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Heuristics for Efficient SAT Solving As implemented in GRASP, Chaff and GSAT.

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Why SAT? Fundamental problem from theoretical point of view –Cook theorem, 1971: the first NP-complete problem. Numerous applications: –Solving any NP problem... –Verification: Model Checking, theorem-proving,... –AI: Planning, automated deduction,... –Design and analysis: CAD, VLSI –Physics: statistical mechanics (models for spin-glass material)

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SAT made some progress…

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The SAT competitions

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Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction.More techniques: decision heuristics, deduction Stochastic SAT solvers: the GSAT approach

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The K-Coloring problem: Given an undirected graph G(V,E) and a natural number k, is there an assignment color: Formulation of famous problems as SAT: k- Coloring (1/2)

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Formulation of famous problems as SAT: k- Coloring (2/2) x i,j = node i is assigned the ‘color’ j (1 i n, 1 j k) Constraints: i) At least one color to each node: (x 1,1 x 1,2 … x 1,k …) ii) At most one color to each node: iii) Coloring constraints: for each i, j such that ( i, j ) 2 E:

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Given a property p : (e.g. “always signal_a = signal_b”) Is there a state reachable within k cycles, which satisfies p ?... s0s0 s1s1 s2s2 s k-1 sksk pp p pp p Formulation of famous problems as SAT: Bounded Model Checking

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The reachable states in k steps are captured by: The property p fails in one of the cycles 1.. k : Formulation of famous problems as SAT: Bounded Model Checking

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The safety property p is valid up to cycle k iff k is unsatisfiable:... s0s0 s1s1 s2s2 s k-1 sksk pp p pp p Formulation of famous problems as SAT: Bounded Model Checking

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Example: a two bit counter Property: G ( l r ). 00 01 10 11 For k = 2, k is unsatisfiable. For k = 4 k is satisfiable Initial state: I : : l Æ : r Transition: R : l ’ = ( l r ) Æ r ’ = : r Formulation of famous problems as SAT: Bounded Model Checking

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Bounded Model Checking k = 0 BMC(M, ,k) yes k++ k ¸ ?k ¸ ? no

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Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

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What is SAT? SATisfying assignment! Given a propositional formula in CNF, find an assignment to Boolean variables that makes the formula true: 1 = (x 2 x 3 ) 2 = ( x 1 x 4 ) 3 = ( x 2 x 4 ) A = {x 1 =0, x 2 =1, x 3 =0, x 4 =1} 1 = (x 2 x 3 ) 2 = ( x 1 x 4 ) 3 = ( x 2 x 4 ) A = {x 1 =0, x 2 =1, x 3 =0, x 4 =1}

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CNF-SAT Conjunctive Normal Form: Conjunction of disjunction of literals. Example: ( : x 1 Ç : x 2 ) Æ (x 2 Ç x 4 Ç : x 1 ) Æ... Experience shows that CNF-SAT solving is faster than solving a general propositional formula. There exists a polynomial transformation due to Tseitin (1970) of a general propositional formula to CNF, with addition of | | variables.

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Tseitin’s encoding by example = x 1 Ç : ( x 2 Æ x 3 ) ’ = a 0 Æ (a 0 $ x 1 Ç a 1 ) Æ (a 1 $ : a 2 ) Æ (a 2 $ x 2 Æ x 3 ) It is left to transform ’ to CNF. x1x1 x2x2 x3x3 Æ : Ç a0a0 a1a1 a2a2

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Tseitin’s encoding: CNF encodings of gates And gate. e.g., for a i $ x 1 Æ x 2 add to S (a i Ç : x 1 Ç : x 2 ), ( : a i Ç x 1 ), ( : a i Ç x 2 ) Or gate. e.g. for a i $ x 1 Ç x 2 add to S ( : a i Ç x 1 Ç x 2 ), (a i Ç : x 1 ), (a i Ç : x 2 ) Not gate. e.g. for a i $ : x 1 add to S ( : a i Ç : x 1 ), (a i Ç x 1 )

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Tseitin’s encoding For each Boolean gate instance g i in , add a new auxiliary variable a i, and add to a stack S the CNF clauses encoding a i $ g i. Let a 0 denote the auxiliary variable encoding the main operator of . Let Theorem (Tseitin): is satisfiable iff ’ is satisfiable.

