Grid-Based SAT Solving with Iterative Partitioning and Clause - PowerPoint PPT Presentation

Grid-Based SAT Solving with Iterative Partitioning and Clause Learning Antti E. J. Hyv¨ arinen, Tommi Junttila, and Ilkka Niemel¨ a Department of Information and Computer Science Aalto University, School of Science Finland September 16, 2011

Motivation ◮ We study solving of challenging SAT instance ◮ Using distributed computing with approx. 64 cores and 1h time limit on each running job ◮ In a setting which adapts straightforwardly to conflict driven clause learning SAT solvers Part-Tree-Learn time (in s) 20000 15000 10000 5000 0 0 5000 10000 15000 20000 MiniSat 2.2.0 time (in s) ◮ System is based on MiniSat 2.2.0 ◮ × — sat, � — unsat September 16, 2011

Randomness in SAT Solver Runtimes 1 0.9 0.8 probability P ( T ≤ t ) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 100 1000 10000 time t (in seconds) Modern SAT solvers exhibit high variance in solving a SAT instance. The random distributions for two instances (green and red) are shown above. September 16, 2011

Algorithm Portfolios Many modern parallel SAT solvers are based on having several algorithms compete on solving the same instance. 100 Speedup (simulated) 10 1 50 100 150 200 250 300 Number of CPUs N The obtained speedup depends heavily on the formula (no learned clause sharing in the simulated results above) September 16, 2011

Divide-and-conquer approaches ◮ Algorithm portfolios do not always scale ◮ Dividing the search space has been suggested as an alternative ◮ A formula φ can be divided to n derived formulas φ ∧ Co 1 , . . . , φ ∧ Co n s.t. φ ≡ ( φ ∧ Co 1 ) ∨ . . . ∨ ( φ ∧ Co n ) . ◮ For example, pick log 2 n variables from φ and conjoin them in all polarities with φ . If n = 2, then φ Co 1 = ( x 1 ∧ x 2 ) Co 2 = ( x 1 ∧ ¬ x 2 ) Co 3 = ( ¬ x 1 ∧ x 2 ) φ ∧ Co 1 φ ∧ Co 3 φ ∧ Co 2 φ ∧ Co 4 Co 4 = ( ¬ x 1 ∧ ¬ x 2 ) Formulas φ ∧ Co i can be solved independently in parallel. September 16, 2011

Risks in Divide-and-Conquer In particular for unsat instances the divide-and-conquer approach might result in worse speedup than the algorithm portfolio approach. Suppose that a formula φ with run time distribution q ( t ) is divided to n derived formulas φ i having run time distribution q ( n α t ) , 0 ≤ α ≤ 1. 180 1 probability P ( T ≤ t ) E T 0.9 α =0.5 160 0.8 α =0.6 0.7 α =0.7 140 Expected run time (in s) 0.6 α =0.8 0.5 α =0.9 120 0.4 0.3 100 0.2 0.1 80 0 60 100 1000 10000 40 time t (in seconds) 20 original portfolio 0 div&conq q ( t/n ) 10 20 30 40 50 60 70 80 90 100 experimental div&conq Number of derived formulas n ◮ In pathological cases the run time increases (left) ◮ In practice, divide-and-conquer is no better than portfolio (right) September 16, 2011

On the Iterative Partitioning and Learning The increase in run time can be avoided by running the original formula and the derived formulas simultaneously. This can be generalized recursively. 4 t/o 1 unsat unsat 2 3 unsat ◮ Initially, say, 8 formulas are running simultaneously ◮ As unsat or timeout results are obtained, it would be nice to collect also the learned clauses to further constrain subsequent derived formulas, as indicated by the arrows (right tree). September 16, 2011

Learned Clauses As the constraints used in divide-and-conquer are not (necessarily) logical consequences of the original formula, the clauses learned by the solvers might not be either. ◮ Let φ = ( x 1 ∨ x 2 ∨ x 3 ) ∧ ( x 2 ∨ ¬ x 3 ) and Co 1 = ¬ x 1 . ◮ Then a modern SAT solver simplifies φ ∧ ¬ x 1 to ( x 2 ∨ x 3 ) ∧ ( x 2 ∨ ¬ x 3 ) ◮ Now branching on ¬ x 2 results in learned clause ( x 2 ) which is not a logical consequence of φ as x 1 , ¬ x 2 , ¬ x 3 satisfies φ . x 3 ¬ x 2 @1 λ ¬ x 3 September 16, 2011

