0
$\begingroup$

Suppose you have a connected graph and want to remove k nodes such that the result is still connected. How could you do this efficiently?

It occurs to me that you could find any spanning tree, say by a tree search of any kind. Identify all leaves in the spanning tree, all of these can be removed without disconnecting the remaining vertices. If you have more than k leaves then you're done, but in any tree you're only guaranteed 2 leaves. So you may need to reiterate the process until you've removed k vertices.

That implies O(k) runs of a tree search. Does a more efficient algorithm exist? I don't think you can just look for articulation points or bridge edges because removing a single vertex may suddenly make other vertices which weren't articulation points now turn into articulation points.

$\endgroup$
1
$\begingroup$

Let $G$ be your graph. Compute any spanning tree $T$ of $G$. Perform a postorder visit of $T$, and keep track of the set $D$ of the first $k$ vertices visited ($T$ and $D$ can be computed by the same DFS visit).

The graph $G'$ obtained by deleting the vertices in $D$ from $G$ is still connected (and $T-D$ is a spanning tree of $G'$). To prove this notice that if $v \in D$, then all descendants of $v$ in $T$ must belong to $D$ as well.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.