An Introduction to Enumeration Schemes Lara Pudwell Valparaiso - - PowerPoint PPT Presentation

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An Introduction to Enumeration Schemes Lara Pudwell Valparaiso - - PowerPoint PPT Presentation

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Pattern-Avoiding Permutations Enumeration Schemes Summary An Introduction to Enumeration Schemes Lara Pudwell Valparaiso University Lara.Pudwell@valpo.edu AWM Workshop Washington, DC January 8, 2009 Pattern-Avoiding Permutations

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Pattern-Avoiding Permutations Enumeration Schemes Summary

An Introduction to Enumeration Schemes

Lara Pudwell Valparaiso University Lara.Pudwell@valpo.edu AWM Workshop Washington, DC January 8, 2009

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Pattern-Avoiding Permutations Enumeration Schemes Summary Definitions

Reduction Given a string of numbers p = p1 · · · pn, the reduction of p is the string obtained by replacing the ith smallest number

  • f p with i.

For example, the reduction of 26745 is

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Pattern-Avoiding Permutations Enumeration Schemes Summary Definitions

Reduction Given a string of numbers p = p1 · · · pn, the reduction of p is the string obtained by replacing the ith smallest number

  • f p with i.

For example, the reduction of 26745 is 1••••.

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SLIDE 4

Pattern-Avoiding Permutations Enumeration Schemes Summary Definitions

Reduction Given a string of numbers p = p1 · · · pn, the reduction of p is the string obtained by replacing the ith smallest number

  • f p with i.

For example, the reduction of 26745 is 1••2•.

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Pattern-Avoiding Permutations Enumeration Schemes Summary Definitions

Reduction Given a string of numbers p = p1 · · · pn, the reduction of p is the string obtained by replacing the ith smallest number

  • f p with i.

For example, the reduction of 26745 is 1••23.

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Pattern-Avoiding Permutations Enumeration Schemes Summary Definitions

Reduction Given a string of numbers p = p1 · · · pn, the reduction of p is the string obtained by replacing the ith smallest number

  • f p with i.

For example, the reduction of 26745 is 14•23.

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SLIDE 7

Pattern-Avoiding Permutations Enumeration Schemes Summary Definitions

Reduction Given a string of numbers p = p1 · · · pn, the reduction of p is the string obtained by replacing the ith smallest number

  • f p with i.

For example, the reduction of 26745 is 14523.

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Pattern-Avoiding Permutations Enumeration Schemes Summary Definitions

Pattern Avoidance in Permutations Given p ∈ Sn and q ∈ Sm, we say p contains q if there are 1 ≤ i1 < · · · < im ≤ n such that pi1 · · · pim reduces to q. Otherwise, p avoids q. p = 21354 contains 132. (since 21354 reduces to 132.) p = 21354 avoids 321. (since p has no decreasing subsequence of length 3.)

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

Two Questions Easy: Given p ∈ Sn, what patterns does p contain? Hard: Given q ∈ Sm, Let Sn(q) = {p ∈ Sn | p avoids q}. Find an expression for |Sn(q)|.

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

Patterns of length 1, 2, and 3 |Sn(1)| =

  • 1

n = 0 n > 0

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

Patterns of length 1, 2, and 3 |Sn(1)| =

  • 1

n = 0 n > 0 |Sn(12)| = |Sn(21)| = 1, n ≥ 0

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

Patterns of length 1, 2, and 3 |Sn(1)| =

  • 1

n = 0 n > 0 |Sn(12)| = |Sn(21)| = 1, n ≥ 0 Graph of p ∈ Sn(132)

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

Patterns of length 1, 2, and 3 |Sn(1)| =

  • 1

n = 0 n > 0 |Sn(12)| = |Sn(21)| = 1, n ≥ 0 Graph of p ∈ Sn(132)

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

Patterns of length 1, 2, and 3 |Sn(1)| =

  • 1

n = 0 n > 0 |Sn(12)| = |Sn(21)| = 1, n ≥ 0 Graph of p ∈ Sn(132) So, |Sn(132)| = n

i=1 |Si−1(132)| · |Sn−i(132)|

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

Patterns of length 1, 2, and 3 |Sn(1)| =

  • 1

n = 0 n > 0 |Sn(12)| = |Sn(21)| = 1, n ≥ 0 Graph of p ∈ Sn(132) So, |Sn(132)| = n

i=1 |Si−1(132)| · |Sn−i(132)| =

2n

n

  • n + 1 = Cn

In fact, |Sn(q)| = Cn if q is any permutation of length 3.

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

Patterns of Length 4 There are 24 patterns of length 4. Using several clever bijections, we can narrow our work to 3 cases: Sn(1342), Sn(1234), and Sn(1324).

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

Patterns of Length 4 There are 24 patterns of length 4. Using several clever bijections, we can narrow our work to 3 cases: Sn(1342), Sn(1234), and Sn(1324).

1 2 3 4 5 6 7 8 |Sn(1342)| 1 2 6 23 103 512 2740 15485 ∼ 8n |Sn(1234)| 1 2 6 23 103 513 2761 15767 ∼ 9n |Sn(1324)| 1 2 6 23 103 513 2762 15793 ∼ 9.3n

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Pattern-Avoiding Permutations Enumeration Schemes Summary Counting Results

For Permutations Most techniques studying |Sn(q)| finds formulas for a specific q. 1998: Zeilberger’s prefix enumeration schemes, i.e. a system of recurrences to count |Sn(q)|. 2005: Vatter’s modified schemes automate the enumeration of |Sn(q)| for even more patterns q.

