chapter 11 - graph csnb 143 discrete mathematical structures

Chapter 11 - Graph

CSNB 143 Discrete Mathematical

Structures

OBJECTIVESStudent should be able to identify graphs

and its components. Students should be able to understand all

the important things about graphs. Students should know how to use certain

ways to solve problems in graphs.

GraphA graph G consist of a finite set V of

objects called vertices (bucu-bucu), a finite set E of objects called edges (sisi-sisi) and a function , that assigns two vertices to each edge (may be the same vertex).

We will write G = (V, E, ). If e is an edge, and (e) = {v, w), that are

vertices for e, we say e is an edge between v and w. The vertices v and w are called the end points of e.

Ex 1: Let say V = {1, 2, 3, 4} and E = {e1, e2, e3, e4, e5}. Let defined by

(e1) = (e5) = {1, 2}; (e2) = {4, 3}; (e3) = {1, 3}; (e4) = {2, 4}

Then G = (V, E, ) is a graph. Usually, graphs are represented by pictures, using a point for each vertex and a line for each edge, as below:

1 2

4 3

Graphs are often used to record information about relationships or connections. It is because the connections are the most important information, and not the pictures.

Ex 2: From Ex 1, other pictures for graph G: 1 2 1 2 3

4

3 4

The degree of a vertex is the number of edges connected to it.

Any edge that connects from a vertex to the same vertex, such as loop, contributes 2 to the degree of a vertex.

A vertex with degree 0 is called an isolated vertex.

A pair of vertices that determine an edge, or sharing the same edge, are adjacent (bersebelahan) vertices.

Ex 3: Consider graphs below:

A B .a .b 2 5

3 4

E C .c 1 .d .e

6 D

(a) (b) (c)

In (a), vertex A has degree 2, vertex B has degree 4, vertex C has degree 1, vertex D has degree 3, vertex E has degree 2.

Find the degrees for all vertices in (b) and (c).

In (b), vertex e is an isolated vertex. Vertices a and b are adjacent. Vertices a and d are not adjacent.

A path in a graph G consists of a vertex sequence of : v1, v2, …, vk, each one is adjacent to the next, and no edge occurs more than once.

In short, path is a journey from vertex v1 and travel through the edges to vk without using any edge twice.

A circuit is a path that begins and ends at the same vertex. (circuit = cycle).

A path is called simple if no vertex appears more than once in the vertex sequence.

Ex 4: Consider Ex 3:In (a), path 1: A, B, E, D, D; path 2: D, E, B, C and

path 3: A, B, A. Note that in 3, we do not know which path between A and B has been used first.

In (b), path 4: a, b, c, a; and 5: d, c, a, a, b. Path 4 is a cycle. Path 6: c, a, b, c, d is not simple.

In (c), sequence 1, 2, 3, 2 is not a path, because the single edge between 2 and 3 has been traveled twice. Path 7: 4, 5, 6, 4 is simple and it is a cycle.

A graph is called connected if there is a path from any vertex to any vertex in the graph.

If not, the graph is called disconnected. If the graph is disconnected, the various connected pieces are called the components of the graph.

Ex 5: Consider Ex 3:Graph (a) is connected. Graphs (b) and (c)

are disconnected graphs. Graph (c) has two components.

Sub graphLet say G = (V, E, ) is a graph. Choose a

subset E1 from E and subset V1 from V so that V1 contains all the end points of edges in E1. Then H = (V1, E1, 1) is also a graph where 1 is restricted to edges in E1.

H is called the sub graph of G.

Ex 6: Consider Ex 3. All of these below are the sub graphs of (a).

A B A B B

E E C E C

D D

(d) (e) (f)

Euler Path and CircuitA path in a graph G is called an Euler path if it

includes EVERY EDGE exactly once. An Euler circuit is an Euler path that is a

circuit. Ex 7: Consider the following graph: E 2 1 D 3 B C 4 5

A (a) (b)

Euler path exist in (a); E, D, B, A, C, D, but Euler circuit is not exist.

Euler circuit exist in (b); 5, 3, 2, 1, 3, 4, 5.

Ex 8: Consider the floor plan as below. Each room is connected to every room and to the outside. Is it possible to begin in a room or outside and take a walk that goes through each door exactly once?

Room A Room B

Room C D outside

D is labeled as outside area. The answer would be easier to find if we

change it into graph.

A B D

CWhen we are looking at a graph, is it

possible to determine whether an Euler path or Euler circuit exists without actually finding it?

Theorem 1 (to identify a circuit)If a graph G has a vertex of odd degree,

there can be no Euler circuit in G. If G is a connected graph and every vertex

has even degree, then there is an Euler circuit in G.

Theorem 2 (to identify a path)If a graph G has more than two vertices

of odd degree, then there can be no Euler path in G.

If G is connected and has exactly two vertices of odd degree, there is an Euler path in G. The path must begin at one vertex of odd degree and end at the other.

So the previous graph will be:

Room A Room B

Room C D outside

So, Euler path exist. Use the Theorem given above to prove that.

Degrees for the four vertices are: A = 4, B = 4, C = 5, D = 7.

From Theorem 1, there is odd degree vertex, so, there is no Euler circuit.

From Theorem 2, there are exactly two odd degree vertices, so there is an Euler path. It must begin from C/D and ends at D/C.

The path is: C, D, C, A, D, A, B, D, B, C, D

 Exercise: Consider the graphs below, and find an Euler path or circuit.

(a)

(c) (b)

(d)

Hamilton Path and CircuitA Hamilton path is a path that contains EACH

VERTEX exactly once. Hamilton circuit is a circuit that contains each vertex

exactly once except for the first vertex, which is also the last.

Ex 9: Consider the following graphs: d c e 2 5 b 3 4 a 1 6 (a) (b)

A B A B

D C E C

(c) D

(d)

In (a), path c, d, e, b, a is a Hamilton path because it contains each vertex exactly once. There is no Hamilton circuit.

In (b), there is no Hamilton path or circuit because it is a non connected graph.

In (c), path A, D, C, B is a Hamilton circuit.

In (d), there is no Hamilton path or Hamilton circuit.

When we saw a graph, is it possible to determine whether a Hamilton path or circuit exists or not?

Theorem 1: Let G be a connected graph with n vertices,

n > 2 and no loops or multiple edges, G has a Hamilton circuit if for any two vertices u and v of G that are not adjacent, the degree of u plus the degree of v is greater that or equal to n.

D(u) + D(v) the number of vertices.Corollary 1: G has a Hamilton circuit if each vertex has degree

greater than or equal to n/2.

Reminder: If Theorem 1 failed, we cannot say that the graph has no Hamilton circuit. We have to check the circuit or path manually.

Ex 10: Consider the following: B C D

A E

H G F

Take two non adjacent vertices; let say A and E. The sum of their degrees are (2 + 2) = 4 < 8 (number of vertices). But if we check properly in manual, we can have the Hamilton circuit that is A, B, C, D, E, F, G, H, A

Exercise: Consider graphs from Euler exercise. Use Theorem 1 first before manual checking.