Exponential formula presentation

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.

For Example:

If Aegf←→ {an}∞0 then

A =

∞∑n=0

ann!xn



What is an EGF?



For Example:


A =

∞∑n=0

ann!xn



What is an EGF?



For Example:


A =

∞∑n=0

ann!xn



Multiplying EGFs

Given two EGFs A and B, what EGF would

A ·B be?

In other words, if( ∞∑n=0

ann!x

n

)( ∞∑n=0

bnn!x

n

)=∞∑n=0

cnn!x

n

then what would cn be in terms of an and bn?



Multiplying EGFs

Given two EGFs A and B, what EGF would

A ·B be?

In other words, if( ∞∑n=0

ann!x

n

)( ∞∑n=0

bnn!x

n

)=∞∑n=0

cnn!x

n

then what would cn be in terms of an and bn?



Multiplying EGFs, cont.

Let’s take an example. To calculate c44! , we

simply take the sum

a0b40!4! +

a1b31!3! +

a2b22!2! +

a3b13!1! +

a4b04!0!

Which is equal to

a0b44! +

4!1!3!a1b3

4! +4!2!2!a2b2

4! +4!3!1!a3b1

4! + a4b04!




Let’s take an example. To calculate c44! , we

simply take the sum

a0b40!4! +

a1b31!3! +

a2b22!2! +

a3b13!1! +

a4b04!0!

Which is equal to

a0b44! +

4!1!3!a1b3

4! +4!2!2!a2b2

4! +4!3!1!a3b1

4! + a4b04!




Therefore, c4 is equal to the sum:(40

)a0b4 +

(41

)a1b3 +

(42

)a2b2 +

(43

)a3b1 +

(44

)a4b0

Generally, cn will be equal ton∑

m=0

(nm

)ambn−m




Therefore, c4 is equal to the sum:(40

)a0b4 +

(41

)a1b3 +

(42

)a2b2 +

(43

)a3b1 +

(44

)a4b0

Generally, cn will be equal ton∑

m=0

(nm

)ambn−m



The Exponential Function

Just as

11−x ←→ 1 + x + x2 + ...

What f (x)egf←→ 1 + x

1! +x2

2! + ...?



The Exponential Function

Correct!

exegf←→ 1 + x

1! +x2

2! + ...



Graph Theory

A few definitions:

A graph:

a set of vertices and relationsbetween the vertices.

A connected component of a graph: a set ofvertices within a graph such that any twovertices are connected to each other by paths,and which is connected to no additional verticesin the supergraph.

A picture: a connected graph such that the setof vertices is the set {1, 2, ..., n}, where n is thenumber of vertices of the graph.



Graph Theory

A few definitions:

A graph: a set of vertices and relationsbetween the vertices.





Graph Theory

A few definitions:


A connected component of a graph:

a set ofvertices within a graph such that any twovertices are connected to each other by paths,and which is connected to no additional verticesin the supergraph.




Graph Theory

A few definitions:






Graph Theory

A few definitions:



A picture:

a connected graph such that the setof vertices is the set {1, 2, ..., n}, where n is thenumber of vertices of the graph.



Graph Theory

A few definitions:






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 withn vertices and k connected components?



Million Dollar Question

Our Question is:




Poker Playing: The Basic Unit

A card C(S, p):

A non-empty ’label set’ S

A ’picture’ p

|S| = number of vertices in p.

Each card represents a connected component

of a graph.




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

A ’picture’ p



of a graph.





A ’picture’ p



of a graph.





A ’picture’ p



of a graph.





A ’picture’ p



of a graph.



Card Example

A card is called standard if the label set S is

the set {1, 2, ..., n} for some n.



Card Example





Card Example





Card Example





Card Example


the set {1, 2, ..., n} for some n.The Exponential Formula


Hands

If a connected component can be represented

by a card, what would a graph be?

A hand H is a set of cards whose label sets

form a partition of {1, 2, ..., n} for some n.

A hand is usually created with cards from a

specified Exponential Family, we’ll get to

that in a bit



Hands







that in a bit



Hands






specified Exponential Family,

we’ll get to

that in a bit



Hands







that in a bit



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.



Hand Example






Hand Example






Hand Example






Notation!

We denote h(n, k) to be the number of

hands with weight n and k cards.

What would h(3,2) be? 3




Notation!



What would h(3,2) be?

3




Notation!







Notation!




What would h(2,3) be?

0



Notation!







Notation!







Million Dollar Question Revisited

Our Question:


Can be rephrased to:

How many hands are there with weight n and kcards?

or what is h(n, k)?




Our Question:




or what is h(n, k)?




Our Question:




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.



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, ...

In an exponential family, let dn be defined as

the number of cards in deck Dn



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 D′n was the deck with all cards of weight

n, then

What would d′1 be? 1



What about d′4? 38





n, then

What would d′1 be?

1








n, then



1







n, then




4






n, then




What about d′4?

38





n, then







Generating Functions pt. 1

We can finally start building generating

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.





functions! Yay!


