Independent set on cubic triangle-free graphs

Question

I know that maximum independent set on cubic triangle-free graphs is NP-complete.

Is it still NP-complete in case we require the independent set to be of size exactly $|V|/2$?

Basiclly, YES instance of independent set problem on cubic triangle-free graphs problem must have exactly $|V|/2$ nodes. NO instance has an independent set of size less than $|V|/2$.

http://cs.stackexchange.com/questions/1176/half-clique-np-complete-problem may be relevant. — Louis, Feb 07 '13 at 12:30
@YuvalFilmus He's asking the problem is $\alpha(G) = |G| / 2$ where $|G|$ is the order of the graph. It should be possible to pad some isolated vertices onto the graph to boost the independence number. Mohammad, do you know the reduction? Is it not possible to add $n/2 - k$ isolated vertices to get the wanted reduction? — Pål GD, Feb 07 '13 at 16:32
@PålGD Does the reduction you suggested produce cubic triangle-free graph? — Mohammad Al-Turkistany, Feb 07 '13 at 18:21
@PålGD The reduction wouldn't work, since the usual problem asks whether $\alpha(G) \geq k$ rather than $\alpha(G) = k$. In fact, it's not even clear that the problem is in NP. — Yuval Filmus, Feb 07 '13 at 20:58
@YuvalFilmus I guess the problem is in $NP$ since the maximum independent set in cubic triangle-free graphs can not be more than $|V|/2$. So, for YES instance you just need to check that the size of the independent set is $|V|/2$. — Mohammad Al-Turkistany, Feb 08 '13 at 00:06

score 7 · Accepted Answer · answered Feb 08 '13 at 04:43

Let's start by proving that the maximum independent set is of size at most $|V|/2$. Let $I$ be an independent set. For each vertex $v$, let $\alpha(v)$ be the number of its neighbors in $I$. If $\alpha(v) \geq 1$, then we know that $v \notin I$. Since the graph is cubic, $\sum_v \alpha(v) = 3|I|$. Since $\alpha(v) \leq 3$, the number of vertices such that $\alpha(v) \geq 1$ is at least $|I|$. Hence $|I| \leq |V|/2$.

When can we have equality? We must have $\alpha(v) \in \{0,3\}$, so for each vertex not in $I$, all its neighbors must be in $I$. The converse is also true - for each vertex in $I$, all its neighbors are not in $I$. In other words, the graph must be bipartite. This can be checked in polynomial time.

YuvalFilmus Thanks a lot. Does this give polynomial time algorithm for my problem? — Mohammad Al-Turkistany, Feb 08 '13 at 12:14

Independent set on cubic triangle-free graphs

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