Peter Winkler, Dartmouth

1. The Worm Order and its Applications FPSAC, San Francisco State University, August 2010 Peter Winkler, Dartmouth joint work in part with Graham Brightwell (LSE) and in part with Lizz Moseman (NIST)

2. Put a b if you can schedule them so as to keep a below b pairwise. For example: Lemma: (i) a a b = <0,2,3,-1,0,2> (ii) a b c implies a c (iii) a b implies min a min b andmax a max b. a = <0,3,0,2,-1,2,1,0> Put a b if a b a and mod out by . A pre-order among words on an ordered alphabet Theorem: There is a unique shortest word (called a worm) in each equivalence class.

3. A typical worm 6 0 -6 the worm<0,2,-1,5,-2,6,-5,4,-3,2,-1,1>

4. 2 212 202 12 21 121 1202 2021 12021 102 201 1201 20 1021 1 02 021 120 10201 1020 0201 101 020 01 10 1 n+1 (4 -3n-4) 010 9 0 The lattice of worms of range{0,1,2} Number of worms with givenn-element range:

5. You need to upgrade SFSU’s computer system, one component at a time, without allowing performance to fall below a certain level at any time. Some Scheduling Problems to be Solved… Can it ever be right to downgrade a component during the process?

6. Sweeping a Forest You need to find a lost child in a portion of the Cascades, using a line of searchers which cannot be stretched beyond a certain total length. Assuming you can do it, can you do it without ever covering the same point twice? In other words, can you do it without retreating?

7. Planar s-t Networks: Peter Doyle’s problem 3 2 4 2 2 2 2 2 s 3 t 3 2 3 2 3 2 2 3 2 3 3 Suppose you can make your way from the southernmost s-t path to the northernmost, crossing one face at a time, without using a path of length more than some bound B. Does it follow that you can do it moving only north, that is, crossing each face only once?

8. 4 7 3 6 1 3 7 10 6 9 4 6 For example: 2 5 1 4 -1 a = <1,4,0,3,-2> 1 5 8 4 7 2 4 b = <-1,4,1,6,3> 0 3 -1 2 -3 -1 1 4 0 3 -2 Scheduling Real Words Given two words (finite strings of reals), how hard is it to schedule them to minimize their maximum pairwise sum? The schedule constitutes a lattice path and yields a new word c=a+b. Can we do this efficiently with many words?

9. A lattice is a poset in which, for every x and y , there is a least upper bound x y and a greatest lower bound x y . If finite it thus has a least element 0 and a greatest element 1 . A lattice is distributive if x (y z) = (x y) (x z) ; equivalently, if it is the lattice of ideals (downward-closed subsets) of some poset. These sublattices are forbidden in a distributive lattice. A function f from a lattice to the reals is submodular if f(x y) + f(x y) f(x) + f(y) . A distributive lattice together with a submodular function is called a submodular system. A Digression to Lattices

10. C Theorem: In any submodular system (L,f) there is a maximal chain C such that f(C) f(P) (in the worm order) for any path P from 0 to 1 in L . P A General Result Corollary: For any two words a and b there is an optimal way of adding them such that the word a+b precedes the result of any other schedule, in the worm order. Consequence for adding many words to minimize the maximum sum: You can add the words pairwise in any order, if you use the canonical sum above. This provides a fast algorithm for adding multiple words.

11. S a a a b a b b b a b a b A “bone” A Useful (and Easily Proved) Corollary: 1 If S is a boneless subset of a distributive lattice, and it contains a path from 0 to 1, it contains a maximal chain. 0

12. If some y is not in S, j j i j i Letx … x be a shortest path from 0 to 1 in S. Put y = x . 0 m y y j j z x x Putu = x z. x z j j i i i Then u is in S,elsex and z constitute a bone. i x z i i x z i Sketch of a proof:

13. C P Corollary 2: In any submodular system (L,f) there is a maximal chain C such that max{f(x): x in C} max{f(x): x in P} for any path P from 0 to 1 in L . Proof: For any bound b the set of points x in L where f(x)b is boneless. A 2nd Corollary and a Consequence Consequence for upgrading your computer system: If performance is submodular (a reasonable assumption), then indeed you can upgrade the whole system one component at a time without any backtracking.

14. Sweeping a Forest (with detachments) Again you need to find a lost child in a forest, without stretching the line of searchers beyond a fixed length. But this time you needn’t start all the searchers in the same place. Now, you are guaranteed that if you can do it at all, then you can you do it without any retreating.

15. Planar s-t Networks: A Counterexample 1 1 10 There is only one way to migrate from the southernmost s-t path to the northernmost, crossing one face at a time, without using a path of length more than 40. 1 1 20 30 t s 1 20 1 1 1 To do it, you must cross and recross the rightmost face several times.

16. 3 2 4 2 2 2 2 2 3 3 2 3 2 3 2 3 2 3 3 Migrating Through a Directed Network s t 2 Again, we wish to migrate from the southernmost s-t path to the northernmost, crossing one face at a time, without using any path of length greater than L. But this time, we have a distributive lattice and submodular function,so we never need to cross any face more than once.

17. A new version of graph searching Vertices of a graph are contaminated, and we wish to clear the vertices but must use c “blocking material” along an edge of capacity c to keep contamination from spreading. Now, the “slipchain” version of our theorem (cf. Thm 3.2 of Graph Minors X) shows that if we can decontaminate with a total of b material, we never need to decontaminate any vertex more than once.

18. .4 .7 .3 .6 .3 .3 .8 1.1 .7 1.0 .7 .7 .2 .5 .1 .4 .1 .1 .5 .8 .4 .7 .4 .4 Then: as a function of t, the escape probability is: .2 .5 .1 .4 .1 .1 1 .1 .4 0 .3 0 2 2 1 t t t Consequences for Coordinate Percolation Suppose: uniform random reals from [0,1] are assigned to the coordinates; each grid point inherits the sum of its coordinate reals; and any vertex whose sum exceeds some bound t is deleted. (t=.75 in figure.) …as is believed for independent percolation.

19. Proof To do this, let a be the value associated with the ith column and b the value for the jth row. If we define c:=t-b then we have a +b < t iff a < c . i j j j j i j i Note that in a region with no open lines (i.e. each a and b exceeds 1–t), all a’s and c’s are now independent draws from [1-t,1] . This reduces the problem to: given two uniformly random words a and c , what is the probability that a precedes c in the worm order? j i We make use of the fact that independent draws from any continuous distribution fall in a uniformly random linear order.

20. We use a time-varying Markov chain to get… 1 1 Eigenvalues: 1 +(- 2), 1 +(- 2) n n Eigenvectors: (, 1,); (, 1,); (0, 0, 1) . ThenQ(n):= Pr(survive nsteps) n n = ( + 1) - ( + 1) 1 (-1) (-1) - 2 - 2 1+ = 1 - 1 2 5 5 5 ( ( ) ) n n where= and .

21. t t Calculating the probability of percolation 3+ 2 Now, ifQ(n)= Pr(survive nsteps), then n = (t-1)(2-t) Q(n) 3- 5 5 2 n 5 + 3 3 - 5 (t-1) (t-1) - = 10 10 giving critical exponent 2-~ .382 5 5 as well as showing that is continuous everywhere, analytic except at t=1 and t=2, and C except at the critical point t=1. 1

22. References: Moseman & W., On a form of coordinate percolation, CPC 17 #6 (Nov ’08), 837-845. Thank you for your attention! Brightwell & W., Submodular percolation, SIDMA 23 #3 (2009), 1149-1178. Brightwell & W., Forward processes and a `lost child’ theorem, in preparation.