Start with a puzzle

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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 ?

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Packing Densities of Permutations
Graph Theory With AltitudeDenver, May 17, 2005

• Walter Stromquist
• Bryn Mawr College / Swarthmore College

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Outline

• History
• Example ? 132
• Definitions
• Layered patterns
• Reid Bartons proof
• Results for ? in S3 and S4 and open
  problems
• Connection to partially ordered sets
• and more open problems

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History

• 1992 Wilfs 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 Prices thesis
• 2002-2005 Many (?5) papers in Electronic
  Journal of Combinatorics
• 2004 Reid Bartons Morgan Prize paper (EJC 11)

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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

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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.
• Were 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.

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Definitions

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

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Definitions

• In this talk, the pattern is always called ?
  and always has length m.
• The permutation ? always has length n.
• Well always assume that n gt m.

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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 thats the largest packing density for
  any ? of
• length 5, we also say that

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Definitions

• The packing density of ? in ? is
• Clearly,
• Were 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 ?.

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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 ? ?

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Example ? 132

• What can we do with longer sequences ? ?
• For n 9, try ? 123 987654
• ? 123 987654
• In fact, ?9( 132 ) 46 / 84.

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Example ? 132

• In general, heres the best we can do for large n

Now So the packing density of ? 132
is
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Another Example 123

• Now let ? 123.
• If ? 1234n, 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.

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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

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Theorem and Proof

• Theorem (Galvin) The limit
  always exists.

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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 (nm) of the ?is.

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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 (nm) of the ?is.

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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 (nm) of the ?is.

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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 (nm) of the ?is.
• It follows (with a bit of algebra) that
• So ?( ?, ? ) cant exceed the largest of the
  ?( ?, ?i )s.

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Theorem and Proof

• ...
• So ?( ?, ? ) cant 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. //

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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.

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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 ?.

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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.

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Permutations in S3 Symmetry

• Symmetry

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Permutations in S3 Symmetry

• Reversal

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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 )

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1324

• Let ? 1324.
• Price Optimal ratios are
  .39 .19 .07
• and ?(1324) ? 0.244.

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1342

• Let ? 1342.
• This optimizer gives a
• lower bound. If you think
• its the best you can do,
• then
• ?(1342) ? 0.1966.

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1342

• If the lower bound holds
• ?(1342) ? 0.1966...
• (Batayev)
• ?(1342) ?(1432) ?(132)

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Partially Ordered Sets

• A (finite) partially ordered set is a finite set
  together with a relation lt such that
• (a) It is never true that x lt x
• (b) It is never true that both x lt y and y
  lt x and
• (c) If x lt y and y lt z, then x lt 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 lt y as x is below
  y ).
• Diagrams

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Partially Ordered Sets

• Example Consider a finite set of vectors (x,
  y) in R2. Say that
• (x1, y1) lt (x2, y2 )
• if
• x1 lt x2 and y1 lt 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.
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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.

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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.

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Posets arent 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.

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Reid Bartons 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.

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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.

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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 isnt an optimizer.
As a result, theres an extra occurrence of the
pattern in Q2. If Q were an optimizer, the
theorem would force ?(P,Q)?(P,Q1)?(P,Q2).
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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. //

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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 isnt just a statement about
   permutations, in disguise ?
6. Whats the packing density of this poset ?