MMaximum Overhang
Mike Paterson ∗ Yuval Peres † Mikkel Thorup ‡ Peter Winkler § Uri Zwick ¶ February 1, 2008
Abstract
How far can a stack of n identical blocks be made to hang over the edge of a table? The questiondates back to at least the middle of the 19th century and the answer to it was widely believed to beof order log n . Recently, Paterson and Zwick constructed n -block stacks with overhangs of order n / ,exponentially better than previously thought possible. We show here that order n / is indeed bestpossible, resolving the long-standing overhang problem up to a constant factor. The problem of stacking n blocks on a table so as to achieve maximum overhang has a long history. Itappears in physics and engineering textbooks from as early as the mid 19th century (see, e.g., [P1850],[W1855], [M1907]). The problem was apparently first brought to the attention of the mathematical com-munity in 1923 when J.G. Coffin posed it in the “Problems and Solutions” section of the American Math-ematical Monthly [C1923]; no solution was presented there.Figure 1: Optimal stacks with 3 and 4 blocks, compared to the corresponding harmonic stacks. The 4block solution is from [A1979]. Like the harmonic stacks it can be made stable by minute displacements.The problem recurred from time to time over subsequent years, e.g., [S1953, S1954, S1955, J1955, GS1958,E1959, G1964, G1971, A1979, D1981, GKP1988, H2005], achieving much added notoriety from its appear-ance in 1964 in Martin Gardner’s “Mathematical Games” column of Scientific American [G1964, G1971]. ∗ Department of Computer Science, University of Warwick, Coventry CV4 7AL, UK. E-mail: [email protected] † Department of Statistics, University of California, Berkeley, California 94710, USA. E-mail: [email protected] ‡ AT&T Labs - Research, 180 Park Avenue, Florham Park, NJ 07932, USA. E-mail: [email protected] § Department of Mathematics, Dartmouth College, Hanover, NH 03755-3551, USA. E-mail: [email protected] ¶ School of Computer Science, Tel Aviv University, Tel Aviv 69978, Israel. E-mail: [email protected] a r X i v : . [ m a t h . HO ] J u l locks = 20overhang = 2.32014 blocks = 30overhang = 2.70909 Figure 2: Optimal stacks with 20 and 30 blocks from [PZ2006] with corresponding harmonic stacks in thebackground.Most of the references mentioned above describe the now classical harmonic stacks in which n unit-lengthblocks are placed one on top of the other, with the i th block from the top extending by i beyond the blockbelow it. The overhang achieved by such stacks is H n = (cid:80) ni =1 1 i ∼ ln n . The cases n = 3 and n = 4are illustrated at the top of Figure 1 above, and the cases n = 20 and n = 30 are shown in the backgroundof Figure 2. Verifying that harmonic stacks are balanced and can be made stable (see definitions in thenext section) by minute displacements is an easy exercise. (This is the form in which the problem appearsin [P1850], [W1855], [M1907].) Harmonic stacks show that arbitrarily large overhangs can be achieved ifsufficiently many blocks are available. They have been used by countless teachers as an introduction torecurrence relations, the harmonic series and simple optimization problems (see, e.g., [GKP1988]). Many readers of the above mentioned references were led to believe that H n ( ∼ ln n ), the overhangachieved by harmonic stacks, is the maximum overhang that can be achieved using n blocks. This isindeed the case under the restriction, explicit or implicit in some of these references, that the blocks shouldbe stacked in a one-on-one fashion, with each block resting on at most one other block. It has been knownfor some time, however, that larger overhangs may be obtained if the one-on-one restriction is lifted. Threeblocks, for example, can easily be used to obtain an overhang of 1. Ainley [A1979] found that four blockscan be used to obtained an overhang of about 1.16789, as shown at the bottom right of Figure 1, and this ismore than 10% larger than the overhang of the corresponding harmonic stack. Using computers, Patersonand Zwick [PZ2006] found the optimal stacks with a given limited number of blocks. Their solutions with20 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 by more than a constant factor, or is over-hang of order log n the best that can be achieved? In a recent cover article in the American Journal ofPhysics , Hall [H2005] observes that the addition of counterbalancing blocks to one-on-one stacks can dou-ble (asymptotically) the overhang obtainable by harmonic stacks. However, he then incorrectly concludesthat no further improvement is possible, thus perpetuating the order log n “mythology”.Recently, however, Paterson and Zwick [PZ2006] discovered that the modest improvements gained for smallvalues of n by using layers with multiple blocks mushroom into an exponential improvement for large valuesof n , yielding overhang of order n / instead of just log n .2igure 3: A 3-row inverted triangle is unbalanced. But is n / the right answer, or is it just the start of another mythology? In their deservedly popular book Mad About Physics [JP2001], Jargodzki and Potter rashly claim that inverted triangles (such as the oneshown on the left of Figure 3) are balanced. If so, they would achieve overhangs of order n / . It turnsout, however, that already the 3-row inverted triangle is unbalanced, and collapses as shown on the rightof Figure 3, as do all larger inverted triangles.The collapse of the 3-row triangle begins with the lifting of the middle block in the top row. It is temptingto try to avoid this failure by using a diamond shape instead as illustrated in Figure 4. Diamonds wereconsidered by Drummond [D1981], and like the inverted triangle, they would achieve an overhang oforder n / , though with a smaller leading constant. The stability analysis of diamonds is slightly morecomplicated than that of inverted triangles, but it can be shown that d -diamonds, i.e., diamonds that have d blocks in their largest row, are stable if and only if d <
5. In Figure 4 we see a practical demonstrationwith d = 5. Figure 4: The instability of a 5-diamond in theory and practice.It is not hard to show that particular constructions like larger inverted triangles or diamonds are unstable.This instability of inverted triangles and diamonds was already noted in [PZ2006]. However, this does notrule out the possibility of a smarter balanced way of stacking n blocks so as to achieve an overhang oforder n / , and that would be much better than the above mentioned overhang of order n / achieved byPaterson and Zwick [PZ2006]. Paterson and Zwick did consider this general question. They did not ruleout an overhang of order n / , but they proved that no larger overhang would be possible. Thus their workshows that the order of the maximum overhang with n blocks has to be somewhere between n / and n / .3 .3 Our result We show here that an overhang of order n / , as obtained by [PZ2006], is in fact best possible. Morespecifically, we show that any n -block stack with an overhang of at least 6 n / is unbalanced, and musthence collapse. Thus we conclude that the maximum overhang with n blocks is of order n / . The rest of this paper is organized as follows. In the next section we present a precise mathematicaldefinition of the overhang problem, explaining in particular 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 stacksthat achieve an overhang of order n / . In Section 4 we introduce a class of abstract mass movementproblems and explain the connection between these problems and the overhang problem. In Section 5 weobtain bounds for mass movement problems that imply the order n / upper bound on overhang. We endin Section 6 with some concluding remarks and open problems. We briefly state the mathematical definition of the overhang problem. For more details, see [PZ2006]. Asin previous papers, e.g., [H2005], the overhang problem is taken here to be a two-dimensional problem:each block is represented by a frictionless rectangle whose long sides are parallel to the table. Our upperbounds apply, however, in much more general settings, as will be discussed in Section 6.
