graph theory chapter 3 tree

Graph TheoryGraph Theory

Chapter 3 Chapter 3 Tree Tree

大葉大學大葉大學 (Da-Yeh Univ.)(Da-Yeh Univ.) 資訊工程系資訊工程系 (Dept. CSIE)(Dept. CSIE)黃鈴玲黃鈴玲 (Lingling Huang)(Lingling Huang)

Copyright Copyright 黃鈴玲黃鈴玲Ch3-Ch3-22

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3.1 Properties of trees3.1 Properties of trees

3.2 Rooted trees3.2 Rooted trees

3.3 Depth-first search3.3 Depth-first search

3.5 Breadth-first search3.5 Breadth-first search

3.6 The minimum spanning tree 3.6 The minimum spanning tree problem problem



Definition:Definition:treetree:: a connected graph without cycles. a connected graph without cycles.

( ( every edge is a bridge) every edge is a bridge)forestforest:: a graph (connected or a graph (connected or

disconnected) without cycles .disconnected) without cycles .A A spanning treespanning tree of a graph of a graph GG is a is a

spanning subgraph of spanning subgraph of GG that is a tree . that is a tree .

See Figure 3.1.


Thm 3.1:Thm 3.1: Let Let TT be a tree of order be a tree of order pp and size and size qq,, then then qq==pp1 .1 .

Pf.

By strong from of induction on p.

p = 1 : k1 is the only tree of order 1, |E(k1)| = 0. ok!

Assume the result is true for all trees of order p < k

(k 2 , k ).

consider p = k :

Let T be a tree of order p = k, and e be an edge of T.

Let T1, T2 be the two components of T{e}, and |V(T1)| = p1, |V(T2)| = p2 (p1 +p2 = p)


∵∵ pp11,, pp22 << kk

∴∴|E(T|E(T11)| = )| = pp111, |E(T1, |E(T22)| = )| = pp221 by assumption.1 by assumption.

|E(T)| = |E(T)| = pp111 + 1 + pp221 +1 = 1 +1 = pp11++pp22 1 = 1 = pp 1.1.

By induction, the result is true for all trees. By induction, the result is true for all trees. ##


Cor 3.1:Let G be a graph of order p, the following are equivalent : (i) G is a tree. (ii) G is connected and has size p1. (iii) G has size p1 and no cycles.

Pf. (i) (ii) (by Thm 3.1) (ii) (iii) ( 證明 connected + size p1 no cycle) 反證，若有 cycle 可扣掉邊後製造 tree ，

邊數不對 (iii) (i) ( 證明 size p1 + no cycle connected) 反證，每一個 component 本身都是 connected + no cycle tree 邊數總和不對


Thm 3.2:Thm 3.2:

Every nontrivial tree contains at Every nontrivial tree contains at least two end-vertices.least two end-vertices.

Pf.

Suppose T is a tree of order p and size q, ∵ ∵ T is connected and nontrivial. is connected and nontrivial. ∴ ∴ degdeg((vv) ) 1, 1, vv VV((TT).). 222)deg( p - q v

V(T)v

∵

∴ ∴ There are at least two vertices of There are at least two vertices of degree 1.degree 1.##


Thm 3.3:Thm 3.3:

If If TT is a tree, is a tree, uu≠≠vv VV((TT), then there is ), then there is exactly one exactly one uu--vv path. path.

Pf. (Bondy)( 跟書上不同 ) Suppose T has two u-v paths, say P and Q. Since P≠≠Q, there is an edge e=xy E(P) but

eE(Q). Clearly the graph (P∪Q) – e is connected. It therefore contains an x-y path P’. But then P’ + e is a cycle in T, a contradiction.


Note:Note: TT: a tree.: a tree. If If uvuv E( E(TT), then ), then TT++uvuv has exactly one has exactly one

cycle cycle CC. (which must contains the edge . (which must contains the edge uvuv))

If some edge If some edge ee ( ( 未必是未必是 uvuv) is deleted ) is deleted from from CC, a tree results again. , a tree results again.

