[PPT] - Solving the Max2Sat problem Sonia Cafieri LIX, cole Polytechnique PowerPoint Presentation

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Solving the Max2Sat problem

Sonia Cafieri

LIX, École Polytechnique cafieri@lix.polytechnique.fr

Nov 18th 2008

S. Cafieri (LIX)

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Outline

1

Basic concepts and definitions Some definition Problem statement Alternative formulations

2

Graphical representation Implication graph

3

Solving the problem Solution methods Solving MAX2SAT using AMPL

S. Cafieri (LIX)

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Outline

1

Basic concepts and definitions Some definition Problem statement Alternative formulations

2

Graphical representation Implication graph

3

Solving the problem Solution methods Solving MAX2SAT using AMPL

S. Cafieri (LIX)

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Boolean variables

Definition

A boolean variable x is a variable that can assume only two values 0 and 1. 0 is for false, 1 for true. The negation ¯ x of a boolean variable x is the variable that assumes the value 1 − x. Note that ¯ ¯ x = x.

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Literals and Clauses

Definition

A literal l is a variable x or its negation ¯ x. Let X be a set of boolean variables. For every x ∈ X there are 2 literals over x, namely x itself and ¯ x.

Definition

A clause over a set of boolean variables is a disjunction of literals. Example: suppose we have 2 variables x1, x2 we create a clause by using the or operator: x1∨¯ x2.

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Literals and Clauses

Definition

The length of a clause is the number of its literals. Example: x1 ∨ ¯ x2 ∨ ¯ x3 is a clause of length 3 its literals are x1, ¯ x2 and ¯ x3 if x1 = 0, x2 = 0, x3 = 1, then its value is 1.

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CNF formula

Definition

A boolean formula in Conjunctive Normal Form (CNF) Φ is a conjunction of clauses: Φ =

m

i=1

Ci, where Ci are clauses, ∀i ∈ {1, .., m}. Example: suppose we have 3 variables: x1, x2, x3 we can create some clauses by using the or operator: (x1∨¯ x2), (¯ x1∨x2), (¯ x2∨¯ x3), (x1∨x3), (¯ x1∨¯ x2), (¯ x1∨x2) then we can join the clauses by using the and operator: (x1∨¯ x2)∧(¯ x1∨x2)∧(¯ x2∨¯ x3)∧(x1∨x3)∧(¯ x1∨¯ x2)∧(¯ x1∨x2)

Definition

The size of a CNF formula is the sum of the length of all its clauses.

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Assigment of values

Definition

An assigment of values to the set X of variables of a boolean formula is called a truth assignment. An assignment of truth values to the propositional variables: satisfies a literal x if x takes the value 1 satisfies a literal ¯ x if x takes the value 0 satisfies a clause if it satisfies at least one literal of the clause satisfies a CNF formula if it satisfies all the clauses of the formula. An assigmnent for a CNF formula Φ is complete if all the variables occurring in Φ have been assigned, otherwise it is partial.

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Outline

1

Basic concepts and definitions Some definition Problem statement Alternative formulations

2

Graphical representation Implication graph

3

Solving the problem Solution methods Solving MAX2SAT using AMPL

S. Cafieri (LIX)

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The 2-SAT problem

Definition

Given a boolean formula in Conjunctive Normal Form, consisting of a conjunction of m clauses Ci, each of which with 2 literals, the 2-Satisfiability Problem, also called 2-SAT, is the problem of deciding if there is a truth assignment to the literals such that each clause is satisfied (i.e. its value is 1). Depending on whether this is possible or not, we say that the formula is satisfiable or unsatisfiable. Each instance of the 2-SAT problem is a 2CNF formula.

To decide if there exists a truth assignment to the literals such that each clause is satisfied is equivalent to ask whether or not the boolean expression C1 × · · · × Cm can take value 1, where ’×’ is the boolean multiplication: 1 × 1 = 1, 1 × 0 = 0, 0 × 0 = 0.

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The MAX2SAT problem

Two versions: decision version

ptimization version.

