed in the last slot.

Final thoughts

Unless you are a truly extraordinary applicant, if your prospective employer uses the clas-

sic strategy, then you should try to be interviewed last. However, attempting to game the

system is likely to backfire as there is no margin of error. Being the final interviewee gives

one the highest probability of being hired, but being the second-to-last has one of the low-

est! Students going on the job market should put their energy into improving their resumes

rather than strategizing interview timing. We note that a student can do both by general-

izing our work to some of the variants of the secretary problem mentioned earlier—and

publishing their work in future issues of the College Mathematics Journal.

Acknowledgment The author thanks the two students who inspired this question, Brian

Lemak and Paul Smith, as well as his colleagues in the math department at Gettysburg

College and the anonymous referees.

Bibliography

[1] T. S. Ferguson, Who solved the secretary problem? Statist. Sci. 4 (1989) 282–296; available at

dx.doi.org/10.1214/ss/1177012493

[2] M. Gardner, New mathematical diversions, Mathematical Association of America, Washington,

DC, revised edition, 1995.

[3] J. P. Gilbert and F. Mosteller, Recognizing the maximum of a sequence, J. Amer. Statist. Assoc.

61 (1966) 35–73; available at 10.2307/2283044

[4] H. Glickman, A best-choice problem with multiple selectors, J. Appl. Probab. 37 (2000) 718–

735; available at 10.1239/jap/1014842831

[5] Z. Govindarajulu, The secretary problem: optimal selection with interview cost, in Proceeding

of the Symposium on Statistics and Related Topics, Carleton Univ., Ottawa, 1975, 19.

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248 Part VI. Cards and Probability

[6] K. Merow, The view from here: Finding your match mathematically, Math Horizons 17 (2009)

18–20; available at 10.4169/194762109X468346

[7] O. K. Zakusilo, Optimal choice of the best object with possible returning to previously observed,

Theory Stoch. Process. 10 (2004) 142–149.

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34Lake Wobegon Dice

Jorge Moraleda and David G. Stork

Nontransitive dice, discovered by Bradley Efron, but introduced to the public by Martin

Gardner [2], are sets of three dice which if rolled in pairs, die A most frequently beats

die B , B beats C , and C beats A. This is a statistical analogue of the game Rock, paper,

scissors (also called rochambeau, roshambo, and jan-ken-pon), in which each element or

action beats another and is beaten by yet another. In such a set there is no best or dominant

element. Of course the scalar mean values of the dice cannot exhibit nontransitivity; it is

only in the statistical case of rolls of different pairs of dice where nontransitivity can arise.

For an example of nontransitive dice, see Figure 34.1.

2

2

9

9

44

1

1

8

8

66

3

3

7

7

55

A B C

Figure 34.1. A set of nontransitive dice (after Efron), where in fair rolls of pairs of dice, on average

A beats B , B beats C , and C beats A. In this set, P ŒA > B� D P ŒB > C� D P ŒC > A� D 5=9 D55:56%:

Grinstead [3] explored the case where the dominance of dice alternated with successive

rolls. That is, if pips are summed over an even number of rolls, die A dominates die B;

summed over an odd number of rolls, die B dominates die A. Such nontransitive dominance

or paradoxical dominance can occur in preference for artworks and for political candidates,

as well as in dominance among warrior avatars in some computer games, some athletes,

and among teams, for instance the common claim that “Patriots beat Colts, Colts beat

Broncos, and Broncos beat Patriots.” [7, 6]. One experiment showed that in prospective

marriage partners, college students preferred intelligence over looks, looks over wealth,

and wealth over intelligence [5].

We extend Efron’s notion to the case, raised by David Stork, in which on each roll of

the full set of n dice each die is more likely to roll greater than the average of the set on the

roll than to roll less than that average. Informally, where Efron studied the case where each

die need beat merely one other die, in our case each die beats the average of all the dice.

Reprinted from The College Mathematics Journal, Vol. 43, No. 2 (Mar. 2012), pp. 152–159.

249

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250 Part VI. Cards and Probability

In a specific statistical sense, then, each die is “better than the set average.” We call these

dice Lake Wobegon in honor of the children in the mythical Minnesota town chronicled by

humorist Garrison Keillor, “Where all the women are strong, all the men are good looking,

and all the children are above average” [4].

Notation and definitions

We define the Lake Wobegon dominance of a die in a set of n dice as the probability the

die rolls greater than the set average on that roll minus the probability the die rolls less

than that set average. We further define the Lake Wobegon dominance of the set to be the

dominance of the set’s least dominant die. We prove that such paradoxical dominance is

bounded above by .n � 2/=n regardless of the number of sides s on each die and the

maximum number of pips p on a side. A set achieving this bound is called Lake Wobegon

optimal. We give a constructive proof that Lake Wobegon optimal sets exist for all n � 3,

if one is free to choose s and p. We also show how to construct minimal optimal sets, that

is, a set that requires the smallest range in the number of pips on the faces.

We use the terms s-sided die/dice to refer to integer-valued, multinomial random vari-

ables where the probability of each value is of the form k=s, where k and s are natural

numbers and k � s. Furthermore, the integer value on each side—the number of its pips—

is in the set f1; 2; : : : ; pg, and is possibly the same on different sides. We denote the num-

ber of pips on the sides in non-decreasing order, for instance two s D 6; p D 7 dice are:

f1; 2; 3; 4; 4; 7g and f2; 2; 4; 5; 5; 6g. Occasionally we use the alternative notation of listing

the unique number pips followed by the probability with which they occur in parenthesis.

Using this notation, the two dice are

�

1

�1

6

�

; 2

�1

6

�

; 3

�1

6

�

; 4

�1

3

�

; 7

�1

6

��

and

�

2

�1

3

�

; 4

�1

6

�

; 5

�1

3

�

; 6

�1

6

��

:

Since we consider physical dice as random variables, we consider them distinguish-

able if and only if their set of pip values differ; that is, we do not care about the ar-

rangement of pips on the physical die. The number of distinguishable dice, according

to our convention, is the number of different sets of p objects chosen s at a time with

replacement, that is,�sCp�1

s

�

. A roll of a set of dice is an independent uniformly dis-

tributed random selection of the integer pip value of each die in the set. Let Xi be the

pip values of the individual dice in a roll, and let A denote the set average A on a given

roll. For each die Xi in a set, we define its individual Lake Wobegon dominance, Di ,

as the probability that Xi is greater than the average A minus the probability that it is

smaller:Di D P ŒEi > 0� � P ŒEi < 0�;

where

Ei D nXi �nX

j D1

Xj :

Much as with Efron’s nontransitive dice, we do not care by how much a die is above or

below the set mean, just the probability of those events. The dominance of a die may be

negative.

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34. Lake Wobegon Dice 251

We define the Lake Wobegon dominance of a set of dice to be the minimum, D, of the

Lake Wobegon dominance of the individual dice in the set: D D mini Di . It is possible

that the Lake Wobegon dominance of a set of dice is negative.

Finally, a set of dice is said to be Lake Wobegon if its Lake Wobegon dominance D is

greater than zero. Thus, a set of dice is Lake Wobegon if the probability that each die in

the set rolls greater than the set’s average on that roll, A, is greater than the probability that

the die rolls less than A.

A simple example of a Lake Wobegon set is ff1; 2; 2g; f1; 2; 2g; f1; 2; 2gg. Here n D 3,

s D 3, p D 2 and the probability each die (i D 1; 2; 3) exceeds the group average on a roll

is P ŒEi > 0� D 10=27, the probability it is less than the average is P ŒEi < 0� D 8=27 and

the probability it equals the average is P ŒEi D 0� D 9=27 (as can be seen by enumerating

the sn D 27 combinations of faces.) Thus the dominance of each die is Di D 2=27 > 0,

and the dominance of the set is D D 2=27 > 0.

The Lake Wobegon property does not imply nontransitive dominance: the set ff1; 2; 2g,

f1; 2; 2g, f1; 2; 2gg is Lake Wobegon, but clearly not nontransitive, since the dice are iden-

tical. Conversely, a nontransitive set need not be Lake Wobegon, as illustrated by the set

ff1; 1; 6; 6; 8; 8g; f3; 3; 5; 5; 7; 7g; f2; 2; 4; 4; 9; 9gg, shown in Figure 34.1.

If a > 0, it is straightforward to verify that the affine transformation x ! ax C b on

the pip values preserves the Lake Wobegon dominance of each die, as well as the Lake

Wobegon dominance of the set. Such transformations, then, yield a natural partition of

the set of finite sets of multinomial random variables into equivalence classes. Thus, in

the reminder of this paper, and without loss of generality, we demand that a 1 appear

somewhere in every set of dice.

Lake Wobegon optimality

We begin by finding an upper bound for the Lake Wobegon dominance D of a set com-

prised of n dice.

Optimality bound An upper bound for the Lake Wobegon dominance of a set of n

dice is .n � 2/=n.

Proof. In every roll at least one die will be below average, unless all dies equal the average.

(The latter case does not improve the Lake Wobegon dominance of any die above 0, so it

cannot improve the Lake Wobegon dominance of the set above 0 either.) Thus an upper

bound of the Lake Wobegon dominance of the set occurs when in every roll only one die

is below average, n � 1 are above average, and each die is below average in exactly 1=n of

the rolls. This yields an upper bound for Lake Wobegon dominance of

n � 1

n� 1

nD n � 2

n:

A set of dice is Lake Wobegon optimal if the upper bound n�2n

is achieved. We have

proved also that if a set of dice is Lake Wobegon optimal, then exactly one die is below

average on each roll and every die is below average equally often. The next result shows

that Lake Wobegon optimality is achievable.

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252 Part VI. Cards and Probability

Optimality existence For n � 3 the set of n dice:

die 1 W˚

pmax � 1 .1/

die 2 W˚

pmax

�12

�

; pmax � n�

12

�

die 3 W˚

pmax

�23

�

; pmax � n.n � 1/�

13

�

� � �die n W

˚

pmax

�n�1

n

�

; pmax � n.n � 1/n�2�

1n

�

;

(34.1)

where

pmax D 1 C n.n � 1/n�2; (34.2)

is Lake Wobegon optimal.

Proof. We prove every die in the set is below the roll average with probability 1=n and

otherwise above the roll average.

