Claim 6 There exists an assignment that satisfi es at least Σ
k
m
k
(1 − 1/2
k
) clauses. Furthermore, there is an
effi cient algorithm to fi nd one.
CorollaryIf every clause has size 3 (this is sometimes called the MAX-exactly-3-SAT problem), then we
can satisfy at least m(7/8)clauses. So, this is for sure a 7/8-approximation, since the optimal solution satisfi es
at least m clauses.
An interesting fact is—getting a 7/8 + ∈approximation for any constant ∈(like .001) is NP-hard.
Proof of claim Consider a random assignment to the variables. For a clause of size k, the chance it is not
satisfi ed is 1/2
k
. So, the ...
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