SAT SOLVING MAXIMIZATION AND MINIMIZATION PROBLEMS

Student: Victoria Kravchenko

Supervisors: Prof. Yoram Moses

Liat Atsmon

The Project GoalStudy and evaluate the approach of solving optimization problems through SAT reduction, and using SAT solvers to solve the reduced problems.

SAT Boolean satisfiability problem

(x1+x5’+x4) (x1’+x5+x3+x4) (x3’+x4’) NP-Complete CNF (Conjunctive Normal Form) – “AND

of ORs”

MiniSat Developed at 2003 by

Niklas Eén and Niklas Sörensson

Advantages: open-source, successful, small.

The MiniSat input formFor: (x1+x5’+x4) (x1’+x5+x3+x4) (x3’+x4’)

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

The Project PART 1: Equation Checker

PART 2: SAT with Optimization

PART 3: Maximal Acyclic Subgraph

The equation checker 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 MiniSatusing the table:

1

0 0 0 0 0

0 1 0 1 0

1 1 0 0 1

0 0 1 1 0

0 1 1 0 1

1 1 1 1 1

i i i i iX Y C Z C

For the LSB bit:

The equation checker

0

0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

Z

Z X Y X Y X Y

f X Y X Y Z X Y X Y Z

X Y X Y Z X Y X Y Z

X Y Z X Y Z X Y Z X Y Z

0

0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

Z

Z

f X Y Z X Y Z X Y Z X Y Z

f X Y Z X Y Z X Y Z X Y Z

00 0 0

0 0 0 1

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 0

ZX Y Z f

The equation checker

0

0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0C

C X Y

f X Y C X Y C X Y C X Y C X Y C X C Y C

00 0 0

0 0 0 1

0 0 1 0

0 1 0 1

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 1

CX Y C f

0

0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

C

C

f X Y C X Y C X Y C X Y C

f X Y C X Y C

X Y C X Y C

SAT with Optimization Input: Vec – x0,x1,x2,…

Phi – CNF expression Output: maximal value of Vec (as a

binary number) which satisfies Phi

SAT with Optimization Versions:

I. Bit by Bit:

II. MiniSat condition:

MSB1?1 1?0 1?

? ? ?? ?

?1

1 0 01 0

1

0 11 1

Maximal Acyclic Subgraph 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)

Maximal Acyclic Subgraph Output: maximal acyclic subgraph

1 (2)

3 (1)

2 (2)

Example 11

(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

P

Example 21

(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

Example 31

(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

Example 41

(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

Example 51

(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

How Does It Work?For each edge: Define: [node_from] > [node_to]

Condition: Translate the condition to CNF using a tree

(reduction to 3-SAT)

0 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 2 2 ...

X Y

x y x y x y x y x y x y x y x y x y

X Y X Y

How Does It Work?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]

Zfinal ↔ X1*W1 + X2*W2 + … + Xn*Wn

How Does It Work?

Get a result from the MiniSat using only the edges conditions and extract the weight Wres

Demand a new result with a greater weight s.t. Zfinal > Wres

Continue until the conditions can not be saturated

Results

50 100 150 200 250 300 350 400 450 500 550 60012:00:00 AM

12:00:43 AM

12:01:26 AM

12:02:09 AM

12:02:52 AM

12:03:36 AM

12:04:19 AM

12:05:02 AM

12:05:45 AM

12:06:28 AM

12:00:10 AM12:00:16 AM12:00:28 AM

12:00:40 AM12:00:50 AM

12:01:08 AM

12:01:46 AM

12:02:24 AM

12:03:02 AM

12:03:39 AM

12:04:47 AM

12:05:55 AM

Run Time vs. Number of Edges (50 Nodes)

Run Time

Poly (order 4)

Exp

Edges

Tim

e

Results

50 100 150 200 250 30000:00:00

00:01:26

00:02:52

00:04:19

00:05:45

00:07:12

00:08:38

00:10:04

00:11:31

00:03:4800:04:09

00:05:56

00:07:25

00:08:30

00:10:31

Run Time vs. Number of Edges (200 Nodes)

Run Time

Poly (order 4)

Exp

Edges

Tim

e

Results

50 100 150 200 250 30000:00:00

00:01:26

00:02:52

00:04:19

00:05:45

00:07:12

00:08:38

00:10:04

00:11:31

00:12:57

00:00:1000:00:41

00:02:19

00:03:25

00:06:39

00:11:10

Run Time vs. Number of Nodes (50 Edges)

Run Time

Poly (order 3)

Exp

Nodes

Tim

e

Results

50 100 150 200 250 30000:00:00

00:02:52

00:05:45

00:08:38

00:11:31

00:14:24

00:17:16

00:20:09

00:00:40

00:02:01

00:05:05

00:07:25

00:11:39

00:17:12

Run Time vs. Number of Nodes (200 Edges)

Run Time

Poly (order 2)

Exp

Nodes

Tim

e

Conclusions The graphs are polynomial and not

exponential although the problem in NP complete – success!

Nodes have a stronger effect on the run time.

Thank you for your attention!