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and the aim is to nd a satisfying truth assignment In INDEPENDENT SET the input is a graph and a number g, and the problem is to nd a set of g pairwise non-adjacent vertices We must somehow relate Boolean logic with graphs! Let us think To form a satisfying truth assignment we must pick one literal from each clause and give it the value true But our choices must be consistent: if we choose x in one clause, we cannot choose x in another Any consistent choice of literals, one from each clause, speci es a truth assignment (variables for which neither literal has been chosen can take on either value) So, let us represent a clause, say (x y z), by a triangle, with vertices labeled x, y, z Why triangle Because a triangle has its three vertices maximally connected, and thus forces us 244
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Figure 88 The graph corresponding to (x y z) (x y z) (x y z) (x y)
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to pick only one of them for the independent set Repeat this construction for all clauses a clause with two literals will be represented simply by an edge joining the literals (A clause with one literal is silly and can be removed in a preprocessing step, since the value of the variable is determined) In the resulting graph, an independent set has to pick at most one literal from each group (clause) To force exactly one choice from each clause, take the goal g to be the number of clauses; in our example, g = 4 All that is missing now is a way to prevent us from choosing opposite literals (that is, both x and x) in different clauses But this is easy: put an edge between any two vertices that correspond to opposite literals The resulting graph for our example is shown in Figure 88 Let s recap the construction Given an instance I of 3 SAT, we create an instance (G, g) of INDEPENDENT SET as follows Graph G has a triangle for each clause (or just an edge, if the clause has two literals), with vertices labeled by the clause s literals, and has additional edges between any two vertices that represent opposite literals The goal g is set to the number of clauses Clearly, this construction takes polynomial time However, recall that for a reduction we do not just need an ef cient way to map instances of the rst problem to instances of the second (the function f in the diagram on page 241), but also a way to reconstruct a solution to the rst instance from any solution of the second (the function h) As always, there are two things to show 1 Given an independent set S of g vertices in G, it is possible to ef ciently recover a satisfying truth assignment to I For any variable x, the set S cannot contain vertices labeled both x and x, because any such pair of vertices is connected by an edge So assign x a value of true if S contains a vertex labeled x, and a value of false if S contains a vertex labeled x (if S contains neither, then assign either value to x) Since S has g vertices, it must have one vertex per clause; this truth assignment satis es those particular literals, and thus satis es all clauses 2 If graph G has no independent set of size g, then the Boolean formula I is unsatis able 245
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It is usually cleaner to prove the contrapositive, that if I has a satisfying assignment then G has an independent set of size g This is easy: for each clause, pick any literal whose value under the satisfying assignment is true (there must be at least one such literal), and add the corresponding vertex to S Do you see why set S must be independent
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This is an interesting and common kind of reduction, from a problem to a special case of itself We want to show that the problem remains hard even if its inputs are restricted somehow in the present case, even if all clauses are restricted to have 3 literals Such reductions modify the given instance so as to get rid of the forbidden feature (clauses with 4 literals) while keeping the instance essentially the same, in that we can read off a solution to the original instance from any solution of the modi ed one Here s the trick for reducing SAT to 3 SAT: given an instance I of SAT, use exactly the same instance for 3 SAT, except that any clause with more than three literals, (a 1 a2 ak ) (where the ai s are literals and k > 3), is replaced by a set of clauses, (a1 a2 y1 ) (y 1 a3 y2 ) (y 2 a4 y3 ) (y k 3 ak 1 ak ), where the yi s are new variables Call the resulting 3 SAT instance I The conversion from I to I is clearly polynomial time Why does this reduction work I is equivalent to I in terms of satis ability, because for any assignment to the ai s, (a1 a2 ak ) is satis ed there is a setting of the yi s for which (a1 a2 y1 ) (y 1 a3 y2 ) (y k 3 ak 1 ak ) are all satis ed
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