5
$\begingroup$

Let's say I have an optimization problem called $k$-foo which asks for a solution of size $k$ minimizing some quality criterion.

Now the corresponding decision problem $foo(M)$ would be:
Is there a solution to foo with quality at least $M$ of size $k$.

For problems on one parameter (for example vertex cover) it is obvious that solving the optimization problem sovles the decision problem.

But here I do not see such a correspondance between the $k$-foo optimization problem and the $foo(M)$ decision problem. How does for example showing that $foo(M)$ is NP-hard implies that $k$-foo is NP-hard?

The $k$-center problem is an example of such a problem where the decision version takes the radius as input and asks wether a solution of size $k$ exists.

$\endgroup$
  • 1
    $\begingroup$ Only decision problems can be NP-hard. Optimization problems are not languages, they are (partial) functions. $\endgroup$ – Yuval Filmus Apr 27 '13 at 4:42
  • $\begingroup$ NP is a complexity class of languages, not optimization problems. When you say that some optimization problem is in NP, you really mean that the corresponding decision problem is in NP. $\endgroup$ – Yuval Filmus Apr 27 '13 at 4:46
  • 1
    $\begingroup$ [I was missing a not in my previous comment ]For a problem to be NP-hard it does NOT need to be in NP. $\endgroup$ – user695652 Apr 27 '13 at 4:53
  • 1
    $\begingroup$ Thank you for raising this point. There seems to be quite some confusion about this fact. While wikipedia is on my side "NP-hard problems may be of any type: decision problems, search problems, or optimization problems." [en.wikipedia.org/wiki/NP-hard] some lecture notes indeed say that "A language is NP-hard, if...". Do you know of any reference to a well agreed definition? $\endgroup$ – user695652 Apr 27 '13 at 18:39
  • 1
    $\begingroup$ @user695652 The book by Garey & Johnson (which I highly recommend) has a chapter about NP-hardness. They use the term for decision (using many-one-reductions) problems as well as for optimization problems (which they formalize as "string relations") using Turing reductions. Note that these are not two entirely different notions of NP-hardness. The latter is just a generalization of the first. So defining a language to be "NP-hard, if..." is just fine, if you only have to deal with languages. $\endgroup$ – Cornelius Brand Apr 27 '13 at 20:03
2
$\begingroup$

Having more than one input doesn't change anything:

Consider an optimization problem $MinQ$:

Input: $x, k$
Output: find $y$ satisfying $Q(x,k,y)$ which minimizes $f(x,k,y)$: $$g(x,k) = \min_{y : Q(x,k,y)} f(x,k,y)$$

The corresponding decision problem $Q$ is:

Input: $x, k, l$
Query: is there a $y$ satisfying $Q(x,k,l)$ such that $f(x,k,y) \leq l$?

To solve the decision problem, we can solve the optimization problem, then check if the output is less than $l$. I.e. there is a very easy reduction from $Q$ to $MinQ$:

input: $x,k,l$
1. $m = MinQ(x,k)$
2. return $m \leq k$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.