Assumption Tagging One approach is to tag constraints with new assumption literals which will be forced false when solving the derived formula. ◮ Let a be a new variable not occurring in φ , and Co 1 = ( a ∨ ¬ x 1 ) ◮ Simplification can now be avoided, and φ ∧ Co 1 = ( x 1 ∨ x 2 ∨ x 3 ) ∧ ( x 2 ∨ ¬ x 3 ) ∧ ( a ∨ ¬ x 1 ) . ◮ The solver may now learn either ( x 1 ∨ x 2 ) , or ( a ∨ x 2 ) , consequences of φ and φ ∧ Co 1 , respectively. ¬ x 1 ¬ a @1 x 3 ¬ x 2 @2 λ ¬ x 3 September 16, 2011

Results on Assumption Tagging Clause sharing with assumption tagging does not give us speedup! Decisions Run time (in s) MiniSat 2.2.0 assumption tagging MiniSat 2.2.0 assumption tagging 1e+10 100000 1e+09 1e+08 10000 1e+07 1e+06 1000 100000 10000 100 1000 100 10 10 100 100001e+061e+081e+10 10 100 1000 10000 100000 MiniSat 2.2.0 no learned clauses MiniSat 2.2.0 no learned clauses Max. 100 000 literals worth of learned clauses per instance ◮ Left: decisions per derived formula decreases ◮ Right: Run time is roughly equal ◮ Top line: Many memory outs September 16, 2011

Flag Tagging To allow for both the simplification and clause sharing, one can set a flag on the constraints. This flag is then inherited at conflict analysis to the learned clause. ◮ Let φ = ( x 1 ∨ x 2 ∨ x 3 ) ∧ ( x 2 ∨ ¬ x 3 ) and Co 1 = ¬ x 1 . ◮ Then SAT solver simplifies φ ∧ ¬ x 1 to ( x 2 ∨ x 3 ) u ∧ ( x 2 ∨ ¬ x 3 ) , where u stands for “unsafe” ◮ branching on ¬ x 2 results in ( x 2 ) u . x 3 ¬ x 2 @1 λ ¬ x 3 September 16, 2011

Results on Flag Tagging decisions time (in s) 1e+09 100000 MiniSat 2.2.0 flag tagging MiniSat 2.2.0 flag tagging 1e+08 1e+07 10000 1e+06 100000 1000 10000 1000 100 100 10 10 100 10000 1e+06 1e+08 10 100 1000 10000 100000 MiniSat 2.2.0 no learned clauses MiniSat 2.2.0 no learned clauses Again max. 100 000 literals worth of learned clauses ◮ While the number of independent learned clauses is smaller, the approach gives now also speed-up September 16, 2011

Experimental Setup Learned clauses φ | = c 4 t/o U U 1 U unsat U U U U U U U U U unsat 2 3 unsat September 16, 2011

Experimental Setup Learned clauses φ | = c t/o U U U unsat U U U t/o Simplify U U U U U U unsat unsat September 16, 2011

Results (1) Part-Tree-Learn (time in s) Part-Tree-Learn (time in s) 20000 20000 15000 15000 10000 10000 5000 5000 0 0 0 5000 10000 15000 20000 0 5000 10000 15000 20000 Part-Tree (time in s) CL-SDSAT (time in s) ◮ Adding learning to the partition tree approach gives clear speed-up and solves more instances (left fig.) ◮ The approach usually beats a clause-sharing portfolio approach based on the same solver (right fig.) September 16, 2011

Results (2) Unsolved instances from SAT-COMP 2009 Name Type plingeling Part-Tree-Learn Part-Tree 9dlx vliw at b iq8 Unsat 3256.41 — — 9dlx vliw at b iq9 Unsat 5164.00 — — AProVE07-25 Unsat — 9967.24 9986.58 dated-5-19-u Unsat 4465.00 2522.40 4104.30 eq.atree.braun.12.unsat Unsat — 4691.99 5247.13 eq.atree.braun.13.unsat Unsat — 9972.47 12644.24 gss-24-s100 Sat 2929.92 3492.01 1265.33 gss-26-s100 Sat 18173.00 10347.41 16308.65 gus-md5-14 Unsat — 13890.05 13466.18 ndhf xits 09 UNSAT Unsat — 9583.10 11769.23 rbcl xits 09 UNKNOWN Unsat — 9818.59 8643.21 rpoc xits 09 UNSAT Unsat — 8635.29 9319.52 sortnet-8-ipc5-h19-sat Sat 2699.62 4303.93 20699.58 total-10-17-u Unsat 3672.00 4447.26 5952.43 total-5-15-u Unsat — 18670.33 21467.79 September 16, 2011

Conclusions Iterative partitioning is an approach based on search space splitting which solves hard SAT instances, and ◮ Does not suffer from the increased expected run time problem ◮ Is faster than a comparable, efficient portfolio solver ◮ To my knowledge, the first time this has been achieved in “modern history” ◮ Clause sharing improves the approach further, when implemented with care September 16, 2011