Hamiltonian Graph

Question

A complement of a cyclic graph on 5 vertices , has an Hamiltonian circuit . (True/False)

Comments

@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                                                                                         

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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..

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dirac's theorem is only a sufficient condition , it is not necessary condition.
It is a hamiltonian graph.

Answer 1

True it have Hamiltonian ckt