MCA 520: Graph Theory Instructor Neelima Gupta

Table of Contents Graph Isomorphism

Thanks to Anika & Arpana Two graphs G=(V,E) and G’=(V,E) are isomorphic if there is a bijective function f: V(G)  V(G’) such that for all v, w  V: {v,w}  E(G) iff {f(v),f(w)}  E(G’)

X w Y Z C A B D AD CB CB DA is isomorphic to Thanks to Anika & Arpana

Showing Isomorphism using Matrices Thanks to Anika & Arpana Y W X Z D C A B Graph G Graph G’ Let there exist a bijective function f: G  G’ such that f(W) = A f(Y) = B f(Z) = C f(X) = D

Adjacency matrix of graph G Adjacency matrix of graph G’ W X Y Z A B C D W A X B Y C Z D Applying permutation to rows and columns of matrix of graph G such that given bijection holds good. W Y Z X W Y Z X Thanks to Anika & Arpana The resultant matrix corresponds to matrix of graph G’ and hence G and G’ are isomorphic to each other.

Let there exist a bijective function f: G  G’ such that f(W) = C f(Z) = A f(X) = B f(Y) = D Thanks to Anika & Arpana Applying permutation to rows and columns of matrix of graph G such that given bijection holds good. Z X W Y Z X W Y

Isomorphism An Equivalence Relation Let G = (V, E), G’ = (V’, E’) and G’’ = (V’’, E’’) all be graphs. Reflexive: For all graphs G, G = G Symmetric: If G = G’ then G’ = G Transitive: If G = G’ And G’ = G’’ then G = G’’ If f :G G’ i.e for all (u,v) € G (f(u),f(v)) € G’ g : G’ G’’ i.e for all (w,x) € G’ (g(w)g,(x)) € G’’ Then gof : G G’’ Therefore, Isomorphism is an equivalence Relation Thanks to Anika & Arpana ~ ~~ ~~~

 We can divide graphs into equivalence classes. All the graphs within an equivalence class are isomorphic to each other.  Eg All the graphs below are isomorphic to each other Thanks to Anika & Arpana

Isomorphism: as different drawings Members of Isomorphic Class are different drawings of the same graph in a plane where the vertices are fixed with their co-ordinates. So, the vertices are unlabeled in Isomorphic world.

The above graphs represent the drawing of same graph in a 2-D plane. Thanks to Anika & Arpana (2,3 ) (3,1) (5,2) (4,4) (2,3 ) (4,4) (5,2) (3,1) If labels are fixed the graph correspond to a drawing and if labels are removed then these drawings become isomorphic graphs.

Some Common Terms P n : Path with n vertices Eg All the graphs isomorphic to the below graph are P 4. C n : Cycle with n vertices Eg All the graphs isomorphic to below graph are C 4. Thanks to Anika & Arpana

K n : Complete graph with n vertices. Eg K 4 K n,m : Complete Bipartite graph with m vertices on one side and n vertices on other side. There cannot be cycles of odd length in a bipartite graph. eg K 2,3 Thanks to Anika & Arpana

Isomorphic Or Not Check if the following graphs are isomorphic or not. G1 G2 Thus, graphs G1 and G2 are isomorphic. u v w x y z G1 is a bipartite graph. Thanks to Arti, Asmita

G3 G3 is not a bipartite graph as it contains a cycle of odd length. Thus, it is not isomorphic to G1 and G2. — G4 Thus, G4 is isomorphic to G1 and G2. lm n r pq lr q m n p Thanks to Arti, Asmita

There are 2 n C 2 graphs possible for n vertices. For n=3, n C 2 = 3 edges possible => 2 3 = 8 graphs possible. G1 G2 G3 G4 G5 G6 G7 G8 Thus, there are 4 isomorphic classes, {G1, G2, G3}, {G4, G5, G6}, {G7} and {G8}. F or n=4 vertices, 64 graphs are possible with 11 equivalence classes. Thanks to Arti, Asmita

Graph Decomposition A Decomposition of a Graph G is a set of subgraphs H 1, H 2, H 3,……….., H k that partition the edges of G such that E(H 1 ) E(H 2 ) E(H 3 ) ……….. E(H k ) = E(G) and E(H 1 ) ∩ E(H 2 ) ∩ E(H 3 ) ∩ ………..∩ E( H k ) = Ø Thanks to Arti, Asmita

