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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. ASP solving process Graph coloring node(1..6).        edge(1,2). edge(1,3). edge(1,4).     edge(2,4). edge(2,5). edge(2,6).     Problem edge(3,1). edge(3,4). edge(3,5).  instance edge(4,1). edge(4,2).   edge(5,3). edge(5,4). edge(5,6).     edge(6,2). edge(6,3). edge(6,5).         color(r). color(b). color(g).   { assign(N,C) : color(C) } = 1 :- node(N). Problem  encoding :- edge(N,M), assign(N,C), assign(M,C).  Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 66 / 541

  2. ASP solving process Graph coloring node(1..6).        edge(1,2). edge(1,3). edge(1,4).     edge(2,4). edge(2,5). edge(2,6).     Problem edge(3,1). edge(3,4). edge(3,5).  instance edge(4,1). edge(4,2).   edge(5,3). edge(5,4). edge(5,6).     edge(6,2). edge(6,3). edge(6,5).         color(r). color(b). color(g).   { assign(N,C) : color(C) } = 1 :- node(N). Problem  encoding :- edge(N,M), assign(N,C), assign(M,C).  Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 66 / 541

  3. ASP solving process Graph coloring node(1..6).        edge(1,2). edge(1,3). edge(1,4).     edge(2,4). edge(2,5). edge(2,6).     Problem edge(3,1). edge(3,4). edge(3,5).  instance edge(4,1). edge(4,2).   edge(5,3). edge(5,4). edge(5,6).     edge(6,2). edge(6,3). edge(6,5).         color(r). color(b). color(g).   { assign(N,C) : color(C) } = 1 :- node(N). Problem  encoding :- edge(N,M), assign(N,C), assign(M,C).  Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 66 / 541

  4. ASP solving process Graph coloring node(1..6).        edge(1,2). edge(1,3). edge(1,4).     edge(2,4). edge(2,5). edge(2,6).     Problem edge(3,1). edge(3,4). edge(3,5).  instance edge(4,1). edge(4,2).   edge(5,3). edge(5,4). edge(5,6).     edge(6,2). edge(6,3). edge(6,5).         color(r). color(b). color(g).   { assign(N,C) : color(C) } = 1 :- node(N). Problem  encoding :- edge(N,M), assign(N,C), assign(M,C).  Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 66 / 541

  5. ASP solving process Graph coloring node(1..6).        edge(1,2). edge(1,3). edge(1,4).     edge(2,4). edge(2,5). edge(2,6).     Problem edge(3,1). edge(3,4). edge(3,5).  instance edge(4,1). edge(4,2).   edge(5,3). edge(5,4). edge(5,6).     edge(6,2). edge(6,3). edge(6,5).         color(r). color(b). color(g).   { assign(N,C) : color(C) } = 1 :- node(N). Problem  encoding :- edge(N,M), assign(N,C), assign(M,C).  Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 66 / 541

  6. ASP solving process Graph coloring node(1..6).        edge(1,2). edge(1,3). edge(1,4).     edge(2,4). edge(2,5). edge(2,6).     edge(3,1). edge(3,4). edge(3,5).  graph.lp edge(4,1). edge(4,2).   edge(5,3). edge(5,4). edge(5,6).     edge(6,2). edge(6,3). edge(6,5).         color(r). color(b). color(g).   { assign(N,C) : color(C) } = 1 :- node(N).   color.lp :- edge(N,M), assign(N,C), assign(M,C). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 66 / 541

  7. ASP solving process ASP solving process Problem Solution ✻ Modeling Interpreting ❄ Logic ✲ ✲ ✲ Stable Grounder Solver Program Models Solving Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 67 / 541

  8. ASP solving process Graph coloring: Grounding $ gringo --text graph.lp color.lp node(1). node(2). node(3). node(4). node(5). node(6). edge(1,2). edge(2,4). edge(3,1). edge(4,1). edge(5,3). edge(6,2). edge(1,3). edge(2,5). edge(3,4). edge(4,2). edge(5,4). edge(6,3). edge(1,4). edge(2,6). edge(3,5). edge(5,6). edge(6,5). color(r). color(b). color(g). { assign(1,r), assign(1,b), assign(1,g) } = 1. { assign(4,r), assign(4,b), assign(4,g) } = 1. { assign(2,r), assign(2,b), assign(2,g) } = 1. { assign(5,r), assign(5,b), assign(5,g) } = 1. { assign(3,r), assign(3,b), assign(3,g) } = 1. { assign(6,r), assign(6,b), assign(6,g) } = 1. :- assign(1,r), assign(2,r). :- assign(2,r), assign(4,r). [...] :- assign(6,r), assign(2,r). :- assign(1,b), assign(2,b). :- assign(2,b), assign(4,b). :- assign(6,b), assign(2,b). :- assign(1,g), assign(2,g). :- assign(2,g), assign(4,g). :- assign(6,g), assign(2,g). :- assign(1,r), assign(3,r). :- assign(2,r), assign(5,r). :- assign(6,r), assign(3,r). :- assign(1,b), assign(3,b). :- assign(2,b), assign(5,b). :- assign(6,b), assign(3,b). :- assign(1,g), assign(3,g). :- assign(2,g), assign(5,g). :- assign(6,g), assign(3,g). :- assign(1,r), assign(4,r). :- assign(2,r), assign(6,r). :- assign(6,r), assign(5,r). :- assign(1,b), assign(4,b). :- assign(2,b), assign(6,b). :- assign(6,b), assign(5,b). :- assign(1,g), assign(4,g). :- assign(2,g), assign(6,g). :- assign(6,g), assign(5,g). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 68 / 541

