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SAT SolvingPresented by Avi Yadgar
The SAT Problem
Given a Boolean formula , look for assignment A for such that . A is a solution for .
A partial assignment assigns a subset of . CNF representation of :
is a conjunction of clauses: A clause is a disjunction of literals:
( )v( ( ))A v true v
( )v
( )v1 2( ) ... mv cl cl cl
1( ... )i lcl lit lit
v
( )v
The SAT Problem
= (v1 v2 v5) (v1 v7 v9) … (v5 v7)
A satisfies ↔ A satisfies all its clauses. Each clause needs one “true” literal
NP-Complete Many applications Very efficient solvers Search / Proof engine
( )v
Boolean Constraint Propagation Unit Clause: A clause with exactly one
unassigned literal, while all the rest are false. Asserts the value of the unassigned variable.
cl = ( v3)
v1 = 0
v2 = 1
v3 = ?v1 = 0 v2 = 0
v3 = 1 v1 v2
Boolean Constraint Propagation Unit Clause: A clause with exactly one
unassigned literal, while all the rest are false. Asserts the value of the unassigned variable.
cl implies and is its antecedent. v1 and v2 are the antecedent variables of
BCP(): Calculates all the possible implications.
1 2 3( )cl v v v v1 = 0 v2 = 1 v3 = ?
v3 = 1
3v
3v
DPLL: Davis Putnam Logemann Loveland Backtrack Search Choose a decision
variable and value.v6
DPLL: Davis Putnam Logemann Loveland Backtrack Search Choose a decision
variable and value. Run bcp()
v6 ¬v14 v3
DPLL: Davis Putnam Logemann Loveland Backtrack Search Choose a decision
variable and value. Run bcp()
v6 ¬v14 v3
¬v1 v18 v4 ¬v2
v8 ¬v10 v7
DPLL: Davis Putnam Logemann Loveland Backtrack Search Choose a decision
variable and value. Run bcp() If a conflict occurs
Flip the highest decision variable not yet flipped.
v6 ¬v14 v3
¬v1 v18 v4 ¬v2
v8 ¬v10 v7
¬v9 v5 ¬v3
DPLL: Davis Putnam Logemann Loveland Backtrack Search Choose a decision
variable and value. Run bcp() If a conflict occurs
Flip the highest decision variable not yet flipped.
Mark as flipped.
v6 ¬v14 v3
¬v1 v18 v4 ¬v2
v8 ¬v10 v7
v9**
DPLL: Davis Putnam Logemann Loveland Backtrack Search Choose a decision
variable and value. Run bcp() If a conflict occurs
flip the highest decision variable not yet flipped.
Mark as flipped Run bcp().
v6 ¬v14 v3
¬v1 v18 v4 ¬v2
v8 ¬v10 v7
v9** v15 v14
DPLL: Davis Putnam Logemann Loveland Backtrack Search Choose a decision
variable and value. Run bcp() If a conflict occurs
flip the highest decision variable not yet flipped.
Mark as flipped Run bcp().
v6 ¬v14 v3
¬v1 v18 v4 ¬v2
¬v8**
v9** v15 v14
DPLL: Davis Putnam Logemann Loveland Backtrack Search Choose a decision
variable and value. Run bcp() If a conflict occurs
flip the highest decision variable not yet flipped.
Mark as flipped Run bcp().
v6 ¬v14 v3
¬v1 v18 v4 ¬v2
¬v8** v11
DPLL: Davis Putnam Logemann Loveland Backtrack Search
Termination No unassigned variables – SAT No decision variable to flip – un-SAT
Optimized BCP
Solver spends 90% in BCP
2 5 6 9 10( , , , , )v v v v v
1 3 4 6 10( , , , , )v v v v v
10v
1v
3v
2v
4v
cres(c1 , c2) = (v2 v5 v7 v9)
The SAT Problem - Resolution
c1 = (v1 v2 v5) , c2(v1 v7 v9)
cres
cres
v2 v5 v7 v9
SAT Solving - Implication Graph (v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v9
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
¬v8
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v12¬v8
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v12
v7
¬v8
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v12
v7
¬v8
v3
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v4
v12
v7
¬v8
v3
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v4 ¬v1
v12
v7
¬v8
v3
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v4 ¬v1
v12
v7
¬v8
v19
v3
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v4 ¬v1
v12
v7
¬v8
v19
¬v6
v3
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v4 ¬v1
v12
v7
¬v8
v19
¬v6
v3
v15
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v4 ¬v1
v12
v7
¬v8
v19
v3
¬v3 v15
SAT Solving - Implication Graph
¬v6
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v4 ¬v1
v12
v7
¬v8
v19
¬v6
v3
¬v3 v15
Conflict
SAT Solving - Implication Graph
(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v12
v7
¬v8
v3
SAT Solving - Implication Graph
¬v4***
CUT: A bi-partition of the graph. The conflict is on the ‘conflict side’. The decisions are on the ‘reason side’ No edge from the ‘conflict side’ to the ‘reason
side’. The edges which cut the cut represent the
reason to the conflict.
