1. SATISFIABILITY
Eric L. Frederich

2. Overview Definitions K-Satisfiability 3-Satisfiability
Three Colorable Problem
Transformation
Davis – Putnam Algorithm
2-Satisfiability
Algorithm
Summary
References

3. Definitions Satisfiability Product of Sums n m K
An instance of the problem is defined by a Boolean expression written using only AND, OR, NOT, variables, and parentheses. The question is: given the expression, is there some assignment of TRUE and FALSE values to the variables that will make the entire expression true?*
Product of Sums
A way of arranging a boolean expression such that the final output is a result of ‘AND’ing some ‘OR’ clauses where each term within an ‘OR’ clause may or may not be ‘NOT’ed. n
The number of boolean variables found in the expression (with or without a NOT) m
The number of ‘OR’ clauses ‘AND’ed together in the boolean expression K
The maximum number of variables found within each OR clause
* Definition from

4. K-Satisfiability
K is the maximum number of terms allowed per ‘OR’ clause.
For any K > 2, the problem lies in NP-Complete
Solution of order 2n
Special K’s
3SAT
3 Colorable Graph problem
Still NP-Complete
2SAT
Polynomial time solution
K = 5
(A + B’ + C)(B + C’ + E’)(A’ + B + C’+ D’ + E)(B + D)
K = 3 (3SAT)
(X’ + Y)(X + Y’ + Z)(X’ + Z’)(Y + Z’)
K = 2 (2SAT)
(X1 + X2)(X2’ + X3)(X1’ + X2’)(X3 + X4)(X3’ + X5)(X4’ + X5’)(X3’ + X4)

5. 3 – Colorable Transformation
Procedure Transform(G) for each vertex vi in G output(ri + yi + bi) for each edge (vi,vj) in G output(ri’ + rj ’)(yi’ + yj’)(bi’ + bj’) output : boolean expression with K = 3

6. 2(21) = 2,097,152 (R2 + Y2 + B2) (R4’ + R5’)(Y4’ + Y5’)(B4’ + B5’)
( Y1 )( B2)(R )( B4)(R )
( Y6 )( Y7 )
(R1' + R2’)( + Y2’)(B1’ + )(R1’ + )( + Y3’)(B1’ + B3’)
(R1' + R4’)( + Y4’)(B1’ + )(R1’ + )( + Y5’)(B1’ + B5’)
(R2' + )(Y2’ + Y5’)( + B5’)(R2’ + R6’)(Y2’ + )( + B6’)
( + R4’)(Y3’ + Y4’)(B3’ + )(R4’ + )(Y4’ + Y5’)( + B5’)
( + R6’)(Y5’ + )(B5’ + B6’)(R7’ + R4’)( + Y4’)(B7’ + )
(R7' + )( + Y5’)(B7’ + B5’)
(R1 + Y1 + B1)
(R6 + Y6 + B6)
(R2’ + R6’)(Y2’ + Y6’)(B2’ + B6’)
(R5’ + R6’)(Y5’ + Y6’)(B5’ + B6’)
(R5 + Y5 + B5)
(R2’ + R5’)(Y2’ + Y5’)(B2’ + B5’)
(R3 + Y3 + B3)
(R7 + Y7 + B7)
(R1’ + R2’)(Y1’ + Y2’)(B1’ + B2’)
(R1’ + R4’)(Y1’ + Y4’)(B1’ + B4’)
(R4 + Y4 + B4)
(R1’ + R3’)(Y1’ + Y3’)(B1’ + B3’)
(R1’ + R5’)(Y1’ + Y5’)(B1’ + B5’) 1 2 6 3 4 5 7
2(21) = 2,097,152
(R1 + Y1 + B1)(R2 + Y2 + B2)(R3 + Y3 + B3)(R4 + Y4 + B4)(R5 + Y5 + B5)
(R6 + Y6 + B6)(R7 + Y7 + B7)
(R1' + R2’)(Y1’ + Y2’)(B1’ + B2’)(R1’ + R3’)(Y1’ + Y3’)(B1’ + B3’)
(R1' + R4’)(Y1’ + Y4’)(B1’ + B4’)(R1’ + R5’)(Y1’ + Y5’)(B1’ + B5’)
(R2' + R5’)(Y2’ + Y5’)(B2’ + B5’)(R2’ + R6’)(Y2’ + Y6’)(B2’ + B6’)
(R3’ + R4’)(Y3’ + Y4’)(B3’ + B4’)(R4’ + R5’)(Y4’ + Y5’)(B4’ + B5’)
(R5’ + R6’)(Y5’ + Y6’)(B5’ + B6’)(R7’ + R4’)(Y7’ + Y4’)(B7’ + B4’)
(R7' + R5’)(Y7’ + Y5’)(B7’ + B5’)

