DISCRETE MATHEMATICS I
LECTURES CHAPTER 10
Some material adapted from
lecture notes provided by
Dr. Chungsim Han and Dr.
Sam Lomonaco
Dr. Adam Anthony
Spring 2011
2
Section 10.1
Introduction to Graphs
3
Definition: A simple graph G = (V, E) consists of V, a
nonempty set of vertices, and E, a set of unordered
pairs of distinct elements of V called edges.


For each eE, e = {u, v} where u, v  V.

An undirected graph (not simple) may contain:
Loops:
An edge e is a loop if e = {u, u} for some uV
Duplicate
edges: A graph is called a multi-graph if there is
at least one duplicate edge.
Introduction to Graphs
4
Definition: A directed graph G = (V, E) consists of a
set V of vertices and a set E of edges that are ordered
pairs of elements in V.


For each eE, e = (u, v) where u, v  V.

An edge e is a loop if e = (u, u) for some uV.
A simple graph is just like a directed graph, but with
no specified direction of its edges.

Graph Models
5
Example I: How can we represent a network of (bidirectional) railways connecting a set of cities?

We should use a simple graph with an edge {a, b}
indicating a direct train connection between cities a and
b.
Boston
Toronto
Lübeck

New York
Washington
Hamburg
Graph Models
6
Example II: In a round-robin tournament, each team
plays against each other team exactly once. How can
we represent the results of the tournament (which team
beats which other team)?

We should use a directed graph with an edge (a, b)
indicating that team a beats team b.
Bruins
Maple Leafs

Penguins
Lübeck Giants
Exercise 1
7

What might the nodes/edges be if we modeled the
following data? Would the graph be best
undirected or directed?
 Acquaintance
Graph: Identifying people are mutually
acquainted
 Influence Graph: Identifying data where one person
has influence over another.
 Computer network
 Road Map
 The World Wide Web
Graph Terminology
8
Definition: Two vertices u and v in an undirected
graph G are called adjacent (or neighbors) in G if {u,
v} is an edge in G.

If e = {u, v}, the edge e is called incident with the
vertices u and v. The edge e is also said to connect u
and v.

The vertices u and v are called endpoints of the edge
{u, v}.

Graph Terminology
9
Definition: The degree of a vertex in an undirected
graph is the number of edges incident with it, except
that a loop at a vertex contributes twice to the degree
of that vertex.

In other words, you can determine the degree of a
vertex in a displayed graph by counting the lines that
touch it.


The degree of the vertex v is denoted by deg(v).
Graph Terminology
10
A vertex of degree 0 is called isolated, since it is not
adjacent to any vertex.

Note: A vertex with a loop at it has at least degree 2
and, by definition, is not isolated, even if it is not
adjacent to any other vertex.

A vertex of degree 1 is called pendant. It is adjacent
to exactly one other vertex.

Graph Terminology
11
Example: Which vertices in the following graph are
isolated, which are pendant, and what is the maximum
degree? What type of graph is it?
f
h
d
a
e
g
f
b
c
j

Solution: Vertex f is isolated, and vertices a, d and
j are pendant. The maximum degree is deg(g) = 5.
This graph is a non-simple undirected graph.
Graph Terminology
12
Determine the number of its edges and the sum of the
degrees of all its vertices:
f
h
d
a
e
g
f
b
c
j

Result: There are 9 edges, and the sum of all
degrees is 18. This is easy to explain: Each new
edge increases the sum of degrees by exactly two.
Graph Terminology
13
The Handshaking Theorem: Let G = (V, E) be an
undirected graph with e edges. Then


2e = vV deg(v)
Corrolary:
The total degree of any undirected graph is always
even!
Example: How many edges are there in a graph with 10
vertices, each of degree 6?

Solution: The sum of the degrees of the vertices is 610 =
60. According to the Handshaking Theorem, it follows that
2e = 60, so there are 30 edges.

