maximum overhang - mathematical association of americamaximum overhang mike paterson, yuval peres,...

Maximum Overhang

Mike Paterson Yuval Peres Mikkel ThorupPeter Winkler and Uri Zwick

1 INTRODUCTION How far can a stack of n identical blocks be made to hangover the edge of a table The question has a long history and the answer was widelybelieved to be of order log n Recently Paterson and Zwick constructed n-block stackswith overhangs of order n13 exponentially better than previously thought possible Weshow here that order n13 is indeed best possible resolving the long-standing overhangproblem up to a constant factor

This problem appears in physics and engineering textbooks from as early as themid-19th century (see eg [15] [20] [13]) The problem was apparently first broughtto the attention of the mathematical community in 1923 when J G Coffin [2] posed itin the ldquoProblems and Solutionsrdquo section of this MONTHLY no solution was presentedthere The problem recurred from time to time over subsequent years eg [17 1819 12 6 5 7 8 1 4 9 10] achieving much added notoriety from its appearance in1964 in Martin Gardnerrsquos ldquoMathematical Gamesrdquo column of Scientific American [7]and in [8 Limits of Infinite Series p 167]

1

1112 0916667 25

24 104167

15minus4radic

28 116789

Figure 1 Optimal stacks with 3 and 4 blocks compared to the corresponding harmonic stacks The 4-blocksolution is from [1] Like the harmonic stacks it can be made stable by minute displacements

Most of the references mentioned above describe the now-classical harmonic stacksin which n unit-length blocks are placed one on top of the other with the i th blockfrom the top extending by 1

2i beyond the block below it The overhang achieved bysuch stacks is 1

2 Hn = 12

sumni=1

1i sim 1

2 ln n The cases n = 3 and n = 4 are illustrated atthe top of Figure 1 above and the cases n = 20 and n = 30 are shown in the back-ground of Figure 2 Verifying that harmonic stacks are balanced and can be madestable (see definitions in the next section) by minute displacements is an easy exercise(This is the form in which the problem appears in [15 pp 140ndash141] [20 p 183] and[13 p 341]) Harmonic stacks show that arbitrarily large overhangs can be achievedif sufficiently many blocks are available They have been used extensively as an intro-duction to recurrence relations the harmonic series and simple optimization problems(see eg [9])

doi104169000298909X474855

November 2009] MAXIMUM OVERHANG 763

blocks = 20overhang = 232014


Figure 2 Optimal stacks with 20 and 30 blocks from [14] with corresponding harmonic stacks in the back-ground

Building a stack with an overhang is of course also a construction challenge inthe real world Figure 3 shows how the Mayans used such constructions in 900 BC incorbel arches In contrast with a true arch a corbel arch does not use a keystone sothe stacks forming the sides do not benefit from leaning against each other

Figure 3 Mayan corbel arch from Cahal Pech Belize 900 BC (picture by Clark AndersonAquaimages underthe Creative Commons License [3])

11 How far can you go Many readers of the above-mentioned references wereled to believe that 1

2 Hn (sim 12 ln n) the overhang achieved by harmonic stacks is the

maximum overhang that can be achieved using n blocks This is indeed the case underthe restriction explicit or implicit in some of these references that the blocks shouldbe stacked in a one-on-one fashion with at most one block resting on each block It

764 ccopy THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 116

has been known for some time however that larger overhangs may be obtained ifthe one-on-one restriction is lifted Three blocks for example can easily be used toobtain an overhang of 1 Ainley [1] found that four blocks can be used to obtainedan overhang of about 116789 as shown at the bottom right of Figure 1 and this ismore than 10 larger than the overhang of the corresponding harmonic stack Usingcomputers Paterson and Zwick [14] found the optimal stacks with a given limitednumber of blocks Their solutions with 20 and 30 blocks are shown in Figure 2

Now what happens when n grows large Can general stacks not subject to the one-on-one restriction improve upon the overhang achieved by the harmonic stacks bymore than a constant factor or is overhang of order log n the best that can be achievedIn a recent cover article in the American Journal of Physics Hall [10] observes that theaddition of counterbalancing blocks to one-on-one stacks can double (asymptotically)the overhang obtainable by harmonic stacks However he then incorrectly concludesthat no further improvement is possible thus perpetuating the order log n ldquomythologyrdquo

Recently however Paterson and Zwick [14] discovered that the modest improve-ments gained for small values of n by using layers with multiple blocks mushroominto an exponential improvement for large values of n yielding an overhang of ordern13 instead of just log n

12 Can we go further But is n13 the right answer or is it just the start of anothermythology In their deservedly popular book Mad About Physics [11 Challenge 271A staircase to infinity p 246] Jargodzki and Potter rashly claim that inverted triangles(such as the one shown on the left of Figure 4) are balanced If so they would achieveoverhangs of order n12 It turns out however that already the 3-row inverted triangleis unbalanced and collapses as shown on the right of Figure 4 as do all larger invertedtriangles

Figure 4 A 3-row inverted triangle is unbalanced

The collapse of the 3-row triangle begins with the lifting of the middle block in thetop row It is tempting to try to avoid this failure by using a diamond shape insteadas illustrated in Figure 5 Diamonds were considered by Drummond [4] and like theinverted triangle they would achieve an overhang of order n12 though with a smallerleading constant The analysis of the balance of diamonds is slightly more complicatedthan that of inverted triangles but it can be shown that d-diamonds ie diamonds thathave d blocks in their largest row are balanced if and only if d lt 5 In Figure 5 wesee a practical demonstration with d = 5

It is not hard to show that particular constructions like larger inverted triangles ordiamonds are unbalanced This imbalance of inverted triangles and diamonds was al-ready noted in [14] However this does not rule out the possibility of a smarter bal-anced way of stacking n blocks so as to achieve an overhang of order n12 and thatwould be much better than the above mentioned overhang of order n13 achieved byPaterson and Zwick [14] Paterson and Zwick did consider this general question (inthe preliminary SODArsquo06 version) They did not rule out an overhang of order n12but they proved that no larger overhang would be possible Thus their work shows that


Figure 5 The instability of a 5-diamond in theory and practice

the order of the maximum overhang with n blocks has to be somewhere between n13

and n12

13 Our result We show here that an overhang of order n13 as obtained by [14]is in fact best possible More specifically we show that any n-block stack with anoverhang of at least 6n13 is unbalanced and hence must collapse Thus we concludethat the maximum overhang with n blocks is of order n13

14 Contents The rest of this paper is organized as follows In the next section wepresent a precise mathematical definition of the overhang problem explaining in par-ticular when a stack of blocks is said to be balanced (and when it is said to be stable)In Section 3 we briefly review the Paterson-Zwick construction of stacks that achievean overhang of order n13 In Section 4 we introduce a class of abstract ldquomass move-mentrdquo problems and explain the connection between these problems and the overhangproblem In Section 5 we obtain bounds for mass movement problems that imply theorder n13 upper bound on overhang We end in Section 6 with some concluding re-marks and open problems

2 THE MODEL We briefly state the mathematical definition of the overhang prob-lem For more details see [14] As in previous papers eg [10] the overhang problemis taken here to be a two-dimensional problem each block is represented by a fric-tionless rectangle whose long sides are parallel to the table Our upper bounds applyhowever in much more general settings as will be discussed in Section 6

21 Stacks Stacks are composed of blocks that are assumed to be identical homo-geneous frictionless rectangles of unit length unit weight and height h Our resultshere are clearly independent of h and our figures use any convenient height Previousauthors have thought of blocks as cubes books coins playing cards etc

A stack B1 Bn of n blocks resting on a flat table is specified by giving thecoordinates (xi yi ) of the lower left corner of each block Bi We assume that the upperright corner of the table is at (0 0) and that the table extends arbitrarily far to the left


Thus block Bi is identified with the box [xi xi + 1] times [yi yi + h] (its length alignedwith the x-axis) and the table which we conveniently denote by B0 with the region(minusinfin 0] times (minusinfin 0] Two blocks are allowed to touch each other but their interiorsmust be disjoint

We say that block Bi rests on block Bj if Bi cap Bj = empty and yi = y j + h IfBi cap B0 = empty and yi = 0 then Bi rests on the table If Bi rests on Bj we let Ii j =Bi cap Bj = [ai j bi j ] times yi be their contact interval If j ge 1 then ai j = maxxi x j and bi j = minxi +1 x j +1 If j = 0 then ai0 = xi and bi0 = minxi +1 0 (Notethat we do allow single-point contact intervals here and so a block may rest on up tothree other blocks this convention only strengthens our impossibility results In [14]mainly concerned with constructions and stability a more conservative conventionrequired proper contact intervals)