Stacks are composed of blocks that are assumed to be identical, homogeneous, frictionless rectangles ofunit length, unit weight and height h . Our results here are clearly independent of h , and our figures useany convenient height. Previous authors have thought of blocks as cubes, books, coins, playing cards, etc.A stack { B , . . . , B n } of n blocks resting on a flat table is specified by giving the coordinates ( x i , y i ) of thelower left corner of each block B i . We assume that the upper right corner of the table is at (0 ,
0) and thatthe table extends arbitrarily far to the left. Thus block B i is identified with the box [ x i , x i + 1] × [ y i , y i + h ](its length aligned with the x -axis), and the table, which we conveniently denote by B , with the region( −∞ , × ( −∞ , B i rests on block B j , denoted “ B i /B j ”, if and only if B i ∩ B j (cid:54) = ∅ and y i = y j + h . If B i ∩ B (cid:54) = ∅ , then B i /B , i.e., block B i rests on the table. If B i /B j , we let I ij = B i ∩ B j = [ a ij , b ij ] × { y i } be their contact interval . If j ≥
1, then a ij = max { x i , x j } and b ij = min { x i + 1 , x j + 1 } . If j = 0 then a i = x i and b i = min { x i +1 , } .The overhang of a stack is defined to be max ni =1 ( x i +1). Let { B , . . . , B n } be a stack composed of n blocks. If B i rests on B j , then B j may apply an upward force of f ij ≥ B i , in which case B i will reciprocate by applying a downward force of the same magnitude on B j .Since the blocks and table are frictionless, all the forces acting on them are vertical. The force f ij maybe assumed to be applied at a single point ( x ij , y ij ) in the contact interval I ij . A downward gravitationalforce of unit magnitude is applied on B i at its center of gravity ( x i + , y i + h ).4igure 5: Balancing collections of forces within a stack. Definition 2.1 (Equilibrium) . Let B be a homogeneous block of unit length and unit weight, and let a be the x -coordinate of its left edge. Let ( x , f ) , ( x , f ) , . . . , ( x k , f k ) be the positions and the magnitudesof the upward forces applied to B along its bottom edge, and let ( x (cid:48) , f (cid:48) ) , ( x (cid:48) , f (cid:48) ) , . . . , ( x (cid:48) k (cid:48) , f (cid:48) k (cid:48) ) be thepositions and magnitudes of the upward forces applied by B , along its top edge, on other blocks of thestack. Then, B is said to be in equilibrium under these collections of forces if and only if k (cid:88) i =1 f i = 1 + k (cid:48) (cid:88) i =1 f (cid:48) i , k (cid:88) i =1 x i f i = ( a + 12 ) + k (cid:48) (cid:88) i =1 x (cid:48) i f (cid:48) i . The first equation says that the net force applied to B is zero while the second says that the net moment is zero. Definition 2.2 (Balance) . A stack { B , . . . , B n } is said to be balanced if there exists a collection of forcesacting between the blocks along their contact intervals, such that under this collection of forces, and thegravitational forces acting on them, all blocks are in equilibrium.The stacks presented in Figures 1 and 2 are balanced. They are, however, precariously balanced, withsome minute displacement of their blocks leading to imbalance and collapse. A stack can be said to be stable if all stacks obtained by sufficiently small displacements of its blocks are balanced. We do not makethis definition formal as it is not used in the rest of the paper, though we refer to it in some informaldiscussions.A schematic description of a stable stack and a collection of balancing forces acting between its blocks isgiven in Figure 5. Only upward forces are shown in the figure but corresponding downward forces are, ofcourse, present. (We note in passing that balancing forces, when they exist, are in general not uniquelydetermined. This phenomenon is referred to as static indeterminacy .)We usually adopt the convention that the blocks of a balanced stack are numbered consecutively frombottom to top and from left to right. Block B is then the leftmost block in the lowest level while B n isthe rightmost block at the top level. For every 0 ≤ i ≤ n , we let F i be a collection of upward balancingforces applied by blocks in { B , B , . . . , B i } on blocks in { B i +1 , . . . , B n } . (See Figure 5.) We refer to F i asthe collection of forces that cross the i -th slice of the stack.Let us examine the relationship between two consecutive collections F i and F i +1 . The only forces presentin F i but not in F i +1 are upward forces applied to B i , while the only forces present in F i +1 but not in F i areupward forces applied by B i to blocks resting upon it. If we let ( x , f ) , ( x , f ) , . . . , ( x k , f k ) be the positions5nd the magnitudes of the upward forces applied to B i , and ( x (cid:48) , f (cid:48) ) , ( x (cid:48) , f (cid:48) ) , . . . , ( x (cid:48) k (cid:48) , f (cid:48) k (cid:48) ) be the positionsand magnitudes of the upward forces applied by B i , and if we let a be the x -coordinate of the left edgeof B i , we get by Definitions 2.1 and 2.2, that (cid:80) ki =1 f i = 1 + (cid:80) k (cid:48) i =1 f (cid:48) i and (cid:80) ki =1 x i f i = ( a + ) + (cid:80) k (cid:48) i =1 x (cid:48) i f (cid:48) i .Block B i thus rearranges the forces in the interval [ a, a + 1] in a way that preserves the total magnitude ofthe forces and their total moment, when its own weight is taken into account. Note that all forces of F act in non-positive positions, and that if B k is the most overhanging block in a stack and the overhangachieved by it is d , then the total magnitude of the forces in F k − that act at or beyond position d − The natural formulation of the overhang problem is now:What is the maximum overhang achieved by a balanced n -block stack?The main result of this paper is: Theorem 2.3.
