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Answer Set Solving in Practice Torsten Schaub University of Potsdam torsten@cs.uni-potsdam.de Potassco Slide Packages are licensed under a Creative Commons Attribution 3.0 Unported License. Torsten Schaub (KRR@UP) Answer Set Solving in

  1. Tweaking N -Queens A First Refinement At least N queens? Exactly one queen per row and column! queens_0.lp % DOMAIN #const n=4. square(1..n,1..n). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- queen(X,Y), queen(X,Y’), Y < Y’. :- queen(X,Y), queen(X’,Y), X < X’. :- queen(X,Y), queen(X’,Y’), X < X’, X’-X = |Y’-Y|. :- not n { queen(X,Y) }. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 460 / 535

  2. Tweaking N -Queens A First Refinement At least N queens? Exactly one queen per row and column! queens_0.lp % DOMAIN #const n=4. square(1..n,1..n). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- queen(X,Y), queen(X’,Y), X < X’. :- queen(X,Y), queen(X’,Y’), X < X’, X’-X = |Y’-Y|. :- not n { queen(X,Y) }. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 460 / 535

  3. Tweaking N -Queens A First Refinement At least N queens? Exactly one queen per row and column! queens_0.lp % DOMAIN #const n=4. square(1..n,1..n). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- queen(X,Y), queen(X’,Y’), X < X’, X’-X = |Y’-Y|. :- not n { queen(X,Y) }. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 460 / 535

  4. Tweaking N -Queens A First Refinement At least N queens? Exactly one queen per row and column! queens_1.lp % DOMAIN #const n=4. square(1..n,1..n). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- queen(X,Y), queen(X’,Y’), X < X’, X’-X = |Y’-Y|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 460 / 535

  5. Tweaking N -Queens A First Refinement Let’s Place 22 Queens! gringo -c n=22 queens_1.lp | clasp --stats Answer: 1 queen(1,18) queen(2,10) queen(3,21) queen(4,3) queen(5,5) ... SATISFIABLE Models : 1+ Time : 0.113s (Solving: 0.00s 1st Model: 0.00s Unsat: 0.00s) CPU Time : 0.020s Choices : 132 Conflicts : 105 Restarts : 1 Variables : 7238 Constraints : 6710 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 461 / 535

  6. Tweaking N -Queens A First Refinement Let’s Place 22 Queens! gringo -c n=22 queens_1.lp | clasp --stats Answer: 1 queen(1,18) queen(2,10) queen(3,21) queen(4,3) queen(5,5) ... SATISFIABLE Models : 1+ Time : 0.113s (Solving: 0.00s 1st Model: 0.00s Unsat: 0.00s) CPU Time : 0.020s Choices : 132 Conflicts : 105 Restarts : 1 Variables : 7238 Constraints : 6710 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 461 / 535

  7. Tweaking N -Queens A First Refinement Let’s Place 122 Queens! gringo -c n=122 queens_1.lp | clasp --stats Answer: 1 queen(1,24) queen(2,52) queen(3,37) queen(4,60) queen(5,76) ... SATISFIABLE Models : 1+ Time : 79.475s (Solving: 1.06s 1st Model: 1.06s Unsat: 0.00s) CPU Time : 6.930s Choices : 1373 Conflicts : 845 Restarts : 4 Variables : 1211338 Constraints : 1196210 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 462 / 535

  8. Tweaking N -Queens A First Refinement Let’s Place 122 Queens! gringo -c n=122 queens_1.lp | clasp --stats Answer: 1 queen(1,24) queen(2,52) queen(3,37) queen(4,60) queen(5,76) ... SATISFIABLE Models : 1+ Time : 79.475s (Solving: 1.06s 1st Model: 1.06s Unsat: 0.00s) CPU Time : 6.930s Choices : 1373 Conflicts : 845 Restarts : 4 Variables : 1211338 Constraints : 1196210 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 462 / 535

