Carnegie Mellon University

Boolean SatisfiabilityBoolean Satisfiabilitywithwith

Transitivity ConstraintsTransitivity Constraints

http://www.cs.cmu.edu/~bryant

Randal E. BryantMiroslav N. Velev

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Outline

Application DomainApplication Domain Verify correctness of a pipelined processor Based on Burch-Dill correspondence checking

Burch & Dill, CAV ‘94

Verification TaskVerification Task Decide validity of formula in logic of equality with

uninterpreted functions Translate into equational logic

Propositional logic with equations of form vi = vj

Bryant, German & Velev, CAV ’99Goel, Sahid, Zhou, Aziz, & Singhal, CAV ‘98

New ContributionNew Contribution Efficient handling of transitivity constraints

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=

f

T

F

T

F

f T

F

=

e1

e0x0

d0

Decision ProblemLogic of Equality with Uninterpreted Functions (EUF)Logic of Equality with Uninterpreted Functions (EUF)

Truth Values Dashed Lines Model control signals

Domain Values Solid lines Model data words

TaskTask Determine whether formula is universally valid

True for all interpretations of variables and function symbols

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Eliminating Function Applications

Verification TaskVerification Task Prove: x = f(f(x)) x = f(f(f(x))) x = f(x)

Instance of: x = y x = f(y) x = f(x)

Ackermann’s MethodAckermann’s Method Replace: f(x) f1 f( f(x)) f2 f(f( f(x))) f3

Gives: x = f2 x = f3 x = f1

Functional Consistency ConstraintsFunctional Consistency Constraints x = f1 f1 = f2

f1 = f2 f2 = f3

x = f2 f1 = f3

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Eliminating Funct. Apps. (cont.)

Equational FormulaEquational Formula Complement of substituted formula + consistency

constraints Clauses Origin x = f2 x = f3 x f1 [x = f2 x = f3 x

= f1]

(x f1 f1 = f2) x = f1 f1 = f2

(f1 f2 f2 = f3) f1 = f2 f2 = f3 (x f2 f1 = f3) x = f2 f1 = f3

Verification TaskVerification Task Prove that equational formula is not satisfiable

x = f2 x = f3 x f1

(x f1 f1 = f2) (f1 f2 f2 = f3)

(x f2 f1 = f3)

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Solving Equational Formulas

HistoricallyHistorically E.g., Nelson & Oppen ‘80 Create special purpose search engine

Davis-Putnam searchData structure to maintain equivalence classes

QuestionQuestion Can we translate problem into pure propositional logic?

Would enable use of BDDs or SAT checkers

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Replacing Equations by Variables

Relational VariablesRelational Variables Goel, Sahid, Zhou, Aziz, & Singhal, CAV ‘98 Replace vi = vj by propositional variable ei,j

Propositional Formula Propositional Formula FFsatsat

Relabeling: x v1 f1 v2 f2 v3 f3 v4 Clauses Origin e13 e14 e12 x = f2 x = f3 x f1

(e12 e23) (x f1 f1 = f2) (e23 e34) (f1 f2 f2 = f3)

(e13 e24) (x f2 f1 = f3)

e13 e14 e12

(e12 e23) (e23 e34)

(e13 e24)

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Need for Transitivity Constraints

Propositional Formula Propositional Formula FFsatsat

e13 e14 e12

(e12 e23) (e23 e34)

(e13 e24)

SolutionSolutione13 = true e14 = true e12 = false e23 = true e34 = true e24 = true

Transitivity Violation in SolutionTransitivity Violation in Solutione13 = true e23 = true e12 = false

Corresponds to x = f2 and f2 = f1 but x f1

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Handling Transitivity Constraints: Goel, et al., CAV ’98ComplexityComplexity

Finding solution to Fsat that satisfies transitivity constraints is NP-Hard

Even when Fsat represented as OBDD

Their methodTheir method Enumerate implicants of Fsat from OBDD representation Discard any implicant that contains transitivity violation Eventually find solution or run out of implicants

Our ExperimentsOur Experiments Works well for small benchmarks Far too many implicants for larger benchmarks

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Handling Transitivity Constraints: Our MethodIdeaIdea

Generate propositional formula Ftrans expressing transitivity constraints

Satisfy formula Fsat Ftrans

Using OBDDs or SAT checker

Sources of EfficiencySources of Efficiency Equational structure very sparse

Far fewer than n(n-1)/2 relational variablesOnly need to enforce limited set of transitivity constraints

With OBDDs, can reduce set of relational variablesOnly those in true support of Fsat

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Benchmark Circuits

Single Issue Pipeline: Single Issue Pipeline: 1xDLX-C1xDLX-C Analogous to DLX model in Hennessy & Patterson Verified in ‘94 by Burch & Dill

Dual Issue Pipeline #1: Dual Issue Pipeline #1: 2xDLX-CA2xDLX-CA Second pipeline can only handle R-R and R-I instructions Burch (DAC ‘96) required 28 manual case splits, 3

commutative diagrams, and 1800s.

