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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 qr code in c# windows application Using Barcode printer for VS .NET Control to generate, create QR Code image in .NET applications. qr code in c# windows application asp.net vb qr code Using Barcode maker for ASP.NET Control to generate, create QR image in ASP.NET applications. asp.net vb qr code Figure 88 The graph corresponding to (x y z) (x y z) (x y z) (x y) qr code generator vb net Using Barcode printer for .NET framework Control to generate, create QRCode image in VS .NET applications. barcode generator in vb.net 2008 qr code generator vb net qr code generator using vb.net Using Barcode encoder for Visual Studio .NET Control to generate, create QR Code 2d barcode image in Visual Studio .NET applications. qr code generator using vb.net 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 Draw EAN128 In None Using Barcode creation for Software Control to generate, create EAN128 image in Software applications. Universal Product Code Version A Drawer In None Using Barcode printer for Software Control to generate, create UPC A image in Software applications. 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 Generating Barcode In None Using Barcode printer for Software Control to generate, create bar code image in Software applications. Bar Code Maker In None Using Barcode generator for Software Control to generate, create bar code image in Software applications. 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 Painting EAN13 Supplement 5 In None Using Barcode encoder for Software Control to generate, create EAN13 Supplement 5 image in Software applications. 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