The overhang of a stack is defined to be maxni=1(xi +1)

22 Forces equilibrium and balance Let B1 Bn be a stack composed of nblocks If Bi rests on Bj then Bj may apply an upward force of fi j ge 0 on Bi in whichcase Bi will reciprocate by applying a downward force of the same magnitude on Bj Since the blocks and table are frictionless all the forces acting on them are verticalThe force fi j may be assumed to be applied at a single point (xi j yi j ) in the contactinterval Ii j A downward gravitational force of unit magnitude is applied on Bi at itscenter of gravity (xi + 12 yi + h2)

Definition 21 (Equilibrium) Let B be a homogeneous block of unit length and unitweight and let a be the x-coordinate of its left edge Let (x1 f1) (x2 f2) (xk fk)

be the positions and the magnitudes of the upward forces applied to B along its bottomedge and let (x prime

1 f prime1) (x prime


kprime f primekprime) be the positions and magnitudes of the

upward forces applied by B along its top edge on other blocks of the stack Then Bis said to be in equilibrium under these collections of forces if and only if

ksumi=1

fi = 1 +kprimesum

i=1

f primei

ksumi=1

xi fi =(

a + 1

2

)+

kprimesumi=1

x primei f prime

i

The first equation says that the net force applied to B is zero while the second says thatthe net moment is zero

Definition 22 (Balance) A stack B1 Bn is said to be balanced if there existsa collection of forces acting between the blocks along their contact intervals such thatunder this collection of forces and the gravitational forces acting on them all blocksare in equilibrium

The stacks presented in Figures 1 and 2 are balanced They are however precari-ously balanced with some minute displacement of their blocks leading to imbalanceand collapse A stack can be said to be stable if all stacks obtained by sufficiently smalldisplacements of its blocks are balanced We do not make this definition formal as it isnot used in the rest of the paper though we refer to it in some informal discussions

A schematic description of a balanced stack and a collection of balancing forcesacting between its blocks is given in Figure 6 Only upward forces are shown in thefigure but corresponding downward forces are of course present (We note in passingthat balancing forces when they exist are in general not uniquely determined Thisphenomenon is referred to as static indeterminacy)


13

10 11 12

7 8 9

5 6

3 4

2

1

F13

F12

helliphellip

hellip

F9

F6

F4

F2

F3

F1

F0

Figure 6 Balancing collections of forces within a stack

We usually adopt the convention that the blocks of a balanced stack are numberedconsecutively from bottom to top and within each level from left to right Block B1

is then the leftmost block in the lowest level while Bn is the rightmost block at the toplevel For 0 le i le n we let Fi be a collection of upward balancing forces applied byblocks in B0 B1 Bi on blocks in Bi+1 Bn (see Figure 6)

Let us examine the relationship between two consecutive collections Fi and Fi+1The only forces present in Fi but not in Fi+1 are upward forces applied to Bi whilethe only forces present in Fi+1 but not in Fi are upward forces applied by Bi to blocksresting upon it If we let (x1 f1) (x2 f2) (xk fk) be the positions and the mag-nitudes of the upward forces applied to Bi and (x prime



kprime f primekprime) be

the positions and magnitudes of the upward forces applied by Bi and if we let abe the x-coordinate of the left edge of Bi we get by Definitions 21 and 22 thatsumk

i=1 fi = 1 + sumkprimei=1 f prime

i andsumk

i=1 xi fi = (a + 12) + sumkprime

i=1 x primei f prime

i Block Bi thus rear-ranges the forces in the interval [a a + 1] in a way that preserves the total magnitudeof the forces and their total moment when its own weight is taken into account Notethat all forces of F0 act in nonpositive positions and that if Bk is a rightmost blockin a stack and the overhang achieved by it is d then the total magnitude of the forcesin Fkminus1 that act at or beyond position d minus 1 should be at least 1 These simple obser-vations play a central role in the rest of the paper

23 The overhang problem The natural formulation of the overhang problem isnow

What is the maximum overhang achieved by a balanced n-block stack

The main result of this paper is

Theorem 23 The overhang achieved by a balanced n-block stack is at most 6n13

The fact that the stacks in the theorem above are required to be balanced but notnecessarily stable only strengthens our result By the nature of the overhang problemstacks that achieve a maximum overhang are on the verge of collapse and thus unstableIn most cases however overhangs arbitrarily close to the maximum overhang may beobtained using stable stacks (Probably the only counterexample is the case n = 3)


3 THE PATERSON-ZWICK CONSTRUCTION Paterson and Zwick [14]describe a family of balanced n-block stacks that achieve an overhang of about(3n16)13 057n13 More precisely they construct for every integer d ge 1 abalanced stack containing d(d minus 1)(2d minus 1)3 + 1 2d33 blocks that achievesan overhang of d2 Their construction for d = 6 is illustrated in Figure 7 Theconstruction is an example of what they term a brick-wall stack which resemblesthe simple ldquostretcher-bondrdquo pattern in real-life bricklaying In each row the blocksare contiguous with each block centered over the ends of blocks in the row beneathOverall the stack is symmetric and has a roughly parabolic shape with vertical axis atthe table edge

Figure 7 A ldquo6-stackrdquo consisting of 111 blocks and giving an overhang of 3 taken from [14]

The stacks of [14] are constructed in the following simple manner A t-row is a rowof t adjacent blocks symmetrically placed with respect to x = 0 An r-slab has height2r minus 3 and consists of alternating r -rows and (r minus 1)-rows the bottom and top rowsbeing r -rows An r -slab therefore contains r(r minus 1) + (r minus 1)(r minus 2) = 2(r minus 1)2

blocks A 1-stack is a single block balanced at the edge of the table a d-stack isdefined recursively as the result of adding a d-slab symmetrically onto the top of a(d minus 1)-stack The construction itself is just a d-stack and so has overhang d2 its to-tal number of blocks is given by n = 1 + sumd

r=1 2(r minus 1)2 = d(d minus 1)(2d minus 1)3 + 1It is shown in [14] using an inductive argument that d-stacks are balanced for anyd ge 1

Why should a parabolic shape be appropriate Some support for this comes fromconsidering the effect of a block in spreading a single force of f acting from belowinto two forces of almost f2 exerted upwards from its edges This spreading behavioris analogous to a symmetric random walk on a line or to difference equations for theldquoheat-diffusionrdquo process in a linear strip In both cases we see that time of about d2 isneeded for effective spreading to width d corresponding to a parabolic stack profile

Our main result Theorem 23 states that the parabolic stacks of [14] are optimal upto constant factors Better constant factors can probably be obtained however Patersonand Zwick [14] present some numerical evidence to suggest that for large values of nthe overhang achievable using n blocks is at least 102n13 For more on this seeSection 6

4 MASS MOVEMENT PROBLEMS Our upper bound on the maximum achiev-able overhang is obtained by considering mass movement problems that are an abstrac-tion of the way in which balancing forces ldquoflowrdquo though a stack of blocks (See the


discussion at the end of Section 22) In a mass movement problem we are required totransform an initial mass distribution into a mass distribution that satisfies certain con-ditions The key condition is that a specified amount of mass be moved to or beyond acertain position We transform one mass distribution into another by performing localmoves that redistribute mass within a unit interval in a way that preserves the totalmass and the center of mass Our goal is then to show that many moves are required toaccomplish the task As can be seen masses here correspond to forces mass distribu-tions correspond to collections of forces and moves mimic the effects of blocks

The mass movement problems considered are formally defined in Sections 41and 42 The correspondence between the mass movement problems considered andthe overhang problem is established in Section 43 The bounds on mass movementproblems that imply Theorem 23 are then proved in Section 5

41 Distributions

Definition 41 (Distributions and signed distributions) A discrete mass distribu-tion is a set μ = (x1 m1) (x2 m2) (xk mk) where k ge 0 x1 x2 xk arereal numbers and m1 mk gt 0 A signed distribution μ is defined the same waybut without the requirement that m1 m2 mk gt 0

If μ = (x1 m1) (x2 m2) (xk mk) is a (signed) distribution then for any setA sube R we define

μ(A) =sumxi isinA

mi

For brevity we use μ(a) as a shorthand for μ(a) and μx gt a as a shorthand forμ(x | x gt a) (Note that x here is a formal variable that does not represent a specificreal number) We similarly use μx ge a μx lt a μa lt x lt b μ|x | ge a etcwith the expected meaning

We say that a (signed) distribution is on the interval [a b] if μ(x) = 0 for everyx isin [a b]

