Good Evening,

So I am trying to solve this exercise which is a paticular case of maximum flow algorithm. Here the graph must have all even edges and 1 odd edge and it must have a maximum flow that is odd.

Now I need to prove the following two points which state that the single odd edge must have a flow $>1$ and/or that edge must have maximum flow in it with respect to its capacity.

I understand that running certain algorithms to find odd flow in such cases prove that both statements are correct that is there is a flow $>1$ in odd edge and also there is maximum flow always in that odd edge. But I don't know how to prove it formally.

Question text:

Consider a graph $G=(V,E)$, with integer capacities on the edges, such that for all $e\in E\setminus\{e^∗\}$, it holds that $c_e$ is an even number, and $c_{e^∗}$ is odd. Suppose that there is a maximum flow in this graph with odd flow.

a. Prove/Disprove: It must be that in every maximum flow, there is a flow in $e^∗$ (i.e., $f_{e^∗}>0$).

b. Prove/Disprove: It must be that in every maximum flow, there is a full flow in $e^∗$ (i.e.,$f_{e^∗}=c_{e^∗}$).

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  • $\begingroup$ Don't use images as main content of your post. This makes your question impossible to search and inaccessible to the visually impaired; we don't like that. Please transcribe text and mathematics. You can use LaTeX. Don't forget to give proper attribution to your sources! $\endgroup$
    – Nathaniel
    Nov 26, 2022 at 13:07
  • 1
    $\begingroup$ I am sorry, I have added the question in text as well. $\endgroup$
    – ConScience
    Nov 26, 2022 at 13:19

1 Answer 1


The Max-flow/Min-cut theorem can be used to prove this property.

As a reminder, $f$ is a max-flow if and only if there exists a cut $X$ of capacity $|f|$.

Let $f$ be a max-flow of odd value in $G$.

a. Suppose there is no flow through $e^*$. Then in $G[V\setminus\{e^*\}]$, $f$ is still a max-flow. However, since all edges have even capacity, all cuts have even capacity. Contradiction.

b. The same idea can be used, I will let you try to find the details.

  • $\begingroup$ So what I can understand and by running some example network flows., is that the cut must have a one edge as odd capacity. and since there will be odd capacity in the cut, the max flow will be odd since cut flow = max flow. and because odd edge will be one of the edges of the cut, it will have the max flow itself. Is that correct? $\endgroup$
    – ConScience
    Nov 26, 2022 at 18:32

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