1. Student:Victoria Kravchenko Supervisors:Prof. Yoram Moses Liat Atsmon

2. Study and evaluate the approach of solving optimization problems through SAT reduction, and using SAT solvers to solve the reduced problems.

3.  Boolean satisfiability problem (x 1 +x 5 ’+x 4 ) (x 1 ’+x 5 +x 3 +x 4 ) (x 3 ’+x 4 ’)  NP-Complete  CNF (Conjunctive Normal Form) – “AND of ORs”

4.  Developed at 2003 by Niklas Eén and Niklas Sörensson  Advantages: open-source, successful, small.

5. For:(x 1 +x 5 ’+x 4 ) (x 1 ’+x 5 +x 3 +x 4 ) (x 3 ’+x 4 ’) p cnf 5 3 → p cnf number_of_variables number_of_clauses 1 -5 4 0 → 0 is the end of the clause -1 5 3 4 0 -3 -4 0

6.  PART 1: Equation Checker  PART 2: SAT with Optimization  PART 3: Maximal Acyclic Subgraph

7.  Input: X, Y, Z – decimal numbers  Output: whether X+Y=Z or not  Calculates the binary form of the numbers  The program prepares the input for MiniSat using the table:

8.  For the LSB bit:

10.  Input: Vec – x 0,x 1,x 2,… Phi – CNF expression  Output: maximal value of Vec (as a binary number) which satisfies Phi

11.  Versions: I.Bit by Bit: II.MiniSat condition: MSB 1?1 0 ????? ?1?1 1 00 10 1 011 1

12.  Input: graph name nodes – [node_number],[node_weight] edges – [node_from],[node_to] For example: ex1 1,2-2,2-3,1 1,2-2,3-3,1 1 (2) 3 (1) 2 (2)

13.  Output: maximal acyclic subgraph 1 (2) 3 (1) 2 (2)

14. 1 (1) 3 (2) 2 (2) 5 (3) 4 (3) lsg ex1 1,1-2,2-3,2-4,3-5,3 1,2-1,3-2,3-2,4-4,5-5,3

15. 1 (1) 3 (2) 2 (2) 5 (3) 4 (3) lsg ex2 1,1-2,2-3,2-4,3-5,3 1,2-3,1-2,3-2,4-4,5-5,3

16. 1 (2) 3 (2) 2 (1) 5 (3) 4 (3) lsg ex3 1,2-2,1-3,2-4,3-5,3 1,2-3,1-2,3-2,4-4,5-5,3

17. 1 (1) 3 (2) 2 (2) 5 (3) 4 (3) lsg ex4 1,1-2,2-3,2-4,3-5,3 1,2-3,1-2,3-2,4-4,5-5,2-5,3

18. 1 (1) 3 (2) 2 (4) 5 (1) 4 (3) lsg ex5 1,1-2,4-3,2-4,3-5,1 1,2-3,1-2,3-2,4-4,5-5,2-5,3

19. For each edge:  Define: [node_from] > [node_to]  Condition:  Translate the condition to CNF using a tree (reduction to 3-SAT)

20. Building the weighted sum:  [node] x [weight]: Y1 [m vars] ↔ X1 [node number] x W1 [m vars] Y2 [m vars] ↔ X2 [node number] x W2 [m vars]  Sum the intermediate variables as in the Equation Checker: Z1 [m vars] ↔ Y1 [m vars] + Y2 [m vars]  Continue summing: Z2 [m vars] ↔ Z1 [m vars] + Y3 [m vars]  Save the last intermediate variables representing the accumulated sum: Z(n-1) [m vars] ↔ Z(n-2) [m vars] + Yn [m vars]  Z final ↔ X 1 *W 1 + X 2 *W 2 + … + X n *W n

21.  Get a result from the MiniSat using only the edges conditions and extract the weight W res  Demand a new result with a greater weight s.t. Z final > W res  Continue until the conditions can not be saturated

26.  The graphs are polynomial and not exponential although the problem in NP complete – success!  Nodes have a stronger effect on the run time.

27. Thank you for your attention!