  9. ASP solving process Graph coloring: Grounding $ gringo --text graph.lp color.lp node(1). node(2). node(3). node(4). node(5). node(6). edge(1,2). edge(2,4). edge(3,1). edge(4,1). edge(5,3). edge(6,2). edge(1,3). edge(2,5). edge(3,4). edge(4,2). edge(5,4). edge(6,3). edge(1,4). edge(2,6). edge(3,5). edge(5,6). edge(6,5). color(r). color(b). color(g). { assign(1,r), assign(1,b), assign(1,g) } = 1. { assign(4,r), assign(4,b), assign(4,g) } = 1. { assign(2,r), assign(2,b), assign(2,g) } = 1. { assign(5,r), assign(5,b), assign(5,g) } = 1. { assign(3,r), assign(3,b), assign(3,g) } = 1. { assign(6,r), assign(6,b), assign(6,g) } = 1. :- assign(1,r), assign(2,r). :- assign(2,r), assign(4,r). [...] :- assign(6,r), assign(2,r). :- assign(1,b), assign(2,b). :- assign(2,b), assign(4,b). :- assign(6,b), assign(2,b). :- assign(1,g), assign(2,g). :- assign(2,g), assign(4,g). :- assign(6,g), assign(2,g). :- assign(1,r), assign(3,r). :- assign(2,r), assign(5,r). :- assign(6,r), assign(3,r). :- assign(1,b), assign(3,b). :- assign(2,b), assign(5,b). :- assign(6,b), assign(3,b). :- assign(1,g), assign(3,g). :- assign(2,g), assign(5,g). :- assign(6,g), assign(3,g). :- assign(1,r), assign(4,r). :- assign(2,r), assign(6,r). :- assign(6,r), assign(5,r). :- assign(1,b), assign(4,b). :- assign(2,b), assign(6,b). :- assign(6,b), assign(5,b). :- assign(1,g), assign(4,g). :- assign(2,g), assign(6,g). :- assign(6,g), assign(5,g). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 68 / 541

  10. ASP solving process Graph coloring: Grounding $ gringo --text graph.lp color.lp node(1). node(2). node(3). node(4). node(5). node(6). edge(1,2). edge(2,4). edge(3,1). edge(4,1). edge(5,3). edge(6,2). edge(1,3). edge(2,5). edge(3,4). edge(4,2). edge(5,4). edge(6,3). edge(1,4). edge(2,6). edge(3,5). edge(5,6). edge(6,5). color(r). color(b). color(g). { assign(1,r), assign(1,b), assign(1,g) } = 1. { assign(4,r), assign(4,b), assign(4,g) } = 1. { assign(2,r), assign(2,b), assign(2,g) } = 1. { assign(5,r), assign(5,b), assign(5,g) } = 1. { assign(3,r), assign(3,b), assign(3,g) } = 1. { assign(6,r), assign(6,b), assign(6,g) } = 1. :- assign(1,r), assign(2,r). :- assign(2,r), assign(4,r). [...] :- assign(6,r), assign(2,r). :- assign(1,b), assign(2,b). :- assign(2,b), assign(4,b). :- assign(6,b), assign(2,b). :- assign(1,g), assign(2,g). :- assign(2,g), assign(4,g). :- assign(6,g), assign(2,g). :- assign(1,r), assign(3,r). :- assign(2,r), assign(5,r). :- assign(6,r), assign(3,r). :- assign(1,b), assign(3,b). :- assign(2,b), assign(5,b). :- assign(6,b), assign(3,b). :- assign(1,g), assign(3,g). :- assign(2,g), assign(5,g). :- assign(6,g), assign(3,g). :- assign(1,r), assign(4,r). :- assign(2,r), assign(6,r). :- assign(6,r), assign(5,r). :- assign(1,b), assign(4,b). :- assign(2,b), assign(6,b). :- assign(6,b), assign(5,b). :- assign(1,g), assign(4,g). :- assign(2,g), assign(6,g). :- assign(6,g), assign(5,g). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 68 / 541