  9. Tweaking N -Queens A First Refinement Let’s Place 122 Queens! gringo -c n=122 queens_1.lp | clasp --stats Answer: 1 queen(1,24) queen(2,52) queen(3,37) queen(4,60) queen(5,76) ... SATISFIABLE Models : 1+ Time : 79.475s (Solving: 1.06s 1st Model: 1.06s Unsat: 0.00s) CPU Time : 6.930s Choices : 1373 Conflicts : 845 Restarts : 4 Variables : 1211338 Constraints : 1196210 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 462 / 535

  10. Tweaking N -Queens A First Refinement Where Time Has Gone time(gringo -c n=122 queens_1.lp | clasp --stats 1241358 7402724 24950848 real 1m15.468s user 1m15.980s sys 0m0.090s Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 463 / 535

  11. Tweaking N -Queens A First Refinement Where Time Has Gone time(gringo -c n=122 queens_1.lp | wc) 1241358 7402724 24950848 real 1m15.468s user 1m15.980s sys 0m0.090s Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 463 / 535

  12. Tweaking N -Queens A First Refinement Where Time Has Gone time(gringo -c n=122 queens_1.lp | wc) 1241358 7402724 24950848 real 1m15.468s user 1m15.980s sys 0m0.090s Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 463 / 535

  13. Tweaking N -Queens A First Refinement Where Time Has Gone time(gringo -c n=122 queens_1.lp | wc) 1241358 7402724 24950848 real 1m15.468s user 1m15.980s sys 0m0.090s Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 463 / 535

  14. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  15. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  16. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  17. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  18. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  19. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  20. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  21. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  22. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  23. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  24. Tweaking N -Queens A First Refinement Grounding Time ∼ Space queens_1.lp % DOMAIN O( n × n ) #const n=4. square(1..n,1..n). % GENERATE O( n × n ) { queen(X,Y) } :- square(X,Y). % TEST O( n × n ) :- X := 1..n, not 1 #count{ queen(X,Y) } 1. O( n × n ) :- Y := 1..n, not 1 #count{ queen(X,Y) } 1. O( n 2 × n 2 ) :- queen(X1,Y1), queen(X2,Y2), X1 < X2, X2-X1 == |Y2-Y1|. % DISPLAY Diagonals make trouble! #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 464 / 535

  25. Tweaking N -Queens Enumerating Diagonals N = 4 4 0Z0Z 4 0Z0Z 3 Z0Z0 3 Z0Z0 2 0Z0Z 2 0Z0Z 1 Z0Z0 1 Z0Z0 1 2 3 4 1 2 3 4 #diagonal 1 = #diagonal 2 = (#row + #column) − 1 (#row − #column) + N Note For each N , indexes 1 , . . . , (2 ∗ N ) − 1 refer to squares on #diagonal 1 / 2 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 465 / 535

  26. Tweaking N -Queens Enumerating Diagonals N = 4 4 0Z0Z 4 0Z0Z 4 5 6 7 7 6 5 4 3 Z0Z0 3 Z0Z0 3 4 5 6 6 5 4 3 2 0Z0Z 2 0Z0Z 2 3 4 5 5 4 3 2 1 Z0Z0 1 Z0Z0 1 2 3 4 4 3 2 1 1 2 3 4 1 2 3 4 #diagonal 1 = #diagonal 2 = (#row + #column) − 1 (#row − #column) + N Note For each N , indexes 1 , . . . , (2 ∗ N ) − 1 refer to squares on #diagonal 1 / 2 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 465 / 535

  27. Tweaking N -Queens Enumerating Diagonals N = 4 4 0Z0Z 4 0Z0Z 4 5 6 7 7 6 5 4 3 Z0Z0 3 Z0Z0 3 4 5 6 6 5 4 3 2 0Z0Z 2 0Z0Z 2 3 4 5 5 4 3 2 1 Z0Z0 1 Z0Z0 1 2 3 4 4 3 2 1 1 2 3 4 1 2 3 4 #diagonal 1 = #diagonal 2 = (#row + #column) − 1 (#row − #column) + N Note For each N , indexes 1 , . . . , (2 ∗ N ) − 1 refer to squares on #diagonal 1 / 2 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 465 / 535