Since dice are independent random variables, the probability that die i rolls its lowest

side and all die k, k > i roll their highest side is the product of the probabilities of each

die rolling the given side, with the odds given in parentheses in (34.1),

P�

pi D pmax � n.n � 1/i�2 and pk D pmax for k > i�

D 1

i

i

i C 1� � � n � 1

nD 1

n: (34.3)

Let j < i; j ¤ 1 be the die showing the next lowest value in the roll besides die i . (The

special cases when i D 1 or j D 1 are analogous.) Showing that die j is above average

proves that all dice except for i are above average. The largest mean that this roll can have

occurs when all dice except for i and j show their largest side. This average value is:

A D 1

n

�

.pmax � 1/ C�

pmax � n.n � 1/j �2�

C�

pmax � n.n � 1/i�2�

C .n � 3/pmax

�

D pmax � .n � 1/j �2 � .n � 1/i�2 � 1

n:

We subtract A from the pip value of die j , pmax � n.n � 1/j �2, to confirm that die j is

indeed above the average of the roll:

pmax � n.n � 1/j �2 � A D �.n � 1/j �1 C .n � 1/i�2 C 1

n� 1

n:

Thus all n � 1 dice are above average on this roll, except the single die i .

We show next that the Lake Wobegon optimal set just produced is minimal in the sense

that in any optimal set of n dice at least one side shows a value greater than or equal to

pmax.

Minimal pips For fixed n, every Lake Wobegon optimal set has at least one side show-

ing the pip value pmax D 1 C n.n � 1/n�2 or greater.

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34. Lake Wobegon Dice 253

Proof. For die i , let li ; ui ; for i 2 f1; : : : ; ng be the minimum (respectively maximum)

number of pips on a side. Moreover let the dice be sorted so that i < j implies li � lj . For

each die i , we define its spread, �i , as

�i D ln � li C 1: (34.4)

Our proof proceeds as follows: First we show that maxj uj > ln. This proves that the

spread of a die bounds the difference between the maximal number of pips in any side and

the number of pips in the smallest side of die i . Afterwards, proving that �1 � pmax � 1

completes our proof.

To verify maxi ui > ln consider the roll fu1; : : : ; un�1; lng. Since the dice are Lake

Wobegon optimal, only one die can be below the average of the roll. Assume die .j ¤ n/

is that one die. This would imply uj < ln which would, in turn, imply that in any roll, die

n would be higher than die j , and thus die n could never be the sole die below the average.

Such a case is not possible because the set is Lake Wobegon optimal, thus die n must be

the one below the average and u1; : : : ; un�1 > ln, that is

minfu1; : : : ; un�1g � ln C 1: (34.5)

We complete the proof by computing a lower bound on �1 recursively. First we compute a

lower bound on �n�1 then compute a lower bound on �i ; i 2 f1; : : : ; n� 2g as a function

of �iC1.

Initial case. Bound on �n�1

Consider the roll fu1; : : : ; un�2; ln�1; lng. The only die below average is die n � 1 since

ln�1 � ln < u1; : : : ; un�1. Thus, since die n is above average, using (34.5) and (34.4) we

obtain the following:

nln > u1 C � � � C un�2 C ln�1 C ln

> .n � 2/.ln C 1/ C ln�1 C ln D nln C .n � 1/ � .ln � ln�1 C 1/;

hence

�n�1 D ln � ln�1 C 1 > n � 1;

or

�n�1 � n: (34.6)

Recursive step. Bound on �i ; i 2 f1; 2 ; : : : ; n � 2g as a function of �iCi

Consider the roll fu1; : : : ; ui�1; li ; liC1; uiC2; : : : ; ung. The only die below average is die

i since li � liC1 � ln � u1; : : : ; un. Thus since die i C 1 is above average, and ln � un,

using (34.5) and (34.4) yields a lower bound on �i as a function of �iCi :

nliC1 > u1 C � � � C ui�1 C li C liC1 C uiC2 C � � � C un

> .n � 2/.ln C 1/ C li C liC1 D liC1 C .n � 1/.ln C 1/ � .ln � li C 1/;

hence

�i D ln � li C 1 > .n � 1/.ln � liC1 C 1/ D .n � 1/�iC1;

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254 Part VI. Cards and Probability

or

�i � .n � 1/�iC1: (34.7)

Solving the recursion for �i defined by (34.6) and (34.7) yields

�i � n.n � 1/n�i�1;

and thus, by (34.2), we conclude

�1 � n.n � 1/n�2 D pmax � 1:

Figure 34.2 plots n�2n

and pmax. For large n, n�2n

approaches 1, so for large n, every

die in an optimal set is nearly always above average.

n 2

n

n(n – 1)n–2 + 1

3 4 5 6 7 8 90.

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.

100

101

102

103

104

105

106

107

108

109

1010

n

n2

n

pm

ax

Figure 34.2. The maximum value of Lake Wobegon dominance of a set .n� 2/=n (left, linear axis)

and the maximum number of pips on any side in the set, pmax (right, logarithmic axis).

If it is not required that all dice in a set have the same number of sides, then having one

die with one side, another with two sides, up to one die with n sides is sufficient to have a

Lake Wobegon optimal set that is minimal. Another way to obtain the required probabilities

is with rotating game wheels. (See Figure 34.3.) There the angles of the sectors correspond

to probabilities, and the number in each sector is the payoff.

Minimal Lake Wobegon optimality

Figure 34.4 shows the optimal set of dice ff6; 6; 6; 6; 6; 6g,f4; 4; 4; 7; 7; 7g,f1; 1; 7; 7; 7; 7gg.

It is the unique optimum set out of the 115,264,382 distinguishable and non-equivalent sets

of three, six-sided dice with up to seven pips per side. It also has three remarkable minimal

properties:

Property 1: No set with fewer than than three dice is Lake Wobegon (and hence Lake

Wobegon optimal), so n D 3 is minimal.

Proof. With only two variables, n D 2, the Lake Wobegon dominance is at most zero

because on every roll either both variables are equal to the average, or one variable is

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34. Lake Wobegon Dice 255

n D 5

320316

321

241

321

1

321321

301

n D 6

37503745

3751

3721

3751

3601

3751

3001

3751

1

3751

n D 7

54 43254 426

54 433

54 391

54 433

54 181

54 433

52 921

54 433

45 361

54 433

1

54 433

Figure 34.3. Lake Wobegon Optimal game wheels. Each wheel is divided into sectors. The proba-

bilities of the dark-colored sections are 11 ; 1

2 ; 13 ; : : : ; 1

n .

above the average and one variable below the average. Thus the sum of the Lake Wobegon

dominance of both variables is zero, and hence the Lake Wobegon dominance of the set is

non-positive.

Property 2: No set of dice with fewer than seven pips per side is Lake Wobegon optimal,

so p D 7 is minimal.

Proof. The fact that n D 3 is minimal together with the requirement that a side shows at

least pmax pips, show that p D 1 C n.n � 1/n�2 D 7 is minimal.

Property 3: No set of three dice with fewer than six sides is Lake Wobegon optimal, so

s D 6 is minimal.

Proof. If a set of n dice is Lake Wobegon optimal, then exactly 1=n of all roll combinations

have each individual die below the roll average. The total number of roll combinations is

sn. Thus sn � 0 mod n. This means for three Lake Wobegon optimal dice the minimal

number of sides is either three or six. A separate counting argument for each of the 27

possible rolls of three, three-sided dice shows that it is impossible to assign pip values such

that each die is below the roll average for exactly nine rolls.

6

6

6

6

6

7

4

7

7

44

1

1

7

7

776

Figure 34.4. Lake Wobegon optimal set of dice for n D 3; s D 6, and p D 7. This set is minimal in

three distinct senses: n, s and p.

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256 Part VI. Cards and Probability

Concluding remarks

We can invert the pip values on a Lake Wobegon set to produce a new set where the Lake

Wobegon dominance of each variable has opposite sign as in the original set. In particular,

inverting a Lake Wobegon optimal set produces a Lake Wobegon pessimal set where each

die rolls below average with probability .n � 1/=n.

An outstanding problem is to find the minimal optimal sets in the sense of the smallest

number of sides s required for a given n. We conjecture these sets may be those in (34.1)

that are also minimal in the p sense and that require s D LCMŒ1; 2; : : : ; n� sides.

Lake Wobegon dice, and Lake Wobegon distributions more generally, could provide

models for non-cooperative, non-zero-sum games [1]. We can imagine that colluding eco-

nomic agents might engage in policies that each agent properly predicts will provide indi-

vidual benefit above the group average.

Acknowledgment The authors would like to thank Brad Efron for words of encourage-

ment.

Bibliography

[1] D. Fudenberg and J. Tirole, Game theory, MIT Press, Cambridge MA, 1991.

[2] M. Gardner, On the paradoxical situations that arise from non-transitive relations,

Scientific American 231, (1974) 120–125; available at dx.doi.org/10.1038/

scientificamerican0574-120.

[3] C. M. Grinstead, On medians of lattice distributions and a game with two dice, Comb. Probab.

Comput. 6, (1997) 273–294; available at dx.doi.org/10.1017/S0963548397003015.

[4] G. Keillor, Lake Wobegon Days, Penguin, New York NY, 1990.

[5] K. O. May, Intransitivity, utility and the aggregation of preference patterns, Econometrica 22

(1954) 1–13; available at dx.doi.org/10.2307/1909827.

[6] R. P. Savage, Jr., The paradox of nontransitive dice, Amer. Math. Monthly 101, (1994) 429–436;

available at dx.doi.org/10.2307/2974903.

[7] T. L. Tenney and C. C. Foster, Non-transitive dominance, Math. Mag. 49, (1976) 115–120; avail-

able at dx.doi.org/10.2307/2690186.

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35Martin Gardner’s Mistake

Tanya Khovanova

Martin Gardner was amazingly accurate and reliable. That he made a mistake is simply

testimonial to the difficulty of this particular problem, which appeared in 1959 and was

republished in [3]:

Mr. Smith has two children. At least one of them is a boy. What is the proba-

bility that both children are boys?

Mr. Jones has two children. The older child is a girl. What is the probability

that both children are girls?

Mr. Jones has failed to stir any controversy, so we ignore him and his two children [5].

Instead, we concentrate on Mr. Smith. Here is the solution that Martin Gardner published

with the problem:

If Smith has two children, at least one of which is a boy, we have three equally

probable cases: boy-boy, boy-girl, girl-boy. In only one case are both children

boys, so the probability that both are boys is 1=3.

The corrected solution

Later Martin Gardner wrote a column titled “Probability and Ambiguity,” which was also

republished (in [4]). In this column Gardner corrects himself, writing “. . . the answer de-

pends on the procedure by which the information “at least one is a boy” is obtained.”

He suggested two potential procedures.

(i) Pick all the families with two children, one of which is a boy. If Mr. Smith is

chosen randomly from this list, then the answer is 1=3.

(ii) Pick a random family with two children; suppose the father is Mr. Smith. Then

if the family has two boys, Mr. Smith says, “At least one of them is a boy.” If he

has two girls, he says, “At least one of them is a girl.” If he has a boy and a girl

he flips a coin to choose one or another of those two sentences. In this case the

probability that both children are the same sex is 1=2.

Thus, the original problem without a specified procedure is ambiguous.

Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 20–24.

