Chessboards, Hats, and Chinese Poetry :Chessboards, Hats, and Chinese Poetry :

Some Rigorous and Not-So-RigorousSome Rigorous and Not-So-Rigorous

Mathematical ResultsMathematical Results

C. L. LiuC. L. Liu

詩裡有數

List of books published

Introduction to Combinatorial Mathematics, McGraw-Hill Book Company, 1968, (Japanese translation, 1972, Chinese translation, 1982).

Topics in Combinatorial Mathematics, Mathematical Association of America, 1972.

Linear Systems Analysis, with J. W. S. Liu, McGraw-Hill Book Company, 1975.

Elements of Discrete Mathematics, McGraw-Hill Book Company, 1977, (Japanese translation, 1979, Chinese translation, 1981).

Pascal, with G. G. Belford, McGraw-Hill Book Company, 1984.

Elements of Discrete Mathematics, second edition, McGraw-Hill Book Company, 1985. (Chinese translation, 1993, Japanese translation, 1995, Indonesian translation, 1995, Greek, Spanish translation).

Solution of Design Automation Problems by the Method of Simulated Annealing, with D. F. Wong, and H. W. Leong, Kluwer Academic Publishers, 1988.

Fault Covering Problems in Reconfigurable VLSI Systems, with R. Libeskind-Hadas, N. Hasan, J. Cong, and P. McKinley, Kluwer Academic Publishers, 1992.

愛上層樓，天下遠見出版公司、國立清華大學出版社出版， 2002 年。

Hats

ChinesePoetry

0cbxax 2Mathematics

Chessboards

It all begins with a chessboard

Covering a Chessboard

88 chessboard

21 domino

Cover the 88 chessboard with thirty-two 21 dominoes

A Truncated Chessboard

21 domino

Cover the truncated 88 chessboard with thirty-one 21 dominoes

Truncated 88 chessboard

Proof of Impossibility

21 domino

Truncated 88 chessboard

Impossible to cover the truncated 88 chessboard with thirty-one dominoes.

Proof of Impossibility

Impossible to cover the truncated 88 chessboard with thirty-one dominoes. There are thirty-two white squares and thirty black squares. A 2 1 domino always covers a white and a black square.

Proof of Impossibility

Impossible to cover the truncated 88 chessboard with thirty-one dominoes. There are thirty-two white squares and thirty black squares. A 2 1 domino always covers a white and a black square.

01 0

0 0 0 011 1 1

1 1

11 1 10 00

00 011

0

0 1110 0

0

0

0000

00 00

0

0 0

0

0

00

11

1

1

1

1

1

1

1

1

11

1 1

1 11

Modulo-2 Arithmetic

1 2 3 4 5 6 …..

odd even odd even odd even…..

odd even

odd even odd

even odd even

0 1

0 0 1

1 1 0

Coloring the Vertices of a Graph

vertex

edge

2 - Colorability

A necessary and sufficient condition : No circuit of odd length

vertex

edge

vertex

edge

3 - Colorability

3 - Colorability

The problem of determining whether a graph is 3-colorable

is NP-complete. ( At the present time, there is no known

efficient algorithm for determining whether a graph is

3-colorable.)

4 - Colorability : Planar Graphs

All planar graphs are 4-colorable.

How to characterize non-planar graphs ? Genus, Thickness, …

Kuratowski’s subgraphs

A Defective Chessboard

Triomino

Any 88 defective chessboard can be covered with twenty-one triominoes

Defective Chessboards

Any 2n2n defective chessboard can be covered with 1/3(2n2n -1) triominoes

Any 88 defective chessboard can be covered with twenty-one triominoes

Prove by mathematical induction

The first domino falls.

If a domino falls, so will the next domino.

All dominoes will fall !

Proof by Mathematical Induction

Basis : n = 1

Induction step :2 n+1

2 n+1

2 n 2 n

2 n

2 n

Any 2n2n defective chessboard can be covered with 1/3(2n2n -1) triominoes

If there are n wise men wearing white hats, then at the nth hour allthe n wise men will raise their hands.

The Wise Men and the Hats

Basis : n =1 At the 1st hour. The only wise man wearing a white hat will raise his hand.

Induction step : Suppose there are n+1 wise men wearing white hats.

At the nth hour, no wise man raises his hand.

At the n+1th hour, all n+1 wise men raise their hands.

……

Another Hat Problem

Design a strategy so that as few men will die as possible.

No strategy In the worst case, all men were shot.

Strategy 1 In the worst case, half of the men were shot.

Another Hat Problem

x n x n-1 x n-2 x n-3 ……………… x1

………..

x n-1 x n-2 x n-3 ……… x1

x n-2 x n-3 ……… x1

x n-1 x n-3 ……… x1

x n-2

Yet, Another Hat Problem

A person may say, 0, 1, or P(Pass)Winning : No body is wrong, at least one person is rightLosing : One or more is wrong

Strategy 1 : Everybody guesses Probability of winning = 1/8

Strategy 2 : First and second person always says P. Third person guesses Probability of winning = 1/2

Strategy 3 :

observe call

00

01

10

11

1

P

P

0

pattern call

000001010011100101110111

111PP1P1P0PP1PPP0PPP0000

Probability of winning = 3/4

More people ?

