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:
- Every clause is satisfied under the assignement
- 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?
