Integrals don’t have anything to do with
graph theory . . . do they?
Mark Kayll
Department of Mathematical Sciences
University of Montana
Wellesley College — 16 March 2009
c 2009 by P. Mark Kayll
Copyright Integrals
Bipartite Graphs
Rook Polynomials
Outline
Integrals
Bipartite Graphs
Definitions
Counting
Fundamental question
Rook Polynomials
Notation
Building the definition
Example
Main Result
Recall the question (with answer)
Proof
Application
Set-up
Through a bipartite lens
Enter the theorem
Main Result
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Euler’s Gamma Function
Z
∞
Γ(x) =
t x−1 e−t dt
0
• converges for 0 < x < ∞
• extends to C; meromorphic with poles {0, −1, −2, −3, . . .}
∞
• Γ(1)= −e−t = 1
• Γ(1/2) =
√
0
π
• Γ(1/3) = ??
Γ(1/4) = ??
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Euler’s Gamma Function
Z
∞
Γ(x) =
t x−1 e−t dt
0
• converges for 0 < x < ∞
• extends to C; meromorphic with poles {0, −1, −2, −3, . . .}
∞
• Γ(1)= −e−t = 1
• Γ(1/2) =
√
0
π
• Γ(1/3) = ??
Γ(1/4) = ??
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Euler’s Gamma Function
Z
∞
Γ(x) =
t x−1 e−t dt
0
• converges for 0 < x < ∞
• extends to C; meromorphic with poles {0, −1, −2, −3, . . .}
∞
• Γ(1)= −e−t = 1
• Γ(1/2) =
√
0
π
• Γ(1/3) = ??
Γ(1/4) = ??
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Euler’s Gamma Function
Z
∞
Γ(x) =
t x−1 e−t dt
0
• converges for 0 < x < ∞
• extends to C; meromorphic with poles {0, −1, −2, −3, . . .}
∞
• Γ(1)= −e−t = 1
• Γ(1/2) =
√
0
π
• Γ(1/3) = ??
Γ(1/4) = ??
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Euler’s Gamma Function
Z
∞
Γ(x) =
t x−1 e−t dt
0
• converges for 0 < x < ∞
• extends to C; meromorphic with poles {0, −1, −2, −3, . . .}
∞
• Γ(1)= −e−t = 1
• Γ(1/2) =
√
0
π
• Γ(1/3) = ??
Γ(1/4) = ??
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Euler’s Gamma Function
Z
∞
Γ(x) =
t x−1 e−t dt
0
• converges for 0 < x < ∞
• extends to C; meromorphic with poles {0, −1, −2, −3, . . .}
∞
• Γ(1)= −e−t = 1
• Γ(1/2) =
√
0
π
• Γ(1/3) = ??
Γ(1/4) = ??
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Euler’s Gamma Function
Z
∞
Γ(x) =
t x−1 e−t dt
0
• converges for 0 < x < ∞
• extends to C; meromorphic with poles {0, −1, −2, −3, . . .}
∞
• Γ(1)= −e−t = 1
• Γ(1/2) =
√
0
π
• Γ(1/3) = ??
Γ(1/4) = ??
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Euler’s Gamma Function
Z
∞
Γ(x) =
t x−1 e−t dt
0
• converges for 0 < x < ∞
• extends to C; meromorphic with poles {0, −1, −2, −3, . . .}
∞
• Γ(1)= −e−t = 1
• Γ(1/2) =
√
0
π
• Γ(1/3) = ??
Γ(1/4) = ??
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Gamma Function Recurrence
Integrate by parts. . .
Z
Γ(x + 1) =
∞
t x e−t dt
0
Z
∞
+
= −t e x −t t=0
= 0 + x Γ(x)
Corollary
Γ(n + 1) = n!
Corollary
Z
0
∞
t n e−t dt = n!
0
∞
x t x−1 e−t dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Gamma Function Recurrence
Integrate by parts. . .
Z
Γ(x + 1) =
∞
t x e−t dt
0
Z
∞
+
= −t e x −t t=0
= 0 + x Γ(x)
Corollary
Γ(n + 1) = n!
Corollary
Z
0
∞
t n e−t dt = n!
0
∞
x t x−1 e−t dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Gamma Function Recurrence
Integrate by parts. . .
Z
Γ(x + 1) =
∞
t x e−t dt
0
Z
∞
+
= −t e x −t t=0
= 0 + x Γ(x)
Corollary
Γ(n + 1) = n!
Corollary
Z
0
∞
t n e−t dt = n!
0
∞
x t x−1 e−t dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Gamma Function Recurrence
Integrate by parts. . .
Z
Γ(x + 1) =
∞
t x e−t dt
0
Z
∞
+
= −t e x −t t=0
= 0 + x Γ(x)
Corollary
Γ(n + 1) = n!
