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

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

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

3. Problem Description

4. 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

5. Spanning Trees of K2,3 Reference: pg. 5 Problem Description

6. Generating Spanning Trees

7. 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

8. 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

9. “Tree of trees” for K2,3 Reference: pg. 8 Generating Spanning Trees

10. 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

11. Finding the Children of a Tree Reference: pg. 11 Generating Spanning Trees

12. Testing for Isomorphism

13. Rooted Tree Isomorphism We first consider the simpler problem of determining when two rooted trees are isomorphic. Reference: pg. 14 Testing for Isomorphism

14. 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

15. Sample Run of Algorithm for Rooted Trees Reference: pg. 17 Testing for Isomorphism

16. 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

17. 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

18. 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

19. Algorithm for General Tree Isomorphism Reference: pg. 21 Testing for Isomorphism

20. Partitioning Spanning Trees

21. 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

22. Reference: pg. 23 Partitioning Spanning Trees

23. Some Results

24. Finding a Closed Formula for I(Ks,t)

25. 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)

26. 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)

27. 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)

28. Configurations of K2,t Center in 2-set No such tree Reference: pg. 32 Finding a Closed Formula for I(Ks,t)

29. 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)

30. Configurations of K2,t Center is an edge Only one such tree Reference: pg. 33 Finding a Closed Formula for I(Ks,t)

31. 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)

32. 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)

33. Questions?