Partitioning the Labeled Spanning Trees of an Arbitrary Graph into Isomorphism Classes

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Partitioning the Labeled Spanning Trees of an Arbitrary Graph into Isomorphism Classes. Austin Mohr. Outline. Problem Description Generating Spanning Trees Testing for Isomorphism Partitioning Spanning Trees Some Results Finding a Closed Formula for I(K s,t ). Problem Description.

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Partitioning the Labeled Spanning Trees

of an Arbitrary Graph into

Isomorphism Classes

Austin Mohr

Outline

Problem Description

Generating Spanning Trees

Testing for Isomorphism

Partitioning Spanning Trees

Some Results

Finding a Closed Formula for I(Ks,t)

Definitions

Spanning tree T of graph G

T is a tree with E(T)⊆E(G) and V(T)=V(G)

Isomorphic trees T1 and T2

There exists a mapping f where the edge

uv∈T1 if and only if the edge f(u)f(v)∈T2

Reference: pg. 3 - 4

Problem Description

Spanning Trees of K2,3

Reference: pg. 5

Problem Description

Definitions

Index of an edge

“Arbitrary” labeling of the edges of G

T*

Tree induced by the edge-subset {1,2,…,n-1}

top(H)/btm(H)

Edge of H with smallest/largest index

Cut(H,e)

Edges of G connecting the components of H\e

(T)

(T\f)∪g, f = btm(T), g = top(Cut(T,f))

Let G be a graph on n vertices, H⊆G, e bean edge of G,and T be a spanning tree of G.

Reference: pg. 6

Generating Spanning Trees

Regarding (T)

Let T be a spanning tree of G.

Then, (T) is a spanning tree of G.

Let T ≠ T* be a spanning tree of G with

(T) = (T\f)∪g.

Then, g∈T*∌f.

Means iteration of  yields T*

Reference: pg. 7

Generating Spanning Trees

“Tree of trees” for K2,3

Reference: pg. 8

Generating Spanning Trees

Definitions

Pivot edge f of T

An edge such that T`\T = f for some child tree T`

Cycle(T,e)

The set of edges of the unique cycle in T∪e

Let G be a graph on n vertices, e be an edge of G, and T be a spanning tree of G.

Reference: pg. 8

Generating Spanning Trees

Finding the Children of a Tree

Reference: pg. 11

Generating Spanning Trees

Rooted Tree Isomorphism

We first consider the simpler problem of determining when two rooted trees are isomorphic.

Reference: pg. 14

Testing for Isomorphism

Rooted Tree Isomorphism

Given two rooted trees T1 and T2 on n

vertices, a mapping f:V(T1) → V(T2) is an

isomorphism if and only if for every vertex

v∈V(T1), the subtree of T1 rooted at v is isomorphic to the subtree of T2 rooted at f(v).

Means we can start at the bottom of the tree and work recursively toward the root

Reference: pg. 14

Testing for Isomorphism

Sample Run of Algorithm for Rooted Trees

Reference: pg. 17

Testing for Isomorphism

General Tree Isomorphism

To generalize the algorithm, we need a

vertex u∈V(T1) and v∈V(T2) such that

f(u) = v for every isomorphism f.

If found, we root T1 at u, root T2 at v, and use the previous algorithm

The center of each tree is suitable choice

Reference: pg. 18

Testing for Isomorphism

Definitions

d(u,v) (distance)

The number of edges in the shortest uv-path

eccentricity

Let v be a vertex of maximum distance from u. Then, the eccentricity of u is d(u,v).

center

The subgraph of G induced by the vertices of minimum eccentricity

Let u and v be vertices of a graph G.

Reference: pg. 18

Testing for Isomorphism

Finding the Center of a Tree

Theorem (Jordan): The center of a tree is either a vertex or an edge.

Jordan’s proof also shows that we can find the center by successively removing all the leaves from the tree until only a vertex or an edge remains.

Reference: pg. 18 - 19

Testing for Isomorphism

Algorithm for General Tree Isomorphism

Reference: pg. 21

Testing for Isomorphism

Partitioning Spanning Trees

Place T* in a subset S1

For each child T of T*

For each subset Si

If T is isomorphic to a tree in Si, place T in Si

Otherwise, create a new subset for T

Find the children of the children of T* and repeat

Continue until all trees have been partitioned

Reference: pg. 22

Partitioning Spanning Trees

Reference: pg. 23

Partitioning Spanning Trees

Definitions

I(G)

The number of isomorphism classes of the spanning trees of G

pk(n)

The number of partitions of the integer n into at most k parts

Reference: pg. 28

Finding a Closed Formula for I(Ks,t)

Useful Counting Tools

The number of ways to arrange n unlabeled balls into kunlabeled buckets is given by pk(n).

At least two buckets nonempty: pk(n) - 1

The number of ways to arrange n unlabeled balls into klabeled buckets is given by C(n+k-1, n).

At least two buckets nonempty: C(n+k-1, n) - k

Reference: pg. 28 - 29

Finding a Closed Formula for I(Ks,t)

Configurations of Ks,t

A spanning tree of Ks,t belongs to one of three disjoint sets

The center is a vertex in the s-set

The center is a vertex in the t-set

The center is an edge between the two sets

We determine the number of nonisomorphic trees in each set and then sum to find I(Ks,t)

Reference: pg. 29

Finding a Closed Formula for I(Ks,t)

Configurations of K2,t

Center in 2-set

No such tree

Reference: pg. 32

Finding a Closed Formula for I(Ks,t)

Configurations of K2,t

Center in t-set

p2(t-1) – 1 trees

Reference: pg. 32 - 33

Finding a Closed Formula for I(Ks,t)

Configurations of K2,t

Center is an edge

Only one such tree

Reference: pg. 33

Finding a Closed Formula for I(Ks,t)

Summing Across the Sets

Summing across the disjoint sets yields

I(K2,t) = 0 + p2(t-1) – 1 + 1 = p2(t-1), t2.

Similarly, we can find

I(K3,t) = sum{k=2 to t-2}(p2(k)) + p3(t-1) +2, t4.

Reference: pg. 29

Finding a Closed Formula for I(Ks,t)

Nicer Formulas

Using the generating function for pk(n), we can simplify the formulas to:

I(K2,t) = ⌈t/2⌉, t2

I(K3,t) = [1/3(t2 + t + 1)], t4

Reference: pg. 36 - 41

Finding a Closed Formula for I(Ks,t)