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

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

Problem Description

 Spanning tree T of graph G › T is a tree with E(T) ⊆ E(G) and V(T)=V(G)  Isomorphic trees T 1 and T 2 › There exists a mapping f where the edge uv ∈ T 1 if and only if the edge f(u)f(v) ∈ T 2 Problem Description Reference: pg

Problem Description Reference: pg. 5

Generating Spanning Trees

 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 Trees Reference: 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.

 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 Trees Reference: pg. 7

Reference: pg. 8 Generating Spanning Trees

 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 Trees Reference: pg. 8 Let G be a graph on n vertices, e be an edge of G, and T be a spanning tree of G.

Reference: pg. 11 Generating Spanning Trees

Testing for Isomorphism

Reference: pg. 14 We first consider the simpler problem of determining when two rooted trees are isomorphic.

 Given two rooted trees T 1 and T 2 on n vertices, a mapping f: V(T 1 ) → V(T 2 ) is an isomorphism if and only if for every vertex v ∈ V(T 1 ), the subtree of T 1 rooted at v is isomorphic to the subtree of T 2 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

Reference: pg. 17 Testing for Isomorphism

 To generalize the algorithm, we need a vertex u ∈ V(T 1 ) and v ∈ V(T 2 ) such that f(u) = v for every isomorphism f. › If found, we root T 1 at u, root T 2 at v, and use the previous algorithm › The center of each tree is suitable choice Reference: pg. 18 Testing for Isomorphism

 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. 18 Let u and v be vertices of a graph G. Testing for Isomorphism

 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 Testing for Isomorphism

Reference: pg. 21 Testing for Isomorphism

Partitioning Spanning Trees

 Place T* in a subset S 1  For each child T of T* › For each subset S i  If T is isomorphic to a tree in S i, place T in S i  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

Some Results

Finding a Closed Formula for I(K s,t )

 I(G) › The number of isomorphism classes of the spanning trees of G  p k (n) › The number of partitions of the integer n into at most k parts Reference: pg. 28 Finding a Closed Formula for I(K s,t )

 The number of ways to arrange n unlabeled balls into k unlabeled buckets is given by p k (n). › At least two buckets nonempty: p k (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) - k Reference: pg Finding a Closed Formula for I(K s,t )

 A spanning tree of K s,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(K s,t ) Reference: pg. 29 Finding a Closed Formula for I(K s,t )

Reference: pg. 32 Finding a Closed Formula for I(K s,t ) Center in 2-set No such tree

Reference: pg Finding a Closed Formula for I(K s,t ) Center in t-set p 2 (t-1) – 1 trees

Reference: pg. 33 Finding a Closed Formula for I(K s,t ) Center is an edge Only one such tree

 Summing across the disjoint sets yields K 2,t p 2 (t-1) – = p 2 (t-1),. I(K 2,t ) = 0 + p 2 (t-1) – = p 2 (t-1), t  2.  Similarly, we can find K 3,t I(K 3,t ) = sum{k=2 to t-2}(p 2 (k)) + p 3 (t-1) +2, t  4. Reference: pg. 29 Finding a Closed Formula for I(K s,t )

 Using the generating function for p k (n), we can simplify the formulas to: › I(K 2,t ) = ⌈ t/2 ⌉, t  2 › I(K 3,t ) = [1/3(t 2 + t + 1)], t  4 Reference: pg Finding a Closed Formula for I(K s,t )

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