CSE 211Discrete Mathematics

Chapter 8.3Representing Graphs and

Graph Isomorphism

§8.3: Graph Representations & Isomorphism

• Graph representations:– Adjacency lists.– Adjacency matrices.– Incidence matrices.

• Graph isomorphism:– Two graphs are isomorphic iff they are identical

except for their node names.

Adjacency Lists

• A table with 1 row per vertex, listing its adjacent vertices.

a b

dc

f

e

Vertex

Adjacent Vertices

a b

b, c a, c, e, f

c a, b, f d e b f c, b

Directed Adjacency Lists

• 1 row per node, listing the terminal nodes of each edge incident from that node.

0 1

23

4

node Terminal nodes

0 3

1 0, 2, 4

2 1

3

4 0,2

• A simple graph G = (V,E) with n vertices can be represented by its adjacency matrix, A, where the entry aij in row i and column j is:

Adjacency Matrix

otherwise

in edgean is }{ if

0

1 G,vva ji

ij

Adjacency Matrix Example

a b c d e f

abcdef

From

To

W5

db

a

c

e

f

{v1,v2}

row column

0 1 0 0 1 11 0 1 0 0 10 1 0 1 0 10 0 1 0 1 11 0 0 1 0 11 1 1 1 1 0

Adjacency Matrices

• Matrix A=[aij], where aij is 1 if {vi, vj} is an edge of G, 0 otherwise.

ab

c

d

0001

0011

0101

1110a

b

c

d

a b c d

Adjacency Matrices• Example 5: Use an adjacency matrix to represent the

pseudograph shown in Figure 5.

0212

2110

1103

2030

FIGURE A Pseudograph.

8

Incidence Matrix

• Let G = (V,E) be an undirected graph. Suppose v1,v2,v3,…,vn are the vertices and e1,e2,e3,…,em are the edges of G. The incidence matrix w.r.t. this ordering of V and E is the nm matrix M = [mij], where

otherwise

with incident is edge if

0

1 ijij

vem

Incidence Matrix Example

• Represent the graph shown with an incidence matrix.

1 1 0 0 0 00 0 1 1 0 10 0 0 0 1 11 0 1 0 0 00 1 0 1 1 0

v1

v2

v3

v4

v5

vertices

e1 e2 e3 e4 e5 e6 edgesv1 v2 v3

v4 v5

e1e2

e3 e4 e5

e6

Incidence matrices

• Matrix M=[mij], where mij is 1 when edge ej

is incident with vi, 0 otherwise

v1

v2

v3

v4

e1 e2 e3 e4v1

v4

v3

v2

e5

11010

000

000

001

11

11

11e1

e2

e5

e4

e3

Incidence Matrices• Example 7: Represent the pseudograph

shown in Figure 7 using an incidence matrix.

FIGURE 7 A Pseudograph.

12

Isomorphism

• Two simple graphs are isomorphic if:

– there is a one-to one correspondence between the vertices of the two graphs

– the adjacency relationship is preserved

Isomorphism (Cont.)

• The simple graphs G1=(V1,E1) and G2=(V2,E2) are isomorphic if there is a one-to-one and onto function f from V1 to V2 with the property that a and b are adjacent in G1 iff f(a) and f(b) are adjacent in G2, for all a and b in V1.

One to one and onto function

One to one and onto function

Example

u1

u3 u4

u2v1

v3 v4

v2

G H

Are G and H isomorphic?

f(u1) = v1, f(u2) = v4, f(u3) = v3, f(u4) = v2

Isomorphism Example

• If isomorphic, label the 2nd graph to show the isomorphism, else identify difference.

a

b

cd

e

f

b

d

a

efc

Are These Isomorphic?

• If isomorphic, label the 2nd graph to show the isomorphism, else identify difference.

ab

c

d

e

* Same # ofvertices

* Same # ofedges

* Different# of vertices of

degree 2! (1 vs 3)

Example 11: a more general solution

Invariants

• Invariants – properties that two simple graphs must have in common to be isomorphic– Same number of vertices

– Same number of edges

– Degrees of corresponding vertices are the same

– If one is bipartite, the other must be; if one is complete, the other must be; and others …

Graph isomorphism Example

Graph GGraph H An isomorphism

between G and H

ƒ(a) = 1ƒ(b) = 6ƒ(c) = 8ƒ(d) = 3ƒ(g) = 5ƒ(h) = 2ƒ(i) = 4ƒ(j) = 7

The two graphs shown below are isomorphic, despite their different looking drawings.

Example

a

e d

c

G H

b

a

e d

c

b

Are G and H isomorphic?

Example

• Are these two graphs isomorphic?

– They both have 5 vertices– They both have 8 edges– They have the same number of vertices with

the same degrees: 2, 3, 3, 4, 4.

