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

2. 2 Conjunctive Normal Form s x 0 x 1 f 0 0 0 0 1 0 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1

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 5 3 1 -5 4 0 -1 5 3 4 0 -3 -4 0

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 -1 -2 -3 -4 -5 0

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

11. Constaints in CNF

12. DIMACS Format

13. Avionics Example

14. p cnf 50 25 c clauses for valid map forall a exists c AC^c_a 1 2 3 4 5 0 6 7 8 9 10 0 11 12 13 14 15 0 16 17 18 19 20 0 21 22 23 24 25 0 26 27 28 29 30 0 31 32 33 34 35 0 36 37 38 39 40 0 41 42 43 44 45 0 46 47 48 49 50 0

15. Avionics Example c constaints ~AC^c_1 + ~AC^c_2 and ~AC^c_1 + ~AC^c_3 -1 -6 0 -1 -11 0 -2 -7 0 -2 -12 0 -3 -8 0 -3 -13 0 -4 -9 0 -4 -14 0 -5 -10 0 -5 -15 0 c constraint ~AC^c_2 + ~AC^c_3 -6 -11 0 -7 -12 0 -8 -13 0 -9 -14 0 -10 -15 0

16. Avionics Example [jjohnson@tux64-12 Programs]$./MiniSat_v1.14_linux aircraft assignment ==================================[MINISAT]=================================== | Conflicts | ORIGINAL | LEARNT | Progress | | | Clauses Literals | Limit Clauses Literals Lit/Cl | | ============================================================================== | 0 | 25 80 | 8 0 0 nan | 0.000 % | ============================================================================== 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 -1 -2 3 -4 -5 -6 7 -8 -9 -10 11 -12 -13 -14 -15 16 -17 -18 -19 -20 21 -22 -23 -24 -25 26 -27 -28 -29 -30 31 -32 -33 -34 -35 36 -37 -38 -39 -40 41 -42 -43 -44 -45 46 -47 -48 -49 -50 0 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

21. Backtrack

22. N Queens

24. Backtrack

25. N Queens Backtrack

26. N Queens

29. 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-2 00010203 13101112 20212223 33303132

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  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 OperationAssignFormula

38. DPLL Example OperationAssignFormula Deduce1

39. DPLL Example OperationAssignFormula Deduce1

40. DPLL Example OperationAssignFormula Deduce1 Guess

41. DPLL Example OperationAssignFormula Deduce1 Guess Deduce Inconsistency

42. DPLL Example OperationAssignFormula Deduce 11 Guess 3 Deduce 4 Undo 3 Backtrack

43. DPLL Example OperationAssignFormula Deduce 11 Guess 3 Deduce 4 Undo 3 Assignment found