Corollary
Z
0
∞
t n e−t dt = n!
0
∞
x t x−1 e−t dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Gamma Function Recurrence
Integrate by parts. . .
Z
Γ(x + 1) =
∞
t x e−t dt
0
Z
∞
+
= −t e x −t t=0
= 0 + x Γ(x)
Corollary
Γ(n + 1) = n!
Corollary
Z
0
∞
t n e−t dt = n!
0
∞
x t x−1 e−t dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Complete Bipartite Graphs
K 2,2
K 3,3
K 4,4
Skip to bottom line
Integrals
Bipartite Graphs
Rook Polynomials
Matchings in K3,3
Main Result
Application
Integrals
Bipartite Graphs
Rook Polynomials
Matchings in K3,3
Main Result
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Perfect Matchings in K3,3
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting Perfect Matchings in Kn,n
n choices
n 1
n 2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Counting Perfect Matchings in Kn,n : Bottom Line
Proposition
Ξ(Kn,n ) = n!
Proposition
Z
Ξ(Kn,n ) =
0
∞
t n e−t dt
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Counting Perfect Matchings in Kn,n : Bottom Line
Proposition
Ξ(Kn,n ) = n!
Proposition
Z
Ξ(Kn,n ) =
0
∞
t n e−t dt
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Tinkering with Kn,n
∞
Z
t n e−t dt
Ξ(Kn,n ) =
0
Question
If we replace Kn,n by a subgraph G ⊆ Kn,n ,
then how does the integrand change?
Z
Ξ(G) =
0
∞
?
dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Tinkering with Kn,n
∞
Z
t n e−t dt
Ξ(Kn,n ) =
0
Question
If we replace Kn,n by a subgraph G ⊆ Kn,n ,
then how does the integrand change?
Z
Ξ(G) =
0
∞
?
dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Tinkering with Kn,n
∞
Z
t n e−t dt
Ξ(Kn,n ) =
0
Question
If we replace Kn,n by a subgraph G ⊆ Kn,n ,
then how does the integrand change?
Z
Ξ(G) =
0
∞
?
dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Tinkering with Kn,n
∞
Z
t n e−t dt
Ξ(Kn,n ) =
0
Question
If we replace Kn,n by a subgraph G ⊆ Kn,n ,
then how does the integrand change?
Z
Ξ(G) =
0
∞
?
dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Bipartite Complements
G : spanning subgraph of Kn,n
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Bipartite Complements
G : spanning subgraph of Kn,n
e : bipartite complement of G
G
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Notation
Matchings with a fixed number of edges
µG (k ) = number of matchings of G containing exactly k edges
k
µG (k )
0
1
1
6
2
9
3
2
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Notation
Matchings with a fixed number of edges
µG (k ) = number of matchings of G containing exactly k edges
k
µG (k )
0
1
1
6
2
9
3
2
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Notation
Matchings with a fixed number of edges
µG (k ) = number of matchings of G containing exactly k edges
k
µG (k )
0
1
1
6
2
9
3
2
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Notation
Matchings with a fixed number of edges
µG (k ) = number of matchings of G containing exactly k edges
k
µG (k )
0
1
1
6
2
3
2
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Notation
Matchings with a fixed number of edges
µG (k ) = number of matchings of G containing exactly k edges
k
µG (k )
0
1
1
6
2
3
2
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Notation
Matchings with a fixed number of edges
µG (k ) = number of matchings of G containing exactly k edges
k
µG (k )
0
1
1
6
2
3
2
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Notation
Matchings with a fixed number of edges
µG (k ) = number of matchings of G containing exactly k edges
k
µG (k )
0
1
1
6
2
9
3
2
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
The Rook Polynomial
An example
k
µG (k )
0
1
1
6
2
9
3
2
ρG (t) =
+1t 3
−6t 2
+9t
−2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
The Rook Polynomial
An example
k
µG (k )
0
1
1
6
2
9
3
2
ρG (t) =
+1t 3
−6t 2
+9t
−2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
The Rook Polynomial
An example
k
µG (k )
0
1
1
6
2
9
3
2
ρG (t) =
+1t 3
−6t 2
+9t
−2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
The Rook Polynomial
An example
k
µG (k )
0
1
1
6
2
9
3
2
ρG (t) =
+1t 3
