[PPT] - Counting the number of spanning tree Pied Piper Department of PowerPoint Presentation

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Counting the number of spanning tree

Pied Piper Department of Computer Science and Engineering Shanghai Jiao Tong University

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目录 Contents 1

Co Comp mplet lete e Gr Grap aph

2

Proof of the Lemma

3

Arbitrary Graph

4

Proof with Matrics

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Kn : a complete graph with n vertices

Fo For each r each n n ≥2， th the num e number of s ber of spanning panning tr tree ees of s of Kn Kn eq equals n uals nn-2

Cayley’s formula

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vertebrates

De Defini nition ion: Kn : A spanning tree of the complete graph Kn ver erte tebr brate ate : consider a Kn , mark one vertex by a circle, one vertex by a square . : the set of all vertebrates (consider value of n) Ch Chor

rd : a unique path connect and .

A vertebrate on 19 vertices

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From each given spanning tree, we can create n2 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

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Pr Proo

of of th

f of the lemm e lemma. a.

St Starts rts fr from

m W

W

Vertices of the chord

rdered by magnitude :

3 4 7 8 9 14 15 as they appeared in chord: 8 4 14 9 3 7 15 1st

st st

step ep ---

-- define graph P : we make an directed edge from each vertex in 1st line

to the vertex below it in the 2nd line. Vertebrate W 1st step

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2nd

nd st

step ep ---

-- we remove the edges of the chord from W, it splits into components.

Vertebrate W 2nd step

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3rd

rd st

step ep ---

-- we direct the edges of the components point to the vertex of the chord.

Vertebrate W 3rd step

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4th

th st

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

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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

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Starts from mapping f

A mapping f 1st

st 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.

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2nd

nd st

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.

rdered by magnitude :

3 4 7 8 9 14 15 Vertices they map to: 8 4 14 9 3 7 15 A mapping f

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rdered by magnitude :

3 4 7 8 9 14 15 Vertices they map to: 8 4 14 9 3 7 15 3rd

rd st

step ep --- define a undirected graph P P (a path) : with the vertices mentioned in 2nd step, and in order of 2nd line. A mapping f 3rd step

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4th

th st

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

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目录 Contents 1

Complete Graph

2

Proof of the Lemma

3

Ar Arbi bitr trar ary Gr y Grap aph

4

Proof of the theorem

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▪ What’s the number of spanning trees of an arbitrary graph?

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▪ Let G be an arbitrary graph with vertices 1, 2, …, n, n >= 2, and with edges 𝑓1, 𝑓2, …, 𝑓𝑛. ▪ We introduce an n*n matrix Q, called the Laplace matrix of the graph G, whose elements 𝑟𝑗𝑘 are determined by the following formula:

Laplace matrix

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Laplace matrix

3 1 2 2 Graph G Laplace matrix 1 2 3 4 1 2 3 4

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Laplace matrix

3 −1 −1 1 −1 −1 −1 −1 2 −1 −1 2 Graph G Laplace matrix Q 1 2 3 4 1 2 3 4 Ob Obse servat ation ion Th The e su sum m of

f t

the ro e rows s of

f th

the L e Lap aplace ace mat atrix ix is s the zer e zero

vec

ector

r

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▪ Definition

▪ Let 𝑅𝑗𝑘 denote the (n - 1) * (n - 1) matrix arising from the laplace matrix Q by deleting the 𝑗𝑢ℎ row and the 𝑘𝑢ℎ column.

Laplace matrix

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▪ Definition

▪ Let 𝑅𝑗𝑘 denote the (n - 1) * (n - 1) matrix arising from the Laplace matrix Q by deleting the 𝑗𝑢ℎ row and the 𝑘𝑢ℎ column.

Laplace matrix

3 −1 −1 1 −1 −1 −1 −1 2 −1 −1 2 1 2 −1 −1 2 3 −1 −1 −1 −1 −1 2 𝑅 𝑅23 𝑅11

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▪ Theorem

▪ 𝑈 𝐻 is the number of spanning tree in graph G. ▪ For every graph G, we have 𝑈 𝐻 = det 𝑅11

#Spanning Tree

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▪ Theorem ▪ For every graph G, we have 𝑼 𝑯 = 𝐞𝐟𝐮 𝑹𝟐𝟐

▪ 𝐵𝑑𝑢𝑣𝑏𝑚𝑚𝑧, 𝑈 𝐻 = | det 𝑅𝑗𝑘| holds for any 𝑗, 𝑘 ∈ {1, 2, … , 𝑜}. (won’t prove here)

#Spanning Tree

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▪ Theorem ▪ For every graph G, we have 𝑼 𝑯 = 𝐞𝐟𝐮 𝑹𝟐𝟐

▪ 𝑈 𝐻 = | det 𝑅𝑗𝑘| holds for any two indices 𝑗, 𝑘 ∈ {1, 2, … , 𝑜}. (won’t prove here)

▪ Take the figure before as an example

#Spanning Tree

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#Spanning Tree

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#Spanning Tree

3

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#Spanning Tree

3 −1 −1 1 −1 −1 −1 −1 2 −1 −1 2 Q:

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#Spanning Tree

3 −1 −1 1 −1 −1 −1 −1 2 −1 −1 2 𝑅 1 2 −1 −1 2 𝑅11

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#Spanning Tree

3 −1 −1 1 −1 −1 −1 −1 2 −1 −1 2 1 2 −1 −1 2 𝑅11 𝑅 det 𝑅11 = 1 ∗ 2 ∗ 2 − −1 ∗ −1 − 0 + 0 = 3

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目录 Contents 1

Complete Graph

2

Proof of the Lemma

3

Arbitrary Graph

4

Pr Proo

of of th

f of the theo e theorem rem

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