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I was given a full directed connected graph $G$ which it's edges are colored in two colors: red and blue.

I was asked to prove that the subgraph that contains all of the nodes and only one of the colored edges is connected (it can be weakly or strongly connected), but I got stuck in the process. Any help will do!

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closed as off-topic by Morgan Rodgers, Xander Henderson, Saad, Shailesh, Leucippus Jun 2 '18 at 4:27

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  • $\begingroup$ what do you mean by full directed graph? do you take all edges in all directions? $\endgroup$ – Yanko Jun 1 '18 at 19:57
  • $\begingroup$ Weakly or strongly connected? $\endgroup$ – Hagen von Eitzen Jun 1 '18 at 20:00
  • $\begingroup$ If this is a specific coloring of a specific directed graph ("full", I know, but how many vertices?), how can we help answer your question without knowing what the graph or the coloring is? $\endgroup$ – Morgan Rodgers Jun 1 '18 at 20:01
  • $\begingroup$ @HagenvonEitzen what do you mean by weakly? $\endgroup$ – Lola Jun 1 '18 at 20:02
  • $\begingroup$ @MorganRodgers The graph is a full graph- each node is connected to all of the others directly (with either a blue or a red edge) $\endgroup$ – Lola Jun 1 '18 at 20:04
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Suppose the red subgraph is not strongly connected, i.e., there are two vertices $a,b$ such that there is no red path from $a$ to $b$. Then in paricular the edge $a\to b$ is blue. For any other vertex, at least one of $a\to c$, $c\to b$ is blue, hence $c$ is weakly connected to $a$ via blue edges: $c\to b\leftarrow a$ or $c\leftarrow a$. We conclude that the blue subgraph is weakly connected.


If we ask for strongly instead of weakly connected, this need not be the case: In $G$, pick any two vertices $a,b$, colour each edge staring in $a$ and/or ending in $b$ red and each edge starting in $b$ and/or ending in $a$ blue and colour all other edges arbitrarily. Then the red subgraph is not strongly connected because there is no red path $b\to\ldots \to a$, and the blue subgraph is not strongly connected because there is no blue path $a\to\ldots\to b$.

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  • $\begingroup$ What is x exactly?:) $\endgroup$ – Lola Jun 1 '18 at 20:12
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    $\begingroup$ @Lola x is the key to the left of c on my keyboard :) $\endgroup$ – Hagen von Eitzen Jun 1 '18 at 20:15
  • $\begingroup$ Amazing! thank you very much :) $\endgroup$ – Lola Jun 1 '18 at 20:17

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