  11. ASP solving process Graph coloring: Grounding $ gringo --text graph.lp color.lp node(1). node(2). node(3). node(4). node(5). node(6). edge(1,2). edge(2,4). edge(3,1). edge(4,1). edge(5,3). edge(6,2). edge(1,3). edge(2,5). edge(3,4). edge(4,2). edge(5,4). edge(6,3). edge(1,4). edge(2,6). edge(3,5). edge(5,6). edge(6,5). color(r). color(b). color(g). { assign(1,r), assign(1,b), assign(1,g) } = 1. { assign(4,r), assign(4,b), assign(4,g) } = 1. { assign(2,r), assign(2,b), assign(2,g) } = 1. { assign(5,r), assign(5,b), assign(5,g) } = 1. { assign(3,r), assign(3,b), assign(3,g) } = 1. { assign(6,r), assign(6,b), assign(6,g) } = 1. :- assign(1,r), assign(2,r). :- assign(2,r), assign(4,r). [...] :- assign(6,r), assign(2,r). :- assign(1,b), assign(2,b). :- assign(2,b), assign(4,b). :- assign(6,b), assign(2,b). :- assign(1,g), assign(2,g). :- assign(2,g), assign(4,g). :- assign(6,g), assign(2,g). :- assign(1,r), assign(3,r). :- assign(2,r), assign(5,r). :- assign(6,r), assign(3,r). :- assign(1,b), assign(3,b). :- assign(2,b), assign(5,b). :- assign(6,b), assign(3,b). :- assign(1,g), assign(3,g). :- assign(2,g), assign(5,g). :- assign(6,g), assign(3,g). :- assign(1,r), assign(4,r). :- assign(2,r), assign(6,r). :- assign(6,r), assign(5,r). :- assign(1,b), assign(4,b). :- assign(2,b), assign(6,b). :- assign(6,b), assign(5,b). :- assign(1,g), assign(4,g). :- assign(2,g), assign(6,g). :- assign(6,g), assign(5,g). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 68 / 541

  12. ASP solving process Graph coloring: Grounding $ clingo --text graph.lp color.lp node(1). node(2). node(3). node(4). node(5). node(6). edge(1,2). edge(2,4). edge(3,1). edge(4,1). edge(5,3). edge(6,2). edge(1,3). edge(2,5). edge(3,4). edge(4,2). edge(5,4). edge(6,3). edge(1,4). edge(2,6). edge(3,5). edge(5,6). edge(6,5). color(r). color(b). color(g). { assign(1,r), assign(1,b), assign(1,g) } = 1. { assign(4,r), assign(4,b), assign(4,g) } = 1. { assign(2,r), assign(2,b), assign(2,g) } = 1. { assign(5,r), assign(5,b), assign(5,g) } = 1. { assign(3,r), assign(3,b), assign(3,g) } = 1. { assign(6,r), assign(6,b), assign(6,g) } = 1. :- assign(1,r), assign(2,r). :- assign(2,r), assign(4,r). [...] :- assign(6,r), assign(2,r). :- assign(1,b), assign(2,b). :- assign(2,b), assign(4,b). :- assign(6,b), assign(2,b). :- assign(1,g), assign(2,g). :- assign(2,g), assign(4,g). :- assign(6,g), assign(2,g). :- assign(1,r), assign(3,r). :- assign(2,r), assign(5,r). :- assign(6,r), assign(3,r). :- assign(1,b), assign(3,b). :- assign(2,b), assign(5,b). :- assign(6,b), assign(3,b). :- assign(1,g), assign(3,g). :- assign(2,g), assign(5,g). :- assign(6,g), assign(3,g). :- assign(1,r), assign(4,r). :- assign(2,r), assign(6,r). :- assign(6,r), assign(5,r). :- assign(1,b), assign(4,b). :- assign(2,b), assign(6,b). :- assign(6,b), assign(5,b). :- assign(1,g), assign(4,g). :- assign(2,g), assign(6,g). :- assign(6,g), assign(5,g). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 68 / 541

  13. ASP solving process ASP solving process Problem Solution ✻ Modeling Interpreting ❄ Logic ✲ ✲ ✲ Stable Grounder Solver Program Models Solving Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 69 / 541

  14. ASP solving process Graph coloring: Solving $ gringo graph.lp color.lp | clasp 0 clasp version 2.1.0 Reading from stdin Solving... Answer: 1 node(1) [...] assign(6,b) assign(5,g) assign(4,b) assign(3,r) assign(2,r) assign(1,g) Answer: 2 node(1) [...] assign(6,r) assign(5,g) assign(4,r) assign(3,b) assign(2,b) assign(1,g) Answer: 3 node(1) [...] assign(6,g) assign(5,b) assign(4,g) assign(3,r) assign(2,r) assign(1,b) Answer: 4 node(1) [...] assign(6,r) assign(5,b) assign(4,r) assign(3,g) assign(2,g) assign(1,b) Answer: 5 node(1) [...] assign(6,g) assign(5,r) assign(4,g) assign(3,b) assign(2,b) assign(1,r) Answer: 6 node(1) [...] assign(6,b) assign(5,r) assign(4,b) assign(3,g) assign(2,g) assign(1,r) SATISFIABLE Models : 6 Time : 0.002s (Solving: 0.00s 1st Model: 0.00s Unsat: 0.00s) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 70 / 541