257

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258 Part VI. Cards and Probability

More procedures

Let us call procedure (i) “boy-centered,” because from the start we know that we are talking

about boys. Correspondingly, we call (ii) “gender neutral.”

Gardner wanted to emphasize that the problem is ambiguous. For him it was enough

to show two different procedures leading to different answers. However, there are many

procedures. Here are two more that demonstrate the full range of ambiguity.

(iii) Suppose that Mr. Smith wants to brag about his sons and will always mention

as many as he can. In this case the procedure might be the following: If he has

two boys, he says, “I have two boys.” If he has one son, he says “At least one of

them is a boy.” In this case the answer to the problem is 0.

(iv) Suppose Mr. Smith dislikes boys, and wants to de-emphasize the number of

boys he has. In this case the procedure might be the following: If he has two

boys, he says, “At least one of them is a boy.” If he has a boy and a girl, he says,

“I am the proud father of a girl.” In this case the answer is 1.

We challenge the reader to invent a procedure for any given answer between 0 and 1.

Tuesday-child problem

Fast-forward to 2010. In a talk at the 9th Gathering for Gardner, Gary Foshee posed the

following problem:

I have two children. One is a boy born on a Tuesday. What is the probability

that I have two boys?

This is Martin Gardner’s Two-Children problem with an extra twist. Before discussing the

solution, let us agree on some basic assumptions:

� Sons and daughters are equally probable—not exactly true, but a reasonable approx-

imation.

� Twins do not exist. Not only is the proportion of twins in the population small,

but, because they are usually born on the same day, twins might complicate the

calculation.

� All days of the week are equally probable birthdays. While this isn’t actually true—

for example, assisted labors are unlikely to be scheduled for weekends—it is a rea-

sonable approximation.

History repeats itself

Just as occurred with the classical Two-Children problem, many mathematicians have been

fighting for the wrong solution. This is their argument:

Each child can be one of two genders and can be born on one of seven days of the week.

Thus, gender and day present 14 equally probable cases for each child. That, in turn, makes

each two-children family belong to one of 196 equally probable cases. When we restrict

all possible cases to the given information that one child is a boy born on a Tuesday, we

get 27 equally probable cases. We can divide these cases into several groups, as follows:

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35. Martin Gardner’s Mistake 259

� 7 cases, where the first child is a son born on a Tuesday and the second child is a

daughter.

� 7 cases, where the second child is a son born on a Tuesday and the first child is a

daughter.

� 6 cases, where the first child is a son born on a Tuesday and the second child is a son

not born on a Tuesday.

� 6 cases, where the second child is a son born on a Tuesday and the first child is a son

not born on a Tuesday.

� 1 case, where both children are sons born on a Tuesday.

These are a total of 27 equally probable cases, 13 of which correspond to two sons.

Thus the probability must be 13=27.

This incorrect solution was widely published, for example, it appeared in Devlin’s An-

gle [1]. But, like Martin Gardner, Devlin corrected himself in his next column [2].

Procedures

The ambiguity that Martin Gardner found in the Two-Children problem is also present in

the Tuesday-Child problem. To resolve the ambiguity, we need to specify the procedure by

which the information was obtained. Here we present just four procedures. The calculations

are based on 196 equally probable cases for different combinations of gender and the day

of the week.

Gender-neutral, day-of-the-week-neutral procedure In this scenario, a father of two

children is picked at random. He is instructed to choose a child by flipping a coin. Then he

has to provide true information about the chosen child in the following format: “I have a

son/daughter born on a Mon/Tue/Wed/Thu/Fri/Sat/Sun.” If his statement is, “I have a son

born on a Tuesday,” what is the probability that the other child is also a son?

The solution is the following: A father has two daughters in 49 cases. Such a father

will make the above statement with probability zero. A father has a son and a daughter

in 98 cases, and will produce the above statement with probability 1=14: with probability

1=2 the son is chosen over the daughter and with probability 1=7 Tuesday is the birthday.

A father has two sons in 49 cases, and he will make the statement with probability 1=7.

The father of two sons is twice as likely to make the statement as the father of a son and a

daughter, but there are half as many such fathers. Thus the probability is 1=2 that the other

child is also a son. This is the most symmetric scenario and produces the most symmetric

answer.

Boy-centered, day-of-the-week-neutral procedure A father of two children is picked

at random. If he has two daughters, he is sent home and another father is picked until one

is found with at least one son. If he has just one son, he is instructed to provide true infor-

mation on his son’s day of birth. If he has two sons, he has to choose one at random. His

statement will be, “I have a son born on a Mon/Tue/Wed/Thu/Fri/Sat/Sun.” If his statement

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260 Part VI. Cards and Probability

is, “I have a son born on a Tuesday,” what is the probability that the other child is also

a son?

The solution is as follows. A father has a son and a daughter in 98 cases, and will

produce the statement above with probability 1=7. If he has two sons (49 cases), the prob-

ability of the statement above will likewise be 1=7. The father of two sons is exactly as

likely to make the statement as the father of a son and a daughter, but there are half as

many such fathers. Thus the probability is 1=3 that the other child is also a son.

This scenario corresponds to the original procedure leading to the first solution of Mar-

tin Gardner’s Two-Children problem. Unsurprisingly, the answer is the same.

Boy-centered, Tuesday-centered procedure Now let us consider the third scenario. A

father of two children is picked at random. If he doesn’t have a son born on a Tuesday,

he is sent home and another father is picked at random until one who has a son born on a

Tuesday is found. He is instructed to tell you, “I have a son born on a Tuesday.” What is

the probability that the other child is also a son?

Here is the solution. A father of a boy and a girl has a son born on a Tuesday in 14

cases. He will make the statement in question with probability 1. A father of two sons has

a son born on a Tuesday in 13 cases. He too is guaranteed to make the statement. Thus the

probability is 13=27 that the other child is a son.

This procedure corresponds to the procedure many mathematicians assume while solv-

ing the Tuesday-Child problem. This implicit assumption is the source of the erroneous

solution.

Gender-neutral, Tuesday-centered procedure For completeness let us consider a fourth

scenario. A father of two children is picked at random. If he doesn’t have a child who is

born on a Tuesday, he is sent home and another father is picked at random until one who

has a child born on a Tuesday is found. He is instructed to tell you, “I have a son/daughter

born on a Tuesday.” If both of his children were born on Tuesdays, he has to pick one at

random. If his statement is, “I have a son born on a Tuesday,” what is the probability that

the other child is also a son?

Here is the solution. A father of two daughters will have a child born on a Tuesday in

13 cases. He makes the statement in question with probability 0. A father of a boy and a

girl has a child born on a Tuesday in 26 cases. The probability that he makes the statement

is 1=2. A father of two sons will have a son born on a Tuesday in 13 cases. The probability

that he makes the statement is 1. The father of two sons is twice as likely to make the

statement as the father of a son and a daughter, but there are half as many such fathers.

Thus the probability is 1=2 that the other child is also a son.

In this procedure we added an additional constraint on the families with two children—

that a child was born on a Tuesday—that’s independent of gender. Not surprisingly, the

answer is 1=2.

Back to Gardner

Many people I argue with don’t want to listen to me. They refer to Martin Gardner as the

final authority in support of their wrong solution. Gardner was a great thinker; he corrected

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35. Martin Gardner’s Mistake 261

his mistake. I urge those who do not agree with me to trust Martin Gardner. Revisit and

rethink this problem together with him.

Acknowledgment I am grateful to all my friends and colleagues who discussed the prob-

lem with me and supported me in writing about it for my blog [6, 7, 8]. I am especially

grateful to Alexey Radul [9] and Peter Winkler [10] who contributed essays on the subject

to my blog. I also thank Sue Katz and Julie Sussman, P.P.A., for editing.

Bibliography

[1] K. Devlin, Probability can bite, (2010); available at www.maa.org/devlin/

devlin_04_10.html.

[2] K. Devlin, The problem with word problems, (2010); available at www.maa.org/devlin/

devlin_05_10.html.

[3] M. Gardner, Nine more problems in The Second Scientific American Book of Mathematical Puz-

zles and Diversions, Univ. of Chicago Press, 1987, 152–164.

[4] ——, Probability and ambiguity in The Second Scientific American Book of Mathematical Puz-

zles and Diversions, Univ. of Chicago Press, 1987, 220–232.

[5] T. Khovanova, Mr. Jones, (2010); available at blog.tanyakhovanova.com/?p=313.

[6] ——, A son born on Tuesday, (2010); available at blog.tanyakhovanova.com/?p=221.

[7] ——, Sons and Tuesdays, (2010); available at blog.tanyakhovanova.com/?p=233.

[8] ——, A Tuesday quiz, (2010); available at blog.tanyakhovanova.com/?p=247.

[9] A. Radul, Shannon entropy rescues the Tuesday child, (2010); available at

blog.tanyakhovanova.com/?p=254.

[10] P. Winkler, Conditional probability and “He said, she said”, (2010); available at

blog.tanyakhovanova.com/?p=234.

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VIIOther Aspects of Martin

Gardner

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36Against the Odds

Martin Gardner

Luther Washington was a friendly, shy, intelligent boy, the oldest of five children who lived

with their parents in Butterfield, Kansas. He was one of sixteen African Americans who

attended Butterfield Central High. His father owned a small grocery store in the town’s

black district. His mother cooked dinners for one of the town’s bankers.

For some reason, which his parents never understood, Luther was fascinated by num-

bers. One day, when he was ten, he surprised his father by saying: “Dad, I’ve discovered

an easy way to tell if a big number can be divided by four or eight and not have anything

left over.”

The family was having breakfast. “Tell us about it,” said Mr. Washington.

“Well,” said Luther, “if the last two digits”—his grade school teacher had taught him

to distinguish digits from numbers—“can be divided by four with nothing left over, the big

number also can be divided by four. Otherwise, it can’t. And if the big number’s last three

digits are a product of eight, so is the big number.”

“Interesting,” said Mrs. Washington. “But I can’t see how that could be of any use to

anybody.”

“I don’t care whether it’s any use or not. I just think it’s neat.”

“I wish you wouldn’t waste so much time on such things,” said Mr. Washington. “All

the math you’ll ever need in life is knowing how to add, subtract, multiply, and divide, and

how to make change.”

“Maybe so,” said Luther. “But there’s something wonderful and mysterious about num-

bers.”

In high school Luther was quick to learn algebra and geometry even though his teacher,

Miss Perkins, seemed bored by what she taught. The fact is, she really was bored. Her ma-

jor, at a Kansas teacher’s college, was American history, but when one of the school’s two

math teachers quit to take a better paying job as a used car salesman, Miss Perkins was

asked to take over his classes. She reluctantly agreed. By studying the textbook carefully

each night she managed fairly well to stay ahead of her students.