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(CNF) SAT basic definitions: literals A literal is a variable or its negation. Var( l ) is the variable associated with a literal l. A literal is called negative if it is a negated variable, and positive otherwise.

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SAT basic definitions: literals If var( l ) is unassigned, then l is unresolved. Otherwise, l is satisfied by an assignment if (var( l )) = 1 and l is positive, or (var( l )) = 0 and l is negative, and unsatisfied otherwise.

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SAT basic definitions: clauses The state of an n -long clause C under a partial assignment is: Satisfied if at least one of C’s literals is satisfied, Conflicting if all of C’s literals are unsatisfied, Unit if n -1 literals in C are unsatisfied and 1 literal is unresolved, and Unresolved otherwise.

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SAT basic definitions: clauses Example

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SAT basic definitions: the unit clause rule The unit clause rule: in a unit clause the unresolved literal must be satisfied.

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Given in CNF: (x,y,z),(-x,y),(-y,z),(-x,-y,-z) Decide() Deduce() Resolve_Conflict() X XX XX A Basic SAT algorithm

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Basic Backtracking Search Organize the search in the form of a decision tree Each node corresponds to a decision Depth of the node in the decision tree is called the decision level Notation: x = v @ d x is assigned v 2 {0,1} at decision level d

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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 = 0@1 {(x 1,0), (x 2,0), (x 3,1)} x 2 x 2 = 0@2 {(x 1,1), (x 2,0), (x 3,1), (x 4,0)} x 1 = 1@1 x 3 = 1@2 x 4 = 0@1 x 2 = 0@1 x 3 = 1@1 No backtrack in this example, regardless of the decision!

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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 = 0@1 x 2 = 0@1 x 3 = 1@1 conflict {(x 1,0), (x 2,0), (x 3,1)} x 2 x 2 = 0@2 x 3 = 1@2 x 1 = 0@1 x 1 x 1 = 1@1

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While (true) { if (!Decide()) return (SAT); while (!Deduce()) if (!Resolve_Conflict()) return (UNSAT); } Choose the next variable and value. Return False if all variables are assigned Apply unit clause rule. Return False if reached a conflict Backtrack until no conflict. Return False if impossible A Basic SAT algorithm ( DPLL-based)

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Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

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Maintain a counter for each literal: in how many unresolved clauses it appears ? Decide on the literal with the largest counter. Requires O(#literals) queries for each decision. Decision heuristics DLIS (Dynamic Largest Individual Sum)

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Compute for every clause and every literal l : J ( l ) := Choose a variable l that maximizes J ( l ). This gives an exponentially higher weight to literals in shorter clauses. Decision heuristics Jeroslow-Wang method

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Let f *( x ) be the # of unresolved smallest clauses containing x. Choose x that maximizes: (( f *( x ) + f *(! x )) * 2 k + f *( x ) * f *(! x ) k is chosen heuristically. The idea: Give preference to satisfying small clauses. Among those, give preference to balanced variables (e.g. f *( x ) = 3, f *(! x ) = 3 is better than f *( x ) = 1, f *(! x ) = 5). Decision heuristics MOM (Maximum Occurrence of clauses of Minimum size).

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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 learns next.

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Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

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Implication graphs and learning 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 =0@1,x 10 =0@3, x 11 =0@3, x 12 =1@2, x 13 =1@2} Current decision assignment: {x 1 =1@6} 66 66 conflict x 9 =0@1 x 1 =1@6 x 10 =0@3 x 11 =0@3 x 5 =1@6 44 44 55 55 x 6 =1@6 22 22 x 3 =1@6 11 x 2 =1@6 33 33 x 4 =1@6 We learn the conflict clause 10 : ( : x 1 Ç x 9 Ç x 11 Ç x 10 ) and backtrack to the highest (deepest) dec. level in this clause (6).

