Daniel Kroening and Ofer Strichman 1 Decision Procedures An Algorithmic Point of View SAT

Decision Procedures An algorithmic point of view2 Part I Reminders -  What is Logic  Proofs by deduction  Proofs by enumeration  Decidability, Soundness and Completeness  some notes on Propositional Logic Deciding Propositional Logic  SAT tools  BDDs      

Decision Procedures An algorithmic point of view3 Next: Deciding Propositional Formulas SAT solvers Binary Decision Diagrams

Decision Procedures An algorithmic point of view4 Given  in CNF: (x,y,z),(-x,y),(-y,z),(-x,-y,-z) Decide() BCP() Resolve_Conflict()  X XX XX  A Basic SAT algorithm

Decision Procedures An algorithmic point of view5 SAT made some progress…

Decision Procedures An algorithmic point of view6 While (true) { if (!Decide()) return (SAT); while (!BCP()) if (!Resolve_Conflict()) return (UNSAT); } Choose the next variable and value. Return False if all variables are assigned Apply repeatedly the unit clause rule. Return False if reached a conflict Backtrack until no conflict. Return False if impossible A Basic SAT algorithm

Decision Procedures An algorithmic point of view7 Basic Backtracking Search Organize the search in the form of a decision tree  Each node corresponds to a decision  Definition: Decision Level (DL) is the depth of the node in the decision tree.  Notation: x 2 {0,1} is assigned to v at decision level d

Decision Procedures An algorithmic point of view8 Backtracking Search in Action  1 = (x 2  x 3 )  2 = (  x 1   x 4 )  3 = (  x 2  x 4 )  1 = (x 2  x 3 )  2 = (  x 1   x 4 )  3 = (  x 2  x 4 ) x 1 x 1 = {(x 1,0), (x 2,0), (x 3,1)} x 2 x 2 = {(x 1,1), (x 2,0), (x 3,1), (x 4,0)} x 1 =  x 3 =  x 4 =  x 2 =  x 3 = No backtrack in this example, regardless of the decision!

Decision Procedures An algorithmic point of view9 Backtracking Search in Action  1 = (x 2  x 3 )  2 = (  x 1   x 4 )  3 = (  x 2  x 4 )  4 = (  x 1  x 2   x 3 )  1 = (x 2  x 3 )  2 = (  x 1   x 4 )  3 = (  x 2  x 4 )  4 = (  x 1  x 2   x 3 ) Add a clause  x 4 =  x 2 =  x 3 = conflict {(x 1,0), (x 2,0), (x 3,1)} x 2 x 2 =  x 3 = x 1 = x 1 x 1 =

Decision Procedures An algorithmic point of view10 Status of a clause A clause can be  Satisfied: at least one literal is satisfied  Unsatisfied: all literals are assigned but non are satisfied  Unit: all but one literals are assigned but none are satisfied  Unresolved: all other cases Example: C = ( x 1 Ç x 2 Ç x 3 ) x1x1 x2x2 x3x3 C 10Satisfied 000Unsatisfied 00Unit 0Unresolved

Decision Procedures An algorithmic point of view11 For a given variable x :  C x p – # unresolved clauses in which x appears positively  C x n - # unresolved clauses in which x appears negatively  Let x be the literal for which C xp is maximal  Let y be the literal for which C yn is maximal  If C xp > C yn choose x and assign it TRUE  Otherwise choose y and assign it FALSE Requires l (#literals) queries for each decision. DLIS (Dynamic Largest Individual Sum) – choose the assignment that increases the most the number of satisfied clauses Decision heuristics - DLIS

Decision Procedures An algorithmic point of view12 Compute for every clause  and every variable l (in each phase): J ( l ) := Choose a variable l that maximizes J ( l ). This gives an exponentially higher weight to literals in shorter clauses. Decision heuristics - JW Jeroslow-Wang method

Decision Procedures An algorithmic point of view13 Pause... We will see other (more advanced) decision Heuristics soon. These heuristics are integrated with a mechanism called Learning with Conflict-Clauses, which we will learn next.

