Algebraic Dynamic Programming. Solving Satisfiability with ADP

Transcription

1 Algebraic Dynamic Programming Session 12 Solving Satisfiability with ADP Robert Giegerich (Lecture) Stefan Janssen (Exercises) Faculty of Technology Summer

2 Recall the Satisfiability Problem Satisfiability of a Boolean formula A = (x 1 x 2 x 4 ) (x 2 x 3 x 5 ) A is satisfiable if there is a truth value assignment to x 1, x 2,... which satisfies A.

3 Recall the Satisfiability Problem Satisfiability of a Boolean formula A = (x 1 x 2 x 4 ) (x 2 x 3 x 5 ) A is satisfiable if there is a truth value assignment to x 1, x 2,... which satisfies A. x <i> = (T, T, T, T, T ) or x <i> = (F, F, F, T, T ) satisfies A, while x <i> = (F, T, T, T, F) does not.

4 Recall satisfiability (2) (Un)satisfiability of a Boolean formula: A = (x 1 x 2 ) ( x 1 x 2 ) (x 1 x 2 ) ( x 1 x 2 ) is not satisfiable.

5 Satisfiability variants General SAT: arbitrary Boolean formula with Boolean operators. CNF-SAT: Boolean formula in conjunctive normal form B = (x 1 x 2 x 4 ) ( x 2 x 3 x 5 )... with n variables, any number of literals (x or x) in a clause, any number of clauses 3-CNF-SAT: only 3 literals per clause, and any number of clauses. Every SAT problem can be reformulated as 3-CNF-SAT problem in polynomial time.

6 Satisfiability variants General SAT: arbitrary Boolean formula with Boolean operators. CNF-SAT: Boolean formula in conjunctive normal form B = (x 1 x 2 x 4 ) ( x 2 x 3 x 5 )... with n variables, any number of literals (x or x) in a clause, any number of clauses 3-CNF-SAT: only 3 literals per clause, and any number of clauses. Every SAT problem can be reformulated as 3-CNF-SAT problem in polynomial time. In any case: A Boolean formula with n variables creates a search space of size 2 n.

7 Importance of SAT and 3-SAT SAT is one of the most famous problems in computer science many other problems can be expressed as SAT-problems SAT was the first problem to be proved NP-complete, i.e. P = NP if SAT has a polynomial time solution (Cook s Theorem)

8 Proof idea of Cook s theorem (2) Given any decision problem in NP, construct a non-deterministic machine that solves it in polynomial time. Then for each input to that machine, build a Boolean expression which says that the input is passed to the machine, the machine runs correctly, and the machine halts and answers "yes". Then the expression can be satisfied if and only if there is a way for the machine to run correctly and answer "yes", so the satisfiability of the constructed expression is equivalent to asking whether or not the machine will answer "yes".

9 Proof idea of Cook s theorem (3) If decision problem X is in NP, a nondeterministic Turing Machine decides it in time p(n), where n = X and p is a polynomial. The Boolean formula constructed has size O(log(p(n))(p(n)) 2 ), which is still polynomial. For details see s_theorem

10 Ideas Expressing SAT in ADP: The input sequence x encodes the Boolean formula The grammar G constructs 2 n parses (candidates) for each clause, and hence (2 n ) K candidates for a formula with K clauses. These candidates model all 2 n possible truth assignments Evaluation algebra SAT checks whether the assignments in a given candidate satisfy each clause Evaluation algebra EQA checks whether the assignments in a given candidate are consistent.

11 Ideas (2) Expressing SAT in ADP (ctd.): There will be candidates which are consistent but do not satisfy the formula. There will be candidates which satisfy the formula but are not consistent. G(SAT EQA, x) takes care of both and solves the satisfiability problem.

12 Input encoding in ASCII: (x 1 x 2 ) ( x 2 x 3 ) PP-&-NP Structure of the formula: & V V x 1 V V x 2 not x 3 x 2

13 A subtlety How to interpret the empty formula: a formula with 0 clauses? ( TRUE by default) a formula with 1 clause having 0 literals? ( FALSE by default) We choose the first option.

14 String grammar generating encoded formulas This grammar rules that the first clause in a formula must have at least one literal. formula formula & clause literal clause clause ɛ literal clause literal lit lit P N The nonterminal symbol lit appears superfluous...

15 Tree grammar generating candidates from input formula addclause emptyformula literal clause clause formula & clause emptyclause literal clause literal assigntrue assignfalse lit lit lit P N

16 Modeling truth assignments by tree grammar formula addclause emptyformula literal clause clause formula & clause emptyclause literal clause literal assigntrue assignfalse lit lit lit P N The two productions in green represent the assignment of T resp. F to the variable x i in the literal i (if any). All such assignments are made independently.

17 Tree representation of solution candidates addclause & assigntrue assigntrue P assignfalse assignfalse P assignfalse empty clause N assigntrue empty clause P

18 Evaluation of solution candidates addclause & assigntrue assigntrue P assignfalse assignfalse P assignfalse empty clause N assigntrue empty clause P The left clause assigns (T, F, F) and is satisfied. The right clause assigns (T, F, T ) and is satisfied. The candidate does not satisfy the formula, because the two assignments are not consistent.

19 Solving SAT with ADP For the following steps go to 1 Consider algebra PRETTY, which indicates the assignments chosen for each clause in a candidate. 2 Explain why the candidate COUNT for the example is (2 3 ) 2. 3 Use algebras SAT and EQA separately. 4 Study carefully the treatment of missing literals! 5 What can you compute with (SAT COUNT ) and (EQA COUNT )? 6 Use algebra (SAT EQA) and (EQA SAT ). 7 Explain why both solve Satisfiability. 8 Is one of these a case of classified DP?

20 Conclusions Considerations on ADP as a programming method: ADP is nontrivial as it can solve problems in NP. NP-problems can be solved by G(A, x), because grammar G generates an exponential number of candidates. It is typical in DP to have an exponential candidate space, and still to evaluate it in polynomial time. This is because the choice function reduces the candidates to a constant (or polynomial) number. If G(A B, x)) solves an NP-hard problem, then at least one of G(A, x)) or G(B, x)) has exponential runtime. (In our example, this is algebra EQA.) Classified DP may also solve problems in NP, as witnessed not by SAT, but by the HMM path labeling problem discussed earlier.