Graph Terminology
14
Theorem: An undirected graph has an even number of
vertices of odd degree.
Proof: Let V1 and V2 be the set of vertices of even and
odd degrees, respectively (Thus V1  V2 = , and V1
V2 = V).
Then by Handshaking theorem
2|E| = vV deg(v) = vV1 deg(v) + vV2 deg(v)
Since both 2|E| and vV1 deg(v) are even,
vV2 deg(v) must be even.
Since deg(v) is odd for all vV2, |V2| must be even.
QED

Exercise 2
15

Draw a graph with the specified properties or show
that no such graph exists:
A
graph with 6 vertices with the following degrees:
1,1,2,2,3,4
 A graph with 4 vertices of degrees 1,2,3,4
 A simple graph with 4 vertices of degrees 1,2,3,4
Exercise 3
16




A graph has 5 vertices of degrees 1,1,4,4, and 6.
How many edges does the graph have?
Is it possible in a group of 13 people for each to
shake hands with exactly 7 others?
Is it possible to have a graph with 15 edges where
each vertex has degree 4?
Is it possible to have a simple graph with 10 edges
where each vertex has degree 4?
Graph Terminology
17
Definition: When (u, v) is an edge of the graph G with
directed edges, u is said to be adjacent to v, and v is
said to be adjacent from u.

The vertex u is called the initial vertex (or source) of
(u, v), and v is called the terminal vertex (or target) of
(u, v).

The initial vertex and terminal vertex of a loop are the
same.

Graph Terminology
18
Definition: In a graph with directed edges, the indegree of a vertex v, denoted by deg-(v), is the number
of edges with v as their terminal vertex.

The out-degree of v, denoted by deg+(v), is the
number of edges with v as their initial vertex.

Question: How does adding a loop to a vertex
change the in-degree and out-degree of that vertex?

Answer: It increases both the in-degree and the outdegree by one.

Graph Terminology
19
Example: What are the in-degrees and out-degrees
of the vertices a, b, c, d in this graph:
deg-(a) = 1
deg-(b) = 4
a
+
b deg+(b) = 2
deg (a) = 2

deg-(d) = 2
deg+(d) = 1
d
c
deg-(c) = 0
deg+(c) = 2
Graph Terminology
20
Theorem: Let G = (V, E) be a graph with directed
edges. Then:


vV deg-(v) = vV deg+(v) = |E|
This is easy to see, because every new edge increases
both the sum of in-degrees and the sum of out-degrees
by one.

Exercise 4
21


What is the maximum number of edges possible in
a simple graph on n vertices?
What is the maximum number of edges possible in
a directed graph on n vertices (loops included)?
Special Graphs
22
Definition: The complete graph on n vertices, denoted
by Kn, is the simple graph that contains exactly one
edge between each pair of distinct vertices.

K1
K2
K3
K4
K5
Exercise 5
23






What is the degree of each vertex in the complete
graph K9?
What is the total degree of K9?
How many edges are there in K9?
How many edges are there in Kn?
What is the degree of a vertex in Kn?
What is the total degree of Kn?
Special Graphs
24
Definition: The cycle Cn, n  3, consists of n vertices v1,
v2, …, vn and edges {v1, v2}, {v2, v3}, …, {vn-1, vn}, {vn,
v1}.

C3
C4
C5
C6
Special Graphs
25
Definition: We obtain the wheel Wn when we add an
additional vertex to the cycle Cn, for n  3, and connect
this new vertex to each of the n vertices in Cn by adding
new edges.

W3
W4
W5
W6
Special Graphs
27
Definition: A simple graph is called bipartite if its vertex
set V can be partitioned into two disjoint nonempty sets V1
and V2 such that every edge in the graph connects a vertex
in V1 with a vertex in V2 (so that no edge in G connects
either two vertices in V1 or two vertices in V2).

For example, consider a graph that represents each person
in a village by a vertex and each marriage by an edge.