Miss Perkins was a devout Baptist. She felt no hostility toward blacks in general. After

all, they were God’s children and she believed that all persons are equally precious in the

eyes of the Lord. On the other hand, she was convinced that the intelligence of blacks was

considerably below that of whites. It had to be something in their genes.

Reprinted from The College Mathematics Journal, Vol. 32, No. 1 (Jan. 2001), pp. 39–43.

265

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266 Part VII. Other Aspects of Martin Gardner

Among the forty-two students in her geometry class, five were black—three girls and

two boys. They would sit silently at their desks, their dark eyes fixed on hers. They rarely

asked a question. When they did it was usually, in her opinion, a foolish one. If she asked

who had solved a problem, she almost never called on a black student whose hand was

raised.

One morning, after Miss Perkins had demonstrated on the blackboard how Euclid

proved that the three angles of a triangle add to a straight angle, Luther was the only

student to wave a hand. His classmates looked as if they were falling asleep.

“Yes, Luther, do you have a question?”

“No. But I thought of a simpler way to prove the theorem.”

Miss Perkins raised her eyebrow and peered at Luther over her bifocals. “Would you

like to come to the blackboard and explain it to the class?”

Luther stood up and walked to the blackboard. He erased the diagram Miss Perkins had

chalked. After drawing a large triangle with irregular sides, he took a pencil from his shirt

pocket and pressed it against the blackboard along the inside of the triangle’s base.

“As you can see,” he said, “the pencil points to the right.” Luther slid the pencil along

the triangle’s base until its point touched the triangle’s right corner. Keeping the point fixed

at the corner, he swung the pencil clockwise until it was alongside the triangle’s right side.

He then moved the pencil upward along the side until its point touched the triangle’s top

corner. Again he rotated the pencil clockwise until it paralleled the triangle’s left side. He

slid the pencil down to the triangle’s lower left corner, and rotated the pencil once more. It

was now back where it started.

“See,” Luther said as he turned his head to face the class, “the pencil now points left.

Obviously it has rotated a hundred and eighty degrees. This proves that the three angles

add to a straight angle.”

The students were all now wide awake. About a dozen, including the three black girls,

clapped. The others sat silent, waiting for Miss Perkins’ reaction.

Miss Perkins looked a bit flustered. She knew how to demonstrate the theorem by

folding over the corners of paper triangles, but Luther’s method caught her by surprise.

“I’m not sure that’s really a proof,” she said finally.

“Well, it is,” said Luther. “The pencil acts like a vector. The best thing about it is you

can apply it to any polygon, with any number of sides. It proves the inside angles must add

to a multiple of a straight angle. And the same thing is true about any polygon’s exterior

angles.”

Miss Perkins wasn’t sure she grasped what Luther was saying. “What are vectors and

exterior angles?” she thought. Fortunately, the bell rang and the class was over before she

could make any other comments. “That boy is going to be a trouble maker,” she said to

herself. “He’s not nearly as smart as he thinks he is.”

A few days later, during a study period, Miss Perkins walked by Luther’s desk. He

had been doodling on a sheet of paper. Scattered over the sheet were numerous ticktacktoe

patterns, their cells filled with X’s and O’s.

“You’re supposed to be working on your assignment.”

“I’ve already finished it, Miss Perkins. I’m trying to figure out whether the first player

can always win at ticktacktoe if both players do their best. Maybe the game’s a draw. I do

know the second player can never win, no matter how big the field is or how many marks

he has to get in a straight line.”

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36. Against the Odds 267

Miss Perkins snatched up the sheet and crumpled it into a ball. “When you’re in my

class,” she said sternly, “I expect you to work on mathematics and nothing else. If you’ve

finished your assignment, come to my desk and I’ll give you some more problems.”

“Yes, Miss Perkins.”

To Miss Perkins’ chagrin, Luther had near perfect scores on every test she gave. But

she was so annoyed by what she considered his “uppity” way of talking to her that she gave

him a B grade for the semester.

In the middle of each school year Butterfield Central High held contests to decide who

among its seniors excelled in each of the school’s subjects. Winners were given a medal

and made members of an honorary society called the B Club.

Luther entered the math contest. Miss Perkins was annoyed when she learned that he

was the first student in the B Club’s history to solve every problem correctly. The test had

been designed and the papers graded by the school’s principal.

Members of the B Club met each Friday, after school. Miss Perkins was the club’s

sponsor. When Luther showed up at the next meeting, she told him he had been appointed

chairman of the club’s peanut-sacking committee.

“Who appointed me?” Luther asked.

“I did.”

“What am I supposed to do?”

“You’ll be in charge of a group of club members. They meet after school every Friday

to make sure all the peanuts are put in paper sacks to sell at Saturday basketball games.”

“How long will that take?”

“About two hours.”

Luther groaned. Time after school was precious. He had found in the town’s library a

book titled Mathematics and the Imagination. It was an exciting book, and he was eager to

get home to finish it.

“I don’t want to sack peanuts,” Luther said. “Am I allowed to resign from the B Club?”

Miss Perkins was so furious when Luther stalked out of the room that next Monday she

sent him to the principal’s office with a note saying he had refused to sack peanuts, and

that he had been very unpleasant in the way he had spoken to her.

Now it so happened that the principal knew a great deal more about mathematics than

Miss Perkins. Math had been his undergraduate major at the University of Michigan before

he went on to do graduate work in education.

The principal had been impressed by the unusual ways Luther solved the problems

on the test he had prepared. After discussing several mathematical topics with Luther, he

realized he was talking to a young man with deep insights into mathematical structures.

“Are you planning on college?”

“No, sir. My parents can’t afford it. Dad wants me to work in his grocery store. When

he’s too old to run the store, he expects me to take it over.”

“How are your other grades?”

“Not so good. I barely passed Latin with a D. And I got a C in English. I didn’t under-

stand any of Shakespeare’s plays. I was so bored by Ivanhoe that I couldn’t finish it.”

“Hmm.” The principal stared at the ceiling for a while before he asked: “Have you

applied for any scholarships?”

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268 Part VII. Other Aspects of Martin Gardner

Luther shook his head. “My folks don’t want me to go to college. They say it would be

a big waste of time. Dad says if I had a college degree it would only make me overqualified

for any kind of work I could get.”

“What do you make of Miss Perkins as a teacher?”

Luther hesitated for a moment. “She’s okay, I guess. I don’t think she likes mathemat-

ics, though. I know she doesn’t like me.”

Luther told the principal about the time she had grabbed the sheet on which he was

analyzing ticktacktoe, and how she told him he had to work only on math during her study

periods.

The principal laughed, then sighed and looked solemn. “She means well. But she

should have known that ticktacktoe is based on combinatorics and graph theory. Did you

finally solve the game?”

“I did when I got home. I drew a complete game tree. It’s not hard to do if you take

advantage of symmetries like rotations and reflections. The game’s a draw if both sides

make their best moves.”

“That’s right. You may be interested to know that Frank Harary, a well known graph

theorist, generalized the game in an interesting way. Have you head about polyominoes?”

Luther shook his head.

“They’re shapes made by attaching unit squares along their edges. In standard ticktack-

toe the goal is to be the first to form what is called a straight tromino. That’s three little

squares in a row. Harary varied the goal by making it any kind of polyomino on a field of

any size. For example, suppose the goal is to be the first to form the only other tromino,

three squares joined to make a right angle. It’s called the bent tromino. If a bent tromino is

the goal, who wins? Or is it also a draw?”

“I wish I’d thought of that,” said Luther.

“Harary also analyzed reverse games for low-order polyominoes. Who wins if the first

player forced to form a specified polyomino is the loser?”

“I can see how that could get very complicated,” said Luther. “I did think of analyzing

the reverse game for standard ticktacktoe. It’s a draw, too. The first player can’t win but he

can force a draw by going first in the center.”

The principal was interested. “Can you show me how he does that?”

For the next ten minutes Luther and the principal played reverse games of ticktacktoe.

“The chairman of the math department at Stanford University is an old friend,” said the

principal, as he pushed the sheets of paper to one side of his desk. “We were frat brothers at

Michigan. If you don’t mind, I’d like to send him your contest answers. Would you object

if Stanford offered you a full scholarship that included room and board?”

“No.”

“Would your parents object?”

“I don’t know.”

The chairman of Stanford’s math department needed only a glance at Luther’s un-

conventional ways of solving the test problems to know that he wanted Luther in his de-

partment. Even before he had a long phone conversation with the principal of Butterfield

Central High, the chairman had started the process for Luther to be given a full freshman

scholarship.

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36. Against the Odds 269

To Luther’s vast surprise and joy, his mother and father agreed to let him accept the

scholarship. He almost cried when his father embraced him and said how proud he was!

Luther’s scholarship was renewed each year, and on through graduate work until he

got his doctorate. Three years later he was made a full professor at Stanford. By then he

had published a dozen technical papers, and was much in demand as a lecturer at math

conventions.

Fast forward ten years to 1970. Luther was awarded the Fields Medal, the equivalent in

mathematics of a Nobel prize. It was for his brilliant solution of a long-standing conjecture

in Diophantine analysis.

A news release about the award went to the Butterfield Times. The paper ran the story

with Luther’s picture on the front page.

Miss Perkins had retired from teaching after she married Mr. Jones, the school’s bas-

ketball coach. During breakfast, her husband handed her the newspaper’s front section.

“It says here that Professor Washington grew up in Butterfield and attended Central

High. He must have been in one of your classes.”

Miss Perkins adjusted her spectacles and studied Dr. Washington’s picture for a full

minute. “I don’t recognize him,” she said, frowning. “There were lots of black boys in my

classes. They all looked so much alike. I wonder if he really deserved that prize. You know

how affirmative action works these days. Would he have been given that award if he’d been

white?”

Note Frank Harary’s generalization of ticktacktoe to polyomino patterns was the topic

of my Scientific American column, April 1979. The column is reprinted as Chapter 17 in

Fractal Music, Hypercards, and More (W. H. Freeman, 1992).

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37A Modular Miracle

John Stillwell

The following piece is one of a series of results that I called “modular miracles” in a 2001

article in the Monthly. It depends on an earlier miracle, which I must therefore briefly ex-

plain. An 1857 theorem of Kronecker [3] relates the famous modular function, j , to unique

prime factorization in the field Q.p

�D/, where D is a positive integer. Kronecker’s theo-

rem states that unique prime factorization holds for the integers � in Q.p

�D/ only when

j.�/ is an ordinary integer for each such � .

Take, for example, the Gaussian integers, a C bp

�1 for ordinary integers a and b.

These are the integers of the field Q.p

�1/, and they were shown to possess unique prime

factorization by Gauss in 1832. Gauss also found several other values of D for which

Q.p

�D/ has unique prime factorization, the largest of which was D D 163. (It is now

known that no larger such D exists.) The number .1 Cp

�163/=2 is an example of an

integer in this field. It then follows, by Kronecker’s theorem, that j..1 Cp

�163/=2/ is an

ordinary integer, and indeed

j..1 Cp

�163/=2/ D .�640320/3:

The function j is defined by geometric conditions that I will not explain here; it is only

necessary to know that j has a series expansion beginning

j.�/ D q�1 C 744 C 196884q C 21493760q2 C � � � ;

where q D e2i�� .

