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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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slide1

Partitioning the Labeled Spanning Trees

of an Arbitrary Graph into

Isomorphism Classes

Austin Mohr

outline
Outline

Problem Description

Generating Spanning Trees

Testing for Isomorphism

Partitioning Spanning Trees

Some Results

Finding a Closed Formula for I(Ks,t)

definitions
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 k 2 3
Spanning Trees of K2,3

Reference: pg. 5

Problem Description

definitions1
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
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 k 2 3
“Tree of trees” for K2,3

Reference: pg. 8

Generating Spanning Trees

definitions2
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
Finding the Children of a Tree

Reference: pg. 11

Generating Spanning Trees

rooted tree isomorphism
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 isomorphism1
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
Sample Run of Algorithm for Rooted Trees

Reference: pg. 17

Testing for Isomorphism

general tree 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

definitions3
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
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
Algorithm for General Tree Isomorphism

Reference: pg. 21

Testing for Isomorphism

partitioning spanning trees
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

slide22

Reference: pg. 23

Partitioning Spanning Trees

definitions4
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
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 k s 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 k 2 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 k 2 t1
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 k 2 t2
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 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
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)