Chapter 7. Trees

Chapter 7. Trees

Weiqi Luo (骆伟祺 )School of Software

Sun Yat-Sen UniversityEmail ： [email protected] Office ： A309

mailto:[email protected]

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7.1. Trees 7.2. Labeled Trees 7.3. Tree Searching 7.4. Undirected Trees 7.5. Minimal Spanning Trees

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Chapter seven: Trees

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Tree Let A be a set, and let T be a relation on A.

We say that T is a tree if there is a vertex v0 in A with the property that

#1 : there exists a unique path in T from v0 to every other vertex in A

#2 : no path from v0 to v0

Usually, v0 is called the root of the tree, and T is referred to as a rooted tree (denoted by (T, v0))

7.1 Trees

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Theorem 1 Let (T,v0) be a rooted tree. Then

(a) There are no cycle in T

(b) v0 is the only root of T

(c) Each vertex in T, other than v0, has in-degree one, and v0 has in-degree zero.

Proof: (a) Suppose that there is a cycle q in T, beginning and ending at

vertex v. By the definition of a tree, v is not v0 and there is a path from v0 to v. The qop is a different path from v0 to v, which is a contradiction of a tree.

7.1 Trees

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(b) If v0’ is another root of T, there is a path from v0 to v0’ (since v0 is a root) and a path q from v0’ to v0 (since v0’ is a root). Then qop is a cycle from v0 to v0, which is a contradiction of a tree.

(c) Let w1 be a vertex in T other than v0. There is a unique path v0, …, vk, w1, from v0 to w1 in T. Thus (vk, w1) in T, so w1 has in-degree at least one. If the in-degree of w1 is more than 1, then there must be distinct vertices w2,w3 such that (w2,w1) and (w3,w1) both in T.

case #1: If w2 & w3 are not v0, there are two different paths from v0 to w1 (v0w2w1, v0w3w1, contradiction)

case #2: If w2 or w3 is v0, e.g. w2=v0, there also are two different paths from v0 to w1 (v0w1, v0w3w1)

7.1 Trees

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Levels, parent-offspring & siblings

7.1 Trees

6

v4

v0

v1 v2v3

v5 v6 v7 v8 v9

Level 1 vertices: the vertices of the edges beginning at v0 (level 0) .

Level k vertices: the vertices of the edgesbeginning at those level k-1 vertices

Level 0

Level 1

Level 2

Parent-offspring: for all pairs (r1, r2)in T, r1

is called the parent of r2, and r2 is called the offspring of r1

Siblings: the vertices that have the same parent

Height of a tree: the largest level numberof a tree.

Leaves: the vertices that have no offsprings

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Theorem 2 Let (T,v0) be a rooted tree on a set A. Then

(a) T is irreflexive

(b) T is asymmetric

(c) If (a,b) in T and (b,c) in T, then (a,c) is not in T, for all a, b and c in A.

The proof is left as an exercise.

7.1 Trees

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Example 1 Let A be the set consisting of a given woman v0 and all

of her female descendants. We now define the following relation T on A: If v1 and v2 are elements of A, then v1 T v2 if and only if v1 is the mother of v2. The relation T on A is a rooted three with root v0.

7.1 Trees

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Example 2 Let A={v1,v2,…v10} and let T={(v2,v3), (v2,v1), (v4,v5),

(v4,v6), (v5,v8), (v6,v7), (v4,v2), (v7,v9), (v7,v10)}. Show that T is a rooted tree and identify the root.

7.1 Trees

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v4

v6 v2 v5

v7

v9 v10

v1 v3 v8

Root

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n-tree If n is a positive integer, we say that a tree is an n-tree if

every vertex has at most n offspring. In particular, a 2-tree is called a binary tree.

Complete n-tree

If all vertices of T, other than the leaves, have exactly n offspring, we say that T is a complete n-tree. A complete 2-tree is called a completed binary tree.

7.1 Trees

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Let (T,v0) be a rooted tree on the set A, and let v be a vertex of T. Let B be the set consisting of v and all its descendants, i.e., all vertices of T that can be reached by a path beginning at v. Observe that B A. Let ⊆ T(v) be the restriction of the relation T to B, that is T ∩ (B × B) .

Delete all vertices that are not descendants of v and all edges that do not begin and end at any such vertex.

