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Prove that in a complete oriented graph, you can change the orientation of one edge so that the resulting graph is connected.

I tried to use induction. but I dont know how if $n=3$ it's true, then suppose it's working for n let's prove that it will work for $n+1$ I don't know how to prove this. Also I think that if if our primordial graph is connected, it's impossible. But I'm not sure.

Delta Account
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  • Do you know that an oriented complete graph has a path that goes through all the vertices? If you have that, then all you need to make it strongly connected is an edge from the end of the path to the beginning. – bof Jul 08 '20 at 20:08
  • If your tournament (= oriented complete graph) is strongly connected, and if it has more than three vertices, then yhou can always reverse one edge and leave it connected. Namely, the tournament has a cycle that goes through all the vertices, and you can safely change the orientation of any edge which is not part of the cycle. If thee are more than three nodes, the cycle will not contain all the edges of the graph. – bof Jul 08 '20 at 20:11

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It is a famous result of Redei that states every tournament (i.e. directed complete graph) graph has a directed Hamilton path (one that goes thru each vertex). The proof can be read here. Let the Hamilton path start at $u$ and end at $v$. The tournament graph has an edge either from $u$ to $v$ or from $v$ to $u$. If the edge is from $u$ to $v$, then reverse it. If the edge is the other way around, do nothing. (If you have to change an edge, just change an arbitrary edge that isn't the edge from $v$ to $u$ or any of the edges in the Hamilton path.)

In either case, we've created a Hamilton cycle in the graph, so it must be strongly connected now. Note that as bof observed, this proof works whenever the number of vertices is at least four. The cases with one and two vertices are trivial. The case with three vertices is basically as trivial, and can either be proven with brute force or the pigeon hole principle.

paulinho
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