This graph is bipartite, because each edge connects a
vertex in the subset of males with a vertex in the subset of
females (if we think of traditional marriages).

Special Graphs
28

Example I: Is C3 bipartite?
v1
v2
v3
No, because there is no way to
partition the vertices into two sets
so that there are no edges with
both endpoints in the same set.
Example II: Is C6 bipartite?
v1
v6
Yes, because
v2
v5 we can display
C6 like this:
v3
v4
v1
v5
v3
v6
v2
v4
Special Graphs
29
Definition: The complete bipartite graph Km,n is the
graph that has its vertex set partitioned into two subsets
of m and n vertices, respectively. Two vertices are
connected if and only if they are in different subsets.

K3,2
K3,4
Exercise 6
30

Draw the complete bipartite graphs for K2,2, K2,3,
and K3,4
Exercise 8
32

What is the degree of each vertex in the complete
bipartite graph K4,5?
Operations on Graphs
33
Definition: A subgraph of a graph G = (V, E) is a
graph H = (W, F) where WV and FE.

Note: Of course, H is a valid graph, so we cannot
remove any endpoints of remaining edges when
creating H.


Example:
K5
subgraph of K5
Operations on Graphs
34
Definition: The union of two simple graphs G1 =
(V1, E1) and G2 = (V2, E2) is the simple graph with
vertex set V1  V2 and edge set E1  E2.


The union of G1 and G2 is denoted by G1  G2.
G1
G2
G1  G2 = K5
Exercise 9
35

Let G be a simple graph with V = {a,b,c,d,e} and
E = {{a,a},{a,b},{a,c},{b,c},{c,d}}.
 Is
H = (VH,EH) with VH={a,b,c,d} and
EH={{a,c},{b,c},{c,d}} a subgraph of G?
 If so, find a second subgraph L such that H  L = G
36
Section 10.2
Representing Graphs, Walks, Paths and Circuits,
Connectedness
Representing Graphs
37
a
d
b
d
c
Vertex
a
b
c
d
Adjacent
Vertices
b, c, d
a, d
a, d
a, b, c
a
b
c
Initial
Vertex
a
b
c
d
Terminal
Vertices
c
a
a, b, c
Representing Graphs
38
Definition: Let G = (V, E) be a simple graph with |V|
= n. Suppose that the vertices of G are listed in
arbitrary order as v1, v2, …, vn.

The adjacency matrix A (or AG) of G, with respect to
this listing of the vertices, is the nn zero-one matrix with
1 as its (i, j)th entry when vi and vj are adjacent, and 0
otherwise.


In other words, for an adjacency matrix A = [aij],
a =1
aij = 0
 ij
if {vi, vj} is an edge of G,
otherwise.
Representing Graphs
39
Example: What is the adjacency matrix AG for the
following graph G based on the order of vertices a, b,
c, d ?
a
0 1 1 1 
1 0 0 1
b

Solution: AG  
d
1 0 0 1
c


1 1 1 0

Note: Adjacency matrices of undirected graphs
are always symmetric.
Graph Walks
40
Definition: A walk of length n from u to v, where n is a
positive integer, in an undirected graph is a sequence
of edges e1, e2, …, en of the graph such that
e1 = {x0, x1}, e2 = {x1, x2}, …, en = {xn-1, xn}, where
x0 = u and xn = v.

When the graph is simple, we denote this path by its
vertex sequence x0, x1, …, xn, since it uniquely
determines the path.

The path or circuit is said to pass through or traverse
x1, x2, …, xn-1.