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Implication graph, flipped assignment x 1 =0@6 x 11 =0@3 x 10 =0@3 x 9 =0@1 x 7 =1@6 x 12 =1@2 77 77 x 8 =1@6 88 10 99 99 ’’ x 13 =1@2 99 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 ) We learn the conflict clause 11 : ( : x 13 Ç x 9 Ç x 10 Ç x 11 Ç : x 12 ) and backtrack to the highest (deepest) dec. level in this clause (3).

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Non-chronological backtracking Non- chronological backtracking x 1 4 5 6 ’’ Decision level Which assignments caused the conflicts ? x 9 = 0@1 x 10 = 0@3 x 11 = 0@3 x 12 = 1@2 x 13 = 1@2 Backtrack to decision level 3 3 These assignments Are sufficient for Causing a conflict.

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Non-chronological backtracking (option #1) So the rule is: backtrack to the largest decision level in the conflict clause. Q: What if the flipped assignment works ? A: continue to the next decision level, leaving the current one without a decision variable. Backtracking back to this level will lead to another conflict and further backtracking.

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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

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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. 55 55 x 6 =1@6 66 66 conflict x 9 =0@1 x 1 =1@6 x 10 =0@3 x 11 =0@3 x 5 =1@6 44 44 22 22 x 3 =1@6 11 x 2 =1@6 33 33 x 4 =1@6 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 Skip alternative learning

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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” ? The answer requires two definitions: Asserting clauses Unique Implication points (UIP’s)

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Asserting 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. Modern solvers only consider Asserting Clauses.

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Unique Implication Points (UIP’s) Def: 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. The method of choice: an asserting clause that includes the first UIP. In this case ( x 10 Ç : x 4 Ç x 11 ). 55 55 66 66 conflict 44 44 22 22 11 33 33 UIP x 10 =0@3 x 11 =0@3 x 4 =1@6

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Conflict-driven backtracking (option #2) Previous method: backtrack to highest decision level in conflict clause (and erase it). A better method (empirically): backtrack to the second highest decision level in the clause, without erasing it. The asserted literal is implied at that level. In our example: ( x 10 Ç : x 4 Ç x 11 ) Previously we backtracked to decision level 6. Now we backtrack to decision level 3. x 4 = 0@3 is implied. 633

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Conflict-driven 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...

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Conflict-Driven Backtracking So the rule is: backtrack to the second highest decision level dl, but do not erase it. If the conflict clause has a single literal, backtrack to decision level 0. Q: It seems to waste work, since it erases assignments in decision levels higher than dl, unrelated to the conflict. A: indeed. But allows the SAT solver to redirect itself with the new information.

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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

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Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

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Conflict clauses and Resolution The Binary-resolution is a sound inference rule: Example:

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Consider the following example: Conflict clause: c 5 : ( x 2 Ç : x 4 Ç x 10 ) Conflict clauses and resolution

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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

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Applied to our example: Finding the conflict clause: cl is asserting the first UIP

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The Resolution-Graph keeps track of the “inference relation” 11 22 33 44 55 66 10 77 88 99 11 55 55 66 66 conflict 44 44 22 22 11 33 33 77 77 88 10 99 99 ’ conflict 99 Resolution Graph

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The resolution graph What is it good for ? Example: for computing an Unsatisfiable core [Picture Borrowed from Zhang, Malik SAT’03]

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

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Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

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4. Periodically, all the counters are divided by a constant. 3. The unassigned variable with the highest counter is chosen. 2. When a clause is added, the counters are updated. 1. Each variable in each polarity has a counter initialized to 0. (Implemented in Chaff) Decision heuristics VSIDS (Variable State Independent Decaying Sum)

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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 heuristics VSIDS (cont’d)

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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)

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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.