The overhang achieved by a balanced n -block stack is at most n / . The fact that the stacks in the theorem above are required to be balanced, but not necessarily stable,makes our result only stronger. By the nature of the overhang problem, stacks that achieve a maximumoverhang are on the verge of collapse and thus unstable. In most cases, however, overhangs arbitrarilyclose to the maximum overhang may be obtained using stable stacks. (Probably the only counterexampleis the case n = 3.) Paterson and Zwick [PZ2006] describe a family of balanced n -block stacks that achieve an overhang ofabout (3 n/ / (cid:39) . n / . More precisely, they construct for every integer d ≥ d ( d − d − + 1 (cid:39) d / d/
2. Their construction, for d = 6,is illustrated in Figure 6. The construction is an example of what [PZ2006] terms a brick-wall stack,which resembles the simple “stretcher-bond” pattern in real-life bricklaying. In each row the blocks arecontiguous, with each block centered over the ends of blocks in the row beneath. Overall the stack issymmetric and has a roughly parabolic shape, with vertical axis at the table edge.The stacks of [PZ2006] are constructed in the following simple manner. A t -row is a row of t adjacent blocks,symmetrically placed with respect to x = 0. An r -slab has height 2 r − r -rowsand ( r − r -rows. An r -slab therefore contains r ( r − r − r −
2) =2( r − blocks. A 1 -stack is a single block balanced at the edge of the table; a d -stack is defined recursivelyas the result of adding a d -slab symmetrically onto the top of a ( d − d -stack and so has overhang d/
2; its total number of blocks is given by n = 1 + (cid:80) dr =1 r − = d ( d − d − + 1. It is shown in [PZ2006], using an inductive argument, that d -stacks, for any d ≥
1, arebalanced.Why should a parabolic shape be appropriate? Some support for this comes from considering the effect ofa block in spreading a single force of f acting from below into two forces of almost f / d isneeded for effective spreading to width d , corresponding to a parabolic stack profile.Our main result, Theorem 2.3, states that the parabolic stacks of [PZ2006] are optimal, up to constantfactors. Better constant factors can probably be obtained, however. Paterson and Zwick [PZ2006] present6igure 6: A “6-stack” consisting of 111 blocks and giving an overhang of 3, taken from [PZ2006].some numerical evidence that suggests that the overhang that can be achieved using n blocks, for largevalues of n , is at least 1 . n / . For more on this, see Section 6. Our upper bound on the maximum achievable overhang is obtained by considering mass movement prob-lems that are an abstraction of the way in which balancing forces “flow” though a stack of blocks. (Seethe discussion at the end of Section 2.2.)In a mass movement problem we are required to transform an initial mass distribution into a mass distri-bution that satisfies certain conditions. The key condition is that a specified amount of mass be movedto or beyond a certain position. We can transform one mass distribution into another by performing local moves that redistribute mass within a given interval in a way that preserves the total mass and the centerof mass. Our goal is then to show that many moves are required to accomplish the task. As can be seen,masses here correspond to forces, mass distributions correspond to collections of forces, and moves mimicthe effects of blocks.The mass movement problems considered are formally defined in Sections 4.1 and 4.2. The correspondencebetween the mass movement problems considered and the overhang problem is established in Section 4.3.The bounds on mass movement problems that imply Theorem 2.3 are then proved in Section 5.
Definition 4.1 (Distributions and signed distributions) . A discrete mass distribution is a set µ = { ( x , m ) , ( x , m ) , . . . , ( x k , m k ) } , where k > x , x , . . . , x k are real numbers, and m , . . . , m k >
0. A signeddistribution µ is defined the same way, but without the requirement that m , m , . . . , m k > µ = { ( x , m ) , ( x , m ) , . . . , ( x k , m k ) } is a (signed) distribution, then for any set A ⊆ R , we define µ ( A ) = (cid:88) x i ∈ A m i . For brevity, we use µ ( a ) as a shorthand for µ ( { a } ) and µ { x > a } as a shorthand for µ ( { x | x > a } ). (Notethat x here is a formal variable that does not represent a specific real number.) We similarly use µ { x ≥ a } , µ { x < a } , µ { a < x < b } , µ {| x | ≥ a } , etc., with the expected meaning.We say that a (signed) distribution is on the interval [ a, b ] if µ ( x ) = 0, for every x (cid:54)∈ [ a, b ].7or every A ⊆ R , we let µ A be the restriction of µ to A : µ A = { ( x i , m i ) | x i ∈ A } . If µ and µ are two signed distributions, we let µ + µ and µ − µ be the signed distributions for which( µ + µ )( x ) = µ ( x ) + µ ( x ) , for every x ∈ R , ( µ − µ )( x ) = µ ( x ) − µ ( x ) , for every x ∈ R . Definition 4.2 (Moments) . Let µ = { ( x , m ) , ( x , m ) , . . . , ( x k , m k ) } be a signed distribution and let j ≥ j -th moment of µ is defined to be: M j [ µ ] = k (cid:88) i =1 m i x ji . Note that M [ µ ] is the total mass of µ , M [ µ ] is the torque of µ , with respect to the origin, and M [ µ ] isthe moment of inertia of µ , again with respect to the origin. If M [ µ ] (cid:54) = 0, we let C [ µ ] = M [ µ ] /M [ µ ] bethe center of mass of µ .Less standard, but crucial for our analysis, is the following definition. Definition 4.3 (Spread) . The spread of a distribution µ = { ( x , m ) , ( x , m ) , . . . , ( x k , m k ) } is defined asfollows: S [ µ ] = (cid:88) i For any discrete distribution µ we have S [ µ ] ≤ M [ µ ] M [ µ ] . The proof of Lemma 4.4 is given in Section 5.5. Definition 4.5 (Moves) . A move v = ([ a, b ] , δ ) consists of an interval [ a, b ] and a signed distribution δ on [ a, b ] with M [ δ ] = M [ δ ] = 0. A move v can be applied to a distribution µ if the signed distribution µ (cid:48) = µ + δ is a distribution, in which case we denote the result µ (cid:48) of this application by vµ . We refer to a + b as the center of the move. Unless otherwise stated, the moves we consider operate on intervals of length 1,i.e., b − a = 1.Note that v is a move