MaxSATByRandomizedRounding

[M.X. Goemans and D.P. Williamson. New 3 4 -approximation algorithms for MAX SAT. SIAM Journal of Discrete Mathematics, 7, 1994, pp. 656-666]

Max k-sat (1-1/2^k)-approximation algorithm

MAX K-SAT: Given a boolean formula in CNF form, where each clause has K distinct literals, find an assignment to the variables that maximizes the number of clauses satisfied.

Example: (X₁ or X₂ or -X₃) and (-X₁ or -X₂ or X₄) and (X₂ or -X₄ or X₃) and ...

Algorithm:

1. Randomly assign each variable TRUE or FALSE.

thm: This is a (1-1/2^k)-approximation algorithm (in expectation).

proof: Each clause is satisfied with probability 1-1/2^k. By linearity of expectation, expected number of the C clauses satisfied is (1-1/2^k)*C. Since OPT <= C, this is a (1-1/2^k)-approximation algorithm (in expectation). QED

Note: the "method of conditional expectations" can be used to convert the randomized algorithm into a deterministic algorithm. In this case, the resulting algorithm is: Consider the variables in sequence. Assign each variable TRUE or FALSE to maximize the expected number of clauses that would be satisfied at the end if the rest of the assignments were to be made randomly.

Max-sat (1-1/e)-approximation algorithm

Integer linear program:

maximize ∑_i y_i subject to

y_i ≥ 0, y_i ≤ 1 for each clause i

x₁ + x₂ + (1-x₃) ≥ y1 for clause 1: (X₁ or X₂ or -X₃)

(1-x₁) + (1-x₂) + x₄ ≥ y2 for clause 2: (-X₁ or -X₂ or X₄)

... etc. (a constraint for each clause)

0 ≤ x_i ≤ 1 for each clause i.

x_i integer for each clause i.

Relaxation (linear program):

maximize ∑_i y_i subject to

y_i ≥ 0, y_i ≤ 1 for each clause i

x₁ + x₂ + (1-x₃) ≥ y₁ for clause 1: (X₁ or X₂ or -X₃)/) : \((1-x₁) + (1-x₂) + x₄) ≥ y₂ for clause 2: (-X₁ or -X₂ or X₄)

...etc. (a constraint for each clause)

0 ≤ x_i ≤ 1 for each clause i.

Algorithm, given MAX-SAT instance:

1. Formulate and solve the linear program above for the instance.
2. Let (x,y) be the optimal fractional solution.
3. For each variable Xi independently, set the variable TRUE with probability xi.

THM: For instances with at most K literals, the above is a (1-(1-1/k)^k)-approximation algorithm (in expectation).

Note: 1-(1-1/k)^k ≥ (1-1/e).

Proof sketch:

Consider, for example, a clause with 3 literals. WLOG assume the clause is "X_1 or X_2 or X_3".

The probability that the clause is NOT satisfied is (1-x₁)*(1-x₂)*(1-x₃) where x₁+x₂+x₃ ≥ y₁.

claim: 1-(1-x₁)*(1-x₂)*(1-x₃) ≥ 1-(1-y₁/3)3 ≥ y₁(1-(1-1/3)3). (prove using calculus)

Therefore, the ith clause is satisfied with probability at least y_i*[1-(1-1/k)^k].

Therefore, by linearity of expectation, the expected number of satisfied clauses is at least ∑_i y_i*[1-(1-1/k)^k] ≥ OPT * [1-(1-1/k)^k].

3/4-approximation algorithm:

1. Use one of the two previous algorithms, chosen at random.

THM: The above algorithm is a 3/4-approximation algorithm, in expectation.

Proof sketch:

For a clause i with k literals, the probability that it is satisfied is

(1/2)*[1-1/2^k] + (1/2)*y_i[1-(1-1/k)^k] ≥ y_i*{ [1-1/2^k] +[1-(1-1/k)^k] } / 2 ≥ (3/4)y_i.

Therefore, the expected number of satisfied clauses is ≥ ∑_i (3/4)y_i ≥ 3/4 OPT.