Thm 3.4 :

Let TT be a tree of order m, and let G be a graph with δ(G) m1. Then T is isomorphic to some subgraph of G. ( 證明跳過 )


HomeworkHomework

Exercise : 2, 5, 8, 11, 12, 14, 15, Exercise : 2, 5, 8, 11, 12, 14, 15, 16, 1916, 19

Exercise 8. Let T be a tree of order 21 having degree set {1, 3, 5, 6}. Suppose that T has 15 end-vertices and one vertex of degree 6. How many verticesof degree 5 does T have?


Exercise 15. Prove or disprove:(c) Every vertex of a tree is a cut-vertex.(d) A tree of order p3 has more cut-vertices then bridges.(e) There exist exactly two regular trees.


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3.2 Rooted Trees3.2 Rooted Trees

Definition:Definition: A A directed treedirected tree is an is an asymmetricasymmetric

digraph whose underlying graph is a digraph whose underlying graph is a tree.tree.

A directed tree A directed tree TT is called a is called a rooted treerooted tree if there exists a vertex if there exists a vertex rr of of TT, called the , called the rootroot, s. t. , s. t. vertex vertex vv of of TT, there is an , there is an rr--vv path in path in TT..


※※ 為方便起見，將為方便起見，將 treetree 的點分層畫出的點分層畫出 ::

rroot r 在 level 0level

height = 3tree 的 height 即最大的 level 值

0

1

2

3


Thm 3.5 :Thm 3.5 : A directed tree A directed tree TT is a rooted tree iff is a rooted tree iff

TT contains a vertex contains a vertex rr with id with id rr = 0, and = 0, and id id vv = 1 for all other vertices = 1 for all other vertices vv of of TT..

pf : () root 即 r, 每點往上一層只連接到一點 , 否則不是 tree, 故 id id vv = 1 = 1 vv rr. . () ( 證明 u1V(T), r-u1 path)

因 id id vv = 1 = 1 vv rr, 從 u1 開始往回走 , 最後一定停在 r, 否則表示有 cycle, 與 T 是 directed tree 矛盾 .


DefinitioDefinition:n:※※ parent

child

更上層 : ancestor

更下層 : descendant

※※r T

v maximal subtree of T rooted at v ( 由 v 及其子孫構成的 induced subtree)

※※因因 rooted tree rooted tree 中所有中所有 arcarc 都是向下都是向下 ,, 所以可省略方向性所以可省略方向性 ..

See Figure 3.8.


※ ※ leafleaf : a vertex with no children : a vertex with no children

internal vertexinternal vertex: : leaf leaf 之之 vertex.vertex.

※ ※ A rooted tree isA rooted tree is mm-ary -ary if every vertex has if every vertex has at at most most mm children. children.

mm = 2 = 2 binary binary

※ ※ complete complete mm-ary tree-ary tree: every vertex : every vertex has has exactly exactly mm children or no children. children or no children.

left child

right child


Thm 3.6 :Thm 3.6 : A complete A complete mm-ary tree with -ary tree with ii internal internal

vertices has order vertices has order m i m i +1.+1.

Pf.

每個 internal vertex 有 m 個 children ，故共有 m i m i 個個 children ( 每個 child 只有一個 parent ，所以只會被計算一次 ) ， root 沒有 parent ，還沒被計算到

∴∴共有共有 m i m i +1 +1 個點個點


Cor 3.6 :Cor 3.6 : Every Every complete binary treecomplete binary tree with with ii internal internal vertices has vertices has i+i+11 leaves. leaves.

pf. By Thm 3.6, m = 2, i internal vertex ∴ ∴ 點數點數 = 2= 2ii + 1 , + 1 , ii +1 leaves. +1 leaves.

Thm 3.7 :Thm 3.7 :

If If TT is a binary tree of height is a binary tree of height hh and and order order pp,,

then then hh+1 +1 pp 2 2hh+1+1 1. 1.Pf.

