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Let's start here: It is said that all NP problems can be reduced to SAT(boolean satisfiability problem). To be more accurate to Circuit SAT, because all decision problems like NP should end up with answer Yes or No.

But now, if I have a random NP problem, how to build a boolean circuit to test, how to group my input, what kind of gates(AND, NOT, OR etc..) should connect those inputs. So basically, my question how to design boolean Circuit which gives an answer TRUE or FALSE.

Last thing, what that answer means. I understand TRUE as this NP problem can be solved in polynomial time and FALSE cannot, am I correct?

It is huge mess in my mind, don't be really outrageous if I made logical mistakes explaining my question :) I hope you understood it.

Excitingly waiting for answers.

Aivaras Gončarovas
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  • If there was a procedural approach to do this kind of reduction, proving that a problem is NP-hard would be trivial. I can tell from experience that it is often very hard to come up with a reduction, which proves that there is no such procedural approach. – Vincent van der Weele Dec 07 '15 at 15:36
  • You'll probably get a better answer to this sort of question on Computer Science: http://cs.stackexchange.com ; StaclOverflow is really meant for practical programming problems. – m69's been on strike for years Dec 07 '15 at 21:06

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I understand the confusion but your understanding is not quite how it works.

NP-hardness is a qualification of decision problems, that is, a problem with answer is yes or no. If we want to show that a decision problem is NP-hard, we do so by showing that it is as least as hard as a problem of which we know it is NP-hard already, for instance SAT.

How can we show that problem A is at least as hard as problem B? Well, we can phrase that as

if we can solve A, we can also solve B

So, given an instance of problem B, we convert it to an instance of problem A, use our solution to A to solve it, and convert it back to a solution to B. Assuming that both conversations are easy, we know that A cannot be easier than B, since a solution to A is also a solution to B.

Your understanding thus had it backwards. In order to show that some problem is NP-hard, we to show that it is at least as hard as SAT, that is, given an arbitrary instance of SAT, convert it to an instance of your problem, and then solve that problem. If the answer is "yes", then the original SAT problem was satisfiable, otherwise it wasn't.

Now, as I wrote in a comment, there is no standard way to do the conversion. You somehow need to manipulate your problem, such that it "looks like SAT", in order to make the conversion. For some problems that's easier then others, but I'd claim it's the hardest part of the NP-hardness proof.

What people typically do instead is that they look for another problem, which is known to be NP-hard already, but looks a bit more like their own problem. That way, the reduction becomes a bit easier. But still it requires a lot of work and creativity. I recommend you look at some existing proofs to see how others do this.

Vincent van der Weele
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