By Jurisic A.
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Additional info for 1-Homogeneous Graphs with Cocktail Party, mu -Graphs
7) from u to v where vi 2 X; iff i is odd. If there is an edge joining v1 and vi there cannot be an edge joining viÀ1 and v2n ; since H is non-Hamiltonian. Since dðuÞ [ n=2; we find that dðvÞ\n À n2 ; which contradicts the hypothesis. 3 Complement and Self-Complementary Graph " of G is Complement: Let G be a simple graph with n vertices. The complement G defined to be the simple graph with the same vertex set as G and where two vertices u and v are adjacent precisely when they are not adjacent in G.
DðxÞ þ dðyÞ ! n in H as well. If we join the nonadjacent vertices x and y, the resulting graph is Hamiltonian. Hence, in graph H, there is a Hamiltonian path between the vertices x and y. If we write x ¼ v1 and y ¼ vn , this Hamiltonian path can be written as v1 v2 ... vi −1 Fig. 4 A Hamiltonian path from v1 to vn vi vi +1 ... v n −1 vn 30 3 Euler Graphs and Hamiltonian Graphs Suppose the degree of v1 is c in graph H. If there is an edge between v1 and vi in this graph, the existence of an edge between viÀ1 and vn will imply that H is Hamiltonian.
S. 1(Euler Theorem) A connected graph G is Eulerian (Euler graph) iff every vertex has an even degree. Proof Necessary condition: Let the graph G be Eulerian. Ã Let W : u ! u be an Euler tour and v be any internal vertex such that v 6¼ u. Suppose, v appears k times in this Euler tour W: Since every time an edge arrives at v, another edge departs from v, and therefore, dG ðvÞ ¼ 2k (Even). Also, dG ðuÞ is 2, since W starts and ends at u. Hence, the graph G has vertices of all even degree. Sufficient condition: Let us assume G is a non-trivial connected graph such that for all vertex v 2 VðGÞ; dG ðvÞ is even.