3.6. Another Approximation Algorithm for MAxE3SAT 55
Prob[X~l = xl,..., X~ = xd] = Prob[p(il)
E Ah,~,... ,p(id) 9 Ai~,~].
There are IA~I,~I I'..." IAi~,~ I polynomials p fulfilling
p(Q) E AiI,~,... ,p(id) E
Ai~,x~, since p is a polynomial of degree at most d- 1 and, hence, it is uniquely
determined by fixing its value at d different places. Therefore,
Prob[p(Q)
E Ai~,x~,... ,p(id) e A~d,~]
- Prob[X~, = x~]..... Prob[X~ =
xd].
Hence, the X~ are d-wise independent.
We describe how to apply Lemma 3.11 for the derandomization of some ran-
domized algorithm. The random variables X1,..., Xn are the random variables
describing the random choices of the algorithm. We assume that the probabilities
for which X1,..., Xn take the values 0 and 1 are rational numbers. Hence, we
can compute the values p~. We replace
X1,..., Xn
by X~,..., X~n. Since Ill I is
bounded by some polynomial in n, we can consider all Ifll elementary events in
polynomial time or in parallel by a polynomial number of processors. For each
elementary event p we evaluate the random variables X~,..., X~, i.e., we com-
pute X~ (p),..., X~n (p) and run the given randomized algorithm for these values
of the random variables. From the [12] results a best one is selected.
The random variable
X~(p)
can be evaluated in constant time, if d is a constant.
We merely have to evaluate a polynomial of constant degree and to compare the
result with
qp~.
Hence, the evaluation of all X~(p) for all i E {1,... ,n} and all
p E fl is possible by an EREW PRAM with qd+l processors in constant time.
3.6 Another Approximation Algorithm for MAXE3SAT
In this section we show how Lemma 3.11 can be applied in order to derandomize
the algorithm RANDOM EkSAT ASSIGNMENT given in Chapter 2. The derandom-
ization only works for constant k and we only consider the case k = 3, i.e. the
algorithm for the problem MAXE3SAT. Recall that an instance of MAXE3SAT
consists of the set {xl,... ,xn} of variables and the set C = {C1,...
,Cm}
of
clauses. In the randomized algorithm each variable
xl,..., xn
is randomly and
independently chosen to be 0 and 1 with probability 1/2 each. Then the prob-
ability of a clause to be satisfied is 7/8 and the expected number of satisfied
clauses is 7/8. m.
It is easy to see that the same analysis works if xl,..., xn are chosen to be 0
and 1 with probability 1/2 each and if we only assume threewise independence
of Xl,..., xn. Hence, we may apply Lemma 3.11.