7. R1 R2 R3 R4 R5 R6 R7 Y1 Y2 Y3 Y4 Y5 Y6 Y7 B1 B2 B3 B4 B5 B6 B7
3-Colorable
R1' R2' R3' R4' R5' R6' R7' Y1' Y2' Y3' Y4' Y5' Y6' Y7' B1' B2' B3' B4' B5' B6' B7'
(R1 + Y1 + B1)(R2 + Y2 + B2)(R3 + Y3 + B3)(R4 + Y4 + B4)(R5 + Y5 + B5)
(R6 + Y6 + B6)(R7 + Y7 + B7)
(R1' + R2’)(Y1’ + Y2’)(B1’ + B2’)(R1’ + R3’)(Y1’ + Y3’)(B1’ + B3’)
(R1' + R4’)(Y1’ + Y4’)(B1’ + B4’)(R1’ + R5’)(Y1’ + Y5’)(B1’ + B5’)
(R2' + R5’)(Y2’ + Y5’)(B2’ + B5’)(R2’ + R6’)(Y2’ + Y6’)(B2’ + B6’)
(R3’ + R4’)(Y3’ + Y4’)(B3’ + B4’)(R4’ + R5’)(Y4’ + Y5’)(B4’ + B5’)
(R5’ + R6’)(Y5’ + Y6’)(B5’ + B6’)(R7’ + R4’)(Y7’ + Y4’)(B7’ + B4’)
(R7' + R5’)(Y7’ + Y5’)(B7’ + B5’)

8. Davis - Putnam Recursive but still 2n
Prunes off branches which will not yield a result.
Does not examine the entire expression at each guess
Procedure split (E) 1. if E has an empty clause, then return 2. if E has no clauses, then exit with current partial assignment 3. select next unassigned variable, xi, in E 4. split ( E ( xi = 0 ) ) 5. split ( E ( xi = 1 ) )

9. (X1 + X3’ + X4)(X1’ + X2’ + X4’)(X2’ + X3)(X1’ + X2 + X4)
Davis - Putnam
(X1 + X3’ + X4)(X1’ + X2’ + X4’)(X2’ + X3)(X1’ + X2 + X4)
X1 = 0
X1 = 1
(X2’ + X4’)(X2’ + X3)(X2 + X4)
(X3’ + X4)(X2’ + X3)
X2 = 0
X2 = 0
X2 = 1
X2 = 1
(X3’ + X4)
(X3’ + X4)(X3)
(X4)
(X4’)(X3)
X3 = 0
X3 = 1
X3 = 0
X3 = 1
X3 = 0
X3 = 1
X3 = 1
X3 = 0
(X4’)
(X4)
(X4)
(X4)
(X4)
X4 = 0
X4 = 0
X4 = 0
X4 = 0
X4 = 0
X4 = 0
X4 = 0
X4 = 1
X4 = 0
X4 = 0

10. (X1 + X2)(X2’ + X3)(X1’ + X2’)(X3 + X4)(X3’ + X5)(X4’ + X5’)(X3’ + X4)
2- Satisfiability
(X1 + X2)(X2’ + X3)(X1’ + X2’)(X3 + X4)(X3’ + X5)(X4’ + X5’)(X3’ + X4)
Pick an assignment Drop terms that become true Make assignments based on initial assignment. Repeat….

11. Summary
For K-Satisfiability where K > 2 it is an NP-Complete Problem
2-Satisfiability is solvable in polynomial time
Davis-Putnam improves efficiency but does not decrease order of complexity
3-Colorable problem is synonymous to 3-Satisfiability

12. References http://condor.stcloudstate.edu/~pkjha/CSCI504/2SAT.pdf
"The New Turing Omnibus", by A.K. Dewdney A.W.H. Freeman/Owl Book, 2001 New York