7.1 Trees

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v4

v6 v2 v5

v7

v9 v10

v1 v3 v8

v5

v8

T(v5)

v2

v1 v3

T(v2)

v6

v7

v9 v10

T(v6)

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Theorem 3 If (T,v0) is a rooted tree and v in T, then T(v) is also a rooted

tree with root v. We will say that T(v) is the subtree of T beginning at v.

Proof: By the definition of T(v), there is a path from v to every other vertex in T(v). If there is a vertex w in T(v) such that there are two distinct paths q and q’ form v to w, and if p is the path in T from v0 to v, then qop and q’op would be two distinct paths in T form v0 to w (contradiction!). Thus each path from v to another vertex w in T(v) must be unique.

if q is a cycle at v in T(v), then q is also a cycle in T (contradiction!). Thus q cannot exist.

Therefore, T(v) is also a rooted tree with root v.

7.1 Trees

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Homework Ex. 4, Ex. 10, Ex. 14, Ex. 20, Ex. 26, Ex. 27, Ex. 34

7.1 Trees

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Use a tree to denote the following algebraic expression

(3 - ( 2 × x ) ) + ( (x – 2 ) - ( 3 + x ) )

7.2 Labeled Trees

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+

-

3

2 x x 2 3 x

x - +

-

1. represent the vertices simply as dots2. show the label of each vertex next to thedot representing that vertex

Here leaves denote the numbers or variables

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Positional n - tree n-tree: every vertex has at most n offspring.

label the offspring of a given vertex from left to right with numbers 1,2 …n

7.2 Labeled Trees

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2

2

3

3

3

1 1

3

2 31

2 31

L

L

L

L

LL

L

LR

R

R

R R

R

R

R

R

positional 3-tree positional binary tree

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Huffman code trees A variable-length code widely used in lossless data

compression. Base Idea: the more frequently used letters are assigned shorter strings.

7.2 Labeled Trees

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0 1

1

1

1

0

0

0

R C

S

A

E

Decode the string: 0101100

0: E10: A110: S0: E

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Homework Ex. 8, Ex. 14, Ex. 16, Ex. 25, Ex. 27

7.2 Labeled Trees

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Visiting Performing appropriate tasks at a vertex will be called

visiting the vertex.

Tree search

The process of visiting each vertex of a tree in some specific order will be called searching the tree or performing a tree search.

7.3 Tree Searching

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Algorithm PREORDER Step 1: visit v

Step 2: If vL exists, then apply this algorithm to (T(vL), vL).

Step 3: If vR exists, then apply this algorithm to (T(vR), vR)

Recursion: Recursion is the process of repeating items in a self-similar way. Refer to

http://en.wikipedia.org/wiki/Recursion

7.3 Tree Searching

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Example 1

7.3 Tree Searching

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A

B

C

D F

E

G J

I

L

K

H

1

2

3

4

5 6

7

8

9

10

11

A B C D E F G H I J K L

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Example 2

(a – b) × (c + ( d / e) )

7.3 Tree Searching

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×

-

a b

e

/

+

c

d1

2 3

4

5

7 86

× — a b + c / d e

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Prefix or Polish form × － a b + c / d e (a=6, b=4, c=5, d=2, e=2)

1. × － 6 4 + 5 / 2 2

2. × 2 + 5 / 2 2 replacing - 6 4 by 2 since 6-4=2

3. × 2 + 5 1 replacing / 2 2 by 1 since 2/2=1

4. × 2 6 replacing + 5 1 by 6 since 5+1=6

5. 12 replacing×2 6 by12 since 2×6=12

7.3 Tree Searching

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Algorithm INORDER Step 1: Search the left subtree (T(vL), vL), if it exists;

Step 2: Visit the root v;

Step 3: Search the right subtree (T(vR), vR), if it exists.

Algorithm POSTORDER Step 1: Search the left subtree (T(vL), vL), if it exists;

Step 2: Search the left subtree (T(vR), vR), if it exists;

Step 3: Visit the root v.

