Satisfiability and SAT Solvers. CS 270 Math Foundations of CS Jeremy Johnson

Transcription

1 Satisfiability and SAT Solvers CS 270 Math Foundations of CS Jeremy Johnson

2 Conjunctive Normal Form Conjunctive normal form (products of sums) Conjunction of clauses (disjunction of literals) For each row in the truth table where the output is false, write a sum such that the corresponding input not in that row Alternatively use Demorgan s law for the negation of dnf for f (zero rows) E.G. (multiplexor function) (ss + xx 0 + xx 1 ) (ss + xx 0 + xx 1 ) ( ss + xx 0 + xx 1 ) ( ss + xx 0 + xx 1 ) s x 0 x 1 f

3 Satisfiability A formula is satisfiable if there is an assignment to the variables that make the formula true A formula is unsatisfiable if all assignments to variables eval to false A formula is falsifiable if there is an assignment to the variables that make the formula false A formula is valid if all assignments to variables eval to true (a valid formula is a theorem or tautology)

4 Satisfiability Checking to see if a formula f is satisfiable can be done by searching a truth table for a true entry Exponential in the number of variables Does not appear to be a polynomial time algorithm (satisfiability is NP-complete) There are efficient satisfiability checkers that work well on many practical problems Checking whether f is satisfiable can be done by checking if f is not valid An assignment that evaluates to false provides a counter example to validity

5 DNF vs CNF It is easy to determine if a boolean expression in DNF is satisfiable but difficult to determine if it is valid It is easy to determine if a boolean expression in CNF is valid but difficult to determine if it is satisfiable It is possible to convert any boolean expression to DNF or CNF; however, there can be exponential blowup

6 SAT Solvers Input expected in CNF Using DIMACS format One clause per line delimited by 0 Variables encoded by integers, not variable encoded by negating integer We will use MiniSAT (minisat.se)

7 MiniSAT Example (x1 -x5 x4) & (-x1 x5 x3 x4) & (-x3 x4). DIMACS format (c = comment, p cnf = SAT problem in CNF) c SAT problem in CNF with 5 variables and 3 clauses p cnf

8 MiniSAT Example (x1 -x5 x4) & (-x1 x5 x3 x4) & (-x3 x4). This is MiniSat 2.0 beta ============================[ Problem Statistics ]================== Number of variables: 5 Number of clauses: 3 Parsing time: 0.00 s. SATISFIABLE v

9 Avionics Application Aircraft controlled by (real time) software applications (navigation, control, obstacle detection, obstacle avoidance ) Applications run on computers in different cabinets 500 apps 20 cabinets Apps 1, 2 and 3 must run in separate cabinets Problem: Find assignment of apps to cabinets that satisfies constraints

10 Corresponding SAT problem AC is a map from apps to cabinents [indicator variable] AC(app,cab) = t iff AC(app) = cab [Valid Mapping] aa cc AAAA aa cc aa AA cc CC AAAA aa cc [constaints] cc AAAA 1 cc AAAA 2 cc AAAA 3 cc cc AAAA 2 cc AAAA 3 cc cc CC AAAA 1 cc AAAA 2 cc AAAA 3 cc cc CC AAAA 2 cc AAAA 3 cc

11 Constaints in CNF cc CC AAAA 1 cc AAAA 2 cc AAAA 3 cc cc CC AAAA 1 cc AAAA 2 cc AAAA 1 cc AAAA 3 cc cc CC AAAA 2 cc AAAA 3 cc cc CC AAAA 2 cc AAAA 3 cc

12 DIMACS Format Var(AAAA aa cc ) = 20(a-1)+c AAAA cc 1 AAAA cc 2 = -c (20+c) AAAA cc 1 AAAA cc 3 = -c -(40+c) AAAA 1 aa AAAA 20 aa = 20(a-1)+1 20(a-1)

13 Avionics Example 10 apps and 5 cabinets Var(AAAA cc aa ) = 5(a-1)+c 50 variables 25 clauses Valid Map aa=1 10 AAAA aa 1 AAAA aa 5 Constaints cc CC AAAA cc cc 1 AAAA 2 cc CC AAAA cc cc 1 AAAA 3 cc CC AAAA cc cc 2 AAAA 3

14 Avionics Example p cnf c clauses for valid map forall a exists c AC^c_a

15 Avionics Example c constaints ~AC^c_1 + ~AC^c_2 and ~AC^c_1 + ~AC^c_ c constraint ~AC^c_2 + ~AC^c_

16 Avionics Example Programs]$./MiniSat_v1.14_linux aircraft assignment ==================================[MINISAT]=================================== Conflicts ORIGINAL LEARNT Progress Clauses Literals Limit Clauses Literals Lit/Cl ============================================================================== nan % ============================================================================== restarts : 1 conflicts : 0 (nan /sec) decisions : 39 (inf /sec) propagations : 50 (inf /sec) conflict literals : 0 ( nan % deleted) Memory used : 1.67 MB CPU time : 0 s SATISFIABLE