A decision problem is a question, in some formal system, with a answer YES/NO. An optimization problem is concerned to finding the best answer to a particular prob- lem.

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The MAX2SAT problem

Definition

decision version Given a boolean formula in Conjunctive Normal Form, consisting of a conjunction of m clauses Ci, each of which with 2 literals, and given an integer k, the Maximum 2- Satisfiability problem, also called MAX2SAT, is the problem of deciding if there is a truth assignment to the literals such that satisfies at least k clauses.

Definition

ptimization version

Given a boolean formula in Conjunctive Normal Form, consisting of a conjunction

f m clauses Ci, each of which with 2 literals, the Maximum 2-Satisfiability problem,

also called MAX2SAT, is to find a truth assignment to the literals that maximizes the number of satisfied clauses.

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The MAX2SAT problem

Definition

decision version Given a boolean formula in Conjunctive Normal Form, consisting of a conjunction of m clauses Ci, each of which with 2 literals, and given an integer k, the Maximum 2- Satisfiability problem, also called MAX2SAT, is the problem of deciding if there is a truth assignment to the literals such that satisfies at least k clauses.

Definition

ptimization version

Given a boolean formula in Conjunctive Normal Form, consisting of a conjunction

f m clauses Ci, each of which with 2 literals, the Maximum 2-Satisfiability problem,

also called MAX2SAT, is to find a truth assignment to the literals that maximizes the number of satisfied clauses.

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The MAX2SAT problem

Note that: MAX2SAT – decision version – if k = m then MAX2SAT is the same as 2SAT. MAX2SAT – optimization version – asks for the maximum number of satisfied clauses. To maximize the number of satisfied clauses is equivalent to minimize the number

f unsatisfied clauses.
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The MAXSAT problem

The natural generalization of MAX2SAT is the MAXSAT problem.

Definition

The Maximum Satisfiability problem (MAXSAT) asks for the maximum number of clauses which can be satisfied by any assignment. The clauses have not a limit on the number of literals. There are several extensions to MAXSAT: The weighted maximum satisfiability problem (Weighted MAXSAT) asks for the maximum weight which can be satisfied by any assignment, given a set of weighted clauses. The partial maximum satisfiability problem (PMAXSAT) asks for the maximum number of clauses which can be satisfied by any assignment of a given subset of

clauses. The rest of the clauses must be satisfied.
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Applications

SAT and MAXSAT are central problems in Artificial Intelligence, Logic and Compu- tational Complexity. Many important real-world applications: scheduling electronic design automation computer architecture design pattern recognition inference in Bayesian networks

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Outline

1

Basic concepts and definitions Some definition Problem statement Alternative formulations

2

Graphical representation Implication graph

3

Solving the problem Solution methods Solving MAX2SAT using AMPL

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MAXSAT as an Integer Programming problem

A logical variable vj (TRUE or FALSE) is replaced by a corresponding integer variable xj, that takes values 1 or 0. So:

an unnegated literal is simply replaced by xj,
a negated literal is replaced by the expression 1 − xj.

A clause is satisfied if and only if at least one of its nl literals is TRUE. For the integer problem, we sum the corresponding nl expressions. The clause is satisfied if and only if the sum is 1 or more. Example: the clause ¯ v1 ∨ v2 ∨ v5 ∨ ¯ v7 is satisfied if: (1 − x1) + x2 + x5 + (1 − x7) ≥ 1. We must have some way of handling the maximization of satisfied clauses. We do this by adding variables to the problem. A clause is either satisfied or it isn’t:

riginalclause satisfied ∨ originalclause not satisfied.

We use a new variable for each clause: yi = 1 if clause ϕi is satisfied yi = 0 if clause ϕi is not satisfied.

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MAXSAT as an Integer Programming problem

A logical variable vj (TRUE or FALSE) is replaced by a corresponding integer variable xj, that takes values 1 or 0. So:

an unnegated literal is simply replaced by xj,
a negated literal is replaced by the expression 1 − xj.