  28. Tweaking N -Queens Enumerating Diagonals N = 4 4 0Z0Z 4 0Z0Z 4 5 6 7 7 6 5 4 3 Z0Z0 3 Z0Z0 3 4 5 6 6 5 4 3 2 0Z0Z 2 0Z0Z 2 3 4 5 5 4 3 2 1 Z0Z0 1 Z0Z0 1 2 3 4 4 3 2 1 1 2 3 4 1 2 3 4 #diagonal 1 = #diagonal 2 = (#row + #column) − 1 (#row − #column) + N Note For each N , indexes 1 , . . . , (2 ∗ N ) − 1 refer to squares on #diagonal 1 / 2 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 465 / 535

  29. Tweaking N -Queens A Second Refinement Let’s go for Diagonals! queens_1.lp % DOMAIN #const n=4. square(1..n,1..n). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- queen(X,Y), queen(X’,Y’), X < X’, X’-X = |Y’-Y|. % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 466 / 535

  30. Tweaking N -Queens A Second Refinement Let’s go for Diagonals! queens_1.lp % DOMAIN #const n=4. square(1..n,1..n). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- D = 1..2*n-1, 2 { queen(X,Y) : D = (X+Y)-1 }. % Diagonal 1 % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 466 / 535

  31. Tweaking N -Queens A Second Refinement Let’s go for Diagonals! queens_1.lp % DOMAIN #const n=4. square(1..n,1..n). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- D = 1..2*n-1, 2 { queen(X,Y) : D = (X+Y)-1 }. % Diagonal 1 :- D = 1..2*n-1, 2 { queen(X,Y) : D = (X-Y)+n }. % Diagonal 2 % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 466 / 535

  32. Tweaking N -Queens A Second Refinement Let’s go for Diagonals! queens_2.lp % DOMAIN #const n=4. square(1..n,1..n). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- D = 1..2*n-1, 2 { queen(X,Y) : D = (X+Y)-1 }. % Diagonal 1 :- D = 1..2*n-1, 2 { queen(X,Y) : D = (X-Y)+n }. % Diagonal 2 % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 466 / 535

  33. Tweaking N -Queens A Second Refinement Let’s Place 122 Queens! gringo -c n=122 queens_2.lp | clasp --stats Answer: 1 queen(1,98) queen(2,54) queen(3,89) queen(4,83) queen(5,59) ... SATISFIABLE Models : 1+ Time : 2.211s (Solving: 0.13s 1st Model: 0.13s Unsat: 0.00s) CPU Time : 0.210s Choices : 11036 Conflicts : 499 Restarts : 3 Variables : 16098 Constraints : 970 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 467 / 535

  34. Tweaking N -Queens A Second Refinement Let’s Place 122 Queens! gringo -c n=122 queens_2.lp | clasp --stats Answer: 1 queen(1,98) queen(2,54) queen(3,89) queen(4,83) queen(5,59) ... SATISFIABLE Models : 1+ Time : 2.211s (Solving: 0.13s 1st Model: 0.13s Unsat: 0.00s) CPU Time : 0.210s Choices : 11036 Conflicts : 499 Restarts : 3 Variables : 16098 Constraints : 970 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 467 / 535

  35. Tweaking N -Queens A Second Refinement Let’s Place 122 Queens! gringo -c n=122 queens_2.lp | clasp --stats Answer: 1 queen(1,98) queen(2,54) queen(3,89) queen(4,83) queen(5,59) ... SATISFIABLE Models : 1+ Time : 2.211s (Solving: 0.13s 1st Model: 0.13s Unsat: 0.00s) CPU Time : 0.210s Choices : 11036 Conflicts : 499 Restarts : 3 Variables : 16098 Constraints : 970 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 467 / 535

  36. Tweaking N -Queens A Second Refinement Let’s Place 300 Queens! gringo -c n=300 queens_2.lp | clasp --stats Answer: 1 queen(1,62) queen(2,232) queen(3,176) queen(4,241) queen(5,207) ... SATISFIABLE Models : 1+ Time : 35.450s (Solving: 6.69s 1st Model: 6.68s Unsat: 0.00s) CPU Time : 7.250s Choices : 141445 Conflicts : 7488 Restarts : 9 Variables : 92994 Constraints : 2394 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 468 / 535

