Transcript Document

Applied Combinatorics, 4th Ed.
Alan Tucker
Section 3.1
Properties of Trees
Prepared by Joshua Schoenly
and Kathleen McNamara
03/01/2005
Tucker, Sec. 3.1
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 inb
c
degree of 0
e
f
d
j
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g
Tucker, Sec. 3.1
h
i
2
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  y, x 
Children: the vertex y is a child of x if they are connected by an
edge  x, y 
Siblings: two vertices with the same parent
a
Parent of g
and h
b
Level 2
e
h
Children of d
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Level 1
f
d
g
c
i
j
Level 3
k
Siblings of each other
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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.
b
a
d
a
c
Root = a
f
b
d
c
e
f
e
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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
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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
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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,
so there are a total of mi children. Adding the root gives
n = mi + 1
m=3
i=3
n = 10
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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
l  1

Given l , then i 
 m  1
n  1

Given n, then i 
m
ml  1

and n 
 m  1
 m  1 n  1
and l 
m
Note: The proof of the corollary’s formulas follow
directly from n=mi+1 (Theorem 2) and the fact that
l+i=n
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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.
l  56 and m  2
By the corollary of Theorem 2:
Entrants
Matches
l  1

i
 m  1
56  1

i
 55 matches
 2  1
Shortened Graph
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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 ≥ `élogmlù and if the tree is balanced, h = élogmlù
l=5
h=3
m=2
a
b
d
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c
i
e
h
5  23  8
f
g
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3
log 2 5  3
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Prufer Sequence
There exists a sequence (s1, s2,…,sn-2) of length n-2.
This is called a Prufer Sequence.
1
3
5
7
2
6
4
Start with the leaf of the smallest
label (2). Its neighbor is 5. 5 = s1
Delete the edge. Take the next
smallest leaf (4). Its neighbor is 3.
3 = s2 Delete the edge.
Continue like this obtaining,
(5,3,1,7,3,6)
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Note: There is a 1:1 correspondence to
the Prufer Sequence and the tree
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Example 2
Find the graph that has the Prufer Sequence (6, 2, 2, 3, 3, 3)
1
1
4
6
3
8
5
7
2
4
5
2
6
7
3
8
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Theorem 4
There are nn-2 different undirected trees on n items.
1
3
8
6
5
6
3
2
8
4
4
1
7
2
7
5
Two different trees on 8 items.
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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.
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Class Problem
Create a Prufer Sequence from the graph:
2
1
5
8
6
4
7
3
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Solution
Create a Prufer Sequence from the graph:
2
1
5
8
6
4
7
3
(5,6,1,1,5,6)
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