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Patterns, Permutations, and Placements
Jonathan S. Bloom
Dartmouth College
Williams College - February 2014
Permutations
DefinitionA permutation of length n is a rearrangement of the numbers
1, 2, . . . , n.
Notation
I Sn = the set of all permutations of length n.
Example
S3 = {123, 132, 213, 231, 312, 321},
and|Sn| = n!
Permutations
DefinitionA permutation of length n is a rearrangement of the numbers
1, 2, . . . , n.
Notation
I Sn = the set of all permutations of length n.
Example
S3 = {123, 132, 213, 231, 312, 321},
and|Sn| = n!
Permutations
DefinitionA permutation of length n is a rearrangement of the numbers
1, 2, . . . , n.
Notation
I Sn = the set of all permutations of length n.
Example
S3 = {123, 132, 213, 231, 312, 321},
and|Sn| = n!
Permutations
DefinitionA permutation of length n is a rearrangement of the numbers
1, 2, . . . , n.
Notation
I Sn = the set of all permutations of length n.
Example
S3 = {123, 132, 213, 231, 312, 321},
and|Sn| = n!
Permutations
DefinitionA permutation of length n is a rearrangement of the numbers
1, 2, . . . , n.
Notation
I Sn = the set of all permutations of length n.
Example
S3 = {123, 132, 213, 231, 312, 321},
and|Sn| =
n!
Permutations
DefinitionA permutation of length n is a rearrangement of the numbers
1, 2, . . . , n.
Notation
I Sn = the set of all permutations of length n.
Example
S3 = {123, 132, 213, 231, 312, 321},
and|Sn| = n!
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 51 2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 51 2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
3
3
1
1
2
2
5
5
4
4
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
3
3 1
1
2
2
5
5
4
4
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 1
1 2
2
5
5
4
4
3
1
122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11
2
2
5
5
4
4
3
1
1
22
35
44
5
1
2 3 4 51
2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11 2
2 5
5
4
4
3
11
2
2
35
44
5
1
2 3 4 51
2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11 22
5
5
4
4
3
112
2
35
44
5
1 2
3 4 51 2
3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11 22
5
5
4
4
3
1122
3
5
44
5
1 2 3
4 51 2 3
4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11 22 5
5 4
4
3
1122
3
5
44
5
1 2 3
4 51 2 3
4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11 22 55 4
4
3
1122
3
5
4
4
5
1 2 3
4 51 2 3
4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11 22 55 44
3
1122
3
5
4
4
5
1 2 3 4
51 2 3 4
5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4
I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
3
3
1
1
4
4
5
5
2
2
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
3
3 1
1
4
4
5
5
2
2
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 1
1 4
4
5
5
2
2
3
1
14
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 11
4
4
5
5
2
2
3
1
1
4
5
2
1
2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 11 4
4 5
5
2
2
3
11
4
5
2
1
2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 11 44 5
5 2
2
3
11
4
5
2
1
2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 11 44 55 2
2
3
11
4
5
2
1
2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 11 44 55 22
3
11
4
5
2
1 2
5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 11 44 55 22
3
11
4
5
2
1 2 5
4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4
3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2
I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
D. Knuth (1968) defined a sorting algorithm called stack sorting.
Examples
Let α = 3 1 2 5 4I α is stack-sortable
33 11 22 55 44
3
1122
35
44
5
1 2 3 4 5
1 2 3 4 5
Let π = 3 1 4 5 2I π is NOT stack-sortable
33 11 44 55 22
3
114
5
2
1 2 5 4 3
Stack Sorting
QuestionWhy is α = 3 1 2 5 4 stack-sortable, while π = 3 1 4 5 2 is NOT?
Theorem (D. Knuth 1968)
π is NOT stack-sortableiff
π has three entries whose relative ordering is “231”.
Examples
⇒ π contains the pattern 231
α = 3 1 2 5 4 is stack-sortable⇒ α avoids the pattern 231
Stack Sorting
QuestionWhy is α = 3 1 2 5 4 stack-sortable, while π = 3 1 4 5 2 is NOT?