A clause is satisfied if and only if at least one of its nl literals is TRUE. For the integer problem, we sum the corresponding nl expressions. The clause is satisfied if and only if the sum is 1 or more. Example: the clause ¯ v1 ∨ v2 ∨ v5 ∨ ¯ v7 is satisfied if: (1 − x1) + x2 + x5 + (1 − x7) ≥ 1. We must have some way of handling the maximization of satisfied clauses. We do this by adding variables to the problem. A clause is either satisfied or it isn’t:

riginalclause satisfied ∨ originalclause not satisfied.

We use a new variable for each clause: yi = 1 if clause ϕi is satisfied yi = 0 if clause ϕi is not satisfied.

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MAXSAT as an Integer Programming problem

A logical variable vj (TRUE or FALSE) is replaced by a corresponding integer variable xj, that takes values 1 or 0. So:

an unnegated literal is simply replaced by xj,
a negated literal is replaced by the expression 1 − xj.

A clause is satisfied if and only if at least one of its nl literals is TRUE. For the integer problem, we sum the corresponding nl expressions. The clause is satisfied if and only if the sum is 1 or more. Example: the clause ¯ v1 ∨ v2 ∨ v5 ∨ ¯ v7 is satisfied if: (1 − x1) + x2 + x5 + (1 − x7) ≥ 1. We must have some way of handling the maximization of satisfied clauses. We do this by adding variables to the problem. A clause is either satisfied or it isn’t:

riginalclause satisfied ∨ originalclause not satisfied.

We use a new variable for each clause: yi = 1 if clause ϕi is satisfied yi = 0 if clause ϕi is not satisfied.

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IP formulation

Let Φ be a CNF formula: Φ = ϕ1 ∧ · · · ∧ ϕm. For all i ≤ m, let Yi = {j ≤ n | xj ∈ ϕi} and ¯ Yi = {j ≤ n | ¯ xj ∈ ϕi} and x ∈ {0, 1}n, y ∈ {0, 1}m be binary decision variables. Integer Programming formulation for MAXSAT: max

i≤m

yi s.t. ∀i ≤ m

j∈Yi

xj +

j∈¯

Yi

(1 − xj) ≥ yi x ∈ {0, 1}n y ∈ {0, 1}m

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Another formulation

Another formulation for the pure satisfiability SAT problem is the following. min

j≤n

xj(1 − xj) s.t. ∀i ≤ m

j∈Yi

xj +

j∈¯

Yi

(1 − xj) ≥ 1 x ∈ [0, 1]n The SAT problem has a solution if and only if the optimal objective function value is 0, which implies that either xj = 0 or xj = 1 for all j ≤ n.

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Complexity

SAT was the first known NP-complete problem [Cook, 1971].

(NP standing for Nondeterministic Polynomial time – any given solution of the problem p can be verified in polynomial time and every other problem in NP can be reduced to p).

MAX2SAT is shown to be NP-complete by reduction from 3SAT. The more general MAXSAT is also NP-complete. 2SAT is polynomially solvable. MAXSAT problems are difficult problems: the time required to solve the problem increases very quickly as the size of the problem grows.

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Outline

1

Basic concepts and definitions Some definition Problem statement Alternative formulations

2

Graphical representation Implication graph

3

Solving the problem Solution methods Solving MAX2SAT using AMPL

S. Cafieri (LIX)

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Implication graph

Definition

An implication graph is a directed graph G(V, A) (V is the set of verteces and A the set

f directed arcs), where each vertex u ∈ V represents the truth status of a boolean lit-

eral, and each directed arc a(u, v) ∈ A, which connects vertex u and vertex v, represents the implication "if the literal u is true then the literal v is also true". Implication graphs can be used for analyzing complex boolean expressions. An instance of the MAX2SAT problem can be represented using an implication graph.

