Factoring rook polynomials Bruce Sagan Department of Mathematics, - - PowerPoint PPT Presentation

Factoring rook polynomials

Bruce Sagan

Department of Mathematics, Michigan State University East Lansing, MI 48824-1027 www.math.msu.edu/˜sagan

March 14, 2014

Basics The FactorizationTheorem An application Exercises and References

Outline

Basics The FactorizationTheorem An application Exercises and References

Consider tiling the first quadrant of the plane with unit squares: Q = . . . . . .

Consider tiling the first quadrant of the plane with unit squares: Q = . . . . . . Let (c, d) be the square in column c and row d.

Consider tiling the first quadrant of the plane with unit squares: Q = . . . . . . (3, 4) Let (c, d) be the square in column c and row d.

Consider tiling the first quadrant of the plane with unit squares: Q = . . . . . . (3, 4) Let (c, d) be the square in column c and row d. A board is a finite set of squares B ⊆ Q.

Consider tiling the first quadrant of the plane with unit squares: Q = . . . . . . (3, 4) Let (c, d) be the square in column c and row d. A board is a finite set of squares B ⊆ Q.

• Ex. Let Bn be the n × n chess board. For example,

B3 =

Consider tiling the first quadrant of the plane with unit squares: Q = . . . . . . (3, 4) Let (c, d) be the square in column c and row d. A board is a finite set of squares B ⊆ Q.

• Ex. Let Bn be the n × n chess board. For example,

B3 = A placement P of rooks on B is attacking if there is a pair of rooks in the same row or column. Otherwise it is nonattacking.

Consider tiling the first quadrant of the plane with unit squares: Q = . . . . . . (3, 4) Let (c, d) be the square in column c and row d. A board is a finite set of squares B ⊆ Q.

• Ex. Let Bn be the n × n chess board. For example,

B3 = attacking: P = R R R A placement P of rooks on B is attacking if there is a pair of rooks in the same row or column. Otherwise it is nonattacking.

Consider tiling the first quadrant of the plane with unit squares: Q = . . . . . . (3, 4) Let (c, d) be the square in column c and row d. A board is a finite set of squares B ⊆ Q.

• Ex. Let Bn be the n × n chess board. For example,

B3 = attacking: P = R R R nonattacking: P = R R R A placement P of rooks on B is attacking if there is a pair of rooks in the same row or column. Otherwise it is nonattacking.

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B.

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

• Ex. We have

rn(Bn)

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

• Ex. We have

rn(Bn) = (# of ways to place a rook in column 1) ·(# of ways to then place a rook in column 2) · · ·

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

• Ex. We have

rn(Bn) = (# of ways to place a rook in column 1) ·(# of ways to then place a rook in column 2) · · · = n · (n − 1) · · ·

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

• Ex. We have

rn(Bn) = (# of ways to place a rook in column 1) ·(# of ways to then place a rook in column 2) · · · = n · (n − 1) · · · = n!

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

• Ex. We have

rn(Bn) = (# of ways to place a rook in column 1) ·(# of ways to then place a rook in column 2) · · · = n · (n − 1) · · · = n! There is a bijection between placements P counted by rn(Bn) and permutations π in the symmetric group Sn where (c, d) ∈ P if and

• nly if π(c) = d.

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

• Ex. We have

rn(Bn) = (# of ways to place a rook in column 1) ·(# of ways to then place a rook in column 2) · · · = n · (n − 1) · · · = n! There is a bijection between placements P counted by rn(Bn) and permutations π in the symmetric group Sn where (c, d) ∈ P if and

• nly if π(c) = d.
• Ex. Let

Dn = Bn − {(1, 1), (2, 2), . . . , (n, n)}.

