MHB Statement of a theorem due to Bondy.

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The discussion revolves around a misunderstanding of a theorem related to Hamiltonian cycles in graphs, specifically a strengthened form of Mantel's theorem. It highlights that while a non-bipartite graph with at least n^2/4 edges might not necessarily be Hamiltonian, the theorem applies to Hamiltonian graphs. A specific example of a 10-vertex graph with a 9-clique and an isolated vertex is presented to illustrate this point, as it has more edges than the threshold but lacks a Hamiltonian cycle. The participants clarify that the theorem's conditions may have been misinterpreted and confirm that the correct statement includes the Hamiltonian clause. This resolution emphasizes the importance of accurate theorem statements in graph theory.
caffeinemachine
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On this page Turán's theorem - Wikipedia, the free encyclopediaI found that:

"A strengthened form of Mantel's theorem states that any graph with at least $n^2/4$ edges must either be a complete bipartite graph $K_{n/2,n/2}$ or it must be pancyclic, i.e, not only does it contain a triangle, it must also contain cycles of all other possible lengths up to the number of vertices."

Now a special case of this is:

A non-bipartite graph $G$ with $|E(G)|\geq n^2/4$, where $n=|V(G)|$, has a Hamiltonian Cycle.

But if we consider the $10$ vertex graph $G$ which has a $9$-clique and an isolated vertex then we see that $|E(G)|=36>10^2/4$ but $G$ doesn't have a Hamiltonian cycle.

What has gone wrong here? What am I missing?
 
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What I read is:

A strengthened form of Mantel's theorem states that any hamiltonian graph with at least n2/4 edges must either be the complete bipartite graph Kn/2,n/2 or it must be pancyclic: not only does it contain a triangle, it must also contain cycles of all other possible lengths up to the number of vertices in the graph (Bondy 1971).

So the special case is really:

A non-bipartite hamiltonian graph G satisfies $|E(G)| \geq n^2 / 4$ where $n = |V(G)|$.

Which does not imply that a non-bipartite graph satisfying this condition is hamiltonian.

So maybe you got the conditions the other way around. I could be wrong though.. it's late. If not, maybe the theorem only applies to connected graphs.
 
Bacterius said:
What I read is:
So the special case is really:
Which does not imply that a non-bipartite graph satisfying this condition is hamiltonian.

So maybe you got the conditions the other way around. I could be wrong though.. it's late. If not, maybe the theorem only applies to connected graphs.

You are right Bacterius. I found the correct statement on ScienceDirect.com - Journal of Combinatorial Theory, Series B - Pancyclic graphs I and had corrected it on wiki. Wiki didn't have tjhe clause 'Hamiltonian' there.
 
Bacterius said:
Ah, that explains it. Thanks :)
I almost got bald pulling my hair over this one. BUt finally science direct helped. Phewf. Thanks for taking a look at my problem. :)
 
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