Decision Procedures An algorithmic point of view14 55 55 x 6 Implication graphs and learning: option #1  1 = (  x 1  x 2 )  2 = (  x 1  x 3  x 9 )  3 = (  x 2   x 3  x 4 )  4 = (  x 4  x 5  x 10 )  5 = (  x 4  x 6  x 11 )  6 = (  x 5   x 6 )  7 = (x 1  x 7   x 12 )  8 = (x 1  x 8 )  9 = (  x 7   x 8   x 13 )  1 = (  x 1  x 2 )  2 = (  x 1  x 3  x 9 )  3 = (  x 2   x 3  x 4 )  4 = (  x 4  x 5  x 10 )  5 = (  x 4  x 6  x 11 )  6 = (  x 5   x 6 )  7 = (x 1  x 7   x 12 )  8 = (x 1  x 8 )  9 = (  x 7   x 8   x 13 ) Current truth assignment: {x 9 10 x 11 x 12 x 13 Current decision assignment: {x 1 66 66  conflict x 9 x 1 x 10 x 11 x 5 44 44 22 22 x 3 11 x 2 33 33 x 4 We learn the conflict clause  10 : ( : x 1 Ç x 9 Ç x 11 Ç x 10 )

Decision Procedures An algorithmic point of view15 Implication graph, flipped assignment option #1 x 1 x 11 x 10 x 9 x 7 x 12 77 77 x 8 88  10 99 99 ’’ x 13 99 Due to the conflict clause  1 = (  x 1  x 2 )  2 = (  x 1  x 3  x 9 )  3 = (  x 2   x 3  x 4 )  4 = (  x 4  x 5  x 10 )  5 = (  x 4  x 6  x 11 )  6 = (  x 5  x 6 )  7 = (x 1  x 7   x 12 )  8 = (x 1  x 8 )  9 = (  x 7   x 8   x 13 )  10 : ( : x 1 Ç x 9 Ç x 11 Ç x 10 )  1 = (  x 1  x 2 )  2 = (  x 1  x 3  x 9 )  3 = (  x 2   x 3  x 4 )  4 = (  x 4  x 5  x 10 )  5 = (  x 4  x 6  x 11 )  6 = (  x 5  x 6 )  7 = (x 1  x 7   x 12 )  8 = (x 1  x 8 )  9 = (  x 7   x 8   x 13 )  10 : ( : x 1 Ç x 9 Ç x 11 Ç x 10 ) No decision here Another conflict clause:  11 : ( : x 13 Ç : x 12 Ç x 11 Ç x 10 Ç x 9 ) where should we backtrack to now ?

Decision Procedures An algorithmic point of view16 Non-chronological backtracking Non- chronological backtracking x ’’ Decision level Which assignments caused the conflicts ? x 9 = x 10 = x 11 = x 12 = x 13 = Backtrack to DL = 3 3 These assignments Are sufficient for Causing a conflict.

Decision Procedures An algorithmic point of view17 Non-chronological backtracking So the rule is: backtrack to the largest decision level in the conflict clause. This works for both the initial conflict and the conflict after the flip. Q: What if the flipped assignment works? A: Change the decision retroactively.

Decision Procedures An algorithmic point of view18 Non-chronological Backtracking x 1 = 0 x 2 = 0 x 3 = 1 x 4 = 0 x 5 = 0 x 7 = 1 x 9 = 0 x 6 = 0... x 5 = 1 x 9 = 1 x 3 = 0

Decision Procedures An algorithmic point of view19 More Conflict Clauses Def: A Conflict Clause is any clause implied by the formula Let L be a set of literals labeling nodes that form a cut in the implication graph, separating the conflict node from the roots. Claim: Ç l 2 L : l is a Conflict Clause. 55 55 x 6 66 66  conflict x 9 x 1 x 10 x 11 x 5 44 44 22 22 x 3 11 x 2 33 33 x 4 1. (x 10 Ç : x 1 Ç x 9 Ç x 11 ) 2. (x 10 Ç : x 4 Ç x 11 ) 3. (x 10 Ç : x 2 Ç : x 3 Ç x 11 )  1 2 3

Decision Procedures An algorithmic point of view20 Conflict clauses How many clauses should we add ? If not all, then which ones ?  Shorter ones ?  Check their influence on the backtracking level ?  The most “influential” ?

Decision Procedures An algorithmic point of view21 Conflict clauses Def: An Asserting Clause is a Conflict Clause with a single literal from the current decision level. Backtracking (to the right level) makes it a Unit clause. Asserting clauses are those that force an immediate change in the search path. Modern solvers only consider Asserting Clauses.

