Discrete Mathematics

Chapter-10

Trees

Introduction to Tree (§10.1) Def 1. A connected (undirected) graph that contains

no simple circuits is called a tree. Trees are particularly useful in computer science,

where they are employed in a wide range of algorithms. Construct efficient algorithms for locating items in a list. Construct efficient codes saving costs in data transmission

and storage. Study games such as checkers and chess. Model procedure carried out using sequence of decisions.

Nikolaus(1623-1708)

Jacob (1654 -1705)

Nikolaus(1662 - 1716)

Nikolaus (1687 - 1759)

Johann (1667 - 1748 )

Nikolaus (1695 -1726)

Daniel(1700 - 1782)

Johann(1710 - 1790)

Johann(1746-1807)

The Bernoulli Family of Mathematicians

Jacob(1759 - 1789)

EXAMPLE2 Which of the graphs shown below are trees ?

a b

c d

e f

1G 2G 3G 4G

a aa bbb

cc cd

dd

e e ef f f

Tree Tree Not a Tree Not a Tree

Theorem 1 An undirected graph is a tree if and only if there is a

unique simple path between any two of its vertices. Pf: T is a tree. T is connected, no simple circuit.

For any vertices x and y, there is a simple path. If there is a different path from x to y, then two

different paths will form a circuit. This is a contradiction. Thus the path is from x to y is unique.

Assume that there is a unique path between any two vertices in graph T. T is connected. Suppose that T

contains a simple circuit. Then every pair of vertices in this circuit, say x and y, have two different paths from x to

y. That contradicts to the uniqueness of the path. Thus T has no circuit, and then T is a tree.

Def 1. A rooted tree is a tree in which one vertex has been designated as the root and every edge is directed away from the root.

d

f g

b

c

a

e

A root treewith root aA Tree

a

ab

b

c

c

d

de

e

f

f

g

g

A root treewith root c

Terminology

Ancestors Descendants parent Child Siblings Leaf Internal vertex Subtree

Terminology

a

b

c

d e

f

g

hi

j

k l m

A Rooted Tree Ta is the parent of f, b and gj is the parent of l and mf is the child of ae is the child of cf, b and g are siblingse and d are siblings

a, b are ancestors of ce, c are descendants of b

leaves: d, e, f, k, i, l, minternal vertices: others

b

d

c

e

subtree of b

subtree of g

Def 3 A rooted tree is called an m-ary tree if every interval

vertex has no more than m children. The tree is called a full m-ary tree if every interval ver

tex has exactly m children. An m-ary tree with m = 2 is called a binary tree.

Example 3

1T 2T

A full binary tree. A full 3-ary tree.

3T4T

A full 5-ary tree. A m-ary tree. (m3)Not a full m-ary tree

More Terminology An ordered rooted tree is a rooted tree where the chi

ldren of each internal vertex are ordered (from left to right).

In an ordered binary tree (usually called just a binary tree), if an internal vertex has two children, the first child is called the left child and the second one is called the right child.

The tree rooted at the left (right) child of a vertex is called the left (right) subtree of this vertex.

Example 4

a

b c

de

f g j k

h i

l

m

A Binary Tree T

The Left Subtree of the Vertex c.

h

j

The Right Subtree of the Vertex c.

i

kl

m

H

H

H

H

H

H

H

H

H

H

C

C

C

C

H H H

H

HH H

H

H

H

C C C

C

The Two Isomers of ButaneButane

Isobutane

Trees as Models

Representing Organizations

President

VP R&D

VPMarketing

VPFinance

VPServices

DirectorResourch

DirectorSoftware

Development

DirectorSoftware

Development

AVPSales

AVPMarketing

ChiefField

Operations

ChiefField

Operations

DirectorAccounting

DirectorMIS

Theorem 2

A tree with n verices has n1 edges. Pf: Prove by induction. Let P(n) be the statement “A tree with

n verices has n1 edges.”

