10.3 - Restricted Satisfiability (SAT)

Problem

2/19/07

Two Variations of SAT

• 3SAT a set of problems similar to SAT but the

Boolean expressions have a regular form: the logical AND of “clauses”, where each clause is the OR of three variables (could be negated variables)

• CSAT an intermediate step that is used to

show that SAT can be reduced to 3SAT

Purpose of Section 10.3

• To reduce SAT to 3SAT This can be accomplished by converting

each Boolean expression E in SAT to a new Boolean expression F in 3SAT

Hence, F is satisfiable iff E is satisfiable

Important Concepts

• Literal - a variable (could be negated)• Clause - logical OR of one or more

literals• CNF - Conjunctive Normal Form,

logical AND of clauses• Equivalent Boolean Expressions - two

Boolean expressions have the same result on any truth assignment to their variables

Notation for CNF

• OR () = Sum (+)• AND () = Product (* or juxtaposition)• Not = ¬

• Juxtaposition: no operator

Example

• Expression: (x ¬y) (¬x z)• CNF Notation: (x + ¬y)(¬x + z)

k-CNF

• k-Conjunctive Normal Form: an expression that contains a product of clauses. Each clause is the sum of exactly k distinct literals

Example: 2-CNF

• (x + ¬y)(y + ¬z)(z + ¬x)• Why is it 2-CNF?

Each of the clauses has exactly two literals

CSAT and kSAT

• CSAT is a Boolean expression in CNF that is satisfiable

• kSAT is a Boolean expression in k-CNF that is satisfiable

CSAT as a Decision Problem

• CSAT is the problem: Given a Boolean expression in CNF, is it

satisfiable?

• Note: we can also write kSAT as a decision problem: Given a Boolean expression in k-CNF, is

it satisfiable?

NP-Complete?

• CSAT, 3SAT, and kSAT for all k higher than 3 are NP-Complete

• 1SAT and 2SAT have linear-time algorithms

Reduction of SAT to CSAT

• Done in two parts: Part 1: “Push” the negations through the

Boolean expression so that variables are the only thing with negations With good data structure design takes linear

time Part 2: Write the Boolean expressions in

CNF form Done by introducing new variables Takes polynomial time

Helpful Rules for Part 1:

1. ¬(E F) => ¬(E) ¬(F)• One of DeMorgan’s Laws

2. ¬(E F) => ¬(E) ¬(F)• Another one of DeMorgan’s Laws

3. ¬(¬E)) => E• Law of Double Negation

Example

¬((¬(x + y))(¬x + y)) Start

¬(¬(x + y))+¬(¬x + y)

Rule 1

x + y + ¬(¬x + y) Rule 3

x + y + (¬(¬x)) ¬y Rule 2

x + y + x ¬y Rule 3

Theorem 10.12

• Every Boolean expression E is equivalent to an expression F in which the only negations occur in literals.

• Moreover, the length of F is linear in the number of symbols of E, and F can be constructed from E in polynomial time

Proof for Theorem 10.12

• Done using Induction• Accomplished by showing that for

expression E there is an equivalent expression F with ¬’s only in literals

• Also, must show that if E has n>=1 operators, then F has no more than 2n-1 operators

Length of F

• F does not have more than 1 pair of parentheses per operator

• The number of variables in an expression cannot exceed the number of operators by more than 1

• The length of F is linearly proportional to the length of E

• Also, the construction of F is simple and the time to construct F is proportional to the length

Basis:

• E has one operator Possible forms (for variables x, y)

¬x x y x y

E is already in the correct form! Therefore, E = F

Also note: F has at most twice the number of operators as E minus 1 holds

Induction:

• Assume true expressions w/fewer terms than E

• E must be in form: E1 E2

E1 E2

• There are equivalent F1 and F2

• E1 has a operators, E2 has b operators E has a + b + 1 operators

Induction (continued)

• F is formed using F1 and F2 F = F1 F2

F = F1 F2

• F1 has at most 2a-1 operators, F2 has 2b-1 operators F has at most 2a + 2b -1 operators

• F has no more than 2(a+b+1)-1 operators i.e., F has no more than twice the number of

operators of E minus 1

Induction (continued)

• E is of the form ¬E1

• There are three cases: E1 = ¬E2

E1 = E2 E3

E1 = E2 E3

Case 1: E1 = ¬E2

• By the law of double negation, E = E2.

• E2 has fewer operators than E There is an F for E2 that have ¬’s only in

literals

• The number of operators of F is at most twice the number in E2 minus 1 (and the same is true for E)

Case 2: E1 = E2 E3

• By DeMorgan’s Law: E= ¬(E2 E3) = ¬E2 ¬E3

• ¬E2 and ¬E1 have fewer operators than E There is F2 and F3 that have ¬’s only in literal F = F2 F3 = E

• ¬E2 has a+1 operators, ¬E3 has b+1 operators• F2 has at most 2(a+1)-1 operators, F3 has at

most 2(b+1)-1 operators• F has at most 2a+2b+3 operators,

(i.e., twice the number of operators of E minus 1)

Case 3: E1 = E2 E3

• Done using same argument as E1 = E2 E3

Theorem 10.13

• CSAT is NP-Complete• Proof:

Reduce SAT to CSAT in polynomial time Convert an instance of SAT into expression E

using Theorem 10.12 Convert E to CNF expression F in polynomial

time Show that F is satisfiable iff E is satisfiable

Basis:

• If E consists of 1 or 2 symbols, then it is a literal.