For every A sube R we let μA be the restriction of μ to A

μA = (xi mi ) | xi isin AIf μ1 and μ2 are two signed distributions we let μ1 + μ2 and μ1 minus μ2 be the signed

distributions for which

(μ1 + μ2)(x) = μ1(x) + μ2(x) for every x isin R

(μ1 minus μ2)(x) = μ1(x) minus μ2(x) for every x isin R

Definition 42 (Moments) Let μ = (x1 m1) (x2 m2) (xk mk) be a signeddistribution and let j ge 0 be an integer The j th moment of μ is defined to be

M j [μ] =ksum

i=1

mi xj

i

Note that M0[μ] is the total mass of μ M1[μ] is the torque of μ with respect to theorigin and M2[μ] is the moment of inertia of μ again with respect to the origin IfM0[μ] = 0 we let C[μ] = M1[μ]M0[μ] be the center of mass of μ


Less standard but crucial for our analysis is the following definition

Definition 43 (Spread) The spread of a distribution μ = (x1 m1) (x2 m2)

(xk mk) is defined as follows

S[μ] =sumilt j

|xi minus x j | mi m j

If M0[μ] = 1 then μ defines a discrete random variable X for which Pr[X = x] =μ(x) for every x isin R The spread S[μ] is then half the average distance betweentwo independent drawings from μ We also then have M1[μ] = E[X ] and M2[μ] =E[X 2] If M1[μ] = E[X ] = 0 then M2[μ] = E[X 2] = Var[X ] It is also worthwhilenoting that if μ1 and μ2 are two distributions then for any k ge 0 Mk[μ1 + μ2] =Mk[μ1] + Mk[μ2] ie Mk is a linear operator

An inequality that proves very useful in the sequel is the following

Lemma 44 For any discrete distribution μ we have S[μ]2 le 13 M2[μ]M0[μ]3

The proof of Lemma 44 is given in Section 54

42 Mass redistribution moves

Definition 45 (Moves) A move v = ([a b] δ) consists of a unit interval [a b] sob = a + 1 and a signed distribution δ on [a b] with M0[δ] = M1[δ] = 0 A move v

can be applied to a distribution μ if the signed distribution μprime = μ + δ is a distributionin which case we denote the result μprime of this application by vμ We refer to a+b

2 as thecenter of the move

Note that if v is a move and μprime = vμ then M0[μprime] = M0[μ] and M1[μprime] = M1[μ]and consequently C[μprime] = C[μ]

A sequence V = 〈v1 v2 v〉 of moves and an initial distribution μ0 naturallydefine a sequence of distributions μ0 μ1 μ where μi = viμiminus1 for 1 le i le (It is assumed here that vi can indeed be applied to μiminus1) We let V μ0 = μ

Moves and sequences of moves simulate the behavior of weightless blocks andstacks However the blocks that we are interested in have unit weight Instead of ex-plicitly taking into account the weight of the blocks as we briefly do in Section 43 itturns out that it is enough for our purposes to impose a natural restriction on the movesequences considered We start with the following definition

Definition 46 (μmax) If μ0 μ1 μ is a sequence of distributions and a isin R wedefine

μmaxx ge a = max0leile

μi x ge a

Expressions like μmaxx gt a μmaxx le a and μmaxx lt a are defined similarly

Definition 47 (Weight-constrained sequences) A sequence V = 〈v1 v2 v〉of moves that generates a sequence μ0 μ1 μ of distributions is said to beweight-constrained (with respect to μ0) if for every a isin R the number of moves in Vcentered in [a infin) is at most μmaxx ge a


In Section 43 we will show the following relation between stacks and weight-constrained sequences

Lemma 48 If there is a stack composed of n blocks of length 1 and weight 1 thatachieves an overhang of d there is a weight-constrained sequence of moves trans-forming an initial distribution μ with M0[μ] = μx le 0 = n and μx gt 0 = 0 intoa distribution ν with νx ge d minus 1

2 ge 1

The main technical result of this paper is the following result for weight-constrainedsequences

Proposition 49 If a distribution ν is obtained from a distribution μ with μx le 0 =n ge 1 and μx gt 0 = 0 by a weight-constrained move sequence then νx gt

6n13 minus 12 lt 1

Theorem 23 the main result of this paper follows immediately from Proposi-tion 49 and Lemma 48

For general move sequences we have the following almost tight result which mightbe of some independent interest In particular it shows that the weight constraint onlyhas a logarithmic effect on the maximal overhang

Proposition 410 If a distribution ν is obtained from a distribution μ with μx le 0= n ge 1 and μx gt 0 = 0 by a move sequence of length at most n then νx gt

2n13 log2 n lt 1

43 From overhang to mass movement In this subsection we will prove Lemma48 capturing the essential relation between stacks and weight-constrained sequencesThe moves of Definition 45 mimic the effect that a block can have on the collectionsof forces within a stack They fail to take into account however the fact that the weightof a block is ldquoused uprdquo by the move and is then lost To faithfully simulate the forcesbetween blocks we need to introduce the slightly modified definition of lossy movesThese lossy moves are only introduced to prove Lemma 48 and will only be usedhere in Section 43

Definition 411 (Lossy moves) If v = ([a b] δ) is a move then the lossy move vdarrassociated with it is vdarr = ([a b] δdarr) where δdarr = δ minus ( a+b

2 1) A lossy move vdarr

can be applied to a distribution μ if μprime = μ + δdarr is a distribution in which case wedenote the result μprime of this application by vdarrμ

Note that if vdarr = ([a b] δdarr) is a lossy move and μprime = vdarrμ then M0[μprime] =M0[μ] minus 1 and M1[μprime] = M1[μ] minus a+b

2 Hence lossy moves do not preserve totalmass or center of mass

If V = 〈v1 v2 v〉 is a sequence of moves we let V darr = 〈vdarr1 v

darr2 v

darr 〉 be

the corresponding sequence of lossy moves If μ0 is an initial distribution we cannaturally define the sequence of distributions μ0 μ1 μ where μi = v

darri μiminus1 for

1 le i le obtained by applying V darr to μ0A collection of forces Fi may also be viewed as a mass distribution The following

lemma is now a simple formulation of the definitions and the discussion of Section 22

Lemma 412 Let B1 B2 Bn be a balanced stack on the table B0 Let Fi be acollection of balancing forces acting between B0 Bi and Bi+1 Bn for


0 le i lt n Let xi+1 be the x-coordinate of the left edge of Bi+1 Then Fi+1 can beobtained from Fi by a lossy move in the interval [xi+1 xi+1 + 1]

We can now link stacks to sequences of lossy moves

Lemma 413 If there is a stack composed of n blocks of length 1 and weight 1 thatachieves an overhang of d then there is sequence of at most n lossy moves which canbe applied to some distribution μ with M0[μ] = μx le 0 = n and μx gt 0 = 0and which finishes with a lossy move centered at d minus 1

2

Proof Let B1 B2 Bn be a balanced stack and let Bk be a block in it thatachieves an overhang of d As in Lemma 412 we let Fi be a collection of balancingforces acting between B0 Bi and Bi+1 Bn We let μ = F0 and ν = Fk It follows from Lemma 412 that ν may be obtained from μ by a sequence of k lossymoves As all the forces in μ = F0 are forces applied by the table B0 and as the tablesupports the weight of the n blocks of the stack we have M0[μ0] = μx le 0 = n andμx gt 0 = 0 Block Bk corresponds to a lossy move centered at d minus 1

2

The next simple lemma shows that sequences of lossy moves can be easily con-verted into weight-constrained sequences of moves and distributions that ldquodominaterdquothe original sequence

Lemma 414 If μ0 μ1 μ is a sequence of distributions obtained by a sequenceof lossy moves then there exists a sequence of distributions μprime

0 μprime1 μprime

obtainedby a weight-constrained sequence of moves such that μprime

0 = μ0 and μprimei (x) ge μi (x)

for 1 le i le and x isin R Furthermore if vdarr = ([d minus 1 d] δ) is the final lossy move

in the sequence then μprimex ge d minus 1

2 ge 1

Proof The sequence μprime0 μ

prime1 μprime

is obtained by applying the sequence of moveswith which the original sequence of lossy moves is associated More formally if μi =v

darri μiminus1 we let μprime

i = viμprimeiminus1 If vi = ([a minus 1

2 a + 12 ] δ) then μprime

i now has an extra massof size 1 at a This mass is frozen and will not be touched by subsequent movesHence if k moves have their centers at or beyond position a then μprime

maxx ge a geμprime

x ge a ge k as required by the definition of weight-constrained sequences Wenote that the move v leaves a mass of size 1 at d minus 1

