mike paterson
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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
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
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!
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: 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
Applications?
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