Decision Procedures An algorithmic point of view22 Unique Implication Points (UIP’s) Definition: A Unique Implication Point (UIP) is an internal node in the Implication Graph that all paths from the decision to the conflict node go through it. The First-UIP is the closest UIP to the conflict. 55 55 66 66  conflict 44 44 22 22 11 33 33 UIP

Decision Procedures An algorithmic point of view23 Conflict-driven backtracking (option #2) Conflict clause: ( x 10 Ç : x 4 Ç x 11 ) With standard Non-Chronological Backtracking we backtracked to DL = 6. Conflict-driven Backtrack: backtrack to the second highest decision level in the clause (without erasing it). In this case, to DL = 3. Q: why?  conflict x 10 x 11 x 4

Decision Procedures An algorithmic point of view24 Conflict-driven Non-chronological Backtracking x 1 = 0 x 2 = 0 x 3 = 1 x 4 = 0 x 5 = 0 x 5 = 1 x 7 = 1 x 3 = 1 x 9 = 0 x 9 = 1 x 6 = 0...

Decision Procedures An algorithmic point of view25 Decision Conflict Decision Level Time work invested in refuting x =1 (some of it seems wasted) C x =1 Refutation of x =1 C1C1 C5C5 C4C4 C3C3 C2C2 Progress of a SAT solver BCP

Decision Procedures An algorithmic point of view26 Conflict-Driven Backtracking So the rule is: backtrack to the second highest decision level dl, but do not erase it. This way the literal with the currently highest decision level will be implied in DL = dl. Q: what if the conflict clause has a single literal ?  For example, from ( x Ç : y ) Æ ( x Ç y ) and decision x =0, we learn the conflict clause ( x ).

Decision Procedures An algorithmic point of view27 Conflict clauses and Resolution The Binary-resolution is a sound inference rule: Example:

Decision Procedures An algorithmic point of view28 Consider the following example: Conflict clause: c 5 : ( x 2 Ç : x 4 Ç x 10 ) Conflict clauses and resolution

Decision Procedures An algorithmic point of view29 Conflict clause: c 5 : ( x 2 Ç : x 4 Ç x 10 ) Resolution order: x 4, x 5, x 6, x 7  T1 = Res(c 4,c 3,x 7 ) = ( : x 5 Ç : x 6 )  T2 = Res(T1, c 2, x 6 ) = ( : x 4 Ç : x 5 Ç X 10 )  T3 = Res(T2,c 1,x 5 ) = (x 2 Ç : x 4 Ç x 10 ) Conflict clauses and resolution

Decision Procedures An algorithmic point of view30 Applied to our example: Finding the conflict clause: cl is asserting the first UIP

Decision Procedures An algorithmic point of view31 The Resolution-Graph keeps track of the “inference relation” 11 22 33 44 55 66  10 77 88 99  11 77 77 88  10 99 99  ’ conflict 55 55 66 66  conflict 44 44 22 22 11 33 33 99 Resolution Graph

Decision Procedures An algorithmic point of view32 The resolution graph What is it good for ? Example: for computing an Unsatisfiable core [Picture Borrowed from Zhang, Malik SAT’03]

33 Resolution graph: example L :L : Inferred clauses Empty clause Original clauses Unsatisfiable core learning

Decision Procedures An algorithmic point of view34 (Implemented in Chaff) VSIDS (Variable State Independent Decaying Sum) Decision heuristics - VSIDS 1.Each variable in each polarity has a counter initialized to When a clause is added, the counters are updated. 3. The unassigned variable with the highest counter is chosen. 4. Periodically, all the counters are divided by a constant.

Decision Procedures An algorithmic point of view35 Decision heuristics – VSIDS (cont’d) Chaff holds a list of unassigned variables sorted by the counter value. Updates are needed only when adding conflict clauses. Thus - decision is made in constant time.

Decision Procedures An algorithmic point of view36 VSIDS is a ‘quasi-static’ strategy: - static because it doesn’t depend on current assignment - dynamic because it gradually changes. Variables that appear in recent conflicts have higher priority. This strategy is a conflict-driven decision strategy. “..employing this strategy dramatically (i.e. an order of magnitude) improved performance... “ Decision heuristics VSIDS (cont’d)

Decision Procedures An algorithmic point of view37 Decision Heuristics - Berkmin Keep conflict clauses in a stack Choose the first unresolved clause in the stack  If there is no such clause, use VSIDS Choose from this clause a variable + value according to some scoring (e.g. VSIDS) This gives absolute priority to conflicts.

Decision Procedures An algorithmic point of view38 Berkmin heuristic tail- first conflict clause

Decision Procedures An algorithmic point of view39 The SAT competitions

Decision Procedures An algorithmic point of view40 End of SAT (for now) Beginning of Binary Decision Diagrams