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Section 3.1. Properties of Trees. Sarah Graham. root. a. b. c. e. f. d. i. h. g. j. Tree Talk: Vocabulary.

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Section 3.1


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slide1

Section 3.1

  • Properties of Trees

Sarah Graham

slide2

root

a

b

c

e

f

d

i

h

g

j

Tree Talk: Vocabulary

  • Tree: a tree is a special type of graph that contains designated vertex called a root so that there is a unique path from the root to any other vertex in the tree. Equivalently, a tree graph contains no circuits.
  • Rooted Tree: a directed tree graph

Root = the unique vertex with in-degree of 0

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m

f

l

d

h

b

e

j

a

i

k

c

root

g

  • Level Number: the length of the path from the root a to x
  • Parent: for any vertex except the root a parent of x is the vertex y with an edge (y,x)
  • Children: for any vertex say z with a connected edge with a parent (x,z)
  • Siblings: two vertices with the same parent

children

Level # 3

siblings

parent

Level #2

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Theorem 1:

A tree with n vertices has n – 1 edges.

Assume that the tree is rooted. Since each vertex except the root has such a unique incoming edge, there are n – 1 nonroot vertices and hence n- 1 edges.

a

Root = a

Vertices = 6

Edges = 6 – 1 = 5

b

c

d

e

f

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Leaves: vertices with no children

  • Internal Vertices: vertices with children excluding the root
  • M-ary Tree: when each internal vertex of a rooted tree has m children
  • Binary Tree: when m = 2
  • Height of a Rooted Tree: the length of the longest path to the root
  • Level: A set distance form the root (ex: the vertices at level 3 is the set of vertices at distance 3 from the root)
  • Balanced Tree (“good”): if all the leaves are at levels h and h-1

Binary tree

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Theorem 2:

If T is an m-ary tree with n vertices, of which i vertices are internal. Then, n = mi + 1.

Proof

Each vertex in a tree, other than the root, is the child of a unique vertex. Each of the i internal vertices has m children, and there are a total of mi children. Adding the root gives

n = mi + 1

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Corollary

T is a m-ary tree with n vertices, consisting of i internal vertices and l leaves.

  • Given i, then l = (m – 1)i + 1 and n = mi + 1
  • Given l, then i = (l –1)/(m – 1) and n = (ml –1)/(m – 1)
  • Given n, then i = (n – 1)/m and l = [(m – 1)n + 1]/m

n = m + l

n – m = l

m(i) +1 – m = l

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a

b

c

Theorem 3

  • T is a m-ary tree of height h with l leaves.
  • l≤ mh and if all leaves are at height h, l = mh
  • h ≥ [logml] and if the tree is balanced, h = [logml]

5 ≤ 23

3 ≥ [log25]

d

e

f

h

i

g

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Theorem 4

There are nn-2 different undirected trees on n items.

Prufer Sequence:

There exists a sequence (s1, s2,…,sn-2) of length n-2.

1

Start with the smallest leaf (2). Its neighbor is 5. 5= s1Delete the edge. Take the next smallest leaf (4). Its neighbor is 3. 3= s2 Delete the edge.

Continue like this.

3

5

7

2

4

6

8

(5,3,1,7,3,6)

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1

6

3

8

5

7

2

4

Find the graph that goes to the Prufer Sequence (6,2,2,3,3,3)

Set aside

1

4

5

2

6

7

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Show that any tree with more than 1 vertex has at least 2 vertices of degree 1.

By the corollary of theorem 2:

l= ((m-1)n+1)/m

Assume l=1 and n≥2

Then:

l=((m-1)n+1)/m

m=(m-1)n+1

(m-1)=(m-1)n

1=n

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2

1

5

1

3

8

4

6

7

4

3

6

7

5

2

(5,6,1,1,5,6)

8

Create a graph from the Prufer Sequence:

Create a Prufer Sequence from the graph:

(3,3,3,3,3,5)