8 self-learning (special graphs and graph iso morph ism)

8/2/2019 8 Self-Learning (Special Graphs and Graph Iso Morph Ism)

1/26

1

Special Graphs (Self-learning)Complete Graphs

n-Cube

Complete bipartite Graphs


2/26

2

Complete Graphs

Complete Graphs:

The complete graph onn vertices, denoted byKn, is the

simple graph that contains exactly one edge between eachpair of distinct vertices.


3/26

3

Complete Graphs

K1 K2 K3

K4 K5 K6


4/264

n-Cubes

Then-cube, denoted by Qn (n1), is the graph that has vertices

representing the 2n

bit strings of length n. Two vertices are adjacentif and only if the bit strings that they represent differ in exactly one

bit position.

Q2Q3

000 001

011010

100

110 111

101

00 01

10 11Q1

0 1


5/265

Bipartite Graphs

A simple graph G=(V,E) is calledbipartite, ifVcan be

partitioned into two disjoint nonempty sets V1 and V2 (i.e. V=V1V2 and V1V2 = ) such that every edge in the graph

connects a vertex in V1 and a vertex in V2 (so that no edge in G

connects either two vertices in V1 or two vertices in V2)

V1 V2

v1

v2

v3

v4

v5

v6


6/266

Bipartite Graphs

Is C6 bipartite?

Can we partition V={v1, v2, v3, v4, v5, v6}into non-empty sets V1 and V2 such that

every edge in C6 connects a vertex in V1

and a vertex V2 ?

Yes!

v1

v2

v3

v4

v5

v6

V1 V2

v1 v2

v3 v4

v5 v6


7/267

V1 V2

Bipartite Graphs

Is K3 bipartite?

Can we partition V={v1, v2, v3} intonon-empty sets V1 and V2 such that

every edge in K3 connects a vertex in

V1 and a vertex V2 ?

No!

v1 v2

v3

v1

v2v3


8/26

8

Complete Bipartite Graphs

Thecomplete bipartite graph Km,n (m, n 1) is the graph with

m+n vertices that has its vertex set partitioned into two subsets of

m and n vertices, respectively.There is an edge between two vertices if and only ifone vertex is

in the first subset and the other vertex is in the second subset.

K2,3

m

n


9/26

9

Bipartite VS. Complete Bipartite

If a graph can be partitioned into two

parts such that there is no intra-connections in either part, the graph is

bipartite.If every vertex in a bipartite graph hasinterconnections with all the vertices inits counterpart, this graph is completebipartite


10/26

10

Complete Bipartite Graphs

K3,3

K2,6


11/26

11

Graph Isomorphism : Introduction

v1

v2

e1

v1

v2

e1e2v3

e2

v3

e3

v4

e5

e3

v4 e4v5

e4

v5 e5

The same graph can be drawn in many different ways:


12/26

12

Introduction

v1v

2v3v4v5

Vertices

ofG v1v

2v3v4v5

Vertices

ofG

G and G denote the same graph.

Can be shown formally by using a function.

v1

v2v5

v3v4

v1

v3v2

v5v4

G:

G:

One-to-one?

Onto?

Yes!

Yes!


13/26

13

DefinitionThe simple graphs G1=(V1,E1) and G2=(V2,E2) are isomorphic

if there is a one-to-one and onto function ffrom V1 to V2 withthe property that

for all vertices a, b V1:

{a, b} is an edge in G1 {f(a),f(b)} is an edge in G2

Such a functionfis called an isomorphism.

In other words: two simple graphs are isomorphic, if there is

a one-to-one correspondence between the vertices of the two

graphs that preserves the adjacency relationship.


14/26

14

Example 1

u1 u2

v3 v4u3 u4

v1 v2

Show that thesegraphs are

isomorphic:

u1u2

u3u4

v1v2

v3v4

G H

Find one-to-one

correspondencef

between the vertices:


15/26

15

Example 1

0 1 1 0

1 0 0 1

1 0 0 1

0 1 1 0

u1u2

u3u4

u1 u2 u3 u4

AG =

v1v4

v3v2

v1 v4 v3 v2

0 1 1 0

1 0 0 1

1 0 0 10 1 1 0

AH

=

SinceAG =AH, it follows thatfpreserves adjacency, so

u1 u2

v3 v4u3 u4

v1 v2

GH

Rewrite the adjacency

matrix of the secondgraph using the order

of the vertex index of

the preimages of the

functionf,

u1

u2u3u4

v1

v2v3v4


16/26

16

Example 2e.g. Justify if t G and

Hare isomorphic.

