Henc e set sp li t t i n g is in n p ii set sp li t t

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Unformatted text preview: ls e. In the latter c as e, s et zi true. In any c as e, b is s et fals e. On the other hand, if ¶0 has an = as s ignment, y1 ; y2 and y3 c annot all be s et fals e s inc e this may c aus e one of the c laus es in = the = as s ignment to be fals e. Thus , the c orres ponding c laus e (y1 _ y2 _ y3 ) mus t be true. A = s atis fy ing as s ignment of ¶ is then deriv ed. c.) Obv ious ly , = SAT is in NP bec aus e an NTM c an gues s an = as s ignment and c hec k if it = = s atis fies the formula nondeterminis tic ally in poly nomial time. Ac c ording to the reduc tion in b, = SAT is c onc luded to be NP­Complete. = 7.28 i.) SET À SP LI T T I N G 2 N P On input of (S; C ) , where S is a finite s et and C = f C 1 ; C 2 ; :::; C kg is a c ollec tion of s ubs ets of S , for s ome k > 0 , an NTM c ould gues s a c oloring of the elements nondeterminis tic ally and ac c ept if no C i has all its elements c olored with the s ame c olor. Henc e, SET À SP LI T T I N G is in N P : ii.) SET À SP LI T T I N G is NP­hard 3 SAT c an be reduc ed to SET À SP LI T T I N G by c ons truc ting a TM F that c omputes the reduc tion f as follows . F = ”On input of a 3CNF Boolean formula (¶) with l c laus es and...
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