Was hoping if anyone had any way to prove the following claim using proof by contradiction

Let $G = (V, E)$ be a simple graph with at least one vertex, and let $G'$ be the graph formed by adding a new vertex $v$ and making it adjacent to every vertex in $V$.

Claim: $G$ has a Hamiltonian Path if and only if $G'$ has a Hamiltonian cycle.

I tried manipulating the definitions of each of the two (path vs. cycle), but didn't find much luck. Any thoughts?


  • $\begingroup$ What prompted you to want a proof by contradiction for this, or at all? $\endgroup$ – Derek Elkins left SE Oct 31 '18 at 18:32

Contradiction isn't the most natural way to do this, since there's a straightforward constructive proof.

Suppose $G$ has a Hamiltonian path. How can you extend that to a Hamiltonian cycle of $G'$? Conversely, suppose that $G'$ has a Hamiltonian cycle. Cover up $v$ with your finger – what do you see?

It's possible to rewrite the reasoning hinted at above as a proof by contradiction, but it's not very natural. ("Suppose $G$ has a Hamiltonian path but $G'$ has no Hamiltonian cycle. Oh, look, here's a Hamiltonian cycle in $G'$ – contradiction.")

  • $\begingroup$ So, in order for G' to have a Hamiltonian cycle, G has to have a path? That makes sense, since you can't have a cycle without a path (I think). And yeah, the contradiction would be strange, but pretty straightforward as you suggest. Guess I'm still trying to visualize this better. $\endgroup$ – flo_mar Oct 31 '18 at 16:58
  • $\begingroup$ Yes -- a Ham cycle in $G'$ must visit every vertex of $G'$ and, if you forget about the new vertex $v$, then you're left with a path. $\endgroup$ – David Richerby Oct 31 '18 at 17:36

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