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To see this, rst suppose that the clauses on the right are all satis ed Then at least one of the literals a1 , , ak must be true otherwise y1 would have to be true, which would in turn force y2 to be true, and so on, eventually falsifying the last clause But this means (a1 a2 ak ) is also satis ed Conversely, if (a1 a2 ak ) is satis ed, then some ai must be true Set y1 , , yi 2 to true and the rest to false This ensures that the clauses on the right are all satis ed Thus, any instance of SAT can be transformed into an equivalent instance of 3 SAT In fact, 3 SAT remains hard even under the further restriction that no variable appears in more than three clauses To show this, we must somehow get rid of any variable that appears too many times Here s the reduction from 3 SAT to its constrained version Suppose that in the 3 SAT instance, variable x appears in k > 3 clauses Then replace its rst appearance by x 1 , its second appearance by x2 , and so on, replacing each of its k appearances by a different new variable Finally, add the clauses (x1 x2 ) (x2 x3 ) (xk x1 ) And repeat for every variable that appears more than three times 246
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Figure 89 S is a vertex cover if and only if V S is an independent set
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It is easy to see that in the new formula no variable appears more than three times (and in fact, no literal appears more than twice) Furthermore, the extra clauses involving x1 , x2 , , xk constrain these variables to have the same value; do you see why Hence the original instance of 3 SAT is satis able if and only if the constrained instance is satis able
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Some reductions rely on ingenuity to relate two very different problems Others simply record the fact that one problem is a thin disguise of another To reduce INDEPENDENT SET to VERTEX COVER we just need to notice that a set of nodes S is a vertex cover of graph G = (V, E) (that is, S touches every edge in E) if and only if the remaining nodes, V S, are an independent set of G (Figure 89) Therefore, to solve an instance (G, g) of INDEPENDENT SET, simply look for a vertex cover of G with |V | g nodes If such a vertex cover exists, then take all nodes not in it If no such vertex cover exists, then G cannot possibly have an independent set of size g
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I NDEPENDENT SET and CLIQUE are also easy to reduce to one another De ne the complement of a graph G = (V, E) to be G = (V, E), where E contains precisely those unordered pairs of vertices that are not in E Then a set of nodes S is an independent set of G if and only if S is a clique of G To paraphrase, these nodes have no edges between them in G if and only if they have all possible edges between them in G Therefore, we can reduce INDEPENDENT SET to CLIQUE by mapping an instance (G, g) of INDEPENDENT SET to the corresponding instance (G, g) of CLIQUE; the solution to both is identical
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Again, two very different problems We must reduce 3 SAT to the problem of nding, among a set of boy-girl-pet triples, a subset that contains each boy, each girl, and each pet exactly once In short, we must design sets of boy-girl-pet triples that somehow behave like Boolean variables and gates! Consider the following set of four triples, each represented by a triangular node joining a boy, girl, and pet: 247
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