Counting the number of spanning tree. Pied Piper Department of Computer Science and Engineering Shanghai Jiao Tong University

Transcription

1 Counting the number of spanning tree Pied Piper Department of Computer Science and Engineering Shanghai Jiao Tong University

2 目录 Contents 1 Complete Graph 2 Proof of the Lemma 3 Arbitrary Graph 4 Proof with Matrics

3 Cayley s formula Kn : a complete graph with n vertices For each n 2, the number of spanning trees of Kn equals n n-2

4 vertebrates Definition: K n : A spanning tree of the complete graph K n vertebrate : consider a K n, mark one vertex by a circle, one vertex by a square. : the set of all vertebrates (consider value of n) Chord : a unique path connect and. A vertebrate on 19 vertices

5 From each given spanning tree, we can create n 2 vertebrates. Therefore the number of all spanning trees equals Lemma. There exists a bijection F between the of all vertebrates and the set of all mappings of the vertex V to itself. Since = all mappings of an n-element set to itself = according to the lemma, therefore the number of spanning trees is

6 Proof of the lemma. Starts from W Vertices of the chord ordered by magnitude : as they appeared in chord: st step --- define graph P : we make an directed edge from each vertex in 1 st line to the vertex below it in the 2 nd line. Vertebrate W 1 st step

7 2 nd step --- we remove the edges of the chord from W, it splits into components. 2 nd step Vertebrate W

8 3 rd step --- we direct the edges of the components point to the vertex of the chord. 3 rd step Vertebrate W

9 4 th step --- define a graph G : with vertex set V, its edges are all directed edges of the components, plus the edges of the graph P. Step graph G

10 the resulting directed graph G, is a graph of a mapping: For each, we set, where there exists an edge { i, j } in G. graph G

11 Starts from mapping f 1 st step --- draw f into a directed graph( may be disjoint ), for each components, there must exists exactly one circle. Proof : For each vertex in a mapping, there is exactly 1 edges going from it. situation A : if there exists 2 or more circles, at least 1 vertex have more than 1 edges going from it, which is not possible in a mapping. Situation B : if the component is acyclic, at least 1 vertex does not have an edge going from it. A mapping f

12 2 nd step --- we extract each circle from components, and for the vertices related to the circles, we write them in order of magnitude, and write the vertex it maps below it. ordered by magnitude : Vertices they map to: A mapping f

13 ordered by magnitude : Vertices they map to: rd step --- define a undirected graph P (a path) : with the vertices mentioned in 2 nd step, and in order of 2 nd line. 3 rd step A mapping f

14 4 th step --- we add the rest vertices and edges of the mapping into P, change the direct edges into undirected ones.

15 目录 Contents 1 Complete Graph 2 Proof of the Lemma 3 Arbitrary Graph 4 Proof of the theorem

16 What s the number of spanning trees of an arbitrary graph?

17 Laplace matrix Let G be an arbitrary graph with vertices 1, 2,, n, n >= 2, and with edges e 1, e 2,, e m. We introduce an n*n matrix Q, called the Laplace matrix of the graph G, whose elements q ij are determined by the following formula:

18 Laplace matrix Graph G Laplace matrix

19 Laplace matrix Graph G Laplace matrix Q Observation The sum of the rows of the Laplace matrix is the zero vector

20 Laplace matrix Definition Let Q ij denote the (n - 1) * (n - 1) matrix arising from the laplace matrix Q by deleting the i th row and the j th column.

21 Laplace matrix Definition Let Q ij denote the (n - 1) * (n - 1) matrix arising from the Laplace matrix Q by deleting the i th row and the j th column Q Q 11 Q 23

22 #Spanning Tree Theorem T G is the number of spanning tree in graph G. For every graph G, we have T G = det Q 11

23 #Spanning Tree Theorem For every graph G, we have T G = det Q 11 Actually, T G = det Q ij holds for any i, j {1, 2,, n}. (won t prove here)

24 #Spanning Tree Theorem For every graph G, we have T G = det Q 11 T G = det Q ij holds for any two indices i, j {1, 2,, n}. (won t prove here) Take the figure before as an example

25 #Spanning Tree

26 #Spanning Tree 3

27 #Spanning Tree Q:

28 #Spanning Tree Q Q

29 #Spanning Tree Q Q det Q 11 = = 3

30 目录 Contents 1 Complete Graph 2 Proof of the Lemma 3 Arbitrary Graph 4 Proof of the theorem

31 Proofs working with determinants We proceed by induction. To make proofs work, we strengthen the inductive hypothesis and show that the theorem also holds for multigraphs. What does the Laplacian of a multigraph look like? If two vertices u and w are joined by m edges, then quv = m. quu is the degree of the vertex u,

32 Proofs working with determinants New formula to be relied on: Lemma1: T(G) = T(G e) + T(G : e) where e is an arbitrary edge of the graph G.

