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G-Parking Functions, Graph Searching, and Tutte Polynomial Huafei Yan Nankai University andTexas A&M University Joint with Dimitrije Kostic

G-Parking Functions, Graph Searching, and Tutte Polynomial

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Huafei Yan Nankai University andTexas A&M University Joint with Dimitrije Kostic. G-Parking Functions, Graph Searching, and Tutte Polynomial. 1. BFS on a connected graphs. - PowerPoint PPT Presentation

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G-Parking Functions, Graph Searching, and Tutte Polynomial

Huafei YanNankai University andTexas A&M University

Joint with Dimitrije Kostic

1. BFS on a connected graphs

Start a queue which is initially {0}. At each stage we take the vertex x at the head of the queue, remove x from the queue, and add all new neighbors of x to the queue.

--- Spencer: Enumerating Graphs and Brownian --- Spencer: Enumerating Graphs and Brownian MotionMotion, (1997)

BFS on H

0

4

2

3

5

1

BFS on H

0

4

2

3

5

1

t Queue

0 0

BFS on H

0

4

2

3

5

1

t Queue

0 0

1 3,4

BFS on H

0

4

2

3

5

1

t Queue

0 0

1 3,4

2 4,1,2

BFS on H

0

4

2

3

5

1

t Queue

0 0

1 3,4

2 4,1,2

3 1,2

BFS on H

0

4

2

3

5

1

t Queue

0 0

1 3,4

2 4,1,2

3 1,2

4 2,5

BFS on H

0

4

2

3

5

1

t Queue

0 0

1 3,4

2 4,1,2

3 1,2

4 2,5

5 5

6 --

Which H s.t. BFS(H)=T?

0

4

2

3

5

1

t Queue

0 0

1 3,4

2 4,1,2

3 1,2

4 2,5

5 5

6 --

[Spencer] An edge {u,v} can be added to T iff u and v have been present in the queue at the same time.

Ex(T)=set of such edges.

Theorem. Given T. BFS(H)=T iff H [T, T Ex(T) ] .

2. A familiar statistics

Let M(T)=|Ex(T)|. Then number of labeled connected graphs on n+1 vertices with n+k edges is

Let Cn(q)=G q|E(G)|-n,

G: labeled, connected, n+1 vertices.

Mn(q)=T q|Ex(T)|, Then

Cn(q) = Mn(1+q).

Same property holds for external activity of trees inversion of trees level in recurrent configurations of sandpile

model

and (reversed) sum of parking functions…..

Parking function

A PF is a sequence (a1,a2,…, an) such that the number of terms larger than k is less than n-k. n=1. (0 ) n=2. (0,0), (0,1), (1,0) There are (n+1)n-1 many parking functions of

length n.

Reversed sum of PFs

Let a=(a1,a2,…, an) be a PF. The reversed sum rsum(a) is

i (i-1-ai) = n(n-1)/2-i ai

Reversed sum of PFs

Let a=(a1,a2,…, an) be a PF. The reversed sum rsum(a) is

i (i-1-ai) = n(n-1)/2-i ai

rsum(a) has the same distribution as M(T).

rsum(a) has the same distribution as M(T).

3. PF as a vertex function

G=Kn+1 with vertex set {0,1,…,n}

A PF is a function from {1,2,…,n} to non-negative integers with the property:

For each nonempty subset U of {1,2,…,n}, there is a vertex v in U s.t.

a(v) < n-|U|.

An example

0

4/

2/

5/

1/

t Queue

0 0

1 3,4

2 4,1,2

3 1,2

4 2,5

5 5

6 --

3/

a(v) = rank of the parent of v

0

4/ 0

2/ 1

5/ 3

1/ 1

t Queue

0 0

1 3,4

2 4,1,2

3 1,2

4 2,5

5 5

6 --

3/ 0

a(v) = rank of the parent of v

0

4/ 0

2/ 1

5/ 3

1/ 1

t Q

0 0

1 3,4

2 4,1,2

3 1,2

4 2,5

5 5

6 --

3/ 0

M(T) rsum(a) M(T) rsum(a)

4.G-parking functions

Definition. A G-parking function is a function f from {1,2,…,n} to non-negative integers with the property:

For each nonempty subset U of {1,2,…,n}, there is a vertex v in U s.t. the number of edges from v to vertices outside of U is

greater than f(v).

