10.3 - Restricted Satisfiability (SAT) Problem

1. 10.3 - Restricted Satisfiability (SAT) Problem 2/19/07

2. 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

3. 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

4. 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

5. Notation for CNF • OR () = Sum (+) • AND () = Product (* or juxtaposition) • Not = ¬ • Juxtaposition: no operator

6. Example • Expression: (x  ¬y)  (¬x  z) • CNF Notation: (x + ¬y)(¬x + z)

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

8. Example: 2-CNF • (x + ¬y)(y + ¬z)(z + ¬x) • Why is it 2-CNF? • Each of the clauses has exactly two literals

9. CSAT and kSAT • CSAT is a Boolean expression in CNF that is satisfiable • kSAT is a Boolean expression in k-CNF that is satisfiable

10. 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?

11. NP-Complete? • CSAT, 3SAT, and kSAT for all k higher than 3 are NP-Complete • 1SAT and 2SAT have linear-time algorithms

12. 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

13. Helpful Rules for Part 1: • ¬(E  F) => ¬(E)  ¬(F) • One of DeMorgan’s Laws • ¬(E  F) => ¬(E)  ¬(F) • Another one of DeMorgan’s Laws • ¬(¬E)) => E • Law of Double Negation

14. Example

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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

21. Induction (continued) • E is of the form ¬E1 • There are three cases: • E1 = ¬E2 • E1 = E2 E3 • E1 = E2 E3

22. 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)

23. 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 • ¬E2has 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)

24. Case 3: E1 = E2 E3 • Done using same argument as E1 = E2 E3

25. 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

26. 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

27. 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

28. 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 F1and 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

29. 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!

30. 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!

31. 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

32. 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

33. 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

34. Case 2: Only-If Part Cont. • Construct S for T • Rules for S will satisfy F •  variables x in F, S(x) = S1(x) • S(y) = 0 (i.e., all clauses of F derived from F2 true) •  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!

35. 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)

36. 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

37. 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

38. 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)!

39. Theorem 10.15 • 3SAT is NP-Complete • Proof: • 3SAT is NP since SAT is NP • Reduce CSAT to 3SAT

40. 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)

41. 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!