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Applied Combinatorics, 4th Ed. Alan TuckerPowerPoint Presentation

Applied Combinatorics, 4th Ed. Alan Tucker

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Applied Combinatorics, 4th Ed. Alan Tucker

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Applied Combinatorics, 4th Ed.Alan Tucker

Section 3.1

Properties of Trees

Prepared by Joshua Schoenly

and Kathleen McNamara

Tucker, Sec. 3.1

Definitions

- 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 with all edges directed away from the root

a

Root = the unique vertex with in-degree of 0

b

c

e

f

d

i

h

g

j

Tucker, Sec. 3.1

- Level Number: the length of the path from the root a to x
- Parent: the vertex y is a parent of x if they are connected by an edge
- Children: the vertex y is a child of x if they are connected by an edge
- Siblings: two vertices with the same parent

a

Parent of g and h

b

c

Level 1

f

Level 2

d

e

Level 3

j

g

k

i

h

Siblings of each other

Children of d

Tucker, Sec. 3.1

b

a

c

a

b

c

d

f

d

e

f

e

Theorem 1

A tree with n vertices has n – 1 edges.

Proof

Choose a root, and direct all edges away from the root. Since each vertex except the root has a single incoming edge, there are n –1 non-root vertices and hence n –1 edges.

Root = a

Tucker, Sec. 3.1

Definitions

- Leaves: vertices with no children
- Internal Vertices: vertices with children
- m-ary Tree: when each internal vertex of a rooted tree has m children
- Binary Tree: when m = 2

Internal vertices

Leaves

Tucker, Sec. 3.1

Definitions

- Height of a Rooted Tree: the length of the longest path to the root.
- Balanced Tree (“good”): if all the leaves are at levels h and h-1, where h is the height of the tree.

Height (h) is 3

h-1

h

Tucker, Sec. 3.1

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, so there are a total of mi children. Adding the root gives

n = mi + 1

m = 3

i = 3

n = 10

Tucker, Sec. 3.1

Corollary

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

Note: The proof of the corollary’s formulas follow directly from n=mi+1(Theorem 2) and the fact that

l + i = n

Tucker, Sec. 3.1

Example 1

If 56 people sign up for a tennis tournament, how many matches will be played in the tournament?

Setting up as a binary tree, there will be 56 leaves and i matches with two entrants entering a match.

Entrants

Matches

Shortened Graph

Tucker, Sec. 3.1

Theorem3

- 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 ≥ `élogmlù and if the tree is balanced, h = élogmlù

l = 5

h = 3

m = 2

a

b

c

d

i

f

e

h

g

Tucker, Sec. 3.1

Prufer Sequence

There exists a sequence (s1, s2,…,sn-2) of length n-2. This is called a Prufer Sequence.

Start with the leaf of the smallest label (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 obtaining,

1

3

5

7

2

4

6

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

8

Note: There is a 1:1 correspondence to the Prufer Sequence and the tree

Tucker, Sec. 3.1

1

6

3

8

5

7

2

4

Example 2

2,

3,

Find the graph that has the Prufer Sequence

(6,

2,

3,

3)

1

4

5

2

6

7

3

8

Tucker, Sec. 3.1

Theorem 4

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

4

8

1

6

3

5

3

6

2

1

7

4

7

5

2

8

Two different trees on 8 items.

Tucker, Sec. 3.1

Proof of Theorem 4

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

We showed there is a 1-to-1 correspondence between trees on n items and Prufer sequences of length n-2. Count Prufer sequences.

(__, __, __, __, __, __)

…

n choices

n choices

This means there are nn-2 different Prufer sequences. Since each tree has a unique Prufer Sequence, there are nn-2 different trees.

Tucker, Sec. 3.1

2

1

5

8

4

6

7

3

Class Problem

Create a Prufer Sequence from the graph:

Tucker, Sec. 3.1

2

1

5

4

7

3

Solution

Create a Prufer Sequence from the graph:

8

6

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

Tucker, Sec. 3.1