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

• Ex. We have

rn(Bn) = (# of ways to place a rook in column 1) ·(# of ways to then place a rook in column 2) · · · = n · (n − 1) · · · = n! There is a bijection between placements P counted by rn(Bn) and permutations π in the symmetric group Sn where (c, d) ∈ P if and

• nly if π(c) = d.
• Ex. Let

Dn = Bn − {(1, 1), (2, 2), . . . , (n, n)}. Then rn(Dn)

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

• Ex. We have

rn(Bn) = (# of ways to place a rook in column 1) ·(# of ways to then place a rook in column 2) · · · = n · (n − 1) · · · = n! There is a bijection between placements P counted by rn(Bn) and permutations π in the symmetric group Sn where (c, d) ∈ P if and

• nly if π(c) = d.
• Ex. Let

Dn = Bn − {(1, 1), (2, 2), . . . , (n, n)}. Then rn(Dn) = # of permutations π ∈ Sn with π(c) = c for all c

Define the rook numbers of B to be rk(B) = number of ways of placing k nonattacking rooks on B. For any board B we have r0(B) = 1 and r1(B) = |B| (cardinality).

• Ex. We have

rn(Bn) = (# of ways to place a rook in column 1) ·(# of ways to then place a rook in column 2) · · · = n · (n − 1) · · · = n! There is a bijection between placements P counted by rn(Bn) and permutations π in the symmetric group Sn where (c, d) ∈ P if and

• nly if π(c) = d.
• Ex. Let

Dn = Bn − {(1, 1), (2, 2), . . . , (n, n)}. Then rn(Dn) = # of permutations π ∈ Sn with π(c) = c for all c = the nth derangement number.

Outline

Basics The FactorizationTheorem An application Exercises and References

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers.

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j.

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j. Ex. B = (1, 1, 3) =

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j. If x is a variable and n ≥ 0 then the corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1). Ex. B = (1, 1, 3) =

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j. If x is a variable and n ≥ 0 then the corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).

Theorem (Factorization Theorem: Goldman-Joichi-White)

For any Ferrers board B = (b1, . . . , bn) we have

n

• k=0

rk(B)x ↓n−k=

n

• j=1

(x + bj − j + 1). Ex. B = (1, 1, 3) =

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j. If x is a variable and n ≥ 0 then the corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).

Theorem (Factorization Theorem: Goldman-Joichi-White)

For any Ferrers board B = (b1, . . . , bn) we have

n

• k=0

rk(B)x ↓n−k=

n

• j=1

(x + bj − j + 1). Ex. B = (1, 1, 3) = r0(B) = 1, r1(B) = 5, r2(B) = 4, r3(B) = 0.

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j. If x is a variable and n ≥ 0 then the corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).

Theorem (Factorization Theorem: Goldman-Joichi-White)

For any Ferrers board B = (b1, . . . , bn) we have

n

• k=0

rk(B)x ↓n−k=

n

• j=1

(x + bj − j + 1). Ex. B = (1, 1, 3) = r0(B) = 1, r1(B) = 5, r2(B) = 4, r3(B) = 0.

3

• k=0

rk(B)x ↓3−k

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j. If x is a variable and n ≥ 0 then the corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).

Theorem (Factorization Theorem: Goldman-Joichi-White)

For any Ferrers board B = (b1, . . . , bn) we have

n

• k=0

rk(B)x ↓n−k=

n

• j=1

(x + bj − j + 1). Ex. B = (1, 1, 3) = r0(B) = 1, r1(B) = 5, r2(B) = 4, r3(B) = 0.

3

• k=0

rk(B)x ↓3−k = 1 · x ↓3 +5 · x ↓2 +4 · x ↓1

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j. If x is a variable and n ≥ 0 then the corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).

Theorem (Factorization Theorem: Goldman-Joichi-White)

For any Ferrers board B = (b1, . . . , bn) we have

n

• k=0

rk(B)x ↓n−k=

n

• j=1

(x + bj − j + 1). Ex. B = (1, 1, 3) = r0(B) = 1, r1(B) = 5, r2(B) = 4, r3(B) = 0.

3

• k=0

rk(B)x ↓3−k = 1 · x ↓3 +5 · x ↓2 +4 · x ↓1 = x3 + 2x2 + x

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j. If x is a variable and n ≥ 0 then the corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).

