mike paterson uri zwick
DESCRIPTION
Overhang. Mike Paterson Uri Zwick. The overhang problem. How far off the edge of the table can we reach by stacking n identical blocks of length 1 ? J.B. Phear – Elementary Mechanics (1850) J.G. Coffin – Problem 3009, American Mathematical Monthly (1923). No friction. - PowerPoint PPT PresentationTRANSCRIPT
Mike PatersonUri Zwick
Overhang
The overhang problem
How far off the edge of the table can we reach by stacking n identical blocks of length 1?
J.B. Phear – Elementary Mechanics (1850)J.G. Coffin – Problem 3009, American Mathematical Monthly (1923).
“Real-life” 3D version Idealized 2D version
No frictionLength parallel to table
The classical solution
Harmonic Stacks
Using n blocks we can get an overhang of
Is the classical solution optimal?
Obviously not!
Inverted triangles?
Balanced?
???
Inverted triangles?
Balanced?
Inverted triangles?
Unbalanced!
Diamonds?
The 4-diamond is balanced
Diamonds?
The 5-diamond is …
Diamonds?
The 5-diamond is Unbalanced!
What really happens?
What really happens!
Why is this 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
Forces between blocks
Assumption: No friction.All forces are vertical.
Equivalent sets of forces
Balance
Definition: A stack of blocks is balanced iff there is an admissible set of forces under which each block is in equilibrium.
1 1
3
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:
Static indeterminacy:balancing forces, if they exist, are usually not unique!
Balance, Stability and Collapse
Most of the stacks considered are precariously balanced, i.e.,
they are in an unstable equilibrium.
In most cases the stacks can be made stable by small modifications.
The way unbalanced stacks collapse can be determined in polynomial time
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
Assumed to be optimal in:
J.F. Hall, Fun with stacking Blocks, American Journal of Physics 73(12), 1107-1116, 2005.
Loaded vs. standard stacks
1
1
Loaded stacks are slightly more powerful.
Conjecture: The difference is bounded by a constant.
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
Brick-wall constructions
Brick-wall constructions
“Parabolic” constructions
6-stack
Number of blocks: Overhang:
Balanced!
Using n blocks we can get an overhang of (n1/3) !!!
An exponential improvement over the O(log n) overhang of
spinal stacks !!!
“Parabolic” constructions
6-slab
5-slab
4-slab
r-slab
r-slab
r-slab within a (r+1)-slab
“Vases”
Weight = 1151.76
Blocks = 1043
Overhang = 10
“Vases”
Weight = 115467.
Blocks = 112421
Overhang = 50
Forces within “vases”
Unloaded “vases”
“Oil lamps”
Weight = 1112.84
Blocks = 921
Overhang = 10
Forces within “oil lamps”
Brick-by-brick constructions
Is the (n1/3) the final answer?
Mike PatersonYuval Peres
Mikkel ThorupPeter Winkler
Uri Zwick
MaximumOverhangYes!
1
0 1 2 3-3 -2 -1
Splitting game Start with 1 at the origin
How many splits are needed to get, say, a quarter of the mass to
distance n?
At each step, split the mass in a given
position between the two adjacent
positions
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?● What is the gap between loaded stacks
and standard stacks?