Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval Peres Mikkel Thorup

A Crow Problem:

How long does it take to drive off the crow? -n n Simple random walk: about n 2 throws. One way to see that: consider the probability distribution of crow’s location; its variance goes up by 1 after each throw.

A new problem, brought to MSRI in spring ’05 by Zwick: the crow comes back…

…at night! Now what---your first stone will hit the crow and dislodge him, but after that you’re increasingly unsure where he is. You can certainly get him off the wall in order n 3 throws, and you certainly still need at least n 2. Which is the truth?

Spreading the distribution at night Now, you can only push down one bar at a time. -n 0 n The distribution starts with a bar of height 1 at the origin (representing the crow’s center-point landing spot). Each stone splits the distribution at the spot you aimed at.

How can you show necessity of ~n 3 “pushdowns” There seems to be no good potential function when the cost of each pushdown is 1. But we do get a nice potential when the cost of pushing down by an amount h is h, namely, the variance V. And we get another nice potential when the cost of pushing down by an amount h is h 2. This new potential is the expected distance A between two spots on the wall chosen randomly from our distribution. to reach this configuration? -n 0 n

Suppose we make m moves, where the i th move pushes a stack down by h i. From the V potential, we get n 2 /2 ≤ V = ∑ h i. From the A potential, we get n ≥ A = ∑ h i 2. Putting these together: n ≥ ∑ h i 2 ≥ V ∑ (1/m) h i = V ∑ (h i /V) h i (Schwarz inequality) = V 2 /m ≥ n 4 /4m thus m ≥ n 3 /4 and we are done! (from above) (rewriting as weighted sum) (since ∑ h i = V ) (since V ≥ n 2 /2 )

Theorem: Order n 3 throws are necessary. Proof: Uses two different potential functions, each for the wrong problem. An unusual case of two wrongs making a right. -n 0 n

The overhang problem How far off the edge of the table can we reach by stacking n identical blocks of length 1? “Real-life” 3D versionIdealized 2D version

Back in time with the overhang problem… John F. Hall, Fun with Stacking Blocks, Am. J. Physics December Martin Gardner - Scientific American’s “Mathematical Games” column, J.G. Coffin – Problem 3009, American Mathematical Monthly, George M. Minchin, A Treatise on Statics with Applications to Physics, 6th ed. (Clarendon, Oxford, 1907), Vol. 1, p William Walton, A Collection of Problems in Illustration of the Principles of Theoretical Mechanics 2nd ed. (Deighton, Bell, Cambridge, 1855), p J.B. Phear, Elementary Mechanics (MacMillan, Cambridge, 1850), pp. 140–141.

The classical solution “Harmonic Stack” Using n bricks we can get an overhang of

Is the classical solution optimal? Apparently not. How can we improve the construction?

Inverted pyramids? Claimed to be stable in Mad About Physics, by Chris Jargodzki and Franklin Potter, but…

They are unbalanced, when the number of layers exceeds 2.

Diamonds? The 4-diamond is balanced…

But the 5-diamond is …

not.

What really happens?

What really happens!

Why is this unbalanced?

… and this balanced?

Equilibrium F 1 + F 2 + F 3 = F 4 + F 5 x 1 F 1 + x 2 F 2 + x 3 F 3 = x 4 F 4 + x 5 F 5 Force equation Moment equation F1F1 F5F5 F4F4 F3F3 F2F2

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

Balanced Stacks Definition: A stack of bricks is balanced iff there is an admissible set of forces under which each brick is in equilibrium. 11 3

How can we tell if a stack is balanced?

Checking for balance F1F1 F2F2 F3F3 F4F4 F5F5 F6F6 F7F7 F8F8 F9F9 F 10 F 11 F 12 F 13 F 14 F 15 F 16 F 17 F 18 Equivalent to the feasibility of a set of linear inequalities:

Stability and Collapse A feasible solution of the primal system gives a set of balancing forces. A feasible solution of the dual system describes an infinitesimal motion that decreases the potential energy.

Small optimal stacks Overhang = Bricks = 4 Overhang = Bricks = 5 Overhang = Bricks = 6 Overhang = Bricks = 7

Small optimal stacks

Overhang = Bricks = 16 Overhang = Bricks = 17 Overhang = Bricks = 18 Overhang = Bricks = 19

Support and counterweight bricks Support set Counter- weights These examples are “spinal”: support stack has only one brick per level, so overhang increases with height. Spinal stacks can achieve overhang S(n) ~ log n.

Support set Stacks with downward external forces acting on them Loaded stacks Size = number of bricks + sum of external forces.

Support set Stacks in which the support set contains only one brick at each level Spinal stacks Equivalently, each level overhangs further than the one below.

Spinal overhang Let S (n) be the maximal overhang achievable using a spinal stack with n bricks. 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---in agreement with Hall’s simulations and conclusion. Conjecture:

100 bricks example

But are spinal stacks optimal? No! When # bricks reaches Support set is not spinal. Overhang = , slightly exceeding S(20).

Optimal weight 100 construction Overhang = Bricks = 47 Weight = 100

Brick-wall constructions

“Parabolic” construction 5-stack Number of bricks:Overhang: Stable!

Thus: n bricks can achieve an overhang of order n 1/3... an exponential improvement over the order log n overhang of spinal stacks.

Mayan from 900 BC---no keystone

“Parabolic” construction 5-slab 4-slab 3-slab

r-slab: inductive step 5-slab

Yes! Argument is based on the idea that laying bricks is like stoning crows. Each additional brick… spreads forces the same way that throwing a stone (at night) spreads the crow’s probability bar. The Upper Bound Is order n 1/3 best possible??

In particular, a stack of only n bricks cannot overhang by more than 6n 1/3 brick lengths. The parabolic construction gives overhang (3/16) 1/3 n 1/3 ~ n 1/3, so we have the order right but the constant is off by an order of magnitude. Simulations suggest that the constant can be improved by adjusting the shape of the brick wall construction… A generalized version of the “stoning crows” analysis shows that it takes order n 3 bricks to get the stack to lean out by n.

“Vases” Weight = Bricks = 1043 Overhang = 10

“Vases” Weight = Bricks = Overhang = 50

“Oil lamps” Weight = Bricks = 921 Overhang = 10 giving overhang of about 1.02 n 1/3.

How about using the third dimension? Our upper bound proof makes no use of the fact that bricks cannot overlap in space! Hence, the 6n 1/3 bound applies even in 3D, as long as there are no non-vertical forces. However, the constant can be improved in space by skintling, Effectively increasing the brick length to (1+w) 1/2.

Open problems ● What is the correct constant in the maximum overhang, in the rectilinear case? In the general 3-dimensional case? ● What is the asymptotic shape of “vases”? ● What is the asymptotic shape of “oil lamps”? ● What is the gap between brick-wall constructions and general constructions? ● Can the proof be extended to cover non-vertical forces (if, indeed, they are possible for 3D bricks)? ● How much friction is needed to change the 1/3 exponent for overhang?

Thank you for your attention. Happy stacking…