2

Lemma 48 follows immediately from Lemmas 413 and 414

5 BOUNDS ON MASS MOVEMENT PROBLEMS This section is devoted tothe proofs of Propositions 49 and 410 As mentioned Proposition 49 implies Theo-rem 23 which states that an n-block stack can have an overhang of at most 6n13

51 Moves for a large second moment In this subsection we want to show that ittakes a lot of moves to convert an initial distribution μ = (0 1) into a distribution ν

with mass p at distance at least d from 0 that is ν|x | ge d ge p This will later beused to prove that it takes many moves to move a fraction p of the mass out at distanced from the side of the table

Note that M2[μ] = 0 while M2[ν] ge pd2 Our analysis is based on this increase insecond moment The precise result of this subsection is as follows


Lemma 51 If a sequence of moves transforms μ = (0 1) into a distribution ν

then ge (3M2[ν])32 In particular if ν|x | ge d ge p where d gt 0 and 0 lt p lt 1then ge (3p)32d3

In order to prove Lemma 51 we introduce special operations called clearanceswhich are only used here in Section 51 and for the proof of Lemma 53 in Section 54

Definitions 52 (Clearances) Associated with each unit interval [a b] is an operationclearance that can be applied to any distribution μ and produces the distribution μprimewhere μprimea lt x lt b = 0 μprime(x) = μ(x) for every x isin [a b] M0[μ] = M0[μprime] andM1[μ] = M1[μprime] In other words the clearance operation moves all the mass in theinterval [a b] into the endpoints of this interval while maintaining the center of massIf v is a move on an interval [a b] we let v denote the clearance associated with[a b] If V is a sequence of moves we let V denote the corresponding sequence ofclearances

It is a simple observation that for any move v and any distribution μ there is aunique move vprime such that vprimeμ = vμ

Closely related to Lemma 44 is the following lemma

Lemma 53 If μ1 is obtained from μ0 by a clearance then

S[μ1] minus S[μ0] ge 3(M2[μ1] minus M2[μ0])2

The proof of Lemma 53 is deferred to Section 54In the rest of this subsection we will prove the following two lemmas

Lemma 54 If a sequence of moves transforms a distribution μ into a distributionν then there is a corresponding sequence of clearances transforming μ into a dis-tribution ν such that M2[ν] le M2[ν]Lemma 55 If a sequence of clearances transforms μ = (0 1) into a distributionν then ge (3M2[ν])32

Combining the two lemmas we immediately get Lemma 51

511 Clearances and splits We will now prove Lemma 54 by showing that the ob-vious sequence of clearances gives a distribution ν satisfying the inequality The corre-spondence is simple to describe but it is less simple to prove that it can only increasethe second moment

We are going to prove the following specific form of Lemma 54

Lemma 56 If V is a sequence of moves that can be applied to μ then M2[V μ] leM2[V μ]

In order to prove Lemma 56 we define splitting which induces a natural partialorder on distributions This is only used here in Section 511

Definition 57 (Splitting) Let μ and μprime be two distributions We say that μprime is a basicsplit of μ denoted μ 1 μprime if μprime is obtained by taking one of the point masses (xi mi )

of μ and replacing it by a collection (x prime1 m prime

1) (x prime m prime

) of point masses with total


mass mi and center of mass at xi We say that μ splits into μprime denoted μ μprime if μprimecan be obtained from μ by a sequence of zero or more basic splits

To prove Lemma 56 we will show (1) that μ μprime implies M2[μ] le M2[μprime] and(2) that V μ V μ

The following two lemmas summarize simple properties of splits and clearancesthat will be explicitly or implicitly used in this section Their obvious proofs are omit-ted

Lemma 58

(i) If μ μprime and μprime μprimeprime then μ μprimeprime(ii) If μ1 μprime

1 and μ2 μprime2 then μ1 + μ2 μprime

1 + μprime2

(iii) For any distribution μ we have (C[μ] M0[μ]) μ

(iv) If μ = (x1 m1) (x2 m2) and μprime = (x prime1 m prime

1) (x prime2 m prime

2) where x prime1 le x1 le

x2 le x prime2 M0[μ] = M0[μprime] and C[μ] = C[μprime] then μ μprime

Lemma 59

(i) If vμ is defined then vμ vμ

(ii) If v is a clearance then μ vμ

(iii) If v is a clearance then v(μ1 + μ2) = vμ1 + vμ2

The following lemma shows that splitting increases the second moment

Lemma 510 If μ μprime then M2[μ] le M2[μprime]

Proof Due to the linearity of M2 and the fact that is the transitive closure of 1 it isenough to prove the claim when μ = (x m) is composed of a single mass and μprime =(x prime

1 m prime1) (x prime

k m primek) is obtained from μ by a basic split For any distribution ν =

(x1 m1) (xk mk) and any c isin R we define M2[ν c] = sumki=1 mi (xi minus c)2 to

be the second moment of ν about c As M0[μ] = M0[μprime] and M1[μ] = M1[μprime] asimple calculation shows that M2[μprime c] minus M2[μ c] = M2[μprime] minus M2[μ] for any c isinR Choosing c = x and noting that M2[μ x] = 0 while M2[μprime x] ge 0 we get therequired inequality

The next lemma exhibits a relation between clearances and splitting

Lemma 511 If μ μprime and v is a move that can be applied to μ then vμ vμprime

Proof We show that vμ vμ vμprime and use Lemma 58(i) The first relation isjust Lemma 59(i) It remains to show vμ vμprime By Lemma 59(iii) it is enough toprove the claim for μ = (x m) composed of a single mass Let [a b] be the intervalcorresponding to v There are two cases If x isin (a b) then

vμ = μ μprime vμprime

as required using Lemma 59(ii) The more interesting case is when x isin (a b) Letν = vμ = (a m1) (b m2) and ν prime = vμprime Let ν prime

= ν prime(minusinfina] and ν prime

r = ν prime[binfin) As v

leaves no mass in (a b) we get that ν prime = ν prime + ν prime

r Let m = M0[ν prime] mr = M0[ν prime

r ]


x = C[ν prime] and xr = C[ν prime

r ] As x le a lt b le xr we get using Lemma 58(iv) (iii)and (ii) that

ν = (a m1) (b m2) (x m) (xr mr )= (x m) + (xr mr ) ν prime

+ ν primer = ν prime

as required

Using induction we easily obtain the following

Lemma 512 If V is a sequence of moves that can be applied to μ then V μ V μ

Lemma 56 follows directly from Lemma 512 and Lemma 510 and Lemma 56implies Lemma 54

512 Spread vs second moment We will now prove Lemma 55 stating that ifa sequence of clearances transforms μ = (0 1) into a distribution ν then ge(3M2[ν])32 The bound relies heavily on Lemma 44 which relates the spread andsecond moment of a distribution and on Lemma 53 which relates differences inspread to differences in second moments

Let μ = μ0 μ1 μ = ν be the sequence of distributions obtained by the se-quence of clearances Note that M0[μi ] = 1 for all i and that S[μ0] = M2[μ0] = 0

Let hi = M2[μi ] minus M2[μiminus1] for 1 le i le Then

M2[μ] = M2[μ] minus M2[μ0] =sum

i=1

hi

By Lemma 53 we get that

S[μ] = S[μ] minus S[μ0] ge 3sum

i=1

h2i

Using the Cauchy-Schwartz inequality to justify the second inequality below we get

S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2

Moreover by Lemma 44 on arbitrary distributions we have

S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2

S[μ] ge 3M2[μ]2radic13 M2[μ]

= (3M2[μ])32

This completes the proof of Lemma 55

52 Mirroring In Section 51 we showed that a lot of moves are needed to movesome proportion of the mass outwards from the origin To relate this to the overhang


problem we need to transform this result into a lower bound on the moves needed tomove a proportion some distance to the right of the origin To achieve this we use asymmetrization approach showing that for any move sequence which shifts mass to theright there is a sequence of similar length which spreads the initial mass outwards Thisnew sequence corresponds roughly to interleaving the original sequence with a mirrorimage of itself The main technical difficulty comes from making this constructionwork correctly around the origin

The main result of this section is

Lemma 513 Let μ0 μ1 μ be a sequence of distributions obtained by applyinga sequence of moves to an initial distribution μ0 with μ0x gt r = 0 If μmaxx gt r lem and μmaxx ge r + d ge pm where d gt 1 and 0 lt p lt 1 then the sequence ofmoves must contain at least

radic3p32(d minus 1

2 )3 moves whose centers are in (r + 1

2 infin)