  37. Tweaking N -Queens A Second Refinement Let’s Place 300 Queens! gringo -c n=300 queens_2.lp | clasp --stats Answer: 1 queen(1,62) queen(2,232) queen(3,176) queen(4,241) queen(5,207) ... SATISFIABLE Models : 1+ Time : 35.450s (Solving: 6.69s 1st Model: 6.68s Unsat: 0.00s) CPU Time : 7.250s Choices : 141445 Conflicts : 7488 Restarts : 9 Variables : 92994 Constraints : 2394 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 468 / 535

  38. Tweaking N -Queens A Second Refinement Let’s Place 300 Queens! gringo -c n=300 queens_2.lp | clasp --stats Answer: 1 queen(1,62) queen(2,232) queen(3,176) queen(4,241) queen(5,207) ... SATISFIABLE Models : 1+ Time : 35.450s (Solving: 6.69s 1st Model: 6.68s Unsat: 0.00s) CPU Time : 7.250s Choices : 141445 Conflicts : 7488 Restarts : 9 Variables : 92994 Constraints : 2394 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 468 / 535

  39. Tweaking N -Queens A Third Refinement Let’s Precalculate Indexes! queens_2.lp % DOMAIN #const n=4. square(1..n,1..n). diag1(X,Y,(X+Y)-1) :- square(X,Y). diag2(X,Y,(X-Y)+n) :- square(X,Y). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- D = 1..2*n-1, 2 { queen(X,Y) : D = (X+Y)-1 }. % Diagonal 1 :- D = 1..2*n-1, 2 { queen(X,Y) : D = (X-Y)+n }. % Diagonal 2 % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 469 / 535

  40. Tweaking N -Queens A Third Refinement Let’s Precalculate Indexes! queens_2.lp % DOMAIN #const n=4. square(1..n,1..n). diag1(X,Y,(X+Y)-1) :- square(X,Y). diag2(X,Y,(X-Y)+n) :- square(X,Y). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- D = 1..2*n-1, 2 { queen(X,Y) : D = (X+Y)-1 }. % Diagonal 1 :- D = 1..2*n-1, 2 { queen(X,Y) : D = (X-Y)+n }. % Diagonal 2 % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 469 / 535

  41. Tweaking N -Queens A Third Refinement Let’s Precalculate Indexes! queens_2.lp % DOMAIN #const n=4. square(1..n,1..n). diag1(X,Y,(X+Y)-1) :- square(X,Y). diag2(X,Y,(X-Y)+n) :- square(X,Y). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- D = 1..2*n-1, 2 { queen(X,Y) : diag1(X,Y,D) }. % Diagonal 1 :- D = 1..2*n-1, 2 { queen(X,Y) : diag2(X,Y,D) }. % Diagonal 2 % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 469 / 535

  42. Tweaking N -Queens A Third Refinement Let’s Precalculate Indexes! queens_3.lp % DOMAIN #const n=4. square(1..n,1..n). diag1(X,Y,(X+Y)-1) :- square(X,Y). diag2(X,Y,(X-Y)+n) :- square(X,Y). % GENERATE 0 { queen(X,Y) } 1 :- square(X,Y). % TEST :- X = 1..n, not 1 { queen(X,Y) } 1. :- Y = 1..n, not 1 { queen(X,Y) } 1. :- D = 1..2*n-1, 2 { queen(X,Y) : diag1(X,Y,D) }. % Diagonal 1 :- D = 1..2*n-1, 2 { queen(X,Y) : diag2(X,Y,D) }. % Diagonal 2 % DISPLAY #show queen/2. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 469 / 535

  43. Tweaking N -Queens A Third Refinement Let’s Place 300 Queens! gringo -c n=300 queens_3.lp | clasp --stats Answer: 1 queen(1,62) queen(2,232) queen(3,176) queen(4,241) queen(5,207) ... SATISFIABLE Models : 1+ Time : 8.889s (Solving: 6.61s 1st Model: 6.60s Unsat: 0.00s) CPU Time : 7.320s Choices : 141445 Conflicts : 7488 Restarts : 9 Variables : 92994 Constraints : 2394 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 470 / 535

