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

2. Two Variations of SAT 3SAT CSAT
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 ¬((¬(x + y))(¬x + y)) Start ¬(¬(x + y))+¬(¬x + y) Rule 1

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)
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: E must be in form: There are equivalent F1 and F2
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
¬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)

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: There are two cases:
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 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

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: Let:
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!