Basis Step: When n = 1, a tree with one vertex has no edge. P(1) is true.

Inductive Step: Assume that P(k) is true. Suppose that T is a tree with k+1 vertices and v is a leaf. Let w is the parent of v. Remove v and the edge connecting v and w. We’ll get a tree T’ with k vertices and then has k1edges. It follows that T has k edges. Then P(k+1) is true.

Theorem 3

1T 2T

A full binary treem = 2, i = 3, 7 vertices.

A full 3-ary treem = 3, i = 4, 13 vertices

A full m-ary tree with i internal vertices contains mi+1 vertices.

3T4T

A full 5-ary treem = 5, i = 3, 16 vertices Not a full m-ary tree

Not satisfy the theorem

Example 9 A chain letter: Each people sends to 4 people. Some people do

this, but others do not send any letters.

How many people have seen this letter, including the first person, if no one receives more than one letter and if the chain letter ends after there have 100 people who read it but didn’t send it out? How many people sent out the letter?

Sol: This is a 4-ary (m = 4) tree.

Note there are n = mi +1 vertices and l = n i leaves.

100(= l) leaves n = 4(n100) + 1 4001 = 3n There are n = 133 vertices.

There are 133100 = 33 internal vertices.

Theorem 4

A full m-ary tree with

1. n vertices has i = (n 1)/m internal vertices and l = [(m1)n + 1]/m leaves,

2. i interval vertices has n = mi + 1 vertices and l = (m1)i + 1 leaves ,

3. l leaves has n = (ml1)/(m1) vertices and i = (l1)/(m1) internal vertices.

More Terminology The level of a vertex v in a rooted tree is the length o

f the unique path from the root to this vertex v. The level of the root is defined to be zero.

The height of a rooted tree is the maximum of the levels of vertices. That is, the height is the length of the longest path from the

root to any vertex.

A rooted m-ary tree of height h is balanced if all leaves are at levels h or h1.

Example

a

b

c

d

ef

g

h

i

jk

l

m n

Level Vertices 0 a 1 b, j, k 2 c, e, f, l 3 d, g, i, m, n 4 h

The height of this rooted tree is 4.

Example

1T 2T

Balanced Not balanced

3T

Balanced

10.2 Applications of Trees

Introduction How should items in a list be stored so that an item

can be easily located ? What series of decisions should be made to find an

object with a certain property in a collection of objects of a certain type ?

How should a set of characters be efficiently coded by bit strings ?

Binary Search Tree 簡介：二元搜尋樹 是一種二元樹。它可能是空的，若不是

空的，它具有下列特性： ( １ ) 每一個元素有一鍵值，而且每一元素的鍵值都不相同，

即每一個鍵值都是唯一的。　　 ( ２ ) 在非空的左子樹上的鍵值，必小於在該子樹的根節點

中的鍵值。　　 ( ３ ) 在非空的右子樹上的鍵值，必大於在該子樹的根節點

中的鍵值。　　 ( ４ ) 左子樹和右子樹也都是二元搜尋樹。

Example1 Form a binary search tree for the words

mathematic physics geography zoology meteorology geology psychology chemistry (using alphabetical order)

Constructing a Binary Search Tree 按照題目所給的字順序排，如題 Mathematics 就是

root 接下 Physic>Mathematics ，所以在 M 右下加一個

child Geography<Mathematics ，所以在 M 左下加一個 c

hild Zoology>Mathematics 且 >Physics ，所以在 P 右下

加一個 child Meteorology>Mathematics 但 <Physics ，所以在 P

左下加一個 child

Geology<Mathematics 但 >Geography ，所以在 Geography 的右下加一個 child

Psychology>Mathematics 且 >Physics 但 <Zoology所以在 Z 左下加一個 child

Chemistry<Mathematics 且 <Geography ，所以在 Geography 的左下加一個 child

Huffman Codes 簡介： Huffman 編制程式是試圖減少相當數量位元必需

代表標誌串的一個統計技術。 修造 Huffman 樹 ( １ ) 選擇二個 parentless 結點以最低的可能性。 ( ２ ) 創造是二個最低的可能性結點的父母的一個新結點。 ( ３ ) 分配新結點可能性相等與它的孩子的可能性的總和。 ( ４ ) 重覆步驟 1 直到有只一個 parentless 結點剩下