Best possible ?

Generalization : 7 people, Probability of winning = 7/8

Application of Algebraic Coding Theory

A Coin Weighing Problem

Twelve coins, possibly one of them is defective ( too heavyor too light ). Use a balance three times to pick out thedefective coin.

1 2 3 4 5 6 7 8

G 9 10GG 11

12G 109

Step 1

Step 3

Step 2

Balance

Step 3Balance Imbalance

7G

1 2 3 4 5 6 7 8

1 3 452 6

Step 1

Step 2

Imbalance

Step 3Balance

21

Step 3Imbalance

1 2 3 4 5 6 7 8

1 3 452 6

Step 1

Step 2

Imbalance

43

Step 3Imbalance

Another Coin Weighing Problem

Application of Algebraic Coding Theory

• Adaptive Algorithms• Non-adaptive Algorithms

Thirteen coins, possibly one of them is defective ( too heavyor too light ). Use a balance three times to pick out thedefective coin. However, an additional good coin is availablefor use as reference.

Yet, Another Hat Problem

Hats are returned to 10 people at random, what is the probability that no one gets his own hat back ?

Apples and Oranges

ApplesApples OrangesOrangesOrangesOranges

ApplesApples

Take out one fruit from one box to determine the contentsof all three boxes.

Derangements

AA BB CC

a b c

a c b

b a c

b c a

c a b

c b a

Derangement of 10 Objects

Number of derangements of n objects

]!

1)1(....

!3

1

!2

1

!1

11[!

nnd n

n

]!10

1)1(....

!3

1

!2

1

!1

11[!10 10

10 d

Probability !10

1)1(....

!3

1

!2

1

!1

11

!101010 d

36788.01 e

Permutation

1 2 3 4

a

b

c

d

Positions

Objects

Placement of Non-taking Rooks

1 2 3 4

a

b

c

d

Positions

Objects

Permutation with Forbidden Positions

1 2 3 4

a

b

c

d

Positions

Objects1 2 3 4

a

b

c

d

Positions

Objects

Placement of Non-taking Rooks

1 2 3 4

a

b

c

d

Positions

Objects1 2 3 4

a

b

c

d

Positions

Objects

Placement of Non-taking Rooks

1 2 3 4

a

b

c

d

Positions

Objects

Rook Polynomial :

R (C) = r0 + r1 x + r2 x2 + …

ri = number of ways to place i non-taking rooks on chessboard C

R (C) = 1 + 6x + 10x2 + 4x3

At Least One Way to Place Non-taking Rooks

1 2 3 4

a

b

c

d

Positions

Objects1 2 3 4

a

b

c

d

Positions

Objects

Theory of Matching !

怎一個愁字了得

愁

怎一個愁字了得

尋尋覓覓 冷冷清清 淒淒慘慘戚戚 乍暖還寒時候 最難將息 三杯兩盞淡酒 怎敵他晚來風急 雁過也 正傷心 卻是舊時相識 滿地黃花堆積 憔悴損 如今有誰堪摘 守著窗兒 獨自怎生得黑 梧桐更兼細雨 到黃昏點點滴滴這次第 怎一個愁字了得

李清照　＜聲聲慢＞

只恐雙溪舴艋舟，載不動許多愁。

李清照　＜武陵春＞

一斛珠連萬斛愁，關山漂泊腰支細。

吳梅林　＜圓圓曲＞

舟載

斛量

新妝宜面下朱樓，深鎖春光一院愁。劉錫禹　＜春詞＞

白髮三千丈、離愁似個長。 李白　＜秋浦歌＞

問君能有幾多愁，恰似一江春水向東流。李後主　＜虞美人＞

一懷愁緒，幾年離索。陸游　＜釵頭鳳＞

凝眸處，從今又添，一段新愁。 李清照　＜鳳凰台上憶吹簫＞

一點芭蕉一點愁。徐再思　＜雙調水仙子夜雨＞

寂寞深閨，柔腸一寸愁千縷。李清照　＜點絳唇＞

與爾同消萬古愁。

李白　＜將進酒＞

一種相思，兩處閒愁。

李清照　＜一剪梅＞

怎一個愁字了得尋尋覓覓 冷冷清清 淒淒慘慘戚戚 乍暖還寒時候 最難將息 三杯兩盞淡酒 怎敵他晚來風急 雁過也 正傷心 卻是舊時相識 滿地黃花堆積 憔悴損 如今有誰堪摘 守著窗兒 獨自怎生得黑 梧桐更兼細雨 到黃昏點點滴滴這次第 怎一個愁字了得

2 > 1

Conclusion

Mathematics is about finding connections, betweenspecific problems and more general results, and between one concept and another seemingly unrelatedconcept that really is related.

有恆是信真君子無欲為剛大丈夫