−6t 2
+9t
−2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
The Rook Polynomial
An example
k
µG (k )
0
1
1
6
2
9
3
2
ρG (t) =
+1t 3
−6t 2
+9t
−2
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
The Rook Polynomial
Definition (rook polynomial of spanning G ⊆ Kn,n )
ρG (t) =
n
X
k =0
(−1)k µG (k )t n−k
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
The Rook Polynomial
Another example
n
n
µG (k ) =
for 0 ≤ k ≤ n
k
ρG (t) =
n
X
k
(−1) µG (k )t
k =0
n−k
=
n
X
k =0
n n−k
(−1)
t
= (t − 1)n
k
k
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
The Rook Polynomial
Another example
n
n
µG (k ) =
for 0 ≤ k ≤ n
k
ρG (t) =
n
X
k
(−1) µG (k )t
k =0
n−k
=
n
X
k =0
n n−k
(−1)
t
= (t − 1)n
k
k
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
The Rook Polynomial
Another example
n
n
µG (k ) =
for 0 ≤ k ≤ n
k
ρG (t) =
n
X
k
(−1) µG (k )t
k =0
n−k
=
n
X
k =0
n n−k
(−1)
t
= (t − 1)n
k
k
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
The Rook Polynomial
Another example
n
n
µG (k ) =
for 0 ≤ k ≤ n
k
ρG (t) =
n
X
k
(−1) µG (k )t
k =0
n−k
=
n
X
k =0
n n−k
(−1)
t
= (t − 1)n
k
k
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
The Rook Polynomial
Another example
n
n
µG (k ) =
for 0 ≤ k ≤ n
k
ρG (t) =
n
X
k
(−1) µG (k )t
k =0
n−k
=
n
X
k =0
n n−k
(−1)
t
= (t − 1)n
k
k
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
The Rook Polynomial
Another example
n
n
µG (k ) =
for 0 ≤ k ≤ n
k
ρG (t) =
n
X
k
(−1) µG (k )t
k =0
n−k
=
n
X
k =0
n n−k
(−1)
t
= (t − 1)n
k
k
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Tinkering with Kn,n
∞
Z
t n e−t dt
Ξ(Kn,n ) =
0
Question
If we replace Kn,n by a subgraph G ⊆ Kn,n ,
then how does the integrand change?
Z
Ξ(G) =
0
∞
?
dt
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting with Integrals
Proposition
Z
Ξ(Kn,n ) =
∞
t n e−t dt.
0
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Counting with Integrals
Proposition
Z
Ξ(Kn,n ) =
∞
t n e−t dt.
0
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Counting with Integrals
Proposition
Z
Ξ(Kn,n ) =
∞
t n e−t dt.
0
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
See proof
Skip proof
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Set-up
X = set of perfect matchings of Kn,n
e = {1, 2, . . . , m} with m ≥ 0
• E(G)
e
• for i ∈ E(G):
Ai = {M ∈ X : M contains edge i}
•
•
Notice: X −
m
[
i=1
Ai = set of perfect matchings of G
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Set-up
X = set of perfect matchings of Kn,n
e = {1, 2, . . . , m} with m ≥ 0
• E(G)
e
• for i ∈ E(G):
Ai = {M ∈ X : M contains edge i}
•
•
Notice: X −
m
[
i=1
Ai = set of perfect matchings of G
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Set-up
X = set of perfect matchings of Kn,n
e = {1, 2, . . . , m} with m ≥ 0
• E(G)
e
• for i ∈ E(G):
Ai = {M ∈ X : M contains edge i}
•
•
Notice: X −
m
[
i=1
Ai = set of perfect matchings of G
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Set-up
X = set of perfect matchings of Kn,n
e = {1, 2, . . . , m} with m ≥ 0
• E(G)
e
• for i ∈ E(G):
Ai = {M ∈ X : M contains edge i}
•
•
Notice: X −
m
[
i=1
Ai = set of perfect matchings of G
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Set-up
X = set of perfect matchings of Kn,n
e = {1, 2, . . . , m} with m ≥ 0
• E(G)
e
• for i ∈ E(G):
Ai = {M ∈ X : M contains edge i}
•
•
Notice: X −
m
[
i=1
Ai = set of perfect matchings of G
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Set-up
X = set of perfect matchings of Kn,n
e = {1, 2, . . . , m} with m ≥ 0
• E(G)
e
• for i ∈ E(G):
Ai = {M ∈ X : M contains edge i}
•
•
Notice: X −
m
[
Ai = set of perfect matchings of G
i=1
Therefore
m
[
Ξ(G) = X −
Ai i=1
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Proof of Main Result
Inclusion-Exclusion
m
[
Ai Ξ(G) = X −
i=1
Ξ(G) = | X | −
m
X
| Ai | +
i=1
· · · + (−1)
X
| Ai ∩ Aj | − + · · ·
1≤i<j≤m
k
X
| Ai1 ∩ Ai2 ∩ · · · ∩ Aik | + · · ·
i1 <i2 <···<ik
e
Notice: Ai1 ∩ · · · ∩ Aik = ∅ unless {i1 , . . . , ik } is a matching in G.
e =⇒ | Ai ∩ Ai ∩ · · · ∩ Ai | = (n − k )!
