This document discusses parallel SAT solving using instance decomposition. It outlines how incorrect UNSAT answers can arise due to clause sharing and elimination techniques across solvers. The document then presents a formal model for instance decomposition based solvers using label functions to consistently share clauses while allowing certain clause elimination rules. Finally, it describes how unsatisfiability proofs can be generated by recording state transitions in the formal model.

1. An Expressive Model for Instance
Decomposition Based Parallel SAT Solvers
Tobias Philipp
Knowledge Representation and Reasoning Group
Technische Universität Dresden

2. Outline
(1) The Instance Decomposition Approach
(2) Incorrect UNSAT Answers
(3) A Formal Model
(4) Unsatisfiability Proofs
1

3. The Instance Decomposition Approach
F
2

4. The Instance Decomposition Approach
F
2

5. The Instance Decomposition Approach
F
F0
x
F1
x
2

6. The Instance Decomposition Approach
F
F0
x
F1
x
2

7. The Instance Decomposition Approach
F
F0
x
F1
F10
y
F11
y
x
2

8. The Instance Decomposition Approach
Solver1
Solver2
Solver3
Solver4
Solver5
F
F0
F1
F10
F11
F
F0
x
F1
F10
y
F11
y
x
2

9. The Instance Decomposition Approach
Solver1
Solver2
Solver3
Solver4
Solver5
F
F0
F1
F10
F11
SAT
F
F0
x
F1
F10
y
F11
y
x
2

10. The Instance Decomposition Approach
Solver1
Solver2
Solver3
Solver4
Solver5
F
F0
F1
F10
F11
UNSAT
UNSAT
F
F0
x
F1
F10
y
F11
y
x
2

11. The Instance Decomposition Approach
Solver1
Solver2
Solver3
Solver4
Solver5
F
F0
F1
F10
F11
UNSAT
UNSAT
UNSATF
F0
x
F1
F10
y
F11
y
x
2

12. The Instance Decomposition Approach
Solver1
Solver2
Solver3
Solver4
Solver5
F
F0
F1
F10
F11
UNSAT
UNSAT
UNSATF
F0
x
F1
F10
y
F11
y
x
Clause Sharing
Formula Simplifications
2

13. Incorrect UNSAT Answers

14. UNSAT may be Incorrect: Sharing Partition
Constraints
F0 is satisfiable.
p(F0) = (F ∧ x), (F ∧ x)
F0 ∧ x ∧ x is
unsatisfiable.
F0 F0 ∧ x F0 ∧ x
F0 ∧ x F0 ∧ x F0 ∧ x
F0 ∧ x ∧ x F0 ∧ x F0 ∧ x
UNSAT
3

15. Blocked Clauses
C is blocked in F iff there is L ∈ C st
all resolvents of C with resolution candidates in F upon L are
tautologies
F = (x ∨ y) ∧ (x ∨ y) ∧ (x ∨ z) ∧ (y ∨ z)
C = (x ∨ z) is blocked in F by z
D = (y ∨ z) is blocked in F by z
4

16. UNSAT can be Incorrect: Applying
Clause Addition Techniques in Two Solvers
F = (x ∨ y) ∧ (x ∨ y) ∧ (x ∨ z) ∧ (y ∨ z)
C = (x ∨ z) is blocked in F by z
D = (y ∨ z) is blocked in F by z
F is satisfiable:
I = {x, y, z}
J = {x, y, z}
I |= C, J |= D
F ∧ C ∧ D is unsatisfiable
F F
F ∧ C F
F ∧ C F ∧ D
F ∧ C ∧ D F ∧ D
UNSAT
5

17. UNSAT can be Incorrect: Applying Clause
Elimination and Addition Techniques in One
Solver
F = (x ∨ y) ∧ (x ∨ y) ∧ (x ∨ z) ∧ (y ∨ z)
C = (x ∨ z) is blocked in F by z
D = (y ∨ z) is blocked in F by z
F is satisfiable:
I = {x, y, z}
J = {x, y, z}
F ∧ C is satisfiable:
J = {x, y, z}
F ∧ C ∧ D is unsatisfiable, since J |= D
F ∧ C F ∧ C
F F ∧ C
F ∧ D F ∧ C
F ∧ D ∧ C F ∧ C
UNSAT
6

18. Our Formalism
With Clause Elimination and
Label-Based Clause Sharing

19. Label Based Clause Sharing can Handle
Partition Constraints
Label function
C : N × Clauses → 2Labels
F : N → 2Labels
Label based clause sharing
Solveri Solverj
C
C(j, C) ⊆ F(i)
Partition constraints
C(C, i) = {unsafe}, if C is a partition constraint
C(C, i) = ∅, otherwise
F(i) = ∅
7

20. Consistent Label Functions Can Handle
Clause Elimination Techniques
is consistent for F0, . . . , Fn iff
Fi ≡sat Fi ∧
(j,C)⊆ (i),
C∈Fj,
0≤j≤n
C
Proposition: Position Based Tagging and Boolean Tagging are
consistent label functions.
8

