# Does NP-completeness require to find the solution?

In the paper "Computing Equilibria:A Computational Complexity Perspective" by Tim Roughgarden, they consider the problem:

Problem 2.1 (Clique). Given a graph $G = (V, E)$ and an integer $k$:

1. if there is a set $K ⊆ V$ with $|K| = k$ and with $(i, j) ∈ E$ for every distinct $i, j ∈ K$, then output such a set;
2. otherwise, indicate that $G$ has no $k$-clique.

Then they claim that Problem 2.1 is NP-complete in the Theorem 2.7.

My question is about the part (1). I always thought NP-completeness is for yes/no problems and thus they didn't require that the the solution be output if the decision is an yes?

So is this definition of NP-complete problem 2.1 right? Or should part (1) be rephrased as "if there is a set $K \subseteq V$ with $|K| = k$ and with $(i, j) \in E$ for every distinct $i, j \in K$, then output yes;

The Problem 2.1 is a half decision problem and half feasibility problem, what are such problems called?

P.S the paper can be found at http://theory.stanford.edu/~tim/papers/et.pdf

I went through Optimization version of decision problems and still not sure where Problem 2.1 belongs to.

• It's called a search problem and in this case, it's more or less equivalent to the decision variant. – Pål GD Mar 31 '15 at 2:01
• @PålGD Could you please explain what is meant by more or less equivalent? Thanks for the Wikipedia link, that explains the naming. – seek Mar 31 '15 at 2:11
• @PålGD It's clear that the search problem is at least as hard as the decision problem, but the other direction not so much. – Raphael Mar 31 '15 at 7:33

Here the search problem is finding a clique of size $k$ if one exists. We think of the search problem as a two-place predicate $f(\langle G,k \rangle,K)$ which is true if $G$ is a graph and $K$ is a $k$-clique of $G$. This search problem is in FNP if the predicate $f$ is in P. (Usually FNP is defined for functions, i.e. for relations which are functions, but you can also define it for general relations.)