{i1 , i2 , . . . , ik } matching in G
1
2
k
e
There are µGe (k ) such matchings in G.
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Proof of Main Result
Inclusion-Exclusion
m
[
Ai Ξ(G) = X −
i=1
Ξ(G) = | X | −
m
X
| Ai | +
i=1
· · · + (−1)
X
| Ai ∩ Aj | − + · · ·
1≤i<j≤m
k
X
| Ai1 ∩ Ai2 ∩ · · · ∩ Aik | + · · ·
i1 <i2 <···<ik
e
Notice: Ai1 ∩ · · · ∩ Aik = ∅ unless {i1 , . . . , ik } is a matching in G.
e =⇒ | Ai ∩ Ai ∩ · · · ∩ Ai | = (n − k )!
{i1 , i2 , . . . , ik } matching in G
1
2
k
e
There are µGe (k ) such matchings in G.
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Proof of Main Result
Inclusion-Exclusion
m
[
Ai Ξ(G) = X −
i=1
Ξ(G) = | X | −
m
X
| Ai | +
i=1
· · · + (−1)
X
| Ai ∩ Aj | − + · · ·
1≤i<j≤m
k
X
| Ai1 ∩ Ai2 ∩ · · · ∩ Aik | + · · ·
i1 <i2 <···<ik
e
Notice: Ai1 ∩ · · · ∩ Aik = ∅ unless {i1 , . . . , ik } is a matching in G.
e =⇒ | Ai ∩ Ai ∩ · · · ∩ Ai | = (n − k )!
{i1 , i2 , . . . , ik } matching in G
1
2
k
e
There are µGe (k ) such matchings in G.
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Proof of Main Result
Inclusion-Exclusion
m
[
Ai Ξ(G) = X −
i=1
Ξ(G) = | X | −
m
X
| Ai | +
i=1
· · · + (−1)
X
| Ai ∩ Aj | − + · · ·
1≤i<j≤m
k
X
| Ai1 ∩ Ai2 ∩ · · · ∩ Aik | + · · ·
i1 <i2 <···<ik
e
Notice: Ai1 ∩ · · · ∩ Aik = ∅ unless {i1 , . . . , ik } is a matching in G.
e =⇒ | Ai ∩ Ai ∩ · · · ∩ Ai | = (n − k )!
{i1 , i2 , . . . , ik } matching in G
1
2
k
e
There are µGe (k ) such matchings in G.
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Proof of Main Result
Inclusion-Exclusion
m
[
Ai Ξ(G) = X −
i=1
Ξ(G) = | X | −
m
X
| Ai | +
i=1
· · · + (−1)
X
| Ai ∩ Aj | − + · · ·
1≤i<j≤m
k
X
| Ai1 ∩ Ai2 ∩ · · · ∩ Aik | + · · ·
i1 <i2 <···<ik
e
Notice: Ai1 ∩ · · · ∩ Aik = ∅ unless {i1 , . . . , ik } is a matching in G.
e =⇒ | Ai ∩ Ai ∩ · · · ∩ Ai | = (n − k )!
{i1 , i2 , . . . , ik } matching in G
1
2
k
e
There are µGe (k ) such matchings in G.
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Proof of Main Result
Inclusion-Exclusion
m
[
Ai Ξ(G) = X −
i=1
Ξ(G) = | X | −
m
X
| Ai | +
i=1
X
| Ai ∩ Aj | − + · · ·
1≤i<j≤m
· · · + (−1)
X
k
| Ai1 ∩ Ai2 ∩ · · · ∩ Aik | + · · ·
i1 <i2 <···<ik
e
Notice: Ai1 ∩ · · · ∩ Aik = ∅ unless {i1 , . . . , ik } is a matching in G.
e =⇒ | Ai ∩ Ai ∩ · · · ∩ Ai | = (n − k )!
{i1 , i2 , . . . , ik } matching in G
1
2
k
e
There are µGe (k ) such matchings in G.