Paths and Circuits
41
A trivial walk from v to v consists of the single vertex v, and no
edges.
The path is a closed-walk if it begins and ends at the same
vertex, that is, if u = v.
The length of a walk is the number of edges in the walk
A trail is a walk with no repeated edges
A path is a walk with no repeated vertices
A circuit is a closed walk with no repeated edges
A path or circuit is simple if it has no repeated vertices (except
the first and last in a circuit).
A non-trivial simple circuit is called a cycle

Exercise 1
42

1.
2.
3.
4.
5.
6.
7.
8.
In the given graph, determine whether each of the
following is a path, simple path, circuit, or simple circuit
abcfb
abcf
A
F
E
fabfcdf
fabcdf
abfb
B
C
D
cfbbc
bb
e
Connectivity
43

Let us now look at something new:
Definition: An undirected graph is called connected if
there is a walk (or a path, or a simple path) between
every pair of distinct vertices in the graph.

For example, any two computers in a network can
communicate if and only if the graph of this network is
connected.

Note: A graph consisting of only one vertex is always
connected, because it does not contain any pair of
distinct vertices.

Exercise 2: Connected or not?
44
a
b
c
b
c
b
a
e
d
c
d
Yes.
a
e
No.
b
a
e
d
Yes.
c
d
f
No.
e
Connectivity
45
A graph that is not connected is the union of two or
more connected subgraphs, each pair of which has no
vertex in common.

A subgraph H is a connected component of a graph
G if:

H
is connected
H
is not a proper subgraph of any connected subgraph of
G
It follows that a graph is connected  G has only one
connected component

Exercise 3
46
Example: What are the connected components in the
following graph?
g
h
d
a
e

b
c
f
i
j
Solution: The connected components are the
graphs with vertices {a, b, c, d}, {e}, {f}, {g, h, i, j}.
Exercise 4
47

What is the minimum number of edges possible in a
connected graph on 4 vertices?
Mathematical Induction
48


Let p(n) be a statement in an integer variable n.
Suppose:
i.
ii.

p(1) is true [base step]
For all integers k  2, if p(k – 1) is true (induction
hypothesis)
To be shown: Given i and ii above, prove that p(k)
must also be true
Exercise 5
49

Use induction on n to prove that a connected graph
on n vertices has at least n – 1 edges.
50
Section 10.5
Trees
What is a Tree?
51

A graph is called acyclic if it has no non-trivial
circuits
A
tree is an acyclic, connected graph
 A trivial tree is a graph with a single vertex
 A forest is an acyclic, disconnected graph

A tree is a special kind of simple graph
Exercise 1
52

Which of the following is a tree?
Tree Example—Decision Tree
53
Tree Example: Directory Structures
54
C:\
Program Files\
WINDOWS\
Documents and Settings\
apanthon\
My Documents\
Desktop\
katchins\
Tree Example: Parse Tree
55

“The young man caught the ball”
Sentence
Noun Phrase
Verb Phrase
Article
Adjective
noun
verb
Noun Phrase
The
young
man
caught
the
ball
Every Graph Contains A Tree
56

Let G be any graph
 If
G has no cycles, then it is a tree!
 If G does have cycles, for each cycle
 Remove

 The
one edge from the cycle
G will still be connected. Why?
resulting graph G’ will be an acyclic tree
Cyclical Proofs
57

How do we prove a  b?
 Prove
a  b and b  a
a


b
a  b means “a and b are equivalent”
What if we say “a, b, and c are equivalent?
a
bc
a
b
c
a
b
c
Tree Properties
58

Let T = {V,E} be a graph. The following are
equivalent:
a)
b)
c)

T is connected and removing any edge from T
disconnects T into two subgraphs that are trees
(subtrees)
There is a unique path between any two distinct
vertices v and w in T
T is a tree
Use a cyclical proof to show this is true
Proof: ab
59


Prove that if T is connected and removing any single
edge results in two disconnected subgraphs, both of
which are trees, then there is a unique path
between any two vertices u and v
Proof by contraposition:
 Prove
that if there exists at least one alternative path
between u and v, then either T is not connected or
removing any edge will not result in two disconnected
subgraphs.
Proof ab (continued)
60