7.3 Tree Searching

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Traveling the tree using INORDER & POSTORDER

7.3 Tree Searching

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×

-

a b

e

/

+

c

d

INORDER: a - b × c + d / e

(a – b) × (c + ( d / e) )

POSTORDER: a b – c d e / + ×

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Infix notation a - b × c + d / e

(a – b) × (c + ( d / e) ) or a – (b × ((c + d) / e) )

Postfix or reverse polish a b – c d e / + × (if a=2,b=1,c=3,d=4,e=2)

1. 2 1 – 3 4 2 / + ×

2. 1 3 4 2 / + × replacing 2 1 – with 1 since 2-1=1

3. 1 3 2 + × replacing 4 2 / with 2 since 4/2=2

4. 1 5 × replacing 3 2 + with 5 since 3+2=5

5. 5 replacing 1 5× with 5 since 1×5 =5

7.3 Tree Searching

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Let T be any order tree and A be the set of vertices in T

Define a binary positional tree B(T) on A: if v in A, the left offspring vL of v in B(T): the first offspring of v in

T, if exists.

the right offspring vR of v in B(T): the next sibling of v in T, if exists

7.3 Tree Searching

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1

2 3 4

56 7 8 9 10

11 12 13

1

23

45

67 9

10

1112

13

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Homework Ex. 16, Ex. 17, Ex. 18, Ex. 24, Ex. 26, Ex. 32, Ex. 36

7.3 Tree Searching

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Undirected Tree An undirected tree is simply the symmetric closure of a

tree, that is, it is the relation that results from a tree when all the edges are made bidirectional.

7.4 Undirected Trees

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b

c

e

a

f

d

g

An undirected diagram T

a b e f

c

d

g

An ordinary tree T2

Note: T is the symmetric closure of T1 and T2. Thus T is a undirected tree.

ad

g

fe

b

An ordinary tree T1

c

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Simple Let R be a symmetric relation, and let p: v1,v2,…vn be a path

in R. we say p is simple if no two edges of p correspond to the same undirected edge. If , in addition, v1 equals vn, we will call p a simple cycle.

Acyclic A symmetric relation R is acyclic if it contains no simple

cycles.

Connected

A symmetric relation R is connected if there is a path in R from any vertex to others.


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Example 2

Path: a, b, c, e, d

Path: f, e, d, c, d, a

Path: f, e, a, d, b, a, f & d, a, b, d

Path: f, e, d, c, e, f


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a f

be

dc

Simple

Not simple, since d, c & c, d correspond to the same undirected edge.

Simple cycles

Not a simple cycle

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Theorem 1 Let R be a symmetric relation on a set A. Then the

following statements are equivalent.

(a) R is an undirected tree

(b) R is connected and acyclic


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Theorem 2 Let R be a symmetric relation on a set A. Then R is an

undirected tree if and only if either of the following statements is true

(a) R is acyclic, and if any undirected edge is added to R, the new relation will not be acyclic;

(b) R is connected, and if any undirected edge is removed from R, the new relation will not be connected.


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Theorem 3 A tree with n vertices has n-1 edges

Proof: Because a tree is connected, there must be at least n-1 edges

to connect the n vertices.

Support that there are more than n-1 edges. Then either the root has in-degree 1 or some other vertex has in-degree at least 2. But by Theorem 1, Section 7.1, this is impossible. Thus there are exactly n-1 edges.


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Theorem 1 (c) Each vertex in T, other than v0, has in-degree one, and v0 has in-degree zero.

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Spanning Tree If R is a symmetric, connected relation on a set A, we say that

a tree T on A is a spanning tree for R if T is a tree with exactly the same vertices as R and which can be obtained from R by deleting some edges of R. For instance,


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a

b c

d e

f

(a) R

a

b c

d e

f

(b) T’

a

b c

d e

f

(c) T’’

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Undirected spanning Tree An undirected spanning tree is the symmetric closure of

a spanning tree, and it is useful in some applications.


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a

b c

d e

f

(c) T’’

a

b c

d e

f

(c) Undirected Tree

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Example 4


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a

b c

d e

f

(a)

a

b c

d e

f

(b)a

b c

d e

f

(e)

a

b c

d e

f

(c)

a

b c

d e

f

(d)a

b c

d e

f

(f)

Theorem 2(b) & Theorem 3suggests a simple algorithm for finding an undirected spanning tree for a symmetric relation R.

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Merging the vertices Let R be a relation on a set A, and let a, b in A. Let A0=A-

{a, b}, and A’=A0 U {a’}, where a’ is some new element not in A. Define a relation R’ on A’ as follows.

Support u, v in A’, u and v are not a’.

