Is topological sort of an original graph same as post-ordering dfs of its transpose graph
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29-09-2020 - |
Question
I have an intuition that topo-sort of an original graph
A -> B -> C
D -> B
topo-sort is [D, A, B, C] or [A, D, B, C]
If I transpose the graph
C -> B -> A
B -> D
the postordering dfs of this transposed graph also gives [D, A, B, C] or [A, D, B, C]
Please, I can't mathematically prove/disprove it. If not true, an counter example would helpful.
A postordering is a list of the vertices in the order that they were last visited by the algorithm
Solution
Let $G$ be a graph and $G'$ be it's transposed version. Your property follows from the following two facts:
1) The order in which vertices are visited by a postorder visit on $G'$ is the reverse of a topological order $\sigma'$ for $G'$ (in fact, that's a standard way to compute a topological order). You can see that this is true by noticing that if $(u,v) \in G'$ then $v$ must be visited before $u$ can be visited, i.e., $v$ follows $u$ in $\sigma'$.
2) If $\sigma'$ is a topological order for $G'$, then the linear order $\sigma$ obtained by reversing $\sigma'$ is a topological order for $G$. You can see that this is true because, for every edge $(u,v)$ of $G$, $G'$ contains $(v,u)$ and therefore $v$ precedes $u$ in $\sigma'$. This means that $(u,v) \in G \implies u$ precedes $v$ in $\sigma$.