Learning: Cuts
Learning: Cuts
¬v5
¬v9
v4 ¬v1
v12
v7
¬v8
v19
¬v6
v3
¬v3 v15
Conflict
Learning: Conflict Clauses
¬v5
¬v9
v4 ¬v1
v12
v7
¬v8
v19
¬v6
v3
¬v3 v15
Conflict
Conflict Side
Reason Side
v7,¬v5,v15 are the reason for the conflict. Adding the clause will prevent it in
the future.7 5 15(¬v v ¬v )
(v5,¬v7,v3)
(¬v7,¬v15,¬v3)
(¬v7,v5,¬v15)
Learning: Conflict Clausesv1
v2 v2
v3 v3 v3v3
0
00
00
1
1 1
The clause (v2,v3) is
created after a conflict
The search tree is pruned accordingly
Learning: Conflict Clauses Prevent the reason to the conflict.
Consists of the negation to the reason literals. Prunes the search tree.
Different cuts yield different conflict clauses. We choose cuts such that:
Conflict clause includes one variable from the top level.
It is a unit clause after backtracking one level. The new problem is equivalent to the original.
Learning: Implication Graph(v9,¬v5)
(v9,¬v8)
(v9,v12)
(v5,v7)
(v5,¬v7,v3)
(¬v4,¬v1)
(v1,¬v12,v19)
(v1,¬v6)
(v6,¬v19,v8,v15)
(¬v7,¬v15,¬v3)
¬v5
¬v9
v4 ¬v1
v12
v7
¬v8
v19
¬v6
v3
¬v3 v15
Conflict
1 UIP2 UIP3 UIP
Non –Chronological Backtracking Backtrack multiple levels instead of one. Use conflict clause to determine the level
Backtrack to the minimum level where the clause is still asserting.
Emphasis on recent learning.
v8 ¬v10 v3 v7 v14
¬v2 v5 v1
¬v4 v21 ¬v12 ¬v15
v19 v18 v32
¬v6 v16 ¬v9 ¬v14
v8 ¬v10 v3 v7
¬v2 v5 v1 v9
Conflict Clause
(v10,¬v7,v2,v9)
Non –Chronological Backtracking
v8 ¬v10 v3 v7 v14
¬v2 v5 v1
¬v4 v21 ¬v12 ¬v15
v19 v18 v32
¬v6 v16 ¬v9 ¬v14
v8 ¬v10 v3 v7
¬v2 v5 v1 v9
v8
v6
v2
v19
v4
1
0
0
1
1
v8
v2
v9
0
1
1
Branching - VSIDS(Variable State Independent Decaying Sum) Counter for each literal
Updated when clauses are added Branch on the literal with the highest rank Divide counters periodically Favors recent conflict clauses.
Recent gained knowledge
Restarts
After a fixed interval Lengthen interval after each restart
Nullify all branching and implications Keep conflict clauses Leads to a new branching order
New score for the literals Hopefully a better order
Branching for BMC
Add static ordering Forward backward
Branch on latches \ inputs
0 0 0 0 1 1( ... ) ( ) ( , ) ... ( , ) ( )k k k k kv v I v R v v R v v P v
Branching for BMC
Create independent sub-problems
0 1 1 1 1( , ) ... ( , ) ( , ) ... ( , )i i i i k kR v v R v v R v v R v v
un-SAT Core
1 2( , )v v2 3( , )v v 1 3 4( , , )v v v1 3( , )v v 2 3( , )v v1 2 4( , , )v v v 1 2 3( , , )v v v 2 3( , )v v2 3( , )v v
¬v2
¬v3
¬v1
v1
Conflict
2( )v
2 3( , )v v
un-SAT Core
1 2( , )v v2 3( , )v v 1 3 4( , , )v v v1 3( , )v v 2 3( , )v v1 2 4( , , )v v v 1 2 3( , , )v v v 2 3( , )v v2 3( , )v v
v2
¬v3 ¬v2
Conflict
2( )v
( )
2( )v
2 3( , )v v
un-SAT Core
1 2( , )v v2 3( , )v v 1 3 4( , , )v v v1 3( , )v v 2 3( , )v v1 2 4( , , )v v v 1 2 3( , , )v v v 2 3( , )v v2 3( , )v v
2( )v
( )
2( )v
2 3( , )v v
un-SAT Core
1 2( , )v v2 3( , )v v 1 3 4( , , )v v v1 3( , )v v 2 3( , )v v1 2 4( , , )v v v 1 2 3( , , )v v v 2 3( , )v v2 3( , )v v
2( )v
( )
2( )v
2 3( , )v v
Easily obtained from a DPLL SAT solver Requires saving the resolution graph
Branching – BerkMin \HaifaSAT SAT as abstraction / refinement Stack of added conflict clauses
Order clauses chronologically (BerkMin) Order on clauses topologically (HaifaSAT)
Branch on a literal from the top clause
1 2 ... n rescl cl cl cl 1 2 ... ncl cl cl
rescl
The Rise and Fall of Parallel SAT Solving VERY appealing! Obstacles
Dependencies Communication
Parallelized BCP() Centralized Appealing since bcp() is 90% of the runtime Useless since bcp() is only 90% of the runtime
BC
P()
The Rise and Fall of Parallel SAT Solving Parallelize for BMC Create “independent” formulae
Take advantage of symmetry Share learned clauses
Inference is local
0 1 1 1 1( , ) ... ( , ) ( , ) ... ( , )i i i i k kR v v R v v R v v R v v
