1. Presented by Ed Clarke Slides borrowed from P. Chauhan and C. Bartzis
GRASP SAT solver
J. Marques-Silva and K. Sakallah
Presented by Ed Clarke
Slides borrowed from P. Chauhan and C. Bartzis

2. SAT Solvers have made some progress…

3. Conjunctive Normal Form (CNF)
Atomic proposition: x1, x2, x3,…
Literal: xi or xi where xi is atomic
Clause: l1  l2  …  ln where li is a literal
CNF formula: 1  2  …  m where i is a clause
Example:
(x2  x3)  (x1  x4  x3)  (x2  x4)

4. Putting a formula in CNF
Extract from truth table of formula
Put in Negation Normal Form (using De Morgan’s laws) and expand using distributivity of  over  :
p  (q  r)  (p  q)  (p  r)
Put in Negation Normal Form and introduce new variable for each conjunctive subformula. (equi-satisfiable but not logically equivalent)

5. SATisfying assignment!
What is SAT?
Given a propositional formula in CNF, find an assignment to boolean variables that makes the formula true
E.g.
1 = (x2  x3)
2 = (x1  x4)
3 = (x2  x4)
A = {x1=0, x2=1, x3=0, x4=1}
SATisfying assignment!

6. Why is SAT important?
Fundamental problem from theoretical point of view
Numerous applications
CAD, VLSI
Optimization
Model Checking and other kinds of formal verification
AI, planning, automated deduction

7. Terminology CNF formula  Assignment A x1,…, xn: n variables
1,…, m: m clauses
Assignment A
Set of (x,v(x)) pairs
|A| < n  partial assignment {(x1,0), (x2,1), (x4,1)}
|A| = n  complete assignment {(x1,0), (x2,1), (x3,0), (x4,1)}
|A= 0  unsatisfying assignment {(x1,1), (x4,1)}
|A= 1  satisfying assignment {(x1,0), (x2,1), (x4,1)}
|A= X  unresolved {(x1,0), (x2,0), (x4,1)}
1 = (x2  x3)
2 = (x1  x4)
3 = (x2  x4)
A = {x1=0, x2=1, x3=0, x4=1}
Manually write down the examples of all types of assignments
For unsat, same as A but with x1=1

8. Terminology
An assignment partitions the clause database into three classes
Satisfied
Unsatisfied
Unresolved
Free literals: unassigned literals of a clause
Unit clause: unresolved with only one free literal

9. Basic Backtracking Search
Organize the search in the form of a decision tree
Each node is an assignment, called decision assignment
Depth of the node in the decision tree  decision level (x)
 x is assigned to v at decision level d

10. Basic Backtracking Search
Iteration:
Make new decision assignments to explore new regions of search space
Infer implied assignments by a deduction process.
May lead to unsatisfied clauses, conflict. The assignment is called conflicting assignment.
If there is a conflict backtrack

11. DPLL in action    U / /  U /  / / /   conflict x1 = 0@1
1 = (x1  x2  x3  x4)
2 = (x1  x3  x4)
3 = (x1  x2  x3)
4 = (x3  x4)
5 = (x4  x1   x3)
6 = (x2  x4)
7 = (x2  x4 )   U / / x2  U /
x2 =  / / / 
 x3 = 
 x4 =
conflict

12. DPLL in action    / / / U   / / conflict conflict x1 = 0@1
1 = (x1  x2  x3  x4)
2 = (x1  x3  x4)
3 = (x1  x2  x3)
4 = (x3  x4)
5 = (x4  x1   x3)
6 = (x2  x4)
7 = (x2  x4 )   / x2
x2 = /
x2 = / U 
 x3 =
 x4 =  / /
 x4 =
conflict
conflict

13. DPLL in action  / / /       conflict conflict x1 = 0@1 x1 = 1@1
1 = (x1  x2  x3  x4)
2 = (x1  x3  x4)
3 = (x1  x2  x3)
4 = (x3  x4)
5 = (x4  x1   x3)
6 = (x2  x4)
7 = (x2  x4 ) / /
x1 = /   x2 x2 
x2 =
x2 = 
x2 =  x4
 x3 =
 x4 = 
 x4 =
x4 =
conflict
conflict
{(x1,1), (x2,0), (x4,1)}

14. DPLL Algorithm
Deduction
Decision
Backtrack

15. GRASP
GRASP stands for Generalized seaRch Algorithm for the Satisfiability Problem (Silva, Sakallah, ’96)
Features:
Implication graphs for BCP and conflict analysis
Learning of new clauses
Non-chronological backtracking