The lemma follows immediately from the following lemma by shifting coordinatesand renormalizing masses

Lemma 514 Let μ0 μ1 μ be a sequence of distributions obtained by ap-plying a sequence of moves to an initial distribution μ0 with μ0x gt minus 1

2 = 0 Ifμmax

x gt minus 1

2

le 1 and μmaxx ge d ge p where d gt 12 and 0 lt p lt 1 then the

sequence of moves must contain at leastradic

3p32d3 moves whose centers are at strictlypositive positions

Proof We may assume without loss of generality that the first move in the sequencemoves some mass from (minusinfin minus 1

2 ] into (minus 12 infin) and that the last move moves some

mass from (minusinfin d) to [d infin) Hence the center of the first move must be in (minus1 0]and the center of the last move must be at a positive position

We shall show how to transform the sequence of distributions μ0 μ1 μ

into a sequence of distributions μprime0 μ

prime1 μprime

prime obtained by applying a sequenceof prime moves such that μprime

0 = (0 1) μprimeprime |x | ge d ge p and such that the number

of moves prime in the new sequence is at most three times the number + of positivelycentered moves in the original sequence The claim of the lemma would then followimmediately from Lemma 51

The first transformation is ldquonegative truncationrdquo where in each distribution μi weshift mass from the interval (minusinfin minus 1

2) to the point minus 12 Formally the resulting distri-

butionrarrμi is defined by

rarrμi (x) =

⎧⎪⎨⎪⎩

μi (x) if x gt minus 12

1 minus μi x gt minus 12 if x = minus 1

2

0 if x lt minus 12

Note that the total mass of each distribution is 1 and thatrarrμ0 = (minus 1

2 1) Let δi =μi minus μiminus1 be the signed distribution associated with the move that transforms μiminus1

into μi and let [ci minus 12 ci + 1

2 ] be the interval in which it operates For brevity werefer to δi as the move itself with its center ci clear from the context We now compare

the transformed ldquomovesrdquorarrδ i = rarr

μi minus rarrμiminus1 with the original moves δi = μi minus μiminus1

If ci gt 0 then δi acts above minus 12 and

rarrδ i = δi If ci le minus1 then δi acts at or below

minus 12 so

rarrδ i is null and

rarrμi = rarr

μiminus1 In the transformed sequence we skip all such null

moves The remaining case is when the center ci of δi is in (minus1 0] In this caserarrδ i


acts within [minus 12

12 ] and we view it as centered at 0 However typically

rarrδ i does not

define a valid move as it may change the center of mass We call suchrarrδ i semi-moves

Whenever we have two consecutive semi-movesrarrδ i and

rarrδ i+1 we combine them into

a single semi-moverarrδ i + rarr

δ i+1 takingrarrμiminus1 directly to

rarrμi+1 The resulting sequence

we may call the cleaned negative truncation For simplicity we reindex the surviv-ing distributions consecutively so that

rarrμ0

rarrμ1

rarrμprime prime le is the cleaned negative

truncation It contains all the original positively centered moves and now at least ev-ery alternate move is of this type Since the last move in the original sequence waspositively centered we conclude

Claim 515 The cleaned negative truncation is composed of original positively cen-tered moves and semi-moves (acting within [minus 1

2 12 ]) The sequence begins with a semi-

move and at most half of its elements are semi-moves

Next we create a reflected copy of the cleaned negative truncation First we define thereflected copy

larrμi of

rarrμi by

larrμi (x) = rarr

μi (minusx) for every x isin R

and define the reflected (semi-)moveslarrδ i = larr

μi minus larrμiminus1 We can now define the mirrored

distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi

Note thatharrμ0 = rarr

μ0 + larrμ0 = (minus 1

2 1) ( 12 1) The distribution

harrμ2i+1 is obtained from

harrμ2i by the (semi-)move

rarrδ i+1 and the distribution


harrμ2i+1 by the

(semi-)movelarrδ i+1 Now comes a key observation

Claim 516 Ifrarrδ i and

larrδ i are semi-moves on [minus 1

2 12 ] then their sum

harrδ i = rarr

δ i + larrδ i

defines an ordinary move centered at 0 and acting on [minus 12

12 ]

Proof Bothrarrδ i and

larrδ i preserve the total mass As

harrδ i is symmetric about 0 it cannot

change the center of mass

As suggested by the above observation ifrarrδ i and

larrδ i are semi-moves we sum them

into a single ordinary moveharrδ i = rarr

δ i + larrδ i centered at 0 taking us directly from

harrμ2i to

harrμ2i+2

Claim 517 After the above summing of semi-move pairs we have a sequence of atmost 3+ ordinary moves taking us from

harrμ0 to

harrμ2 Here + is the number of positively

centered moves in the original sequence The first move in the sequence is the addedharrδ 0 acting on [minus 1

2 12 ] centered in 0 and taking us to

harrμ2

Proof By Claim 515 before the mirroring we had + ordinary movesrarrδ i and at most

+ semi-moves acting on [minus 12

12 ] For each ordinary move we get a reflected move

larrδ i For each semi-move we get the single merged move

harrδ i By Claim 515

rarrδ 0 is a

semi-move that gets added with its reflection


We now replace the initial distributionharrμ0 = (minus 1

2 1) ( 12 1) by the distribution

μprime0 = (0 2) which has the same center of mass and replace the first move by δprime

1 =harrδ 1 + (minus 1

2 1) (0 minus2) ( 12 1) The distribution after the first move is then again

harrμ2

We have thus obtained a sequence of at most 3+ moves that transforms μprime0 =

(0 2) into a distribution ν prime = harrμ with ν prime|x | ge d ge 2p Scaling these distributions

and moves by a factor of 2 we get by Lemma 51 that 3+ ge (3p)32d3 as claimed

53 Proofs of Propositions 49 and 410 We prove the following lemma which eas-ily implies Proposition 49 from Section 42

Lemma 518 Let μ0 μ1 μ be a sequence of distributions obtained by ap-plying a weight-constrained sequence of moves on an initial distribution μ0 withμ0x gt r = 0 If for some real m ge 15 we have μmaxx gt r le m then μmax

x gt

r + 6m13 minus 12

= 0

Proof We start by showing that the claim of the lemma holds for all m isin [15 1)We then show that for m ge 1 if the claim holds for m5 then it also holds for m Byinduction we get that the claim holds for all m ge 15 (More formally we are showingby induction on k that the claim of the lemma holds for all m isin [15 5k))

If m isin [15 1) then as μmax x gt r le m lt 1 and the sequence is weight-constrained there is no move whose center is greater than r Hence μmax

x gt r + 1

2

= 0 Since 6m13 minus 1

2 ge 6(15)13 minus 12 gt 1

2 we get the claim of the lemma Thisestablishes the base case of the induction

Suppose therefore that μmax x gt r le m where m ge 1 and that the claim of thelemma holds for m5 (We refer to this last condition as the induction hypothesis) Letu be the least number which is at least 2 such that μmax x gt r + u le m5 By theinduction hypothesis with r replaced by r + u we get that

μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0

If u gt 2 we have μmax x ge r + u ge m5 We also have μ0x gt r = 0 andμmaxx gt r le m By Lemma 513 with d = u gt 1 we get that the sequencemust contain at least

radic3(15)32(u minus 1

2 )3 gt 1

7 (u minus 12)

3 moves whose centers are in(r + 1

2 infin) As the sequence of moves is weight-constrained and as μmaxx gt r le mthere can be at most m such moves with centers greater than r ie

1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2

Since m ge 1 this bound is greater than 2 so it also holds when u = 2 Thus

u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)

m13 lt 55m13 le 6m13 minus 1

2

This proves the induction step and completes the proof


Modulo the proofs of Lemmas 44 and 53 which are given in the next sectionthis completes the proof of our main result that the maximum overhang that canbe achieved using n blocks is at most 6n13 It is fairly straightforward to mod-ify the proof of Lemma 518 above so as to obtain the stronger conclusion thatμmax

x ge cn13 minus 1

2

lt 1 for any c gt 552(2 middot 353) 4479 at least for large

enough values of n and hence an improved upper bound on overhang of 45n13say In the modified proof we would choose u to be the least number for whichμmax x ge r + u ge (27125)m (The constant 27125 here is the optimal choice)The proof however becomes slightly messier as several of the inequalities do nothold for small values of m

Next we prove the following lemma which easily implies Proposition 410

Lemma 519 Let μ0 μ1 μn be a sequence of distributions obtained by apply-ing a sequence of n moves to an initial distribution μ0 with μ0x gt 0 = 0 andμ0x le 0 = n Then μnx gt 2n13 log2 n lt 1

