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A graph $G$ has the property that every edge of $G$ joins an odd vertex with an even vertex. Show that $G$ is bipartite and has even size.

Some definitions:

  • A vertex of even degree is called an even vertex, while a vertex of odd degree is an odd vertex.

  • A graph $G$ is a bipartite graph if $V(G)$ (the set of vertices of $G$) can be partitioned into two subsets $U$ and $W$, such that every edge of $G$ joins a vertex of $U$ and a vertex of $W$.

  • Here size refers to the length of the graph, that is, to the number of edges.

What I've tried to do:

So, I've partitioned the set $V(G)$ into two subsets $E$ and $O$, where $E$ consists of the even vertices of $G$ and $O$ consists of the odd vertices of $G$.

By the property that the graph $G$ satisfies it follows that $G$ is bipartite? Do I have to show something else? I'm in doubt because it seems very easy.

Now for the second part of the exercise (G has even size) I know that if $H$ is a bipartite graph of size $m$ with partite sets $U=\{u_1,u_2,\ldots,u_s\}$ and $W=\{w_1,w_2,\ldots, w_t\}$. Since every edge of $H$ joins a vertex of $U$ and a vertex of $W$, it follows that adding the degrees of the vertices in $U$ (or in $W$) gives the number of edges in $H$, that is, $$ \sum_{i=1}^s \mbox{deg}(u_i) = \sum_{j=1}^t \mbox{deg}(w_j)=m, $$ where $\mbox{deg}(u)$ is the degree of a vertex $u$ in $H$.

Since $\mbox{deg}(u_i)$ is even for all $i=1,\ldots,s$ it follows that $$ \sum_{i=1}^s \mbox{deg}(u_i) = m $$ is even. Am I right? I'm confused because it seems to me that any bipartite graph has even size.

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  • $\begingroup$ Why does it seem to you that any bipartite graph has even size? What about the bipartite graphs $K_2$, $K_{1,3}$, $P_4$, $P_6$, $K_{3,3}$, etc. etc. etc.? $\endgroup$ – bof Oct 6 '18 at 22:01
  • $\begingroup$ Oh thanks, I get it now. It would take a long time to explain my confusion here. But I was kind of messing around with a result that says every graph has an even number of odd vertices. Anyway, is my proof correct? $\endgroup$ – Ali Khan Oct 6 '18 at 22:51
  • $\begingroup$ I'm not seeing how the latter lets you conclude that the graph has even size. The numer $m$ is supposed to be the amount of vertices, not edges, right? $\endgroup$ – Guido A. Oct 6 '18 at 23:01
  • $\begingroup$ Your proof looks fine to me. $\endgroup$ – bof Oct 6 '18 at 23:20
  • $\begingroup$ @GuidoA. $m$ is the number of edges. $\endgroup$ – Ali Khan Oct 7 '18 at 0:01
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Proof: Since every edge in $G$ joins an odd vertex with an even vertex, it cannot be the case that two odd/even vertices are adjacent. Hence, $G$ is an $X,Y$-bigraph such that $X$ is the set of even vertices, and $Y$ is the set of odd vertices.

We wish to show that $|E(G)|$ is even. Suppose, on the contrary, that $|E(G)|$ is odd. Thus, there are an odd number of edges that are incident to vertices in $X$. This is a contradiction, as each vertex in $X$ is even.

Therefore, if $G$ is a graph such that every edge connects an even and odd vertex, then $G$ is bipartite, and $|E(G)|$ is even.

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    $\begingroup$ I think we wish to show that the number of edges is even, not the number of vertices, though I am not 100% confident here. $\endgroup$ – Misha Lavrov Oct 7 '18 at 0:14
  • $\begingroup$ @MishaLavrov, thank you for the comment. The proof I had posted was flawed regardless. Ex. $K_{1,2}$ has odd $n(G)$ but satisfies the assumption. $|E(G)|$ being even is a more logical result. $\endgroup$ – Steve Schroeder Oct 7 '18 at 1:19

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