User:Sravan11k/sandbox

Solution Algorithm - 3 SAT => P=NP
3-SAT: Solution Algorithm Count the number of clauses with same variable combination in the Expression(E) having ‘n’ number of clauses. x1 and ~x1 are positive and negatives of x1, hence (x1 OR x2 OR x3) and (~x1 OR ~x2 OR x3) belong to same group, say G1. If the group count for G1, the total number of clauses in G1, is equal to 8 (2^K, where K=3 in 3-SAT) then the decision of G1, say g1 is 1(true) which means that G1 is Unsatisfiable that implies the given expression E is Unsatisfiable. If all possible group decisions are (false)0’s then it’s not Unsatisfiable that implies the expression E is Satisfiable. It’s a generic algorithm, where if we follow the above process it’ll check for all clauses ignoring the previous ones taking n(n+1)/2 steps with an additional n group decisions plus a check in all Group decisions. Total : n(n+1)/2 +n+1 = (n2+3n+2)/2) steps. By the above polynomial expression, we can conclude that the 3-SAT decision problem which is an NP-Complete problem can be solved in polynomial time.

P.S: If we restrict the process by checking for I >=(2^K), where ‘I’ is the number of available clauses to check group decision.