Theorem (D. Knuth 1968)
π is NOT stack-sortableiff
π has three entries whose relative ordering is “231”.
Examples
⇒ π contains the pattern 231
α = 3 1 2 5 4 is stack-sortable⇒ α avoids the pattern 231
Stack Sorting
QuestionWhy is α = 3 1 2 5 4 stack-sortable, while π = 3 1 4 5 2 is NOT?
Theorem (D. Knuth 1968)
π is NOT stack-sortableiff
π has three entries whose relative ordering is “231”.
Examples
π = 3 1 4 5 2 is NOT stack-sortable
⇒ π contains the pattern 231
α = 3 1 2 5 4 is stack-sortable⇒ α avoids the pattern 231
Stack Sorting
QuestionWhy is α = 3 1 2 5 4 stack-sortable, while π = 3 1 4 5 2 is NOT?
Theorem (D. Knuth 1968)
π is NOT stack-sortableiff
π has three entries whose relative ordering is “231”.
Examples
π = 3 1 4 5 2 is NOT stack-sortable
⇒ π contains the pattern 231
α = 3 1 2 5 4 is stack-sortable⇒ α avoids the pattern 231
Stack Sorting
QuestionWhy is α = 3 1 2 5 4 stack-sortable, while π = 3 1 4 5 2 is NOT?
Theorem (D. Knuth 1968)
π is NOT stack-sortableiff
π has three entries whose relative ordering is “231”.
Examples
π = 3 1 4 5 2 is NOT stack-sortable⇒ π contains the pattern 231
α = 3 1 2 5 4 is stack-sortable⇒ α avoids the pattern 231
Stack Sorting
QuestionWhy is α = 3 1 2 5 4 stack-sortable, while π = 3 1 4 5 2 is NOT?
Theorem (D. Knuth 1968)
π is NOT stack-sortableiff
π has three entries whose relative ordering is “231”.
Examples
π = 3 1 4 5 2 is NOT stack-sortable⇒ π contains the pattern 231
α = 3 1 2 5 4 is stack-sortable
⇒ α avoids the pattern 231
Stack Sorting
QuestionWhy is α = 3 1 2 5 4 stack-sortable, while π = 3 1 4 5 2 is NOT?
Theorem (D. Knuth 1968)
π is NOT stack-sortableiff
π has three entries whose relative ordering is “231”.
Examples
π = 3 1 4 5 2 is NOT stack-sortable⇒ π contains the pattern 231
α = 3 1 2 5 4 is stack-sortable⇒ α avoids the pattern 231
Permutation Patterns
It’s easier with pictures!
π = 3 1 4 5 2
×
×
××
×
××××
××
I π contains 132
I π avoids 321
Notation
I Sn(τ) = the set of permutations of length n that avoid τ .
DefinitionTwo patterns σ and τ are Wilf-equivalent if for all n,
|Sn(τ)| = |Sn(σ)|.
Permutation Patterns
It’s easier with pictures!
π = 3 1 4 5 2 ×
×
××
×
××××
××
I π contains 132
I π avoids 321
Notation
I Sn(τ) = the set of permutations of length n that avoid τ .
DefinitionTwo patterns σ and τ are Wilf-equivalent if for all n,
|Sn(τ)| = |Sn(σ)|.
Permutation Patterns
It’s easier with pictures!
π = 3 1 4 5 2 ×
×
××
×
××××
××
I π contains 132
I π avoids 321
Notation
I Sn(τ) = the set of permutations of length n that avoid τ .
DefinitionTwo patterns σ and τ are Wilf-equivalent if for all n,
|Sn(τ)| = |Sn(σ)|.
Permutation Patterns
It’s easier with pictures!
π = 3 1 4 5 2 ×
×
××
××××
×
××
I π contains 132
I π avoids 321
Notation
I Sn(τ) = the set of permutations of length n that avoid τ .
DefinitionTwo patterns σ and τ are Wilf-equivalent if for all n,
|Sn(τ)| = |Sn(σ)|.
Permutation Patterns
It’s easier with pictures!
