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

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

4. Is the classical solution optimal?
Obviously not!

5. Inverted triangles?
Balanced?

6. ???

7. Inverted triangles?
Balanced?

8. Inverted triangles?
Unbalanced!

9. Inverted triangles?
Unbalanced!

10. Diamonds?
Balanced?

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

12. Diamonds?
The 5-diamond is …

13. Diamonds?
… unbalanced!

14. What really happens?

15. What really happens!

16. How do we know this is unbalanced?

17. … and this balanced?

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

19. Checking balance

20. 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

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

22. Small optimal stacks

23. Small optimal stacks

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

25. Support and balancing blocks
Principal block
Balancing set
Support set

26. Support and balancing blocks
Balancing set
Principal block
Support set

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

28. 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

29. Optimality condition:
Optimal spinal stacks …
Optimality condition:

30. 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!

31. Optimal weight 100 loaded spinal stack

32. Optimal 100-block spinal stack

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

34. Optimal 30-block stack
Blocks = 30
Overhang =

35. Optimal (?) weight 100 construction
Blocks = 49
Overhang =

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

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

38. r-slab
r-slab

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

41. 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

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

43. “Vases”
Weight =
Blocks = 1043
Overhang = 10

44. “Vases”
Weight =
Blocks =
Overhang = 50

45. then, asymmetric designs

46. “Oil lamps”
Weight =
Blocks = 921
Overhang = 10

47. 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

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

55. Distributions

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

57. Signed distributions

58. 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.

59. 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

60. 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.

61. Move sequences

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

63. 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

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

65. Main theorem

66. 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

67. Simplified setting
“Integral” distributions
Splitting moves

68. -3 -2 -1 1 2 3

69. 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

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

71. Spread vs. second moment argument

72. 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!

73. 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”.

74. Design the best bridge

75. Design the best seesaw

76. 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!

77. 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.

78. Applications?
The end