(hh+1 +1 pp) ) 當每個當每個 level level 只有一點時只有一點時 ((pp 2 2hh+1+1 –– 1) 1) 當當 tree tree 是是 complete complete 時時


Definition:Definition: A rooted tree A rooted tree TT of height of height hh is is balancedbalanced

if every leaf is at level if every leaf is at level hh or or hh1.1.

Cor 3.7 :Cor 3.7 :

If If TT is a binary tree of height is a binary tree of height hh and order and order pp,,

then then hh log log2 2 ((((pp+1)/2)+1)/2) . .Equality holds if Equality holds if TT is a balanced is a balanced complete binary tree.complete binary tree.

Pf.

pp 2 2hh+1+1 1 1 hh log log2 2 ((((pp+1)/2)+1)/2)


Pf of Cor 3.7: (continued)

If If TT is a balanced complete binary is a balanced complete binary tree,tree, then then pp > 1+2+2 > 1+2+222+…+2+…+2hh11 = 2 = 2hh 11 22hh 1 < 1 < p p 2 2hh+1+1 1 1 22hh11 < < ((pp+1)/2+1)/2 2 2hh

hh = = log log2 2 ((((pp+1)/2)+1)/2) ##


Homework: 4, 5

Exercise 5: A complete m-ary tree T of height h is called a full m-ary tree if every leaf is at the level h. If T has l leaves, show that h = logm l.


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GA spanning tree of a graph G is a tree that is a subgraph of G with vertex set V(G).

T1 T2T3

Some spanning trees of G :

3.3 Depth-first Search3.3 Depth-first Search


※ Many graph algorithms require a systematic method of visiting the vertices of a graph depth-first search (DFS, 深先搜尋 ) bread-first search (BFS, 廣先搜尋 )


Depth-first searchDepth-first search

Gv1 v2

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假設 G 用其 adjacency list 來表示，而 list 中，每個點的neighbor 依下標由小排到大

v11

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Step 1: visit v1


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Gv1 v2

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depth-first search index

dfi(v5)=3

DFS forest


Gv5 v13

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Gv5 v13

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rooted forest

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back edge


Algorithm 3.1 (DFS algorithm)1. dfi(v) 0 vV(G)2. i 1 /* 用來 assign 點的 dfi 值 */3. S /* 用來存放 DFS forest 的所有 arc */4. If dfi(r) 0 r V(G), then output S and stop;

otherwise, let r be the first vertex s. t. dfi(r) = 0 and let w r. /* r 是這個 component 的 tree 的 root */

5. dfi(w) i6. i i + 17. (search) 7.1. If dfi(v) = 0 for some v in the adjacency list of

w, then continue; otherwise, go to Step 7.4.

7.2 S S ∪ {(w, v)} and assign parent(v) w. 7.3 w v and return to Step 5. 7.4 If w r, then w parent(w) and return to Step

7.1; otherwise, return to Step 4.


Thm 3.8.Thm 3.8. Every back edge Every back edge ee of a graph of a graph GG

joins an ancestor and a descendant.joins an ancestor and a descendant.

In particular, if In particular, if ee = = uvuv and and uu is visited is visited before before vv in the depth-first search of in the depth-first search of GG, , then then vv is a descendant of is a descendant of uu..

※ Time complexity of Alg. 3.1: O(max{p, q})


v6

Exercise 1.For the graph G shown below, find thedepth-first search forest. ( 不必為 rooted 形式 )

G

v5 v12

v8v7

v1

v4

v9

v10

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v11


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BFS: visit 過點 v 後，先 visit 完點 v 的每個neighbor ，將之存入 queue 中，再從 queue 中選取第一個點 visit ，依此類推

3.5 Breadth-first 3.5 Breadth-first SearchSearch


Queue

v1

Step 1: visit v1

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v12 v6

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Queue

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Step 2: visit neighbors of v1

Gv5 v13

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v127

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Queue

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Gv5 v13

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v5


Queue

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Step 4: visit v7

v79

v7


Queue

Gv5 v13

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v79v7

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Queue

Gv5 v13

v12 v6

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v22 v65v127

v33v44 v96

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v79

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v14

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v14

v15

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v15v8

12 v8

所有點皆已 visit

BFS forest


v6

Exercise 2.For the graph G shown below, find thebreadth-first search forest.