33 Proofs working with determinants New formula to be relied on: Lemma1: T(G) = T(G e) + T(G : e) where e is an arbitrary edge of the graph G. G e denotes the graph obtained by deleting that edge e in G.

34 Proofs working with determinants New formula to be relied on: Lemma1: T(G) = T(G e) + T(G : e) where e is an arbitrary edge of the graph G. G e denotes the graph obtained by deleting that edge e in G. G : e the one obtained by contracting the edge e.

35 Proofs working with determinants New formula to be relied on: Lemma1: T(G) = T(G e) + T(G : e) where e is an arbitrary edge of the graph G. G e denotes the graph obtained by deleting that edge. G : e the one obtained by contracting the edge e. {1, 2} 5 3 4

36 Proofs working with determinants In order to see why the Lemma holds, we divide the spanning trees of G into two classes. Remember the Lemma1: T(G) = T(G e) + T(G : e)

37 Proofs working with determinants In order to see why the Lemma holds, we divide the spanning trees of G into two classes. The spanning trees that do not contain the edge e. (exactly the spanning trees of the graph G e, T(G e).) Remember the Lemma1: T(G) = T(G e) + T(G : e)

38 Proofs working with determinants In order to see why the Lemma holds, we divide the spanning trees of G into two classes. The spanning trees that do not contain the edge e. (exactly the spanning trees of the graph G e, T(G e).) The spanning trees that do contain the edge e. (one-to-one correspondence with the spanning trees of G : e, as indicated in the following picture, T(G : e).) Remember the Lemma1: T(G) = T(G e) + T(G : e)

39 Proofs working with determinants Here we give some symbols and their meaning: Q : the Laplacian of G e. Q : the Laplacian of G : e Q 11 : the (n 1) (n 1) matrix arising from the matrix Q by deleting the first row and the first column. Q 11 : the (n 1) (n 1) matrix arising from the matrix Q by deleting the first row and the first column. Q 11,22 : the (n 2) (n 2) matrix arising from the matrix Q by deleting the first and second row and the first and second column. Q 11 : the (n 2) (n 2) matrix arising from the matrix Q by deleting the first row and the first column.

40 Proofs working with determinants Assume edge e has endvertices 1 and 2 Observation1: Q 11 arises from Q 11 by subtracting 1 from the element in the upper left corner. Observation2: Q 11 = Q 11,22

41 Proofs working with determinants,,,, = Q 11 = Q 11

42 Proofs working with determinants Now we are ready to show by induction on m that T(G) = detq 11 holds for every multigraph G with at most m edges.

43 Proofs working with determinants If in a multigraph G the vertex number 1 is not incident to any edge, then we have T(G) = 0.

44 Proofs working with determinants If in a multigraph G the vertex number 1 is not incident to any edge, then we have T(G) = 0. And detq 11 =0 Q Q

45 Proofs working with determinants If in a multigraph G the vertex number 1 is not incident to any edge, then we have T(G) = 0. And detq 11 =0 Q Sum Q Sum 0 0 0

46 Proofs working with determinants If in a multigraph G the vertex number 1 is not incident to any edge, then we have T(G) = 0. And detq 11 =0 Q Sum detq 11 =0 T(G) = detq 11 =0

47 Proofs working with determinants If in a multigraph G the vertex number 1 is not incident to any edge, then we have T(G) = 0. (In particular, when m = 0, theorem holds, this is our base case.).

48 Proofs working with determinants If the vertex number 1 is incident to at least one edge, then we fix one such edge, let us call it e, Assume e has endvertices 1 and 2.

49 Proofs working with determinants If the vertex number 1 is incident to at least one edge, then we fix one such edge, call it e. Assume e has endvertices 1 and 2. According to Observation2, we have the following formula : Remember the Observation2 : Q 11 = Q11,22

50 Proofs working with determinants If the vertex number 1 is incident to at least one edge, then we fix one such edge, call it e. Assume e has endvertices 1 and 2. According to Observation2, we have the following formula: the matrix Q 11 arises from Q 11 by adding the vector e1 = (1, 0, 0,..., 0) to the first row. R: vertor e1 as it s first row, the remaning part agrees with Q11 Remember the Observation2 : Q 11 = Q11,22

51 Proofs working with determinants

52 Proofs working with determinants From the three formula above, hence

53 Proofs working with determinants From the three formula above, hence So we have T G = det Q 11 This completes the inductive step and hence we proof it.

54 谢谢! Q & A