;; 2-SAT and P = NP ;; The archetypal NP-complete problem is 3-SAT. Given a set of logical ;; triples that must be enforced, is there a satisfying allocation of ;; variables? ;; To warm up for this, let's try its tractable cousin 2-SAT ;; In 2-SAT, the clauses to be satisfied are all of form either: ;; (a must be true/false) or (b must be true/false) ;; So here's an example problem, with four variables and four clauses ;; Either x1 must be true or x2 must be true ;; Either x3 must be true or x4 must be true ;; Either x1 must be false or x3 must be false ;; Either x2 must be false or x4 must be false ;; A solution to the problem is either a satisfying assignment, or a ;; proof that no such assignment exists. ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;; To make this trial problem more precise: ;; Represent an assignment as a vector of true, false values, then (def test-assignment [ true false true false ]) ;; We can represent the first clause as: (defn satisfies-1 [v] (or (nth v (dec 1) (nth v (dec 2))))) (satisfies-1 test-assignment) ;-> true ;; So far so good ;; The second clause as: (defn satisfies-2 [v] (or (nth v (dec 3)) (nth v (dec 4)))) (satisfies-2 test-assignment) ;-> true ;; Hooray ;; The third clause as: (defn satisfies-3 [v] (or (not (nth v (dec 1))) (not (nth v (dec 3))))) (satisfies-3 test-assignment) ;-> false ;; Darn ;; And finally the fourth clause as: (defn satisfies-4 [v] (or (not (nth v (dec 2))) (not (nth v (dec 4))))) (satisfies-4 test-assignment) ;-> true ;; Three out of four ain't bad ;; So actually we only failed one of the clauses with our guess. ;; What it means for the problem to be in NP is that given a proposed ;; satisfying assignment, we can check all these conditions in ;; polynomial time. (polynomial in n, the length of the input, here ;; take n to be the greater of the number of clauses and variables) ;; In fact it's obvious that it's linear in n. We need to check each ;; clause exactly once on a proposed solution. (defn satisfies? [v] (and (satisfies-1 v) (satisfies-2 v) (satisfies-3 v) (satisfies-4 v))) ;; So 2-SAT is definitely in NP. (satisfies? test-assignment) ;-> false ;; Given speedy checking of proposed solutions, we can find (or rule out) a satisfying solution by brute-force search: (def possible-assignments (for [i [true false] j [true false] k [true false] l [true false]] [i j k l])) possible-assignments ;-> ([true true true true] [true true true false] [true true false true] [true true false false] [true false true true] [true false true false] [true false false true] [true false false false] [false true true true] [false true true false] [false true false true] [false true false false] [false false true true] [false false true false] [false false false true] [false false false false]) (count possible-assignments) ;-> 16 ;; Not too bad, we can just try them all: (for [a possible-assignments :when (satisfies? a)] a) ;-> ([true false false true] [false true true false]) ;; So we've found two possible satisfying assignments to the variables (satisfies? [true false false true]) ;-> true (satisfies? [false true true false]) ;-> true ;; One would have done. And if we'd found none, then that would have ;; shown the problem to be unsatisfiable, which is also a valid answer ;; to this 'decision problem'. ;; Because the space of possible solutions is exponential in the ;; number of variables, which is constrained by the size of the input, ;; the fact that 2-SAT is in NP and so each trial solution can be ;; assessed in polynomial time, guarantees us an exponential-time ;; solution to the problem. ;; The interesting question is 'can we do better than that?' ;; In the case of 2-SAT we can do much better. In fact it will turn ;; out that 2-SAT can be solved in (small) polynomial time, which ;; means that we can solve very large problems of this kind. ;; In the case of 3-SAT, almost certainly not. There's a vast class of ;; problems in NP which are all easier than 3 SAT, in the sense that ;; if we had a polynomial time 3-SAT algorithm then we'd automatically ;; also have polynomial time algorithms for this vast class, and ;; people have been trying to solve them for many years. And nobody ;; has ever come up with a polynomial time algorithm for any of them. ;; This is known as the P=NP problem. It's thought not, but nobody knows. ;; It's probably the most important open maths problem there is.

## Tuesday, October 15, 2013

### 2SAT and P = NP

Subscribe to:
Post Comments (Atom)

## No comments:

## Post a Comment