Start with a puzzle…

1. Start with a puzzle… • There are four occurrences of the pattern  = 132 in the sequence  = 13254: • 1 3 25 4 13 2 5 41 32 5 4 1 3 2 5 4 • Can you find a different sequence  (a different rearrangment of the digits 12345) with more than four occurrences of the pattern  = 132 ?

2. Packing Densities of Permutations Graph Theory With AltitudeDenver, May 17, 2005 Walter Stromquist Bryn Mawr College / Swarthmore College

3. Outline • History • Example:  = 132 • Definitions • Layered patterns • Reid Barton’s proof • Results for  in S3 and S4 — and open problems • Connection to partially ordered sets • — and more open problems

4. History • 1992: Wilf’s address to SIAM • Many results about permutations with no occurrences of  • 1993-1996: • Settle case of 132 (Kleitman, Galvin, WRS) (others?) • Packing densities exist (Galvin) • Layered permutations • 1997: Alkes Price’s thesis • 2002-2005: Many (5) papers in Electronic Journal of Combinatorics • 2004: Reid Barton’s Morgan Prize paper (EJC 11)

5. Example:  = 132 • Let  = 132 and  = 13254. • An occurrence of  in  is a subsequence of  that has the same ordering as  = 132 — that is, low / high / middle. There are 4 such occurrences: • 1 3 25 4 13 2 5 41 32 5 4 1 3 2 5 4

6. Definitions • Two sequences  and  are called order-isomorphic if • For example, 1 3 2 5 4 and 1 2001 2000 5001 5000 • are order-isomorphic. • We’re concerned only with finite sequences of distinct terms. We may as well represent them as permutations of integers • 1, …, n. • The set of permutations of length n is called Sn.

8. Definitions • A pattern is a permutation  in Sm. • An occurrence of  in  is a subsequence of  that is order-isomorphic to . • Let • Clearly,

9. Definitions • In this talk, the pattern is always called  and always has length m. • The permutation  always has length n. • We’ll always assume that n > m.

10. Example:  = 132 • We can do better: If  = 12543, then  • has 6 occurrences of the 132 pattern. • So: • Since there are 10 three-element subsequences • of , we say that the packing density of 132 in  is • …and since that’s the largest packing density for any  of • length 5, we also say that

11. Definitions • The packing density of  in  is • Clearly, • We’re concerned with permutations Sn that maximize the • packing density ( ,  ). So, define: • Any permutation * that achieves this maximum (for a given • size n) is called an optimizer for .

12. Definitions • The packing density of  is the limiting value, • if it exists. • Our problems in this talk are, given , • (1) What are the optimizers for  ? • (2) What is the packing density of  ?

13. Example:  = 132 • What can we do with longer sequences  ? • For n = 9, try  = 123 987654… •  = 123 987654… • In fact, 9( 132 ) = 46 / 84.

14. Example:  = 132 • In general, here’s the best we can do for large n: Now So the packing density of  = 132 is

15. Another Example: 123 • Now let  = 123. • If  = 1234…n, then every 3-term subsequence of  is • order-isomorphic to . So, • The optimizers for 123 are of the form 1234..n, and the • packing density of 123 is 1.

16. Outline • History • Example:  = 132 • Definitions • Packing densities exist • Layered patterns • Results for  in S3 and S4 — and open problems • Connection to partially ordered sets — and more open problems

17. Theorem and Proof • Theorem (Galvin): The limit always exists.

18. 1432 1432 1243 1243 1243 Theorem and Proof • Theorem (Galvin): The limit always exists. • Proof: Let Sn be an optimizer for size n, so that • Now consider its one-point-deleted subsequences 1, 2, …, n. Every occurrence of  in  also appears in exactly (n–m) of the i’s.

19. 1432 1432 1243 1243 1243 Theorem and Proof • Theorem (Galvin): The limit always exists. • Proof: Let Sn be an optimizer for size n, so that • Now consider its one-point-deleted subsequences 1, 2, …, n. Every occurrence of  in  also appears in exactly (n–m) of the i’s.

20. Theorem and Proof • Theorem (Galvin): The limit always exists. • Proof: Let Sn be an optimizer for size n, so that • Now consider its one-point-deleted subsequences 1, 2, …, n. Every occurrence of  in  also appears in exactly (n–m) of the i’s.

21. Theorem and Proof • Theorem (Galvin): The limit always exists. • Proof: Let Sn be an optimizer for size n, so that • Now consider its one-point-deleted subsequences 1, 2, …, n. Every occurrence of  in  also appears in exactly (n–m) of the i’s. • It follows (with a bit of algebra) that • So ( ,  ) can’t exceed the largest of the ( , i )’s.