The Numerical Miracle

Hermite (1859) noticed a curious numerical consequence of Kronecker’s theorem on the

values of j.�/:

e�p

163 D 262537412640768744;

(an integer!) correct to 12 decimal places [2].

This little known discovery of Hermite was exploited by Martin Gardner in an amus-

ing hoax edition of his column in Scientific American. On 1 April 1975 Gardner [1]

Reprinted from The American Mathematical Monthly, Vol. 108, No. 1 (Jan. 2001), pp. 70–76.

271

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272 Part VII. Other Aspects of Martin Gardner

announced—among several other “sensational discoveries that have somehow or another

escaped public attention”—that

e�p

163 D 262537412640768744 exactly. (1)

He gave the announcement an extra coat of varnish by claiming that it settled a conjecture

of Ramanujan, supposedly made in a paper of 1914. The paper cited by Gardner does

indeed discuss near integers of the form e�p

n, but without claiming that they could be

integers, and without mentioning e�p

163. Still, in the pocket calculator days of 1975, it

was pretty hard to decide whether e�p

163 is an integer or not.

Its true value, as Hermite and Ramanujan knew, is the integer in (1) minus a very tiny

number (< 10�12).

In fact, putting � D .1 Cp

�163/=2 in q D e2i�� gives the tiny

q D ei���p

163 D �e��p

163;

and putting this q in

j.�/ D q�1 C 744 C 196884q C 21493760q2 C � � �

gives

j..1 Cp

�163/=2/ D �e�p

163 C 744 � tiny number;

and therefore

e�p

163 D integer � tiny number:

Bibliography

[1] M. Gardner, Mathematical Games, Scientific American 232 (April 1975), p. 126.

[2] C. Hermite, Sur la theorie des equations modulaires, C. R. Acad. Sci. Paris Ser. I Math. XLVIII

(1859) 940. Also in Oeuvres de Charles Hermite, Paris, Gauthier-Villars, 1905–17, vol. 2, pp.

38–82.

[3] L. Kronecker, Uber die elliptischen Functionen fur welche complexe Multiplication stattfindet,

Leopold Kronecker’s Werke, Chelsea Pub. Co., New York, 1968, vol. 4, pp. 179–183.

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38The Golden Ratio—

A Contrary Viewpoint

Clement Falbo

Introduction

Over the past five centuries, a great deal of nonsense has been written about the golden

ratio, ˆ D 1Cp

52

, its geometry, and the Fibonacci sequence. Many authors make claims

that these mathematical entities are ubiquitous in nature, art, architecture, and anatomy.

Gardner [4] has shown that the admiration for this number seems to have been raised to

cult status. Fortunately, however, there have been some recent papers, including Fischler [2]

in 1981, Markowsky [7] in 1992, Steinbach [9] in 1997, and Fowler [3] in 1982, that

are beginning to set the record straight. For example, Markowsky, in his brilliant paper

“Misconceptions about the Golden Ratio,” speaking about ˆ, says:

Generally, its mathematical properties are correctly stated, but much of what

is presented about it in art, architecture, literature and esthetics is false or

seriously misleading. Unfortunately, these statements about the golden ratio

have achieved the status of common knowledge and are widely repeated. Even

current high school geometry textbooks . . . make many incorrect statements

about the golden ratio. It would take a large book to document all the misin-

formation about the golden ratio, much of which is simply repetition of the

same errors by different authors.

It is remarkable that prior to Fischler’s and Markowsky’s papers, there seemed to have

been no set standards for obtaining measurements of artwork. Often, a proponent of the

golden ratio will choose to frame some part of a work of art in an arbitrary way to create

the appearance that the artist made use of an approximation of ˆ. Markowsky shows an

example in which Bergamini [1] arbitrarily circumscribes a golden rectangle about the

figure of St. Jerome in a painting by Leonardo da Vinci, cutting off the poor fellow’s arm

in order to make the picture fit.

It is frequently asserted that the golden ratio occurs in nature as the shape of spi-

rals in sea shells. We can easily test this claim by first providing a protocol for measur-

ing the spirals. One requirement should be to allow for some error in the measurements.

Reprinted from The College Mathematics Journal, Vol. 36, No. 2 (Mar. 2005), pp. 123–134.

273

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274 Part VII. Other Aspects of Martin Gardner

Markowsky [7] suggests an error bar of ˙2%, which seems to be quite adequate. Measur-

ing under this protocol, we find that spirals in sea shells do not generally fit the shape of

the golden ratio. This is true despite the numerous articles on the Internet and elsewhere, in

which pictures apparently have been stretched to fit the ˆ ratio—“stretching the truth”—so

to speak.

The golden ratio is associated with the Fibonacci sequence in a very simple way. The

sequence is an example of a quadratic recursive equation sometimes used to describe var-

ious scientific and natural phenomena such as age-structured population growth. In order

to define the general quadratic recursive formula, let x0, x1, p, and q be fixed positive

numbers, and for any integer n � 2, define xn as

xn D pxn�1 C qxn�2: (38.1)

Murthy [8] provides a number of theorems for this general recursive equation. It is clear

that many of the features that are proclaimed to be unique to the Fibonacci sequence are,

indeed, common to all second-order recursive equations. For example,

limn!1

xnC1

xn

D r; (38.2)

where r is the positive root of the quadratic equation

x2 � px � q D 0; (38.3)

obtained by assuming that xn D xn is a solution to (38.1). One of our objectives in this

paper is to show that if q D 1 and r is the limit in (38.2), then the pair .r; p/ has all of

the geometric and algebraic properties that are often ascribed as being unique to the pair

.ˆ; 1/. For example, we have r � p D 1=r corresponding to the property ˆ � 1 D 1=ˆ:

For equation (38.1) to be useful in describing aged-structured population growth in

plants and animals, the coefficients p and q must be determined by some niche or fecundity

properties of the organism being studied. In other applications, such as phyllotaxy, we may

use a second-order recursive equation such as (38.1) to predict and explain the evolution

of leaf placement on a stem in terms of maximizing the gathering of sunlight. However,

we should not expect the complexities of natural systems to yield to the easy-to-compute

Fibonacci sequence, and there seems to be no unbiased evidence favoring the Fibonacci

sequences over all other possible sequences. If one expends great effort in looking only

for this special sequence, then it may be perceived, whether or not it is there. This is

succinctly illustrated in terms of statistical analysis by Fischler [2], who shows that careless

computations and misused formulas produce the golden number when it isn’t there.

In a popular new book, Livio [6] draws upon the information developed in Markowsky

and others to discuss the protocol violations that are the source of claims that ˆ occurs in

classical architecture, such as the Parthenon and the Egyptian pyramids. Livio presents a

well-written explanation of misleading claims concerning these classics, as well as various

paintings and other art work. Livio calls the advocates for these claims the “golden num-

berists.” It seems, however, that he believes that certain constructions (such as Kepler’s

triangle) or arithmetic equations (such as a continued fraction representation of ˆ) are sig-

nificant and unique enough for him to subtitle his book “The Story of Phi, the World’s

Most Astonishing Number.”

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38. The Golden Ratio—A Contrary Viewpoint 275

Origins of the golden ratio

The golden ratio is the solution to a problem given by Euclid (c. 300 BC) in his Elements,

Book VI, Proposition 30:

To cut a given finite line in extreme and mean ratio.

That is, given a segment AE, find the point B for which AE=AB D AB=BE. Euclid had

already solved this in a previous theorem (Book II, Proposition 11). We show his construc-

tion in Figure 38.1 in order to generalize it later.

D C F

A M B E

1

1

1_2

1_2

Figure 38.1. The mean proportion AE W AB WW AB W BE.

Start with the unit square ABCD and let M be the midpoint of AB. Construct the line

segment MD. Draw a circle with center at M and radius MD so that it cuts�!AB at the point

E . So, MD D ME. In modern notation, we have the lengths,

ME Dp

5

2; MB D

1

2; and BE D

p5 � 1

2:

Now, since AE D AB C BE, or AE D 1 C BE, we can write: AE D .1 Cp

5/=2. Thus, we

can easily show that AE=AB D AB=BE.

The length AE, .1 Cp

5/=2, is denoted by ˆ, and is called the golden ratio, or the

divine proportion. In the above figure, the rectangle AEFD is called the golden rectangle.

The history of the golden ratio pre-dates Euclid. As early as 540 BC, the Pythagoreans had

studied it in their work with the pentagon. We discuss some of the ratios that appear in the

pentagon and all other odd polygons later.

The golden ratio ˆ can be used to construct a beautiful logarithmic spiral, shown in

Figure 38.2. This graph can be obtained by fitting the polar equation � D bec� to selected

points on the golden rectangle. We can, however, also plot a logarithmic spiral inscribed in

a rectangle with a ratio other than ˆ. Figure 38.3 shows a spiral with a 1.33 to 1 ratio.

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276 Part VII. Other Aspects of Martin Gardner

0.8

–0.2

0.2

0.4

0.6

–1 –0.5 0.5

(1.6182 = Ratio)

Figure 38.2. A spiral with the golden ratio.

0.8

–0.2

0.2

0.4

0.6

–0.5 0.5

–0.4

–0.75 –0.25 0.25 0.75 1

(1.333 = Ratio)

Figure 38.3. A spiral with a 1:33 to 1 ratio.

Spirals in sea shells

Basically, the two types of sea shells that many of us are familiar with are the cephalopod

(head-foot) and the gastropod (stomach-foot), or with apologies to my biologist friends,

the octopus and the snail. The nautilus, a cephalopod, is an octopus in a shell that consists

of a series of chambers, each sealed from the previous one. This animal lives in the latest

chamber with its eight tentacles sticking out. In Figure 38.4 is a photo that I took of a

longitudinal section of a nautilus.

The nautilus is definitely not in the shape of the golden ratio. Anyone with access to

such a shell can see immediately that the ratio is somewhere around 4 to 3. In 1999, I

measured shells of Nautilus pompilius, the chambered nautilus, in the collection at the

California Academy of Sciences in San Francisco. The measurements were taken to the

nearest millimeter, which gives them error bars of ˙1 mm. The ratios ranged from 1:24 to

1:43, and the average was 1:33, not ˆ (which is approximately 1:618). Using Markowsky’s

˙2% allowance for ˆ to be as small as 1:59, we see that 1:33 is quite far from this ex-

Figure 38.4. A digital photograph of a longitudinal section of a nautilus.