G

v5 v12

v8v7

v1

v4

v9

v10

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3.6 The minimum 3.6 The minimum spanning tree spanning tree problemproblemDefinition:Definition: (1) (1) weighted graphweighted graph: graph 中每個 edge 上有 weight (2) weight w(H) of a subgraph H of a weighted graph: the sum of the weights of the edges of H

The Minimum Spanning Tree Problem :The Minimum Spanning Tree Problem : Find a minimum spanning tree in all Find a minimum spanning tree in all

connected weighted graph.connected weighted graph.


Algorithm 3.3 (KruskalAlgorithm 3.3 (Kruskal’’s s Algorithm)Algorithm)

1. 1. SS ← ← ..

2. Let 2. Let ee be an edge of minimum be an edge of minimum weight such that weight such that eeSS and < and <SS∪{∪{ee}> }> is acyclic, and let is acyclic, and let SS ← ← SS∪{∪{ee}.}.

3. If |3. If |SS| = | = pp – 1, then output – 1, then output SS; ; otherwise, return to Step 2.otherwise, return to Step 2.

建議：在 step2 執行前先將 G 的 edge 依 weight 由小至大排列O(p2)


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Constructing the tree by Kruskal algorithmConstructing the tree by Kruskal algorithm

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FIGURE 3-18FIGURE 3-18

6

G

Sort the edges of G: uv, xy, vw, uw, ux, vx, vy, uy, wx

Step 1 Step 2

S={uv} S={uv, xy}


Constructing the treeConstructing the tree

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Sort the edges of G: uv, xy, vw, uw, ux, vx, vy, uy, wx

Step 3 y

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Step 4|S|=p-1 stop!

S={uv, xy, vw} S={uv, xy, vw, ux}


Theorem Theorem 3.12.3.12.

KruskalKruskal’’s algorithm produces a minimum s algorithm produces a minimum

spanning tree in a nontrivial connected spanning tree in a nontrivial connected

weighted graph.weighted graph.

Pf : Let G be a nontrivial connected weighted graph of order p, and let T be a subgraph produced by Kruskal’s algorithm. Certainly, T is a spanning tree of G (and so has size p – 1), say

E(T) = { e1, e2, …, ep1 }where w(e1) w(e2) … w(ey1).Therefore, the weight of T is

1

1

).()(p

iiewTw


Suppose, to the contrary, that Suppose, to the contrary, that TT is not a is not a minimum spanning tree. minimum spanning tree.

Then, among the minimum spanning trees of Then, among the minimum spanning trees of GG, , choose one, call it choose one, call it HH, which has a maximum , which has a maximum number of edges in common with number of edges in common with TT..

Now Now HH TT; so ; so TT has at least one edge that has at least one edge that does not belong to does not belong to HH..

Let Let eeii(1(1 ii pp1)1) be the first edge with be the first edge with eeii EE((TT)),, eeii EE((H).H). { {e1, e2, …, ei1} EE((H).H).

Define Define GG00 = = HH++eeii. .

Then Then GG00 has exactly one cycle has exactly one cycle CC..

Since Since TT has no cycles, there is an edge has no cycles, there is an edge ee0 0 of of CC that is not in that is not in TT. . ee00 EE((T),T), ee00 EE((H).H).

The graph The graph TT00 = = GG00 ee00 is also a spanning tree of is also a spanning tree of GG and and ww((TT00) = ) = ww((HH) + ) + ww((eeii) ) ww((ee00).).

Proof of Thm 3.12 (continued):


Since Since ww((HH) ) ww((TT00) ) ww((ee00) ) ww((eeii).).