π = 3 1 4 5 2 ×
×
××
×
×××
×
××
I π contains 132
I π avoids 321
Notation
I Sn(τ) = the set of permutations of length n that avoid τ .
DefinitionTwo patterns σ and τ are Wilf-equivalent if for all n,
|Sn(τ)| = |Sn(σ)|.
Permutation Patterns
It’s easier with pictures!
π = 3 1 4 5 2 ×
×
××
×
××××
××
I π contains 132
I π avoids 321
Notation
I Sn(τ) = the set of permutations of length n that avoid τ .
DefinitionTwo patterns σ and τ are Wilf-equivalent if for all n,
|Sn(τ)| = |Sn(σ)|.
Permutation Patterns
It’s easier with pictures!
π = 3 1 4 5 2 ×
×
××
×
××××
××
I π contains 132
I π avoids 321
Notation
I Sn(τ) = the set of permutations of length n that avoid τ .
DefinitionTwo patterns σ and τ are Wilf-equivalent if for all n,
|Sn(τ)| = |Sn(σ)|.
Permutation Patterns
It’s easier with pictures!
π = 3 1 4 5 2 ×
×
××
×
××××
××
I π contains 132
I π avoids 321
Notation
I Sn(τ) = the set of permutations of length n that avoid τ .
DefinitionTwo patterns σ and τ are Wilf-equivalent if for all n,
|Sn(τ)| = |Sn(σ)|.
Permutation Patterns
Example (Patterns of length 2)
In general,Sn(21) = {123 . . . n}.
Sn(12) = {n . . . 321}.
⇒ 12 is Wilf-equivalent to 21
Permutation Patterns
Example (Patterns of length 2)
In general,Sn(21) =
{123 . . . n}.
Sn(12) = {n . . . 321}.
⇒ 12 is Wilf-equivalent to 21
Permutation Patterns
Example (Patterns of length 2)
In general,Sn(21) = {123 . . . n}.
Sn(12) = {n . . . 321}.
⇒ 12 is Wilf-equivalent to 21
Permutation Patterns
Example (Patterns of length 2)
In general,Sn(21) = {123 . . . n}.
Sn(12) =
{n . . . 321}.
⇒ 12 is Wilf-equivalent to 21
Permutation Patterns
Example (Patterns of length 2)
In general,Sn(21) = {123 . . . n}.
Sn(12) = {n . . . 321}.
⇒ 12 is Wilf-equivalent to 21
Permutation Patterns
Example (Patterns of length 2)
In general,Sn(21) = {123 . . . n}.
Sn(12) = {n . . . 321}.
⇒ 12 is Wilf-equivalent to 21
Permutation Patterns
Patterns of length 3If τ is any pattern of length 3, then
|S3(τ)| = 5
|S4(τ)| = 14
|S5(τ)| = 42
|S6(τ)| = 132
...
|Sn(τ)| =1
n + 1
(2n
n
)= nth Catalan number
I ALL length 3 patterns are Wilf-equivalent
Permutation Patterns
Patterns of length 3If τ is any pattern of length 3, then
|S3(τ)| = 5
|S4(τ)| = 14
|S5(τ)| = 42
|S6(τ)| = 132
...
|Sn(τ)| =1
n + 1
(2n
n
)= nth Catalan number
I ALL length 3 patterns are Wilf-equivalent
Permutation Patterns
Patterns of length 3If τ is any pattern of length 3, then
|S3(τ)| = 5
|S4(τ)| = 14
|S5(τ)| = 42
|S6(τ)| = 132
...
|Sn(τ)| =1
n + 1
(2n
n
)= nth Catalan number
I ALL length 3 patterns are Wilf-equivalent
Permutation Patterns
Patterns of length 3If τ is any pattern of length 3, then
|S3(τ)| = 5
|S4(τ)| = 14
|S5(τ)| = 42
|S6(τ)| = 132
...
|Sn(τ)| =1
n + 1
(2n
n
)= nth Catalan number
I ALL length 3 patterns are Wilf-equivalent
Permutation PatternsPatterns of length 4
n = 5 6 7 8 9
|Sn(2314)| 103 512 2740 15485 91245
|Sn(1234)| 103 513 2761 15767 94359
|Sn(1324)| 103 513 2762 15793 94776
⇒ NOT all patterns of length 4 are Wilf-equivalent.