Theorem (Factorization Theorem: Goldman-Joichi-White)

For any Ferrers board B = (b1, . . . , bn) we have

n

• k=0

rk(B)x ↓n−k=

n

• j=1

(x + bj − j + 1). Ex. B = (1, 1, 3) = r0(B) = 1, r1(B) = 5, r2(B) = 4, r3(B) = 0.

3

• k=0

rk(B)x ↓3−k = 1 · x ↓3 +5 · x ↓2 +4 · x ↓1 = x3 + 2x2 + x = (x + 1)x(x + 1)

A partition is a weakly increasing sequence (b1, . . . , bn) of nonnegative integers. A Ferrers board is B = (b1, . . . , bn) consisting of the lowest bj squares in column j of Q for all j. If x is a variable and n ≥ 0 then the corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).

Theorem (Factorization Theorem: Goldman-Joichi-White)

For any Ferrers board B = (b1, . . . , bn) we have

n

• k=0

rk(B)x ↓n−k=

n

• j=1

(x + bj − j + 1). Ex. B = (1, 1, 3) = r0(B) = 1, r1(B) = 5, r2(B) = 4, r3(B) = 0.

3

• k=0

rk(B)x ↓3−k = 1 · x ↓3 +5 · x ↓2 +4 · x ↓1 = x3 + 2x2 + x = (x + 1)x(x + 1) = (x + b1)(x + b2 − 1)(x + b3 − 2).

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer.

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer. Consider

n x B R Bx =

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer. Consider

n x B R Bx = Claim: both sides of (1) equal rn(Bx).

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer. Consider

n x B R Bx = Claim: both sides of (1) equal rn(Bx). Placing rooks left to right

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer. Consider

n x B R Bx = Claim: both sides of (1) equal rn(Bx). Placing rooks left to right rn(Bx) =

n

• j=1

(# of unattacked squares in column j)

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer. Consider

n x B R Bx = Claim: both sides of (1) equal rn(Bx). Placing rooks left to right rn(Bx) =

n

• j=1

(# of unattacked squares in column j) = (x + b1)(x + b2 − 1) . . .

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer. Consider

n x B R Bx = Claim: both sides of (1) equal rn(Bx). Placing rooks left to right rn(Bx) =

n

• j=1

(# of unattacked squares in column j) = (x + b1)(x + b2 − 1) . . . = RHS of (1).

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer. Consider

n x B R Bx = Claim: both sides of (1) equal rn(Bx). Placing rooks left to right rn(Bx) =

n

• j=1

(# of unattacked squares in column j) = (x + b1)(x + b2 − 1) . . . = RHS of (1). rn(Bx) =

n

• k=0

(# of ways to put k rooks on B and n − k on R)

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer. Consider

n x B R Bx = Claim: both sides of (1) equal rn(Bx). Placing rooks left to right rn(Bx) =

n

• j=1

(# of unattacked squares in column j) = (x + b1)(x + b2 − 1) . . . = RHS of (1). rn(Bx) =

n

• k=0

(# of ways to put k rooks on B and n − k on R) =

n

• k=0

rk(B) · x(x − 1) . . . (x − n + k + 1)

n

• k=0

rk(b1, . . . , bn)x ↓n−k=

n

• j=1

(x + bj − j + 1).

1 2

(1)

• Proof. It suffices to prove (1) for x a positive integer. Consider

n x B R Bx = Claim: both sides of (1) equal rn(Bx). Placing rooks left to right rn(Bx) =

n

• j=1

(# of unattacked squares in column j) = (x + b1)(x + b2 − 1) . . . = RHS of (1). rn(Bx) =

n

• k=0

(# of ways to put k rooks on B and n − k on R) =

n

• k=0

rk(B) · x(x − 1) . . . (x − n + k + 1) = LHS of (1).

Outline

Basics The FactorizationTheorem An application Exercises and References

Call boards B and B′ rook equivalent, B ≡ B′, if rk(B) = rk(B′) for all k ≥ 0.

Call boards B and B′ rook equivalent, B ≡ B′, if rk(B) = rk(B′) for all k ≥ 0. Note that B ≡ B′ implies |B| = r1(B) = r1(B′) = |B′|.

