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This paper discusses the generation of minimum transitivity constraints for deciding equality logic in polynomial time, providing an improvement over previous methods. It explores different approaches and their effectiveness, and presents results from experiments and benchmarks.
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Generating minimum transitivity constraints in P-time for deciding Equality Logic Ofer Strichman and Mirron Rozanov Technion, Haifa, Israel Technion
Deciding Equality Logic (TE) • The eager approach: TE! Pr • Bryant & Velev [BV-CAV’00] – Boolean satisfiability with transitivity constraints. • Meir and Strichman [MS-CAV’05] – Yet another decision procedure for equality logic. • This work: a ‘closure’ on [MS-CAV’05] Technion
Basic notions E: x = yÆy = zÆzx (non-polar) Equality Graph: y x z Technion
From Equality to Propositional Logic[BV-CAV'00] – the Sparse method x1 E :x1 = x2Æx2 = x3Æx1x3 sk : e1,2 Æe2,3Æ:e1,3 • Encode all edges with Boolean variables • Add transitivity constraints e1,2 e1,3 x2 e2,3 x3 Technion
From Equality to Propositional Logic[BV-CAV'00] – the Sparse method x1 E :x1 = x2Æx2 = x3Æx1x3 sk : e1,2 Æe2,3Æ:e1,3 • Transitivity Constraints: For each cycle of size n, forbid a true assignment to n-1 edges T S = (e1,2Æe2,3!e1,3) Æ (e1,2Æe1,3!e2,3) Æ (e1,3Æe2,3!e1,2) Check: skÆTS e1,2 e1,3 x2 e2,3 x3 Technion
From Equality to Propositional Logic[BV-CAV'00] – the Sparse method • Thm-1: It is sufficient to constrain chord-free simple cycles • There can be an exponential number of chord-free simple cycles… e2 e5 e1 e3 e4 Technion
From Equality to Propositional Logic[BV-CAV'00] – the Sparse method • Make the graph ‘chordal’. • In a chordal graph, it is sufficient to constrain only triangles. • Polynomial # of edges and constraints. • # constraints = 3 £ #triangles Technion
An improvement[MS-CAV’05] – the RTC method • So far we did not consider the polarity of the edges. • Assuming E is in Negation Normal Form E: x = yÆy = zÆzx y (polar) Equality Graph: = = x z Technion
e3 e1 e2 An improvementReduced Transitivity Constraints (RTC) • Here, T R = e3Æe2!e1 is sufficient • This is only true because of monotonicity of NNF z Allowing e.g. :x = z, x = y, zy F T ’:x = z, x = y, z = y T = = y x T Technion
Definitions • Dfn-1: A contradictory cycle is a cycle with exactly one disequality edge. • Dfn-2: A contradictory Cycle C is constrainedunder T if T does not allow such an assignment. T T C = T T F Technion
Main theorem[MS-CAV’05] • LetT Rbe a conjunction of transitivity constraints. • IfT Rconstrains all simple contradictory cycles thenE is satisfiable iff skÆT R is satisfiable The Equality Formula Technion
Transitivity: 5 constraints RTC: 0 constraints T Transitivity: 5 constraints RTC: 1 constraint T T T F Technion
Applying RTC • How can we use the theorem without enumerating contradictory cycles ? • Answer: • Consider the chordal graph. • Still – which triangles ? which constraints? Technion
The RTC solution [MS-CAV’05] • 1) Exp # cycles to traverse 2) Not all cycles are simple. • Solution to 1): Stop before adding an existing constraint • Solution to 2): Explore only simple cycles • These solutions cannot be combined. x2 x0 x4 cache: e0,2 Æe1,2 e0,1 e1,3 Æe2,3 e1,2 e2,4 Æe3,4 e2,3 e0,2 Æe0,4 e2,4 x1 x3 Technion
x7 Constraining simple contradictory cycles • Focus on each solid edge es separately • - (find its dashed Bi-connected component) 2. Make the graph chordal x2 x0 x4 es x1 x5 x3 x6 Do we need: e5,6Æ e3,6! e3,5 ? Do we need: e3,5Æ e3,6! e5,6 ? Technion
Constraining simple contradictory cycles 3.Remove a vertexxkthat leans on an edge(xi,xj) 4.Is(xi,xj)on a simple cycle withes?O(|E|) 5.If yes, add(ek,iÆ ek,j! ei,j) e5,6 Æe3,6 e3,5 x2 x0 x4 es x1 x5 x3 x6 Technion
Constraining simple contradictory cycles • Remove a vertex vkthat leans on an edge (vi,vj) • Does (vi,vj) on the same simple cycle with es? • If yes, add (ek,iÆ ek,j! ei,j) e5,6 Æe3,6 e3,5 x2 x0 x4 es x1 x5 x3 x6 Technion
Correctness • The set of generated constraints is sufficient. • The set of generated constraints is necessary. Technion
Results – random graphs V=200, E=800, 16 random topologies # constraints: reduction of 17% Run time: reduction of 32% Technion
Results – random graphs V=200, E=800, 16 random topologies # constraints: reduction of 17% Run time: reduction of 32% Technion
