2 2 votes A complement of a cyclic graph on 5 vertices , has an Hamiltonian circuit . (True/False) Mathematical Logic graph-theory discrete-mathematics + – VS 2.3k views answer comment Share Follow Print See all 9 Comments 9 9 Comments reply joshi_nitish commented Jan 16, 2018 reply Follow flag yes, true! cycle graph with 5 vertices is self complementary, therefore complement of $C_5$ is also $C_5$ and therefore it will also have Hamiltonian cycle. 0 0 replyShare Ajay Jadhav commented Jan 16, 2018 reply Follow flag I think it will be K5 0 0 replyShare joshi_nitish commented Jan 16, 2018 reply Follow flag can complement of $C_5$ be ever $K_5$ ? 0 0 replyShare Ajay Jadhav commented Jan 16, 2018 reply Follow flag yeah,you are right,my mistake 0 0 replyShare sid1221 commented Jan 16, 2018 reply Follow flag can i say like complement of c5 has vertex degree 2 ( max degree - degree of cycle graph with 5 vertex =4-2) which is even , from theorem its hamilton ckt 0 0 replyShare VS commented Jan 16, 2018 i edited by VS Jan 16, 2018 reply Follow flag @joshi_nitish I am not talking about Cycle graph, I am asking Cyclic Graph. A cyclic graph is a graph containing at least one graph cycle. 2 2 replyShare VS commented Jan 16, 2018 reply Follow flag @Pawan Kumar 2 This graph is Eulerian , but NOT Hamiltonian. Here, deg(v) >= floor(n/2) = 2 , But , it is not hamiltonian. In formula for Dirac's theorem , we don't have floor. Dirac's Theorem Let G be a simple graph with n vertices where n ≥ 3 If deg(v) ≥ n/2 for each vertex v, then G is Hamiltonian. Ref: http://personal.kent.edu/~rmuhamma/GraphTheory/MyGraphTheory/eulerGraph.htm 2 2 replyShare Pawan Kumar 2 commented Jan 16, 2018 i edited by Pawan Kumar 2 Jan 16, 2018 reply Follow flag yes mam I've corrected it now thanks .. In hamiltonian ckt , every vertex is traversed once and traversal starts and ends at same node .. but thats not the case with hamiltonian path....But still Dirac's and Ore's theorem used for Hamiltonian Ckt are sufficient but not necessary conditions.. 0 0 replyShare K_Nishant commented Aug 13, 2018 reply Follow flag dirac's theorem is only a sufficient condition , it is not necessary condition. It is a hamiltonian graph. 0 0 replyShare Please log in or register to add a comment.
0 0 votes True it have Hamiltonian ckt Anup dogrial answered Dec 11, 2019 Anup dogrial comment Share Follow 0 reply Please log in or register to add a comment.