and µ (cid:48) = vµ , then M [ µ (cid:48) ] = M [ µ ], M [ µ (cid:48) ] = M [ µ ] and consequently C [ µ (cid:48) ] = C [ µ ].A sequence V = (cid:104) v , v , . . . , v (cid:96) (cid:105) of moves and an initial distribution µ naturally define a sequence ofdistributions µ , µ , . . . , µ (cid:96) , where µ i = v i µ i − for 1 ≤ i ≤ (cid:96) . (It is assumed here that v i can indeed beapplied to µ i − .) We let V µ = µ (cid:96) .Moves and sequences of moves simulate the behavior of weightless blocks and stacks. However, the blocksthat we are interested in have unit weight. Instead of explicitly taking into account the weight of theblocks, as we briefly do in Section 4.3, it turns out that it is enough for our purposes to impose a naturalrestriction on the move sequences considered. We start with the following definition:8 efinition 4.6 ( µ max ) . If µ , µ , . . . , µ (cid:96) is a sequence of distributions, and a ∈ R , we define µ max { x > a } = max ≤ i ≤ (cid:96) µ i { x > a } . Expressions like µ max { x ≥ a } , µ max { x < a } and µ max { x ≤ a } are defined similarly. Definition 4.7 (Weight-constrained sequences) . A sequence V = (cid:104) v , v , . . . , v (cid:96) (cid:105) of moves that generatesa sequence µ , µ , . . . , µ (cid:96) of distributions is said to be weight-constrained , with respect to µ if, for every a ∈ R , the number of moves in V centered in ( a, ∞ ) is at most µ max { x > a } .The two main technical results of this paper are the following theorems. Theorem 4.8. If a distribution ν is obtained from a distribution µ with µ { x ≤ } ≤ n and µ { x > } = 0 ,where n ≥ , by a weight-constrained move sequence, then ν { x ≥ n / − } = 0 . For general move sequences we have the following almost tight result, which might be of some independentinterest. In particular, it shows that the weight constraint only has a logarithmic effect on the maximaloverhang. Theorem 4.9. If a distribution ν is obtained from a distribution µ with µ { x ≤ } ≤ n and µ { x > } = 0 ,where n ≥ , by a move sequence of length at most n , then ν { x ≥ n / log n } < . We show next that Theorem 4.8 does indeed imply Theorem 2.3, the main result of this paper. The moves of Definition 4.5 capture the essential effect that a block can have on the collections of forceswithin a stack. They fail to take into account, however, the fact that the weight of a block is “used up” bythe move and is then lost. To faithfully simulate the effect of unit weight blocks we introduce the slightlymodified definition of lossy moves : Definition 4.10 (Lossy moves) . If v = ([ a, b ] , δ ) is a move, then the lossy move v ↓ associated with it is v ↓ = ([ a, b ] , δ ↓ ), where δ ↓ = δ − { ( a + b , } . A lossy move v ↓ can be applied to a distribution µ if µ (cid:48) = µ + δ ↓ is a distribution, in which case we denote the result µ (cid:48) of this application by v ↓ µ .Note that if v ↓ = ([ a, b ] , δ ↓ ) is a lossy move and µ (cid:48) = v ↓ µ , then M [ µ (cid:48) ] = M [ µ ] − M [ µ (cid:48) ] = M [ µ ] − a + b .Hence, lossy moves do not preserve total mass or center of mass.If V = (cid:104) v , v , . . . , v (cid:96) (cid:105) is a sequence of moves, we let V ↓ = (cid:104) v ↓ , v ↓ , . . . , v ↓ (cid:96) (cid:105) be the corresponding sequenceof lossy moves. If µ is an initial distribution, we can naturally define the sequence of distributions µ , µ , . . . , µ (cid:96) , where µ i = v ↓ i µ i − for 1 ≤ i ≤ (cid:96) , obtained by applying V ↓ to µ .A collection of forces F i may also be viewed as mass distribution. The following lemma is now a simpleformulation of the definitions and the discussion of Section 2.2: Lemma 4.11. Let { B , B , . . . , B n } be a balanced stack. Let F i be a collection of balancing forces actingbetween { B , . . . , B i } and { B i +1 , . . . , B n } , for ≤ i ≤ n . Let x i be the x -coordinate of the left edge of B i .Then, F i +1 can be obtained from F i by a lossy move in the interval [ x i , x i +1] . As an immediate corollary, we get: Lemma 4.12. If there is a stack composed of n blocks of length and weight that achieves an overhangof d , then there is sequence of at most n − lossy moves that transforms a distribution µ with M [ µ ] = µ { x ≤ } = n and µ { x > } = 0 into a distribution µ (cid:48) with µ (cid:48) { x ≥ d − } ≥ . roof. Let { B , B , . . . , B n } be a balanced stack and let B k be a block in it that achieves an overhang of d .As before, we let F i be a collection of balancing forces acting between { B , . . . , B i } and { B i +1 , . . . , B n } .We let µ = F and µ (cid:48) = F k − . It follows from Lemma 4.11 that µ (cid:48) may be obtained from µ by a sequenceof k − µ = F are forces applied by the table B , and as the table supportsthe weight of the n blocks of the stack, we have M [ µ ] = µ { x ≤ } = n and µ { x > } = 0. As the forcesin µ (cid:48) = F k − must at least support the weight of B k , we have µ (cid:48) { d − ≤ x ≤ d } ≥ Lemma 4.13. If µ , µ , . . . , µ (cid:96) is a sequence of distributions obtained by a sequence of lossy moves, thenthere exists a sequence of distributions µ (cid:48) , µ (cid:48) , . . . , µ (cid:48) (cid:96) obtained by a weight-constrained sequence of movessuch that µ (cid:48) = µ , and µ (cid:48) i ( x ) ≥ µ i ( x ) , for every ≤ i ≤ (cid:96) and x ∈ R .Proof. The sequence µ (cid:48) , µ (cid:48) , . . . , µ (cid:48) (cid:96) is obtained by performing exactly the same moves used to obtain thesequence µ , µ , . . . , µ (cid:96) , treating them now as moves rather than lossy moves. More formally, if µ i = v ↓ i µ i − ,we let µ (cid:48) i = v i µ (cid:48) i − . If v i = ([ a − , a + ] , δ ), then µ (cid:48) i now has an extra mass of size 1 at a . This mass is frozen , and will not be touched by subsequent moves. Hence, if k moves have their center beyond position a ,then µ (cid:48) max { x > a } ≥ µ (cid:48) (cid:96) { x > a } ≥ k , as required by the definition of weight-constrained sequences.It is now easy to see that Theorem 4.8 together with Lemmas 4.11 and 4.12 imply Theorem 2.3. This section is devoted to the proofs of Theorems 4.8 and 4.9. As mentioned, Theorem 4.8 implies Theo-rem 2.3, which states that an n -block stack can have an overhang of at most 6 n / . We