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# SAT (Boolean Formula Satisfiability Problem) Computer Science Engineering (CSE) Notes | EduRev

## Computer Science Engineering (CSE) : SAT (Boolean Formula Satisfiability Problem) Computer Science Engineering (CSE) Notes | EduRev

The document SAT (Boolean Formula Satisfiability Problem) Computer Science Engineering (CSE) Notes | EduRev is a part of the Computer Science Engineering (CSE) Course Theory of Computation.
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Part IV. SAT (Boolean Formula Satisfiability Problem)

7 SAT

Satisfiability
An example for a boolean formula is
(¬x1 ∪ x2) ∩ (x1 ∪ ¬x2) ∩ (¬x2 ∪ x3) ∩ (¬x∪ ¬x2 ∪ x3).
This formula is in conjunctive normal form (CNF) (intersection of unions).
This boolean formula is satisfiable if there is some assignment of the values 0 and 1 to its variables that causes it to
evaluate to 1 (true).
If we put, x= 0, x2 = 0, x3 = 1, we can see that the above formula evaluates to 1 (true).
This means above formula is satisfiable.

Consider the boolean formula,
(x1 ∪ x2) ∩ (¬x1 ∪ ¬x2)
If we put x1 = 0, x2 = 0, this formula evaluates to 0. If we put x1 = 0, x2 = 1, this formula evaluates to 0. If we put
x1 = 1, x= 0, this formula evaluates to 0. If we put x1 = 1, x2 = 1, this formula evaluates to 0. This means for any values assigned to the variables the above boolean formula always evaluate to 0 (false).
This means the formula is not satisfiable.

SAT Problem
SAT (Boolean Formula Satisfiability) problem is defined as:
Is there a set of values for the boolean variables that assigns the value 1 (true) to the boolean expression given in CNF?

8 Cook’s Theorem
Cook’s theorem states that
satisfiability of boolean formulas (SAT problem) is NP-Complete.

SAT Problem is NP - Complete
We know that a problem is in class NP-Complete if: it is in NP and it is NP Hard.

Theorem:
Satisfiability of boolean formulas is NP-Complete.

Proof:
Step 1: Prove that SAT problem is in class NP.
A boolean variable can have two possible values. Hence, for an n variable boolean expression in CNF, there are 2n possible combinations of values of these variables.
If we want to check whether a boolean formula is satisfiable, we need to apply each combination to the formula until we get a 1 as the result of evaluation. In the worst case, we need to apply all the 2possible combinations.
Consider the formula, (¬x1 ∪ x2) ∩ (x1 ∪ ¬x2) ∩ (¬x2 ∪ x3) ∩ (¬x1 ∪ ¬x2 ∪ x3). There are 3 variables. Each variable can have two possible values, 0 or 1. There are 23 combinations. So the time complexity of this problem is Θ(2n). (not polynomial time) If we are given a certificate containing a satisfiable assignment for a formula, then we can easily verify it in polynomial time.
For example, for the above formula, if we are given a certificate saying that if x1 = 0, x2 = 0, x= 1, then just applying these values to the formula, we get,
(¬x1 ∪ x2) ∩ (x1 ∪ ¬x2) ∩ (¬x2 ∪ x3) ∩ (¬x1 ∪ ¬x2 ∪ x3)
=(¬0 ∪ 0) ∩ (0 ∪ ¬0) ∩ (¬0 ∪ 1) ∩ (¬0 ∪ ¬0 ∪ 1)
=(1 ∪ 0) ∩ (0 ∪ 1) ∩ (1 ∪ 1) ∩ (1 ∪ 1 ∪ 1)
=1 ∩ 1 ∩ 1 ∩ 1=1
This can be done in linear time.
So the SAT problem is in class NP.