Example 5 Use Huffman coding to encode the fallowing symbols

with the frequencies listed: A:0.08, B:0.10, C:0.12, D:0.15, E:0.20, F:0.35. What is the average number of bits used to encode a character ?

Example 5 於題目中找尋最小兩點→ A:0.08 和 B:0.10(A 於右下 B 於

左下∵ B>A) 接著把 A+B 視為新的一點 0.18 ，再找最小兩點→ C:0.12

和 D:0.15 如同 AB 方法接起來 C+D 也視為新的一點 0.27 再比大小，之後依此方法作完

全部的點 而全圖於右下角的支線都為 1, 左下角都之線都為 0 因此 A by 111,B by 110,C by 011,D by 010, E by 10, F by 00. So the average number of bits used to encode a symbol using t

his encoding is 3×0.08+3×0.10+3×0.12+3×0.15+2×0.20+2×0.35=2.45

9.3 Tree Traversal

Universal address system

We will describe one way to order totally thevertices of an ordered rooted tree. To produce thisordering, we must first label all the vertices. We dothis recursively :1. Label the root with the integer 0. Then label its k children (at label 1) from left to right with 1,2,...,k .

2. For each vertex v at level n with label A, label its

k v children, as they are drawn from left to right,

with A.1,A.2,…A.k v .

Example 1we display the labelings of the universal address system next to the vertices in the ordered rooted tree shown in

Figure 1.

Sol. The lexicographic ordering of the labelings is0 < 1 < 1.1 < 1.2 < 1.3 < 2 < 3 < 3.1 < 3.1.1 < 3.1.2 < 3.1.2.1 < 3.1.2.2 < 3.1.2.3< 3.1.2.4 < 3.1.3 < 3.2 <4 < 4.1 < 5 < 5.1 < 5.1.1 < 5.2 < 5.3

0

1 2 43 5

1.1 1.2 1.33.1

3.2 4.1 5.1 5.2 5.3

3.1.33.1.2

3.1.1

3.1.2.1 3.1.2.2 3.1.2.3 3.1.2.4

5.1.1

Def. 1

Let T be an order rooted tree with root r . If T consists

only of r , then r is the preorder traversal of T .

Otherwise , suppose that T1 , T2 , … , Tn are the subtree

at r from left to right in T . The preorder traversal begins by visiting r . It continues by traversing T1 in preorder , then T2 in preorder , and so on , until Tn is traversed in preorder.

Example 2 In which order does a preorder traversal visit the vertices in the ordered rooted tree T shown below.

a

bc

d

ef

g

h i

jk

l m

n o p

a b c de f g h i

j

k

l m

n o p

a k n o p f c d g l m h ijb e

The preorder traversal of T is a , b , e , j , k , n , o , p , f , c , d , g , l , m , h , i .

Let T be an ordered rooted tree with root r . If T consists Only of r , then r is the inorder traversal of T. Otherwise , suppose that T1 , T2 , … , Tn are the subtree at r from left to right in T . The inorder traversal begins by traversing T1 in inorder , then visiting r . It continues by traversing T2 in inorder , then T3 in inorder ,…, and Finally Tn in inorder.

Def. 2

Example 3In which order does a inorder traversal visit the vertices in the ordered rooted tree T shown below .

a

bc

d

ef

g

h i

jk

l m

n o p

ab c d

f

g

h i

l m

e

jk

n o p

ab c df g h il mej k

n o p

The preorder traversal of T is

a , b , e , j , k , n , o , p , f , c , d , g , l , m , h , i .