What’s Known?
I Every pattern of length 4 is Wilf-equivalent to one of:
2314 1234 1324
I I. Gessel (1990) gave a formula for |Sn(1234)|I M. Bona (1997) gave a formula for |Sn(2314)|
Open Problem
Find a formula for |Sn(1324)|.
Permutation PatternsPatterns of length 4
n = 5 6 7 8 9
|Sn(2314)| 103 512 2740 15485 91245
|Sn(1234)| 103 513 2761 15767 94359
|Sn(1324)| 103 513 2762 15793 94776
⇒ NOT all patterns of length 4 are Wilf-equivalent.
What’s Known?
I Every pattern of length 4 is Wilf-equivalent to one of:
2314 1234 1324
I I. Gessel (1990) gave a formula for |Sn(1234)|I M. Bona (1997) gave a formula for |Sn(2314)|
Open Problem
Find a formula for |Sn(1324)|.
Permutation PatternsPatterns of length 4
n = 5 6 7 8 9
|Sn(2314)| 103 512 2740 15485 91245
|Sn(1234)| 103 513 2761 15767 94359
|Sn(1324)| 103 513 2762 15793 94776
⇒ NOT all patterns of length 4 are Wilf-equivalent.
What’s Known?
I Every pattern of length 4 is Wilf-equivalent to one of:
2314 1234 1324
I I. Gessel (1990) gave a formula for |Sn(1234)|I M. Bona (1997) gave a formula for |Sn(2314)|
Open Problem
Find a formula for |Sn(1324)|.
Permutation PatternsPatterns of length 4
n = 5 6 7 8 9
|Sn(2314)| 103 512 2740 15485 91245
|Sn(1234)| 103 513 2761 15767 94359
|Sn(1324)| 103 513 2762 15793 94776
⇒ NOT all patterns of length 4 are Wilf-equivalent.
What’s Known?
I Every pattern of length 4 is Wilf-equivalent to one of:
2314 1234 1324
I I. Gessel (1990) gave a formula for |Sn(1234)|I M. Bona (1997) gave a formula for |Sn(2314)|
Open Problem
Find a formula for |Sn(1324)|.
Permutation PatternsPatterns of length 4
n = 5 6 7 8 9
|Sn(2314)| 103 512 2740 15485 91245
|Sn(1234)| 103 513 2761 15767 94359
|Sn(1324)| 103 513 2762 15793 94776
⇒ NOT all patterns of length 4 are Wilf-equivalent.
What’s Known?
I Every pattern of length 4 is Wilf-equivalent to one of:
2314 1234 1324
I I. Gessel (1990) gave a formula for |Sn(1234)|I M. Bona (1997) gave a formula for |Sn(2314)|
Open Problem
Find a formula for |Sn(1324)|.
Permutation PatternsPatterns of length 4
n = 5 6 7 8 9
|Sn(2314)| 103 512 2740 15485 91245
|Sn(1234)| 103 513 2761 15767 94359
|Sn(1324)| 103 513 2762 15793 94776
⇒ NOT all patterns of length 4 are Wilf-equivalent.
What’s Known?
I Every pattern of length 4 is Wilf-equivalent to one of:
2314 1234 1324
I I. Gessel (1990) gave a formula for |Sn(1234)|
I M. Bona (1997) gave a formula for |Sn(2314)|
Open Problem
Find a formula for |Sn(1324)|.
Permutation PatternsPatterns of length 4
n = 5 6 7 8 9
|Sn(2314)| 103 512 2740 15485 91245
|Sn(1234)| 103 513 2761 15767 94359
|Sn(1324)| 103 513 2762 15793 94776
⇒ NOT all patterns of length 4 are Wilf-equivalent.
What’s Known?
I Every pattern of length 4 is Wilf-equivalent to one of:
2314 1234 1324
I I. Gessel (1990) gave a formula for |Sn(1234)|I M. Bona (1997) gave a formula for |Sn(2314)|
Open Problem
Find a formula for |Sn(1324)|.