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Berkmin heuristic tail- first conflict clause

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Deduction() allocates new implied variables and conflicts. How can this be done efficiently ? Observation: More than 90% of the time SAT solvers perform Deduction(). More engineering aspects of SAT solvers

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Hold 2 counters for each clause : val1 ( ) - # of negative literals assigned 0 in + # of positive literals assigned 1 in . val0 ( ) - # of negative literals assigned 1 in + # of positive literals assigned 0 in . Grasp implements Deduction() with counters

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Every assignment to a variable x results in updating the counters for all the clauses that contain x. Grasp implements Deduction() with counters is satisfied iff val1( ) > 0 is unsatisfied iff val0( ) = | | is unit iff val1( ) = 0 val0( ) = | | - 1 is unresolved iff val1( ) = 0 val0( ) < | | - 1. Backtracking: Same complexity.

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Observation: during Deduction(), we are only interested in newly implied variables and conflicts. These occur only when the number of literals in with value ‘ false ’ is greater than | | - 2 Conclusion: no need to visit a clause unless (val0( ) > | | - 2) How can this be implemented ? Chaff implements Deduction() with a pair of observers

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Define two ‘ observers ’ : O1( ), O2( ). O1( ) and O2( ) point to two distinct literals which are not ‘ false ’. becomes unit if updating one observer leads to O1( ) = O2( ). Visit clause only if O1( ) or O2( ) become ‘ false ’. Chaff implements Deduction() with a pair of observers

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Both observers of an implied clause are on the highest decision level present in the clause. Therefore, backtracking will un-assign them first. Conclusion: when backtracking, observers stay in place. V[1]=0 V[5]=0, v[4]= 0 V[2]=0 O1O2 Unit clause Backtrack v[4] = v[5]= X v[1] = 1 Backtracking: No updating. Complexity = constant. Chaff implements Deduction() with a pair of observers

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The choice of observing literals is important. Best strategy is - the least frequently updated variables. The observers method has a learning curve in this respect: 1. The initial observers are chosen arbitrarily. 2. The process shifts the observers away from variables that were recently updated (these variables will most probably be reassigned in a short time). In our example: the next time v[5] is updated, it will point to a significantly smaller set of clauses. Chaff implements Deduction() with a pair of observers

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Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

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GSAT: stochastic SAT solving for i = 1 to max_tries { T := randomly generated truth assignment for j = 1 to max_flips { if T satisfies return TRUE choose v s.t. flipping v’s value gives largest increase in the # of satisfied clauses (break ties randomly). T := T with v’s assignment flipped. } } Given a CNF formula , choose max_tries and max_flips Many alternative heuristics

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Numerous progressing heuristics Hill-climbing Tabu-list Simulated-annealing Random-Walk Min-conflicts...

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Improvement # 1: clause weights Initial weight of each clause: 1 Increase by k the weight of unsatisfied clauses. Choose v according to max increase in weight Clause weights is another example of conflict-driven decision strategy.

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Improvement # 2: Averaging-in Q: Can we reuse information gathered in previous tries in order to speed up the search ? A: Yes! Rather than choosing T randomly each time, repeat ‘good assignments’ and choose randomly the rest.

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Let X1, X2 and X3 be equally wide bit vectors. Define a function bit_average : X1 X2 X3 as follows: b 1 i b 1 i = b 2 i randomotherwise (where b j i is the i-th bit in Xj, j {1,2,3}) Improvement # 2: Averaging-in (cont’d) b 3 i :=

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Improvement # 2: Averaging-in (cont’d) Let T i init be the initial assignment (T) in cycle i. Let T i best be the assignment with highest # of satisfied clauses in cycle i. T 1 init := random assignment. T 2 init := random assignment. i > 2, T i init := bit_average(T i-1 best, T i-2 best )

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