Title: PRESENTATION NAME
1Partitioning the Labeled Spanning Trees of an
Arbitrary Graph into Isomorphism Classes
Austin Mohr
2Outline
- Problem Description
- Generating Spanning Trees
- Testing for Isomorphism
- Partitioning Spanning Trees
- Some Results
- Finding a Closed Formula for I(Ks,t)
3Problem Description
4Definitions
- 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
Problem Description
Reference pg. 3 - 4
5Spanning Trees of K2,3
Problem Description
Reference pg. 5
6Generating Spanning Trees
7Definitions
Let G be a graph on n vertices, H?G, e be an edge
of G, and T be a spanning tree of G.
- 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
8Regarding ?(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 Trees
Reference pg. 7
9Tree of trees for K2,3
Reference pg. 8
Generating Spanning Trees
10Definitions
Let G be a graph on n vertices, e be an edge of
G, and T be a spanning tree of G.
- 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
11Finding the Children of a Tree
Reference pg. 11
Generating Spanning Trees
12Testing for Isomorphism
13Rooted Tree Isomorphism
We first consider the simpler problem of
determining when two rooted trees are isomorphic.
Testing for Isomorphism
Reference pg. 14
14Rooted 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
15Sample Run of Algorithm for Rooted Trees
Reference pg. 17
Testing for Isomorphism
16General 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
17Definitions
Let u and v be vertices of a graph G.
- 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
Testing for Isomorphism
18Finding the Center of a Tree
- Theorem (Jordan) The center of a tree is either
a vertex or an edge. - Jordans 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
19Algorithm for General Tree Isomorphism
Reference pg. 21
Testing for Isomorphism
20Partitioning Spanning Trees
21Partitioning 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
22Reference pg. 23
Partitioning Spanning Trees
23Some Results
24Finding a Closed Formula for I(Ks,t)
25Definitions
- 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)
26Useful 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(nk-1, n). - At least two buckets nonempty C(nk-1, n) - k
Reference pg. 28 - 29
Finding a Closed Formula for I(Ks,t)
27Configurations 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)
28Configurations of K2,t
Center in 2-set No such tree
Reference pg. 32
Finding a Closed Formula for I(Ks,t)
29Configurations of K2,t
Center in t-set p2(t-1) 1 trees
Reference pg. 32 - 33
Finding a Closed Formula for I(Ks,t)
30Configurations of K2,t
Center is an edge Only one such tree
Reference pg. 33
Finding a Closed Formula for I(Ks,t)
31Summing 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) sumk2 to t-2(p2(k)) p3(t-1) 2,
t?4.
Reference pg. 29
Finding a Closed Formula for I(Ks,t)
32Nicer 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)
33Questions?