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This question already has an answer here:

I read a general principal of graph theory which says "A graph that can be traversed without lifting the pencil from the paper , while tracing each edge exactly once, can have not more than 2 odd vertices." Can I have any proof of it? I am new to graph theory

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marked as duplicate by hardmath, user147263, Leucippus, graydad, user99914 Sep 10 '15 at 5:19

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Let's say a circuit is a path in a graph that ends where it starts. Such a circuit graph has even order at every vertex.

Now consider removing any edge.

Now, is the above argument reversible?

... and we're done!

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think in your way you every times that have entered an odd vertice you have to be able to go out. thus for doing this goal you only can atmost have two odd vertice, using them in start and finish point.(I hope you understand me, sorry for weak english)

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  • $\begingroup$ welcome, dear :) $\endgroup$ – R.N Sep 9 '15 at 15:52

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