Mike Paterson

Mike Paterson

Overhang bounds

Joint work with Uri Zwick, Yuval Peres, Mikkel Thorup

and Peter Winkler

The classical solution

Harmonic Stacks

Using n blocks we can get an overhang of

1

2+

14

+16

+L +12n

=

1

21+

1

2+

1

3+L +

1

n⎛⎝⎜

⎞⎠⎟

≈1

2loge n

Is the classical solution optimal?

Obviously not!

Inverted triangles?

Balanced?

???

Inverted triangles?

Balanced?

Inverted triangles?

Unbalanced!

Inverted triangles?

Unbalanced!

Diamonds?

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

F1 + F2 + F3 = F4 + F5

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

Force equation

Moment equation

F1

F5F4

F3

F2

Checking balance

Checking balance

F1F2 F3 F4 F5 F6

F7F8 F9 F10

F11 F12

F13F14 F15 F16

F17 F18

Equivalent to the feasibilityof a set of linear inequalities:

Small optimal stacks

Overhang = 1.16789Blocks = 4




Small optimal stacks





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

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 weight 100 loaded spinal stack

Optimal 100-block spinal stack

Are spinal stacks optimal?

No!

Support set is not spinal!


Tiny gap

Optimal 30-block stack


Optimal (?) weight 100 construction


Weight = 100

“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 !!!

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

Ωn1is 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

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: (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!

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;

But …

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.

The end

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Mike Paterson