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Special Graphs (Self-learning)Complete Graphs
n-Cube
Complete bipartite Graphs
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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.
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Complete Graphs
K1 K2 K3
K4 K5 K6
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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
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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
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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
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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
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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
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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
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Complete Bipartite Graphs
K3,3
K2,6
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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:
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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!
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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.
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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:
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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
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Example 2e.g. Justify if t G and
Hare isomorphic.
G
e d
a c
b
H
e d
a c
b
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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
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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
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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.
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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
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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=
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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.!
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Exercise 2
u1
u2 u3
u4
u5u6
v1
v2 v3
v4
v5v6
Are these graphs isomorphic?
G H
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Exercise 3Are these Graphs Isomorphic?
u1
u2
u3
u5 u4
G H
v1
v2
v3v4
v5
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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
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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