G H

u1 u2

u5 u3

u4

v1 v2

v5

v3

v4

Example (Cont.)

u1 u2 u3 u4 u5

u1 0 1 0 1 1u2 1 0 1 1 1u3 0 1 0 1 0u4 1 1 1 0 1u5 1 1 0 1 0

v1 v2 v3 v4 v5

v1 0 0 1 1 1v2 0 0 1 0 1v3 1 1 0 1 1v4 1 0 1 0 1v5 1 1 1 1 0

– G and H don’t appear to be isomorphic.

– However, we haven’t tried mapping vertices from G onto H yet.

v1 v3 v2 v5 v4

v1

v3

v2

v5

v4

G H H G?

Example (Cont.)

• Start with the vertices of degree 2 since each graph only has one:

deg(u3) = deg(v2) = 2 therefore f(u3) = v2

G H

u1 u2

u5 u3

u4

v1 v2

v5

v3

v4

Example (Cont.)• Now consider vertices of degree 3

deg(u1) = deg(u5) = deg(v1) = deg(v4) = 3

therefore we must have either one of

f(u1) = v1 and f(u5) = v4

f(u1) = v4 and f(u5) = v1

G H

u1 u2

u5 u3

u4

v1 v2

v5

v3

v4

Example (Cont.)

• Now try vertices of degree 4:deg(u2) = deg(u4) = deg(v3) = deg(v5) = 4

therefore we must have one of:f(u2) = v3 and f(u4) = v5 or

f(u2) = v5 and f(u4) = v3

G H

u1 u2

u5 u3

u4

v1 v2

v5

v3

v4

Example (Cont.)

• There are four possibilities (this can get messy!)f(u1) = v1, f(u2) = v3, f(u3) = v2, f(u4) = v5, f(u5) = v4

f(u1) = v4, f(u2) = v3, f(u3) = v2, f(u4) = v5, f(u5) = v1

f(u1) = v1, f(u2) = v5, f(u3) = v2, f(u4) = v3, f(u5) = v4

f(u1) = v4, f(u2) = v5, f(u3) = v2, f(u4) = v3, f(u5) = v1

G H

u1 u2

u5 u3

u4

v1 v2

v5

v3

v4

Example (Cont.)

u1 u2 u3 u4 u5

u1 0 1 0 1 1u2 1 0 1 1 1u3 0 1 0 1 0u4 1 1 1 0 1u5 1 1 0 1 0

v1 v2 v3 v4 v5

v1 0 0 1 1 1v2 0 0 1 0 1v3 1 1 0 1 1v4 1 0 1 0 1v5 1 1 1 1 0

We permute the adjacency matrix of H (per function choices above) to see if we get the adjacency of G. Let’s try:

f(u1) = v1, f(u2) = v3, f(u3) = v2, f(u4) = v5, f(u5) = v4

Does G = H’? Yes!

v1 v3 v2 v5 v4

v1 0 1 0 1 1v3 1 0 1 1 1v2 0 1 0 1 0v5 1 1 1 0 1v4 1 1 0 1 0

G H H’

Example (Cont.)

u1 u2 u3 u4 u5

u1 0 1 0 1 1u2 1 0 1 1 1u3 0 1 0 1 0u4 1 1 1 0 1u5 1 1 0 1 0

v1 v2 v3 v4 v5

v1 0 0 1 1 1v2 0 0 1 0 1v3 1 1 0 1 1v4 1 0 1 0 1v5 1 1 1 1 0

It turns out that

f(u1) = v4, f(u2) = v3, f(u3) = v2, f(u4) = v5, f(u5) = v1

also works.

v4 v3 v2 v5 v1

v4 0 1 0 1 1v3 1 0 1 1 1v2 0 1 0 1 0v5 1 1 1 0 1v1 1 1 0 1 0

G H H’

Isomorphism of Graphs• Example 10: Determine whether the graphs

shown in below are isomorphic.•

FIGURE 10 The Graphs G and H. 31

Isomorphism of Graphs

FIGURE 11 The Subgraphs of G and H Made Up of Vertices of

Degree Three and the Edges Connecting Them.

32

Isomorphism of Graphs• Example 11: Determine whether the graphs

G and H displayed in below are isomorphic.

FIGURE 12 Graphs G and H. 33

CSE 211Discrete Mathematics

Chapter 8.4Connectivity

Paths in Undirected Graphs

• There is a path from vertex v0 to vertex vn if there is a sequence of edges from v0 to vn

– This path is labeled as v0,v1,v2,…,vn and has a length of n.

• The path is a circuit if the path begins and ends with the same vertex.

• A path is simple if it does not contain the same edge more than once.

