wake forest university - february 2014patterns, permutations, and placements jonathan s. bloom...

Patterns, Permutations, and Placements

Jonathan S. Bloom

Dartmouth College

Wake Forest University - February 2014

Permutations

DefinitionA permutation of length n is a rearrangement of the numbers

1, 2, . . . , n.

NotationLet Sn denote the set of all permutations of length n.

Example

S3 = {123, 132, 213, 231, 312, 321},

and|Sn| = n!

Permutations

1, 2, . . . , n.

Example

S3 = {123, 132, 213, 231, 312, 321},

and|Sn| = n!

Permutations

1, 2, . . . , n.

Example

S3 = {123, 132, 213, 231, 312, 321},

and|Sn| = n!

Permutations

1, 2, . . . , n.

Example

S3 = {123, 132, 213, 231, 312, 321},

and|Sn| = n!

Permutations

1, 2, . . . , n.

Example

S3 = {123, 132, 213, 231, 312, 321},

and|Sn| =

Permutations

1, 2, . . . , 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

1 2 3 4 5

Let π = 3 1 4 5 2I π is NOT stack-sortable

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

33 11 22 55 44

1 2 3 4 5

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

1 2 3 4 5

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

1 2 3 4 5

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

1 2 3 4 5

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

2 3 4 5

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

33 11 2

2 3 4 5

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

33 11 22

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

33 11 22

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

33 11 22 5

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

33 11 22 55 4

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

33 11 22 55 44

1 2 3 4

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4

33 11 22 55 44

1 2 3 4 5

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

Let α = 3 1 2 5 4I α is stack-sortable

33 11 22 55 44

1 2 3 4 5

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

I π is NOT stack-sortable

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

1 2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

1 2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

1 2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

33 11 4

2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

33 11 44 5

2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

33 11 44 55 2

2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

33 11 44 55 22

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

33 11 44 55 22

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

33 11 44 55 22

1 2 5 4

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

Let π = 3 1 4 5 2

33 11 44 55 22

1 2 5 4 3

Stack Sorting

Examples

33 11 22 55 44

1 2 3 4 5

33 11 44 55 22

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-sortable ⇐⇒ π has three entries whose relativeordering is “231”.

Examples

⇒ π contains the pattern 231

α = 3 1 2 5 4 is stack-sortable⇒ α avoid the pattern 231

Stack Sorting

Examples

Stack Sorting

Examples

π = 3 1 4 5 2 is NOT stack-sortable

Stack Sorting

Examples

π = 3 1 4 5 2 is NOT stack-sortable⇒ π contains the pattern 231

Stack Sorting

Examples

α = 3 1 2 5 4 is stack-sortable

⇒ α avoid the pattern 231

Stack Sorting

Examples

Permutation Patterns

Its easier with pictures!

π = 31452

×××

×××××

× I π contains 123

I π contains 213

I π avoids 321

NotationLet Sn(τ) be the set of permutations of length n that avoid τ .

π = 31452×

×××

×××××

I π contains 213

I π avoids 321

π = 31452×

×××

×××××

I π contains 123

I π contains 213

I π avoids 321

π = 31452×

××××

×××××

I π contains 123

I π contains 213

I π avoids 321

π = 31452×

×××

I π contains 123

I π contains 213

I π avoids 321

π = 31452×

×××

×××××

I π contains 123

I π contains 213

I π avoids 321

π = 31452×

×××

I π contains 213

I π avoids 321

π = 31452×

×××

×××××

I π contains 123

I π contains 213

I π avoids 321

π = 31452×

×××

×××××

I π contains 123

I π contains 213

I π avoids 321

DefinitionWe say two patterns τ, σ ∈ Sk are Wilf-equivalent provided

|Sn(τ)| = |Sn(σ)|

for all n.

Example (Patterns of length 2)

What permutations avoid 21?

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) = {n . . . 321}.

⇒ 12 is Wilf-equivalent to 21

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) = {n . . . 321}.

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) = {n . . . 321}.

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) = {n . . . 321}.

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) = {n . . . 321}.

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) = {n . . . 321}.

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) =

{123 . . . n}.

Sn(12) = {n . . . 321}.

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) = {n . . . 321}.

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) =

{n . . . 321}.

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) = {n . . . 321}.

S2 = {12, 21}

⇒ S2(21) = {12}.

S3 = {123, 132, 213, 231, 312, 321}

⇒ S3(21) = {123}.

In general,Sn(21) = {123 . . . n}.

Sn(12) = {n . . . 321}.