Proof The result is immediate when n = 1 and it is easy to verify whenever n lt

2n13 log2 n ie for 2 le n le 31 From now on we assume just that n ge 8Let k = log2 n + 1 gt log2 n Define u0 = 0 and for i = 1 k let ui be the least

number which is at least uiminus1 + 2 and such that μmaxx gt ui le n2i If ui gt uiminus1 + 2we have μmaxx ge ui ge n2i ge μmaxx gt uiminus12 By Lemma 513 applied withr = uiminus1 d = ui minus uiminus1 gt 1 and p = 1

2 and noting that the total number of moves isn we find that

radic3p32

(d minus 1

2

)3

le n

ie

ui minus uiminus1 leradic

2

316n13 + 1

2lt

3

2n13

since n ge 4 This bound on ui minus uiminus1 also holds if ui minus uiminus1 = 2 so we conclude thatui minus uiminus1 lt 3

2 n13 for all i Then

uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n

since n ge 8 Because μmaxx gt uk le n2k lt 1 this completes the proof

As before the constants in the above proof are not optimized We believe that astronger version of the lemma which states under the same conditions that μnx gt

cn13(log2 n)23 lt 1 for some c gt 0 actually holds This would match an examplesupplied by Johan Hastad Lemma 519 (and Proposition 410) imply an almost tightbound on an interesting variant of the overhang problem that involves weightlessblocks as discussed in Section 6

54 Proof of spread vs second moment inequalities

Lemma 44 (The proof was deferred from Section 41) For any discrete distributionμ

S[μ]2 le 1

3M2[μ]M0[μ]3


The method of proof used here was suggested to us by Benjy Weiss and resulted ina much improved and simplified presentation The lemma is essentially the case n = 2of a more general result proved by Plackett [16]

Proof Suppose that μ = (x1 m1) (xk mk) where x1 lt x2 lt middot middot middot lt xk We first transform the coordinates into a form which will be more convenient for

applying the Cauchy-Schwartz inequality Since the statement of the lemma is invari-ant under scaling of the masses we may assume that M0[μ] = 1 Also since S isinvariant under translation and M2 is minimized by a translation which moves C[μ] tothe origin we may assume without loss of generality that C[μ] = 0

Define a function g(t) for minus 12 lt t le 1

2 by

g(t) = xi whereiminus1sumr=1

mr lt t + 1

2le

isumr=1

mr

and define g(minus 12 ) = x1 Thus the value of g is x1 on [minus 1

2 minus 12 + m1] x2 on (minus 1

2 + m1

minus 12 + m1 + m2] etcNow we have that

M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j

mi m j (x j minus xi ) =int 12

t=minus12

int t

s=minus12(g(t) minus g(s)) ds dt

Above it may seem that the integral should have been restricted to the case whereg(s) lt g(t) However if g(t) = g(s) the integrand is zero so this case does not con-tribute to the value of the integral

Recall that C[μ] = 0 so that M1[μ] = int 12t=minus12 g(t) dt = 0 Therefore

int 12

t=minus12

int t

s=minus12g(t) ds dt =

int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t

s=minus12g(s) ds dt =

int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12

t=minus12tg(t) dt (dagger)


Using the Cauchy-Schwartz inequality

S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12

t=minus12g(t)2 dt middot

int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]

Lemma 53 (The proof was deferred from Section 42) If μ1 is obtained from μ0 bya clearance then


Proof For any move the resulting changes in spread and second moment are invariantunder linear translation of the coordinates so we may assume that the interval of themove is [minus 1

2 12 ] These differences are also independent of any weight outside the

interval of the move so we may assume that μ0 has all its support in [minus 12

12 ]

Since our move is a clearance the addition of an extra point mass at either minus 12 or

12 leaves the differences in spread and second moment invariant We may thereforeadd such a mass to bring the center of mass of μ0 to 0 Finally since the statementof the lemma is invariant under scaling of the masses we may further assume thatM0[μ0] = 1

Since the clearance pushes all the mass to the endpoints of the interval we get

μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4

We define g(t) for minus 12 le t le 1

2 just as in the proof of Lemma 44 but now corre-

sponding to the distribution μ0 As before M j [μ0] = int 12minus12 g(t) j dt for j ge 0 and we

recall as in (dagger) that S[μ0] = 2int 12

minus12 tg(t) dt

We have M1[μ0] = int 12minus12 g(t) dt = 0 Let c = M2[μ0] = int 12

minus12 g(t)2 dt and s =S[μ0] By Lemma 44 we have s2 le c

3 If c le 112 then s le radic

c3 le 1

6 and the resultfollows immediately as

S[μ1] minus S[μ0] minus 3(M2[μ1] minus M2[μ0])2 = 1

4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0

We next claim that if c = M2[μ0] gt 112 then s = S[μ0] le 1

4 minus a2

12 where a =32 minus 6c lt 1 To prove this claim we define a function h(t) as follows

h(t) =

ta if |t | le a

2 12 sgn(t) otherwise


We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12

minus12t h(t) dt = 1

8minus a2

24

By the Cauchy-Schwartz inequality

(int 12

minus12h(t)g(t) dt

)2

leint 12

minus12h(t)2 dt middot

int 12

minus12g(t)2 dt = c2

and so int 12

minus12h(t)g(t) dt le c =

int 12

minus12h(t)2 dt (lowast)

We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)

)h(t) dt (lowastlowast)

since h(t) minus g(t) le 0 and ta minus h(t) le 0 for t lt minus a

2 h(t) minus g(t) ge 0 and ta minus h(t) ge 0

for t gt a2 and t

a minus h(t) = 0 for |t | le a2 Adding inequalities (lowast) and (lowastlowast) and multi-

plying by 2a gives

S[μ0] = 2int 12

minus12t g(t) dt le 2

int 12


4minus a2

12

Finally

S[μ1] minus S[μ0] ge 1

4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

This completes the proof

We end the section by noting that although the inequalities of Lemmas 44 and 53are only claimed for discrete distributions which is all we need in this paper ourproofs can easily be modified to show that they also hold for general continuous distri-butions In fact for nontrivial discrete distributions the inequalities in the two lemmasare always strict In the continuous case the inequalities are satisfied with equality byappropriately chosen uniform distributions In particular the constant factors 1

3 and 3appearing in the two lemmas cannot be improved

6 CONCLUDING REMARKS AND OPEN PROBLEMS We have shown thatthe maximum overhang achieved using n homogeneous frictionless blocks of unitlength is at most 6n13 Thus the constructions of [14] cannot be improved by morethan a constant factor establishing order n13 as the asymptotic answer to the age-oldoverhang problem

The discussions and results presented so far all referred to the standard two-dimensional version of the overhang problem Our results hold however in greatergenerality We briefly discuss some natural generalizations and variants of the over-hang problem for which our bounds still apply


In Section 2 we stipulated that all blocks have a given height h It is easy to seehowever that all our results remain valid even if blocks have different heights but stillhave unit length and unit weight In particular blocks are allowed to degenerate intosticks ie have height 0 Also even though we required blocks not to overlap we didnot use this condition in any of our proofs

Loaded stacks introduced in [14] are stacks composed of standard unit length andunit weight blocks and point weights that can have arbitrary weight (Point weightsmay be considered to be blocks of zero height and length but nonzero weight) Ourresults with essentially no change imply that loaded stacks of total weight n can havean overhang of at most 6n13

What happens when we are allowed to use blocks of different lengths and weightsOur results can be generalized in a fairly straightforward way to show that if a block oflength has weight proportional to 3 as would be the case if all blocks were similarthree-dimensional cuboids then the overhang of a stack of total weight n is again oforder at most n13 It is amusing to note that in this case an overhang of order n13 canbe obtained by stacking n unit-length blocks as in the construction of [14] or simplyby balancing a single block of length n13 and weight n at the edge of the table

In case the weights are not strictly proportional to the lengths cubed we coulddefine c to be the smallest constant such that blocks of weight w have length at mostcw13 If the total weight of the stack is W then its overhang is bounded by 6cW 13

Proposition 410 supplies an almost tight upper bound for the following variant ofthe overhang problem how far away from the edge of a table can a mass of weight 1be supported using n weightless blocks of length 1 and a collection of point weights oftotal weight n The overhang in this case beats the classical one by a factor of betweenlog23 n and log n