  44. Tweaking N -Queens A Third Refinement Let’s Place 300 Queens! gringo -c n=300 queens_3.lp | clasp --stats Answer: 1 queen(1,62) queen(2,232) queen(3,176) queen(4,241) queen(5,207) ... SATISFIABLE Models : 1+ Time : 8.889s (Solving: 6.61s 1st Model: 6.60s Unsat: 0.00s) CPU Time : 7.320s Choices : 141445 Conflicts : 7488 Restarts : 9 Variables : 92994 Constraints : 2394 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 470 / 535

  45. Tweaking N -Queens A Third Refinement Let’s Place 300 Queens! gringo -c n=300 queens_3.lp | clasp --stats Answer: 1 queen(1,62) queen(2,232) queen(3,176) queen(4,241) queen(5,207) ... SATISFIABLE Models : 1+ Time : 8.889s (Solving: 6.61s 1st Model: 6.60s Unsat: 0.00s) CPU Time : 7.320s Choices : 141445 Conflicts : 7488 Restarts : 9 Variables : 92994 Constraints : 2394 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 470 / 535

  46. Tweaking N -Queens A Third Refinement Let’s Place 600 Queens! gringo -c n=600 queens_3.lp | clasp --stats Answer: 1 queen(1,477) queen(2,365) queen(3,455) queen(4,470) queen(5,237) ... SATISFIABLE Models : 1+ Time : 76.798s (Solving: 65.81s 1st Model: 65.75s Unsat: 0.00s) CPU Time : 68.620s Choices : 869379 Conflicts : 25746 Restarts : 12 Variables : 365994 Constraints : 4794 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 471 / 535

  47. Tweaking N -Queens A Third Refinement Let’s Place 600 Queens! gringo -c n=600 queens_3.lp | clasp --stats Answer: 1 queen(1,477) queen(2,365) queen(3,455) queen(4,470) queen(5,237) ... SATISFIABLE Models : 1+ Time : 76.798s (Solving: 65.81s 1st Model: 65.75s Unsat: 0.00s) CPU Time : 68.620s Choices : 869379 Conflicts : 25746 Restarts : 12 Variables : 365994 Constraints : 4794 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 471 / 535

  48. Tweaking N -Queens A Case for Oracles Let’s Place 600 Queens! gringo -c n=600 queens_3.lp | clasp --stats --heuristic=vsids --trans-ext=dynamic Answer: 1 queen(1,477) queen(2,365) queen(3,455) queen(4,470) queen(5,237) ... SATISFIABLE Models : 1+ Time : 76.798s (Solving: 65.81s 1st Model: 65.75s Unsat: 0.00s) CPU Time : 68.620s Choices : 869379 Conflicts : 25746 Restarts : 12 Variables : 365994 Constraints : 4794 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 472 / 535

  49. Tweaking N -Queens A Case for Oracles Let’s Place 600 Queens! gringo -c n=600 queens_3.lp | clasp --stats --heuristic=vsids --trans-ext=dynamic Answer: 1 queen(1,477) queen(2,365) queen(3,455) queen(4,470) queen(5,237) ... SATISFIABLE Models : 1+ Time : 76.798s (Solving: 65.81s 1st Model: 65.75s Unsat: 0.00s) CPU Time : 68.620s Choices : 869379 Conflicts : 25746 Restarts : 12 Variables : 365994 Constraints : 4794 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 472 / 535

  50. Tweaking N -Queens A Case for Oracles Let’s Place 600 Queens! gringo -c n=600 queens_3.lp | clasp --stats --heuristic=vsids --trans-ext=dynamic Answer: 1 queen(1,422) queen(2,458) queen(3,224) queen(4,408) queen(5,405) ... SATISFIABLE Models : 1+ Time : 37.454s (Solving: 26.38s 1st Model: 26.26s Unsat: 0.00s) CPU Time : 29.580s Choices : 961315 Conflicts : 3222 Restarts : 7 Variables : 365994 Constraints : 4794 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 472 / 535