begin by considering an important class of moves: Definition 5.1 (Extreme moves) . An extreme move ¯ v is defined solely as an interval [ a, b ]. An extrememove ¯ v can be applied to any distribution µ resulting in the distribution µ (cid:48) = ¯ vµ such that µ (cid:48) { a Definition 5.3 (Splitting) . Let µ and µ (cid:48) be two distributions. We say that µ (cid:48) is a basic split of µ , denoted µ (cid:22) µ (cid:48) , if µ (cid:48) is obtained by taking one of the point masses ( x i , m i ) of µ and replacing it by a collection { ( x (cid:48) , m (cid:48) ) , . . . , ( x (cid:48) (cid:96) , m (cid:48) (cid:96) ) } of point masses with total mass m i and center of mass at x i . We say that µ (cid:48) splitsinto µ , denoted µ (cid:22) µ (cid:48) , if µ (cid:48) can be obtained from µ by a sequence of zero or more basic splits.10he following two lemmas summarize simple properties of splits and extreme moves that will be explicitlyor implicitly used in this section. Their obvious proofs are omitted. Lemma 5.4. (i) If µ (cid:22) µ (cid:48) and µ (cid:48) (cid:22) µ (cid:48)(cid:48) , then µ (cid:22) µ (cid:48)(cid:48) .(ii) If µ (cid:22) µ (cid:48) and µ (cid:22) µ (cid:48) , then µ + µ (cid:22) µ (cid:48) + µ (cid:48) .(iii) For any distribution µ we have { ( C [ µ ] , M [ µ [) } (cid:22) µ .(iv) If µ = { ( x , m ) , ( x , m ) } and µ (cid:48) = { ( x (cid:48) , m (cid:48) ) , ( x (cid:48) , m (cid:48) ) } , where x (cid:48) ≤ x ≤ x ≤ x (cid:48) , M [ µ ] = M [ µ (cid:48) ] and C [ µ ] = C [ µ (cid:48) ] , then µ (cid:22) µ (cid:48) . Lemma 5.5. (i) If vµ is defined then vµ (cid:22) ¯ vµ .(ii) If ¯ v is an extreme move then µ (cid:22) ¯ vµ .(iii) If ¯ v is an extreme move then ¯ v ( µ + µ ) = ¯ vµ + ¯ vµ . The following lemma shows that splitting increases the second moment. Lemma 5.6. If µ (cid:22) µ (cid:48) then M [ µ ] ≤ M [ µ (cid:48) ] .Proof. Due to the linearity of M and the fact that (cid:22) is the transitive closure of (cid:22) , it is enough toprove the claim when µ = { ( x, m ) } is composed of a single mass and µ (cid:48) = { ( x (cid:48) , m (cid:48) ) , . . . , ( x (cid:48) k , m (cid:48) k ) } isobtained from µ by a basic split. For any distribution ν = { ( x , m ) , . . . , ( x k , m k ) } and any c ∈ R wedefine M [ ν, c ] = (cid:80) ki =1 m i ( x i − c ) to be the second moment of ν about c . As M [ µ ] = M [ µ (cid:48) ] and M [ µ ] = M [ µ (cid:48) ], a simple calculation shows that M [ µ (cid:48) , c ] − M [ µ, c ] = M [ µ (cid:48) ] − M [ µ ], for any c ∈ R .Choosing c = x and noting that M [ µ, x ] = 0 while M [ µ (cid:48) , x ] ≥ 0, we get the required inequality.The next lemma exhibits a relation between extreme moves and splitting. Lemma 5.7. If µ (cid:22) µ (cid:48) and v is a move that can be applied to µ , then vµ (cid:22) ¯ vµ (cid:48) .Proof. We show that vµ (cid:22) ¯ vµ (cid:22) ¯ vµ (cid:48) , and use Lemma 5.4( i ). The first relation is just Lemma 5.5( i ). Itremains to show ¯ vµ (cid:22) ¯ vµ (cid:48) . By Lemma 5.5( iii ), it is enough to prove the claim for µ = { ( x, m ) } composedof a single mass. Let [ a, b ] be the interval corresponding to ¯ v . There are two cases. If x (cid:54)∈ [ a, b ], then¯ vµ = µ (cid:22) µ (cid:48) (cid:22) ¯ vµ (cid:48) , as required. The more interesting case is when x ∈ [ a, b ]. Let ν = ¯ vµ = { ( a, m ) , ( b, m ) } and ν (cid:48) = ¯ vµ (cid:48) .Let ν (cid:48) (cid:96) = µ ( −∞ ,a ] and ν (cid:48) r = µ [ b, ∞ ) . As ¯ v leaves no mass in ( a, b ), we get that ν (cid:48) = ν (cid:48) (cid:96) + µ (cid:48) r . Let ¯ m (cid:96) = M [ ν (cid:48) (cid:96) ],¯ m r = M [ ν (cid:48) r ], ¯ x (cid:96) = C [ ν (cid:48) (cid:96) ] and ¯ x r = C [ ν (cid:48) r ]. As ¯ x (cid:96) ≤ a < b ≤ ¯ x r , we get using Lemma 5.4( ii ) and ( iii ) that ν = { ( a, m ) , ( b, m ) } (cid:22) { (¯ x (cid:96) , ¯ m (cid:96) ) , (¯ x r , ¯ m r ) } = { (¯ x (cid:96) , ¯ m (cid:96) ) } + { (¯ x r , ¯ m r ) } (cid:22) ν (cid:48) (cid:96) + ν (cid:48) r = ν (cid:48) , as required.Using induction we easily obtain: Theorem 5.8. If V is a sequence of moves that can be applied to µ , then V µ (cid:22) ¯ V µ . Combining Theorem 5.8 and Lemma 5.6 we get the following immediate corollary. Corollary 5.9. If V is a sequence of moves that can be applied to µ , then M [ V µ ] ≤ M [ ¯ V µ ] . .2 Spread vs. second moment We now obtain our first bound for mass movement problems. The bound relies heavily on Lemma 4.4 thatrelates the spread and second moment of a distribution, on Lemma 5.2 that relates differences in spread todifferences in second moments, and finally, on Corollary 5.9 that states that converting moves to extrememoves can only increase the second moment. Lemma 5.10. Any sequence of moves that transforms the distribution µ = { (0 , } into a distribution ν with ν {| x | ≥ d } ≥ p , where d > and < p < , must contain at least (3 p ) / d moves.Proof. Let µ , µ , . . . , µ (cid:96) be the sequence of distributions obtained by applying a sequence V of (cid:96) movesto µ = { (0 , } , and suppose that µ (cid:96) {| x | ≥ d } ≥ p . By the definition of the second moment we have M [ µ (cid:96) ] ≥ pd .Let ¯ µ , ¯ µ , . . . , ¯ µ (cid:96) be the sequence of distributions obtained by applying the sequence ¯ V of the extrememoves corresponding to the moves of V on ¯ µ = µ = { (0 , } . By Corollary 5.9, we get that M [¯ µ (cid:96) ] ≥ M [ µ (cid:96) ] ≥ p d . By Lemma 4.4 we have M [¯ µ (cid:96) ] S [¯ µ (cid:96) ] = (cid:18) M [¯ µ (cid:96) ] M [¯ µ (cid:96) ] S [¯ µ (cid:96) ] (cid:19) / M [¯ µ (cid:96) ] / ≥ √ M [¯ µ (cid:96) ] / ≥ √ p / d . Let h i = M [¯ µ i ] − M [¯ µ i − ], for 1 ≤ i ≤ (cid:96) . As M [¯ µ ] = 0, we clearly have, M [¯ µ (cid:96) ] = (cid:96) (cid:88) i =1 h i . By Lemma 5.2, we get that S [¯ µ (cid:96) ] ≥ (cid:96) (cid:88) i =1 h i . Using the Cauchy-Schwartz inequality to justify the second inequality below, we get: S [¯ µ (cid:96) ] ≥ (cid:96) (cid:88) i =1 h i ≥ (cid:80) (cid:96)i =1 h i ) (cid:96) = 3 M [¯ µ (cid:96) ] (cid:96) . Thus, as claimed, (cid:96) ≥ M [¯ µ (cid:96) ] S [¯ µ (cid:96) ] ≥ (3 p ) / d . The main result of this section is: Theorem 5.11. Let µ , µ , . . . , µ (cid:96) be a sequence of distributions obtained by applying a sequence of movesto an initial distribution µ with µ { x > r } = 0 . If µ max { x > r } ≤ m and µ max { x ≥ r + d } ≥ pm , where d > and < p < , then the sequence of moves must contain at least √ p / ( d − ) moves whose centersare in ( r + , ∞ ) . The theorem follows immediately from the following lemma by shifting coordinates and renormalizingmasses. 