Let T be an ordered rooted tree with root r . If T Consists only of r , then r is the postrder traversal of

T. Otherwise , suppose that T1 , T2 , … , Tn are the subtreesat r from left to right in T . The postorder

traversal begins by traversing T1 in postorder , then T2 in inorder , then T3 in inorder ,…, and ends by

visiting r .

Def. 3

Example 4

a

bc

d

ef

g

h i

jk

l m

n o p

In which order does a postorder traversal visit the vertices in the ordered rooted tree T shown below .

ab c d

f

g

h i

l m

e

jk

n o p

ab c df g h il mej k

n o p

The preorder traversal of T is

a , b , e , j , k , n , o , p , f , c , d , g , l , m , h , i .

Infix, Prefix, and Postfix notation

Example 5 what is the ordered rooted tree that represents the

expression((x+y)↑2)+((x-4)/3)?

+

x y x

_

4

↑

+ 2

x y x

_

4

3

/ ↑

+ 2

x y x

_

4

3

/

+

↑

+ 2

x y x

_

4

3

/

+

Example 6 what is the prefix from for ((x+y)↑2)+((x-4)/3)? Example 8 what is the postfix from for ((x+y)↑2)+((x-4)/3)?

This produces the prefix expression : +↑+ x y 2 / - x 4 3This produces the postefix expression : x y + 2 ↑ x 4 - 3 / +

+ - * 2 3 5 / ↑ 2 3 4

+ - * 2 3 5 / 8 4

+ - * 2 3 5 2

+ - 6 5 2

+ 1 2

3

Example 7 what is the value of the prefix expression + - * 2 3 5 / ↑ 2 3 4 ?

7 2 3 * - 4 ↑ 9 3 / +

7 6 - 4 ↑ 9 3 / +

1 4 ↑ 9 3 / +

1 9 3 / +

1 3 +

4

Example 9 what is the value of the postfix expression 7 2 3 * - 4 ↑9 3 / + ?

Example 10 Find the ordered rooted tree representing the compound proposition ( ┐ (p ^ q)) (┐p ˇ ┐q) . Then use this rooted tree to find the prefix , postfix and infix firms of this expression.

^

p q

┐ ┐

p q

^

p q

┐

┐ ┐

p q

ˇ

^

p q

┐

┐ ┐

p q

ˇ

preorder : ┐ ^ p q ˇ ┐p ┐qpostorder : p q ^ ┐p ┐q ˇinorder : (( ┐ (p ^ q)) ((┐p) ˇ (┐q)))

10.4 Spanning Trees (§9.4)

Let G be a simple graph. A spanning tree of G is a subgraph of G that is a tree containing every vertex of G.

Spanning Tree of a graph is not unique.

Theorem A simple graph is connected if and only if it has a

spanning tree. Pf: Suppose that a simple graph G has a spanning tree T. T

contains every vertex of G. There is a path in T between any two of its vertices. Thus there is a path in G between any two of its vertices. Hence, G is connected.

Suppose that G is connected. If G is not a tree, it must contain a simple circuit. Removing an edge from this circuit obtains a subgraph of G, say G’. If G’ is a tree, than it is a spanning tree of G. Otherwise, we continue to remove edge from a circuit until we get a tree subgraph. This tree is a spanning tree of G.

Depth-First Search

Arbitrary choose a vertex as a root. Successively adding vertices and edges where each new edge

is incident with the last vertex in the path and a vertex not already in the path. Let this path as long as possible.

Move back to the next to last vertex in the path.

a

bc

d

ef

gh

i j

k

f

ec

d

a b

gh

kj

i

Algorithm: Depth-First Searchprocedure DFS(G; connected graph with vertices v1, v2, …, vn)T := tree consisting only of the vertex vi

visit(vi)procedure visit(v: vertex of G)for each vertex w adjacent to v and not yet in Tbegin

add vertex w and edge {v, w} to Tvisit(w)

end

The procedure DFS constructs a spanning tree using O(e), or O(n2).