16. GRASP search template
Decide: simple one, choose the one with most # of sat. clauses
Deduce: do BCP
Diagnose: do learning
Erase: erase assignment at current decision level
Search tree is maintained implicitly in procedure call stack

17. GRASP Decision Heuristics
Procedure decide()
Which variable to split on
What value to assign
Default heuristic in GRASP:
Choose the variable and assignment that directly satisfies the largest number of clauses
Other possibilities exist

18. GRASP Deduction
Boolean Constraint Propagation using implication graphs
E.g. for the clause  = (x  y), if y=1, then we must have x=1
For a variable x occuring in a clause, assignment 0 to all other literals is called antecedent assignment A(x)
E.g. for  = (x  y  z),
A(x) = {(y,0), (z,1)}, A(y) = {(x,0),(z,1)}, A(z) = {(x,0), (y,0)}
Variables directly responsible for forcing the value of x
Antecedent assignment of a decision variable is empty

19. (x) = max{(y)|(y,v(y))A(x)}
Implication Graphs
Nodes are variable assignments x=v(x)
(decision or implied)
Predecessors of x are antecedent assignments A(x)
No predecessors for decision assignments!
Special conflict vertices  have A() = assignments to variables in the unsatisfied clause
Decision level for an implied assignment is
(x) = max{(y)|(y,v(y))A(x)}
Tell them it is assigned to each node in the search tree

20. Example Implication Graph
Current truth assignment:
Current decision assignment:
1 = (x1  x2)
2 = (x1  x3  x9)
3 = (x2  x3  x4)
4 = (x4  x5  x10)
5 = (x4  x6  x11)
6 = (x5  x6)
7 = (x1  x7  x12)
8 = (x1 x8)
9 = (x7  x8   x13) 4 1 3 6 
conflict 2 5

21. GRASP Deduction Process

22. GRASP Conflict Analysis
After a conflict arises, analyze the implication graph at current decision level
Add new clauses that would prevent the occurrence of the same conflict in the future  Learning
Determine decision level to backtrack to, might not be the immediate one  Non-chronological backtracking

23. Learning Determine the assignment that caused conflict 
Backward traversal of the IG, find the roots of the IG in the transitive fanin of 
This assignment is necessary condition for 
Negation of this assignment is called conflict induced clause C()
Adding C() to the clause database will prevent the occurrence of  again

24. Learning C() = (x1  x9  x10  x11) conflict x10=0@3 x5=1@6 4 13 6 
conflict 2 5
C() = (x1  x9  x10  x11)

25. Learning For any node of an IG x, partition A(x) into
(x) = {(y,v(y)) A(x)|(y)<(x)}
(x) = {(y,v(y)) A(x)|(y)=(x)}
Conflicting assignment AC() = causesof(), where
(x,v(x)) if A(x) = 
(x)  [  causesof(y) ]o/w
(y,v(y))  (x)
causesof(x) =

26. Learning Learning of new clauses increases clause database size
Increase may be exponential
Heuristically delete clauses based on a user provided parameter
If size of learned clause > parameter, don’t include it

27. Backtracking Failure driven assertions (FDA):
If C() involves current decision variable, a different assignment for the current variable is immediately tried.
In our IG, after erasing the assignment at level 6, C() becomes a unit clause x1
This immediately implies x1=0

28. Example Implication Graph
Current truth assignment:
Current decision assignment:
1 = (x1  x2)
2 = (x1  x3  x9)
3 = (x2  x3  x4)
4 = (x4  x5  x10)
5 = (x4  x6  x11)
6 = (x5   x6)
7 = (x1  x7  x12)
8 = (x1 x8)
9 = (x7  x8   x13)
C()=(x1  x9  x10  x11) 4 1 3 6 
conflict 2 5

29. Example Implication Graph
Current truth assignment:
Current decision assignment:
1 = (x1  x2)
2 = (x1  x3  x9)
3 = (x2  x3  x4)
4 = (x4  x5  x10)
5 = (x4  x6  x11)
6 = (x5  x6)
7 = (x1  x7  x12)
8 = (x1 x8)
9 = (x7  x8   x13)
C() = (x1  x9  x10  x11)
C()
C()
C()

30. Non-chronological backtracking
Decision level 4 5 x1 6  ' 3
C() 8 9 ’
C() 9 9
C() 7 7
AC(’) = {x9 x10 = x11 =
C() = (x9  x10  x11   x12   x13)

31. Backtracking Backtrack level is given by
 = d-1 chronological backtrack
 < d-1 non-chronological backtrack
 = max{(x)|(x,v(x))AC(’)}

32. Procedure Diagnose()

33. Chaff What’s next? Reduce overhead for constraint propagation
Better decision heuristics
Better learning, problem specific
Better engineering
Chaff