Suppose T is any graph, such that there are two
distinct paths between some vertices u and v.
 If
there are two distinct paths, then there must be a
cycle in the graph, found by following the first path
from u to v, then reversing the second path to get back
from v to u.
 We have already observed that removing an edge in a
cycle will not disconnect a graph. Therefore, there is at
least one edge which can be removed from this graph
that will not disconnect T, which is what was to be
shown.
Proof bc
61


Prove that if there exists a unique path between all
pairs of vertices in T, then T must be a tree.
Proof by contraposition that if T is not a tree, then
there is not a unique path between all pairs of
vertices in T
 There
T
are two cases:
is not connected
 T has a cycle
Proof of bc (continued)
62

Case 1: T is not connected
 Then
there must not be a path between at least two
vertices u and v, which is what was to be shown.

Case 2: T has a cycle
 Pick
any edge {u,v} on the cycle. This represents a path
from u to v. Now follow a path the ‘long way around’
the cycle, starting at v and ending at u. This path is
distinct from the first, which is what was to be shown.
Proof of ca
63


Prove that if T is a tree, then T is connected and
removing any edge results in two disconnected
subgraphs that are themselves trees.
Suppose T is a graph that is a tree
Then T is, by definition, connected and has no cycles
 Remove any one edge {u,v} from T. Since T has no cycles,
then this edge represents a unique path from u to v because
if there were another path, then {u,v} would complete a
cycle
 Therefore, removing {u,v} disconnects the graph into two
distinct components.
 The two subgraphs have no cycles, and since only one edge
was removed, they are connected. Thus they are trees.

More tree properties!
66


A tree T with n vertices has n-1 edges.
Proof (by induction):
A tree with n = 1 vertex has n – 1 = 0 edges.
 For a graph with n  2 vertices, suppose that any graph with
n – 1 or fewer vertices has one less edge than it does
vertices.

Take a tree with n vertices and remove a single edge from the tree.
This results in two smaller subtrees T1 = (V1,E1) and T2 = (V2 E2).
 n = |V| = |V1| + |V2|
 |E| = |E1| + |E2| + 1
 By the induction hypothesis, |E1| = |V1| - 1 and |E2| = |V2| - 1
 Then |E| = (|V1| - 1) + (|V2| - 1) + 1 = |V1| + |V2| - 1
 = |V| - 1, which is what was to be shown

Exercise 2
67

Prove by contradiction that a finite tree with more
than 1 vertex has at least 1 vertex of degree 1
Exercise 3
68

Is a graph with 12 vertices and 12 edges a tree?

Is any graph with 5 vertices and 4 edges a tree?

Is any connected graph with 5 vertices and 4 edges a
tree?
Exercise 4
69

Let T be a graph on n vertices. Prove that the
following are all equivalent:
a)
b)
c)
T is a tree
T is connected and has n – 1 edges
T is acyclic and has n – 1 edges
70
Section 10.6
Rooted Trees and Tree Traversals
What is a rooted tree?
71
Parent
Root(level = 0)
Child
Internal Vertex
(level = 1)
Leaf(level = 2)
Tree Definitions
72







The root is any vertex in a tree that is selected to be the
root
Level(v) = #edges it takes to reach v from the root
height(T) = the maximum level of any vertices in T
Children(v)  all vertices adjacent to v whose level is
Level(v) + 1
Parent(v) The unique vertex that is adjacent to v whose
level is Level(v) – 1
A leaf is a node with no children.
An ancestor v is any vertex w that lies on the path from the
root to v. v would be considered a descendant of all such
vertices.
Exercise 1
73
Find the:
a) Level of e
b) Height of the tree
c) Children of t
d) Parent of t
e) Ancestors of g
f) Descendants of t a
g) Leaves of the tree
r
t
s
v
w
c
b
f
g
x
u
y
z
e
d
h
i
Exercise 2
74

Design a tree to represent the table of contents of a
book:
C1
S1.1
S1.2
S1.3
C2
S2.1
S2.2
S2.2.1
S2.2.2
C3
S3.1
S3.2
Binary Trees
75


A Binary Tree is a rooted tree in which each internal vertex
has at most two children
Since there are only two, they get special names:




Subtrees get special names too!