Permutation PatternsPatterns of length 4
n = 5 6 7 8 9
|Sn(2314)| 103 512 2740 15485 91245
|Sn(1234)| 103 513 2761 15767 94359
|Sn(1324)| 103 513 2762 15793 94776
⇒ NOT all patterns of length 4 are Wilf-equivalent.
What’s Known?
I Every pattern of length 4 is Wilf-equivalent to one of:
2314 1234 1324
I I. Gessel (1990) gave a formula for |Sn(1234)|I M. Bona (1997) gave a formula for |Sn(2314)|
Open Problem
Find a formula for |Sn(1324)|.
Rook Placements
A Ferrers Board F is a square array of boxes with a “bite” takenout of the northeast corner.
F =
××
××
×
××
A full rook placement (f.r.p.) on F is a placement of markerswith EXACTLY one in each row and column.
Notation
I RF = set of all f.r.p.’s on the Ferrers board F
Rook Placements
A Ferrers Board F is a square array of boxes with a “bite” takenout of the northeast corner.
F =
××
××
×
××
A full rook placement (f.r.p.) on F is a placement of markerswith EXACTLY one in each row and column.
Notation
I RF = set of all f.r.p.’s on the Ferrers board F
Rook Placements
A Ferrers Board F is a square array of boxes with a “bite” takenout of the northeast corner.
F =
××
××
×
××
A full rook placement (f.r.p.) on F is a placement of markerswith EXACTLY one in each row and column.
Notation
I RF = set of all f.r.p.’s on the Ferrers board F
Rook Placements
A Ferrers Board F is a square array of boxes with a “bite” takenout of the northeast corner.
F =
××
××
×
××
A full rook placement (f.r.p.) on F is a placement of markerswith EXACTLY one in each row and column.
Notation
I RF = set of all f.r.p.’s on the Ferrers board F
Rook Placements
A Ferrers Board F is a square array of boxes with a “bite” takenout of the northeast corner.
F =
××
××
×
××
A full rook placement (f.r.p.) on F is a placement of markerswith EXACTLY one in each row and column.
Notation
I RF = set of all f.r.p.’s on the Ferrers board F
Rook Placements
A Ferrers Board F is a square array of boxes with a “bite” takenout of the northeast corner.
F =
××
××
×
××
A full rook placement (f.r.p.) on F is a placement of markerswith EXACTLY one in each row and column.
Notation
I RF = set of all f.r.p.’s on the Ferrers board F
Rook Placements
Patterns?
××
××
×
××
×
××××
×
I contains the pattern 312
I avoids the pattern 231
Notation
I RF (τ) = set of all f.r.p. on F that avoid τ .
Rook Placements
Patterns?
××
××
×
××
×
××××
×
I contains the pattern 312
I avoids the pattern 231
Notation
I RF (τ) = set of all f.r.p. on F that avoid τ .
Rook Placements
Patterns?
××
××
×
××
×
××
××
×
I contains the pattern 312
I avoids the pattern 231
Notation
I RF (τ) = set of all f.r.p. on F that avoid τ .
Rook Placements
Patterns?
××
××
×
××
×
××××
×
I contains the pattern 312
I avoids the pattern 231
Notation
I RF (τ) = set of all f.r.p. on F that avoid τ .
Rook Placements
Patterns?
××
××
×
××
×
××
××
×
I contains the pattern 312
I avoids the pattern 231
Notation
I RF (τ) = set of all f.r.p. on F that avoid τ .
Rook Placements
Patterns?
××
××
×
××
×
××××
×
I contains the pattern 312
I avoids the pattern 231
Notation
I RF (τ) = set of all f.r.p. on F that avoid τ .
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof ⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof ⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof ⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof ⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)
I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof ⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof ⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof
⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof ⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof ⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Rook Placements
DefinitionTwo patterns σ and τ are shape-Wilf-equivalent if for every F ,
|RF (σ)| = |RF (τ)|.
In this case we write σ ∼ τ .
Observe
shape-Wilf-equivalence ⇒Wilf-equivalence.
What’s Known?