Call boards B and B′ rook equivalent, B ≡ B′, if rk(B) = rk(B′) for all k ≥ 0. Note that B ≡ B′ implies |B| = r1(B) = r1(B′) = |B′|. Ex. B = (1, 1, 3) = B′ = (2, 3) =

Call boards B and B′ rook equivalent, B ≡ B′, if rk(B) = rk(B′) for all k ≥ 0. Note that B ≡ B′ implies |B| = r1(B) = r1(B′) = |B′|. Ex. B = (1, 1, 3) = B′ = (2, 3) = For B, B′: r0 = 1, r1 = 5, r2 = 4, rk = 0 for k ≥ 3

Call boards B and B′ rook equivalent, B ≡ B′, if rk(B) = rk(B′) for all k ≥ 0. Note that B ≡ B′ implies |B| = r1(B) = r1(B′) = |B′|. Ex. B = (1, 1, 3) = B′ = (2, 3) = For B, B′: r0 = 1, r1 = 5, r2 = 4, rk = 0 for k ≥ 3 so B ≡ B′.

Call boards B and B′ rook equivalent, B ≡ B′, if rk(B) = rk(B′) for all k ≥ 0. Note that B ≡ B′ implies |B| = r1(B) = r1(B′) = |B′|. Ex. B = (1, 1, 3) = B′ = (2, 3) = For B, B′: r0 = 1, r1 = 5, r2 = 4, rk = 0 for k ≥ 3 so B ≡ B′. A Ferrers board B = (b1, . . . , bn) is increasing if b1 < · · · < bn.

Call boards B and B′ rook equivalent, B ≡ B′, if rk(B) = rk(B′) for all k ≥ 0. Note that B ≡ B′ implies |B| = r1(B) = r1(B′) = |B′|. Ex. B = (1, 1, 3) = B′ = (2, 3) = For B, B′: r0 = 1, r1 = 5, r2 = 4, rk = 0 for k ≥ 3 so B ≡ B′. A Ferrers board B = (b1, . . . , bn) is increasing if b1 < · · · < bn. In the example above, B′ is increasing but B is not.

Call boards B and B′ rook equivalent, B ≡ B′, if rk(B) = rk(B′) for all k ≥ 0. Note that B ≡ B′ implies |B| = r1(B) = r1(B′) = |B′|. Ex. B = (1, 1, 3) = B′ = (2, 3) = For B, B′: r0 = 1, r1 = 5, r2 = 4, rk = 0 for k ≥ 3 so B ≡ B′. A Ferrers board B = (b1, . . . , bn) is increasing if b1 < · · · < bn. In the example above, B′ is increasing but B is not.

Theorem (Foata-Sch¨ utzenberger)

Every Ferrers board is rook equivalent to a unique increasing board.

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn)

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn)

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn)

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

B′ = (0, 2, 3) so ζ(B′) = (0, 1, 2)−(0, 2, 3) = (0, −1, −1) ∴ B ≡ B′.

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

B′ = (0, 2, 3) so ζ(B′) = (0, 1, 2)−(0, 2, 3) = (0, −1, −1) ∴ B ≡ B′. Every Ferrers board B is rook equivalent to a unique increasing board.

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

B′ = (0, 2, 3) so ζ(B′) = (0, 1, 2)−(0, 2, 3) = (0, −1, −1) ∴ B ≡ B′. Every Ferrers board B is rook equivalent to a unique increasing board. Proof sketch. Pad B with zeros so that ζ = ζ(B) starts with 0 and has all entries ≥ 0.

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

B′ = (0, 2, 3) so ζ(B′) = (0, 1, 2)−(0, 2, 3) = (0, −1, −1) ∴ B ≡ B′. Every Ferrers board B is rook equivalent to a unique increasing board. Proof sketch. Pad B with zeros so that ζ = ζ(B) starts with 0 and has all entries ≥ 0. Let m = max ζ(B).