Therefore
Ξ(G) =
X
(−1)k µGe (k )(n − k )!
k ≥0
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Conclusion
Ξ(G) =
n
X
(−1)k µGe (k )(n − k )!
k =0
Ξ(G) =
=
n
X
(−1)k µGe (k )
k =0
Z ∞ X
n
0
Z
∞
t n−k e−t dt
0
(−1)k µGe (k )t n−k e−t dt
k =0
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Conclusion
Ξ(G) =
n
X
(−1)k µGe (k )(n − k )!
k =0
Ξ(G) =
=
n
X
(−1)k µGe (k )
k =0
Z ∞ X
n
0
Z
∞
t n−k e−t dt
0
(−1)k µGe (k )t n−k e−t dt
k =0
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Conclusion
Ξ(G) =
n
X
(−1)k µGe (k )(n − k )!
k =0
Ξ(G) =
=
n
X
(−1)k µGe (k )
k =0
Z ∞ X
n
0
Z
∞
t n−k e−t dt
0
(−1)k µGe (k )t n−k e−t dt
k =0
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Conclusion
Ξ(G) =
n
X
(−1)k µGe (k )(n − k )!
k =0
Ξ(G) =
=
n
X
(−1)k µGe (k )
k =0
Z ∞ X
n
0
Z
∞
t n−k e−t dt
0
(−1)k µGe (k )t n−k e−t dt
k =0
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Conclusion
Ξ(G) =
n
X
(−1)k µGe (k )(n − k )!
k =0
Ξ(G) =
=
n
X
(−1)k µGe (k )
k =0
Z ∞ X
n
0
Z
∞
t n−k e−t dt
0
(−1)k µGe (k )t n−k e−t dt
k =0
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Conclusion
Ξ(G) =
n
X
(−1)k µGe (k )(n − k )!
k =0
Ξ(G) =
=
n
X
(−1)k µGe (k )
k =0
Z ∞ X
n
0
Z
∞
t n−k e−t dt
0
(−1)k µGe (k )t n−k e−t dt
k =0
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Proof of Main Result
Conclusion
Ξ(G) =
n
X
(−1)k µGe (k )(n − k )!
k =0
Ξ(G) =
=
n
X
(−1)k µGe (k )
k =0
Z ∞ X
n
0
Z
∞
t n−k e−t dt
0
(−1)k µGe (k )t n−k e−t dt
k =0
We [Emerson and the speaker] have re-proved:
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Proof of Main Result
Conclusion
Ξ(G) =
n
X
(−1)k µGe (k )(n − k )!
k =0
Ξ(G) =
=
n
X
(−1)k µGe (k )
k =0
Z ∞ X
n
0
Z
∞
t n−k e−t dt
0
(−1)k µGe (k )t n−k e−t dt
k =0
We [Emerson and the speaker] have re-proved:
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
To Appendix
Wrap-up
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: introduction
Definition
A derangement σ of a set S is a permutation of S admitting no
fixed points; i.e., σ : S → S is a bijection such that σ(x) 6= x for
each x ∈ S.
Notation
dn = number of derangements of {1, 2, . . . , n}
Example
n
1
2
3
4
5
6
dn
0
1
2
9
44
265
derangements of {1, 2, . . . , n}
∅
{21}
{231, 312}
{2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321}
{31452, . . .}
{534621, . . .}
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: introduction
Definition
A derangement σ of a set S is a permutation of S admitting no
fixed points; i.e., σ : S → S is a bijection such that σ(x) 6= x for
each x ∈ S.
Notation
dn = number of derangements of {1, 2, . . . , n}
Example
n
1
2
3
4
5
6
dn
0
1
2
9
44
265
derangements of {1, 2, . . . , n}
∅
{21}
{231, 312}
{2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321}
{31452, . . .}
{534621, . . .}
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: introduction
Definition
A derangement σ of a set S is a permutation of S admitting no
fixed points; i.e., σ : S → S is a bijection such that σ(x) 6= x for
each x ∈ S.
Notation
dn = number of derangements of {1, 2, . . . , n}
Example
n
1
2
3
4
5
6
dn
0
1
2
9
44
265
derangements of {1, 2, . . . , n}
∅
{21}
{231, 312}
{2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321}
{31452, . . .}
{534621, . . .}
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: enumeration
Question
What is the value of dn ?
Answer
dn = n!
n
X
(−1)k
k =0
k!
Proof technique.
Via generating functions or inclusion-exclusion.
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: enumeration
Question
What is the value of dn ?
Answer
dn = n!
n
X
(−1)k
k =0
k!
Proof technique.
Via generating functions or inclusion-exclusion.
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: enumeration
Question
What is the value of dn ?
Answer
dn = n!
n
X
(−1)k
k =0
k!
Proof technique.
Via generating functions or inclusion-exclusion.
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: through a bipartite lens
1
2
3
4
1
2
3
4
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: through a bipartite lens
1
2
3
4
1
2
3
4
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: through a bipartite lens
1
2
3
4
1
2
3
4
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: through a bipartite lens
1
2
3
4
1
2
3
4
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: through a bipartite lens
1
2
3
4
1
2
3
4
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: through a bipartite lens
1
2
3
4
1
2
3
4
2413
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: through a bipartite lens
1
2
3
4
1
2
3
4
dn = Ξ(Kn,n − M)
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Recall the Main Result
Theorem (Joni & Rota, 1980; Godsil, 1981)
If G is a spanning subgraph of Kn,n , then
Z ∞
Ξ(G) =
ρGe (t)e−t dt.