  51. Tweaking N -Queens A Case for Oracles Let’s Place 600 Queens! gringo -c n=600 queens_3.lp | clasp --stats --heuristic=vsids --trans-ext=dynamic Answer: 1 queen(1,422) queen(2,458) queen(3,224) queen(4,408) queen(5,405) ... SATISFIABLE Models : 1+ Time : 37.454s (Solving: 26.38s 1st Model: 26.26s Unsat: 0.00s) CPU Time : 29.580s Choices : 961315 Conflicts : 3222 Restarts : 7 Variables : 365994 Constraints : 4794 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 472 / 535

  52. Tweaking N -Queens A Case for Oracles Let’s Place 600 Queens! gringo -c n=600 queens_3.lp | clasp --stats --heuristic=vsids --trans-ext=dynamic Answer: 1 queen(1,90) queen(2,452) queen(3,494) queen(4,145) queen(5,84) ... SATISFIABLE Models : 1+ Time : 22.654s (Solving: 10.53s 1st Model: 10.47s Unsat: 0.00s) CPU Time : 15.750s Choices : 1058729 Conflicts : 2128 Restarts : 6 Variables : 403123 Constraints : 49636 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 472 / 535

  53. Do’s and Dont’s Outline 1 Tweaking N -Queens 2 Do’s and Dont’s 3 Hints Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 473 / 535

  54. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pro(asparagus,fresh). pro(cucumber,fresh). pro(asparagus,tasty). pro(cucumber,tasty). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  55. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pro(asparagus,fresh). pro(cucumber,fresh). pro(asparagus,tasty). pro(cucumber,tasty). buy(X) :- veg(X), pro(X,cheap), pro(X,fresh), pro(X,tasty). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  56. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pro(asparagus,fresh). pro(cucumber,fresh). pro(asparagus,tasty). pro(cucumber,tasty). pro(asparagus,clean). buy(X) :- veg(X), pro(X,cheap), pro(X,fresh), pro(X,tasty), pro(X,clean). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  57. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pro(asparagus,fresh). pro(cucumber,fresh). pro(asparagus,tasty). pro(cucumber,tasty). pro(asparagus,clean). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  58. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pre(cheap). pro(asparagus,fresh). pro(cucumber,fresh). pre(fresh). pro(asparagus,tasty). pro(cucumber,tasty). pre(tasty). pro(asparagus,clean). buy(X) :- veg(X), pro(X,P) : pre(P). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  59. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pre(cheap). pro(asparagus,fresh). pro(cucumber,fresh). pre(fresh). pro(asparagus,tasty). pro(cucumber,tasty). pre(tasty). pro(asparagus,clean). pre(clean). buy(X) :- veg(X), pro(X,P) : pre(P). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  60. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pre(cheap). pro(asparagus,fresh). pro(cucumber,fresh). pre(fresh). pro(asparagus,tasty). pro(cucumber,tasty). pre(tasty). pro(asparagus,clean). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  61. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pre(cheap). pro(asparagus,fresh). pro(cucumber,fresh). pre(fresh). pro(asparagus,tasty). pro(cucumber,tasty). pre(tasty). pro(asparagus,clean). buy(X) :- veg(X), not bye(X). bye(X) :- veg(X), pre(P), not pro(X,P). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  62. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pre(cheap). pro(asparagus,fresh). pro(cucumber,fresh). pre(fresh). pro(asparagus,tasty). pro(cucumber,tasty). pre(tasty). pro(asparagus,clean). pre(clean). buy(X) :- veg(X), not bye(X). bye(X) :- veg(X), pre(P), not pro(X,P). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  63. Do’s and Dont’s Implementing Universal Quantification Goal: identify objects such that ALL properties from a “list” hold 1 check all properties explicitly . . . obsolete if properties change ✘ 2 use variable-sized conjunction (via ‘ : ’) . . . adapts to changing facts ✔ 3 use negation of complement . . . adapts to changing facts ✔ Example: vegetables to buy veg(asparagus). veg(cucumber). pro(asparagus,cheap). pro(cucumber,cheap). pre(cheap). pro(asparagus,fresh). pro(cucumber,fresh). pre(fresh). pro(asparagus,tasty). pro(cucumber,tasty). pre(tasty). pro(asparagus,clean). pre(clean). buy(X) :- veg(X), not bye(X). bye(X) :- veg(X), pre(P), not pro(X,P). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 474 / 535

