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Mike Paterson Yuval Peres Mikkel Thorup Peter Winkler Uri Zwick

Extremal Combinatorics in Banana-Shampoo, November 2006. Random Walk with Planted Coins. Mike Paterson Yuval Peres Mikkel Thorup Peter Winkler Uri Zwick. The Problem.

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Mike Paterson Yuval Peres Mikkel Thorup Peter Winkler Uri Zwick

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  1. Extremal Combinatorics in Banana-Shampoo, November 2006 Random Walk with Planted Coins Mike Paterson Yuval Peres Mikkel Thorup Peter WinklerUri Zwick

  2. The Problem Suppose Tina Token is taking a simple random walk on the number line and needs to achieve distance n from her start point, with probability at least (say) ½. But she cannot proceed from any node until a coin appears there for her to flip. The coins must be planted in advance in space-time, and are wasted if Tina is not there when they appear. n -n 0

  3. 2 There must be at least n coins, because it takes that many steps (on average) for Tina to get to n. 3 On the other hand n coins are clearly enough, because you can plant each of n coins at every even (or odd) point between n and -n. 2 3 So what’s right? n or n ? Or something in between? And why should we care? 2 The Obvious

  4. Some Observations (1) There’s no need to place two coins at the same instant of time (we don’t care how long Tina takes to get there, only how many coins are used). (2) Each planted coin changes the probability distribution of Tina’s location by “pushing down” the bar at one location, and sending its material half to the left, half to the right.

  5. -3 -2 -1 0 1 2 3 Example: Tinamoves 3 places

  6. Wanted: a function f(x) such that (for example) f(0…010…0)=0, f(any winningconfiguration) = n , and f never goes up by more than one after a single push-down. 3 Potential Functions Problem: No such function has chosen to manifest itself to us. But suppose we change the cost function, charging y (instead of 1) to push a stack down by an amount y ?

  7. 2 V(x) = x i (variance) is a perfect potential: i V(x) goes up by exactly y when any bar is pushed down by y; V(0…010…0) = 0; and V(W) is at least n /2 for any winning configuration W. 2 Conclusion: the total amount of “pushing down” must be at least n /2. 2 With that cost function…

  8. Here’s another differential cost: charge 2ydy for pushing down by dy, so the total cost of pushing a bar down from y to 0 is y . 2 Why does this work? Because “pushing down by y “ commutes while “pushing all the way down” does not… In turn, because pushing down by y is a differential operation with marginal cost dy.

  9. Let S(x) = i – j x x (expected distance). Then S(x) increases by exactly y when a bar of height y is pushed down; and S(0…010) = 0. j i A new potential function For this cost function the potential was (for us) hard to find---but easy to prove. The key here is that S(W) is at most n.

  10. Suppose we have a successful pushing strategy which takes m steps, pushing down a bar of height y(t) at time t , and ends at a configuration of variance V = y(t) . 2 Then n y(t) = V (y(t)/V)y(t) V (1/m)y(t) = V /m n /4m 2 4 3 by Cauchy-Schwarz, and thus m n /4. And now for some sleight of hand… But what on earth was all this for?

  11. 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 version Idealized 2D version

  12. Back in time with the overhang problem… John F. Hall, Fun with Stacking Blocks, Am. J. Physics December 2005. Martin Gardner - Scientific American’s “Mathematical Games” column, 1969. J.G. Coffin – Problem 3009, American Mathematical Monthly, 1923. George M. Minchin, A Treatise on Statics with Applications to Physics, 6th ed. (Clarendon, Oxford, 1907), Vol. 1, p. 341.William Walton, A Collection of Problems in Illustration of the Principles of Theoretical Mechanics 2nd ed. (Deighton, Bell, Cambridge, 1855), p. 183.J.B. Phear, Elementary Mechanics (MacMillan, Cambridge, 1850), pp. 140–141.

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

  14. 1/3 gives overhang of order n . Paterson and Zwick’s parabolic stretcher-bond construction But can we prove this is best possible?

  15. You can get a better constant… Weight = 1112.84 Bricks = 921 Overhang = 10 But that’s all.

  16. Random walk vs. distribution of forces -n 0 n Think of the distribution of forces at some level of the stack as the probability distribution of Tina’s location at some particular time. The distribution starts with a bar of height 1 at the origin (representing Tina at the start, or the force exerted by a single brick on the bottom of the stack). Each added brick (or flipped coin) spreads out the force (probability) in both directions.

  17. The Planted Coin Solution and (substantial) generalizations thereof show that in fact order n is the best overhang you can get. 1/3

  18. Open problems • What is the correct constant in the maximum overhang (n1/3), in the rectilinear case? In the general 3-dimensional case (with “skintling”)? • What is the asymptotic shape of the “oil lamp”? • What is the gap between brick-wall constructionsand general constructions? • What is the correct constant in the Tina token problem? • Is the greedy algorithm for moving Tina essentially optimal?

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