I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)I 231 ∼ 312 (Z. Stankova and J. West, 2002)
- Complicated proof ⇒ can’t count things
I We give a simple proof that 231 ∼ 312
- Can count things!
Dyck Paths
1
2
3
4
1 2 3 4 5 6 7 8 9 10 11 12 13 14
A Dyck path of size n is a path that:
I starts at the origin
I ends at the point (2n, 0)
I never goes below the x-axis
It is well known that
# Dyck paths of size n =1
n + 1
(2n
n
)
Dyck Paths
1
2
3
4
1 2 3 4 5 6 7 8 9 10 11 12 13 14
A Dyck path of size n is a path that:
I starts at the origin
I ends at the point (2n, 0)
I never goes below the x-axis
It is well known that
# Dyck paths of size n =1
n + 1
(2n
n
)
Dyck Paths
1
2
3
4
1 2 3 4 5 6 7 8 9 10 11 12 13 14
A Dyck path of size n is a path that:
I starts at the origin
I ends at the point (2n, 0)
I never goes below the x-axis
It is well known that
# Dyck paths of size n =1
n + 1
(2n
n
)
Dyck Paths
1
2
3
4
1 2 3 4 5 6 7 8 9 10 11 12 13 14
A Dyck path of size n is a path that:
I starts at the origin
I ends at the point (2n, 0)
I never goes below the x-axis
It is well known that
# Dyck paths of size n =1
n + 1
(2n
n
)
Dyck Paths
1
2
3
4
1 2 3 4 5 6 7 8 9 10 11 12 13 14
A Dyck path of size n is a path that:
I starts at the origin
I ends at the point (2n, 0)
I never goes below the x-axis
It is well known that
# Dyck paths of size n =1
n + 1
(2n
n
)
Dyck Paths
1
2
3
4
1 2 3 4 5 6 7 8 9 10 11 12 13 14
A Dyck path of size n is a path that:
I starts at the origin
I ends at the point (2n, 0)
I never goes below the x-axis
It is well known that
# Dyck paths of size n =1
n + 1
(2n
n
)
Dyck Paths
1
2
3
4
1 2 3 4 5 6 7 8 9 10 11 12 13 14
A Dyck path of size n is a path that:
I starts at the origin
I ends at the point (2n, 0)
I never goes below the x-axis
It is well known that
# Dyck paths of size n =1
n + 1
(2n
n
)
Labeled Dyck paths
0
1
1
2
3
2
2
1
1
1
1
2
2
1
0
We label the Dyck path so that:
I Monotonicity
- +1/0 up step and −1/0 down step
I Zero Condition
- All zeros lie precisely on the x-axis
I Tunnel Property
- “Left” ≤ “Right”
Labeled Dyck paths
0
1
1
2
3
2
2
1
1
1
1
2
2
1
0
We label the Dyck path so that:I Monotonicity
- +1/0 up step and −1/0 down step
I Zero Condition
- All zeros lie precisely on the x-axis
I Tunnel Property
- “Left” ≤ “Right”
Labeled Dyck paths
0
1
1
2
3
2
2
1
1
1
1
2
2
1
0
We label the Dyck path so that:I Monotonicity
- +1/0 up step and −1/0 down step
I Zero Condition
- All zeros lie precisely on the x-axis
I Tunnel Property
- “Left” ≤ “Right”
Labeled Dyck paths
0
1
1
2
3
2
2
1
1
1
1
2
2
1
0
We label the Dyck path so that:I Monotonicity
- +1/0 up step and −1/0 down step
I Zero Condition
- All zeros lie precisely on the x-axis
I Tunnel Property
- “Left” ≤ “Right”
Our proof of 231 ∼ 312
An outline
1. 231-avoiding rook placement 7→ Tunnel property
2. Tunnel Property 7→ Reverse Tunnel Property
3. Reverse Tunnel Property 7→ 312-avoiding rook placement