  15. ASP solving process Graph coloring: Solving $ gringo graph.lp color.lp | clasp 0 clasp version 2.1.0 Reading from stdin Solving... Answer: 1 node(1) [...] assign(6,b) assign(5,g) assign(4,b) assign(3,r) assign(2,r) assign(1,g) Answer: 2 node(1) [...] assign(6,r) assign(5,g) assign(4,r) assign(3,b) assign(2,b) assign(1,g) Answer: 3 node(1) [...] assign(6,g) assign(5,b) assign(4,g) assign(3,r) assign(2,r) assign(1,b) Answer: 4 node(1) [...] assign(6,r) assign(5,b) assign(4,r) assign(3,g) assign(2,g) assign(1,b) Answer: 5 node(1) [...] assign(6,g) assign(5,r) assign(4,g) assign(3,b) assign(2,b) assign(1,r) Answer: 6 node(1) [...] assign(6,b) assign(5,r) assign(4,b) assign(3,g) assign(2,g) assign(1,r) SATISFIABLE Models : 6 Time : 0.002s (Solving: 0.00s 1st Model: 0.00s Unsat: 0.00s) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 70 / 541

  16. ASP solving process Graph coloring: Solving $ clingo graph.lp color.lp 0 clasp version 2.1.0 Reading from stdin Solving... Answer: 1 node(1) [...] assign(6,b) assign(5,g) assign(4,b) assign(3,r) assign(2,r) assign(1,g) Answer: 2 node(1) [...] assign(6,r) assign(5,g) assign(4,r) assign(3,b) assign(2,b) assign(1,g) Answer: 3 node(1) [...] assign(6,g) assign(5,b) assign(4,g) assign(3,r) assign(2,r) assign(1,b) Answer: 4 node(1) [...] assign(6,r) assign(5,b) assign(4,r) assign(3,g) assign(2,g) assign(1,b) Answer: 5 node(1) [...] assign(6,g) assign(5,r) assign(4,g) assign(3,b) assign(2,b) assign(1,r) Answer: 6 node(1) [...] assign(6,b) assign(5,r) assign(4,b) assign(3,g) assign(2,g) assign(1,r) SATISFIABLE Models : 6 Time : 0.002s (Solving: 0.00s 1st Model: 0.00s Unsat: 0.00s) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 70 / 541

  17. ASP solving process ASP solving process Problem Solution ✻ Modeling Interpreting ❄ Logic ✲ ✲ ✲ Stable Grounder Solver Program Models Solving Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 71 / 541

  18. ASP solving process A coloring Answer: 6 node(1) [...] \ assign(6,b) assign(5,r) assign(4,b) assign(3,g) assign(2,g) assign(1,r) 3 5 6 1 2 4 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 72 / 541

  19. ASP solving process A coloring Answer: 6 node(1) [...] \ assign(6,b) assign(5,r) assign(4,b) assign(3,g) assign(2,g) assign(1,r) 3 5 6 1 2 4 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 72 / 541

  20. Methodology Outline 1 Elaboration tolerance 2 ASP solving process 3 Methodology 4 Case studies Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 73 / 541

  21. Methodology Basic methodology Methodology Generate and Test (or: Guess and Check) Generator Generate potential stable model candidates (typically through non-deterministic constructs) Tester Eliminate invalid candidates (typically through integrity constraints) Nutshell Logic program = Data + Generator + Tester ( + Optimizer) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 74 / 541

  22. Methodology Basic methodology Methodology Generate and Test (or: Guess and Check) Generator Generate potential stable model candidates (typically through non-deterministic constructs) Tester Eliminate invalid candidates (typically through integrity constraints) Nutshell Logic program = Data + Generator + Tester ( + Optimizer) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 74 / 541

  23. Methodology Graph coloring node(1..6).        edge(1,2). edge(1,3). edge(1,4).     edge(2,4). edge(2,5). edge(2,6).     Problem edge(3,1). edge(3,4). edge(3,5).  instance edge(4,1). edge(4,2).   edge(5,3). edge(5,4). edge(5,6).     edge(6,2). edge(6,3). edge(6,5).         color(r). color(b). color(g).   { assign(N,C) : color(C) } = 1 :- node(N). Problem  encoding :- edge(N,M), assign(N,C), assign(M,C).  Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 75 / 541

  24. Methodology Graph coloring node(1..6).        edge(1,2). edge(1,3). edge(1,4).     edge(2,4). edge(2,5). edge(2,6).     edge(3,1). edge(3,4). edge(3,5).  Data edge(4,1). edge(4,2).   edge(5,3). edge(5,4). edge(5,6).     edge(6,2). edge(6,3). edge(6,5).         color(r). color(b). color(g).   { assign(N,C) : color(C) } = 1 :- node(N). Problem  encoding :- edge(N,M), assign(N,C), assign(M,C).  Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 75 / 541

  25. Methodology Graph coloring node(1..6).        edge(1,2). edge(1,3). edge(1,4).     edge(2,4). edge(2,5). edge(2,6).     edge(3,1). edge(3,4). edge(3,5).  Data edge(4,1). edge(4,2).   edge(5,3). edge(5,4). edge(5,6).     edge(6,2). edge(6,3). edge(6,5).         color(r). color(b). color(g).   { assign(N,C) : color(C) } = 1 :- node(N). Generator  Tester :- edge(N,M), assign(N,C), assign(M,C).  Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 75 / 541

  26. Case studies Outline 1 Elaboration tolerance 2 ASP solving process 3 Methodology 4 Case studies Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 76 / 541