  64. Do’s and Dont’s Running Example: Latin Square Given: an N × N board Wanted: assignment of 1 , . . . , N 1 1 1 2 3 4 5 6 2 2 2 3 4 5 6 1 3 3 3 4 5 6 1 2 4 4 4 5 6 1 2 3 5 5 5 6 1 2 3 4 6 6 6 1 2 3 4 5 1 2 3 4 5 6 1 2 3 4 5 6 represented by facts: represented by atoms: square(1,1). ... square(1,6). num(1,1,1) num(1,2,2) ... num(1,6,6) square(2,1). ... square(2,6). num(2,1,2) num(2,2,3) ... num(2,6,1) square(3,1). ... square(3,6). num(3,1,3) num(3,2,4) ... num(3,6,2) square(4,1). ... square(4,6). num(4,1,4) num(4,2,5) ... num(4,6,3) square(5,1). ... square(5,6). num(5,1,5) num(5,2,6) ... num(5,6,4) square(6,1). ... square(6,6). num(6,1,6) num(6,2,1) ... num(6,6.5) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 475 / 535

  65. Do’s and Dont’s Running Example: Latin Square Given: an N × N board Wanted: assignment of 1 , . . . , N 1 1 1 2 3 4 5 6 2 2 2 3 4 5 6 1 3 3 3 4 5 6 1 2 4 4 4 5 6 1 2 3 5 5 5 6 1 2 3 4 6 6 6 1 2 3 4 5 1 2 3 4 5 6 1 2 3 4 5 6 represented by facts: represented by atoms: square(1,1). ... square(1,6). num(1,1,1) num(1,2,2) ... num(1,6,6) square(2,1). ... square(2,6). num(2,1,2) num(2,2,3) ... num(2,6,1) square(3,1). ... square(3,6). num(3,1,3) num(3,2,4) ... num(3,6,2) square(4,1). ... square(4,6). num(4,1,4) num(4,2,5) ... num(4,6,3) square(5,1). ... square(5,6). num(5,1,5) num(5,2,6) ... num(5,6,4) square(6,1). ... square(6,6). num(6,1,6) num(6,2,1) ... num(6,6.5) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 475 / 535

  66. Do’s and Dont’s Projecting Irrelevant Details Out A Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- square(X,Y), N = 1..n, not num(X,Y’,N) : square(X,Y’). :- square(X,Y), N = 1..n, not num(X’,Y,N) : square(X’,Y). Note unreused “singleton variables” gringo latin_0.lp | wc gringo latin_1.lp | wc 105480 2558984 14005258 42056 273672 1690522 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 476 / 535

  67. Do’s and Dont’s Projecting Irrelevant Details Out A Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- square(X,Y), N = 1..n, not num(X,Y’,N) : square(X,Y’). :- square(X,Y), N = 1..n, not num(X’,Y,N) : square(X’,Y). Note unreused “singleton variables” gringo latin_0.lp | wc gringo latin_1.lp | wc 105480 2558984 14005258 42056 273672 1690522 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 476 / 535

  68. Do’s and Dont’s Projecting Irrelevant Details Out A Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- square(X,Y), N = 1..n, not num(X,Y’,N) : square(X,Y’). :- square(X,Y), N = 1..n, not num(X’,Y,N) : square(X’,Y). Note unreused “singleton variables” gringo latin_0.lp | wc gringo latin_1.lp | wc 105480 2558984 14005258 42056 273672 1690522 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 476 / 535

  69. Do’s and Dont’s Projecting Irrelevant Details Out A Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). squareX(X) :- square(X,Y). squareY(Y) :- square(X,Y). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- squareX(X), N = 1..n, not num(X,Y’,N) : square(X,Y’). :- squareY(Y), N = 1..n, not num(X’,Y,N) : square(X’,Y). Note unreused “singleton variables” gringo latin_0.lp | wc gringo latin_1.lp | wc 105480 2558984 14005258 42056 273672 1690522 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 476 / 535