Our proof of 231 ∼ 312
An outline
1. 231-avoiding rook placement 7→ Tunnel property
2. Tunnel Property 7→ Reverse Tunnel Property
3. Reverse Tunnel Property 7→ 312-avoiding rook placement
Our proof of 231 ∼ 312
An outline
1. 231-avoiding rook placement 7→ Tunnel property
2. Tunnel Property 7→ Reverse Tunnel Property
3. Reverse Tunnel Property 7→ 312-avoiding rook placement
Our proof of 231 ∼ 312
An outline
1. 231-avoiding rook placement 7→ Tunnel property
2. Tunnel Property 7→ Reverse Tunnel Property
3. Reverse Tunnel Property 7→ 312-avoiding rook placement
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
1. 231-avoiding f.r.p. 7→ Tunnel property
××
××
×
××
RF (231)
1
2
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
Our proof of 231 ∼ 312
2. Tunnel property 7→ Reverse tunnel property
0
2
3
5
3
2
3
2
3
4
3
2
0
−4 44 4
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
1 2
1 2
=0
1
2
3
1
1
2
1
1
2
1
1
0
≥
3 2
Our proof of 231 ∼ 312
2. Tunnel property 7→ Reverse tunnel property
0
2
3
5
3
2
3
2
3
4
3
2
0
−4 4
4 4
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
1 2
1 2
=0
1
2
3
1
1
2
1
1
2
1
1
0
≥
3 2
Our proof of 231 ∼ 312
2. Tunnel property 7→ Reverse tunnel property
0
2
3
5
3
2
3
2
3
4
3
2
0
−4 4
4 4
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
1 2
1 2
=0
1
2
3
1
1
2
1
1
2
1
1
0
≥
3 2
Our proof of 231 ∼ 312
2. Tunnel property 7→ Reverse tunnel property
0
2
3
5
3
2
3
2
3
4
3
2
0
−
4 4
4 4
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
1 2
1 2
=0
1
2
3
1
1
2
1
1
2
1
1
0
≥
3 2
Our proof of 231 ∼ 312
2. Tunnel property 7→ Reverse tunnel property
0
2
3
5
3
2
3
2
3
4
3
2
0
−
4 4
4 4
0
1
1
2
2
1
1
1
2
2
2
1
0
≤
1 2
1 2
=0
1
2
3
1
1
2
1
1
2
1
1
0
≥
3 2
Our proof of 231 ∼ 312
3. Reverse tunnel property 7→ 312-avoiding f.r.p.
0
1
2
3
3
2
1
1
2
1
1
2
1
1
0
≥
×××
×
×
×
×
RF (312)
Theorem (Bloom–Saracino ’11)
This mapping is a bijection between RF (231) and RF (312).⇒ 231 and 312 are shape-Wilf-equivalent.
Our proof of 231 ∼ 312
3. Reverse tunnel property 7→ 312-avoiding f.r.p.
0
1
2
3
3
2
1
1
2
1
1
2
1
1
0
≥
×××
×
×
×
×
RF (312)
Theorem (Bloom–Saracino ’11)
This mapping is a bijection between RF (231) and RF (312).⇒ 231 and 312 are shape-Wilf-equivalent.
Our proof of 231 ∼ 312
3. Reverse tunnel property 7→ 312-avoiding f.r.p.
0
1
2
3
3
2
1
1
2
1
1
2
1
1
0
≥
×××
×
×
×
×
RF (312)
Theorem (Bloom–Saracino ’11)
This mapping is a bijection between RF (231) and RF (312).⇒ 231 and 312 are shape-Wilf-equivalent.
Enumerative Results: 2314-Avoiding Permutations
In 1997, Bona proved the celebrated result:
|Sn(2314)|
= (−1)n
[−7n2 + 3n + 2
2+ 6
n∑i=2
(−2)i(2i − 4)!
i !(i − 2)!
(n − i + 2
2
)]
Our Proof
6257413
Sn(2314)
×
×
×
×
×
×
×
×
×
×
×
×
×
×
RF (231)
0
1
1 2
1
2
1
1
2
1
1
0
≤
23
2
Enumerative Results: 2314-Avoiding Permutations
In 1997, Bona proved the celebrated result:
|Sn(2314)|
= (−1)n
[−7n2 + 3n + 2
2+ 6
n∑i=2
(−2)i(2i − 4)!
i !(i − 2)!