12 emma 5.12. Let µ , µ , . . . , µ (cid:96) be a sequence of distributions obtained by applying a sequence of movesto an initial distribution µ with µ { x > − } = 0 . If µ max (cid:8) x > − (cid:9) ≤ and µ max { x ≥ d } ≥ p , where d > and < p < , then the sequence of moves must contain at least √ p / d moves whose centers areat strictly positive positions.Proof. We may assume, without loss of generality, that the first move in the sequence moves some massfrom ( −∞ , ] into ( , ∞ ) and that the last move moves some mass from ( −∞ , d ) to [ d, ∞ ). Hence, thecenter of the first move must be in ( − , 0] and the center of the last move must be at a positive position.We shall show how to transform the sequence of distributions µ , µ , . . . , µ (cid:96) into a sequence of distributions µ (cid:48) , µ (cid:48) , . . . , µ (cid:48) (cid:96) (cid:48) , obtained by applying a sequence of (cid:96) (cid:48) moves, such that µ (cid:48) = { (0 , } , µ (cid:48) (cid:96) (cid:48) {| x | ≥ d } ≥ p , andsuch that the number of moves (cid:96) (cid:48) in the new sequence is at most three times the number (cid:96) + of positivelycentered move in the original sequence. The claim of the lemma would then follow immediately fromLemma 5.10.The first transformation is “negative truncation”, where in each distribution µ i , we shift mass from theinterval ( −∞ , − ) to the point − . Formally the resulting distribution → µ i is defined by → µ i ( x ) = µ i ( x ) if x > − − µ i { x > − } if x = − x < − . Note that the total mass of each distribution is 1 and that → µ = { ( − , } . Let δ i = µ i − µ i − be the signeddistribution associated with the move that transforms µ i − into µ i and let [ c i − , c i + ] be the interval inwhich it operates. For brevity, we refer to δ i as the move itself, with its center c i clear from the context. Wenow compare the transformed “moves” → δ i = → µ i − → µ i − with the original moves δ i = µ i − µ i − . If c i > δ i acts above − and → δ i = δ i . If c i ≤ − 1, then δ i acts at or below − , so → δ i is null and → µ i = → µ i − .In the transformed sequence, we skip all such null moves. The remaining case is when the center c i of δ i is in ( − , → δ i acts within [ − , ], and we view it as centered at 0. However, typically → δ i does not define a valid move as it may change the center of gravity. We call these → δ i semi-moves . Ifwe have two consecutive semi-moves → δ i and → δ i +1 , we combine them into a single semi-move → δ i + → δ i +1 ,taking → µ i − directly to → µ i +1 . In the resulting negatively truncated and simplified sequence, we know thatat least every alternate move is an original, positively centered, move. Since the last move in the originalsequence was positively centered we conclude: Claim 5.13. The sequence obtained by the negative truncation transformation and the subsequent clean-upis composed of original positively centered moves and semi-moves (acting within [ − , ] ). The sequencebegins with a semi-move and at most half of its elements are semi-moves. Next, we create a reflected copy of the negatively truncated distributions. The reflected copy ← µ i of → µ i isdefined by ← µ i ( x ) = → µ i ( − x ) , for every x ∈ R . We similarly define the reflected (semi-)moves ← δ i = ← µ i − ← µ i − . We can now define the mirrored distributions ↔ µ i = → µ i + ← µ i . Note that ↔ µ = → µ + ← µ = { ( − , , ( , } . The distribution ↔ µ i may be obtained from ↔ µ i − by applyingthe (semi-)move → δ i , resulting in the distribution → µ i + ← µ i − , and then the (semi-)move ← δ i , resulting in → µ i + ← µ i = ↔ µ i . The ↔ µ i sequence is therefore obtained by interleaving the (semi-)moves → δ i with theirreflections ← δ i . Now comes a key observation: 13 laim 5.14. If → δ i and ← δ i are semi-moves, then their sum ↔ δ i = → δ i + ← δ i defines an ordinary move centeredat and acting on [ − , ] .Proof. Both → δ i and ← δ i preserve the total mass. As ↔ δ i is symmetric about 0, it cannot change the centerof mass.As suggested by the above observation, if → δ i and ← δ i are semi-moves, we combine them into a singleordinary move ↔ δ i centered at 0. We thus obtain a sequence of at most 3 (cid:96) + moves, where (cid:96) + is the numberof positively centered moves in the original sequence, that transforms ↔ µ to ↔ µ (cid:96) .Recall from Claim 5.13 that the first “move” → δ in the negatively truncated sequence is a semi-move. Thefirst move ↔ δ , obtained by combining → δ and ← δ , is therefore a move acting on [ − , ]. We now replacethe initial distribution ↔ µ = { ( − , , ( , } by the distribution µ (cid:48) = { (0 , } , which has the same centerof gravity, and replace the first move by δ (cid:48) = ↔ δ + { ( − , , (0 , − , ( , } . The distribution after the firstmove is then again ↔ µ .We have thus obtained a sequence of at most 3 (cid:96) + moves that transforms µ (cid:48) = { (0 , } into a distribution ν (cid:48) = ↔ µ (cid:96) with ν (cid:48) {| x | ≥ d } ≥ p . Scaling these distribution and moves by a factor of 2, we get, byLemma 5.10, that 3 (cid:96) + ≥ (3 p ) / d , as claimed. We prove the following theorem which easily implies Theorem 4.8. Theorem 5.15. Let µ , µ , . . . , µ (cid:96) be a sequence of distributions obtained by applying a constrained sequenceof moves on an initial distribution µ with µ { x > r } = 0 . If µ max { x > r } ≤ n , where n ≥ , then µ max (cid:8) x > r + 6 n / − (cid:9) = 0 .Proof. The proof is by induction on n . If n < r and hence µ max (cid:8) x > r + (cid:9) = 0. Since 6( ) / − > / 2, the result clearly holds.Suppose now that µ max { x > r } = n and that the result holds for all ≤ n (cid:48) < n . Let u be the largest numberfor which µ max { x ≥ r + u } > n . As the distributions µ i are discrete, it follows that µ max { x > r + u } ≤ n .As u ≥ 0, we have µ { x > r + u } = 0. By the induction hypothesis with r replaced by r + u , we