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38. The Golden Ratio—A Contrary Viewpoint 277

panded value of ˆ: It seems highly unlikely that there exists any nautilus shell that is within

2% of the golden ratio, and even if one were to be found, I think it would be rare rather

than typical.

A ratio of different color,p

2

A spiral of a different ratio is in the shape of the “silver ratio”p

2 to 1. (Some authors call

1 Cp

2 the silver ratio.) A spiral based onp

2 is considerably closer to the shape of the

nautilus than ˆ is (see Figure 38.5).

0.8

–0.2

0.2

0.4

0.6

–0.4

(1.41421 = Ratio)

–1 –0.5 0.5 1

Figure 38.5. A spiral with the ratiop

2 to 1.

It is well known that the golden rectangle can be subdivided into subrectangles (each

similar to the original) forming a kind of “spiral” of smaller and smaller similar rectangles.

This interesting property happens to be shared by every rectangle (except the square). For

example, if you take ap

2-by-1 rectangle and cut it in two (in the natural way), you get

two rectangles with sides in the same ratio. Continuing this, we get similar rectangles as

shown in Figure 38.6.

Figure 38.6. A rectangle and subrectangles withp

2 to 1.

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278 Part VII. Other Aspects of Martin Gardner

(This rectangle was actually used in a field study in biology. In an article in Science,

Harte, Kinzig, and Green [5] used Figure 38.6 when they wanted to determine a relation be-

tween species distribution and geographical area. They were interested in studying regions

of varying sizes, but they wanted to maintain similarity in the shape of the region so as not

to introduce a bias against rare species. One of their protocols was to start with a rectangle

of known dimensions and then to generate a sequence of smaller similar subrectangles in

which to collect samples. They chose the rectangle with the ratio ofp

2 to 1:)

The extreme mean ratio, generalized

Figure 38.6 illustrates the fact that the golden rectangle is not the only one that can be sub-

divided into similar subrectangles. Indeed, Fowler shows that we can construct an infinite

sequence of extreme mean ratios. What he does is to generalize the standard procedure for

constructing ˆ. He starts with a unit square ABCD and for any positive integer n, takes

the point Mn on�!AB, at a distance of n=2 from A. Then the segment MnD has lengthp

n2 C 4=2. Now rotating MnD about Mn he gets a circular arc that cuts�!AB at a point

E . (Mn is between A and E .) The length of AE is n Cp

n2 C 4=2, which he calls the

noem, or nth order extreme mean. That is, .MnE C AMn/=AB D AB=.MnE � AMn/.

If n D 1, the noem is ˆ, or the first order extreme mean, while if n D 2, the noem

is 1 Cp

2: In Figure 38.7a, n D 3, AM3 D 32

, and G is a point so that M3 is the

midpoint of the segment AG. AE D .3 Cp

13/=2, (the third order extreme mean) and

GE D .�3 Cp

13/=2, so AE=AB D AB=GE. One property of the nth order extreme mean

is that noem � 1=noem D n.

D C F

A M3 E 1 G B

noem

1

p_2

D C F

A M E B

p

r

p_2

(a) (b)

Figure 38.7. (a) The noem, with n D 3; (b) The poem, for any p > 0.

We can obtain the same result for any positive real number p, not just for an integer n,

as follows. After Fowler, we call our ratio the pth order extreme mean, or the poem. Its

construction is shown in Figure 38.7b. Start with any p-by-1 rectangle ABCD, AB D p.

Select the midpoint M of AB and construct the segment MD; its length isp

p2 C 4=2.

Rotate the radius MD about M until it cuts the line�!AB at the point E , giving the segment

AE with length .p

p2 C 4 C p/=2, denoted by r . Then BE D AE � AB; this length is

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38. The Golden Ratio—A Contrary Viewpoint 279

.�p Cp

p2 C 4/=2, which is 1=r: Thus, r � p D 1=r: The number r is the poem. Notice

that r is a root of the quadratic equation (38.3) when q D 1. Completing the rectangle

BEFC, as shown, we have that AE=EF D FE=BE.

Self-similar decompositions of rectangles of any ratio

The poem, defined above, has a simple interpretation. For any r > 1, and any rectangle R

with a ratio of r to 1, we can divide R into two smaller rectangles, one of which has the

same ratio r to 1, and this in turn can be so divided etc., as in Figure 38.8. The coordinates

of the vertices A4, A5, A6, are given below.

A4 D�

r � 1

rC 1

r3; 0

�

A5 D�

r � 1

r;

1

r2� 1

r4

�

A6 D�

r � 1

rC 1

r3� 1

r5;

1

r2

�

The vertices A1; A2; A3; : : : converge to a “center-point,”

�r3

1 C r2;

1

1 C r2

�

:

In order to show this, we look at the abscissas of the corners. First notice that the corner

points alternate in returning to a previous x value before moving on to a new x value. So,

skipping those that return to a previous abscissa, we have:

Point A2 A4 A6 . . .

Abscissa r � 1r

r � 1r

C 1

r3 r � 1r

C 1

r3 � 1

r5 . . .

The alternating geometric series

r � 1

rC 1

r3� 1

r5C � � � converges to

r3

1 C r2:

Similarly, the corner points return to a previous ordinate in an alternating pattern and they

converge to 1=.1 C r2/. Notice that, for any r (including ˆ, of course) the center point

determined by these infinite series is the intersection of the line through .0; 1/ and .r; 0/

and the line through the points .r � 1=r; 0/ and .r; 1/.

Other geometric properties

Here are some other common geometric properties that are not unique to ˆ.

Spirals An equiangular spiral can be drawn through the points A1; A2; A3; : : : in Figure

38.8. Just use the center point�

r3

1 C r2;

1

1 C r2

�

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280 Part VII. Other Aspects of Martin Gardner

(0,1)

A1=(0,0) A4

A6

A5

A2=(r–1/r,1)(r, 1)

(r, 0)

A3=r ,1_

r2

Figure 38.8. A rectangle of dimensions r to 1, with similar subrectangles.

and fit a polar equation for the logarithmic spiral through any two of these points. It is a

general property of all such spirals that the tangents to the spiral at any point make a fixed

angle with the rays from the center point.

Sum of the areas of all of the subrectangles If r > 1 and p D r � 1=r , then the sum

of the areas,

r C 1

rC 1

r3C � � � ;

of all the rectangles in Figure 38.8 is r2=p, and in the special case in which r D ˆ, the

sum is ˆ2 because p D 1.

Kepler’s triangle 4ACD in Figure 38.9 is called Kepler’s triangle; here is how it is

constructed. Let r > 1 and again let p D r � 1=r ; then r2 D rp C 1: (As before, p D 1

when r D ˆ:) Now construct a semicircle with rp C 1 as the diameter AC. Mark the point

B so that AB D rp and BC D 1. Erect a line at B that intersects the circle at D. Then

the length of DC (shown as x here) is r . Apparently this was thought to be amazing when

Kepler did it, but it is not so amazing—it is simply the Euclidean construction for finding

D

A C rp B 1

x

Figure 38.9. Kepler’s triangle.

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38. The Golden Ratio—A Contrary Viewpoint 281

the square root of a segment AB, that is DB D prp. The hypotenuse of the right triangle

DBC isp

rp C 1 D r .

Pentagons, heptagons, nonagons, . . .

Regular polygons with an odd number of sides have some interesting properties related

to the ratios we have been discussing. Steinbach makes the case for other polygons in his

paper; he opens with this:

One of the best-kept secrets in plane geometry is the family of ratios of di-

agonals to sides in the regular polygons. So much attention has been given to

one member of this family, the golden ratio ˆ in the pentagonal case, that the

others live in undeserved obscurity. But the wealth of material that pours from

the pentagon—proportional sections, recursive sequences, and quasiperiodic

systems—can be matched wonder-for-wonder by any other polygon : : : :

It is well known that the golden ratio occurs in the regular pentagon with unit sides. In

Figure 38.10, the length AC D 2 cos.�=5/, which is ˆ:

1.6

18

108˚ B

A

E

C D

1

1

Figure 38.10. The pentagon and the golden ratio, AC � AB D 1=AC.

Much of the publicity given to ˆ has come from the special place that the pentagon had

among the Pythagoreans. In this case, the equation AC � AB D 1=AC says that ˆ � 1 D1=ˆ. But, if we look at other regular odd polygons, we see that some pair of diagonals also

satisfy the same relationship. Indeed, it is an interesting exercise to prove the following

result.

Let P be any regular odd polygon with n sides, n � 5. If r is the length of a longest

diagonal of P and p is the length of a second longest diagonal, then r � p D 1=r .

Example Behold the heptagon! See Figure 38.11. The diagonals AD and AC have the

relationship described:

AD D 4 cos2.�=7/ � 1 � 2:247 and AC D 2 cos.�=7/ � 1:802

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282 Part VII. Other Aspects of Martin Gardner

2.2

47

102.86˚

B

A

EC

D

1

128.57˚

G

F

1

11.8

02

Figure 38.11. The heptagon and two related ratios. If AD D r , then AC D r � 1=r .

That is, AD � AC D 1=AD:

Algebraic properties of xn D pxn�1 C xn�2

We want to consider a modest generalization of the Fibonacci sequence, equation (38.4)

below. The paper by Murthy presents substantially more general recursive equations that

include (38.4).

Again, let r > 1 and p D r � 1r

. Let x0 D 1, x1 D 1, and for n � 2, let

xn D pxn�1 C xn�2: (38.4)

Then the general term of the sequence is

xn D c1rn C c2

�

�1

r

�n

; (38.5)

where

r D p Cp

p2 C 4

2;

�1

rD p �

p

p2 C 4

2;

and

c1 D 2 � p Cp

p2 C 4

2p

p2 C 4; c2 D p � 2 C

p

p2 C 4

2p

p2 C 4:

Because of the way that p is defined, r and �1r

are the roots of the quadratic equation

x2 � px � 1 D 0: (38.6)

A few terms of this sequence, obtainable from either equation (38.4) or equation (38.5),

are

x2 D p C 1;

x3 D p2 C p C 1;

x4 D p3 C p2 C 2p C 1:

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38. The Golden Ratio—A Contrary Viewpoint 283

If p ¤ 1, this is not Fibonacci. We now point out several properties that any such

solution r of (38.6) would have and that the quadratic equation is sufficient to imbue r

with these properties.

Subtracting a number to get the reciprocal From the fact that ˆ is the solution to

equation (38.6) when p D 1, ˆ � 1 D 1=ˆ: Thus, ˆ is the only number whose reciprocal

can be found by subtracting 1. But for any other positive number p, if r is the positive

root of (38.6), then r � p D 1=r , and r is the only number whose reciprocal can be

found by subtracting p. This is simply a relation between the coefficients and the sum of

the roots of any quadratic equation. For example, if p D 12

, and r is the positive root

of x2 � 12x � 1 D 0, then r D .1 C

p17/=4. Subtracting 1

2gives .�1 C

p17/=4, the

reciprocal of .1 Cp

17/=4:

Square of nth term It is easy to prove by induction that for any three consecutive terms

of the sequence in equation (38.4), the square of the middle term is p more or p less than

the product of the other two terms. The Fibonacci special case says that given any three

consecutive terms, the square of the middle one is 1 more or 1 less than the product of the

other two.

