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# vaziraniKSAT1 - 16 Maximum Satisﬁability The maximum...

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Unformatted text preview: 16 Maximum Satisﬁability The maximum satisﬁability problem has been a classical problem in approx- imation algorithms. More recently, its study has led to crucial insights in the area of hardness of approximation [see Chapter 29). In this chapter, we will use LP-rounding, with randomization, to obtain a 3/4 factor approxi- mation algorithm. We will derandomizc this algorithm using the method of conditional expectation. Problem 16.1 (Maximum satisﬁability (MAX—SAT” Given a con- junctive normal form formula f on Boolean variables :::1, . . . , an, and non— negative weights, me, for each clause 0 of f, find a truth assignment to the Boolean variables that maximizes the total weight of satisﬁed clauses. Let C represent the set of clauses of f, i.e., f = And} c. Each clause is a disjunction of literals; each literal being either a Boolean variable or its negation. Let sizo(c) denote the size of clause e, i.e., the number of literals in it. We will assume that the sizes of clauses in f are arbitrary. For any positive integer k, we will denote by M A X-kSAT the restriction of MAX-SAT to instances in which each clause is of size at most is. MAX- SAT is N P-hard; in fact, even MAX-28AT is NP-hard (in contrast, ZSAT is in P). We will ﬁrst present two approximation algorithms for MAX—SAT, having guarantees of 1/2 and 1 1 fa, respectively. The first performs better if the clause sizes are large, and the seconds performs better if they are small. We will then show how an appropriate combination of the two algorithms achieves the promised approximation guarantee. In the interest of mininﬂzing notation, let us introduce common terminol- ogy for all three algorithms. Random variable W will denote the total weight. of satisfied clauses. For each clause c, random variable l/Vc denotes the weight. contributed by clause at to W. Thus, W ; Zed} WC and E[Wc] 2 inc - Pr[c is satisﬁed]. . (Strictly speaking, this is abuse of notation, since the randomization used by the three algorithms is different.) 132 16 Maximum Satisﬁability 16.1 Dealing with large clauses The first algorithm is straightforward. Set each Boolean variable to be True independently with probability 1/2 and output the resulting truth assign- ment, say T. For k _>_ 1, deﬁne at —- l —2' k. Lemma 16.2 If sizo(c) : k, then E[W¢] ; (Ik‘mc. Proof: Clause e is not satisﬁed by T iff all its literals are set to False. The probability of this event is 2 "7. III For It 2 l, at 2 1/2. By linearity of expectation. 1 1 _ E[W] = Z awn :_> iZwt 2 50131", CEC ‘ CEC where we have used a trivial upper bound on OPT the total weight of claimes in C. Instead of converting this into a high probability statement, with a. cor- responding lose in guarantee, we show how to derandomize this procedure- The resulting algorithm deterministieally computes; a truth assigmnent such that the weight of satisﬁed clauses is _>_ E[W] 2 OPT/2. Observe that out increases with h; and the guarantee of this algorithm is 3/ 4 if each clause has two or more literals. (The next algorithm is designed to deal with unit clauses more effectively.) 16.2 Derandomizing via the method of conditional expectation We will critically use the sell-reducibility of SAT (see Section A5). Consider the selilreducibility tree T for formula ft Each internal node at level '5 corre- sponds to a setting for Boolean variables :31, . . . , 331;, and each leaf represents a corriplete truth assigmnent to the n variables. Let us label each node of 7‘ with its conditional expectation as follows. Let a1,...,a,- be a truth as- signment to 2:1, . . . ,1}. The node corresponding to this assignment will be labeled with E[W[:rl = a1, . . ‘ ,m = oi]. Ift' --: n, this is a leaf node and its conditional expectation is simply the total weight of clauses satisﬁed by its truth assignment. Lemma 16.3 The conditional expectation of any node in T can be competed in polynomial time. Proof: Consider a node at = o1, . . . ,mi -— at. Let (b be the Boolean formula, on variables n+1, . . ., an,“ obtained for this node via self—reducibility. Clearly, 16.2 Derandomizing via the method of conditional expectation 133 the expected weight of satisﬁed clauses of I!) under a random truth assignment to the variables m: H,. ,:1:,, can be computed in polynomial time Adding to this the total weight of clauses of f already satisiied by the partial assigmncnt :1:1:c,,...,:1:-—c,g1vestheanswerEl Theorem 16.4 We can compete, in polynomial time, a path. from the root to a. leaf such that the conditional expectation of each. node on this path. is 2 ElW]. Proof: The conditional expectation of a node is the average of the condi— tional expectations of its two children, i.e., ElWle = (1.1," 211,] =ElWlQ‘1 -—-— c1,...,:12, =' 04,335.14 = T1110“? *l' EIWICB} :-. (11, "”131: = can“ = Falsel/Z The reason, of course, is that n+1 is equally likely to be set to True or False. As a result, the child with the larger value has a conditional expectation at least as large as that. of the parent. This establishes the existence of the desired path. As a consequence of Lemma 16.3, it can be computed in polynomial time. D The deterministic algorithm follows as a corollary of Theorem 16.4. We simply output the truth assignment on the leaf node of the path computed. The total weight of clauses satisﬁed by it is 2 E[W]. Let us show that the technique outlined above can, in principle, be used to derandomize more complex randomized algorithms. Suppose the algorithm does not set the Boolean variables independently of each other (for instance, see Remark 16.6). Now, ElWIIEi = a}, ...,.’II{ -= as] = ElWlm1 = a1, ...,:1:,, = chm,“ = True] - Pr[:r.;+1 ='I‘rue|:1:1 = (1.1. ...,:1:,- _— a.,-l+ ElWlw; = 111,...,:r¢ = one,“ = False] -Pr[m,+1 = Falselm = 0.1, ...,:c,- = (1,]. The sum of the two conditional probabilities is again 1, since the two events are exhaustive. So, the conditional expectation of the parent is still a convex combination of the conditional expectations of the twochildren. If we can determine, in polynomial time, which of the two children has a larger value, we can again deranclomize the algorithm. However, computing the con—- clitional expectations may not be 'easy. Observe how critically independence was used in the proof of Lemma 16.3. it was because of independence that we could assume a random truth assignment on Boolean variables \$14.1, . . . , 33,, and thereby compute the expected weight of satisﬁed clauses of qt. In general, a randomized algorithm may pick from a larger set of choices and not necessarin with equal probability. But once again a conVeX combina- tion of the conditional eXpectations of these choices, given by the probabilities 134 16 Maximum Satisﬁability of picking them, equals the conditional expectation of the parent. Hence there must be a choice that has at least as large a conditional expectation as the parent" ...
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vaziraniKSAT1 - 16 Maximum Satisﬁability The maximum...

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