Start with a puzzle… There are four occurrences of the pattern  = 132 in the sequence  = 13254: Can you find a different sequence  (a different rearrangment of the digits 12345) with more than four occurrences of the pattern  = 132 ?

Packing Densities of Permutations Walter Stromquist Bryn Mawr College / Swarthmore College Graph Theory With Altitude Denver, May 17, 2005

Outline History Example:  = 132 Definitions Layered patterns Reid Barton’s proof Results for  in S 3 and S 4 — and open problems Connection to partially ordered sets — and more open problems

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

Example:  = 132 Let  = 132 and  = 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:

Definitions Two sequences  and  are called order-isomorphic if For example, and 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 S n.

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

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.

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

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

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  ?

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

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

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 n, and the packing density of 123 is 1.

Outline History Example:  = 132 Definitions Packing densities exist Layered patterns Results for  in S 3 and S 4 — and open problems Connection to partially ordered sets — and more open problems

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

Theorem and Proof Theorem (Galvin): The limit always exists. Proof: Let  S n 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

Theorem and Proof Theorem (Galvin): The limit always exists. Proof: Let  S n 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

Theorem and Proof Theorem (Galvin): The limit always exists. Proof: Let  S n 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.

Theorem and Proof Theorem (Galvin): The limit always exists. Proof: Let  S n 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.

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. //

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: but the following are not layered:

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 .

Permutations in S 3 Here are the permutations  in S 3 :  (  )=1  (  )=.464  (  )=1 The rest of these cases can be resolved by symmetry.

Permutations in S 3 — Symmetry Symmetry:

Permutations in S 3 — Symmetry Reversal:

Permutations in S 4 Permutations in S 4 : Layered permutations, by symmetry class: 1234 (two variations) - packing density (four variations) - packing density (Price) 1243 (four variations) - packing density 3/ (two variations) - packing density 3/ (two variations) - approximately (Price) Unlayered permutations: 1342 (eight variations) - unknown ( lower bound ) 2413 (two variations) - unknown ( bounds 51/511, 2/9 )

1324 Let  = Price: Optimal ratios are… … and  (1324) 

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

1342 If the lower bound holds…  (1342)  (Batayev)  (1342) =  (1432)  (132)

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:

Partially Ordered Sets Example: Consider a finite set of vectors (x, y) in R 2. Say that (x 1, y 1 ) < (x 2, y 2 ) if x 1 < x 2 and y 1 < y 2. This construction can also be done in R 3, or in R n. Fact: Every finite partially ordered set is isomorphic to a poset constructed in this way. The smallest n for which R n suffices is called the dimension of the poset.

Partially Ordered Sets Posets that can be represented in R 2 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.

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.

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.

Reid Barton’s Proof of the Layered 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 Q 1 by replacing v with an incomparable copy u’ of u. Form Q 2 by replacing u with an incomparable copy v’ of v.

Proof, continued… Then every occurrence of P in Q appears… once each in Q 1, Q 2 if it omits both u and v twice in Q 1 if it includes u but not v twice in Q 2 if it inlcudes v but not u once each in Q 1, Q 2 if it includes both u and v (in the last case, because P is layered). So, But Q is an optimizer, so and Q 1 and Q 2 are both optimizers.

Pattern: Actually, in this example Q isn’t an optimizer. As a result, there’s an extra occurrence of the pattern in Q 2. If Q were an optimizer, the theorem would force (P,Q)= (P,Q 1 )= (P,Q 2 ). Q u v (P,Q)=2 Q1Q1 u u’ (P,Q 1 )=2 v’ Q2Q2 v (P,Q 2 )=3 Every occurrence of P in Q recurs once each in Q 1 and Q 2, or twice in Q 1, or twice in Q 2.

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. //

Open Problems 1.Find a better way to compute  ( 1324 ). 2.What is  ( 1342 ) ? More generally, can you say anything useful about recursively layered permutations ? 3.What is  ( 2413 ) ? 4.Find any general way of attacking non-layered permutations. 5.Can you say anything about packing densities of posets that isn’t just a statement about permutations, in disguise ? 6.What’s the packing density of this poset ?