Examples K n can be decomposed in K n-1, K n-1, 1 Decomposition of K 4 contains three copies of P 3. K 7 can be decomposed into 7 copies of K 3 K 6 can be decomposed into 5 copies of P 4

K n can be decomposed into K n-1, K n-1,1 K n-1 Thanks: Gayatri Koli Roll No. 10 Gunjan Sawhney Roll no. 11 (MCA 2014)

K n = Uk n-1,1 Bi-clique is not necessarily a complete graph. K n can be constructed from k n-1. K n-1 Thanks: Gayatri Koli Roll No. 10 Gunjan Sawhney Roll no. 11 (MCA 2014)

K 7 can be decomposed into 7 copies of k 3 (wrong figure, pls correct it) Pick one edge from each type of edge then rotate. In the given graph there are 3 types of edges- 1. circumference (black) 2. radial (blue) 3. dotted (red) Thanks: Gayatri Koli Roll No. 10 Gunjan Sawhney Roll no. 11 (MCA 2014)

K 6 can be decomposed into 5 copies of P 4. Pick one edge from each type of edge then rotate. In the given graph there are 3 types of edges- 1. circumference (black) 2. radial (blue) 3. dotted (red) Thanks: Gayatri Koli Roll No. 10 Gunjan Sawhney Roll no. 11 (MCA 2014)

Graph Complement Complement of a Graph : Let G = (V, E) be a simple graph and let K consist of all 2-element subsets of V. Then H = (V, K \ E) is the complement of G. The Complement or Inverse of a graph G is a graph H on the same vertices such that two distinct vertices are adjacent in H if and only if they are not adjacent in G. Thanks to Arti, Asmita

Self-complementary Graph Self-Complementary Graph : A graph which is isomorphic to its complement. An n-vertex Graph H is Self-Complementary if and only if K n has a decomposition consisting of two copies of H. Note:- Two copies mean two graphs isomorphic to each other. Thanks to Arti, Asmita

Example :C 5 is self complementary = + G GBGB GAGA Graph G is decomposed into G A and G B where G A is complement of G B and also, G A is isomorphic to G B. Thanks to Arti, Asmita

Example :C 5 is self complementary. K 5 = C 5 U C 5 c Thanks: Gayatri Koli Roll No. 10 Gunjan Sawhney Roll no. 11 (MCA 2014)

Formal Proof Given : G = G A G B such that E(G A ) ∩ E(G B ) = Ø and G A is isomorphic to G B. To Prove : G A and G B are Self-Complementary Graphs. Proof : Given G = G A G B …………….(1) & E(G A ) ∩ E(G B ) = Ø …………….(2) Therefore, by (1) and (2) G A is complement of G B and vice versa i.e G A = G B c and G B = G a c …………….(3) Also given that, G A is isomorphic to G B …………….(4) By (3) and (4) Hence, proved that G A and G B are SELF-COMPLEMENTARY. Thanks to Arti, Asmita

Some Special Graphs And Their Complements Triangle Claw Triangle’s complement Claw’s complement Thanks: Gayatri Koli Roll No. 10 Gunjan Sawhney Roll no. 11 (MCA 2014)

Paw Kite Paw’s complement Kite’s complement Thanks: Gayatri Koli Roll No. 10 Gunjan Sawhney Roll no. 11 (MCA 2014)

House Bull House’s complement Bull’s complement Thanks: Gayatri Koli Roll No. 10 Gunjan Sawhney Roll no. 11 (MCA 2014)

Bow Arrow Bow’s complement Arrow’s complement Thanks: Gayatri Koli Roll No. 10 Gunjan Sawhney Roll no. 11 (MCA 2014)

Petersen Graph Definition: A Simple graph whose vertices are the 2-element subsets of a 5-element set and there is an edge between {a,b} and{c,d} iff they are disjoint.

Girth A girth of a graph with a cycle is the length of its shortest cycle. A girth of a graph without a cycle is infinity.

Girth of Petersen Graph Girth of Petersen Graph is 5.

Automorphism Defn A graph G is vertex transitive if for every pair of vertices u, v in G, there is an automorphism that maps u to v. K m,n has m! n! automorphisms. K n,n has 2 (n!) 2 automorphisms.