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Implication graph

A 2CNF formula (i.e. an instance of MAX2SAT or 2SAT) can be transformed into an implication graph by replacing each of its disjunctions by a pair of implications and then representing these implications graphically as directed arcs in a graph. Given two literals a and b, the disjunction a ∨ b is equivalent to: ¯ a ⇒ b ¯ b ⇒ a Note: for 1 clause in the 2CNF formula, we have 2 arcs in the graph.

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Implication graph

Example 1: (¯ x1 ∨ x2) ∧ (x1 ∨ ¯ x2) ∧ (¯ x2 ∨ x3) ∧ (¯ x1 ∨ ¯ x3)

clause 1 ¯

x1 ∨ x2 gives: x1 ⇒ x2 and ¯ x2 ⇒ ¯ x1

clause 2 x1 ∨ ¯

x2 gives: ¯ x1 ⇒ ¯ x2 and x2 ⇒ x1

clause 3 ¯

x2 ∨ x3 gives: x2 ⇒ x3 and ¯ x3 ⇒ ¯ x2

clause 4 ¯

x1 ∨ ¯ x3 gives: x1 ⇒ ¯ x3 and x3 ⇒ ¯ x1

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Implication graph

Example 2: (x1 ∨ ¯ x2) ∧ (¯ x1 ∨ x2) ∧ (¯ x2 ∨ ¯ x3) ∧ (x1 ∨ x3) ∧ (¯ x1 ∨ ¯ x2) ∧ (x1 ∨ x2) ∧ (x2 ∨ x3) ∧ (x2 ∨ ¯ x3)

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Implication graph

Example 3: a more complicated instance 2CNF formula with m = 200 clauses n = 60 variables

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Outline

1

Basic concepts and definitions Some definition Problem statement Alternative formulations

2

Graphical representation Implication graph

3

Solving the problem Solution methods Solving MAX2SAT using AMPL

S. Cafieri (LIX)

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How can we solve the problem?

A very simple algorithm: exhaustive search. Enumerate all possible 1 and 0 configurations, and verify each one. As the number of variables becomes large, enumerating all possible combinations becomes impossible. For n variables, this search can require 2n tests. Ex.: n = 5, there are 25 = 32 possible configurations: 00000, 00001, 00010, 00100, . . . , 11111.

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Solution methods

Two classes of methods are used to solve SAT and MAXSAT problems: exact methods and approximate methods. Exact methods

guarantee finding an optimal solution for an instance or, if there is no solution, proving its unsatisfiability perform exhaustive searches do not check every possible configuration (they usually use strategies to speed up the search) their performance can be limited for large instances are generally based on the exploration of a binary search tree, building incrementally truth assignments.

Approximate methods

do not guarantee finding an optimal solution are quickly in finding a solution (if they find it) are mainly based on local search and evolutionary algorithms.

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Solution methods

Exact methods variants of Davis-Putnam-Logemann-Loveland (DPLL) procedure

(designed for SAT - deterministic tree-search with backtracking) [Davis et al., 1962]

Branch and Bound (BB) algorithms [Borchers and Furman, 1999]. Approximate methods GSAT

(greedy local search – starts with a randomly generated assignment, then repeatedly changes (’flips’) the assignment of the variable that leads to the largest decrease in the total number

f unsatisfied clauses).

WalkSAT

(starts with randomly generated assignment, then repeatdly picks a clause which is unsat- isfied by the current assignment and flips a variable within that clause).

. . . From now on, we will focus on BB algorithms.

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Branch and Bound algorithm

Branch and Bound algorithms are usually applied for solving general Mixed Integer Linear Programming (MILP) problems. They work by recursively partitioning the search space and computing a lower and an upper bound to the optimal value of the objective function in each subregion. For a minimization problem, the lower bound is usually computed by constructing and solving a linear relaxation, and the upper bound can simply be a local optimum found by a local MILP solver.

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Branch and Bound algorithm

Let’s try to use a BB algorithm for solving a MAX2SAT problem. Basic concepts: The space of all possible assignments for a CNF formula can be represented as a search tree, where internal nodes represent partial assignments and leaf nodes represent complete assignments. A Branch and Bound algorithm for MAXSAT explores the search tree in a depth- first manner. Let’s examine these concepts in details.