(n − i + 2
2
)]
Our Proof
6257413
Sn(2314)
×
×
×
×
×
×
×
×
×
×
×
×
×
×
RF (231)
0
1
1 2
1
2
1
1
2
1
1
0
≤
23
2
Enumerative Results: 2314-Avoiding Permutations
In 1997, Bona proved the celebrated result:
|Sn(2314)|
= (−1)n
[−7n2 + 3n + 2
2+ 6
n∑i=2
(−2)i(2i − 4)!
i !(i − 2)!
(n − i + 2
2
)]
Our Proof
6257413
Sn(2314)
×
×
×
×
×
×
×
×
×
×
×
×
×
×
RF (231)
0
1
1 2
1
2
1
1
2
1
1
0
≤
23
2
Enumerative Results: 2314-Avoiding Permutations
In 1997, Bona proved the celebrated result:
|Sn(2314)|
= (−1)n
[−7n2 + 3n + 2
2+ 6
n∑i=2
(−2)i(2i − 4)!
i !(i − 2)!
(n − i + 2
2
)]
Our Proof
6257413
Sn(2314)
×
×
×
×
×
×
×
×
×
×
×
×
×
×
RF (231)
0
1
1 2
1
2
1
1
2
1
1
0
≤
23
2
Enumerative Results: 2314-Avoiding Permutations
In 1997, Bona proved the celebrated result:
|Sn(2314)|
= (−1)n
[−7n2 + 3n + 2
2+ 6
n∑i=2
(−2)i(2i − 4)!
i !(i − 2)!
(n − i + 2
2
)]
Our Proof
6257413
Sn(2314)
×
×
×
×
×
×
×
×
×
×
×
×
×
×
RF (231)
0
1
1
2
3
2
2
1
2
1
1
2
1
1
0
≤
23
2
Enumerative Results: 2314-Avoiding Permutations
In 1997, Bona proved the celebrated result:
|Sn(2314)|
= (−1)n
[−7n2 + 3n + 2
2+ 6
n∑i=2
(−2)i(2i − 4)!
i !(i − 2)!
(n − i + 2
2
)]
Our Proof
6257413
Sn(2314)
×
×
×
×
×
×
×
×
×
×
×
×
×
×
RF (231)
0
1
1 2
1
2
1
1
2
1
1
0
≤
23
2
Thank you!
Appendix: New Enumerative Results
I In 2012, D. Callan and V. Kotesovec conjectured that
∞∑n=0
|Sn(2314, 1234)|zn =1
1− C (zC (z))
= 1 + z + 2z + 6z2 + 22z3 + · · ·
where C (z) is the generating function for the Catalannumbers.
I All 231-avoiding f.r.p. are counted by
54z
1 + 36z − (1− 12z)3/2= 1 + z + 3z2 + 14z3 + 83z4 + · · ·
I New enumerative results in the theory of perfect matchingsand set partitions.
Appendix: New Enumerative Results
I In 2012, D. Callan and V. Kotesovec conjectured that
∞∑n=0
|Sn(2314, 1234)|zn =1
1− C (zC (z))
= 1 + z + 2z + 6z2 + 22z3 + · · ·
where C (z) is the generating function for the Catalannumbers.
I All 231-avoiding f.r.p. are counted by
54z
1 + 36z − (1− 12z)3/2= 1 + z + 3z2 + 14z3 + 83z4 + · · ·
I New enumerative results in the theory of perfect matchingsand set partitions.
Appendix: New Enumerative Results
I In 2012, D. Callan and V. Kotesovec conjectured that
∞∑n=0
|Sn(2314, 1234)|zn =1
1− C (zC (z))
= 1 + z + 2z + 6z2 + 22z3 + · · ·
where C (z) is the generating function for the Catalannumbers.
I All 231-avoiding f.r.p. are counted by
54z
1 + 36z − (1− 12z)3/2= 1 + z + 3z2 + 14z3 + 83z4 + · · ·
I New enumerative results in the theory of perfect matchingsand set partitions.
