3. Satisfiability Checking 3.1 SAT-Checking Procedures - PowerPoint PPT Presentation

Fachgebiet RechnerSysteme Technische Universität Verification Technology Darmstadt 3. Satisfiability Checking Computer Systems Lab 1 3. Satisfiability Checking 3 3. Satisfiability Checking 3.1 SAT-Checking Procedures Verification Technology  Satisfiability (SAT): Content Content  A Boolean function f is satisfiable, if there exists an  A Boolean function f is satisfiable, if there exists an assignment A of Boolean constants to variables so that f = 1 3.1 Satisfiability-(SAT)-Checking — Example: f = xy + xz + r 3.2 Variants of SAT-Checking is satisfied by the assignment A: (x = 1, y = 0)  Boolean functions are tautologies ( = 1), are satisfiable (SAT), or are unsatisfiable (UNSAT, = 0)  SAT checker  Rather than to demonstrate the tautology f = 1, show that f is unsatisfiable      f a b a b 1 ?      f ( a b a b )     a b ( a b ) 3. Satisfiability Checking 2 3. Satisfiability Checking 4 3.1 SAT-checking procedures — Statement x  y = 1 and original circuit in  SAT checkers work on conjunctive normal forms (cnf's) combination: — Example: (a + b + c)·(b + c + d)·(a + d)  The sum-terms are called (or)-clauses ( )  Satisfying a cnf means that all clauses have to be satisfied ! a x & For a tautology, the b output must always be 1 y  1  1 This output must Implication always be 0 circuit (unsatisfiable)!

Fachgebiet RechnerSysteme Technische Universität Verification Technology Darmstadt 3. Satisfiability Checking 5 3. Satisfiability Checking 7 3.1 SAT-checking procedures 3.1 Satisfiability checking 3.1 SAT-checking procedures  Example:  Resolution (Davis/Putnam 1960)  Idea: (a + c)(b + c)(a + b)(a + c)(a + c) — Assume that there are two clauses with opposing pp g b: (a + c)(c + a)(a + c)(a + c) b ( + )( + )( + )( + ) values of some variable x, e.g., (x + a)(x + b) a: cc — We have generally (x + a)(x + b) = (x + a)(x + b)(a + b) — Selecting x as a decision variable, we will delete c: 0 (x + a) and (x + b) from our set of clauses and will continue with ( (a + b) b)  (a + b)=0 will also make (x + a)(x + b)=0  Systematic application to all variables  Unsatisfiability: 0 can be derived 3. Satisfiability Checking 6 3. Satisfiability Checking 8 3.1 SAT-checking procedures 3.1 SAT-checking procedures 3.1 Satisfiability checking  A different explanation of the resolution step:  Given an assignment A of Boolean constants to variables, each clause is either:  In order to prove the unsatisfiability of f, we apply  Satisfied (= 1) Boole's expansion theorem to f for some variable x: f f = x*f(0) + x*f(1) *f(0) + *f(1)  Unsatisfied (= 0)  f = 0 iff f(0)+f(1) = 0  Unresolved (can not be reduced to a constant)  Let f be in the form f = Z*(x + a)*(x + b) where Z stands for — Example: (b + e)·(b + c + d)·(a + d) the rest of the cnf. Then f(0) = Z*a and f(1) = Z*b. Thus, unresolved satisfied unsatisfied f(0) + f(1) = Z*(a + b). with the assignment A: (a = 1, b = 1, d = 0)  Satisfiability checking of a cnf g:  Find an assignment so that all clauses of g are satisfied  If this is infeasible then g is equal 0 Problem: how to organize "finding an assignment"?

Fachgebiet RechnerSysteme Technische Universität Verification Technology Darmstadt 3. Satisfiability Checking 9 3. Satisfiability Checking 11 3.1 Satisfiability checking 3.1 SAT-checking procedures 3.1 Satisfiability checking 3.1 SAT-checking procedures — Example: (b + e)·(b + c + d)·(a + d) a 0 1  With the new assignment A: (a = 1, b = 1, d = 1) the first clause remains unresolved c  The unresolved clause (b + e) is a unit clause b 1 1 0 0  A unit clause is an unresolved clause which has exactly one unassigned literal (e in the example) c c b b  Unit clauses should be used to determine the next 0 1 0 1 0 1 0 1 variable to be assigned a 1 backtrack + d d d d d d d d d d d d d d d d reverse b 1 decision 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 d with b = 1, b + e can 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 clause 0 1 only be satisfied unsatisfied, with e = 1 UNSAT conflicting e assignment 1 3. Satisfiability Checking 10 3. Satisfiability Checking 12 3.1 Satisfiability checking 3.1 SAT-checking procedures 3.1 Satisfiability checking 3.1 SAT-checking procedures  Finding an assignment can be organized as a decision tree  In the example, a satisfying assignment (Davis/Logemann/Loveland 1962, DLL algorithm) A: (a = 1, b = 1, d = 1, e = 1) was found — Example: (b + e)·(b + c + d)·(a + d)   with the assignment A: (a = 1, b = 1, d = 0) ith th i t A ( 1 b 1 d 0) If all branches terminate in conflicting assignments then If ll b h t i t i fli ti i t th the function is unsatisfiable  If we arrive at an assignment which results in at least  In the worst case, an exponential number of steps has to one unsatisfied clause ( a conflicting assignment) then be executed the last decision has to be reversed and a backtracking step is performed  Practically, the selection of decision variables on the a basis of detected unit clauses results in an important 1 backtrack + improvement of efficiency improvement of efficiency reverse b 1 decision d 1 clause 0 1 unsatisfied, conflicting assignment