Continued fraction and endless square root Two other relations between r and p are

r D p C 1

p C 1

pC 1pC���

and r Dr

1 C p

q

1 C pp

1 C p : : : ;

which specialize when p D 1 to

ˆ D 1 C1

1 C 1

1C 11C���

and ˆ D

r

1 Cq

1 Cp

1 C � � � :

A word in favor of ˆ

We do recognize that the number ˆ, like e and � , has many interesting properties. The

sense in which this is true comes from the fact that ˆ shows up when we try to simplify

certain formulas, such as setting p D 1 in the continued fraction given above. This is

the same sense in which the number e is interesting. If you want to find the derivative of

logb.x/, b > 0 and b ¤ 1; from the definition

limh!0

logb.x C h/ � logb.x/

h;

after some work, you get

1

xlogb

limh!0

�

1 C h

x

�x=h!

:

The indicated limit is e; so we have 1x

logb.e/: Now, it is natural to select b D e; so, the

desired derivative is 1x

: Using e for the logarithm base just makes the formula simpler.

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284 Part VII. Other Aspects of Martin Gardner

Conclusions

We have shown that in certain instances, the claim that the golden ratio has a special place

among numbers as a valid description of nature is unsupported. Furthermore, it has been

well refuted that this ratio is somehow exceptionally pleasing and that it occurs frequently

in art and architecture. For example, from taking measurements, we find that there is no

basis for saying that ˆ is the ratio that naturally occurs in sea shells. In particular, there is

no basis for the assertion that it occurs in the nautilus. I disagree with Livio’s implication

that ˆ is “The World’s Most Astonishing Number.” The interesting properties of ˆ as a

positive root to the quadratic equation (38.6) (with p D 1) are matched by the positive

root to (38.6) for any positive number, p ¤ 1. The wonderful geometric properties of ˆ

as an extreme mean are matched by the other means discussed here. Also, finding ˆ, as

2 cos.�=5/, in the pentagon is exciting, but no more so than finding 4 cos2.�=7/ � 1 in the

heptagon. Perhaps, ˆ deserves to be included in a list along with e, � , and other numbers

because it can be used to simplify certain formulas. In that sense, it is interesting, perhaps

even very interesting, but not entirely astonishing.

Acknowledgment I would like to thank my wife, Jean, for her help with the many rough

draft versions of this paper. Also, I appreciate the amount of time and energy expended by

the editor of this journal in assisting me in the development of this paper.

Bibliography

[1] D. Bergamini, Mathematics, Time Incorporated, 1963.

[2] R. Fischler, How to find the golden number without really trying, Fibonacci Quart. 19 (1981)

406–410.

[3] D. H. Fowler, A generalization of the golden section, Fibonacci Quart. 20 (1982) 146–158.

[4] M. Gardner, The cult of the golden ratio, Skeptical Inquirer 18 (1994) 243–247.

[5] J. Harte, A. Kinzig, and J. Green, Self-Similarity in the distribution of abundance of species,

Science (April 9, 1999) 334–339.

[6] M. Livio, The Golden Ratio: The Story of Phi, the World’s Most Astonishing Number, Broadway

Books, 2002.

[7] G. Markowsky, Misconceptions about the golden ratio, College Math. J. 23 (1992) 2–19.

[8] P. V. S. Murthy, Generalizations of some problems on Fibonacci numbers, Fibonacci Quart. 20

(1982) 65–66.

[9] P. Steinbach, Golden fields: A case for the heptagon, Math. Mag. 70 (1997) 22–31.

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39Review of The Mysterious Mr.

Ammann by Marjorie Senechal

Philip D. Straffin

In 1975, Martin Gardner wrote in his Mathematical Games column in Scientific American

about Roger Penrose’s discovery of a set of tiles that do not tile the plane periodically,

but do tile it nonperiodically. Gardner did not show pictures because Penrose was waiting

for a patent. In response to that column, Gardner received a letter from Robert Ammann,

who wrote, “I am also interested in nonperiodic tiling, and have discovered both a set of

two polygons which tile the plane only nonperiodically and a set of four solids which fill

space only nonperiodically.” Ammann included pictures: he had independently discovered

Penrose tilings and generalized them to three dimensions. He went on to find five more sets

of planar tilings, discover the general organizing principle now known as Ammann bars,

and make many other contributions to the theory of nonperiodic tiling.

No one in the tiling community knew who Ammann was. He described himself as “an

amateur doodler” and he declined all invitations to meetings and conferences. In 1987,

Marjorie Senechal finally succeeded in tracking him down, arranging for him to meet Pen-

rose, Donald Coxeter, and John Conway, and even getting him to speak at two conferences.

After Ammann’s early death from a heart attack in 1994, she learned more about his life.

At age three, he was precociously brilliant. At age four, he stopped talking and withdrew

into his own world. He struggled through childhood and adolescence, “off the charts in-

tellectually but impossible emotionally.” During the years of his tiling discoveries he was

living in a motel and supporting himself by sorting mail in a post office. His mother said

that the mathematical meetings were the high point of his life: “No one else reached out

to him.” Perhaps we should draw the moral that it is important to encourage mathemati-

cal talent wherever we find it, and to welcome into the mathematical community all able

contributors.

Reprinted from The College Mathematics Journal, Vol. 36, No. 2 (Mar. 2005), pp. 167–168.

285

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40Review of PopCo by Scarlett Thomas

Martin Gardner

This is the only novel I know of that is saturated with mathematics. Alice Butler, the 29-

year-old narrator of the story, works for the world’s third-largest toy company, PopCo. Her

job is to create ideas for toys and games, especially products that will appeal to teen-age

girls.

Orphaned when young, Alice is raised by loving and loved grandparents, both of whom

are first-rate mathematicians. Her grandfather writes a monthly column on recreational

math, said to be similar to the one I wrote for Scientific American. He is obsessed with

trying to decode the notorious Voynich manuscript, now widely believed to be a hoax.

Although he fails in this project, he succeeds in decoding a cipher-text that reveals the

location of a buried treasure. The treasure is on a bird sanctuary, and because digging it up

would disturb the birds, he never reveals its location. He does, however, leave cipher clues

to Alice. The grandmother’s obsession on the other hand is trying to solve the Riemann

hypothesis.

Alice has been strongly influenced by her grandfather’s fondness for mathematical

play, and she constantly interrupts her narrative with accurate descriptions of a variety

of mathematical topics. These include Conway’s computer game of Life, Newcomb’s fa-

mous paradox involving decision theory, trapdoor ciphers, and the Monty Hall problem

of the three doors. In addition to recreational topics, Alice provides excellent accounts of

Cantor’s alephs, prime factorization, Godel’s proofs, and even the Riemann hypothesis!

In addition to being extremely knowledgeable about math, Alice plays both Go and

chess well. There is an amusing flashback to her secondary school days when she partici-

pated in a tournament run by her loathed sexist math teacher—the winner of the tournament

gets to play him. Alice of course wins, and when the teacher and Alice sit down to play,

he pompously tells her that the game probably won’t last five minutes. Alice recognizes

his opening as one used by Kasparov in a game that she had once studied intensively (in

fact, her grandfather had written one of his columns about it), so she knows exactly how to

meet the attack. Ten minutes later the teacher is defeated. He is so humiliated that he takes

extensive sick leave.

The novel concludes with a double climax. Expert in cipher-breaking—one of Al-

ice’s toys is a code-breaking kit that includes Thomas Jefferson’s cylinder with rotating

wheels—Alice cracks her grandfather’s cipher about the buried treasure. The treasure,

worth billions, is unearthed, and Alice generously gives it all to the bird sanctuary.

Reprinted from The College Mathematics Journal, Vol. 38, No. 3 (May 2007), pp. 241–242.

287

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288 Part VII. Other Aspects of Martin Gardner

The second climax, much more improbable, concerns Alice’s recruitment into a myste-

rious secret organization with a revolutionary agenda—but I’ll not reveal that here in case

a reader is moved to read the book.

Although I enjoyed the story very much, there were a couple of things that bothered

me. One was the frequency with which the characters, especially Alice’s lover Ben, seem

unable to speak without peppering their conversation with meaningless f-words. Do em-

ployees in British toy firms actually talk that way?

The other was Alice’s apparent low opinion of mainline doctors. I can understand how

she might be a vegan, avoiding all food coming from animals, but her devotion to home-

opathy is much harder to comprehend. One wonders whether this reflects the author’s opin-

ion or is just written into her lead character. Surely Alice would know that homeopathic

remedies dilute a drug so thoroughly that at most two or three molecules, if any, remain.

(Somehow, the water is supposed to “remember” the drug’s properties.)

Other features of the book include a table for writing a Vigenere cipher, a postscript

listing the primes less than 1000, a clever crossword puzzle, and a recipe for a cake. Part 1

opens with a quotation from a paper by psychologist Stanley Milgram, Part 2 with an brief

excerpt from Paul Hoffman’s biography of Paul Erdos, and Part 3 starts with one of Piet

Hein’s best “grooks”:

A bit beyond perception’s reach

I sometimes believe I see

That life is two locked boxes, each

Containing the other’s key.

Scarlett Thomas writes captivatingly; her novel is not easy to put down.

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41Superstrings and Thelma

Martin Gardner

Several years ago I was a graduate student at the University of Chicago. I was working on

my doctorate in physics, about possible ways to test superstring theory, when my brother

in Tulsa died suddenly from a heart attack. Both parents had earlier passed away. After

the funeral I drove past my past, marveling at the enormous changes that had taken place

since I grew up there. I drove by the red brick building, now an enormous warehouse, that

had once been Tulsa Central High. My grades in history, Latin, and English lit were low,

but I was good in math and had a great physics teacher. He was mainly responsible for my

majoring in physics after a scholarship took me to the University of Chicago.

While I was having dinner at a popular restaurant on the corner of Main and Sixth

streets, the waitress stared at me with a look of surprise. “Are you Michael Brown?”

“That’s me,” I said. She smiled and held out a hand. “I’m Thelma O’Keefe. We were

in the same algebra 101 class.”

We shook hands.

“You won’t remember me,” she said. “I was fat in those days, and shy, and not very

pretty.”

“That’s hard to believe,’ I said. “You look gorgeous now.”

“Well, thank you, kind sir,” she said, smiling. “You were a whiz at algebra. Do you

remember when you caught Mr. Miller in a mistake he made on the blackboard, and how

embarrassed he was?”

“I remember. He was a miserable teacher. I think he hated math.”