0
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: calculations
∞
Z
dn = Ξ(Kn,n − M) =
ρM (t)e
−t
0
Z
∞
n −t
(t − 1) e
=
x =t −1
dt =
(t − 1)n e−t dt
1
dt +
(t − 1)n e−t dt
0
Z
∞
n −(x+1)
x e
=
Z
dx +
0
= e−1 Γ(n + 1) + En
=
∞
0
Z
1
Z
n!
+ En
e
0
1
(t − 1)n e−t dt
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: calculations
∞
Z
dn = Ξ(Kn,n − M) =
ρM (t)e
−t
0
Z
∞
n −t
(t − 1) e
=
x =t −1
dt =
(t − 1)n e−t dt
1
dt +
(t − 1)n e−t dt
0
Z
∞
n −(x+1)
x e
=
Z
dx +
0
= e−1 Γ(n + 1) + En
=
∞
0
Z
1
Z
n!
+ En
e
0
1
(t − 1)n e−t dt
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: calculations
∞
Z
dn = Ξ(Kn,n − M) =
ρM (t)e
−t
0
Z
∞
n −t
(t − 1) e
=
x =t −1
dt =
(t − 1)n e−t dt
1
dt +
(t − 1)n e−t dt
0
Z
∞
n −(x+1)
x e
=
Z
dx +
0
= e−1 Γ(n + 1) + En
=
∞
0
Z
1
Z
n!
+ En
e
0
1
(t − 1)n e−t dt
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: calculations
∞
Z
dn = Ξ(Kn,n − M) =
ρM (t)e
−t
0
Z
∞
n −t
(t − 1) e
=
x =t −1
dt =
(t − 1)n e−t dt
1
dt +
(t − 1)n e−t dt
0
Z
∞
n −(x+1)
x e
=
Z
dx +
0
= e−1 Γ(n + 1) + En
=
∞
0
Z
1
Z
n!
+ En
e
0
1
(t − 1)n e−t dt
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: calculations
∞
Z
dn = Ξ(Kn,n − M) =
ρM (t)e
−t
0
Z
∞
n −t
(t − 1) e
=
x =t −1
dt =
(t − 1)n e−t dt
1
dt +
(t − 1)n e−t dt
0
Z
∞
n −(x+1)
x e
=
Z
dx +
0
= e−1 Γ(n + 1) + En
=
∞
0
Z
1
Z
n!
+ En
e
0
1
(t − 1)n e−t dt
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: calculations
∞
Z
dn = Ξ(Kn,n − M) =
ρM (t)e
−t
0
Z
∞
n −t
(t − 1) e
=
x =t −1
dt =
(t − 1)n e−t dt
1
dt +
(t − 1)n e−t dt
0
Z
∞
n −(x+1)
x e
=
Z
dx +
0
= e−1 Γ(n + 1) + En
=
∞
0
Z
1
Z
n!
+ En
e
0
1
(t − 1)n e−t dt
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: calculations
∞
Z
dn = Ξ(Kn,n − M) =
ρM (t)e
−t
0
Z
∞
n −t
(t − 1) e
=
x =t −1
dt =
(t − 1)n e−t dt
1
dt +
(t − 1)n e−t dt
0
Z
∞
n −(x+1)
x e
=
Z
dx +
0
= e−1 Γ(n + 1) + En
=
∞
0
Z
1
Z
n!
+ En
e
0
1
(t − 1)n e−t dt
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: calculations
∞
Z
dn = Ξ(Kn,n − M) =
ρM (t)e
−t
Z
dt =
0
∞
Z
n −t
(t − 1) e
=
x =t −1
1
dt +
(t − 1)n e−t dt
0
Z
∞
n −(x+1)
x e
=
Z
dx +
0
1
(t − 1)n e−t dt
0
−1
= e Γ(n + 1) + En
=
(t − 1)n e−t dt
0
Z
1
∞
n!
+ En
e
dn =
n!
+ En
e
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: denouement
n!
+ En
e
dn =
Z
n −t
(t − 1) e
En =
1
dt
0
e−t < 1 on the interval (0, 1)
1
Z
(t − 1)n e−t dt <
|En | ≤
0
|En | <
Z
1
(1 − t)n dt =
0
1
for each n ≥ 1
2
1
n+1
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: denouement
n!
+ En
e
dn =
Z
n −t
(t − 1) e
En =
1
dt
0
e−t < 1 on the interval (0, 1)
1
Z
(t − 1)n e−t dt <
|En | ≤
0
|En | <
Z
1
(1 − t)n dt =
0
1
for each n ≥ 1
2
1
n+1
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: denouement
n!