22. Theorem and Proof • ... • So ( ,  ) can’t exceed the largest of the ( , i )’s. • So: • So the sequence { n(  ) } is non-increasing. Since it is bounded below by zero, it must have a limit. //

23. Layered Permutations • A permutation is layered if it consists of one or more blocks, such that the symbols are increasing between blocks and decreasing within blocks. • Examples: The following are layered: • 132 123 1432 2143 • but the following are not layered: • 312 1342.

24. Layered Permutations • Theorem: If  is layered, then its optimizers  are layered. • More precisely: For every n, • This means that to find the packing density of a layered • pattern , we need only consider layered permutations .

25. Permutations in S3 • Here are the permutations  in S3: • 123 132 213 231 312 321 • ()=1 ()=.464 ()=1 • The rest of these cases can be resolved by symmetry.

26. Permutations in S3 — Symmetry • Symmetry:

27. Permutations in S3 — Symmetry • Reversal:

28. Permutations in S4 • Permutations in S4: • Layered permutations, by symmetry class: • 1234 (two variations) - packing density 1 • 1432 (four variations) - packing density 0.4236 (Price) • 1243 (four variations) - packing density 3/8 • 2143 (two variations) - packing density 3/8 • 1324 (two variations) - approximately 0.244 (Price) • Unlayered permutations: • 1342 (eight variations) - unknown ( lower bound 0.1966 ) • 2413 (two variations) - unknown ( bounds 51/511, 2/9 )

29. 1324 • Let  = 1324. • Price: Optimal ratios are… .39 .19 .07… • and (1324)  0.244.

30. 1342 • Let  = 1342. • This optimizer gives a • lower bound. If you think • it’s the best you can do, • then • (1342)  0.1966.

31. 1342 • If the lower bound holds… • (1342)  0.1966... • (Batayev) • (1342) = (1432) (132)

32. Partially Ordered Sets • A (finite) partially ordered set is a finite set together with a relation < such that • (a) It is never true that x < x; • (b) It is never true that both x < y and y < x; and • (c) If x < y and y < z, then x < z (transitivity). • A partially ordered set is also called a poset. We use the terms “above” and “below” to describe the relation (that is, read x < y as “x is below y” ). • Diagrams:

33. Partially Ordered Sets • Example: Consider a finite set of vectors (x, y) in R2. Say that • (x1, y1) < (x2, y2 ) • if • x1 < x2and y1 < y2. This construction can also be done in R3, or in Rn. Fact: Every finite partially ordered set is isomorphic to a poset constructed in this way. The smallest n for which Rn suffices is called the dimension of the poset.

34. Partially Ordered Sets • Posets that can be represented in R2 have graphs like those of permutations: • Match each such poset with the permutation that has the same graph. • This matching is not 1-to-1, nor does it cover all posets. But, it is a bijection for layered posets — that is, the ones that correspond to layered permutations.

35. Partially Ordered Sets • Packing densities of posets: • Theorem: Layered posets have layered optimizers. • The theory for layered posets is exactly like that for layered permutations.

36. Posets aren’t exactly like permutations • Example: •  = = • These are the same poset, but different permutations. • So, = 0 (as permutations) • but = 1 (if we think of them as posets). • If  is layered, then is the same in both worlds.

37. Reid Barton’s Proof of theLayered Poset Theorem • Theorem: If P is a layered poset, and n  |P|, then P has an optimizer Q’ of size n such that Q’ is a layered poset. • Proof: Let Q be any optimizer of size n for P, and let u and v be any two incomparable elements of Q. • Form Q1 by replacing v with an incomparable copy u’ of u. • Form Q2 by replacing u with an incomparable copy v’ of v.

38. Proof, continued… • Then every occurrence of P in Q appears… • once each in Q1, Q2 if it omits both u and v • twice in Q1 if it includes u but not v • twice in Q2 if it inlcudes v but not u • once each in Q1, Q2 if it includes both u and v • (in the last case, because P is layered). • So, • But Q is an optimizer, so • and Q1 and Q2 are both optimizers.

39. u Q v (P,Q)=2 v’ u’ u v Q1 Q2 (P,Q1)=2 (P,Q2)=3 Pattern: Every occurrence of P in Q recurs once each in Q1 and Q2, or twice in Q1, or twice in Q2. Actually, in this example Q isn’t an optimizer. As a result, there’s an extra occurrence of the pattern in Q2. If Q were an optimizer, the theorem would force (P,Q)=(P,Q1)=(P,Q2).

40. Proof, concluded • So in general, we can freely modifiy any modifier by replacing elements incomparable to u with incomparable copies of u… • that is, by moving them into a layer with u. • Ultimately, any optimizer can be altered until it becomes a layered optimizer. //

41. Open Problems • Find a better way to compute  ( 1324 ). • What is  ( 1342 ) ? More generally, can you say anything useful about recursively layered permutations ? • What is  ( 2413 ) ? • Find any general way of attacking non-layered permutations. • Can you say anything about packing densities of posets that isn’t just a statement about permutations, in disguise ? • What’s the packing density of this poset ?