Dual Issue Pipeline #2: Dual Issue Pipeline #2: 2xDLX-CC2xDLX-CC Second pipeline can also handle all instructions

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Verifying Original Benchmarks

None Require Transitivity ConstraintsNone Require Transitivity Constraints Fsat is unsatisfiable in every case Circuits don’t make use of transitivity in forwarding or stall

decisions

PerformancePerformanceCircuit OBDD Secs. FGRASP Secs.1xDLX-C 0.2 32xDLX-CA 11. 1762xDLX-CC 29. 5,035

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Transitivity Benchmarks

Modified, but Correct CircuitsModified, but Correct Circuits Modify forwarding logicESrc1=MDest

ESrc1=MDest (ESrc1=ESrc2 ESrc2=MDest)Equivalent under transitivity

Circuit names 1xDLX-Ct, 2xDLX-CAt, 2xDLX-CCt

Buggy CircuitsBuggy Circuits 100 buggy versions of 2xDLX-CC

Each contains single modification of control logic Must ensure that counterexample satisfies transitivity

constraints

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1xDLX-C Equation Structure

VerticesVertices For each vi

13 different register identifiers

EdgesEdges For each equation Control stalling and

forwarding logic 27 relational variables

Out of 78 possible

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2DLX-CCt Equation Structure

EquationsEquations Between 25

different register identifiers

143 relational variables

Out of 300 possible

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Graph Interpretation of Transitivity

Transitivity ViolationTransitivity Violation Cycle in graph Exactly one edge has ei,j = false

== ==

==

==

====

==

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Exploiting Chords

ChordChord Edge connecting two non-

adjacent vertices in cyclePropertyProperty

Sufficient to enforce transitivity constraints for all chord-free cycles

If transitivity holds for all chord-free cycles, then holds for arbitrary cycles

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Enumerating Chord-Free Cycles

StrategyStrategy Enumerate chord-free cycles in graph Each cycle of length k yields k transitivity constraints

• • •

1 2 k• • •

ProblemProblem Potentially exponential number of chord-free cycles

2k+k chord-free cycles

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Adding Chords

StrategyStrategy Add edges to graph to reduce number of chord-free cycles

• • •

1 2 k• • • 2k+k chord-free cycles

2k+1 chord-free cycles

Trade-OffTrade-Off Reduces formula size Increases number of relational variables

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Chordal Graph

DefinitionDefinition Every cycle of length > 3 has a

chord

GoalGoal Add minimum number of edges

to make graph chordal

Relation to Sparse Gaussian Relation to Sparse Gaussian EliminationElimination

Choose pivot ordering that minimizes fill-in

NP-hard Simple heuristics effective

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Adding Chordal Edges to 1xDLX-C

OriginalOriginal 27 relational variables 286 cycles 858 clauses

AugmentedAugmented 33 relational

variables 40 cycles 120 clauses

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Adding Chordal Edges to 2xDLX-CCtOriginalOriginal

143 relational variables

2,136 cycles 8,364 clauses

AugmentedAugmented 193 relational

variables 858 cycles 2,574 clauses

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SAT Checker on Good Circuits

StrategyStrategy Run on clauses encoding Fsat and Ftrans

FGRASP Performance (Secs.)FGRASP Performance (Secs.)Circuit Fsat Fsat Ftrans

1xDLX-C 3 41xDLX-Ct --- 92xDLX-CA 176 1,2752xDLX-CAt --- 8962xDLX-CC 5,035 9,9322xDLX-CCt --- 15,003

ObservationObservation Much more challenging with transitivity constraints imposed

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SAT Checker on Buggy Circuits

Performance Penalty with Transitivity ConstraintsPerformance Penalty with Transitivity Constraints Geometric average slowdown = 2.3X

1

10

100

1000

10000

100000

1 10 100 1000 10000 100000

Without Constraints

With

Con

stra

ints

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Using OBDDs

Possible StrategyPossible Strategy Build OBDDs for Fsat and Ftrans

Compute Fsat Ftrans Find satisfying solution

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Limitation of OBDDsOBDD for OBDD for FFtranstrans can be of exponential size can be of exponential size

Regardless of variable ordering Formal result

Relational variables forming k X k mesh OBDD representation has (2k/4) nodes

Experimental ResultsExperimental Results Unable to build OBDD of Ftrans for large benchmarks

6 X 6 mesh6 X 6 mesh

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Better Use of OBDDsStrategyStrategy

Build OBDD for Fsat Determine relational variables in true support

Easy with OBDD

Generate Ftrans for these variables Compute conjunction and find satisfying solution

PerformancePerformance When Fsat unsatisfiable, no further steps required

For other benchmarks, yields tractable Ftrans

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2DLX-CCt Reduced Constraints

Relational variablesRelational variables 46 original 6 chordal

OBDD RepresentationOBDD Representation 7,168 nodes

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Reduced Constraints: Average-Case Buggy CircuitRelational VariablesRelational Variables

17 original 3 chordal

OBDD RepresentationOBDD Representation 70 nodes

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Reduced Constraints: Worst-Case Buggy CircuitRelational variablesRelational variables

52 original 16 chordal

OBDD RepresentationOBDD Representation 93,937 nodes

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OBDDs on Good Circuits

CUDD Performance (Secs.)CUDD Performance (Secs.)Circuit Time1xDLX-C 0.21xDLX-Ct 22xDLX-CA 112xDLX-CAt 1092xDLX-CC 292xDLX-CCt 441

ObservationObservation Significantly more effort with transitivity constraints Better performance than FGRASP

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1

10

100

1000

10000

1 10 100 1000 10000

Without Constraints

With

Con

stra

ints

OBDDs on Buggy Circuits

Performance Penalty with Transitivity ConstraintsPerformance Penalty with Transitivity Constraints Geometric average slowdown = 1.01X

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Conclusion

Equational Formulas can be Solved by Propositional Equational Formulas can be Solved by Propositional MethodsMethods Exploit sparse structure of equations

Reduces number of variablesReduces formula size

With OBDDs, can identify essential relational variables In true support of Fsat

Can use either SAT checker or OBDDsOBDDs do best for unsatisfiable formulas

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Extension

Formulas with Ordering ConstraintsFormulas with Ordering Constraints Constraints of form vi vj

Symbolic SolutionSymbolic Solution Introduce variables ai,j and aj,i for each constraint vi rel vj

ai,j true when vi vj Solution defines partial ordering

ApplicationApplication Scheduling problems