Patterns of length 3

S3 = {123, 132, 213, 231, 312, 321}

I S3(321) = {123, 132, 213, 231, 312}

|S3(321)| = 5

|S4(321)| = 14

|S5(321)| = 42

|Sn(321)| =1

)= nth Catalan number

In fact, this is true for ALL length 3 patterns!!

I ALL length 3 patterns are Wilf-equivalent

S3 = {123, 132, 213, 231, 312, 321}

I S3(321) = {123, 132, 213, 231, 312}

|S3(321)| = 5

|S4(321)| = 14

|S5(321)| = 42

|Sn(321)| =1

S3 = {123, 132, 213, 231, 312, 321}

I S3(321) = {123, 132, 213, 231, 312}

|S3(321)| = 5

|S4(321)| = 14

|S5(321)| = 42

|Sn(321)| =1

S3 = {123, 132, 213, 231, 312, 321}

I S3(321) = {123, 132, 213, 231, 312}

|S3(321)| = 5

|S4(321)| = 14

|S5(321)| = 42

|Sn(321)| =1

S3 = {123, 132, 213, 231, 312, 321}

I S3(321) = {123, 132, 213, 231, 312}

|S3(321)| = 5

|S4(321)| = 14

|S5(321)| = 42

|Sn(321)| =1

= nth Catalan number

S3 = {123, 132, 213, 231, 312, 321}

I S3(321) = {123, 132, 213, 231, 312}

|S3(321)| = 5

|S4(321)| = 14

|S5(321)| = 42

|Sn(321)| =1

S3 = {123, 132, 213, 231, 312, 321}

I S3(321) = {123, 132, 213, 231, 312}

|S3(321)| = 5

|S4(321)| = 14

|S5(321)| = 42

|Sn(321)| =1

S3 = {123, 132, 213, 231, 312, 321}

I S3(321) = {123, 132, 213, 231, 312}

|S3(321)| = 5

|S4(321)| = 14

|S5(321)| = 42

|Sn(321)| =1

Patterns of length 4Things get messy...

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.

In fact, every pattern of length 4 is Wilf-equivalent to one of:

2314 1234 1324

What is known?

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)|.

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

2314 1234 1324

What is known?

Open Problem

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

2314 1234 1324

What is known?

Open Problem

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

2314 1234 1324

What is known?

Open Problem

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

2314 1234 1324

What is known?

Open Problem

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

2314 1234 1324

What is known?

I I. Gessel (1990) gave a formula for |Sn(1234)|

I M. Bona (1997) gave a formula for |Sn(2314)|

Open Problem

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

2314 1234 1324

What is known?

Open Problem

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

2314 1234 1324

What is known?

Open Problem

Rook Placements

DefinitionA Ferrers Board F is a square array of boxes with a “bite” takenout of the northeast corner.

A full rook placement (f.r.p.) on F is a placement of markerswith EXACTLY one in each row and column.

NotationRF = set of all f.r.p.’s on the fixed board F

Rook Placements

×××

××××

×I This f.r.p. contains the pattern 312

I This f.r.p. avoids the pattern 231

Notation

I RF (τ) = set of all f.r.p. on F that avoid τ .

Rook Placements

×××

××××

I This f.r.p. contains the pattern 312

Notation

Rook Placements

××××

Notation

Rook Placements

Notation

Rook Placements

×××

××××

Notation

Rook Placements

×××

×I This f.r.p. contains the pattern 312

Notation

Rook Placements

×××

××××

Notation

Rook Placements

DefinitionWe say two patterns σ, τ ∈ Sk are shape-Wilf-equivalent andwrite σ ∼ τ if for every Ferrers board F

|RF (σ)| = |RF (τ)|.

Note: shape-Wilf equivalence ⇒Wilf-equivalence.

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

|RF (σ)| = |RF (τ)|.

- Can count things!

Rook Placements

|RF (σ)| = |RF (τ)|.

I 123 . . . k ∼ k . . . 321 (J. Backlin, J. West, and G. Xin, 2000)

I 231 ∼ 312 (Z. Stankova and J. West, 2002)

- Can count things!

Rook Placements

|RF (σ)| = |RF (τ)|.

- Can count things!

Rook Placements

|RF (σ)| = |RF (τ)|.

- Complicated proof

⇒ can’t count things

- Can count things!

Rook Placements

|RF (σ)| = |RF (τ)|.

- Can count things!

Rook Placements

|RF (σ)| = |RF (τ)|.

- Can count things!

Rook Placements

|RF (σ)| = |RF (τ)|.

- Can count things!