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Branch and Bound algorithm

SAT problem

First consider the SAT problem. Basic algorithm: Select a variable and set its value to 0 or 1. Note: different 0/1 assignments create branches of the search tree. Replace the selected variable with the selected value in the formula. If the selected value satisfies all the clauses, then assign a value to the next vari- able, and so on. Else, if the selected value fails to satisfy one of the clauses, prune the current branch and backtrack the search. Keep performing this process until all clauses are satisfied.

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Search tree

Example of binary search tree

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Branch and Bound algorithm

SAT problem

Example (¯ x2 ∨ ¯ x5) ∧ (x1 ∨ ¯ x3) ∧ (x4 ∨ ¯ x5) ∧ (x1 ∨ x2) Start setting x1 = 0 and substitute this value into the formula. We get a 0 in two clauses, but we don’t know yet if the formula is satisfied or not. Go to the next variable in the tree, x2. Set x2 = 0. The 4-th clause is unsatisfied, because has both x1 and x2 as 0. Therefore, we don’t continue exploring that branch, since we cannot satisfy the formula. Try the other branch for x2: set x2 to 1. The 4-th clauses is now satisfied. Set x3 = 0. The 2-nd clauses is satisfied. Set x4 = 0. Set x5 = 0. Then all clauses are satisfied: we solved the SAT problem.

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Branch and Bound algorithm

SAT problem

Observation

Notice that we didn’t need to go through every branch of the tree to find the solution.

Observation

If an assigned variable fails to satisfy one of the clauses, then we prune the correspond- ing branch of the search tree. What does it happen when we consider MAXSAT?

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Branch and Bound algorithm

MAXSAT problem

Consider now the MAXSAT problem. We want to satisfy the maximum number of clauses, not all the clauses. So, if we find an unsatisfied clause when assigning a value to a variable, then we have not to stop at that point and go back to the other branch. If a certain 0/1 value for a variable fails to satisfy a clauses, it could be that the same value satisfies many other clauses: it is possible that continuing down on that branch we get maximum satisfiability. We have to explore more branches! When can we stop exploring a branch?

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BB algorithm for MAX2SAT

Consider the IP formulation of the MAX2SAT problem. Algorithm: At each node of the search tree solve the Linear Programming (LP) relaxation

btained by replacing the integrality requirements by the simple bounds:

∀j ∈ {1, . . ., n} 0 ≤ xj ≤ 1, ∀i ∈ {1, . . . , m} 0 ≤ yi ≤ 1. If the solution to the LP is infeasible, backtrack to a higher level in the search tree. If the solution to the LP is integral, compare it to the best integral solution found so far. Backtrack to a higher level in the search tree. Else (if the solution has one or more fractional variables) branch on one of the fractional variables and repeat the procedure.

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BB algorithm for MAX2SAT

Some observation: Branching allows to extend the current partial assignment by instantiating a new

variable. This leads to the creation of two new branches of the tree from the

current branch: the left branch corresponds to instantiate the new variable to 0, the right branch corresponds to instantiate the new variable to 1. Backtracking is used to go back to a higher level in the search tree and continue exploring the tree along others branches. When we find an assignment of values such that the number of satisfied clauses is greater than the number we previously found, we store the values we have found. The corresponding value of the objective function is the current bound and it represents the current best value of the objective function.

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BB algorithm for MAX2SAT

For the sake of simplicity, we only consider the basic BB algorithm for MAX2SAT, but usually suitable strategies are used to reduce the computational complexity.

Observation

BB algorithms for MAX2SAT usually implement methods to reduce the input formula: the formula is simplified according to transformation rules without changing the max- imum number of satisfiable clauses

[Gramm, PhD thesis, 1999].