thereforeget that µ max (cid:110) x > r + u + 6( n / − (cid:111) = 0 . As µ { x > r } = 0, µ max { x > r } ≤ n and µ max { x ≥ r + u } > n , we get by Theorem 5.11 that the sequencemust contain at least √ ) / ( u − ) > ( u − ) moves whose centers are positive. As the sequence ofmoves is constrained, and as µ max { x > r } ≤ n , there can be at most n such moves with centers greaterthan r , i.e., 17 (cid:18) u − (cid:19) ≤ n . Hence u ≤ (7 n ) / + 12 , and so u + 6( n / − ≤ (7 / + 6 · ( 15 ) / ) n / − < . n / − ≤ n / − , for n ≥ n blocks in at most 6 n / . It is fairlystraightforward to modify the proof of Theorem 5.15 above so as to obtain the stronger conclusion that µ max (cid:8) x > cn / − (cid:9) = 0, for any c > / · / (cid:39) . n , and hence animproved upper bound on overhang of, say, 4 . n / . This is done by choosing u to be the largest number forwhich µ max { x ≥ u } > n . (The constant here is the optimal choice.) The proof, however, becomesslightly messier, as several of the inequalities do not hold for small values of n .Next, we prove the following theorem which easily implies Theorem 4.9. Theorem 5.16. Let µ , µ , . . . , µ n be a sequence of distributions obtained by applying a sequence of n moves to an initial distribution µ with µ { x ≤ } ≤ n and µ { x > } = 0 , where n ≥ . Then µ n { x > n / log n } < .Proof. Suppose that 2 k ≤ n < k +1 , where k ≥ 0. For 1 ≤ i ≤ k , let u i be the largest number for which µ max { x ≥ u i } ≥ n i . By the discreteness of the distributions we again have µ max { x > u i } < n i . Let u = 0.Assume, for the sake of contradiction, that µ n { x > n / log n } ≥ 1. Then, u k ≥ kn / . There is then atleast one value of i for which u i − u i − ≥ n / . By Theorem 5.11, applied with r = u i − and d = u i − u i − ,we conclude that the sequence must contain more than n moves, a contradiction.As before, the constants in the above proof are not optimized. We believe that a stronger version ofthe theorem, which states under the same conditions that µ n { x > cn / (log n ) / } < 1, for some c > Lemma 4.4. (The proof was deferred from Section 4.1.) For any discrete distribution µ , S [ µ ] ≤ M [ µ ] M [ µ ] . The method of proof used here was suggested to us by Benjy Weiss, and resulted in a much improvedand simplified presentation. The lemma is essentially the case n = 2 of a more general result proved byPlackett [P1947]. Proof. Suppose that µ = { ( x , m ) , ..., ( x k , m k ) } where x < x < · · · < x k .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 invariant under scaling of the masses, we mayassume that M [ µ ] = 1.Define a function g ( t ) for − ≤ t ≤ by g ( t ) = x i , where i − (cid:88) r =1 m r < t + 12 ≤ i (cid:88) r =1 m r , and define g ( − ) = x .Now we have that M j [ µ ] = k (cid:88) i =1 x ji m i = (cid:90) t = − g ( t ) j dt j ≥ 0, and S [ µ ] = (cid:88) i 112 = 13 M [ µ ] . Lemma 5.2. (The proof was deferred from Section 4.2.) If µ is obtained from µ by an extreme move(in an interval of length 1) then S [ µ ] − S [ µ ] ≥ M [ µ ] − M [ µ ]) . Proof. Since the statement of the lemma is invariant under linear translation of the coordinates, we mayassume that the interval of the move is [ − , ]. Let ν = ( µ ) [ − , ] , i.e., the restriction of µ to [ − , ].Note that the lemma relates the difference in spread and the difference in second moment resulting from theextreme move. Since the addition of an extra point mass at either − or leaves each of these differencesinvariant, we may add such a mass as will bring the center of mass of ν to 0, and continue the proofunder this assumption. Since the statement of the lemma is invariant under scaling of the masses, we mayfurther assume that M [ ν ] = 1.If ν is the result within the interval [ − , ] of the extreme move, then: ν = (cid:26)(cid:18) − , (cid:19) , (cid:18) , (cid:19)(cid:27) and M [ ν ] = S [ ν ] = 14 . We define g ( t ) for − ≤ t ≤ just as in the proof of Lemma 4.4 but now corresponding to the distribu-tion ν , and so − ≤ g ( t ) ≤ , for − ≤ t ≤ . As before, M j [ ν ] = (cid:82) − g ( t ) j dt for j ≥ 0, and we recallas in ( † ) that S [ ν ] = 2 (cid:82) − tg ( t ) dt . 16e have M [ ν ] = (cid:82) − g ( t ) dt = 0. Let c = M [ ν ] = (cid:82) − g ( t ) dt and s = S [ ν ]. By Lemma 4.4 we have s ≤ c . If c ≤ then s ≤ (cid:112) c ≤ and the result follows immediately as S [ ν ] − S [ ν ] − M [ ν ] − M [ ν ]) = − s − (cid:0) − c (cid:1) ≥ − s − (cid:0) − s (cid:1) = (1+2 s )(1 − s ) ≥ . We next claim that if c = M [ ν ] > , then s = S [ ν ] ≤ − a , where a = − c < 1. To prove thisclaim, we define a function h ( t ) as follows: h ( t ) = (cid:26) ta if | t | ≤ a , and sgn( t ) otherwise.We may verify that (cid:90) − h ( t ) dt = 14 − a c and (cid:90) − t h ( t ) dt = 18 − a . By the Cauchy-Schwartz inequality, (cid:32)(cid:90) − h ( t ) g ( t ) dt (cid:33) ≤ (cid:90) − h ( t ) dt · (cid:90) − g ( t ) dt = c , and so (cid:90) − h ( t ) g ( t ) dt ≤ c = (cid:90) − h ( t ) dt . ( ∗ )We also have (cid:90) − (cid:18) ta − h ( t ) (cid:19) g ( t ) dt ≤ (cid:90) − (cid:18) ta − h ( t ) (cid:19) h ( t ) dt , ( ∗∗ )since h ( t ) − g ( t ) ≤ ta − h ( t ) ≤ t < − a , and h ( t ) − g ( t ) ≥ ta − h ( t ) ≥ t > a , and ta − h ( t ) = 0 for | t | ≤ a . Adding inequalities (*) and (**), and multiplying by 2 a , gives S [ ν ] = 2 (cid:90) − t g ( t ) dt ≤ (cid:90) − t h ( t ) dt = 14 − a . Finally, S [ ν ] − S [ ν ] ≥ − (cid:18) − a (cid:19) = a 12 = 3 (cid:18) − c (cid:19) = 3( M [ ν ] − M [ ν ]) . This completes the proof.We end the section by noting that although the inequalities of Lemmas 4.4 and 5.2 are only claimed fordiscrete distributions, which is all we need in this paper, our proofs can be easily modified to show that theyhold also for general continuous distributions. In fact, for non-trivial discrete distributions, the inequalitiesin the two lemmas are always strict . In the continuous case, the inequalities are satisfied with equality byappropriately chosen uniform distributions. In particular, the constant factors and 3 appearing in thetwo lemmas cannot be improved. 