Fachgebiet RechnerSysteme Technische Universität Verification Technology Darmstadt 3. Satisfiability Checking 13 3. Satisfiability Checking 15 3.1 SAT-checking procedures 3.1 Satisfiability checking 3.1 Satisfiability checking 3.1 SAT-checking procedures  GRASP (Silva, Sakallah '95)  There are direct and indirect implications of assignments http://portal.acm.org/citation.cfm?id=244560  If a = 0 was decided previously then the decision  Systematic investigation of the implications of c = 0@i at a certain decision level i implies d = 1@i at assignments i t the same level i for clause (a + c + d) for satisfiability th l l i f l ( + + d) f ti fi bilit  "Learning"  d = 1@i has e = 0@i as an indirect implication if we have also clause (d + e)  Non-chronological backtracking  And so on ...  The detection of all implications is called Boolean Constraint Propagation (BCP) 3. Satisfiability Checking 14 3. Satisfiability Checking 16 3.1 Satisfiability checking 3.1 SAT-checking procedures 3.1 Satisfiability checking 3.1 SAT-checking procedures  The ordering of decisions is recorded by means of a  The implications of a decision are represented by means decision level associated with each decision of an implication graph  We refer to a variable assignment at a certain decision  Assume n clauses k 1 , ... , k n l level by means of "@", e.g., a = 1@1, b = 1@ 2, etc. l b f "@" 1@1 b 1@ 2 t  "Variable assignment va 1 implies variable assignment va 2 due to clause k i " is represented by k i k i d=1@i va 1 va 2 e.g. c=0@i decision a 1 level #1 decision b 1 level #2 decision d level #3 0

Fachgebiet RechnerSysteme Technische Universität Verification Technology Darmstadt 3. Satisfiability Checking 17 3. Satisfiability Checking 19 3.1 SAT-checking procedures 3.1 Satisfiability checking 3.1 Satisfiability checking 3.1 SAT-checking procedures  Example:  We now know that the assignment a = 0@1, b = 0@2, c = 0 0 0@5 leads to a conflict resulting in a backtracking step c = 1 (c + d)(d + e)(e + c + f + b)(d + a + f)...  If the procedure makes the same decision later in the k 1 k 2 k 3 k 4 1 2 3 4 graph then the complex detection of all implications has h th th l d t ti f ll i li ti h Assume the following previous decisions: to be repeated a = 0@1, b = 0@2, ... @3, ... @4 Now assume the decision c = 0@5 at level 5. The implication graph becomes: (c + d)(d + e)(e + c + f + b)(d + a + f)... a = 0@1 k 4 k 4 f = 0@5 f = 0@5 decision decision k 4 a d = 0@5 0 level #1 k 1 k 2 decision b c = 0@5 0 e = 1@5 conflict! level #2 ... k 3 k 3 decision c c decision level #5 k 3 0 level #i b = 0@2 f = 1@5 0 3. Satisfiability Checking 18 3. Satisfiability Checking 20 3.1 Satisfiability checking 3.1 SAT-checking procedures 3.1 SAT-checking procedures 3.1 Satisfiability checking  We may also view this procedure as a resolution step  We now know that the assignment a = 0@1, b = 0@2, c = after a partial variable assignment. 0@5 leads to a conflict resulting in a backtracking step c = 1  If the procedure makes the same decision later in the — In the example we have with the partial variable assignment a=b=c=0: i t b 0 graph then the complex detection of all implications has h th th l d t ti f ll i li ti h to be repeated (c + d)(d + e)(e + c + f + b)(d + a + f)... =  It is possible to "learn" the conflicting assignment by adding the clause (a + b + c): d(d + e)(e + f)(d + f)… (c + d)(d + e)(e + c + f + b)(d + a + f)(a + b + c)... d: d: e(e f)f… e(e + f)f… decision decision a 0 level #1 enforces e: ff c=1 for a=0 decision b 0 and b=0 level #2 ... decision decision c c level #5 0 level #i 0