17 Avionics Assignment SAT True indicator variables: 3 = 5*0 + 3 => AC(1,3) 7 = 5*1 + 2 => AC(2,2) 11 = 5*2 + 1 => AC(3,1) 16 = 5*3+1 => AC(4,1) 21 = 5*4+1 => AC(5,1) 26 = 5*5=1 => AC(6,1) 31 = 5*6+1 => AC(7,1) 36 = 5*7+1 => AC(8,1) 41 = 5*8 + 1 => AC(9,1) 46 = 5*9+1 => AC(10,1)

18 N-Queens Problem Given an N x N chess board Find a placement of N queens such that no two queens can take each other

19 N Queens

20 N Queens

21 N Queens Backtrack

22 N Queens

23 N Queens

24 N Queens Backtrack

25 N Queens Backtrack

26 N Queens

27 N Queens

28 N Queens

29 N Queens Solution Found

30 Recursive Solution to N-Queens Define Queens(board, current, size) Input: board a size x size chess board with placement of current queens in positions without conflict only using the first current columns Output: true if board is a conflict free placement of size queens if (current = size) then return true for row = 0 to size-1 do position := (row,column+1) if ConflictFree(board,position) Update(board,position) done := Queens(board,column+1,size) if done = true return true return false

31 N-Queens as a SAT Problem Introduce variables B ij for 0 i,j < N B ij = T if queen at position (i,j) F otherwise Constraints Exactly one queen per row Row i = B ij, j=0 N-1 Exactly one queen per column Column j = B ij, i=0 N-1 At most one queen on diagonal Diagonal k- = B ij, i-j = k = -N+1,N-1 Diagonal k+ = B ij, i+j = k = 0,2N

32 4-Queens SAT input Exactly one queen in row i B i0 B i1 B i2 B i3 B i0 B i1 B i2 B i3 B i1 B i2 B i3 B i2 B i3

33 4-Queens SAT input Exactly one queen in column j B 0j B 1j B 2j B 3j B 0j B 1j B 2j B 3j B 1j B 2j B 3j B 2j B 3j

34 4-Queens SAT input At most one queen in diagonal k- B 20 B 31 B 00 B 11 B 22 B 33 B 11 B 22 B 33 B 22 B 33 B 02 B 13

35 4-Queens SAT input At most one queen in diagonal k+ B 01 B 10 B 30 B 21 B 12 B 03 B 21 B 12 B 03 B 12 B 03 B 32 B 23

36 DPLL Algorithm Tries to incrementally build a satisfying assignment A: V {T,F} (partial assignment) for a formula ϕ in CNF A is grown by either Deducing a truth value for a literal Whenever all literals except one are F then the remaining literal must be T (unit propagation) Guessing a truth value Backtrack when guess (leads to inconsistency) is wrong

37 DPLL Example Operation Assign Formula 1 2, 2 3 4, 1 2, 1 3 4, 1

38 DPLL Example Operation Assign Formula 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 1 1 2, 2 3 4, 1 2, 1 3 4, 1

39 DPLL Example Operation Assign Formula 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 1 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 2 1 2, 2 3 4, 1 2, 1 3 4, 1

40 DPLL Example Operation Assign Formula 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 1 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 1, 2 1 2, 2 3 4, 1 2, 1 3 4, 1 Guess 1, 2, 3 1 2, 2 3 4, 1 2, 1 3 4, 1

41 DPLL Example Operation Assign Formula 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 1 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 1, 2 1 2, 2 3 4, 1 2, 1 3 4, 1 Guess 1, 2, 3 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 1, 2, 3, 4 1 2, 2 3 4, 1 2, 1 3 4, 1 Inconsistency

42 DPLL Example Operation Assign Formula 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce , 2 3 4, 1 2, 1 3 4, 1 Deduce 2 1, 2 1 2, 2 3 4, 1 2, 1 3 4, 1 Guess 3 1, 2, 3 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 4 1, 2, 3, 4 1 2, 2 3 4, 1 2, 1 3 4, 1 Undo 3 1, 2 1 2, 2 3 4, 1 2, 1 3 4, 1 Backtrack

43 DPLL Example Operation Assign Formula 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce , 2 3 4, 1 2, 1 3 4, 1 Deduce 2 1, 2 1 2, 2 3 4, 1 2, 1 3 4, 1 Guess 3 1, 2, 3 1 2, 2 3 4, 1 2, 1 3 4, 1 Deduce 4 1, 2, 3, 4 1 2, 2 3 4, 1 2, 1 3 4, 1 Undo 3 1, 2 1 2, 2 3 4, 1 2, 1 3 4, 1 Guess 3 1, 2, 3 1 2, 2 3 4, 1 2, 1 3 4, 1 Assignment found