G

e d

a c

b

H

e d

a c

b


17/26

17

Isomorphic InvariantA property P is called isomorphic invariant , if and only if

given any simple graphs G andH, ifG has property P and

His isomorphic to G, thenHhas property P.

Some typical isomorphic invariants:

1. Same number of vertices

2. Same number of edges

3. Same degrees of vertices (that is, a vertex v of degree d

in G must correspond to a vertexf(v) of degree dinH)

4. Simple circuit of length k, where k>2


18/26

18

Example 2e.g. Justify if t G and

Hare isomorphic.

G

e d

a c

b

H

e d

a c

b

Number of vertices: both 5

Number of edges: both 6

Degrees of vertices: deg(e)=1 inH, but G has no such a vertex.

So, NOT isomorphic


19/26

19

How to justify IsomorphismSTEP 1: Check I somorphic I nvariants.

If there is any difference, conclude they are not isomorphic.

STEP 2: Const ruct 1-2-1 Correspondence.If you hardly find such a function, you may need more invariance analyses (see the optional example).

STEP 3: Check New Adj acency Mat r ices.Rewrite the adjacency matrix of the second graph using the order of the index of the vertex preimages of the functionf, to see if

the two adjacency matrices are identical. If yes,they are isomorphic, if no, the difference between the two graphs can beseen from the matrices.


20/26

20

Isomorphism

Show that these

graphs are

isomorphic:

Find one-to-one correspondencefbetween the vertices which

preserves adjacency :

u1u2u3u4

u5

v1

v2v3v4

v5

Tips: Start with a special pair of vertices, consider connections when building up

mapping for the rest of the vertices.

u1

u2

u5

G u3

u4

v2

v4v5

H

v1 v3


21/26

21

Isomorphismu1u2

u3u4u5

v1v2

v3v4v5

Compare adjacency matrices:

u1u2

u5u4

v1

v2

v3

v4v5

G

H

u3

u1

u2u3u4

u5

u1 u2 u3 u4 u5

0 1 1 1 1

1 0 0 0 11 0 0 1 0

1 0 1 0 0

1 1 0 0 0

AG =

v2

v3v1v5

v4

v2 v3 v1 v5 v4

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 0

1 1 0 0 0

AH=


22/26

22

Exercise 1Are these graphs

isomorphic?

u1 u2

u3

u5

u6

u4

v1

v2 v6

v4v5

v3

G H

Find one-to-one correspondence

fbetween the vertices which

preserves adjacency :

u1u2u3u

4u5u6

v1v2v3v

4v5v6

D.I.Y.!


23/26

23

Exercise 2

u1

u2 u3

u4

u5u6

v1

v2 v3

v4

v5v6

Are these graphs isomorphic?

G H


24/26

24

Exercise 3Are these Graphs Isomorphic?

u1

u2

u3

u5 u4

G H

v1

v2

v3v4

v5


25/26

25

Are these

GraphsIsomorphic?

u1 u2 u3 u5 u6 u8

u4 u7

v1 v2 v4 v5 v6 v8

v3 v7

Check typical Invariants:

Number of vertices: both 8

Number of edges: both 7

Degrees of vertices: 4 deg=1, 2 deg=2, 2 deg=3

G H

More Invariance Analyses (Optional)

Simple circuits: both 0


26/26

26

More Invariance Analyses (Optional)

u1 u2 u3 u5 u6 u8

u4 u7

v1 v2 v4 v5 v6

v3 v7

v8

since deg(u2)=2 in G, u2 must correspond to v4 or v5 inH,

since these are the vertices of degree 2 inH

Although, invariants are the same, there is still a difference!

G H

u2 is adjacent u1, a vertex of degree 1

but neither v4 or v5 is adjacent to a vertex of degree 1 inH

deg(u1)=1

8 self-learning (special graphs and graph iso morph ism)

Documents