By KruskalBy Kruskal’’s algorithm, s algorithm, eeii is an edge of is an edge of minimum weight such that <{minimum weight such that <{ee11,,ee22,,……, , eeiii-i-

11}∪{}∪{eeiiii}> is acyclic.}> is acyclic.

However, <{However, <{ee11,,ee22,,……, , eeiii-1i-1, , ee00}> is a subgraph of }> is a subgraph of H H and is therefore acyclic, so and is therefore acyclic, so ww((eeii) = ) = ww((ee00).).

Thus, Thus, ww((TT00) = ) = ww((HH),), which implies that which implies that TT00 is also a is also a minimum spanning tree of minimum spanning tree of GG. .

But But TT00 has more edges in common with has more edges in common with TT than than H H has with has with TT, which contradicts our assumption., which contradicts our assumption.

Proof of Thm 3.12 (continued):


Algorithm 3.3 (PrimAlgorithm 3.3 (Prim’’s s Algorithm)Algorithm)

1. Let 1. Let uu be an arbitrary vertex of be an arbitrary vertex of GG and and TT ← ← uu..

2. Let 2. Let ee be an edge of minimum be an edge of minimum weight joining a vertex of weight joining a vertex of TT and a and a vertex not in vertex not in TT, and , and TT ← ← TT∪{∪{ee}.}.

3. If |3. If |EE((TT))| = | = pp – 1, then output – 1, then output TT; ; otherwise, return to Step 2.otherwise, return to Step 2.

O(p3)


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Constructing the tree by Prim’s algorithmConstructing the tree by Prim’s algorithm

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Step 2

Step 3

V(T)={u, v}, E(T)={u v}

V(T)={u, v,w}, E(T)={uv, uw}

Step 1V(T)={u}

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Constructing the tree by Prim’s algorithmConstructing the tree by Prim’s algorithm

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Step 5V(T)=V(G) E(T)={uv, uw, ux, xy}

Step 4

V(T)={u, v, w, x},

E(T)={uv, uw, ux}

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|E(T)|=p-1 stop!


Algorithm 3.5 (BorAlgorithm 3.5 (Borůůvkavka’’s s Algorithm)Algorithm)

1. 1. FF ← ← KKpp..

2. For each component 2. For each component F’F’ of of FF, join a , join a vertex of vertex of F’F’ to a vertex of some to a vertex of some other component of other component of FF by an edge by an edge of minimum weight. Denote this of minimum weight. Denote this set of edges by set of edges by SS, and , and FF ← ← FF + + SS..

3. If 3. If ||EE((FF)| = )| = pp – 1 – 1, then output , then output EE((FF);); otherwise, return to Step 2.otherwise, return to Step 2.

If G is a connected weighted graph whose edges have distinct weights:


F1:

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Constructing the tree by Constructing the tree by BorBorůůvka’svka’s algorithm: algorithm:Step 1 Step 2

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F3:

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Constructing the tree by Constructing the tree by BorBorůůvka’svka’s algorithm: algorithm:Step 3

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Spanning treeSpanning tree

Every graph can be considered a Every graph can be considered a weighted graph in which every weighted graph in which every edge has weight 1.edge has weight 1.

Kruskal, Prim, DFS, and BFSKruskal, Prim, DFS, and BFS

can be applied to producecan be applied to produce a spanning tree a spanning tree(see Figure 3-21)(see Figure 3-21)


Theorem 3.14 Theorem 3.14 (Cayley(Cayley’’s Tree Formula)s Tree Formula)

Let Let GG be a complete labeled graph of be a complete labeled graph of

order order pp. Then there are . Then there are pppp22 unequal unequal

spanning trees of spanning trees of GG..

Note : Cayley’s Tree Formula implies that if V is a set of cardinality p1, then there are pp-2

unequal trees with vertex set V.(see Figure 3-22, 16 unequal trees with 4 vertices.)(see Figure 3-22, 16 unequal trees with 4 vertices.)


HomeworkHomework

Exercise 3.6: Exercise 3.6: 1, 9 1, 9

graph theory chapter 3 tree

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