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BB algorithm for MAX2SAT

Input a 2CNF instance while (termination test not satisfied) do solve the continuous relaxation if (the solution is infeasible) then backtracking else if (the solution is integer) then if (the current value of the obj. func. is greater than the one previously found) then store the current solution endif backtracking else branching endif endif endwhile The output is the maximum number of satisfied clauses.

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BB algorithm for MAX2SAT: problem relaxation

The linear continuous relaxation is obtained by relaxing the integrality constraints on the variables. Let Φ be a CNF formula: Φ = ϕ1 ∧ · · · ∧ ϕm. For all i ≤ m, let Yi = {j ≤ n | xj ∈ ϕi} and ¯ Yi = {j ≤ n | ¯ xj ∈ ϕi} and x ∈ Rn, y ∈ Rm be decision variables. max

i≤m

yi s.t. ∀i ≤ m

j∈Yi

xj +

j∈¯

Yi

(1 − xj) ≥ yi 0 ≤ x ≤ 1 0 ≤ y ≤ 1

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BB algorithm for MAX2SAT: branching rule

We branch if we find a solution containing still fractional values. A lot of branching rules are possible. What rule can we use? A very simple branching rule: branch on the first fractional variable xi that we find, for i ∈ {1, . . . , n}. Using this criterion, branching index is: i = min {1, . . . , n} | xi / ∈ {0, 1}.

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Outline

1

Basic concepts and definitions Some definition Problem statement Alternative formulations

2

Graphical representation Implication graph

3

Solving the problem Solution methods Solving MAX2SAT using AMPL

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Using AMPL to implement the algorithm

Can we use AMPL for solving the MAX2SAT problem by means of the Branch and Bound algorithm? Of course we can! We can write an AMPL model for the LP relaxation of MAX2SAT. We can solve at each node of the search tree the LP problem using the solver CPLEX. We can implement the BB algorithm, incorporating the above solution, using the AMPL language: scripts of AMPL commands can be set up to run as programs that perform repeti- tive actions. A script can solve a series of instances, by including a solve statement inside a loop [http://www.ampl.com/].

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MAX2SAT problem (relaxation): AMPL model

## parameters param m, integer; # number of clauses param n, integer; # number of variables (2n is the number of literals) set M := 1..m; set N := 1..n; # phi[i,j] = 1 if x_j appears in clause i # = -1 if bar{x}_j appears in clause i # = 0

therwise

param phi{M,N} integer, default 0; ## variables # x = 1 if literal x is true and 0 if false # y = 1 if clause is satisfied ## continuous relaxation var x{ i in N} >= 0, <=1; var y{ i in M} >= 0, <=1; ## objective maximize clauses: sum{i in M} y[i]; ## constraints subject to clause {i in M} : sum{j in N : phi[i,j] = 1} x[j] + sum{j in N : phi[i,j] = -1} (1-x[j]) >= y[i];

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Encoding an instance: an AMPL data file

param m := 4; # number of clauses param n := 3; # number of variables # phi[i,j] = 1 if x_j appears in clause i # = -1 if bar{x}_j appears in clause i # = 0

therwise

param phi := 1 1

1

1 2 1 2 1 1 2 2

1

3 2

1

3 3 1 4 1

1

4 3

1

;

This corresponds to the following instance: (¯ x1 ∨ x2) ∧ (x1 ∨ ¯ x2) ∧ (¯ x2 ∨ x3) ∧ (¯ x1 ∨ ¯ x3)

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Solving MAX2SAT: an AMPL .run file

# max2satBB.run

ption solver_msg 0;

# model model max2sat.mod; # data file data max2sat.dat; # branch and bound include BB.run; display xstar; printf "max clauses = %d\n", fstar;

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Solving MAX2SAT: an AMPL .run file

Observations

Our .run file use an "include" command. The AMPL command

include filename

cause input to be taken from a file named filename: the command is replaced by the content of filename. We include BB.run, containing the implementation of the BB algorithm. All the ampl commands contained in this file will be executed.

BB.run is an AMPL script.