17 Concluding remarks and open problems We have shown that the maximum overhang achieved using n homogeneous, frictionless blocks of unitlength is at most 6 n / . Thus, the constructions of [PZ2006] cannot be improved by more than a constantfactor, establishing order n / as the asymptotic answer to the age-old overhang problem.The discussions and results presented so far all referred to the standard two-dimensional version of theoverhang problem. Our results hold, however, in greater generality. We briefly discuss some naturalgeneralizations and variants of the overhang problem for which our bounds still apply.In Section 2 we stipulated that all blocks have a given height h . It is easy to see, however, that all ourresults remain valid even if blocks have different heights, but still have unit length and unit weight. Inparticular, blocks are allowed to degenerate into sticks , i.e., have height 0. Also, even though we requiredblocks not to overlap, we did not use this condition in any of our proofs. Loaded stacks , introduced in [PZ2006], are stacks composed of standard unit length and unit weight blocks,and point weights that can have arbitrary weight. (Point weights may be considered to be blocks of zeroheight and length, but nonzero weight.) Our results, with essentially no change, imply that loaded stacksof total weight n can have an overhang of at most 6 n / .What happens when we are allowed to use blocks of different lengths and weights? Our results can begeneralized in a fairly straightforward way to show that if a block of length (cid:96) has weight proportional to (cid:96) ,as would be the case if all blocks were similar three-dimensional cuboids, then the overhang of a stackof total weight n is again of order at most n / . It is amusing to note that in this case an overhang oforder n / can be obtained by stacking n unit-length blocks as in the construction of [PZ2006], or simplyby balancing a single block of length n / and weight n at the edge of the table! Might this mean thatthere is some physical principle that could have told us, without all the calculations, that the right answerto the original overhang problem had to be of order n / ?Theorem 4.9 supplies an almost tight upper bound for the following variant of the overhang problem: Howfar away from the edge of a table can a mass of weight 1 be supported using n weightless blocks of length 1,and a collection of point weights of total weight n ? The overhang in this case beats the classical one by afactor of between log / n and log n .In all variants considered so far, blocks were assumed to have their largest faces parallel to the table’ssurface and perpendicular to its edge. The assumption of no friction then immediately implied that allforces within a stack are vertical, and our analysis, which assumes that there are no horizontal forces,was applicable. A nice argument, communicated to us by Harry Paterson, shows that in the frictionlesstwo-dimensional case, horizontal forces cannot be present even if some of the blocks are tilted. Our resultsthus apply also in this case. Figure 7: A “skintled” 4-diamond.We believe that our bounds apply, with slightly adjusted constants, also in three dimensions, but proving18o remains an open problem. Overhang larger by a factor of √ w may be obtained with 1 × w × h blocks, where h ≤ w ≤ 1, using a technique called skintling (see Figure 7). In skintling (a term we learnedfrom an edifying conversation with John H. Conway about brick-laying) each block is rotated about itsvertical axis, so that—in our case—the diagonal of its bottom face is perpendicular to the edge of the table.With suitably adjusted notion of length, however, our bounds apply to any three-dimensional constructionthat can be balanced using vertical forces. It is an interesting open problem whether there exist three-dimensional stacks composed of frictionless, possibly tilted, blocks that can only be balanced with the aidof some non-vertical forces. (We know that this is possible if the blocks are nonhomogeneous and are ofdifferent sizes.) As mentioned, we believe that our bounds do apply in three dimensions, even if it turnsout that non-vertical forces are sometimes useful, but proving this requires some additional arguments.We end by commenting on the tightness of the analysis presented in this paper. Our main result isa 6 n / upper bound on the overhang that may be obtained using n blocks. As mentioned after the proofof Theorem 5.15, this bound can be easily improved to about 4 . n / , for sufficiently large values of n .Various other small improvements in the constants are possible. For example, a careful examination of ourproofs reveals that whenever we apply Lemma 5.2, the distribution µ contains at most three masses inthe interval acted upon by the move that produces µ . (This follows from the fact that a block can restupon at most three other blocks.) The constant 3 appearing in Lemma 5.2 can then be improved, thoughit is optimal when no assumption regarding the distribution µ is made. We believe, however, that newideas would be needed to reduce the upper bound to below, say, 3 n / . weight (cid:1) (cid:1) (cid:1) Figure 8: An “oil-lamp”-shaped stackAs mentioned, Paterson and Zwick [PZ2006] describe simple balanced n -block stacks that achieve anoverhang of about 0 . n / . They also present some numerical evidence that suggests that the overhangthat can be achieved using n blocks, for large values of n , is at least 1 . n / . These larger overhangs areobtained using stacks that are shaped like the “oil-lamp” depicted in Figure 8. For more details on thefigure and on “oil-lamp” constructions, see [PZ2006]. (The stack shown in the figure is actually a loadedstack, as defined above, with the external forces shown representing the point weights.)A small gap still remains between the best upper and lower bounds currently available for the overhangproblem, though they are both of order n / . Determining a constant c such that the maximum overhangachievable using n blocks is asymptotically cn / is a challenging open problem. Acknowledgements We would like to thank John H. Conway, Johan H˚astad, Harry Paterson, Anders Thorup and Benjy Weissfor useful discussions observations, some of which appear with due credit within the paper.19 eferences [A1979] S. 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