Joint work with Yuval Peres, Mikkel Thorup, Peter Winkler and Uri Zwick Overhang Bounds Mike...

Joint work with Yuval Peres, Mikkel Thorup, Peter Winkler and Uri Zwick

Overhang Bounds

Mike Paterson

DIMAP & Dept of Computer Science

University of Warwick

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The classical solution

Harmonic Stacks

Using n blocks we can get an overhang of

Diamonds

The 4-diamond is balanced

Diamonds

The 5-diamond is …

Diamonds?

… unbalanced!

What really happens?

What really happens!

Small optimal stacks

Overhang = 1.16789Blocks = 4

Overhang = 1.30455Blocks = 5

Overhang = 1.4367Blocks = 6

Overhang = 1.53005Blocks = 7

Small optimal stacks

Overhang = 2.14384Blocks = 16

Overhang = 2.1909Blocks = 17

Overhang = 2.23457Blocks = 18

Overhang = 2.27713Blocks = 19

Note “spinality”

Support and balancing blocks

Principalblock

Support set

Balancing

set

Support and balancing blocks

Principalblock

Support set

Balancing

set

Principalblock

Support set

Stacks with downward external

forces acting on them

Loaded stacks

Size =

number of blocks

+ sum of external

forces.

Principalblock

Support set

Stacks in which the support set contains

only one block at each level

Spinal stacks

Assumed to be optimal in:

J.F. Hall, Fun with stacking Blocks, American Journal of Physics 73(12), 1107-1116, 2005.

Optimal spinal stacks

…

Optimality condition:

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:

A factor of 2 improvement over harmonic stacks!

Conjecture:

Optimal 100-block spinal stack

Spine

Shield

Towers

Optimal weight 100 loaded spinal stack

Loaded spinal stack + shield

spinal stack + shield + towers

Are spinal stacks optimal?

No!

Support set is not spinal!

Overhang = 2.32014Blocks = 20

Tiny gap

Optimal 30-block stack

Overhang = 2.70909Blocks = 30

Optimal (?) weight 100 construction

Overhang = 4.2390Blocks = 49

Weight = 100

“Parabolic” constructions

6-stack

Number of blocks in d-stack: Overhang:Balanced!

“Parabolic” constructions

6-slab

5-slab

4-slab

r-slab

r-slab

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

(r-1)-slab

Nested inductions

“Smooth” parabola?

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Stacks with monotonic right contour can achieve only about ln n overhang [theorem above]

No good!

“Vases”

Weight = 1151.76

Blocks = 1043

Overhang = 10

“Oil lamps”

Weight = 1112.84

Blocks = 921

Overhang = 10

What about an upper bound?

Ωnis a lower bound

for overhang with n blocks

Can we do better?

Equilibrium

F1 + F2 + F3 = F4 + F5

x1 F1+ x2 F2+ x3 F3 = x4 F4+ x5 F5

Force equation

Moment equation

F1

F5F4

F3

F2

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

MovesA 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

F1 + F2 + F3 = F4 + F5

x1 F1+ x2 F2+ x3 F3 = x4 F4+ x5 F5

Force equation

Moment equation

F1

F5F4

F3

F2

Recall!

MovesA 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 movementIf there is an n-block stack that achieves an overhang of d, then

n–1 lossy moves

Main theorem

Four stepsShift half mass outside interval Shift half mass across interval

Shift some mass across intervaland no further

Shift some mass across interval

Simplified setting

“Integral” distributions

Splitting moves

0 1 2 3-3 -2 -1

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: Ω (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!

Can we extend the proof to the general case, with general distributions and moves?

Can we get improved boundsfor small values of p?

Can moves beyond position d help?

But …

We did not yet use the lossy nature of moves.

Spread vs. second moment argument

Spread vs. second moment argument

Spread vs. second moment inequalities

If 1 is obtained from 0 by an extreme move, then

Plackett (1947):

Spread vs. second moment argument(for extreme moves)

Splitting

“Basic” splitting move

A single mass is split into arbitrarily many

parts, maintaining the total and center of mass

if 1 is obtained from 0 by a sequence of splitting moves

Def:

Splitting and extreme moves

If V is a sequence of moves, we let V* be the corresponding sequence of

extreme moves

Lemma:

Corollary:

Spread vs. second moment argument(for general moves)

extreme

Notation

An extended bound

An almost tight bound

An almost tight bound - Proof

An asymptotically tight bound

lossy moves

An asymptotically tight bound - Proof

lossy

Our paper was in SODA’08 this week

An early version is at http://arXiv.org/pdf/0707.0093

Some open questions

What shape gives optimal overhang?

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.

The end

Applications!

The end