Counting the number of spanning tree Pied Piper Department of Computer Science and Engineering Shanghai Jiao Tong University
目录 Contents Co Comp mplet lete e Gr Grap aph 1 2 Proof of the Lemma Arbitrary Graph 3 4 Proof with Matrics
Cayley’s formula Kn : a complete graph with n vertices n ≥ 2 ， th For each Fo r each n the num e number of s ber of spanning panning uals n n-2 tr tree ees of s of Kn Kn eq equals n
vertebrates De Defini nition ion: K n : A spanning tree of the complete graph K n ver erte tebr brate ate : consider a K n , mark one vertex by a circle, one vertex by a square . : the set of all vertebrates (consider value of n) Ch Chor ord : a unique path connect and . A vertebrate on 19 vertices
From each given spanning tree, we can create n 2 vertebrates. Therefore the number of all spanning trees equals Lem emma. a. 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
Pr Proo oof of th f of the lemm e lemma. a. St Starts rts fr from om W W Vertices of the chord ordered by magnitude : 3 4 7 8 9 14 15 as they appeared in chord: 8 4 14 9 3 7 15 st st --- define graph P : we make an directed edge from each vertex in 1 st line 1 st step ep --- to the vertex below it in the 2 nd line. 1 st step Vertebrate W
nd st 2 nd step ep --- --- we remove the edges of the chord from W, it splits into components. 2 nd step Vertebrate W
rd st 3 rd step ep --- --- we direct the edges of the components point to the vertex of the chord. 3 rd step Vertebrate W
th st 4 th step ep --- --- 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 4 --- graph G
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
Starts from mapping f st st 1 st step ep --- 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
nd st 2 nd step ep --- 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 : 3 4 7 8 9 14 15 Vertices they map to: 8 4 14 9 3 7 15 A mapping f
ordered by magnitude : 3 4 7 8 9 14 15 Vertices they map to: 8 4 14 9 3 7 15 rd st 3 rd step ep --- define a undirected graph P P (a path) : with the vertices mentioned in 2 nd step, and in order of 2 nd line. 3 rd step A mapping f
th st 4 th step ep --- we add the rest vertices and edges of the mapping into P, change the direct edges into undirected ones.
目录 Contents Complete Graph 1 2 Proof of the Lemma Ar Arbi bitr trar ary Gr y Grap aph 3 4 Proof of the theorem
▪ What’s the number of spanning trees of an arbitrary graph?
Laplace matrix ▪ Let G be an arbitrary graph with vertices 1, 2, … , n, n >= 2, and with edges 𝑓 1 , 𝑓 2 , … , 𝑓 𝑛 . ▪ We introduce an n*n matrix Q, called the Laplace matrix of the graph G, whose elements 𝑟 𝑗𝑘 are determined by the following formula:
Laplace matrix 1 2 3 4 1 3 2 1 3 2 4 2 Graph G Laplace matrix
Laplace matrix 1 2 3 4 1 3 −1 −1 −1 2 0 0 −1 1 −1 0 2 −1 3 −1 0 −1 2 4 Graph G Laplace matrix Q Obse Ob servat ation ion Th The e su sum m of of t the ro e rows s of of th the L e Lap aplace ace mat atrix ix is s the zer e zero o vec ector or
Laplace matrix ▪ Definition ▪ Let 𝑅 𝑗𝑘 denote the (n - 1) * (n - 1) matrix arising from the laplace matrix Q by deleting the 𝑗 𝑢ℎ row and the 𝑘 𝑢ℎ column.
Laplace matrix ▪ Definition ▪ Let 𝑅 𝑗𝑘 denote the (n - 1) * (n - 1) matrix arising from the Laplace matrix Q by deleting the 𝑗 𝑢ℎ row and the 𝑘 𝑢ℎ column. 3 −1 −1 −1 3 −1 −1 −1 1 0 0 1 0 0 −1 0 2 −1 2 −1 −1 0 0 −1 −1 0 −1 2 −1 2 −1 0 0 2 𝑅 𝑅 11 𝑅 23
#Spanning Tree ▪ Theorem ▪ 𝑈 𝐻 is the number of spanning tree in graph G. ▪ For every graph G, we have 𝑈 𝐻 = det 𝑅 11
#Spanning Tree ▪ Theorem ▪ For every graph G, we have 𝑼 𝑯 = 𝐞𝐟𝐮 𝑹 𝟐𝟐 ▪ 𝐵𝑑𝑢𝑣𝑏𝑚𝑚𝑧, 𝑈 𝐻 = | det 𝑅 𝑗𝑘 | holds for any 𝑗, 𝑘 ∈ {1, 2, … , 𝑜} . (won’t prove here)
#Spanning Tree ▪ Theorem ▪ For every graph G, we have 𝑼 𝑯 = 𝐞𝐟𝐮 𝑹 𝟐𝟐 ▪ 𝑈 𝐻 = | det 𝑅 𝑗𝑘 | holds for any two indices 𝑗, 𝑘 ∈ {1, 2, … , 𝑜} . (won’t prove here) ▪ Take the figure before as an example
#Spanning Tree
#Spanning Tree 3
#Spanning Tree 3 −1 −1 −1 Q: −1 1 0 0 −1 0 2 −1 −1 0 −1 2
#Spanning Tree 3 −1 −1 −1 𝑅 −1 1 0 0 −1 0 2 −1 −1 0 −1 2 𝑅 11 1 0 0 0 2 −1 0 −1 2
#Spanning Tree 3 −1 −1 −1 𝑅 −1 1 0 0 −1 0 2 −1 −1 0 −1 2 𝑅 11 1 0 0 0 2 −1 0 −1 2 det 𝑅 11 = 1 ∗ 2 ∗ 2 − −1 ∗ −1 − 0 + 0 = 3
目录 Contents Complete Graph 1 2 Proof of the Lemma Arbitrary Graph 3 Pr Proo oof of th f of the theo e theorem rem 4
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 q uv = − m . • q uu is the degree of the vertex u ,
Proofs working with determinants ▪ New formula to be relied on: a1: T(G) = T(G − e) + T(G : e) Le Lemma1 ▪ where e is an arbitrary edge of the graph G.
Proofs working with determinants ▪ New formula to be relied on: a1: T(G) = T(G − e) + T(G : e) Le Lemma1 ▪ where e is an arbitrary edge of the graph G. ▪ G − e denotes the graph obtained by deleting that edge e in G.
Proofs working with determinants ▪ New formula to be relied on: a1: T(G) = T(G − e) + T(G : e) Le Lemma1 ▪ 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.
Proofs working with determinants ▪ New formula to be relied on: a1: T(G) = T(G − e) + T(G : e) Le Lemma1 ▪ 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
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)
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)
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)
Proofs working with determinants ▪ Here we give some symbols and their meaning : : the Laplacian of G − e. ▪ Q ′ ▪ 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.
Proofs working with determinants Assume edge e has endvertices 1 and 2 ▪ Ob Obse serva rvation1 tion1: Q ′ 11 arises from Q 11 by subtracting 1 from the element in the upper left corner. ▪ Ob Obse serva rvation2 tion2: Q ′ ′ 11 = Q 11,22
Proofs working with determinants , , , , = 𝑅′ 11 = 𝑅′′ 11
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.
Recommend
More recommend