  27. Case studies Satisfiability Outline 1 Elaboration tolerance 2 ASP solving process 3 Methodology 4 Case studies Satisfiability Queens Traveling salesperson Reviewer Assignment Planning Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 77 / 541

  28. Case studies Satisfiability Satisfiability testing Problem Instance A propositional formula φ in CNF Problem Class Is there an assignment of propositional variables to true and false such that a given formula φ is true Example: Consider formula ( a ∨ ¬ b ) ∧ ( ¬ a ∨ b ) Logic Program Generator Tester Stable models { a } ← ← ∼ a , b X 1 = { a , b } { b } ← ← a , ∼ b X 2 = {} Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 78 / 541

  29. Case studies Satisfiability Satisfiability testing Problem Instance A propositional formula φ in CNF Problem Class Is there an assignment of propositional variables to true and false such that a given formula φ is true Example: Consider formula ( a ∨ ¬ b ) ∧ ( ¬ a ∨ b ) Logic Program Generator Tester Stable models { a } ← ← ∼ a , b X 1 = { a , b } { b } ← ← a , ∼ b X 2 = {} Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 78 / 541

  30. Case studies Satisfiability Satisfiability testing Problem Instance A propositional formula φ in CNF Problem Class Is there an assignment of propositional variables to true and false such that a given formula φ is true Example: Consider formula ( a ∨ ¬ b ) ∧ ( ¬ a ∨ b ) Logic Program Generator Tester Stable models { a } ← ← ∼ a , b X 1 = { a , b } { b } ← ← a , ∼ b X 2 = {} Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 78 / 541

  31. Case studies Satisfiability Satisfiability testing Problem Instance A propositional formula φ in CNF Problem Class Is there an assignment of propositional variables to true and false such that a given formula φ is true Example: Consider formula ( a ∨ ¬ b ) ∧ ( ¬ a ∨ b ) Logic Program Generator Tester Stable models { a } ← ← ∼ a , b X 1 = { a , b } { b } ← ← a , ∼ b X 2 = {} Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 78 / 541

  32. Case studies Satisfiability Satisfiability testing Problem Instance A propositional formula φ in CNF Problem Class Is there an assignment of propositional variables to true and false such that a given formula φ is true Example: Consider formula ( a ∨ ¬ b ) ∧ ( ¬ a ∨ b ) Logic Program Generator Tester Stable models { a } ← ← ∼ a , b X 1 = { a , b } { b } ← ← a , ∼ b X 2 = {} Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 78 / 541

  33. Case studies Queens Outline 1 Elaboration tolerance 2 ASP solving process 3 Methodology 4 Case studies Satisfiability Queens Traveling salesperson Reviewer Assignment Planning Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 79 / 541

  34. Case studies Queens The n-queens problem 5 Z0Z0Z 4 0Z0Z0 Place n queens on an n × n chess board Queens must not attack one another 3 Z0Z0Z 2 0Z0Z0 Q Q Q 1 Z0Z0Z Q Q 1 2 3 4 5 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 80 / 541

  35. Case studies Queens Defining the field queens.lp row(1..n). col(1..n). Create file queens.lp Define the field n rows n columns Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 81 / 541

  36. Case studies Queens Defining the field Running . . . $ clingo queens.lp --const n=5 Answer: 1 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) SATISFIABLE Models : 1 Time : 0.000 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 82 / 541

  37. Case studies Queens Placing some queens queens.lp row(1..n). col(1..n). { queen(I,J) : row(I), col(J) } . Guess a solution candidate by placing some queens on the board Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 83 / 541

  38. Case studies Queens Placing some queens Running . . . $ clingo queens.lp --const n=5 3 Answer: 1 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) Answer: 2 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) queen(1,1) Answer: 3 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) queen(2,1) SATISFIABLE Models : 3+ Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 84 / 541

  39. Case studies Queens Placing some queens Answer: 1 5 Z0Z0Z 4 0Z0Z0 3 Z0Z0Z 2 0Z0Z0 Answer: 1 row(1) row(2) row(3) row(4) row(5) \ 1 Z0Z0Z col(1) col(2) col(3) col(4) col(5) 1 2 3 4 5 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 85 / 541

  40. Case studies Queens Placing some queens Answer: 2 5 Z0Z0Z 4 0Z0Z0 3 Z0Z0Z 2 0Z0Z0 Answer: 2 row(1) row(2) row(3) row(4) row(5) \ 1 L0Z0Z col(1) col(2) col(3) col(4) col(5) \ queen(1,1) 1 2 3 4 5 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 86 / 541

  41. Case studies Queens Placing some queens Answer: 3 5 Z0Z0Z 4 0Z0Z0 3 Z0Z0Z 2 QZ0Z0 Answer: 3 row(1) row(2) row(3) row(4) row(5) \ 1 Z0Z0Z col(1) col(2) col(3) col(4) col(5) \ queen(2,1) 1 2 3 4 5 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 87 / 541

  42. Case studies Queens Placing n queens queens.lp row(1..n). col(1..n). { queen(I,J) : row(I), col(J) }. :- { queen(I,J) } != n. Place exactly n queens on the board Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 88 / 541

