Partitioning the Labeled Spanning TreesPartitioning the Labeled Spanning Treesof an Arbitrary Graph into of an Arbitrary Graph into Isomorphism ClassesIsomorphism ClassesAustin Mohr

Outline Problem Description Generating Spanning Trees Testing for Isomorphism Partitioning Spanning Trees Some Results Finding a Closed Formula for I(Ks,t)

Problem DescriptionProblem Description

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

Problem DescriptionProblem DescriptionReference: pg. 3 - 4Reference: pg. 3 - 4

Spanning Trees of K2,3

Problem DescriptionProblem DescriptionReference: pg. 5Reference: pg. 5

Generating Spanning TreesGenerating Spanning Trees

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))

Generating Spanning TreesGenerating Spanning TreesReference: pg. 6Reference: pg. 6

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

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*

Generating Spanning TreesGenerating Spanning TreesReference: pg. 7Reference: pg. 7

“Tree of trees” for K2,3

Reference: pg. 8Reference: pg. 8 Generating Spanning TreesGenerating 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

Generating Spanning TreesGenerating Spanning TreesReference: pg. 8Reference: pg. 8

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

Finding the Children of a Tree

Reference: pg. 11Reference: pg. 11 Generating Spanning TreesGenerating Spanning Trees

Testing for IsomorphismTesting for Isomorphism

Rooted Tree Isomorphism

Testing for IsomorphismTesting for IsomorphismReference: pg. 14Reference: pg. 14

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

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. 14Reference: pg. 14 Testing for IsomorphismTesting for Isomorphism

Sample Run of Algorithm for Rooted Trees

Reference: pg. 17Reference: pg. 17 Testing for IsomorphismTesting 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. 18Reference: pg. 18 Testing for IsomorphismTesting 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

Reference: pg. 18Reference: pg. 18

Let u and v be vertices of a graph G.

Testing for IsomorphismTesting 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 - 19Reference: pg. 18 - 19 Testing for IsomorphismTesting for Isomorphism

Algorithm for General Tree Isomorphism

Reference: pg. 21Reference: pg. 21 Testing for IsomorphismTesting for Isomorphism

Partitioning Spanning TreesPartitioning 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

partitionedReference: pg. 22Reference: pg. 22 Partitioning Spanning TreesPartitioning Spanning Trees

Reference: pg. 23Reference: pg. 23 Partitioning Spanning TreesPartitioning Spanning Trees

Some ResultsSome Results

Finding a Closed Formula for Finding a Closed Formula for I(KI(Ks,ts,t))

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. 28Reference: pg. 28 Finding a Closed Formula for Finding a Closed Formula for I(KI(Ks,ts,t))

Useful Counting Tools The number of ways to arrange n

unlabeled balls into k unlabeled buckets is given by pk(n).

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

The number of ways to arrange n unlabeled balls into k labeled buckets is given by C(n+k-1, n).› At least two buckets nonempty: C(n+k-1, n) -

kReference: pg. 28 - 29Reference: pg. 28 - 29 Finding a Closed Formula for Finding a Closed Formula for I(KI(Ks,ts,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. 29Reference: pg. 29 Finding a Closed Formula for Finding a Closed Formula for I(KI(Ks,ts,t))

Configurations of K2,t

Reference: pg. 32Reference: pg. 32 Finding a Closed Formula for Finding a Closed Formula for I(KI(Ks,ts,t))

Center in 2-setCenter in 2-setNo such treeNo such tree

Configurations of K2,t

Reference: pg. 32 - 33Reference: pg. 32 - 33 Finding a Closed Formula for Finding a Closed Formula for I(KI(Ks,ts,t))

Center in Center in tt-set-setpp22(t(t-1-1)) – 1 trees – 1 trees

Configurations of K2,t

Reference: pg. 33Reference: pg. 33 Finding a Closed Formula for Finding a Closed Formula for I(KI(Ks,ts,t))

Center is an edgeCenter is an edgeOnly one such treeOnly one such tree

Summing Across the Sets Summing across the disjoint sets yields

I(KK2,t2,t) = 0 + pp22(t(t-1-1)) – 1 + 1 = – 1 + 1 = pp22(t(t-1-1), ), t2.. Similarly, we can find

I(KK3,t3,t) = sum{k=2 to t-2}(p2(k)) + p3(t-1) +2, t4.

Reference: pg. 29Reference: pg. 29 Finding a Closed Formula for Finding a Closed Formula for I(KI(Ks,ts,t))

Nicer Formulas Using the generating function for pk(n),

we can simplify the formulas to:› I(K2,t) = ⌈t/2⌉, t2› I(K3,t) = [1/3(t2 + t + 1)], t4

Reference: pg. 36 - 41Reference: pg. 36 - 41 Finding a Closed Formula for Finding a Closed Formula for I(KI(Ks,ts,t))

Questions?Questions?