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This is the problem that I have been given:

Consider the Independent-Set problem, in which the input is an undirected graph $G = (V,E)$ and a parameter $k$, and the goal is to determine if $G$ has an independent set of size $k$. Suppose we have an oracle $O$ for solving this decision version of independent set (think of it as a library function that takes input a graph $G$ and $k$ and answers YES/NO). Prove that there exists an algorithm that can find an independent set of size $k$, if one exists, using a polynomial number of calls to the oracle $O$, and possibly a polynomial amount of computation of its own.

My first question is: Is there any way to prove that an algorithm exists without just giving a specific algorithm?

I'm sort of taking a crash course in computer science, so this is not my strongest subject. Any hints as to what direction to take this would be appreciated!

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  • $\begingroup$ Isn't this related to "families of k-independent sets" ? I have no clue over your particular problem. But maybe finding/constructing a family of sets $F$ and then calling the oracle to every element would suffice. $\endgroup$
    – seteropere
    Oct 18, 2016 at 6:54

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Here's a hint.

Try running the decision problem subroutine on the original graph $G = (V, E)$. If the answer is NO, clearly you have your answer. If the answer is YES, pick some arbitrary vertex $v \in V$ and remove it from $G$ to produce a new graph $G'$. Now run the decision problem subroutine on $G'$. If the answer is still YES, what did you learn? (This is the "easy" case.) If the answer is now NO, what did you learn? (This case needs a bit more thinking, but not too much.) How many times might you need to repeat this procedure?

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If the problem has the property that the decision version can be converted to the search version, it is self reducibility. You can show that all NP-complete problems are self reducible. The idea is to build the instance with the prefix of the verifier for NP-complete problem, and solve it based on the decision version. This is kinda well-known several lines proof.

I am not aware of any other methods to show the fact you asked for.

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  • $\begingroup$ I am pretty confused on how you would actually go about this: "The idea is to build the instance with the prefix of the verifier for NP-complete problem, and solve it based on the decision version." $\endgroup$
    – flapdoodle
    Oct 19, 2016 at 3:43
  • $\begingroup$ The first link in Google for the given topic gave me the following : cs.umd.edu/~jkatz/complexity/f11/lecture3.pdf I hope this will help. You are interested in section 2. $\endgroup$
    – Eugene
    Oct 19, 2016 at 8:43

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