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Does there exist a simple graph that has five vertices each of degree three?

I say it does not because co-ordinate to the rule $2m =$ sum of degrees, where $m$ are the edges. The sum of the degrees is $15$, then $2m = 15$. Then, the number of edges $= \frac{xv}{two}$, which is not an integer.

asked Apr 3, 2014 at 3:46

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  • $\begingroup$ Right! $\ \ $ $\endgroup$

    Apr 3, 2014 at iv:27

  • $\begingroup$ Well, actually yes there is, but it must have at least 6 vertices in total :). $\endgroup$

    Jan 10, 2015 at i:53

2 Answers 2

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As David mentions in the comments, you are correct. The dominion you refer to is often chosen the degree sum formula or the handshaking lemma. The same rule can be used to bear witness the following more than general statement:

Any graph must have an fifty-fifty number of odd vertices.

Here a vertex is called odd if it has odd degree.

answered January 10, 2015 at 1:47

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really information technology does non exit.because according to handshaking theorem twice the edges is the degree.but v vertices of degree 3 which is equal to 3+3+iii+3+three=15.information technology should be an even number and fifteen is not an even number and besides the number of odd degree vertices in an undirected graph must be an fifty-fifty count.

answered Oct 23, 2018 at 12:30

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