Step 2: Prove that SAT problem is NP-hard.
We know that a problem is NP-Hard, if every problem in NP can be reduced to this problem in polynomial time. A lot of details are involved in this proof. [This proof is beyond the scope of this class].
[Refer the text book ’Introduction to the Theory of computation’ by Michael Sipser - Chapter 7- Theorem 7.37]

3-CNF Satisfiability Problem (3-CNF-SAT)

3-CNF
We know that a boolean formula is in conjunctive normal form(CNF) if it is expressed as an AND of clauses, each clause is the OR of one or more variables.
Then, a formula is in 3-conjunctive normal form (3-CNF) is each clause has exactly three distinct literals.
For example, the following boolean formula is in 3-CNF.
(x1 ∪ ¬x1 ∪ ¬x2) ∩ (x3 ∪ x2 ∪ x4) ∩ (¬x1 ∪ ¬x3 ∪ ¬x4) ∩ (x2 ∪ x1 ∪ ¬x3)
The clauses in the above formula contain exactly 3 literals. The formula is in 3-CNF.

3-CNF Satisfiability Problem
3-CNF-SAT problem is defined as:
Is there a set of values for the boolean variables that assigns the value 1 (true) to the boolean expression given in 3-CNF?
3-CNF Satisfiability Problem is NP-complete
We know that a problem is in class NP-Complete if: it is in NP and it is NP Hard.

Theorem:
Satisfiability of boolean formulas in 3-CNF is NP-complete.

Proof:
Step 1: Prove that 3-CNF-SAT is in class NP.
A boolean variable can have two possible values. Hence, for an n variable boolean expression in CNF, there are 2n possible combinations of values of these variables.
If we want to check whether a boolean formula is satisfiable, we need to apply each combination to the formula until we
get a 1 as the result of evaluation. In the worst case, we need to apply all the 2n possible combinations.
Consider the 3-CNF formula, (x1 ∪ ¬x1 ∪ ¬x2) ∩ (x3 ∪ x2 ∪ x4) ∩ (¬x1 ∪ ¬x3 ∪ ¬x4) ∩ (x2 ∪ x1 ∪ ¬x3). There are 4 variables. Each variable can have two possible values, 0 or 1. There are 24 combinations.So the time complexity of this problem is Θ(2n). (not polynomial time) If we are given a certificate containing a satisfiable assignment for a formula, then we can easily verify it in polynomial time.
For example, for the above formula, if we are given a certificate saying that if x1 = 0, x2 = 1, x3 = 0, x4 = 1, then
just applying these values to the formula, we get,

(x1 ∪ ¬x1 ∪ ¬x2) ∩ (x3 ∪ x2 ∪ x4) ∩ (¬x1 ∪ ¬x3 ∪ ¬x4) ∩ (x2 ∪ x1 ∪ ¬x3)
=(0 ∪ ¬0 ∪ ¬1) ∩ (0 ∪ 1 ∪ 1) ∩ (¬0 ∪ ¬0 ∪ ¬1) ∩ (1 ∪ 0 ∪ ¬0)
=(0 ∪ 1 ∪ 0) ∩ (0 ∪ 1 ∪ 1) ∩ (1 ∪ 1 ∪ 0) ∩ (1 ∪ 0 ∪ 1)
=1 ∩ 1 ∩ 1 ∩ 1 ∩ 1=1
This can be done in linear time.
So the 3-CNF-SAT problem is in class NP.

Step 2: Prove that 3-CNF-SAT problem is NP-hard.
We know that a problem is NP-Hard, if every problem in NP can be reduced to this problem in polynomial time.
The approach we used is as follows:
In the previous section, we already proved that all NP problems can be reduced to SAT. Then, if we can reduce SAT to 3-CNF-SAT in polynomial time, we can say that 3-CNF-SAT is NP-hard. We learned in the subject, ’logic design’ that any boolean formula can be expressed in CNF. Then, a boolean formula
in CNF can be rewritten in 3-CNF.
Consider a boolean formula as follows:
a ∪ b ∪ c ∪ d
This can be converted to 3-CNF as,
(a ∪ b ∪ x) ∩ (c ∪ d ∪ x)
Thus any boolean formula can be transformed to a 3-CNF boolean formula. This can be done using De Morgan’s laws.
That means SAT problem can be reduced to 3-CNF-SAT problem. That is, 3-CNF-SAT is NP-hard.

In step 1, we proved that 3-CNF-SAT is in NP.
In step 2, we proved that 3-CNF-SAT is NP-hard.
So, 3-CNF-SAT is an NP-complete problem.

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