Question

Computational Complexity Question

Show that the 3-SAT-3 problem is NP-complete by reducing SAT to 3-SAT-3.

Answer 1

Definition and understandability basic: fact that of every problem in NP can be polynomial-time reducible to a problem 'A' ->(NP-Hard)

P=NP ->cheked quickly can be solved quickly. in polynomial time. Thee are a lot of NP-Complete problems that can't be solved quickly.

EG: shortest path and longest path between two points in a graph. So the shortest path problems given points as source and destination required to find the path in polynomial time. there is an algorithm Dijkstra's algorithm helping to find shortest path. the longest path between points.

shortest path-NP Complete - O(V^)

NP-Hard -longest path

EG: Satisfiability problem - If a boolean formula is given in the conjunctive normal form, assignment to a variable in formula leads to true. its satisfiable problem.

CNF->(a v b v c v ~d) ^ (a v b) ^ (a v c v ~b) //these are clauses., and variables literals, clauses are conjunction.

Among these clauses all literals are disjunction or the boolean are of expression. Literals are in normal form or. A Boolean formula in conjunctive normal form as shown above

3Sat:- The cnf formulas are restricted to 3 literals per clause

(a v b v ~c) ^ (a v b v c) ^ (~a v ~b v ~c)

Any cnf can have exactly 3 literals, just 3 cnf if existes any assignment to varabe to true

2sat restricted to 2 cnf formula. for every clause there will be exact 2 literals for 2 cnf and if assignment exists it to true then its a 2 saat problem.

Here you can see 2 sat and 3 sat are quite similar as the only difference is number of literals per clause. The final procedure quite similar in both of problem still 2 sat problem is polynimia time solvable and P Clause, 3 sat is np complete.

3 SAT NP Hard

1. Theorem : SAT where all clauses have length 2 and varibales occur 3 times, is satisfaible.
2. If there is a pair of clauses (a v b) ^ (a-  v b-),replace occurence(there wil be onlllllyl one) of b to a-. Remove that pair of clauses. This will result in two occurences of a and no occurences of b.
3. Take unstable 3 SAT and performing reduction where each variable occurs at most three times. That is, for each variable A that occcurs k>3 times, replaces the occurences respectively and add new clauses (A-1 V A2) ^ (A-2 V A3) ^ ..... ^ (Ak-1-  V Ak) ^ (A-k V A1)
   A ... gtx has same acalre is saisting assignment Havina A, VA TO clause ATV A. but not wo 1 2 use n ce an use M: Aq' and N=A. so delete first class and replace any instance of to with A and A with Ai. (HIVA) ^ (A, VA) A... ^(A ki vax) a C CHEV A₂), (A, 7A,) (A, → Ag) 1.1 (Ak-yAx) ^ CA >A). Apply transformation again to clause Aj v Az, (AIVA „) ^ (A VAJA ....(Ani vd x) ^ CAEVA), - Now Az is not present and take any value. ane