Start with a puzzle…

Start with a puzzle…

There are four occurrences of the pattern = 132 in the sequence = 13254:

1 3 2 5 4 1 3 2 5 4 1 3 2 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 ?

Packing Densities of Permutations

Walter StromquistBryn Mawr College / Swarthmore

College

Graph Theory With AltitudeDenver, May 17, 2005

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

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)

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 2 5 4 1 3 2 5 4 1 3 2 5 4 1 3 2 5 4

Definitions

Two sequences and are called order-isomorphic if

For example, 1 3 2 5 4 and 1 2001 2000 5001 5000are order-isomorphic.

We’re concerned only with finite sequences of distinct terms. We may as well represent them as permutations of integers1, …, n.

The set of permutations of length n is called Sn.

whenever (and vice versa).i j i j

Definitions

A pattern is a permutation in Sm.

An occurrence of in is a subsequence of that is order-isomorphic to .

Let

Clearly,

( , ) the number of occurrences

of in .

n0 ( , ) .

m

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

6(132, ) .

10

5

6(132) .

10

(132, ) 6.

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 .

n

m

( , )( , ) .

0 ( , ) 1.

nn

S( ) max ( , ).

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 ?

n

n( ) lim ( )

Example: = 132

What can we do with longer sequences ?

For n = 9, try = 123 987654… = 123 987654…

In fact, 9( 132 ) = 46 / 84.

46(132, ) 0.54884

45(132, ) 0.53684

Example: = 132

In general, here’s the best we can do for large n:

ratio 1: 3

Now

So the packing density of = 132 is

n(132) 2 3 3 .464 .

( ) 2 3 3.

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.

n( , ) , and

m

( , ) 1.

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

Theorem and Proof

Theorem (Galvin): The limit always exists.

nn

( ) lim ( )

Theorem and Proof


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.

n( ) ( , ).

1432 1432 1243 1243 1243

n

n( ) lim ( )

Theorem and Proof




n( ) ( , ).

1432 1432 1243 1243 1243

n

n( ) lim ( )

Theorem and Proof




n( ) ( , ).

n

n( ) lim ( )

Theorem and Proof




It follows (with a bit of algebra) that

So ( , ) can’t exceed the largest of the ( , i )’s.

n

n( ) lim ( )

n( ) ( , ).

ii 1,...,n

( , ) average ( , ) .

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. //

i n 1n i i n 1

S( ) ( , ) max ( , ) max ( , ) ( ).

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.

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 .

n nS S ,

layered

max ( , ) max ( , ).

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.

Permutations in S3 — Symmetry

Symmetry:

Permutations in S3 — Symmetry

Reversal:

Permutations in S4

Permutations in S4:

Layered permutations, by symmetry class:

1234 (two variations) - packing density 11432 (four variations) - packing density 0.4236 (Price)1243 (four variations) - packing density 3/82143 (two variations) - packing density 3/81324 (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 )

1324

Let = 1324.

Price: Optimal ratios are… .39 .19 .07…

and (1324) 0.244.

1342

Let = 1342.

This optimizer gives alower bound. If you thinkit’s the best you can do,then

(1342) 0.1966.

1342

If the lower bound holds…

(1342) 0.1966...

(Batayev)

(1342) = (1432) (132)

33 3

43 3

4 2 2 4 2 224 8 3

1 3(1342)

1 4 2 2 4 2 2 1

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:


Example: Consider a finite set of vectors (x, y) in R2. Say that

(x1, y1) < (x2, y2 )

if

x1 < x2 and 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.


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.


Packing densities of posets:

Theorem: Layered posets have layered optimizers.

The theory for layered posets is exactly like that for layered permutations.

(P,Q) the number of subsets of Q that

are isomorphic to P

n

m

nQ n

nn

(P,Q)(P,Q)

(P) max (P,Q)

( ) lim (P)

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 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.

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.

1 2( , ) ( , ) 2 ( , ).P Q P Q P Q

1 2( , ) ( , ) ( , )P Q P Q P Q

Pattern:

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).

Q u

v

(P,Q)=2

Q1

u u’

(P,Q1)=2

v’

Q2

v

(P,Q2)=3

Every occurrence of P in Q recurs once each in Q1 and Q2, or twice in Q1, or twice in Q2.

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 ?

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