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 May 23 accepted Proving a simple graph is a connected graph May 23 accepted Proofs involving subtrees of a tree May 23 accepted Constructing a 3-regular graph with no 3-cycles May 23 accepted Counting the number of graphs with a certain property May 23 accepted How to show a graph is not Hamiltonian? Apr 10 comment Counting the number of graphs with a certain property I found something in the text that says if a graph has an odd cycle, then $\chi (G) \geq 3$. But how do I check all simple graphs with odd cycles to see which follow the property above? Apr 10 asked Counting the number of graphs with a certain property Apr 3 comment How to show a graph is not Hamiltonian? Thank you very much for the clear answer! I am curious. Does a proof exist that does not require referencing the Petersen graph? (Even if it is the only graph satisfying such requirements) That is, is there some general criteria that would force a graph to be non-Hamiltonian based solely on the number of vertices/girth/regularness? Apr 3 comment How to show a graph is not Hamiltonian? $v_G$ is the number of vertices of $G$. I have edited my post to add this explanation! Apr 3 revised How to show a graph is not Hamiltonian? added 45 characters in body Apr 3 comment How to show a graph is not Hamiltonian? Oops, upon reading my post again I realize I forgot to add the property that $G$ was 3-regular. My bad! But nice counterexample for the original post. Apr 3 revised How to show a graph is not Hamiltonian? added 24 characters in body Apr 3 asked How to show a graph is not Hamiltonian? Mar 20 comment Proofs involving subtrees of a tree Sorry, I didn't notice the typo. Indeed it was a nonempty intersection, and I've corrected the post. Mar 20 revised Proofs involving subtrees of a tree added 3 characters in body Mar 20 asked Proofs involving subtrees of a tree Mar 13 asked Proving a simple graph is a connected graph Mar 13 comment Constructing a 3-regular graph with no 3-cycles No, there cannot, but I'm having trouble writing this "formally". Do I simply write that this is because $k+1$, say, would only have edges between $k$, $k+2$, and $k+(n+1)$, and so $k+n$ is not included? Mar 13 revised Constructing a 3-regular graph with no 3-cycles added 1 characters in body Mar 13 comment Constructing a 3-regular graph with no 3-cycles Oops, yes, my bad!