Graph Theoretic Foundations

• Paths and Cycles:

• Walk in a graph G is v0, e1, v1, …, vl-1, el, vl

Graph Theoretic Foundations

• Paths and Cycles:

• IF v0, v1, …, vl are distinct (except possible v0,vl) then the walk is a path.

• Denoted by v0, v1, …, vl or e1, e2, …, el

• Length of path is l

Paths, Cycles, and Trails

• A trail is a walk with no repeated edge.

Graph Theoretic Foundations

• Paths and Cycles:

• A path or walk is closed if v0 = vl

• A closed path containing at least one edge is called a cycle.

Paths in Undirected Graphs

• A path or circuit is said to pass through the vertices v0, v1, v2, …, vn or traverse the

edges e1, e2, …, en.

Example

• u1, u4, u2, u3

– Is it simple? – yes

– What is the length? – 3

– Does it have any circuits? – no

u1 u2

u5 u4 u3

Example

• u1, u5, u4, u1, u2, u3

– Is it simple?– yes

– What is the length?– 5

– Does it have any circuits?– Yes; u1, u5, u4, u1

u1 u2

u5 u3

u4

Example

• u1, u2, u5, u4, u3

– Is it simple?– yes

– What is the length?– 4

– Does it have any circuits?– no

u1 u2

u5 u3

u4

Connectedness

• An undirected graph is called connected if there is a path between every pair of distinct vertices of the graph.

• There is a simple path between every pair of distinct vertices of a connected undirected graph.

Example

Are the following graphs connected?

c

e

f

a

d

b

g

e

c

a

f

b

d

Yes No

Connectedness (Cont.)

• A graph that is not connected is the union of two or more disjoint connected subgraphs (called the connected components of the graph).

Example

• What are the connected components of the following graph?

b

a c

d e

h g

f

Example

• What are the connected components of the following graph?

b

a c

d e

h g

f

{a, b, c}, {d, e}, {f, g, h}

Cut edges and vertices

• If one can remove a vertex (and all incident edges) and produce a graph with more connected components, the vertex is called a cut vertex.

• If removal of an edge creates more connected components the edge is called a cut edge or bridge.

Example

• Find the cut vertices and cut edges in the following graph.

a

b c

d f

e h

g

Example

• Find the cut vertices and cut edges in the following graph.

a

b c

d f

e h

g

Cut vertices: c and eCut edge: (c, e)

Graph Theoretic Foundations• Connectivity:

• Connectivity (G) of graph G is …

• G is k connected if (G) k

• Separator or vertex-cut Cut vertex

Separation pair

Graph Theoretic Foundations

• Trees and Forests:

• Tree – connected graph without any cycle

• Forest – a graph without any cycle

Connectedness in Directed Graphs

• A directed graph is strongly connected if there is a directed path between every pair of vertices.

• A directed graph is weakly connected if there is a path between every pair of vertices in the underlying undirected graph.

Example

• Is the following graph strongly connected? Is it weakly connected?

a b

c

e d

This graph is strongly connected. Why? Because there is a directed path between every pair of vertices.

If a directed graph is strongly connected, then it must also be weakly connected.

Example

• Is the following graph strongly connected? Is it weakly connected?

a b

c

e d

This graph is not strongly connected. Why not? Because there is no directed path between a and b, a and e, etc.

However, it is weakly connected. (Imagine this graph as an undirected graph.)

Connectedness in Directed Graphs

• The subgraphs of a directed graph G that are strongly connected but not contained in larger strongly connected subgraphs (the maximal strongly connected subgraphs) are called the strongly connected components or strong components of G.

Example

• What are the strongly connected components of the following graph?a b

c

e d

This graph has three strongly connected components:• The vertex a• The vertex e• The graph consisting of V = {b, c, d} and E = { (b, c), (c, d), (d, b)}

Consider a connected graph G. The distance between vertices u and v in G, written d(u, v), is the length of the shortest path between u and v. The distance of G, written diam(G), is the maximum distance between any two points in G. For example, in Fig. 1-8(a), d(A,F) = 2 and diam(G) = 3, whereas in Fig. 1-8(b), d(A, F) = 3 and diam(G) = 4

Distance in Trees and Graphs

Distance in Trees and Graphs• If G has a u, v-path, then the distance from

u to v, written dG(u, v) or simply d(u, v), is the least length of u, v-path.

• The diameter (diam G) is maxu,vV(G)d(u,v)

• The eccentricity [(u)] of a vertex u is maxvV(G)d(u,v)

• The radius of G (rad G) is minuV(G)(u)

• A vertex with minimum eccentricity in G is called a center of G.

Distance in Trees and Graphs

• Find the diameter, eccentricity, radius and center of the given G.