  70. Do’s and Dont’s Projecting Irrelevant Details Out A Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). squareX(X) :- square(X,Y). squareY(Y) :- square(X,Y). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- squareX(X), N = 1..n, not num(X,Y’,N) : square(X,Y’). :- squareY(Y), N = 1..n, not num(X’,Y,N) : square(X’,Y). Note unreused “singleton variables” gringo latin_0.lp | wc gringo latin_1.lp | wc 105480 2558984 14005258 42056 273672 1690522 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 476 / 535

  71. Do’s and Dont’s Unraveling Symmetric Inequalities Another Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- num(X,Y,N), num(X,Y’,N), Y != Y’. :- num(X,Y,N), num(X’,Y,N), X != X’. Note duplicate ground rules (swapping Y / Y’ and X / X’ gives the “same”) gringo latin_2.lp | wc gringo latin_3.lp | wc 2071560 12389384 40906946 1055752 6294536 21099558 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 477 / 535

  72. Do’s and Dont’s Unraveling Symmetric Inequalities Another Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- num(X,Y,N), num(X,Y’,N), Y != Y’. :- num(X,Y,N), num(X’,Y,N), X != X’. Note duplicate ground rules (swapping Y / Y’ and X / X’ gives the “same”) gringo latin_2.lp | wc gringo latin_3.lp | wc 2071560 12389384 40906946 1055752 6294536 21099558 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 477 / 535

  73. Do’s and Dont’s Unraveling Symmetric Inequalities Another Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- num(X,Y,N), num(X,Y’,N), Y != Y’. :- num(X,Y,N), num(X’,Y,N), X != X’. Note duplicate ground rules (swapping Y / Y’ and X / X’ gives the “same”) gringo latin_2.lp | wc gringo latin_3.lp | wc 2071560 12389384 40906946 1055752 6294536 21099558 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 477 / 535

  74. Do’s and Dont’s Unraveling Symmetric Inequalities Another Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- num(X,Y,N), num(X,Y’,N), Y < Y’. :- num(X,Y,N), num(X’,Y,N), X < X’. Note duplicate ground rules (swapping Y / Y’ and X / X’ gives the “same”) gringo latin_2.lp | wc gringo latin_3.lp | wc 2071560 12389384 40906946 1055752 6294536 21099558 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 477 / 535

  75. Do’s and Dont’s Unraveling Symmetric Inequalities Another Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- num(X,Y,N), num(X,Y’,N), Y < Y’. :- num(X,Y,N), num(X’,Y,N), X < X’. Note duplicate ground rules (swapping Y / Y’ and X / X’ gives the “same”) gringo latin_2.lp | wc gringo latin_3.lp | wc 2071560 12389384 40906946 1055752 6294536 21099558 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 477 / 535

  76. Do’s and Dont’s Linearizing Existence Tests Still another Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- num(X,Y,N), num(X,Y’,N), Y < Y’. :- num(X,Y,N), num(X’,Y,N), X < X’. Note uniqueness of N in a row/column checked by enumerating pairs! gringo latin_3.lp | wc gringo latin_4.lp | wc 1055752 6294536 21099558 228360 1205256 4780744 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 478 / 535

  77. Do’s and Dont’s Linearizing Existence Tests Still another Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- num(X,Y,N), num(X,Y’,N), Y < Y’. :- num(X,Y,N), num(X’,Y,N), X < X’. Note uniqueness of N in a row/column checked by enumerating pairs! gringo latin_3.lp | wc gringo latin_4.lp | wc 1055752 6294536 21099558 228360 1205256 4780744 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 478 / 535

  78. Do’s and Dont’s Linearizing Existence Tests Still another Latin square encoding % DOMAIN #const n=32. square(1..n,1..n). % GENERATE 1 { num(X,Y,N) : N = 1..n } 1 :- square(X,Y). % TEST :- num(X,Y,N), num(X,Y’,N), Y < Y’. :- num(X,Y,N), num(X’,Y,N), X < X’. Note uniqueness of N in a row/column checked by enumerating pairs! gringo latin_3.lp | wc gringo latin_4.lp | wc 1055752 6294536 21099558 228360 1205256 4780744 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 478 / 535

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