  43. Case studies Queens Placing n queens queens.lp row(1..n). col(1..n). { queen(I,J) : row(I), col(J) }. :- not { queen(I,J) } = n. Place exactly n queens on the board Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 88 / 541

  44. Case studies Queens Placing n queens Running . . . $ clingo queens.lp --const n=5 2 Answer: 1 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) \ queen(5,1) queen(4,1) queen(3,1) queen(2,1) queen(1,1) Answer: 2 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) \ queen(1,2) queen(4,1) queen(3,1) queen(2,1) queen(1,1) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 89 / 541

  45. Case studies Queens Placing n queens Answer: 1 5 L0Z0Z 4 QZ0Z0 3 L0Z0Z Answer: 1 2 QZ0Z0 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) \ 1 L0Z0Z queen(5,1) queen(4,1) queen(3,1) queen(2,1) queen(1,1) 1 2 3 4 5 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 90 / 541

  46. Case studies Queens Placing n queens Answer: 2 5 Z0Z0Z 4 QZ0Z0 3 L0Z0Z Answer: 2 2 QZ0Z0 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) \ 1 LQZ0Z queen(1,2) queen(4,1) queen(3,1) queen(2,1) queen(1,1) 1 2 3 4 5 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 91 / 541

  47. Case studies Queens Horizontal and vertical attack queens.lp row(1..n). col(1..n). { queen(I,J) : row(I), col(J) }. :- { queen(I,J) } != n. :- queen(I,J), queen(I,J’), J != J’. :- queen(I,J), queen(I’,J), I != I’. Forbid horizontal attacks Forbid vertical attacks Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 92 / 541

  48. Case studies Queens Horizontal and vertical attack queens.lp row(1..n). col(1..n). { queen(I,J) : row(I), col(J) }. :- { queen(I,J) } != n. :- queen(I,J), queen(I,J’), J != J’. :- queen(I,J), queen(I’,J), I != I’. Forbid horizontal attacks Forbid vertical attacks Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 92 / 541

  49. Case studies Queens Horizontal and vertical attack Running . . . $ clingo queens.lp --const n=5 Answer: 1 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) \ queen(5,5) queen(4,4) queen(3,3) queen(2,2) queen(1,1) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 93 / 541

  50. Case studies Queens Horizontal and vertical attack Answer: 1 5 Z0Z0L 4 0Z0L0 3 Z0L0Z Answer: 1 2 0L0Z0 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) \ 1 L0Z0Z queen(5,5) queen(4,4) queen(3,3) queen(2,2) queen(1,1) 1 2 3 4 5 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 94 / 541

  51. Case studies Queens Diagonal attack queens.lp row(1..n). col(1..n). { queen(I,J) : row(I), col(J) }. :- { queen(I,J) } != n. :- queen(I,J), queen(I,J’), J != J’. :- queen(I,J), queen(I’,J), I != I’. :- queen(I,J), queen(I’,J’), (I,J) != (I’,J’), I-J == I’-J’. :- queen(I,J), queen(I’,J’), (I,J) != (I’,J’), I+J == I’+J’. Forbid diagonal attacks Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 95 / 541

  52. Case studies Queens Diagonal attack Running . . . $ clingo queens.lp --const n=5 Answer: 1 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) \ queen(4,5) queen(1,4) queen(3,3) queen(5,2) queen(2,1) SATISFIABLE Models : 1+ Time : 0.000 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 96 / 541

  53. Case studies Queens Diagonal attack Answer: 1 5 ZQZ0Z 4 0Z0ZQ 3 Z0L0Z Answer: 1 2 QZ0Z0 row(1) row(2) row(3) row(4) row(5) \ col(1) col(2) col(3) col(4) col(5) \ 1 Z0ZQZ queen(4,5) queen(1,4) queen(3,3) queen(5,2) queen(2,1) 1 2 3 4 5 Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 97 / 541

  54. Case studies Queens Optimizing queens-opt.lp { queen(I,1..n) } = 1 :- I = 1..n. { queen(1..n,J) } = 1 :- J = 1..n. :- { queen(D-J,J) } > 1, D = 2..2*n. :- { queen(D+J,J) } > 1, D = 1-n..n-1. Encoding can be optimized Much faster to solve Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 98 / 541