+ En
e
dn =
Z
n −t
(t − 1) e
En =
1
dt
0
e−t < 1 on the interval (0, 1)
1
Z
(t − 1)n e−t dt <
|En | ≤
0
|En | <
Z
1
(1 − t)n dt =
0
1
for each n ≥ 1
2
1
n+1
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: denouement
n!
+ En
e
dn =
Z
n −t
(t − 1) e
En =
1
dt
0
e−t < 1 on the interval (0, 1)
1
Z
(t − 1)n e−t dt <
|En | ≤
0
|En | <
Z
1
(1 − t)n dt =
0
1
for each n ≥ 1
2
1
n+1
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: denouement
n!
+ En
e
dn =
Z
n −t
(t − 1) e
En =
1
dt
0
e−t < 1 on the interval (0, 1)
1
Z
(t − 1)n e−t dt <
|En | ≤
0
|En | <
Z
1
(1 − t)n dt =
0
1
for each n ≥ 1
2
1
n+1
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: denouement
n!
+ En
e
dn =
Z
n −t
(t − 1) e
En =
1
dt
0
e−t < 1 on the interval (0, 1)
1
Z
(t − 1)n e−t dt <
|En | ≤
0
|En | <
Z
1
(1 − t)n dt =
0
1
for each n ≥ 1
2
1
n+1
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: denouement
n!
+ En
e
dn =
Z
n −t
(t − 1) e
En =
1
dt
0
e−t < 1 on the interval (0, 1)
1
Z
(t − 1)n e−t dt <
|En | ≤
0
|En | <
Z
1
(1 − t)n dt =
0
1
for each n ≥ 1
2
1
n+1
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: denouement
n!
+ En
e
dn =
Z
n −t
(t − 1) e
En =
1
dt
0
e−t < 1 on the interval (0, 1)
1
Z
(t − 1)n e−t dt <
|En | ≤
0
|En | <
Z
1
(1 − t)n dt =
0
1
for each n ≥ 1
2
1
n+1
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: summary
dn =
|En | <
n!
+ En
e
1
for each n ≥ 1
2
dn = integer closest to
dn = n!
n
X
(−1)k
k =0
n!
e
k!
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: summary
dn =
|En | <
n!
+ En
e
1
for each n ≥ 1
2
dn = integer closest to
dn = n!
n
X
(−1)k
k =0
n!
e
k!
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Derangements: summary
dn =
|En | <
n!
+ En
e
1
for each n ≥ 1
2
dn = integer closest to
dn = n!
n
X
(−1)k
k =0
n!
e
k!
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Application
Derangements: summary
dn =
|En | <
n!
+ En
e
1
for each n ≥ 1
2
dn = integer closest to
dn = n!
n
X
(−1)k
k =0
n!
e
To Appendix
Wrap-up
k!
Back to proof
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Take-Home Messages
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Take-Home Messages
•
Gamma function (and other integrals) form a bridge
between discrete and continuous mathematics.
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Take-Home Messages
•
Gamma function (and other integrals) form a bridge
between discrete and continuous mathematics.
•
Connection manifests nicely in counting matchings.
Application
Integrals
Bipartite Graphs
Rook Polynomials
Main Result
Take-Home Messages
•
Gamma function (and other integrals) form a bridge
between discrete and continuous mathematics.
•
Connection manifests nicely in counting matchings.
•
Ties can be fruitful, not merely beautiful.
Application
Integrals
Bipartite Graphs
Rook Polynomials
fin
applause (optional)
Main Result
Application
Appendix
More Counting with Integrals
Proposition
1
Ξ(Kn ) = √
2π
Z
∞
t n e−t
2 /2
dt.
−∞
Theorem (Godsil, 1981)
If G is a spanning subgraph of Kn , then
Z ∞
1
2
Ξ(G) = √
α G (t)e−t /2 dt.
2 π −∞
Appendix
More Counting with Integrals
Proposition
1
Ξ(Kn ) = √
2π
Z
∞
t n e−t
2 /2
dt.
−∞
Theorem (Godsil, 1981)
If G is a spanning subgraph of Kn , then
Z ∞
1
2
Ξ(G) = √
α G (t)e−t /2 dt.