In all variants considered so far blocks were assumed to have their largest faces par-allel to the tablersquos surface and perpendicular to its edge The assumption of no frictionthen immediately implied that all forces within a stack are vertical and our analy-sis which assumes that there are no horizontal components was applicable A niceargument communicated to us by Harry Paterson shows that in the frictionless two-dimensional case no blocks can lean against each other inducing horizontal forces Wecould have a block balanced on its corner but this would not create nonvertical forcesOur results thus apply also to this general two-dimensional case where the blocks maybe stacked arbitrarily

We believe that our bounds also apply with slightly adjusted constants in threedimensions but proving so remains an open problem Overhang larger by a factor of

Figure 8 A ldquoskintledrdquo 4-diamond


radic1 + w2 may be obtained with 1 times w times h blocks where h le w le 1 using a tech-

nique called skintling (see Figure 8) In skintling (a term we learned from an edifyingconversation with John H Conway about brick-laying) each block is rotated aboutits vertical axis so thatmdashin our casemdashthe diagonal of its bottom face is perpendic-ular to the edge of the table With length defined by the projection on the directionof the overhang our bounds apply to any three-dimensional construction that can bebalanced using vertical forces only It is an interesting open problem whether thereexist three-dimensional stacks composed of frictionless possibly tilted homogeneousrectangular blocks that can only be balanced with the aid of some nonvertical forcesFigure 9 shows that this is possible with some convex homogeneous blocks of dif-ferent shapes and sizes As mentioned we believe that our bounds do apply in threedimensions for regular blocks even if it turns out that nonvertical forces are sometimesuseful but proving this requires some additional arguments

Figure 9 A balanced stack of convex objects with nonvertical forces

We end by commenting on the tightness of the analysis presented in this paperOur main result is a 6n13 upper bound on the overhang that may be obtained using nblocks As mentioned after the proof of Lemma 518 this bound can easily be im-proved to about 45n13 for sufficiently large values of n Various other small improve-ments in the constants are possible For example a careful examination of our proofsreveals that whenever we apply Lemma 53 the distribution μ0 contains at most threemasses in the interval acted upon by the move that produces μ1 (This follows from thefact that a block can rest upon at most three other blocks) The constant 3 appearing inLemma 53 can then be improved though it is optimal when no assumption regardingthe distribution μ0 is made Finally we note our bound only counts blocks with centersstrictly over the side of the table that is if there are at most n such blocks over theside then the overhang is proved to be at most 6n13 We believe however that newideas would be needed to reduce the upper bound to below 3n13 say

As mentioned Paterson and Zwick [14] describe simple balanced n-block stacksthat achieve an overhang of about 057n13 They also present some numerical evidencethat suggests that the overhang achievable using n blocks is at least 102n13 for largevalues of n These larger overhangs are obtained using stacks that are shaped like theldquooil lamprdquo depicted in Figure 10 For more details on the figure and on ldquooil-lamprdquoconstructions see [14] (The stack shown in the figure is actually a loaded stack asdefined above with the external forces shown representing the point weights)

A small gap still remains between the best upper and lower bounds currently avail-able for the overhang problem though they are both of order n13 Determining a con-stant c such that the maximum overhang achievable using n blocks is asymptoticallycn13 is a challenging open problem


Figure 10 An ldquooil-lamprdquo loaded stack with overhang 10 having 921 blocks and total weight 111288

ACKNOWLEDGMENTS We would like to thank John H Conway Johan Hastad Harry Paterson AndersThorup and Benjy Weiss for useful discussions and observations some of which appear with due credit withinthe paper

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1 S Ainley Finely balanced Math Gaz 63 (1979) 272 doi10230736180492 J G Coffin Problem 3009 this MONTHLY 30 (1923) 76 doi10230722984903 Creative Commons License (CC-BY-SA-25) httpcreativecommonsorglicensesby-sa2

54 J E Drummond On stacking bricks to achieve a large overhang (Note 658) Math Gaz 65 (1981)

40ndash42 doi10230736179375 L Eisner Leaning tower of the Physical Review Amer J Phys 27 (1959) 121 doi1011191

19347716 G Gamow and M Stern Puzzle-Math Viking New York 19587 M Gardner Mathematical games Some paradoxes and puzzles involving infinite series and the concept

of limit Scientific American (Nov 1964) 126ndash1338 Martin Gardnerrsquos Sixth Book of Mathematical Games from Scientific American WH Freeman

San Francisco 19719 R L Graham D E Knuth and O Patashnik Concrete Mathematics Addison-Wesley Longman Read-

ing MA 198810 J F Hall Fun with stacking blocks Amer J Phys 73 (2005) 1107ndash1116 doi1011191207400711 C P Jargodzki and F Potter Mad About Physics Braintwisters Paradoxes and Curiosities John Wiley

New York 200112 P B Johnson Leaning tower of lire Amer J Phys 23 (1955) 240 doi1011191193395713 G M Minchin A Treatise on Statics With Applications to Physics 6th ed Clarendon Oxford 190714 M Paterson and U Zwick Overhang this MONTHLY 116 (2009) 19ndash44 A preliminary version appeared

in Proceedings of the 17th Annual ACM-SIAM Symposium on Discrete Algorithms (SODArsquo06) Societyfor Industrial and Applied Mathematics Philadelphia 2006 231ndash240

15 J B Phear Elementary Mechanics Macmillan Cambridge 185016 R L Plackett Limits of the ratio of mean range to standard deviation Biometrika 34 (1947) 120ndash12217 R T Sharp Problem 52 Pi Mu Epsilon J 1 (1953) 32218 Problem 52 Pi Mu Epsilon J 2 (1954) 41119 R Sutton A problem of balancing Amer J Phys 23 (1955) 547 doi1011191193409420 W Walton A Collection of Problems in Illustration of the Principles of Theoretical Mechanics 2nd ed

Deighton Bell Cambridge 1855


MIKE PATERSON (PhD FRS) took degrees in mathematics at Cambridge and during this time rose tofame as the co-inventor with John Conway of Sprouts He evolved from president of the Trinity MathematicalSociety to president of the European Association for Theoretical Computer Science and migrated throughMIT to the University of Warwick where he has been in the Computer Science department for 38 yearsDepartment of Computer Science University of Warwick Coventry CV4 7AL United Kingdommspdcswarwickacuk

YUVAL PERES is the manager of the Theory Group at Microsoft Research Redmond and an Affiliate Pro-fessor at the University of Washington Seattle Previously he taught at the Hebrew University Jerusalem andthe University of California Berkeley He is most proud of his former students and his favorite quote is fromhis son Alon who at age six was heard asking a friend ldquoLeo do you have a religion You know a religionlike Jewish or Christian or Mathematics rdquoMicrosoft Research One Microsoft Way Redmond WA 98052-6399peresmicrosoftcom

MIKKEL THORUP has a DPhil from Oxford University from 1993 From 1993 to 1998 he was at theUniversity of Copenhagen Since then he has been at ATampT LabsndashResearch He is also a Fellow of the ACMand a Member of the Royal Danish Academy of Sciences and Letters His main work is in algorithms anddata structures and he is the editor of this area for the Journal of the ACM One of his best-known results is alinear-time algorithm for the single-source shortest paths problem in undirected graphs Mikkel prefers to seekhis mathematical inspiration in nature combining the quest with his hobbies of bird watching and mushroompickingATampT LabsndashResearch Florham Park NJ 07932mthorupresearchattcom

PETER WINKLER is Professor of Mathematics and Computer Science and Albert Bradley Third CenturyProfessor in the Sciences at Dartmouth College His research is in discrete math and the theory of computingwith forays into statistical physics He has also published two collections of mathematical puzzles and in somecircles is best known for his invention of cryptologic techniques for the game of bridgeDepartment of Mathematics Dartmouth College Hanover NH 03755peterwinklerdartmouthedu

URI ZWICK received his BSc in Computer Science from the Technion Israel Institute of Technology andhis MSc and PhD in Computer Science from Tel Aviv University where he is currently a professor ofComputer Science His main research interests are algorithms and complexity combinatorial optimizationmathematical games and recreational mathematics In an early collaboration with Mike Paterson he deter-mined the optimal strategy for playing the popular Memory Game known also as Concentration His favoriteHawaiian musician is Israel KamakawiworsquooleSchool of Computer Science Tel Aviv University Tel Aviv 69978 Israelzwickcstauacil




Figure 2 Optimal stacks with 20 and 30 blocks from [14] with corresponding harmonic stacks in the back-ground

Building a stack with an overhang is of course also a construction challenge inthe real world Figure 3 shows how the Mayans used such constructions in 900 BC incorbel arches In contrast with a true arch a corbel arch does not use a keystone sothe stacks forming the sides do not benefit from leaning against each other

