Overhang bounds Mike Paterson Joint work with Uri Zwick, Yuval Peres, Mikkel Thorup and Peter Winkler

The classical solution Using n blocks we can get an overhang of Harmonic Stacks

Is the classical solution optimal? Obviously not!

Inverted triangles? Balanced?

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Inverted triangles? Balanced?

Inverted triangles? Unbalanced!

Inverted triangles? Unbalanced!

Diamonds? Balanced?

The 4-diamond is balanced Diamonds? The 4-diamond is balanced

Diamonds? The 5-diamond is …

Diamonds? … unbalanced!

What really happens?

What really happens!

How do we know this is unbalanced?

… and this balanced?

Equilibrium Force equation Moment equation F1 F2 F3 F4 F5 x1 F1+ x2 F2+ x3 F3 = x4 F4+ x5 F5

Checking balance

Equivalent to the feasibility of a set of linear inequalities: Checking balance F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F12 F13 F14 F15 F16 Equivalent to the feasibility of a set of linear inequalities: F17 F18

Small optimal stacks Overhang = 1.16789 Blocks = 4 Overhang = 1.30455

Small optimal stacks

Small optimal stacks

Small optimal stacks Overhang = 2.14384 Blocks = 16 Overhang = 2.1909

Support and balancing blocks Principal block Balancing set Support set

Support and balancing blocks Balancing set Principal block Support set

Loaded stacks Stacks with downward external forces acting on them Principal block Size = number of blocks + sum of external forces Support set

Stacks in which the support set contains only one block at each level Spinal stacks Stacks in which the support set contains only one block at each level Principal block Support set

Optimality condition: Optimal spinal stacks … Optimality condition:

A factor of 2 improvement over harmonic stacks! Spinal overhang Let S (n) be the maximal overhang achievable using a spinal stack with n blocks. Let S*(n) be the maximal overhang achievable using a loaded spinal stack on total weight n. Theorem: Conjecture: A factor of 2 improvement over harmonic stacks!

Optimal weight 100 loaded spinal stack

Optimal 100-block spinal stack

Are spinal stacks optimal? No! Support set is not spinal! Blocks = 20 Overhang = 2.32014 Tiny gap

Optimal 30-block stack Blocks = 30 Overhang = 2.70909

Optimal (?) weight 100 construction Blocks = 49 Overhang = 4.2390

“Parabolic” constructions 6-stack Number of blocks: Overhang: Balanced!

“Parabolic” constructions 6-slab 5-slab 4-slab

r-slab r-slab

r-slab within an (r +1)-slab

An exponential improvement over the ln n overhang of spinal stacks !!! So with n blocks we can get an overhang of c n1/3 for some constant c !!! An exponential improvement over the ln n overhang of spinal stacks !!! Note: c n1/3 ~ e1/3 ln n Overhang, Paterson & Zwick, American Math. Monthly Jan 2009

What is really the best design? Some experimental results with optimised “brick-wall” constructions Firstly, symmetric designs

“Vases” Weight = 1151.76 Blocks = 1043 Overhang = 10

“Vases” Weight = 115467. Blocks = 112421 Overhang = 50

then, asymmetric designs

“Oil lamps” Weight = 1112.84 Blocks = 921 Overhang = 10

Theorem: Maximum overhang is less than C n1/3 for some constant C Ωn is a lower bound for overhang with n blocks? Can we do better? Not much! Theorem: Maximum overhang is less than C n1/3 for some constant C Maximum overhang, Paterson, Perez, Thorup, Winkler, Zwick, American Math. Monthly, Nov 2009

Forces between blocks Assumption: No friction. All forces are vertical. Equivalent sets of forces

Distributions

Moments and spread j-th moment Center of mass Spread NB important measure

Signed distributions

Moves A move is a signed distribution  with M0[ ] = M1[ ] = 0 whose support is contained in an interval of length 1 A move is applied by adding it to a distribution. A move can be applied only if the resulting signed distribution is a distribution.

Equilibrium Recall! F1 F2 F3 F4 F5 F1 + F2 + F3 = F4 + F5 Force equation x1 F1+ x2 F2+ x3 F3 = x4 F4+ x5 F5 Moment equation

Moves A move is a signed distribution  with M0[ ] = M1[ ] = 0 whose support is contained in an interval of length 1 A move is applied by adding it to a distribution. A move can be applied only if the resulting signed distribution is a distribution.

Move sequences

Extreme moves Moves all the mass within the interval to the endpoints

Lossy moves If  is a move in [c-½,c+½] then A lossy move removes one unit of mass from position c Alternatively, a lossy move freezes one unit of mass at position c

Overhang and mass movement If there is an n-block stack that achieves an overhang of d, then n–1 lossy moves

Main theorem

Four steps Shift half mass outside interval Shift half mass across interval Shift some mass across interval and no further Shift some mass across interval

Simplified setting “Integral” distributions Splitting moves

-3 -2 -1 1 2 3

Basic challenge Suppose that we start with a mass of 1 at the origin. How many splits are needed to get, say, half of the mass to distance d ? Reminiscent of a random walk on the line O(d3) splits are “clearly” sufficient To prove: (d3) splits are required

Effect of a split Note that such split moves here have associated interval of length 2.

Spread vs. second moment argument

That’s a start! But … That’s another talk! we have to extend the proof to the general case, with general distributions and moves; we need to get improved bounds for small values of p; we have to show that moves beyond position d cannot help; we did not yet use the lossy nature of moves. That’s another talk!

Open problems What is the asymptotic shape of “vases”? What is the asymptotic shape of “oil lamps”? What is the gap between brick-wall stacks and general stacks? Other games! “Bridges” and “seesaws”.

Design the best bridge

Design the best seesaw

A big open area We only consider frictionless 2D constructions here. This implies no horizontal forces, so, even if blocks are tilted, our results still hold. What happens in the frictionless 3D case? With friction, everything changes!

With friction With enough friction we can get overhang greater than 1 with only 2 blocks! With enough friction, all diamonds are balanced, so we get Ω(n1/2) overhang. Probably we can get Ω(n1/2) overhang with arbitrarily small friction. With enough friction, there are possibilities to get exponents greater than 1/2. In 3D, I think that when the coefficient of friction is greater than 1 we can get Ω(n) overhang.

Applications? The end