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# Sat solving maximization and minimization problems PowerPoint PPT Presentation

Student:Victoria Kravchenko Supervisors:P rof. Yoram Moses Liat Atsmon. Sat solving maximization and minimization problems. The Project Goal. Study and evaluate the approach of solving optimization problems through SAT reduction, and using SAT solvers to solve the reduced problems. SAT.

Sat solving maximization and minimization problems

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#### Presentation Transcript

Student:Victoria Kravchenko

Supervisors:Prof. Yoram Moses

LiatAtsmon

## Sat solving maximization and minimization problems

### The Project Goal

Study 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 byNiklasEén and NiklasSörensson

### The MiniSat input form

For:(x1+x5’+x4) (x1’+x5+x3+x4) (x3’+x4’)

p cnf 5 3 → p cnfnumber_of_variablesnumber_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 preparesthe input for MiniSatusing the table:

### The equation checker

• For the LSB bit:

### 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:

• Bit by Bit:

• MiniSat condition:

MSB

1?

1

1?

0

1?

1

?1

?

1

?

0

1

1

?

0

0

?

1

0

?

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)

2(2)

3(1)

### Maximal Acyclic Subgraph

• Output: maximal acyclic subgraph

1(2)

2(2)

3(1)

### Example 1

P

1 (1)

2 (2)

3 (2)

4 (3)

5 (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

### Example 2

1 (1)

2 (2)

3 (2)

4 (3)

5 (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 3

1 (2)

2 (1)

3 (2)

4 (3)

5 (3)

lsgex3 1,2-2,1-3,2-4,3-5,3 1,2-3,1-2,3-2,4-4,5-5,3

### Example 4

1 (1)

2 (2)

3 (2)

4 (3)

5 (3)

lsgex4 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 5

1 (1)

2 (4)

3 (2)

4 (3)

5(1)

lsgex5 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)

### 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

### Conclusions

• The graphs are polynomial and not exponential although the problem in NP complete – success!

• Nodes have a stronger effect on the run time.