• A literal is a clause, therefore E is already in CNF

Induction:

• Assume: Every expression shorter than E can be

converted to a product of clauses Conversion takes at most cn2 time on

expressions of length n

• There are two cases: E = E1 E2

E = E1 E2

Case 1: E = E1 E2

• By the inductive hypothesis: All and only the satisfying assignments of

E1 can be extended to a satisfying assignment for F1

E2 can be extended to a satisfying assignment for F2

Assume: F1 and F2 are distinct variables Let F = F1 F2 where:

F1 F2 is a CNF expression if F1 and F2 are CNF expressions

• Need to show: Truth assignment T for E can be extended to a satisfying assignment for F iff T satisfies E

Case 1: If Part

• Suppose: T satisfies E Let: T1 be T restricted to apply to only

variables in E1

Let: T2 be T restricted to apply to only variables in E2

T1 and T2 can be extended to assignments S1 and S2 that satisfy F1 and F2

Let S agree with S1 and S2

S is an extension of T that satisfies F!

Case 1: Only-If Part

• Suppose: T has extension S that satisfies F Let: T1 be T restricted to apply to only variables in E1

Let: T2 be T restricted to apply to only variables in E2

Let: S1 be S restricted to the variables of F1

Let: S2 be S restricted to the variables of F2

Then: S1 is an extension of T1 and S2 is an extension of T2

F = F1 F2, then S1 satisfies F1 and S2 satisfies F2

T1 must satisfy E1 and T2 must satisfy E2 => T satisfies E!

Case 2: E = E1 E2

• By the inductive hypothesis: CNF expressions F1 and F2 have the properties: A truth assignment for E1 satisfies E1 iff it can

extended to a satisfying assignment for F1

A truth assignment for E2 satisfies E2 iff it can extended to a satisfying assignment for F2

F1 and F2 are disjoint except for those variables that appear in E

F1 and F2 are in CNF

• Note: A simple OR of F1 and F2 to construct F would not result in an expression in CNF. A more complicated construction will be needed

Case 2: Continued

• Suppose: (where gi and hi are clauses) F1 = g1 g2 … gp

F2 = h1 h2 … hp

• Let: (where y is a new variable) F = (y+g1) * (y+g2) *… * (y+gp) *

(¬y+h1) * (¬y+h2) *… * (¬y+hq)

• Show: A truth assignment T for E satisfies E iff T can be extended to a truth assignment S that satisfies F

Case 2: Only-If Part

• Assume: T satisfies E Let: T1 be T restricted to apply to only

variables in E1

Let: T2 be T restricted to apply to only variables in E2

E = E1 E2 then either: T satisfies E1 or T satisfies E2

W.L.O.G Assume: T satisfies E1

T1 can be extended to S1 which satisfies F1

Case 2: Only-If Part Cont.

• Construct S for T Rules for S will satisfy F

1) variables x in F, S(x) = S1(x)

2) S(y) = 0 (i.e., all clauses of F derived from F2 true)

3) variables x in F2 but not in F1, S(x) is T(x) if T(x) is defined (otherwise either 0 or 1)

• S makes all clauses derived from the g’s true by Rule 1

• S makes all clauses derived from the h’s true by Rule 2

• S satisfies F!

Case 2: If Part

• Suppose: Truth assignment T for E is extended to truth assignment S for F and S satisfies F

• T satisfies E whenever S satisfies F• Show: Time to construct F from E is

at most quadratic in n (where n is the length of E)

Case 2: If Part Cont.

• Takes linear (in size of E) time to: Splitting E into E1 and E2

Constructing F from F1 and F2

• Let: dn = upper bound on time to construct

E1 and E2 plus construction of F n = the length of E

Case 2: If Part Cont.

• T(n): (a recurrence equation) time to construct F from any E T(1) = T(2) ≤ e (for some constant e) T(n) ≤ dn + cmax0<i<n-1(T(i)+T(n-1-i))

For n≥3 where c is a constant such that it can be shown that T(n) ≤ cn2

• T(1) & T(2): If E has 1 or 2 symbols, no recursion and process takes time e

• Otherwise, conversion from E to F is: E to E1 and E2, F to F1 and F2 => takes dn 2 recursive conversions: E1 to F1 and E2 to F2

Case 2: If Part Cont.

• There is a constant c such that for all n, T(n) ≤ cn2

• Thus, the construction of F from E takes O(n2)!

Theorem 10.15

• 3SAT is NP-Complete• Proof:

3SAT is NP since SAT is NP Reduce CSAT to 3SAT

Proof:

• Given a CNF expression E = e1 … ek, replace each clause ei to create a new expression F• ei = (x), introduce new variables u and v

• Replace with (x+u+v)(x+u+ ¬v)(x+¬u+v)(x+¬u+¬v)

• ei = (x+y), introduce new variable z• Replace with (x+y+z)(x+y+¬z)

• ei = sum of 3 literals, no replacement needed• ei = (x1+…+xm) for some m>=4, introduce new

variables y1, …, ym-3

• Replace with (x1+x2+y1)(x3+¬y1+y2)(x4+¬y2+y3)…(xm-

2+¬ym-4+ym-3)(xm-1+xm+ym-3)

Proof Cont.

• Thus, each instance of E of CSAT can be reduced to an instance of F of 3SAT such that F is satisfiable iff E is satisfiable

• Construction time is linear in length of E

• Since CSAT is NP-Complete, 3SAT is NP-complete!