Enumerator

H(x, y) =∑n,k≥0

h(n, k)xn

n!yk






functions! Yay!


Enumerator

H(x, y) =∑n,k≥0

h(n, k)xn

n!yk





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,

we can merge them together to

create a new exponential family F such that

For any D′n and D′′n in F ′ and F ′′

respectively, Dn in F is the union of the card

sets of D′n and D′′n.



Merging Families


pictures, we can merge them together to







Merging Families


pictures, we can merge them together to







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

F = F ′ ⊗ F ′′

Then

H(x,y) = H’(x,y)H”(x,y)



A Quick Lemma





F = F ′ ⊗ F ′′

Then

H(x,y) = H’(x,y)H”(x,y)



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.

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

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(nr

)ways, the second card in

(n−rr

)ways, ... the

last card can be picked in(n−(k−1)r

r

)= 1 way.

Additionally, the order of the k cards doesn’t

matter so we divide out another k!.

h(kr, k) = 1k!

n!r!k



The Initial Case



(n−rr

)ways, ... the


r

)= 1 way.



h(kr, k) = 1k!

n!r!k



The Initial Case



(n−rr

)ways, ... the


r

)= 1 way.



h(kr, k) = 1k!

n!r!k



The Initial Case

If h(kr, k) = 1k!

n!r!k

then the hand enumerator

of this family is

H(x, y) =∑n,k

h(n, k)xnyk/n!

=∑k

xkryk

k!r!k

=∑k

(xryr!

)k/k!

= exp{yxr

r! }



The Initial Case

If h(kr, k) = 1k!

n!r!k


of this family is

H(x, y) =∑n,k

h(n, k)xnyk/n!

=∑k

xkryk

k!r!k

=∑k

(xryr!

)k/k!

= exp{yxr

r! }



The Initial Case

If h(kr, k) = 1k!

n!r!k


of this family is

H(x, y) =∑n,k

h(n, k)xnyk/n!

=∑k

xkryk

k!r!k

=∑k

(xryr!

)k/k!

= exp{yxr

r! }



The Initial Case

If h(kr, k) = 1k!

n!r!k


of this family is

H(x, y) =∑n,k

h(n, k)xnyk/n!

=∑k

xkryk

k!r!k

=∑k

(xryr!

)k/k!

= exp{yxr

r! }



The Initial Case

If h(kr, k) = 1k!

n!r!k


of this family is

H(x, y) =∑n,k

h(n, k)xnyk/n!

=∑k

xkryk

k!r!k

=∑k

(xryr!

)k/k!

= exp{yxr

r! }



Induction

Given an exponential family F and positive

integer r such that

every deck besides Dr is

empty, we claim its hand enumerator is:

H(x, y) = exp{ydrxr

r! }



Induction


integer r such that every deck besides Dr is

empty,

we claim its hand enumerator is:

H(x, y) = exp{ydrxr

r! }



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.

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.

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



Induction



in its rth deck. Then F is the result of






Induction






exp{y(m− 1)xr/r!}

exp{yxr/r!}= exp{ymxr/r!}



Induction






exp{y(m− 1)xr/r!}exp{yxr/r!}

= exp{ymxr/r!}



Induction









Multi-deck Induction

Theorem (The Exponential Formula)Let F be an exponential family whose deck andhand enumerators are D(x) and H(x, y).Then

H(x, y) = eyD(x)

andh(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! + yd2x2

2! + ...}

= eyD(x)






= exp{

yd1x1

1!

}exp{

yd2x2

2!

}...

= exp{

yd1x1

1! + yd2x2

2! + ...}

= eyD(x)






= exp{

yd1x1

1!

}exp{

yd2x2

2!

}...

= exp{

yd1x1

1! + yd2x2

2! + ...}

= eyD(x)






= 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 nvertices and k connected components?is just

h(n, k) = H(x, y)[xn

n! yk]

= eyD(x)[xn

n! yk]

=[xn

n! yk] {ykD(x)k

k!

}∞0

=[xn

n!

] {D(x)k

k!

}∞0






n! yk]

= eyD(x)[xn

n! yk]

=[xn

n! yk] {ykD(x)k

k!

}∞0

=[xn

n!

] {D(x)k

k!

}∞0






n! yk]

= eyD(x)[xn

n! yk]

=[xn

n! yk] {ykD(x)k

k!

}∞0

=[xn

n!

] {D(x)k

k!

}∞0






n! yk]

= eyD(x)[xn

n! yk]

=[xn

n! yk] {ykD(x)k

k!

}∞0

=[xn

n!

] {D(x)k

k!

}∞0






n! yk]

= eyD(x)[xn

n! yk]

=[xn

n! yk] {ykD(x)k

k!

}∞0

=[xn

n!

] {D(x)k

k!

}∞0



Thanks For Watching!

Bibliography:

Generatingfunctionology by Herbert S. Wilfhttps://www.math.upenn.edu/∼wilf/gfologyLinked2.pdf


Education

Exponential formula presentation