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

B′ = (0, 2, 3) so ζ(B′) = (0, 1, 2)−(0, 2, 3) = (0, −1, −1) ∴ B ≡ B′. Every Ferrers board B is rook equivalent to a unique increasing board. Proof sketch. Pad B with zeros so that ζ = ζ(B) starts with 0 and has all entries ≥ 0. Let m = max ζ(B). Rearrange ζ to form ζ′ = (0, 1, 2, . . . , m, ζ′

m+1, . . . , ζ′ n)

where ζ′

m+1 ≥ · · · ≥ ζ′ n.

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

B′ = (0, 2, 3) so ζ(B′) = (0, 1, 2)−(0, 2, 3) = (0, −1, −1) ∴ B ≡ B′. Every Ferrers board B is rook equivalent to a unique increasing board. Proof sketch. Pad B with zeros so that ζ = ζ(B) starts with 0 and has all entries ≥ 0. Let m = max ζ(B). Rearrange ζ to form ζ′ = (0, 1, 2, . . . , m, ζ′

m+1, . . . , ζ′ n)

where ζ′

m+1 ≥ · · · ≥ ζ′

• n. Then ∃ increasing B′ with ζ(B′) = ζ′.

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

B′ = (0, 2, 3) so ζ(B′) = (0, 1, 2)−(0, 2, 3) = (0, −1, −1) ∴ B ≡ B′. Every Ferrers board B is rook equivalent to a unique increasing board. Proof sketch. Pad B with zeros so that ζ = ζ(B) starts with 0 and has all entries ≥ 0. Let m = max ζ(B). Rearrange ζ to form ζ′ = (0, 1, 2, . . . , m, ζ′

m+1, . . . , ζ′ n)

where ζ′

m+1 ≥ · · · ≥ ζ′

• n. Then ∃ increasing B′ with ζ(B′) = ζ′.
• Ex. B = (0, 1, 1, 3) so ζ(B) = (0, 1, 2, 3) − (0, 1, 1, 3) = (0, 0, 1, 0).

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

B′ = (0, 2, 3) so ζ(B′) = (0, 1, 2)−(0, 2, 3) = (0, −1, −1) ∴ B ≡ B′. Every Ferrers board B is rook equivalent to a unique increasing board. Proof sketch. Pad B with zeros so that ζ = ζ(B) starts with 0 and has all entries ≥ 0. Let m = max ζ(B). Rearrange ζ to form ζ′ = (0, 1, 2, . . . , m, ζ′

m+1, . . . , ζ′ n)

where ζ′

m+1 ≥ · · · ≥ ζ′

• n. Then ∃ increasing B′ with ζ(B′) = ζ′.
• Ex. B = (0, 1, 1, 3) so ζ(B) = (0, 1, 2, 3) − (0, 1, 1, 3) = (0, 0, 1, 0).

Now ζ′ = (0, 1, 0, 0)

The root vector of B = (b1, . . . , bn) is ζ(B) = (0−b1, 1−b2, . . . , n−1−bn) = (0, 1, . . . , n−1)−(b1, b2, . . . , bn) The entries of ζ(B) are exactly the zeros of

k rk(B)x ↓n−k.

So if B = (b1, . . . , bn) and B′ = (b′

1, . . . , b′ n) then

B ≡ B′ ⇐ ⇒ ζ(B) is a rearrangement of ζ(B′).

• Ex. B = (1, 1, 3) so ζ(B) = (0, 1, 2) − (1, 1, 3) = (−1, 0, −1).

B′ = (0, 2, 3) so ζ(B′) = (0, 1, 2)−(0, 2, 3) = (0, −1, −1) ∴ B ≡ B′. Every Ferrers board B is rook equivalent to a unique increasing board. Proof sketch. Pad B with zeros so that ζ = ζ(B) starts with 0 and has all entries ≥ 0. Let m = max ζ(B). Rearrange ζ to form ζ′ = (0, 1, 2, . . . , m, ζ′

m+1, . . . , ζ′ n)

where ζ′

m+1 ≥ · · · ≥ ζ′

• n. Then ∃ increasing B′ with ζ(B′) = ζ′.
• Ex. B = (0, 1, 1, 3) so ζ(B) = (0, 1, 2, 3) − (0, 1, 1, 3) = (0, 0, 1, 0).