DFS is also called backtracking,

since the algorithm returns to vertices previously visited to add paths.

Example How can backtracking be used to decide whether a given

graph can be colored using 3 colors?

a b c

de a 1

b

a b c d e

Spanning Tree

2

c c

d

1

3

eno way

3

d 1

e 3 Done

Breadth-First Search

Arbitrary choose a root. Add all edges (and vertices) incident to this vertex. Order

these vertices (in level 1) arbitrarily. Continue this step for the vertices we obtaining in previous

step as long as it does not produce a simple circuit till all vertices have been added.

a b c

d ef

g

h ij

k

l

m

f

e c

i

e j g

b dl

a hk

m

Algorithm: Breadth-First Searchprocedure BFS(G; connected graph with vertices v1, v2, …, vn)

T := tree consisting only of the vertex vi

L := empty list

put v1 in the list L of unprocessed verticeswhile L is not emptybegin

remove the first vertex , v, from Lfor each neighbor w of v if w is not in L and not in T then begin

add w to the end of the list Ladd w and edge {v, w} to T

endend

The procedure BFS constructs a spanning tree using O(e), or O(n2).

DFS in Directed Graphsa b c d

e f gh

i j k l

a

bc

gf

e

d

h

l

k

j

i

The output of DFS of a directed graphis not a tree, but a spanning forest.

10.5 Minimum Spanning Trees A minimum spanning tree in a (weighted) connected

graph is a spanning tree that has the smallest possible sum of weights of its edges.

The first algorithm was given by Robert Prim in 1957. Choose any edge with smallest weight. Successively add to the tree edges of minimum weight that

are incident to the vertex already in the tree and not forming a simple circuit.

Stop when n 1 edges have been added.

Algorithm Prim’s Algorithmprocedure Prim(G: weighted connected undirected graph with n vertices)

T := a minimum-weight edge

for i := 1 to n 2begin

e := an edge of minimum weight incident to a vertex in T and not forming a simple circuit in T if added to T

T := T with e added

end (T is a minimum spanning tree of G)

Example

San Francisco

Chicago New York

Atlanta

Denver

$2000

$1200

$1000

$2200

$900

$1400

$130

0

$800

$700$1600

700 (Chicago, Atlanta)800 (Atlanta, New York)

900 (Denver, San Francisco)

1200 (San Francisco, Chicago)

Total cost = $3600

Chicago

Atlanta S F

New York Denver

Examplea b c d

e f gh

i j k l

2 3

3 3

33

3

32

4

4

2

4

5

1

1

1

1 (c, d)

1 (b, f)

1 (k, l)

2 (c, g)

2 (a, b)2 (f, j)

3 (b, c)

3 (a, e)3 (i, j)3 (g, h)3 (h, l)

Total = 24

Algorithm Kruskal’s Algorithmprocedure Kruskal(G: weighted connected undirected graph with n vertices)

T := empty tree

for i := 1 to n 1begin

e := an edge in G with smallest weight that does not form a simple circuit when added to T

T := T with e added

end (T is a minimum spanning tree of G)

Example

San Francisco

Chicago New York

Atlanta

Denver

$2000

$1200

$1000

$2200

$900

$1400

$130

0

$800

$700$1600

700 (Chicago, Atlanta)800 (Atlanta, New York)900 (Denver, San Francisco)

1200 (San Francisco, Chicago) Total cost = $3600

Examplea b c d

e f gh

i j k l

2 3

3 3

33

3

32

4

4

2

4

5

1

1

1

1 (c, d)1 (b, f)1 (k, l)2 (a, b)2 (c, g)2 (f, j)3 (b, c)3 (g, h)3 (a, e)3 (h, l)3 (i, j)

Total = 24