Let G be a simple connected graph. Let a, b, c, d be four distinct vertices of G.
Is there a way to partition the above four vertices to two pairs, so that the two shortest paths between the vertices of each pair are edge-disjoint?
1
-
$\begingroup$ Yes, it does seem to be possible in all cases. You might want to add some context to this question; right now, I'm not sure why this is an interesting problem. $\endgroup$– Darsh RanjanCommented Oct 31, 2009 at 21:56
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Yes. Just pick the two paths (not necessarily edge disjoint) in G of shortest total length which together join the four vertices into two pairs. If they contained a common edge, you could remove that edge from both paths (changing which vertices are connected to which) to obtain a pair of paths of shorter total length. The resulting paths must be shortest paths between the pairs of vertices they connect. (I assume I'm allowed to pick any shortest path between two vertices, if there's more than one.) Any pair of shortest paths between the same pairs of vertices has the same total length, so by the same argument the two paths of the pair must be edge-disjoint.
answered Oct 31, 2009 at 22:10