Now ζ′ = (0, 1, 0, 0) so B′ = (0, 1, 2, 3) − (0, 1, 0, 0) = (0, 0, 2, 3).

Outline

Basics The FactorizationTheorem An application Exercises and References

• 1. Let Bn be the n × n Ferrers board.

(a) Compute rk(Bn) for any 0 ≤ k ≤ n.

• 1. Let Bn be the n × n Ferrers board.

(a) Compute rk(Bn) for any 0 ≤ k ≤ n. (b) Factor n

k=0 rk(Bn)x ↓n−k.

• 1. Let Bn be the n × n Ferrers board.

(a) Compute rk(Bn) for any 0 ≤ k ≤ n. (b) Factor n

k=0 rk(Bn)x ↓n−k.

(c) Find the unique increasing board equivalent to Bn.

• 1. Let Bn be the n × n Ferrers board.

(a) Compute rk(Bn) for any 0 ≤ k ≤ n. (b) Factor n

k=0 rk(Bn)x ↓n−k.

(c) Find the unique increasing board equivalent to Bn.

• 2. Let Tn = (0, 1, 2, . . . , n − 1).

(a) Show that for 0 ≤ k ≤ n, rk(Tn) equals the number of partitions of {1, . . . , n} into n − k subsets. This number is called a Stirling number of the second kind.

• 1. Let Bn be the n × n Ferrers board.

(a) Compute rk(Bn) for any 0 ≤ k ≤ n. (b) Factor n

k=0 rk(Bn)x ↓n−k.

(c) Find the unique increasing board equivalent to Bn.

• 2. Let Tn = (0, 1, 2, . . . , n − 1).

(a) Show that for 0 ≤ k ≤ n, rk(Tn) equals the number of partitions of {1, . . . , n} into n − k subsets. This number is called a Stirling number of the second kind. (b) Factor n

k=0 rk(Tn)x ↓n−k.

• 1. Let Bn be the n × n Ferrers board.

(a) Compute rk(Bn) for any 0 ≤ k ≤ n. (b) Factor n

k=0 rk(Bn)x ↓n−k.

(c) Find the unique increasing board equivalent to Bn.

• 2. Let Tn = (0, 1, 2, . . . , n − 1).

(a) Show that for 0 ≤ k ≤ n, rk(Tn) equals the number of partitions of {1, . . . , n} into n − k subsets. This number is called a Stirling number of the second kind. (b) Factor n

k=0 rk(Tn)x ↓n−k.

(c) Give a second proof of the identity in (b) by counting the number of functions f : {1, . . . , n} → {1, . . . , x} (where x is a positive integer) in two different ways.

• 1. Let Bn be the n × n Ferrers board.

(a) Compute rk(Bn) for any 0 ≤ k ≤ n. (b) Factor n

k=0 rk(Bn)x ↓n−k.

(c) Find the unique increasing board equivalent to Bn.

• 2. Let Tn = (0, 1, 2, . . . , n − 1).

(a) Show that for 0 ≤ k ≤ n, rk(Tn) equals the number of partitions of {1, . . . , n} into n − k subsets. This number is called a Stirling number of the second kind. (b) Factor n

k=0 rk(Tn)x ↓n−k.

(c) Give a second proof of the identity in (b) by counting the number of functions f : {1, . . . , n} → {1, . . . , x} (where x is a positive integer) in two different ways. (d) Find the unique increasing board equivalent to Tn.

References.

• 1. D. Foata and M. P. Sch¨

utzenberger, On the rook polynomials

• f Ferrers relations, in Combinatorial theory and its

applications, II (Proc. Colloq., Balatonf¨ ured, 1969), pages 413–436, North-Holland, Amsterdam, 1970.

• 2. Jay R. Goldman, J. T. Joichi, and Dennis E. White, Rook
• theory. I. Rook equivalence of Ferrers boards, Proc. Amer.
• Math. Soc., 52:485–492, 1975.

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