Left child
Right child
If there is only one, call it the left child
The left subtree of a vertex v is the subtree rooted at the left child
of v
The right subtree of a vertex v is the subtree rooted at the right
child of v
A full (or complete) binary tree is a binary tree in which each
internal vertex has exactly two children
Exercise 3
76

Use a complete binary tree to represent the
following mathematical expressions:
a
+b
 (a + b)  ((c * d) – e)
M-ary Trees
77



An m-ary tree is a rooted tree in which each
internal vertex has at most m children
A full m-ary tree is an m-ary tree in which each
internal vertex has exactly m children
m = 2 (binary tree), m = 3 (ternary tree)
Theorem 1
78

Let T be a full m-ary tree with n vertices, i internal
vertices, and L leaves. Then each of the following is
true:
a) n  m i  1
b) L  i (m  1)  1
L 1 n 1
c) i 

m 1
m
Corollary to Theorem 1
79

Let T be a full binary tree with n vertices, i internal
vertices, and L leaves. Then each of the following is
true:
a) n  2i  1
n 1
b) L  i  1 
2
n 1
c) i  L  1 
2
Exercise 4
80




How many vertices does a full ternary tree with 11
leaves have?
Is there a full binary tree with 12 vertices?
How many edges does a full 5-ary tree with 100
internal vertices have?
Is there a full binary tree that has 10 internal vertices
and 13 leaves?
Exercise 5
81

The Wimbledon tennis championship is a singleelimination tournament in which a player is
eliminated after a single loss. If 31 women
compete in the championship, how many matches
must be played to determine the champion?
Exercise 6
82

Suppose someone starts a chain letter. Each person
who receives the letter is asked to send it to 5 other
people.
 If
everyone who receives the letter follows the
instructions, how many people can be reached in a tree
of height 2?
 If 125 people received the letter, but did not send it,
determine the following:
 How
many people sent the letter?
 How many people in total have seen the letter, including the
person who started it?
Exercise 7
83

A computer lab has a single wall socket with 6
outlets in it. Using power strips with 6 connections
each, how many extension cords do we need to
power 46 all-in-one computers?
Balanced Trees
84


An m-ary tree of height h is balanced if every leaf
is at level h or h – 1.
Which of the following trees are balanced?
85
Number of Vertices in a Binary Tree
Based on Height

If we have binary a tree of height h, what is the
maximum number of vertices n we can have in that
tree?
 Hint:
it’s when every internal vertex has as many
children as possible!

Given the solution to the above, how many leaves
are in the tree, given the height?
86
Minimum Binary Tree Height Based on
Number of Vertices


If a binary tree has n vertices, what is the maximum
height?
More interesting, what is the minimum height?
 n < 21, h  0
 21  n < 22, h  1
 22  n < 23, h  2
 23  n < 24, h  3
 2k  n < 2k+1, h  k (Now let’s solve this for k!)
 h  log(n)
 20
M-Ary Tree Version
87

In an m-ary tree of height h, what is the maximum number of
vertices n?






#V at Level 0: m0,
#V at Level 1: m1,
#V at Level 2: m2,
#V at Level 3: m3,
#V at Level k: mk,
n1
n  m + 1 = (m2 – 1)/(m – 1)
n  m2 + m + 1 = (m3 – 1)/(m – 1)
n  m3 + m2 + m + 1 = (m4 – 1)/(m – 1)
n  mk + … + m2 + m + 1 = (mk – 1)/(m – 1)
Back to the chain-letter example, now how many vertices would
there be if the letter was sent by everybody for 10 levels?
Theorem
88


Prove by induction that if T is an m-ary tree of
height h, then the number of leaves l  mh
Then prove the corrolary that if T is an m-ary tree
with l leaves, then logm l  h
Exercise 8
89

