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Define: $$L=\{(G,k) : G\text{ has a vertex cover of size at most $k$, and a clique of size at least $k$}\}$$

I need to determine whether $L\in \mathrm{NPC}$ or $L\in \mathrm{P}$. I suspect that $L\in \mathrm{NPC}$, been trying to prove it by looking for a reduction from Vertex Cover, but couldn't find one.

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The graphs with clique size at least $k$ and VC at most $k$ have a particular structure: They can be partitioned into three sets, $C$, $I$ and a singleton $\{s\}$; $G[C]$ is a clique, $G[I]$ is an independent set and $s$ is just a vertex. $G[C \cup I]$ is what's called a split graph.

Proof: If a graph $G$ contains a clique of size $k$, then that clique uses up (at least) $k-1$ of your vertex cover budget. Let $G,k$ be a yes instance and $C$ any $k$-clique. Observe that $G - C$ is a star with possibly isolated vertices.

There are two cases: $G-C$ has edges, and $G-C$ does not have edges. The latter case is trivial, since we can let $C$ be the solution. In the former case, $G-C$ is the star graph with isolates. Let $s$ be the vertex with non-zero degree.

Both these cases lead to polynomial recognition algorithm. The one in which $G-C$ is an independent set, is the case where $G$ is a split graph. The other case has one vertex $s$ for which $G-s$ is a split graph. Guess $s$, and you're back at case 1.

As xskxzr pointed out, there is not necessarily a unique "split partition", that is, a partitioning of a split graph into the vertex sets $C$ and $I$, however, there are at most $n$ such partitions, and you can try all. This leads in the end to an $O(n^3m)$ algorithm for VC∩C, which I'm sure can be improved.

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    $\begingroup$ If I understand you correctly, the reduction would be to add a clique of size $k$. but then if I use $k'=2k-1$ I would need a clique of size $2k-1$, and that's a problem. $\endgroup$ – John Apr 22 at 13:35
  • $\begingroup$ I havn't heard of split graphs. Is there a simple polynomial algorithm to determine if a graph is a split graph? and if so, how can I use this information to solve my problem? $\endgroup$ – John Apr 22 at 16:05

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