SMT benchmarks • Never really finished the implementation… • Our 4-5 experiments with them showed that • We still have a small advantage comparing to the Sparse method. • Yet Yices is much better…. • A result of the Uninterpreted functions. • Are there formulas for which the eager approach still wins? • Generating meaningful equality formulas is hard… Technion
A crafted example 2n assignments satisfysk. None satisfy the theory. Technion
Thank you Technion
ResultsUclid benchmarks* (all unsat) * Results strongly depend on the reduction method of Uninterpreted Functions. Technion
Possible refutations of CNF’s generated by Sparse Transitivity constraints Boolean Encoding Æ T S B B T R P3 P2 P0 P4 T S – T R P1 P2 Constraints of the form e1Æ e2! e3 A P3 proof exists according to the main theorem. Hypothesis: (T S – T R) clauses hardly participate in the proof Thm: B is satisfiable !B Æ (T S – T R) is satisfiable Technion
B T R Average on: 10 graphs, ~890K clauses All Unsat Sparse: ~ 22 sec. RTC: ~ 12 Sec. T S- T R B T R B – Boolean encoding T R –RTC constraints T S –Sparse constraints T S- T R Technion
Summary • The RTC method is ~dominant over the Sparse method. • Open issue: find a P-time algorithm that exploits the full power of the main theorem. Technion
Example: Circuit Transformations Stage 1 • A pipeline processes data in stages • Data is processed in parallel – as in an assembly line • Formal Model: Stage 2 Stage 3 Technion
Example: Circuit Transformations • The maximum clock frequency depends on the longest path between two latches • Note that the output of g is usedas input to k • We want to speed up the design by postponing k to the third stage Technion
Validating Circuit Transformations ? = Technion
Source program z= (x1+y1) (x2+y2); Target program u1=x1+y1;u2=x2+y2;z=u1u2 ; Validating a compilation process Compilation • Need to prove that:(u1=x1+y1 u2=x2+y2 z=u1u2) $z= (x1+y1) (x2+y2) Source Target Technion
Validating a compilation process • Target program u1=x1+y1;u2=x2+y2;z=u1u2 ; • Source program z= (x1+y1) (x2+y2); Compilation • Need to prove that:(u1=x1+y1 u2=x2+y2 z=u1u2) $z= (x1+y1) (x2+y2) g1 f1 f2 f1 f2 g2 Technion
Need to prove that:(u1=x1+y1 u2=x2+y2 z=u1u2) $z= (x1+y1) (x2+y2) g1 f1 f2 f1 f2 g2 Validating a compilation process • Instead, prove: under functional consistency: for every uninterpreted function fx = y!f(x) = f(y) • Which translates to (via Ackermann’s reduction): Technion
Definitions for the proof… • A Violating cycle under an assignment R: • This assignment violates T S but not necessarily T R Either dashed or solid eT1 F T eF T eT2 Technion
More definitions for the proof… • An edge e = (vi,vj) is equal under an assignment iff there is an equality path between vi and vj all assigned T under . Denote: v3 T F T v1 v2 T T Technion
More definitions for the proof… • An edge e = (vi,vj) is disequal under an assignment iff there is a disequality path between vi and vj in which the solid edge is the only one assigned false by . Denote: v3 T F T v1 v2 T T Technion
v3 F T T v1 v2 Proof… • Observation 1:The combinationis impossible if = R(recall:R²T R) • Observation 2: if (v1,v3) is solid, then Technion
Type 1: It is not the case that Assign S (e23) = F Type 2: Otherwise it is not the case that Assign (e13) = T ReConstructing S v3 v3 F F T T T F T T v1 v1 v2 v2 In all other casesS = R Technion
ReConstructing S • Starting from R, repeat until convergence: • (eT) := F in all Type 1 cycles • (eF) := T in all Type 2 cycles • All Type 1 and Type 2 triangles now satisfy T S • B is still satisfied (monotonicity of NNF) • Left to prove: all contradictory cycles are still satisfied Technion
T T Proof… • Invariant: contradictory cycles are not violating throughout the reconstruction. • contradicts the precondition to make this assignment… v3 F T F T v1 v2 Technion
T F Proof… • Invariant: contradictory cycles are not violating throughout the reconstruction. • contradicts the precondition to make this assignment… v3 F T T T v1 v2 Technion
Constraining simple contradictory cycles The constraint e3,6 Æe3,5 e5,6is not added cache: … e5,6 Æe4,6 e4,5 x2 x0 x4 x1 x5 x3 x6 Open problem: constrain simple contradictory cycles in P time Technion
Constraining simple contradictory cycles the constraint e3,6 Æe3,5 e5,6is not added, though needed Suppose the graph has 3 more edges Here we will stop, although … cache: … e5,6 Æe4,6 e4,5 x2 x0 x4 x1 x5 x3 x6 Open problem: constrain simple contradictory cycles in P time Technion