Exponential Generating Functions (EGFs

Exponential Generating Functions
The Exponential Formula
Topics in Algebraic Combinatorics
Oliver Zhang • Proof School • 9th Grade, Block 1, 2016

What is an EGF?
An EGF or Exponential Generating Function
is a generating function over a factorial.

What is an EGF?
For Example:
If A
egf
←→ {an}∞
0 then

What is an EGF?
For Example:
If A
egf
←→ {an}∞
0 then
A =
∞
n=0
an
n!
xn

Multiplying EGFs
Given two EGFs A and B, what EGF would
A · B be?

Multiplying EGFs
Given two EGFs A and B, what EGF would
A · B be?
In other words, if
∞
n=0
an
n! xn
∞
n=0
bn
n! xn
=
∞
n=0
cn
n! xn
then what would cn be in terms of an and bn?

Multiplying EGFs, cont.
Let’s take an example. To calculate c4
4! , we
simply take the sum
a0b4
0!4! + a1b3
1!3! + a2b2
2!2! + a3b1
3!1! + a4b0
4!0!

Let’s take an example. To calculate c4
4! , we
simply take the sum
a0b4
0!4! + a1b3
1!3! + a2b2
2!2! + a3b1
3!1! + a4b0
4!0!
Which is equal to
a0b4
4! +
4!
1!3!a1b3
4! +
4!
2!2!a2b2
4! +
4!
3!1!a3b1
4! + a4b0
4!

Therefore, c4 is equal to the sum:
4
0 a0b4 + 4
1 a1b3 + 4
2 a2b2 + 4
3 a3b1 + 4
4 a4b0

Therefore, c4 is equal to the sum:
4
0 a0b4 + 4
1 a1b3 + 4
2 a2b2 + 4
3 a3b1 + 4
4 a4b0
Generally, cn will be equal to
n
m=0
n
m ambn−m

The Exponential Function
Just as
1
1−x ←→ 1 + x + x2
+ ...
What f(x)
egf
←→ 1 + x
1! + x2
2! + ...?

The Exponential Function
Correct!
ex egf
←→ 1 + x
1! + x2
2! + ...

Graph Theory
A few deﬁnitions:
A graph:

Graph Theory
A few deﬁnitions:
A graph: a set of vertices and relations
between the vertices.

Graph Theory
A few deﬁnitions:
A connected component of a graph:

Graph Theory
A few deﬁnitions:
A connected component of a graph: a set of
vertices within a graph such that any two
vertices are connected to each other by paths,
and which is connected to no additional vertices
in the supergraph.

Graph Theory
A few deﬁnitions:
in the supergraph.
A picture:

Graph Theory
A few deﬁnitions:
in the supergraph.
A picture: a connected graph such that the set
of vertices is the set {1, 2, ..., n}, where n is the
number of vertices of the graph.

Graph Theory
Note: These two pictures are considered
distinct because the labeling of their vertices
are diﬀerent.

Million Dollar Question
Our Question is:
How many ways are there to build a graph with
n vertices and k connected components?

Poker Playing: The Basic Unit
A card C(S, p):

A card C(S, p):
A non-empty ’label set’ S

A card C(S, p):
A ’picture’ p

A card C(S, p):
A ’picture’ p
|S| = number of vertices in p.

A card C(S, p):
A ’picture’ p
|S| = number of vertices in p.
Each card represents a connected component
of a graph.

Card Example

Card Example
A card is called standard if the label set S is
the set {1, 2, ..., n} for some n.

Hands
If a connected component can be represented
by a card, what would a graph be?

Hands
A hand H is a set of cards whose label sets
form a partition of {1, 2, ..., n} for some n.

Hands
A hand is usually created with cards from a
speciﬁed Exponential Family,

Hands
A hand is usually created with cards from a
speciﬁed Exponential Family, we’ll get to
that in a bit

Hand Example

Hand Example
The weight of a card is the size of the label
set. The weight of a hand is the sum of the
weight of the cards.

Notation!
We denote h(n, k) to be the number of
hands with weight n and k cards.

Notation!
What would h(3,2) be?

Notation!
What would h(3,2) be? 3

Notation!
What would h(2,3) be?

Notation!

Million Dollar Question Revisited
Our Question:

Our Question:
Can be rephrased to:
How many hands are there with weight n and k
cards?

Our Question:
Can be rephrased to:
How many hands are there with weight n and k
cards?
or what is h(n, k)?

Decks
A deck Dn is a ﬁnite set of standard cards
whose weights are all n and whose pictures
are all diﬀerent.

Exponential Families
An exponential family [EF] F is deﬁned to
be a collection of decks with weights 1, 2, ...

