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                    CS 577 (Intro to Algorithms)
Lec 23 (11/28/06)                                       Shuchi Chawla
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In these lecture notes we will prove that the Max-2-SAT problem is
NP-complete. 

Max-2-SAT is defined as follows. We are given a 2-CNF formula and a
bound k, and asked to find an assignment to the variables that
satisfies at least k of the clauses.

For example, suppose the formula is 

(x V y) A (not y V x) A (not x V not y) A (not x)

then, if k=3, this is a Yes instance, because we can satisfy the first
three clauses by setting x=True and y=False. But if k=4, then this is
a No instance because no assignment will make all of the clauses true.


Max-2-SAT is closely related to the 2-SAT problem, where we are
supposed to find an assignment for a 2-CNF formula that satisfies
*all* of the clauses. Recall that 2-SAT is in P, that is, there is a
polynomial time algorithm for it. On the other hand we will show that
Max-2-SAT is NP-complete.

Theorem: Max-2-SAT is NP-complete

Proof: 

First we will show that Max-2-SAT is in NP. This is easy. A proof for
a Yes instance of the problem is simply an assignment to the
variables. In polynomial time we can verify that this assignment does
indeed satisfy at least k of the clauses.

In order to prove NP-hardness, we will reduce the 3-SAT problem to
Max-2-SAT.

Recall that in 3-SAT we are given a 3-CNF formula and asked to find a
satisfying assignment for it. We will convert this formula into a
2-CNF formula clause by clause.

So consider a clause (x V y V z) in the given 3-CNF formula. We
introduce a new variable for this clause, call it w. We then replace
this clause by the following 10 clauses:

      (x)        (y)         (z)        (w)

(not x V not y)  (not y V not z)  (not x V not z)

  (x V not w)      (y V not w)      (z V not w)

We replace each clause in the 3-CNF formula by 10 clauses in this
manner, introducing a fresh variable w for every clause. Suppose that
the original formular has m clauses, then we end up with a 2-CNF
formula with 10m clauses.

We now claim that:

1. If an assignment satisfies (x V y V z), then there is an assignment
   for w, that together with the assignment for x, y and z, satisfies
   exactly 7 of the 10 clauses above.

2. If an assignment does not satisfy (x V y V z), then every
   assignment for w, together with the assignment for x, y and z,
   satisfies at most 6 of the 10 clauses above.


These claims imply that if there is a satisfying assignment for the
original 3-CNF formula, then there is an assignment for the new 2-CNF
formula that satisfies at least 7m of the clauses. On the other hand,
if no assignment satisfies the 3-CNF formula, then at least for one of
the original clauses, we will only be able to satisfy 6 of the 10 new
clauses, and therefore, the number of clauses satisfied in the 2-CNF
formula is strictly less than 7m. Therefore, we can decided whether or
not the given 3-CNF formula is satisfiable by determining whether or
not it is possible to satisfy 7m of the clauses in the 2-CNF formula.


It remains to prove the above claims. We do this by case analysis.

Case 1: x=y=z=True. Then by setting w=True, we can satisfy the entire
        first and third rows, or 7 clauses in all.

Case 2: x=y=True z=False. Then by setting w=True, we can satisfy three
        clauses from the first row, two from the second row, and two
        from the third row, or 7 clauses in all.

Case 3: x=True, y=z=False. Then by setting w=False, we can satisfy one
        clause on the first row, and the entire second and third rows,
        which is again 7 clauses in all.

Case 4: x=y=z=False. Note that (x V y V z) is not satisfied. If we set
        w=True, this only satisfies one clause on the first row, and
        all clauses on the second row; a total of 4 clauses. On the
        other hand if we set w=False, then we only satisfy all clauses
        on the second and third rows, a total of 6 clauses.

Other cases are analogous to the above.

QED.


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