00

1

234 334

2/ 2

1/ 0 4/ 1

3/ 2

0

Tutte polynomial of G

To count connected subgraphs of G by the number of excess edges, use

Tutte polynomial tG(x,y)

Theorem.

tG(1+x,1+y) = H xc(H)-1 y|E(H)|+c(H)-n-1

where H is over all spanning subgraphs.

General picture

Tutte polynomial of GG-parking functions

Spanning trees of G

BFSbijections

5. BFS to subgraphs of G

Theorem. Given G and a spanning tree T. Then BFS(H)=T iff

H 2 [T, T[ (Ex(T) \ G) ]

Corollary.

tG(1, y) = T y|Ex(T) in G|

where T ranges over all spanning trees of G.

BFS to subgraphs of G

Theorem. Given G and a spanning tree T. Then BFS(H)=T iff

H 2 [T, T[ (Ex(T) \ G) ]

Corollary.

tG(1, y) = T y|Ex(T) in G|

where T ranges over all spanning trees of G.

6. From T to G-parking function

Given T in G, apply BFS on T.

Define

f(v) = number of edges {w,v} in G such that w is processed before the parent of v in the queue.

An example

0

4

2

5

3

1

An example

0

4/

2/

5/

1/

t Queue

0 0

1 3,4

2 4,1,2

3 1,2

4 2,5

5 5

6 --

3/

f(v)={ (u,v) in E(G): rank(u)<rank(parent of v) }

0

4/ 0

2/ 1

5/ 2

1/ 0

t Queue

0 0

1 3,4

2 4,1,2

3 1,2

4 2,5

5 5

6 --

3/ 0

7.From G-parking function to tree

BFS with a value function. Initially, val_0(v)=f(v)

Run BFS on G and update the value function

At each stage, add new neighbors only if the value is -1.

Example

0

4/ 0

2/ 1

5/ 2

1/ 0

t Queue

0 03/ 0

Example

0

4/ -1

2/ 0

5/ 1

1/ 0

t Q

0 0

1 3,4

3/ -1

Example

0

4/ -2

2/ -1

5/ 1

1/ -1

t Q

0 0

1 3,4

2 4,1,2

3/-1

Example

0

4/ -2

2/ -2

5/ 0

1/ -2

t Q

0 0

1 3,4

2 4,1,2

3/-1

3 1,2

Example

0

4/ -2

2/ -3

5/ -1

1/ -2

t Q

0 0

1 3,4

2 4,1,2

3/-1

3 1,2

4 2,5

Example

0

4/ -2

2/ -3

5/ -2

1/ -2

t Q

0 0

1 3,4

2 4,1,2

3/-1

3 1,2

4 2,5

5 5

Example

0

4/ -2

2/ -3

5/ -2

1/ -2

t Q

0 0

1 3,4

2 4,1,2

3/-1

3 1,2

4 2,5

5 56 --

The ending value records the number of “extra edges”.

|E(G)|= v f(v) +|E(T)| +|Ex(T) in G|

Example

0

4/ -2

2/ -3

5/ -2

1/ -2

t Q

0 0

1 3,4

2 4,1,2

3/-1

3 1,2

4 2,5

5 56 --

Conclusion

Let rsum(f) = |E(G)|-n-v f(v)

One can get the full tG(x,y) by allowing multiroots for the

G-parking function.

Theorem.

tG(1,y) = f yrsum(f)

Theorem.

tG(1,y) = f yrsum(f)

8 Multiparking functions

Definition. A G-multiparking function is a function f from {1,2,…,n} to non-negative integers and (*) with the property:

For each nonempty subset U of {1,2,…,n}, either (i) f(v)=* where v=min(U), or (ii) there is a vertex v in U s.t. the number of edges from v to vertices outside of U is greater than f(v).

General formula

Let r(f)=number of v s.t. f(v)=*

Theorem.

tG(1+x,y) = yE(G)-n+1f (xy)r(f)-1y-sum(f)-Rec(f) ,

where Rec(f) is the number of edges incident to roots.

Theorem.

tG(1+x,y) = yE(G)-n+1f (xy)r(f)-1y-sum(f)-Rec(f) ,

where Rec(f) is the number of edges incident to roots.

A corollary

In a parking function (a1a2…an), a term ai=j is critical if

(i) no other term =j;

(ii) There are j terms <j, and n-j-1 terms >j.

Let p(a1…an) = #{ j: j is critical, and a left-to-right maximal}

Theorem

TKn+1(x,y)=f 2 P(n) xp(f)yn(n-1)/2-sum(f).

Example: n=2

(0,1), (1,0), (0,0)

TK3 = x2 + x + y.