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I'm given a more restricted version of 3-SAT called 3-SAT-M:

Problem: 3-SAT-M

INPUT: A set of clauses C {c1,...,ck} over n boolean variables {x1,...,xn}, where every clause contains exactly three literals (as in 3-SAT).

OUTPUT: YES if there is a truth assignment to the boolean variables such that the following constraints are satisfied:

  1. Every clause is satisfied under the assignement
  2. There are at least k/2 clauses in which all 3 literals are set to true under the assignment.

NO otherwise.

So from what I understand it is the same as 3-SAT apart from the second constraint which demands that at least half of the clauses have all 3 variables set to true under the assignment.

I want to prove that 3-SAT-M is NP-Complete by showing how 3-SAT can be reduced to 3-SAT-M in polynomial time.

I'm having a hard time coming up with such a transformation that can show the reduction. I've tried by introducing some extra helper variables but it doesn't seem to be working out.

Any help would be appreciated.

EDIT: What I have tried:

Ok, so say for my 3SAT problem instance I have k clauses each with 3 variables. If I add k new clauses and 3k new variables (which are all set to true). So new k = 2* (old) k.

This means that I now satisfy the 2nd constraint of 3-SAT-M as at least k/2 of the clauses have all variables set to TRUE.

Is this correct?

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  • $\begingroup$ This looks like a standard exercise (and a good test of your understanding of the concepts). I suggest you take a look at our reference question on reductions. In your case, which of the following kinds of transformations have you tried? (a) Adding extra variables. (b) Adding extra clauses. Try each of them, and then edit the question to show what you've come up with and where specifically you are stuck. $\endgroup$ – D.W. Oct 5 '15 at 18:49
  • $\begingroup$ @D.W. Thanks. I've had a thought about it and have updated my question. Am I on track or is it completely wrong? $\endgroup$ – DSF Oct 5 '15 at 20:28
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    $\begingroup$ You've made a great start. Now can you formally prove your reduction? Given any instance of 3-SAT, the reduction must create a 3-SAT-M instance (e.g. the way you have described in your edit) such that the correct output on your instance is YES if the original 3-SAT instance is satisfiable and NO otherwise. $\endgroup$ – usul Oct 5 '15 at 21:46
  • $\begingroup$ @usul Thanks for your feedback. I'm a bit rusty when it comes to proofs and have been told that I am too informal when it comes to proving stuff. Is there a similar proof that you could suggest I look at before writing out a proof for this problem ? $\endgroup$ – DSF Oct 6 '15 at 10:28
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    $\begingroup$ @D.Singh many textbooks would have them. As an example let me prove that the problem SAT is NP-hard by reduction from 3SAT. The reduction just outputs the input instance again without changing it. The output is a valid SAT instance because 3SAT is just a special case. Now, if the output is satisfiable as a SAT instance, then the input is satisfiable as a 3SAT instance. If it is not satisfiable as a SAT instance, then it is not satisfiable as a 3SAT instance. Therefore solving SAT would enable us to solve 3SAT, so SAT is NP-hard. $\endgroup$ – usul Oct 6 '15 at 13:46

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