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Pattern-Avoiding Permutations Enumeration Schemes Summary Divide

Refinement Notation Goal: Divide Sn(q) into subsets. Sn

  • q; p1 · · · pl
  • :=
  • π ∈ Sn
  • π avoids q

π has prefix p1 · · · pl

  • and

Sn

  • q; p1 · · · pl

i1 · · · il

  • :=

  π ∈ Sn

  • π avoids q

π has prefix p1 · · · pl π = i1 · · · ilπl+1 · · · πn    If p1 · · · pl−1 reduces to p∗

1 · · · p∗ l−1, then p = p1 · · · pl is called a

refinement if p∗. e.g p = 2413 is a refinement of p∗ = 231.

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Pattern-Avoiding Permutations Enumeration Schemes Summary Divide

Refinement Example For any pattern q, Sn(q) = Sn(q; 1) = Sn(q; 12) ∪ Sn(q; 21).

  • r graphically:

∅ 1 12 21

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Pattern-Avoiding Permutations Enumeration Schemes Summary Conquer

Reversibly Deletable Positions Reversibly Deletable Positions Given a pattern q and a prefix p, position r is reversibly deletable if Deleting pr from π ∈ Sn(q; p1 · · · pl) produces a q-avoiding permutation of length n − 1, and Inserting pr into π ∈ Sn−1(q; p1 · · · pr−1pr+1 · · · pl) produces a q avoiding permutation of length n. In other words, deleting and re-inserting pr gives a bijection between Sn(q; p1 · · · pl) and Sn−1(q; p1 · · · pr−1pr+1 · · · pl), and |Sn(q; p1 · · · pl)| = |Sn−1(q; p1 · · · pr−1pr+1 · · · pl)| .

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Pattern-Avoiding Permutations Enumeration Schemes Summary Conquer

Reversibly Deletable Example Graphically: ∅ 1 12 21

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Pattern-Avoiding Permutations Enumeration Schemes Summary Conquer

Reversibly Deletable Example Graphically: ∅ 1 12 21

  • d1
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Pattern-Avoiding Permutations Enumeration Schemes Summary Conquer

Gap Vectors Gap vectors give a condition for which choices of i1, . . . , il yield

  • Sn
  • q; p1 · · · pr · · · pl

i1 · · · ir · · · il

  • = 0.

Graphically: ∅ 1 12 21

  • d1
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Pattern-Avoiding Permutations Enumeration Schemes Summary Conquer

Gap Vectors Gap vectors give a condition for which choices of i1, . . . , il yield

  • Sn
  • q; p1 · · · pr · · · pl

i1 · · · ir · · · il

  • = 0.

Graphically: ∅ 1 12 21 ≥ (0, 1, 0)

  • d1
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Pattern-Avoiding Permutations Enumeration Schemes Summary Conquer

Gap Vectors Gap vectors give a condition for which choices of i1, . . . , il yield

  • Sn
  • q; p1 · · · pr · · · pl

i1 · · · ir · · · il

  • = 0.

Graphically: ∅ 1 12 21 ≥ (0, 1, 0)

  • d1
  • d2
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Pattern-Avoiding Permutations Enumeration Schemes Summary Enumeration Schemes

Enumeration Scheme Definition An enumeration scheme is a set of triples [pi, Ri, Gi] such that for each triple pi is a reduced prefix of length n Ri a subset of {1, . . . , n} Gi is a set of vectors of length n + 1 and either Ri is non-empty or all refinements of pi are also in the scheme.

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Pattern-Avoiding Permutations Enumeration Schemes Summary Enumeration Schemes

Enumeration Scheme Definition An enumeration scheme is a set of triples [pi, Ri, Gi] such that for each triple pi is a reduced prefix of length n (prefix) Ri a subset of {1, . . . , n} (reversibly deletable positions) Gi is a set of vectors of length n + 1 (gap vectors) and either Ri is non-empty or all refinements of pi are also in the scheme.

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Pattern-Avoiding Permutations Enumeration Schemes Summary Enumeration Schemes

Avoid(12) and Avoid(123) Sn(12) ∅ 1 ≥ (0, 1) Sn(123) ∅ 1 12 ≥ (0, 0, 1) 21

  • d1
  • d2
  • d1
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Pattern-Avoiding Permutations Enumeration Schemes Summary Enumeration Schemes

Avoid(1234) Sn(1234) ∅ 1 12 21 132 123 ≥ (0, 0, 0, 1) 231 2413 3412 3421 2314 ≥ (0, 0, 0, 0, 1)

  • d1
  • d2
  • d3
  • d1,2
  • d1,2
  • d3
  • d4
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Pattern-Avoiding Permutations Enumeration Schemes Summary Summary

Summary There are few techniques to count large classes of pattern-avoiding permutations. Extending Zeilberger’s and Vatter’s schemes gives a good success rate for counting the elements of Sn(q). Enumeration schemes have also been successfully used to count:

pattern-avoiding words (strings with repeated letters) permutations avoiding barred patterns (permutations that avoid a particular pattern unless that pattern is part of an even larger specified pattern)

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Pattern-Avoiding Permutations Enumeration Schemes Summary Summary

Contact Info Lara Pudwell Lara.Pudwell@valpo.edu http://faculty.valpo.edu/lpudwell