Exponential Families
An exponential family [EF] F is deﬁned to
be a collection of decks with weights 1, 2, ...
In an exponential family, let dn be deﬁned as
the number of cards in deck Dn

Exponential Families Example
If Dn was the deck with all cards of weight
n, then

n, then
What would d1 be?

n, then
What would d1 be? 1
What would d2 be?

n, then
What would d1 be? 1
What would d2 be? 1
What would d3 be?

n, then
What would d1 be? 1
What would d2 be? 1
What would d3 be? 4
What about d4?

n, then
What would d1 be? 1
What would d2 be? 1
What would d3 be? 4
What about d4? 38

Generating Functions pt. 1
We can ﬁnally start building generating
functions! Yay!

functions! Yay!
We introduce the 2-variable Hand
Enumerator
H(x, y) =
n,k≥0
h(n, k)
xn
n!
yk

functions! Yay!
We introduce the 2-variable Hand
Enumerator
H(x, y) =
n,k≥0
h(n, k)
xn
n!
yk
This generating function is a half-blood.

The deck enumerator is the egf of the
sequence {dn}∞
1 and is denoted as D(x).

Merging Families
Given two EFs F and F with no shared
pictures,

Merging Families
pictures, we can merge them together to
create a new exponential family F such that

Merging Families
pictures, we can merge them together to
create a new exponential family F such that
For any Dn and Dn in F and F
respectively, Dn in F is the union of the card
sets of Dn and Dn.

A Quick Lemma
Lemma (Label Counting)
Let F, F , F be three exponential families
and let H(x, y), H (x, y), H (x, y) be their
respective 2-variable hand enumerators. If

A Quick Lemma
F = F ⊗ F

A Quick Lemma
F = F ⊗ F
Then
H(x,y) = H’(x,y)H”(x,y)

The Initial Case
Given a ﬁxed r, let all decks besides Dr be
empty. Additionally, let Dr only contain one
card with r vertices.

The Initial Case
Given a ﬁxed r, let all decks besides Dr be
empty. Additionally, let Dr only contain one
card with r vertices.
h(n, k) for this deck is zero unless n can be
represented by kr for some integer k.
But then how many hands are there?

The Initial Case
The ﬁrst card labels can be picked in n
r
ways, the second card in n−r
r ways, ... the
last card can be picked in n−(k−1)r
r = 1 way.

The Initial Case
r
r ways, ... the
r = 1 way.
Additionally, the order of the k cards doesn’t
matter so we divide out another k!.

The Initial Case
r
r ways, ... the
r = 1 way.
Additionally, the order of the k cards doesn’t
matter so we divide out another k!.
h(kr, k) = 1
k!
n!
r!k

The Initial Case
If h(kr, k) = 1
k!
n!
r!k then the hand enumerator
of this family is

The Initial Case
If h(kr, k) = 1
k!
n!
of this family is
H(x, y) =
n,k
h(n, k)xn
yk
/n!

The Initial Case
If h(kr, k) = 1
k!
n!
of this family is
H(x, y) =
n,k
h(n, k)xn
yk
/n!
=
k
xkryk
k!r!k

The Initial Case
If h(kr, k) = 1
k!
n!
of this family is
H(x, y) =
n,k
h(n, k)xn
yk
/n!
=
k
xkryk
k!r!k
=
k
xry
r!
k
/k!

The Initial Case
If h(kr, k) = 1
k!
n!
of this family is
H(x, y) =
n,k
h(n, k)xn
yk
/n!
=
k
xkryk
k!r!k
=
k
xry
r!
k
/k!
= exp{yxr
r! }

Induction
Given an exponential family F and positive
integer r such that

Induction
integer r such that every deck besides Dr is
empty,

Induction
integer r such that every deck besides Dr is
empty, we claim its hand enumerator is:
H(x, y) = exp{ydrxr
r! }

Induction
Suppose the claim is true for
dr = 1, 2, ...m − 1 and let F have m cards
in its rth deck.

Induction
in its rth deck. Then F is the result of
merging a family with m − 1 cards in the rth
deck and a family with 1 card in that deck.

Induction
exp{y(m − 1)xr
/r!}

Induction
exp{y(m − 1)xr
/r!}exp{yxr
/r!}

Induction
exp{y(m − 1)xr
/r!}exp{yxr
/r!}= exp{ymxr
/r!}

Multi-deck Induction
Theorem (The Exponential Formula)
Let F be an exponential family whose deck and
hand enumerators are D(x) and H(x, y).
Then
H(x, y) = eyD(x)
and
h(n, k) = xn
n!
D(x)k
k!

By merging a sequence of decks D1, D2, ...
we can obtain any exponential family F.

= exp yd1x1
1! exp yd2x2
2! ...

= exp yd1x1
1! exp yd2x2
2! ...
= exp yd1x1
1! + yd2x2
2! + ...

= exp yd1x1
1! exp yd2x2
2! ...
= exp yd1x1
1! + yd2x2
2! + ...
= eyD(x)

Solving the question
The number of ways to build a graph with n
vertices and k connected components?
is just

is just
h(n, k) = H(x, y) xn
n! yk

is just
n! yk
= eyD(x) xn
n! yk

is just
n! yk
= eyD(x) xn
n! yk
= xn
n! yk yk
D(x)k
k!
∞
0

is just
n! yk
= eyD(x) xn
n! yk
= xn
n! yk yk
D(x)k
k!
∞
0
= xn
n!
D(x)k
k!
∞
0

Thanks For Watching!
Bibliography:
Generatingfunctionology by Herbert S. Wilf
https://www.math.upenn.edu/∼wilf/gfologyLinked2.pdf

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