Dyck Paths

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

Dyck Paths

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Dyck Paths

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Dyck Paths

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Dyck Paths

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Dyck Paths

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Labeled Dyck paths

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

We label the Dyck path so that:I Monotonicity

I Zero Condition

I Tunnel Property

Labeled Dyck paths

I Zero Condition

I Tunnel Property

Labeled Dyck paths

I Zero Condition

I Tunnel Property

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

An outline

1. 231-avoiding f.r.p. ⇒ Tunnel property

RF (231)

2. Tunnel property ⇒ Reverse tunnel property

−4 44 4

−4 4

3. Reverse tunnel property ⇒ 312-avoiding f.r.p.

×××

RF (312)

Theorem (Bloom–Saracino ’11)

This mapping is a bijection between RF (231) and RF (312).⇒ 231 and 312 are shape-Wilf-equivalent.

×××

RF (312)

×××

RF (312)

Generating FunctionsThe generating function for a sequence of integers

a0, a1, a2, a3, . . .

is the “formal” series∞∑n=0

I “A generating function is a clothesline on which we hang up asequence of numbers for display” - H. Wilf

I We do not worry about convergence!

Example

Let Dn be the set of Dyck paths with length n.

C (z) =∞∑n=0

|Dn|zn

=1−√

1− 4z

= 1 + z + 2z2 + 5z3 + 14z4 + 42z5 + · · ·

a0, a1, a2, a3, . . .

Example

C (z) =∞∑n=0

|Dn|zn

=1−√

1− 4z

= 1 + z + 2z2 + 5z3 + 14z4 + 42z5 + · · ·

a0, a1, a2, a3, . . .

Example

C (z) =∞∑n=0

|Dn|zn

=1−√

1− 4z

= 1 + z + 2z2 + 5z3 + 14z4 + 42z5 + · · ·

a0, a1, a2, a3, . . .

Example

C (z) =∞∑n=0

|Dn|zn

=1−√

1− 4z

= 1 + z + 2z2 + 5z3 + 14z4 + 42z5 + · · ·

a0, a1, a2, a3, . . .

Example

C (z) =∞∑n=0

|Dn|zn =1−√

1− 4z

= 1 + z + 2z2 + 5z3 + 14z4 + 42z5 + · · ·

a0, a1, a2, a3, . . .

Example

C (z) =∞∑n=0

|Dn|zn =1−√

1− 4z

= 1 + z + 2z2 + 5z3 + 14z4 + 42z5 + · · ·

Enumerative Results: 2314-Avoiding Permutations

In 1990 Bona proved the following celebrated result

∞∑n=0

|Sn(2314)|zn =32z

1 + 20z − 8z2 − (1− 8z)3/2.

Our Proof

6257413

Sn(2314)

RF (231)

∞∑n=0

|Sn(2314)|zn =32z

1 + 20z − 8z2 − (1− 8z)3/2.

Our Proof

6257413

Sn(2314)

RF (231)

∞∑n=0

|Sn(2314)|zn =32z

1 + 20z − 8z2 − (1− 8z)3/2.

Our Proof

6257413

Sn(2314)

RF (231)

∞∑n=0

|Sn(2314)|zn =32z

1 + 20z − 8z2 − (1− 8z)3/2.

Our Proof

6257413

Sn(2314)

RF (231)

∞∑n=0

|Sn(2314)|zn =32z

1 + 20z − 8z2 − (1− 8z)3/2.

Our Proof

6257413

Sn(2314)

RF (231)

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

1 + 36z − (1− 12z)3/2= 1 + z + 3z2 + 14z3 + 83z4 + · · ·

I New enumerative results in the theory of perfect matchingsand set partitions.

∞∑n=0

|Sn(2314, 1234)|zn =1

1− C (zC (z))

= 1 + z + 2z + 6z2 + 22z3 + · · ·

1 + 36z − (1− 12z)3/2= 1 + z + 3z2 + 14z3 + 83z4 + · · ·

∞∑n=0

|Sn(2314, 1234)|zn =1

1− C (zC (z))

= 1 + z + 2z + 6z2 + 22z3 + · · ·

1 + 36z − (1− 12z)3/2= 1 + z + 3z2 + 14z3 + 83z4 + · · ·

∞∑n=0

|Sn(2314, 1234)|zn =1

1− C (zC (z))

= 1 + z + 2z + 6z2 + 22z3 + · · ·

1 + 36z − (1− 12z)3/2= 1 + z + 3z2 + 14z3 + 83z4 + · · ·

Thank you!

wake forest university - february 2014patterns, permutations, and placements jonathan s. bloom...

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