21. Instance Decomposition Based Solvers can
be Described As State Transition Systems
Partition function pn(F0) = (F1, F2, . . . , Fn)
F0 ≡ F1 ∨ F2 ∨ . . . ∨ Fn.
Cooperation tree E E ⊆ {0, . . . , n} × {0, . . . , n}
F0 is satisfiable, if Fi is satisfiable,
Fi is unsatisfiable, if Fj is unsatisfiable for all children j of i.
Label functions:
9

22. The States in the
Instance Decomposition Model
States
F0 F1
. . . Fn E
Initial state initpn, ,E(F0) = (F0, . . . , Fn, , E)
pn(F0) = (F1, . . . , Fn) s.t. F0 ≡ F1 ∨ . . . ∨ Fn.
is a consistent label function for F0, . . . , Fn.
E is a cooperation tree for F0, . . . , Fn, where F0 is the root.
Terminal states SAT, UNSAT
10

23. SAT Termination Rule
F0 F1
. . . Fi
. . . Fn E
SAT
some Fi is satisfiable
11

24. UNSAT Termination Rule
F0 F1
. . . Fi
. . . Fn E
UNSAT
∅ ∈ F0
12

25. LOCUNSAT Rule
F0 F1
. . . Fi
. . . Fn E
F0 F1
. . . Fi ∧ ∅ . . . Fn E
all children of Fi in E contain ∅,
(i, ∅) := (i)
13

26. Labeled Resolution Rule
F0 F1
. . . Fi
. . . Fn E
F0 F1
. . . Fi ∧ D . . . Fn E
D = C ⊗ C ,
C, C ∈ Fi,
(i, D) := (i, C) ∪ (i, C )
14

27. Clause Deletion Rule
F0 F1
. . . Fi
. . . Fn E
F0 F1
. . . Fi
. . . Fn E
Fi ⊂ Fi and Fi ≡sat Fi
15

28. Label Based Clause Sharing Rule
F0 F1
. . . Fi
. . . Fn E
F0 F1
. . . Fi ∧ C . . . Fn E
C ∈ Fj and (j, C) ⊆ (i),
(i, C) := (j, C)
16

29. Properties of the
Instance Decomposition Model
Clause Elimination Techniques:
Blocked Clause Elimination
Variable Elimination
Instances: a restricted form of PCASSO
Key Invariant: If initpn,E, (F0)
∗
; (F0, . . . , Fn, , E):
F0 ≡sat F0,
is consistent for F0, . . . , Fn,
E is a cooperation tree for F0, . . . , Fn.
Theorem: The Instance Decomposition Model is sound.
We can use clause elimination techniques in PCASSO.
17

30. Unsatisfiability Proof
F SAT Solver
SAT, I
UNSAT, P
Checker
Checker
Execution traces correspond to unsatisfiability results
Record changes in the formulas
18

31. Conclusion
Contributions
A sound formalism for instance decomposition based
parallel SAT solvers
consistent label functions
A proof format
Future Work
How can we use clause elimination and addition methods in
all solver incarnations?
How can we construct clausal proofs?
19

32. An Expressive Model for Instance
Decomposition Based Parallel SAT Solvers
Tobias Philipp
Knowledge Representation and Reasoning Group
Technische Universität Dresden
Thank you for your attention.

34. Position-Based Tagging
F
F0
x
F1
F10
y
F11
y
x
Position-based label function
F( ) = { }
F(0) = { , 0} C(0, {x}) = {0}
F(1) = { , 1} C(w, {x}) = {1} for w ∈ {1, 10, 11}
F(10) = { , 1, 10} C(10, {y}) = {10}
F(11) = { , 1, 11} C(11, {y}) = {11}
20

35. F = ˙{{x, y}˙}
F0 = ˙{{x, y}, {x}˙}
F1 = ˙{{x, y}, {x}˙}
F10 = ˙{{x, y}, {x}, {y}˙}
F11 = ˙{{x, y}, {x}, {y}˙}
F
F0
x
F1
F10
y
F11
y
x
21

36. Proof Format
Position-based tagging
Labeled and extended resolution derivation of Cn in F
(Ci | 1 ≤ i ≤ n) such
Ci ∈ F,
Ci is a labeled resolvent from two previous clauses Cj and Ck
where j < i and k < n, or
Ci is the empty clause with label u and for every a ∈ Σ there
is the empty clause labeled with ua.
Proof Checking
Check correct partition constraints and labels in F
Check derivation
Contains empty clause with no labels attached
22

37. Solveri Solverj
23

38. Partition Functions
Partition function pf (F) = (F1, . . . , Fn)
F ≡ F1 ∨ . . . ∨ Fn
F = ˙{{x, y}˙}
F0 = ˙{{x, y}, {x}˙}
F1 = ˙{{x, y}, {x}˙}
F10 = ˙{{x, y}, {x}, {y}˙}
F11 = ˙{{x, y}, {x}, {y}˙}
F
F0
x
F1
F10
y
F11
y
x
24

39. Clause Addition Techniques can make
Formulas Unsatisfiable
F0 ≡ (F0 ∧ x) ∨ (F0 ∧ x)
Assume F0 ≡sat F0 ∧ x.
F0 ∧ x ∧ x is unsatisfiable.
Result: There is no consistent label
function that allows to share x.
F F
F F
F ∧ F F
UNSAT
25