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Does a (simple, undirected) graph have an Euler cycle if it has at least one vertex of odd degree, or is that statement false? Since it has at least one vertex of odd degree, it has at least two, right? So it can but it mustn't, right?

Zap
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    Have you drawn out any examples at all? – Randall Sep 25 '19 at 13:39
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    Does a simple graph even exist with precisely one vertex of odd degree? I suppose if we are being pedantic, the statement "If $G$ is a simple undirected graph with precisely one vertex of odd degree then $0=1$, it has an Euler cycle, and the moon is made of cheese" is a true statement... so the statement you are asking the truth of needs to be clarified. – JMoravitz Sep 25 '19 at 13:42
  • @JMoravitz thank you for pointing that out, I typed the question falsely. – Zap Sep 25 '19 at 13:48

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Use Equivalence condition for a graph to be Eulerian (Using degree of vertices). look here

sabeelmsk
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