  55. Case studies Queens And sometimes it rocks $ clingo -c n=5000 queens-opt-diag.lp --config=jumpy -q --stats=2 clingo version 4.1.0 Solving... SATISFIABLE Models : 1+ Time : 3758.143s (Solving: 1905.22s 1st Model: 1896.20s Unsat: 0.00s) CPU Time : 3758.320s Choices : 288594554 Conflicts : 3442 (Analyzed: 3442) Restarts : 17 (Average: 202.47 Last: 3442) Model-Level : 7594728.0 Problems : 1 (Average Length: 0.00 Splits: 0) Lemmas : 3442 (Deleted: 0) Binary : 0 (Ratio: 0.00%) Ternary : 0 (Ratio: 0.00%) Conflict : 3442 (Average Length: 229056.5 Ratio: 100.00%) Loop : 0 (Average Length: 0.0 Ratio: 0.00%) Other : 0 (Average Length: 0.0 Ratio: 0.00%) Atoms : 75084857 (Original: 75069989 Auxiliary: 14868) Rules : 100129956 (1: 50059992/100090100 2: 39990/29856 3: 10000/10000) Bodies : 25090103 Equivalences : 125029999 (Atom=Atom: 50009999 Body=Body: 0 Other: 75020000) Tight : Yes Variables : 25024868 (Eliminated: 11781 Frozen: 25000000) Constraints : 66664 (Binary: 35.6% Ternary: 0.0% Other: 64.4%) Backjumps : 3442 (Average: 681.19 Max: 169512 Sum: 2344658) Executed : 3442 (Average: 681.19 Max: 169512 Sum: 2344658 Ratio: 100.00%) Bounded : 0 (Average: 0.00 Max: 0 Sum: 0 Ratio: 0.00%) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 99 / 541

  56. Case studies Queens And sometimes it rocks $ clingo -c n=5000 queens-opt-diag.lp --config=jumpy -q --stats=2 clingo version 4.1.0 Solving... SATISFIABLE Models : 1+ Time : 3758.143s (Solving: 1905.22s 1st Model: 1896.20s Unsat: 0.00s) CPU Time : 3758.320s Choices : 288594554 Conflicts : 3442 (Analyzed: 3442) Restarts : 17 (Average: 202.47 Last: 3442) Model-Level : 7594728.0 Problems : 1 (Average Length: 0.00 Splits: 0) Lemmas : 3442 (Deleted: 0) Binary : 0 (Ratio: 0.00%) Ternary : 0 (Ratio: 0.00%) Conflict : 3442 (Average Length: 229056.5 Ratio: 100.00%) Loop : 0 (Average Length: 0.0 Ratio: 0.00%) Other : 0 (Average Length: 0.0 Ratio: 0.00%) Atoms : 75084857 (Original: 75069989 Auxiliary: 14868) Rules : 100129956 (1: 50059992/100090100 2: 39990/29856 3: 10000/10000) Bodies : 25090103 Equivalences : 125029999 (Atom=Atom: 50009999 Body=Body: 0 Other: 75020000) Tight : Yes Variables : 25024868 (Eliminated: 11781 Frozen: 25000000) Constraints : 66664 (Binary: 35.6% Ternary: 0.0% Other: 64.4%) Backjumps : 3442 (Average: 681.19 Max: 169512 Sum: 2344658) Executed : 3442 (Average: 681.19 Max: 169512 Sum: 2344658 Ratio: 100.00%) Bounded : 0 (Average: 0.00 Max: 0 Sum: 0 Ratio: 0.00%) Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 99 / 541

  57. Case studies Traveling salesperson Outline 1 Elaboration tolerance 2 ASP solving process 3 Methodology 4 Case studies Satisfiability Queens Traveling salesperson Reviewer Assignment Planning Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 100 / 541

  58. Case studies Traveling salesperson The traveling salesperson problem (TSP) Problem Instance A set of cities and distances among them, or simply a weighted graph Problem Class What is the shortest possible route visiting each city and returning to the city of origin? Note TSP extends the Hamiltonian cycle problem: Is there a cycle in a graph visiting each node exactly once TSP is relevant to applications in logistics, planning, chip design, and the core of the vehicle routing problem Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 101 / 541

  59. Case studies Traveling salesperson The traveling salesperson problem (TSP) Problem Instance A set of cities and distances among them, or simply a weighted graph Problem Class What is the shortest possible route visiting each city and returning to the city of origin? Note TSP extends the Hamiltonian cycle problem: Is there a cycle in a graph visiting each node exactly once TSP is relevant to applications in logistics, planning, chip design, and the core of the vehicle routing problem Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 101 / 541

  60. Case studies Traveling salesperson Traveling salesperson node(1..6). edge(1,(2;3;4)). edge(2,(4;5;6)). edge(3,(1;4;5)). edge(4,(1;2)). edge(5,(3;4;6)). edge(6,(2;3;5)). cost(1,2,2). cost(1,3,3). cost(1,4,1). cost(2,4,2). cost(2,5,2). cost(2,6,4). cost(3,1,3). cost(3,4,2). cost(3,5,2). cost(4,1,1). cost(4,2,2). cost(5,3,2). cost(5,4,2). cost(5,6,1). cost(6,2,4). cost(6,3,3). cost(6,5,1). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 102 / 541

  61. Case studies Traveling salesperson Traveling salesperson node(1..6). edge(1,(2;3;4)). edge(2,(4;5;6)). edge(3,(1;4;5)). edge(4,(1;2)). edge(5,(3;4;6)). edge(6,(2;3;5)). cost(1,2,2). cost(1,3,3). cost(1,4,1). cost(2,4,2). cost(2,5,2). cost(2,6,4). cost(3,1,3). cost(3,4,2). cost(3,5,2). cost(4,1,1). cost(4,2,2). cost(5,3,2). cost(5,4,2). cost(5,6,1). cost(6,2,4). cost(6,3,3). cost(6,5,1). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 102 / 541