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AMPL scripts

AMPL scripts can contain any AMPL command and may include programming lan- guage constructs like for, repeat and if to repeat statements and perform them conditionally

[Fourer, Gay, Kernighan, AMPL, 2nd edition].

Iterating over a set: the for statement. Iterating subject to a condition: the repeat statement. Testing a condition: the if-then-else statement.

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AMPL scripts

for: executes commands for each member of a set. Example:

for {i in 1..4} { let v[i] := i; }

repeat: continues iterating as long as a logical condition is satisfied. Example:

repeat while (i <= 4) { let v[i] := i; }

if: evaluates a condition and executes some specified command if the condition is true. Example:

if (i <> 4) then { let v[i] := i; }

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SLIDE 55

BB for MAX2SAT: AMPL script framework

# param definitions ... repeat while (termination = 0) { # retrieve x’ from list # fix x’ (to 0 or 1) # solve LP relaxation

ption solver cplex;

solve > /dev/null; if (solve_result = "infeasible") then { # no branching } else { # check if all xi are binary (integ = 1) if (integ = 1) then { if (clauses > fstar) then { let fstar := clauses; let {i in N} xstar[i] := x[i]; } # no branching } else { # branching # choice of the branching variable # update list } } # check for termination }

S. Cafieri (LIX)

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SLIDE 56

BB for MAX2SAT: use of AMPL iteration and conditional statements

Main loop:

repeat while (termination = 0) { ..... }

The parameter termination is initially set to 0. The loop causes performing itera- tively the commands specified therein until termination is changed to 1: this hap- pens when all possible configurations have been tested.

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SLIDE 57

BB for MAX2SAT: use of AMPL iteration and conditional statements

Another use of the repeat statement: choice of the branching variable. Iterations are performed until the flag parameter is changed to 1, that is when the first non-integer variable is encountered, or all the variables have been examined (so the counter ii is qual to n.

# choice of branching variable let ii := 1; let flag := 0; repeat while (ii <= n and flag = 0) { if (x[ii] <> 0 and x[ii] <> 1) then { let l := ii; let flag := 1; } let ii := ii+1; }

Note also the use of the if statement.

S. Cafieri (LIX)

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SLIDE 58

BB for MAX2SAT: use of AMPL iteration and conditional statements

Use of the for statement: in order to check integrality of the variables, we use a loop

n N and test the values of all variables.

Use of the if ...then statement: at each iteration i, the if statement evaluates the condition xi = 0 and xi = 1 and, if it is satisfied (i.e. variable xi is not binary), then changes the value of the parameter integ to 0.

# check if all xi are binary let integ := 1; for {i in N} { if (x[i] <> 0 and x[i] <> 1) then { let integ := 0; } }

S. Cafieri (LIX)

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SLIDE 59

BB for MAX2SAT: use of AMPL iteration and conditional statements

Nested if statements: if all variables are integer, then check if the computed value

f the objective function (clauses) is greater than the current best value and, if so,

update the current best value and the current solution.

if (integ = 1) then { if (clauses > fstar) then { let fstar := clauses; let {i in N} xstar[i] := x[i]; } let branch := 0; } else { .....

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SLIDE 60

Summary

We have learned: what satisfiability and maximum satisfiability problems are; how MAX2SAT can be solved by a Branch and Bound algorithm; how MAX2SAT can be solved using AMPL.

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SLIDE 61

Bibliography

J. Gramm.

Exact algorithms for Max2Sat and their applications. PhD thesis, 1999.

M. Davis, G. Logemann, D. Loveland.

A machine program for theorem-proving. Communications of the ACM, 5:394–397, 1962.

T. Alsiniet, F. Manyá, J. Planes.

Improved Branch and Bound algorithms for Max-Sat. Proceedings of the 6th International Conference on the Theory and Applications of Satisfiability Testing, 2003.

S. Joy, J.E. Mitchell, B. Borchers.

Solving MAX-SAT and Weighted MAX-SAT Problems Using Branch-and-Cut. http://www.rpi.edu/ mitchj/papers.html, 1998.

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