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
1 1 1
1
  L 
2 4 6
2n
1
1 1
1
1
  1    L    log e n
2
2 3
n
2
Harmonic Stacks
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
F4
F3
F1 + F2 + F3 = F4 + F5
x1 F1+ x2 F2+ x3 F3 = x4 F4+ x5 F5
F5
Force equation
Moment equation
Checking balance
Checking balance
F2
F1
F3
F4
F5
F11
F7
F8
F9
F10
F13
F14
F15
F16
F6
F12
Equivalent to the feasibility
of a set of linear inequalities:
F17
F18
Small optimal stacks
Blocks = 4
Overhang = 1.16789
Blocks = 5
Overhang = 1.30455
Blocks = 6
Overhang = 1.4367
Blocks = 7
Overhang = 1.53005
Small optimal stacks
Blocks = 16
Overhang = 2.14384
Blocks = 17
Overhang = 2.1909
Blocks = 18
Overhang = 2.23457
Blocks = 19
Overhang = 2.27713
Support and balancing blocks
Balancing
set
Principal
block
Support
set
Support and balancing blocks
Balancing
set
Principal
block
Support
set
Loaded stacks
Stacks with
downward external
forces acting on them
Size
=
Principal
block
number of
blocks
+
sum of external
forces
Support
set
Spinal stacks
Stacks in which the
support set contains
only one block
at each level
Principal
block
Support
set
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:
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
Weight = 100
Blocks = 49
Overhang = 4.2390
“Parabolic” constructions
6-stack
Number of blocks:
Balanced!
Overhang:
“Parabolic” constructions
6-slab
5-slab
4-slab
r-slab
r-slab
r-slab within an (r +1)-slab
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
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.
Recall!
Equilibrium
F1
F2
F4
F3
F1 + F2 + F3 = F4 + F5
x1 F1+ x2 F2+ x3 F3 = x4 F4+ x5 F5
F5
Force equation
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 some mass across interval
and no further
Shift half mass across interval
Shift some mass across interval
Simplified setting
“Integral” distributions
Splitting moves
-3
-2
-1
0
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 …
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