Let (a’, u) in R’ iff (a,u) in R or (b,u) in R

(u, a’) in R’ iff (u,a) in R or (u,b) in R

(u, v) in R’ iff (u,v) in R

We say that R’ is a result of merging the vertices a and b


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Example 5


38

v0

v1 v2

(a) R

v6

v5

v4

v3

v’0

v2

(b) Merging v1 & v0 into v’0

v6

v5

v4

v3

v’’0

v6

(c) Merging v’0 & v2 into v’’0

v5

v4

v3

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Suppose that vertices a and b of a relation R are merged into a new vertex a’ that replaces a and b to obtain the relation R’. To determine the matrix of R’, we proceed as follows.

Step 1: Let row i represent vertex a and row j represent vertex b. Replace row i by the join of rows i and j.

Note: The join of two n-tuples of 0’s and 1’s has 1 in some position exactly when either of those two n-tuples has a 1 in that position.

Step 2: Replace column i by the join of columns i and j Step 3: Restore the main diagonal to its original values in R Step 4: Delete row j and column j.


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Example 6


40

0110010

1001100

1000011

0100000

0100000

1010000

0010000

v0

v1

v2

v3

v4

v5

v6

v0 v1 v2 v3 v4 v5 v6

011110

100011

100000

100000

110000

010000

v’0

v2

v3

v4

v5

v6

v’0 v2 v3 v4 v5 v6

01111

10000

10000

10000

10000

v’’0

v3

v4

v5

v6

v’’0 v3 v4 v5 v6

(a) (b) (c)

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The Prim’s algorithm for finding a spanning tree for a symmetric, connected relation R on A={v1,v2,…,vn}

Step 1: Choose v1 of R, and arrange the matrix of R so that the first row corresponds to v1

Step 2: Choose a vertex v2 of R such that (v1,v2) in R, merge v1 and v2 into a new vertex v’1, representing {v1,v2}, and replace v1 by v’1. compute the matrix of R’. Call the vertex v’1 a merged vertex.

Step 3: Repeat Step 1& 2 on R’ until a relation with a single vertex is obtained. At each stage, keep a record of the set of original vertices that represented by each merged vertex.

Step 4: Construct the spanning tree. At each stage, when merging vertices a & b, select an edge in R from one of the original vertices represented by a to one of the original vertices represented by b.


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Example 7


42

a b

c d

ba

c d

Matrix Original & merged vertices

New vertexto be merged

0011

0011

1100

1100

a b c d

a

b

c

d

011

101

110

a’

b

d

a’ b d

01

10

a’’ da’’

d

c

a’ {a, c} b

da’’ {a, c, b}

0a’’’

a’’’ a’’’ {a, c, d, b}

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Homework

Ex. 6, Ex. 12, Ex. 14, Ex. 16, Ex. 21, Ex. 22


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Weighted graph A weighted graph is a graph for which each edges is

labeled with a numerical value called its weight.

For instance

vertices: towns

edges: the distance between two connected towns

7.5 Minimal Spanning Trees

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B D

CA

F G

H

E

6 2

6

2

3 2

34

5

5

4

3

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Example 2


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BC

IA

H

G

JF

E

D

2.64.22.1 3.6

2.8

2.4

2.7

3.4

1.7

1.8

3.2 5.3

2.5

2.1

2.2

3.3

2.9Vertices: relay stations

Edges: the cost for upgrading4.4

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Distance between vertices The weight of an edge (vi,vj) is sometimes referred to as

the distance between vertices vi and vj.

Nearest neighbor A vertex u is nearest neighbor of vertex v

if u and v are adjacent and no other vertex is joined to v by an edge of lesser weight than (u, v)

(Note: may have more than one nearest neighbor)


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Example 3


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B D

CA

F G

H

E

6 2

6

2

3 2

34

5

5

4

The nearest neighbor of A is C

The nearest neighbor of F is E & G

3

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The nearest neighbor of a set of vertices A vertex v is a nearest neighbor of a set vertices V={v1,v2,

…,vk} in a graph if v is adjacent to some member vi of V and no other vertex adjacent to a member of V is joined by an edge of lesser weight than (v,vi).

Note: This vertex v may belong to V


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The nearest neighbor of {C,E,J} is D.