2 π −∞
Appendix
Ties to Orthogonal Polynomials I
Rook polynomials
ρG (t) =
n
X
(−1)k µG (k )t n−k
k =0
Laguerre polynomials
d n −t n Ln (t) = (−1) e n e t
dt
n t
Appendix
Ties to Orthogonal Polynomials I
Rook polynomials
ρG (t) =
n
X
(−1)k µG (k )t n−k
k =0
Laguerre polynomials
d n −t n Ln (t) = (−1) e n e t
dt
n t
Appendix
Ties to Orthogonal Polynomials I
Rook polynomials
ρG (t) =
n
X
(−1)k µG (k )t n−k
k =0
Laguerre polynomials
d n −t n Ln (t) = (−1) e n e t
dt
n t
Appendix
Ties to Orthogonal Polynomials I
Rook polynomials
ρG (t) =
n
X
(−1)k µG (k )t n−k
k =0
Laguerre polynomials
d n −t n Ln (t) = (−1) e n e t
dt
n t
Proposition
ρKn,n (t) = Ln (t)
Appendix
Ties to Orthogonal Polynomials II
Matchings polynomials
bn/2c
αG (t) =
X
(−1)k µG (k )t n−2k
k =0
Hermite polynomials
n t 2 /2
Hn (t) = (−1) e
d n −t 2 /2 e
dt n
Appendix
Ties to Orthogonal Polynomials II
Matchings polynomials
bn/2c
αG (t) =
X
(−1)k µG (k )t n−2k
k =0
Hermite polynomials
n t 2 /2
Hn (t) = (−1) e
d n −t 2 /2 e
dt n
Appendix
Ties to Orthogonal Polynomials II
Matchings polynomials
bn/2c
αG (t) =
X
(−1)k µG (k )t n−2k
k =0
Hermite polynomials
n t 2 /2
Hn (t) = (−1) e
d n −t 2 /2 e
dt n
Appendix
Ties to Orthogonal Polynomials II
Matchings polynomials
bn/2c
αG (t) =
X
(−1)k µG (k )t n−2k
k =0
Hermite polynomials
n t 2 /2
Hn (t) = (−1) e
d n −t 2 /2 e
dt n
Proposition
αKn (t) = Hn (t)
Appendix
Further Generalizations
Proposition
∞
Z
Ξ(Kn,n ) =
t n e−t dt.
0
Proposition
1
Ξ(Kn ) = √
2π
Z
∞
t n e−t
2 /2
dt.
−∞
)
Meta Theorem (
Other combinatorially interesting sequences take the form
Z
t n dµ
Ω
for some measure µ and space Ω.
Appendix
Further Generalizations
Proposition
∞
Z
Ξ(Kn,n ) =
t n e−t dt.
0
Proposition
1
Ξ(Kn ) = √
2π
Z
∞
t n e−t
2 /2
dt.
−∞
)
Meta Theorem (
Other combinatorially interesting sequences take the form
Z
t n dµ
Ω
for some measure µ and space Ω.
Appendix
Further Generalizations
Proposition
∞
Z
Ξ(Kn,n ) =
t n e−t dt.
0
Proposition
1
Ξ(Kn ) = √
2π
Z
∞
t n e−t
2 /2
dt.
−∞
Meta Theorem (
)
Other combinatorially interesting sequences take the form
Z
t n dµ
Ω
for some measure µ and space Ω.
Appendix
Further Generalizations
Proposition
∞
Z
Ξ(Kn,n ) =
t n e−t dt.
0
Proposition
1
Ξ(Kn ) = √
2π
Z
∞
t n e−t
2 /2
dt.
−∞
Meta Theorem (see Godsil, 1993, Chapter 9)
Other combinatorially interesting sequences take the form
Z
t n dµ
Ω
for some measure µ and space Ω.
Appendix
Further Generalizations
Proposition
∞
Z
Ξ(Kn,n ) =
t n e−t dt.
0
Proposition
1
Ξ(Kn ) = √
2π
Z
∞
t n e−t
2 /2
dt.
−∞
Meta Theorem (see Godsil, 1993, Chapter 9)
Other combinatorially interesting sequences take the form
Z
t n dµ
Ω
for some measure µ and space Ω.
Wrap-up
The End
Appendix
Further Reading
Books
E. Artin.
The Gamma Function.
Holt, Rinehart and Winston, New York, 1964.
C.D. Godsil.
Algebraic Combinatorics.
Chapman & Hall, New York, 1993.
J. Riordan.
An Introduction to Combinatorial Analysis.
John Wiley & Sons, Inc., New York, 1958.
Appendix
Further Reading
Articles
E.E. Emerson and P.M. Kayll.
A short proof of the Joni-Rota-Godsil integral formula for
counting bipartite matchings.
Submitted for publication, 2009.
C.D. Godsil.
Hermite polynomials and a duality relation for matching
polynomials.
Combinatorica 1 (1981), 257–262.
S.A. Joni and G.-C. Rota.
A vector space analog of permutations with restricted
position.
J. Combin. Theory Ser. A 29 (1980), 59–73.
Appendix
Really fin
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