Figure 3 Mayan corbel arch from Cahal Pech Belize 900 BC (picture by Clark AndersonAquaimages underthe Creative Commons License [3])

11 How far can you go Many readers of the above-mentioned references wereled to believe that 1

2 Hn (sim 12 ln n) the overhang achieved by harmonic stacks is the

maximum overhang that can be achieved using n blocks This is indeed the case underthe restriction explicit or implicit in some of these references that the blocks shouldbe stacked in a one-on-one fashion with at most one block resting on each block It












and n12

















ksumi=1

fi = 1 +kprimesum

i=1

f primei

ksumi=1

xi fi =(

a + 1

2

)+

kprimesumi=1

x primei f prime

i






13

10 11 12

7 8 9

5 6

3 4

2

1

F13

F12

helliphellip

hellip

F9

F6

F4

F2

F3

F1

F0










i andsumk





















41 Distributions



μ(A) =sumxi isinA

mi









M j [μ] =ksum

i=1

mi xj

i






S[μ] =sumilt j















μi x ge a






2 ge 1



6n13 minus 12 lt 1




2n13 log2 n lt 1








darr2 v

darr 〉 be


darri μiminus1 for








2


2



0 μprime1 μprime





2 ge 1


prime1 μprime






maxx ge a geμprime


2
























1) (x prime m prime






Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES





























and n12

















ksumi=1

fi = 1 +kprimesum

i=1

f primei

ksumi=1

xi fi =(

a + 1

2

)+

kprimesumi=1

x primei f prime

i






13

10 11 12

7 8 9

5 6

3 4

2

1

F13

F12

helliphellip

hellip

F9

F6

F4

F2

F3

F1

F0










i andsumk





















41 Distributions



μ(A) =sumxi isinA

mi









M j [μ] =ksum

i=1

mi xj

i






S[μ] =sumilt j















μi x ge a






2 ge 1



6n13 minus 12 lt 1




2n13 log2 n lt 1








darr2 v

darr 〉 be


darri μiminus1 for








2


2



0 μprime1 μprime





2 ge 1


prime1 μprime






maxx ge a geμprime


2
























1) (x prime m prime






Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES





























ksumi=1

fi = 1 +kprimesum

i=1

f primei

ksumi=1

xi fi =(

a + 1

2

)+

kprimesumi=1

x primei f prime

i






13

10 11 12

7 8 9

5 6

3 4

2

1

F13

F12

helliphellip

hellip

F9

F6

F4

F2

F3

F1

F0










i andsumk





















41 Distributions



μ(A) =sumxi isinA

mi









M j [μ] =ksum

i=1

mi xj

i






S[μ] =sumilt j















μi x ge a






2 ge 1



6n13 minus 12 lt 1




2n13 log2 n lt 1








darr2 v

darr 〉 be


darri μiminus1 for








2


2



0 μprime1 μprime





2 ge 1


prime1 μprime






maxx ge a geμprime


2
























1) (x prime m prime






Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES



















13

10 11 12

7 8 9

5 6

3 4

2

1

F13

F12

helliphellip

hellip

F9

F6

F4

F2

F3

F1

F0










i andsumk





















41 Distributions



μ(A) =sumxi isinA

mi









M j [μ] =ksum

i=1

mi xj

i






S[μ] =sumilt j















μi x ge a






2 ge 1



6n13 minus 12 lt 1




2n13 log2 n lt 1








darr2 v

darr 〉 be


darri μiminus1 for








2


2



0 μprime1 μprime





2 ge 1


prime1 μprime






maxx ge a geμprime


2
























1) (x prime m prime






Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES






























41 Distributions



μ(A) =sumxi isinA

mi









M j [μ] =ksum

i=1

mi xj

i






S[μ] =sumilt j















μi x ge a






2 ge 1



6n13 minus 12 lt 1




2n13 log2 n lt 1








darr2 v

darr 〉 be


darri μiminus1 for








2


2



0 μprime1 μprime





2 ge 1


prime1 μprime






maxx ge a geμprime


2
























1) (x prime m prime






Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES






















S[μ] =sumilt j















μi x ge a






2 ge 1



6n13 minus 12 lt 1




2n13 log2 n lt 1








darr2 v

darr 〉 be


darri μiminus1 for








2


2



0 μprime1 μprime





2 ge 1


prime1 μprime






maxx ge a geμprime


2
























1) (x prime m prime






Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES





















2 ge 1



6n13 minus 12 lt 1




2n13 log2 n lt 1








darr2 v

darr 〉 be


darri μiminus1 for








2


2



0 μprime1 μprime





2 ge 1


prime1 μprime






maxx ge a geμprime


2
























1) (x prime m prime






Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES






















2


2



0 μprime1 μprime





2 ge 1


prime1 μprime






maxx ge a geμprime


2
























1) (x prime m prime






Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES




































1) (x prime m prime






Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES






















Lemma 58



1 + μprime2






Lemma 59




















r ]






as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES























as required








i=1

hi



i=1

h2i


S[μ] ge 3sum

i=1

h2i ge 3

(sum

i=1 hi )2

= 3M2[μ]2


S[μ] leradic

1

3M2[μ]

Hence

ge 3M2[μ]2


= (3M2[μ])32







radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES






















radic3p32(d minus 1


2 infin)



2 = 0 Ifμmax

x gt minus 1

2









prime1 μprime







rarrμi (x) =

⎧⎪⎨⎪⎩



2

0 if x lt minus 12









minus 12 so


rarrμi = rarr






rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES





















rarrδ i does not








rarrμ0

rarrμ1







larrμi of

rarrμi by

larrμi (x) = rarr




distributions

harrμ2i = rarr

μi + larrμi

harrμ2i+1 = rarr

μi+1 + larrμi








harrμ2i+1 by the





harrδ i = rarr

δ i + larrδ i


12 ]









harrμ2i to

harrμ2i+2


harrμ0 to




harrμ2






rarrδ 0 is a








harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES
























harrμ2






x gt

r + 6m13 minus 12

= 0



x gt r + 1

2


2 ge 6(15)13 minus 12 gt 1



μmax

x gt r + u + 6

(m

5

)13 minus 1

2

= 0



2 )3 gt 1

7 (u minus 12)



1

7

(u minus 1

2

)3

le m

Hence

u le (7m)13 + 1

2


u + 6(m

5

)13 minus 1

2le

(713 + 6 middot

(1

5

)13)


2




x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES




















x ge cn13 minus 1

2









radic3p32

(d minus 1

2

)3

le n

ie


2

316n13 + 1

2lt

3

2n13



uk lt3kn13

2= 3n13

2(log2 n + 1) le 2n13 log2 n






S[μ]2 le 1

3M2[μ]M0[μ]3






2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES























2 by


mr lt t + 1

2le

isumr=1

mr



2 + m1


M j [μ] =ksum

i=1

x ji mi =

int 12

t=minus12g(t) j dt

for j ge 0 and

S[μ] =sumilt j


t=minus12

int t




int 12

t=minus12

int t


int 12

t=minus12

(t + 1

2

)g(t) dt =

int 12

t=minus12tg(t) dt

while

int 12

t=minus12

int t


int 12

s=minus12

int 12

t=sg(s) dt ds

=int 12

s=minus12

(1

2minus s

)g(s) ds = minus

int 12

s=minus12sg(s) ds

So

S[μ] = 2int 12




S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES




















S[μ]2 = 4

(int 12

t=minus12tg(t) dt

)2

le 4int 12


int 12

t=minus12t2 dt

= 4M2[μ] middot 1

12= 1

3M2[μ]






12 ]




μ1 =(

minus1

2

1

2

)

(1

2

1

2

)and M2[μ1] = S[μ1] = 1

4





minus12 tg(t) dt




c3 le 1



4minus s minus 3

(1

4minus c

)2

ge 1

4minus s minus 3

(1

4minus 3s2

)2

= (1 + 2s)(1 minus 6s)3

16ge 0


4 minus a2


h(t) =

ta if |t | le a



We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES



















We may verify that

int 12

minus12h(t)2 dt = 1

4minus a

6= c and

int 12


8minus a2

24


(int 12

minus12h(t)g(t) dt

)2

leint 12


int 12


and so int 12


int 12


We also have int 12

minus12

(t

aminus h(t)

)g(t) dt le

int 12

minus12

(t

aminus h(t)




for t gt a2 and t


plying by 2a gives

S[μ0] = 2int 12


int 12


4minus a2

12

Finally


4minus

(1

4minus a2

12

)= a2

12= 3

(1

4minus c

)2

= 3(M2[μ1] minus M2[μ0])2

























REFERENCES





































REFERENCES



















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