# Finding maximum-cardinality independent set with a particular oracle

We suppose we have a polynomial algorithm which receives a graph $G$ (any graph) and returns a stable set of $G, SA(G)$ with the following property:

$\alpha(G) − |SA(G)| \leq k$ , for every natural $k$ ($k$ is a constant)

$\alpha(G)$ is the stability number

I need to show that this algorithm can be used to find(in polynomial time) a stable set (of maximum cardinality) in a graph.

I discovered that if I give $k=0$ then we get to this relation. But does that mean that the algorithm finds the stable set with maximum cardinal.

• I don't understand the stated property. Since $SA(G)$ is stable, we know that $|SA(G)|\leq\alpha(G)$. But the fact that $\alpha(G)-|SA(G)|\leq k$ for every natural number $k$ means, in particular, that $\alpha(G)-|SA(G)|\leq 0$, so we deduce that $|SA(G)|=\alpha(G)$ and we're done. Did you mean to give some other property? – David Richerby Nov 7 '16 at 15:59
• Also, what did you try? Where did you get stuck? We're happy to help with conceptual questions but just answering homework-style exercises for you is unlikely to really help you. – David Richerby Nov 7 '16 at 16:00
• Welcome to Computer Science! What have you tried? Where did you get stuck? We do not want to just hand you the solution; we want you to gain understanding. However, as it is we do not know what your underlying problem is, so we can not begin to help. See here for tips on asking questions about exercise problems. If you are uncertain how to improve your question, why not ask around in Computer Science Chat? – Raphael Nov 7 '16 at 16:05
• The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! – Raphael Nov 7 '16 at 16:05
• Voting to close as unclear: the property given still implies that the oracle returns a set of size max-0=max. – David Richerby Apr 26 '17 at 17:58

## 1 Answer

If $k = 0$ then your algorithm outputs a stable set of maximum size. In the general case, try duplicating each vertex $k+1$ times, and see if it helps.