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We have the bipartite graph $G=K_{5,9}$.

We construct a new graph $G'$ by adding a new vertex $u$ that is connected with each vertex of $G$.

Then $G'$ has an Euler circuit, because every vertex has an even degree (the degree of $u$ is $5+9=14$, the degrees of the old vertices in the new graph $G'$ are $9+1=10$ and $5+1=6$ respectively) and $G'$ is connected.

Is this correct? Or doesn't $G'$ have an Euler circuit?

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Yes, it's correct. A graph has an Euler circuit if and only if it satisfies the following two conditions: every vertex has even degree, and there is only one non-empty component. This graph is clearly connected, and the degrees are even as you say.

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  • $\begingroup$ Great!! Thank you so much!! :-) $\endgroup$ – Mary Star Aug 29 '18 at 16:07

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