1. Digitaalsüsteemide verifitseerimise kursus1 Formal verification: SAT SAT applied in equivalence checking

2. Digitaalsüsteemide verifitseerimise kursus2 Equivalence Checking Two principal approaches: Transform implementation and spec ( reference implementation) to a canonical form Search for an input assignment that would distinguish the responses of the implementation and reference implementation. SAT applied for the latter

3. Digitaalsüsteemide verifitseerimise kursus3 Satisfiability aka SAT SAT: Boolean function is satisfiable if there exists a variable assignment for which the function is TRUE

4. Digitaalsüsteemide verifitseerimise kursus4 Equivalence checking with SAT Equivalence Checking can be reduced to SAT: d = f  g Iff d is satisfiable then f and g are NOT equivalent.

5. Digitaalsüsteemide verifitseerimise kursus5 Miter circuit + 1 + Specification (reference implementation) Implementation SAT?

6. Digitaalsüsteemide verifitseerimise kursus6 Satisfiability aka SAT SAT is transformed to CNF (i.e. product of sums). Sums are called terms. If terms have max 2 literals then 2-SAT 2-SAT solved in a polynomial time  3-SAT is an NP complete task

7. Digitaalsüsteemide verifitseerimise kursus7 Satisfiability aka SAT Is this CNF satisfiable? Yes: a = 1, b = 0, c = 0! Worst case: 2 n combinations to try

8. Digitaalsüsteemide verifitseerimise kursus8 Some terminology If x in the formula alwaysin one phase (i.e. always inverted or always noninverted then x is unate. If x in the formula in both phases then x is binate. Term having just one literal called unit term.

9. Digitaalsüsteemide verifitseerimise kursus9 Resolvent-algorithm Resolvent: f = (x+A)(¬x+B) = (x +A)(¬x+B)(A+B) Consensus: f = xC + ¬xD = xC + ¬xD + CD Since SAT is in CNF we use resolvent.

10. Digitaalsüsteemide verifitseerimise kursus10 Resolvent-algorithm 1.Choose another variable x. 2.If x is unate, apply unate rule. 3.If x is unit term, apply unit term rule. 4.If x is unate, solve resolvent of x. 5.Repeat the steps until all resolvents solved. 6.If the result is 1, then function satisfiable; otherwise not satisfiable (unit term).

11. Digitaalsüsteemide verifitseerimise kursus11 Resolvent-algorithm example a binate terms resolvent solved

12. Digitaalsüsteemide verifitseerimise kursus12 Resolvent-algorithm: summary Resolvent-algorithm mathematically elegant but...... Designed for small SAT problems In the worst case 2 n resolvents to solve In order to solve complex SAT instances, search based algorithms needed

13. Digitaalsüsteemide verifitseerimise kursus13 Search-based SAT

14. Digitaalsüsteemide verifitseerimise kursus14 SolveSAT() input: a formula output: SAT or UNSAT forever { state = select_branch(); // choose and assign a variable if (state == EXHAUSTED) return UNSAT; result = infer(); // infer variable values if ( result == SAT) return SAT; else if (result == UNSAT) backtrack(); // backtrack to a prior decision else // result == INDETERMINATE continue; // need further assignment } Search-based SAT

15. Digitaalsüsteemide verifitseerimise kursus15 Implication Graph Directed acyclic graph: Nodes labeled by variable names, followed by the rank of the decision Variables preceded by minus were assigned 0, not preceded by minus were assigned 1 Directed arcs show from which assignments what new assignments imply Decision nodes (grey) and implication nodes (white)

16. Digitaalsüsteemide verifitseerimise kursus16 decisions: k = 1, j = 1, a = 0, b = 1. reach a conflict: x = 1 ja x = 0! learning: add a new term (¬e + h + ¬d) Implication Graph

17. Digitaalsüsteemide verifitseerimise kursus17 It implies that c = 0; the function is simplified: Since e is a unit term then e = 1; first decision: a = 1 If we choose b=1, then conflict! Two possibilities to handle this: 1) Invert the last decision (backtrack) 2) Add a new term (learning): Implication Graph Example

18. Digitaalsüsteemide verifitseerimise kursus18 Equivalence checking with SAT Equivalence Checking can be reduced to SAT: d = f  g Iff d is satisfiable then f and g are NOT equivalent.

19. Digitaalsüsteemide verifitseerimise kursus19 Miter circuit + 1 + Specification (reference implementation) Implementation SAT?

20. Digitaalsüsteemide verifitseerimise kursus20 SAT for schematics: characteristic formula Build CNFs corresponding to logic gates using logic implication: a  b = ¬a + b ab abab 001 011 100 111

21. Digitaalsüsteemide verifitseerimise kursus21 Implications for describing the AND gate: ¬a  ¬c & ¬b  ¬c & ¬c  ¬a  ¬b Characteristic formula for AND in CNF: (a+ ¬c) (b+ ¬c) (c+ ¬a+ ¬b) & a b c SAT for schematics: characteristic formula

22. Digitaalsüsteemide verifitseerimise kursus22 Implications for describing the OR-gate: a  c & b  c & c  a  b Characteristic formula for OR in CNF: (¬a + c) (¬b + c) (¬c + a + b) 1 a b c SAT for schematics: characteristic formula

23. Digitaalsüsteemide verifitseerimise kursus23 Characteristic formula for a schematic: (a+¬d)(b+¬d)(d+¬a+¬b)(¬c+¬e)(c+e)(¬d+f)(¬e+f)(¬f+d+e) 1 c e f & a b d SAT for schematics: characteristic formula