“I know I hated it,” Thelma said.

“I’m sorry to hear that. Math can be exciting and beautiful when it’s taught properly.”

After Thelma brought my receipt and credit card, I said, “Any chance I could see you

after work? Maybe you could steer me to a late night bar where we could chat about old

times?”

“I’m free at eleven,” she said.

I followed Thelma’s car to a pleasant little bar on the outskirts of town, near where she

lived. She was divorced, she told me, and had a boy of ten who was probably asleep in her

apartment. The bar served only beer. She said she didn’t drink anything with more alcohol

than beer. Her ex, she added, had an alcohol problem. I didn’t press her for details. Instead,

I fear I talked too much about myself, and mainly about superstrings.

Reprinted from Math Horizons, Vol. 18, No. 1 (Sept. 2010), pp. 6–9.

289

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290 Part VII. Other Aspects of Martin Gardner

I did my best to explain that strings were inconceivably tiny loops, like rubber bands,

that vibrated at different rates. Their frequencies generated all the properties of the fun-

damental particles, such as electrons and quarks. The simplest string vibration produces

gravitons, conjectured particles that carry gravity waves.

“You have nice dark eyes,” she interrupted.

“Thanks,” I said. “Your eyes aren’t so bad either.”

I tried hard to explain how a famous physicist named Ed Witten had generalized strings

to what he called M-theory. The M stands for membranes, or branes for short. A superstring

is a brane of one dimension. Other branes have higher dimensions. Our universe, I said,

has ten or eleven dimensions, of which six or seven are squeezed into compact tiny spheres

that are attached to every point in our spacetime.

“I didn’t understand a word you spoke,” Thelma said. “It sounds nutty to me. Do you

believe all that?”

“Mostly. I think strings are for real, but I’m not so sure about Witten’s membranes.”

“Is everything made of strings?” Thelma asked.

“Everything.”

“And what are the strings made of?”

“Nothing. They’re just pure mathematical structures.”

“If the universe isn’t made of anything,” she said, “how come it exists?”

“Good question. Nobody knows.”

“Well, maybe God knows,” she said.

Outside the bar, standing by our cars, Thelma invited me to her apartment for some

coffee.

“No,” I said. “I really can’t stay another minute. I have to be up early to catch a plane

to Chicago. It was great getting to know you.”

“Will I see you again?”

“It would be a pleasure,” I said.

Like a fool I failed to ask for her address and phone number. We shook hands. She said

goodbye, then startled me with a quick kiss on the mouth.

Almost a year drifted by. My thesis was published as a book by the University of

Chicago Press. My suggestions for testing string theory were taken seriously by most

stringers. There was hope that some of the tests might actually be made by a new atom

cruncher under construction in Switzerland. There were vague rumors about a Nobel prize.

I was unable to get Thelma out of my head. I kept thinking of her wonderful smile, and

how good she smelled. It wasn’t perfume. Was it her hair? I thought about her more than I

thought about superstrings!

The University of Oklahoma, at Norman, hired me as an assistant professor. A suburb

of Oklahoma City, Norman is only a few hours drive from Tulsa.

None of the waitresses at the restaurant where Thelma had worked knew what had

happened to her. She left her job six months before, and they hadn’t heard from her since.

No Thelma O’Keefe was in the Tulsa phone book. I drove back to Norman feeling

sad and frustrated. Should I hire a detective? The Norman yellow book had a long list of

“investigators” and two detective agencies.

I was planning to call one of the agencies when my telephone rang. It was Thelma!

“I heard you were asking about me,” she said.

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41. Superstrings and Thelma 291

“Yes. How did you get my phone number?”

“It’s on the Internet. How are strings?”

“Not so good. It didn’t predict dark matter. It didn’t predict dark energy. It even failed

to pass one of my tests. Lots of stringers are starting to have doubts, including me.”

“If we meet again,” said Thelma, “don’t tell me about it.”

Further Reading

Two recent books attacking string/M theory as pseudoscience are Not Even Wrong, by

mathematician Peter Woit (Basic Books, 2006), and The Trouble with Physics, by Lee

Smolin (Mariner Books, 2007). See Chapter 18, “Is String Theory in Trouble?” in my

book, The Jinn from Hyperspace (Prometheus Books, 2008).

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IndexAbe, Gakuho, 163

Adamatzky, Andrew, 208

Aitken, A. C., 39

Almkvist, Gert, 188

Ammann, Robert, 285

Ascher, Marcia, 45

Ball, W. W. Rouse, 223

Bankoff, Leon, 3, 7

Bernhart, Frank, 164

Beverley, William, 154

Bodlaender, Hans, 73

Bojanov, Borislav, 188

Bolt, Brian, 159

Bottomley, Henry, 46

Brams, Steven, 73

Brandt, Jørgen, 188

Brooks, R. L., 143

Carroll, Lewis, vi, 58, 66

Carver, W. B., 26

Catalan, Eugene Charles, 119

Christ, Henry, 227

Chu, I. Ping, 131

Cipra, Barry, 176

Coeter, H. S. M., 3

Conway, John, 207, 285

Coxeter, H. S. M., 224, 285

da Vinci, Leonardo, 273

de Vasa, H. E., 155

Denef, Yann, 154

diophantine, 203, 269

Dodgson, Charles, 58

Dudeney, Henry, 162, 167

Efron, Bradley, 249

Eggleton, Roger, 159

Erdos, Paul, 4, 288

Eriksson, Henrik, 188

Euler, Leonhard, 20

Feynman, Richard, 87, 110

Fibonacci numbers, 51, 143, 274

Fields Medal, 269

Ford, L. R., 26

Foshee, Gary, 258

Four Color Theorem, 67, 73

Frederickson, Greg, 135

Friedman, Erich, 16

Fulves, Karl, 181, 224

Gale, David, 221

games

Go, 78, 207

KenKen, 173

Life, 207, 287

map coloring, 73

Nim, 45

sudoku, 173

ticktacktoe, 266

two-player Bulgarian solitaire, 191

two-player Monty, 236

Wythoff, 213

Garnett, F. M., 26

Gauss, Carl Friederich, 271

Gessel, Ira, 114

Gilbreath, Norman, 221

Gilks, Joe, 159

golden ratio, 123, 273

Golomb, Solomon, 38, 128, 136, 140, 150

Gomory, Ralph, 127

Gosper, Bill, 208

Grunbaum, Branko, 66, 150

Grabarchuk, Serhiy, 139

Graham, Ron, 188

Gutenmacher, Victor, 188

293

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294 Index

Guthrie, Francis, 65

Hall, Monty, 223, 231, 287

Hamilton circuit, 68, 191

Harary, Frank, 268

Hardy, G. H., 54

Hein, Piet, 288

Hermite, Charles, 271

Hickerson, Dean, 208

High, Robert, 74

Hirayama, Akira, 163

Hoffman, Paul, 288

Hoggatt, Verner, Jr., 150

Huber, Greg, 33

Jaenisch, C. F., 154

Jefferson, Thomas, 287

Jelliss, George, 154

Jensen, Christopher, 132

Johnsonbaugh, Richard, 131

Jones, Kate, 132

Karatsuba, Anatolii Alexeevich, 188

Karp, Richard, 222

Keillor, Garrison, 250

Khodulev, Andrei, 16

Kim, Scott, 19

Kirstead, Friend, Jr., 225

Klamkin, Murray, 4

Knuth, Donald, 188, 222

Kronecker, Leopold, 271

Kumar, Awani, 155

Ligocki, Terry, 33

Lindgren, Harry, 135

Liouville, Joseph, 37

Loyd, Sam, 201

Mackay, Hughues, 154

magic square, 153, 160, 225

Mahler, Kurt, 38

Marlow, T. W., 154

Maxwell, Brian, 226

Mayrignac, Jean-Charles, 154

McIntosh, Harold, 99

McLean, Bruce, 103

Milgram, Stanely, 288

Miyamoto, Tetsuya, 173

Moseteller, Fred, 231

Murray, H. J. R., 154

Nelsen, Roger, 128

Nieuwland, Pieter, 25

Nobel Prize, 290

O’Beirne, T. H., 13

Ollerenshaw, Kathleen, 224

Online Encyclopedia of Integer Sequences,

46, 107, 217

Oskolkov, Konstantin, 188

Polya Award, 3

Palmatelli-Palmarini, Massimo, 232

partitions, 51, 53, 188

Penrose, Roger, 285

Petkov, Milko, 188

Pomerance, Carl, 150

Poniachik, Jaime, 38

Putnam Competition, 3

Rademacher, Hans, 54

Ramanujan, Srinivasa, 54, 272

Ransom, William, 202

Reiter, Harold, 163

Rendell, Paul, 208

Ritchie, David, 164

Roberts, T. S., 154

Robinson, Raphael, 11

Schattschneider, Doris, 11

Scherer, Karl, 150

Schreck, D. J. E., 25

Selvin, Steve, 231

Senechal, Marjorie, 285

Shapley, Lloyd, 74

Shasha, Dennis, 199

Sherman, Scott, 111

shuffle, 170, 221

Silver, Stephen, 208

Siu, Man-Keung, 184

Smith, C. A. B., 143

Smith, John Maynard, 231

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Index 295

Starr, Norton, 133

state diagram, 182, 189, 197

Stertenbrink, Guenter, 154

Stone, Arthur, 87, 103, 109, 113, 143

Stork, David, 249

Susco, Barbara, 19

Theobald, Gavin, 14

Thiel, Von J. Christian, 162

Thomas, Scarlett, 287

Toom, Andrei, 188

tree

Calkin Wilf, 215

Monty Hall, 233

pat, 115

Stern Brocot, 217

ticktacktoe, 268

Tucker, Bryant, 87

Tukey, John, 87, 110

Tutte, William, 143

vos Savant, Marilyn, 232

Wainwright, Robert, 208

Wallis, John, 25

Wenzelides, Karl, 154

Willcocks, T. H., 155

Witten, Ed, 290

Wythoff, Willem Abraham, 213

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About the Editors

Michael Henle is a professor of mathematics at Oberlin College in Oberlin, Ohio. He is

the author of several previous books including Which Numbers are Real? which was just

published by the MAA in 2012. Trained as a functional analyst, he has written as well on

combinatorial subjects and geometry. He is serving as editor of The College Mathematics

Journal through 2013.

Brian Hopkins is a professor of mathematics at Saint Peter’s University in Jersey City,

New Jersey. He won, with Robin Wilson, the 2005 George Polya Award, edited the 2008

MAA Notes volume Resources for Teaching Discrete Mathematics, and was given the

2011 MAA New Jersey Section Award for Distinguished College or University Teaching

of Mathematics. Much of his research stems from Bulgarian Solitaire, a topic popularized

by Martin Gardner. Hopkins will be the editor of The College Mathematics Journal from

2014 to 2018.

297