On a random walk strategy for the Q2SAT problem K. Subramani

Topics to be covered 1.Definitions of 2SAT and Q2SAT problems 2.2 player games 3.Motivation 4.Related Work 5.The random walk algorithm for 2SAT 6.Analysis of Papadimitriou`s algorithm 7.Prerequisites for coin-flipping to work 8.How Q2SAT meets those prerequisites 9. Conclusion

2SAT and Q2SAT Given a Boolean formula where each clause has at most 2 literals from the variable set {x 1, x 2, …x n }, the 2SAT problem is to check whether this formula has a satisfying assignment.

2SAT and Q2SAT (contd.) Given a Boolean formula and a quantifier string Q(x,y) where each clause has at most 2 literals from the variable set {x 1, x 2, …x n, y 1,y 2,…y n }, the Q2SAT problem is to check whether the formula Q(x,y) has a satisfying model

2 player games 1.Let X be the Existential player and Y be the Universal player. 2. Assume without loss of generality that all x variables are existentially quantified and the y variables are universa- lly quantified. 3. Each player makes a sequence of moves non-deterministi- cally in the order specified by Q(x,y). 4. A move is a true/false guess. 5. If at the end of the moves, the expression evaluates to true, we say that X has a winning strategy and this strategy is called a model for Q(x,y)

2 Player games (contd.) So the QBF problem in general, is to decide whether the existential player has a winning strategy.

Motivation 1.Algorithmic elegance. 2.How much can a simple strategy net us? 3.Deeper understanding of game theory. 4.First use of randomization to study QBF feasibility. 5.Papadimitriou got a FOCS paper out of his 2SAT random walk algorithm! [Pap91]

Related Work 1. [Pap91] introduced the random walk algorithm for simple 2SAT. 2. [WS02] applies the random walk technique to 3SAT problems with good empirical results. 3. Even for HornSAT (which is in P), the random walk algorithm does not converge in a polynomial number of steps (analytically). (For a pathological example, see [Ros00].) 4. [APT79] solves both 2SAT and Q2SAT in linear time, using connected components of the implication graph.

Related Work (contd.) 5. I am working on strategies which enhance simple random walk with “greed”, i.e., selecting the unsatisified clause with the fewest number of literals to fix. It works for HornSAT, but HornSAT has an extremely simple deterministic algorithm.

The random walk algorithm for 2SAT 1.Start with a random assignment for. 2.If all clauses are satisfied stop. 3.Pick a clause which is not satisfied. 4.Pick one of the 2 literals in this clause at random and `flip´ it. 5.Repeat steps (2)-(5) for 2n 2 steps. 6.Say that the formula is `probably unsatisfiable.´

Analysis Let us say that the given formula is satisfiable and focus on a specific satisfying assignment T. Let the initial random assignment be T´, which does not satisfy Now consider a clause C which is violated by T´. At least one of the 2 literals in C must be true in T. Therefore flipping a literal at random results in T´getting closer to T, with probability at least one half. Likewise, T´could be moving away from T, with probability one half.

Analysis (contd.) Let X i denote the number of flips required by the above algorithm, assuming that the current assignment differs from T in exactly i variables. We use the probability law: E[X] = E[E[X| Y]] (see [Ros00]) to conclude that E[X i ]= ½ (E[X i-1 ]+1)+ ½ (E[X i+1 ]+1) Also note that E[X 0 ]=0 and E[X n ]=1+E[X n-1 ], since if by chance we start off with the correct assignment, no flips are required, whereas if all variables are incorrect, then a flip must move us in the right direction!

Analysis (contd.) The above system can be solved to get E[X n ] = n 2. Now applying Markov´s law, we can conclude that the probability that more than 2.n 2 flips will be required is less than ½.

Prerequisites for flipping to work 1.A literal must be flippable, i.e., there are exactly 2 values to assign to it. (Y 1 ) 2.When a literal is flipped with probability at least one half, we move in the right direction, and with probability at most one half, we move in the wrong direction. (Y 2 ) Y 1 is trivially met in existential first order logic, by all SAT formulas, since every variable is either true or false. However, Y 2 fails for even HornSAT!

Why Q2SAT meets the prerequisites First observe that if Y 1 is met by Q2SAT then Y 2 is trivially met, since under the current assignment both literals have been set incorrectly (assuming that both variables are existentially quantified) and with probability at least ½, we will move in the right direction by choosing one at random.

Why Q2SAT meets the prerequisites I have provided a detailed explanation of Page 4 and Page 5 of my paper of how Y 1 is met by Q2CNF formulas. The main idea is that Q2SAT problems are guaranteed to have simple models, if satisfiable, i.e., a model in which each existentially quantified variable is either true, false, or y i or the complement of y i, where y i is a universally quantified variable preceding it.

Why Q2SAT meets the prerequisites A linear scan of the clause set cuts these choices to precisely two. We are now free to use Papadimitriou´s algorithm and the accompanying analysis!

Conclusion A number of interesting problems are currently open, insofar as randomized approaches are concerned: (a)Boolean equivalence, i.e., given 2 2CNF formulas, do they have the same set of satisfying assignments? (b) UNIQUE2SAT – Does a 2CNF formula have exactly one satisfying assignment? (c) CRITICAL2SAT – Is a 2CNF formula unsatisfiable, but knocking off any clause makes it satisfiable?