A full binary tree has 500 leaves
 What

is the minimum height of the tree?
Is there a ternary tree of height 4 with 100 leaves?
90
Binary Tree Traversals
YES THIS MATERIAL WILL BE ON THE EXAM!!!
Tree Traversals
91




In mathematics, we’re always studying tree properties,
and they are useful
In computing, we’re not just studying trees, we are
storing and processing trees
A tree traversal is a method for efficiently retrieving
information from a tree
There are three different traversals that have different
applications:
Pre-order
 In-order
 Post-order

Preorder Traversal of a Binary Tree
92


Let T be a rooted binary tree with root R, and left
subtree TL and right subtree TR.
The Preorder traversal of T is as follows:
 ‘Visit’
R (print value, perform computation, etc.)
 Perform a preorder traversal on TL
 Perform a preorder traversal on TR
A
B
ABC
C
About Recursion
93


Recursion is weird…but really cool!
The pre-order procedure is very easy to define
because it uses itself as part of the definition!
 ‘Visit’
R (print value, perform computation, etc.)
 Perform a preorder traversal on TL
 Perform a preorder traversal on TR

If we follow this precisely, then it will gradually take
us throughout the entire tree!
Exercise 1
94

Find the pre-order traversal of the following tree (ROOT-L-R):
Your Turn! Finish the
traversal by
performing a preorder traversal of the
right subtree.
r
a
b
c
f
j
n
Result:
d
g
e
h
k
o
i
l
p
m
q
s
Post-Order and In-Order Traversals
95


As before, let T be a rooted binary tree with root R,
and left subtree TL and right subtree TR.
The post-order traversal of T is:
 Perform
a post-order traversal of TL
 Perform a post-order traversal of TR
 ‘Visit’ R

A
BCA
B
C
The in-order traversal of T is:
 Perform
 ‘Visit’
a post-order traversal of TL
R
B
 Perform a post-order traversal of TR
A
BAC
C
Exercise 2
96

Find the Post-Order (L-R-Root) traversal:
r
a
b
c
f
j
n
d
g
e
h
k
o
i
l
p
m
q
s
Exercise 3
97

Find the in-order (L-Root-R) traversal:
r
a
b
c
f
j
n
d
g
e
h
k
o
i
l
p
m
q
s
Arithmetic Trees, Revisited
98

Consider the tree for ((a + b) * (c – d)):
*
+
a


b
c
d
Perform an in-order (L-Root-R) traversal of the tree
How can we get the parentheses back?
Arithmetic Trees, Revisited
99

Consider the tree for ((a + b) * (c – d)):
*
+
a


b
c
d
Perform a pre-order traversal
Such an expression is called the prefix form of an
arithmetic computation
Arithmetic Trees, Revisited
100

Consider the tree for ((a + b) * (c – d)):
*
+
a


b
c
d
Perform a post-order traversal
Such an expression is called the postfix form of an
arithmetic computation
Why pre-fix, post-fix matter
101


You don’t need parentheses!!!
Consider the prefix notation:
*
+ab–cd
 Scan rightleft.
 When you reach an operator, apply it to the two values
to the right
 Then
 Try
*
replace the three items with the single answer
it!
+ - 4 6 8  16 4
Computing Post-fix Expressions
102

Similarly, compute post-fix expressions from leftright:
 You’ll
always have two values, then an operand
 Apply
the operand to the values, then replace all three with
the answer you get.

Try again!
4
6 – 8 + 16 4  *
Exercise 4
103

Try some more!
 Pre-fix:
+432^8*
 Post-fix: 3 5 * 2 + 4 2 ^ 8  –

How about logic (works similarly)?
 Let
a = TRUE, b = FALSE
 Prefix:    a b   a  b
 Postfix: a b  a  ~ b

Thought exercise: postfix is easier for humans, but
prefix is easier for computers. Why?