What is the nearest neighbor of {D,E,F}? Answer: D & F

BC

IA

H

G

JF

E

D

2.64.22.1 3.6

2.8

2.4

2.7

3.4

1.7

1.8

3.2 5.3

2.52.1

2.2

3.3

2.9

4.4

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Minimal spanning tree

An undirected spanning tree in which the total weight of the edges is as small as possible, is called a minimal spanning tree.

Prim’s algorithm (Section 7.4, p. 293) can easily be adapted to produce a minimal spanning tree for a weighted graph by using the greedy algorithm.

Greedy algorithm makes the locally optimal choice at each stage, and hopes to find the global optimum.

http://en.wikipedia.org/wiki/Greedy_algorithm


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Algorithm: Prim’s ALGORITHM Step 1: Choose a vertex v1 of R. Let V={v1} and E={};

Step 2: Choose a nearest neighbor vi of V that is adjacent to vj, vj in V, and for which the edge (vi, vj) does not form a cycle with members of E. Add vi to V and add (vi, vj) to E.

Step 3: Repeat Step 2 until |E|=n-1. Then V contains all n vertices of R, and E contains the edges of a minimal spanning tree for R.


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Theorem 1 Prim’s algorithm produces a minimal tree for the

relation.

Proof

…


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Example 5


52

B D

CA

F G

H

E

6 2

6

2

3

2

34

5

5

4

3

B D

CA

F G

H

E

6 2

6

2

3

2

34

5

5

4

3

21 Kilometers 21 Kilometers

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Example 6


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BC

IA

H

G

JF

E

D

2.6

4.22.1 3.6

2.8

2.4

2.7

3.4

1.7

1.8

3.2 5.3

2.5

2.1

2.2

3.3

2.9

4.4

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Weight graph & Matrix representation


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A B

C

D

0110

1011

1101

0110

4

33

5

2

4

4

3

3 2

5

5 3

3

2

A B C D

A

B

C

D

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Algorithm PRIM’S ALGORITHM (Matrix version) Let R be a symmetric, connected relation with n vertices and

M be the associated matrix of weights. Step 1 Choose the smallest entry in M, say mij. Let a be the

vertex that is represented by row I and b the vertex represented by column j.

Step 2 Merge a with b as follows:

Replace row i with


55

0

0

min( , ) 0 0

0 0

ik jk

jk ik

ikik jk ik jk

ik jk

m if m

m if mm

m m if m m

if m m

1 k n

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Replace column i with

Replace the main diagonal with the original entries of M.

Delete row j and column j. Call the resulting matrix M’. Step 3 Repeat Step 1 & 2 on M’ and subsequent matrices

until a single vertex is obtained. At each stage, keep a record of the set of original vertices that is represented by each merged vertex.


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0

0

min( , ) 0 0

0 0

ki kj

kj ki

kiki kj ki kj

ki kj

m if m

m if mm

m m if m m

if m m

1 k n

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Step 4 Construct the minimal spanning tree as follows: At each stage, when merging vertices a and b, select the edge represented by the minimal weight from one of the original vertices represented by a to one of the original vertices represented by b.


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Example 7


58

Either 2 may be selected as mij. We choose m3,4 and merge C and D, This produces

0430

4053

3502

0320

M

A

A

B

C

D

B C D

043

' 403

330

M

A

A

B

C’

C’B

With C’ {C,D} and the first edge (C,D), Repeat 1 and 2On M’ using m1,3=3. This gives

03''

30M

A’

A’

B

B

With A’ {A,C,D} and the edge (A,C)

A B

C

D

4

33

5

2

A final merge yields the edge (B,D)

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Algorithm KRUSKAL’S ALGORITHM Let R be a symmetric, connected relation with n vertices and

let S={e1,e2,…,ek} be the set of weighted edges of R.

Step 1 Choose an edge e1 in S of least weight. Let E={e1}. Replace S with S-{e1}

Step 2 Select an edge ei in S of least weight that will not make a cycle with members of E. Replace E with E U {ei} and S with S-{ei}

Step 3 Repeat Step 2 until |E|=n-1


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Example 8


60

B D

CA

F G

H

E

6 2

6

2

3

2

34

5

5

4

3

21 Kilometers

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Example 9


61

B

A

F

C

E

20

D

G

12

12

1410

12

15

15

20

16

10

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Homework Ex. 4, Ex. 8, Ex. 13, Ex. 15, Ex. 20, Ex. 21


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Chapter 7. Trees