  62. Case studies Traveling salesperson Traveling salesperson node(1..6). edge(1,(2;3;4)). edge(2,(4;5;6)). edge(3,(1;4;5)). edge(4,(1;2)). edge(5,(3;4;6)). edge(6,(2;3;5)). cost(1,2,2). cost(1,3,3). cost(1,4,1). cost(2,4,2). cost(2,5,2). cost(2,6,4). cost(3,1,3). cost(3,4,2). cost(3,5,2). cost(4,1,1). cost(4,2,2). cost(5,3,2). cost(5,4,2). cost(5,6,1). cost(6,2,4). cost(6,3,3). cost(6,5,1). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 102 / 541

  63. Case studies Traveling salesperson Traveling salesperson node(1..6). edge(1,(2;3;4)). edge(2,(4;5;6)). edge(3,(1;4;5)). edge(4,(1;2)). edge(5,(3;4;6)). edge(6,(2;3;5)). cost(1,2,2). cost(1,3,3). cost(1,4,1). cost(2,4,2). cost(2,5,2). cost(2,6,4). cost(3,1,3). cost(3,4,2). cost(3,5,2). cost(4,1,1). cost(4,2,2). cost(5,3,2). cost(5,4,2). cost(5,6,1). cost(6,2,4). cost(6,3,3). cost(6,5,1). edge(X,Y) :- cost(X,Y,_). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 102 / 541

  64. Case studies Traveling salesperson Traveling salesperson node(1..6). edge(1,(2;3;4)). edge(2,(4;5;6)). edge(3,(1;4;5)). edge(4,(1;2)). edge(5,(3;4;6)). edge(6,(2;3;5)). cost(1,2,2). cost(1,3,3). cost(1,4,1). cost(2,4,2). cost(2,5,2). cost(2,6,4). cost(3,1,3). cost(3,4,2). cost(3,5,2). cost(4,1,1). cost(4,2,2). cost(5,3,2). cost(5,4,2). cost(5,6,1). cost(6,2,4). cost(6,3,3). cost(6,5,1). edge(X,Y) :- cost(X,Y,_). node(X) :- cost(X,_,_). node(Y) :- cost(_,Y,_). Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 102 / 541

  65. Case studies Traveling salesperson Traveling salesperson { cycle(X,Y) : edge(X,Y) } = 1 :- node(X). { cycle(X,Y) : edge(X,Y) } = 1 :- node(Y). reached(Y) :- cycle(1,Y). reached(Y) :- cycle(X,Y), reached(X). :- node(Y), not reached(Y). #minimize { C,X,Y : cycle(X,Y), cost(X,Y,C) }. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 103 / 541

  66. Case studies Traveling salesperson Traveling salesperson { cycle(X,Y) : edge(X,Y) } = 1 :- node(X). { cycle(X,Y) : edge(X,Y) } = 1 :- node(Y). reached(Y) :- cycle(1,Y). reached(Y) :- cycle(X,Y), reached(X). :- node(Y), not reached(Y). #minimize { C,X,Y : cycle(X,Y), cost(X,Y,C) }. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 103 / 541

  67. Case studies Traveling salesperson Traveling salesperson { cycle(X,Y) : edge(X,Y) } = 1 :- node(X). { cycle(X,Y) : edge(X,Y) } = 1 :- node(Y). reached(Y) :- cycle(1,Y). reached(Y) :- cycle(X,Y), reached(X). :- node(Y), not reached(Y). #minimize { C,X,Y : cycle(X,Y), cost(X,Y,C) }. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 103 / 541

  68. Case studies Traveling salesperson Traveling salesperson { cycle(X,Y) : edge(X,Y) } = 1 :- node(X). { cycle(X,Y) : edge(X,Y) } = 1 :- node(Y). reached(Y) :- cycle(1,Y). reached(Y) :- cycle(X,Y), reached(X). :- node(Y), not reached(Y). #minimize { C,X,Y : cycle(X,Y), cost(X,Y,C) }. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 103 / 541

  69. Case studies Reviewer Assignment Outline 1 Elaboration tolerance 2 ASP solving process 3 Methodology 4 Case studies Satisfiability Queens Traveling salesperson Reviewer Assignment Planning Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 104 / 541

  70. Case studies Reviewer Assignment Reviewer Assignment Problem Instance A set of papers and a set of reviewers along with their first and second choices of papers and conflict of interests Problem Class A nice assignment of three reviewers to each paper Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 105 / 541

  71. Case studies Reviewer Assignment Reviewer Assignment Problem Instance A set of papers and a set of reviewers along with their first and second choices of papers and conflict of interests Problem Class A “nice” assignment of three reviewers to each paper Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 105 / 541

  72. Case studies Reviewer Assignment Reviewer Assignment by Ilkka Niemel¨ a paper(p1). reviewer(r1). classA(r1,p1). classB(r1,p2). coi(r1,p3). paper(p2). reviewer(r2). classA(r1,p3). classB(r1,p4). coi(r1,p6). [...] { assigned(P,R) : reviewer(R) } = 3 :- paper(P). :- assigned(P,R), coi(R,P). :- assigned(P,R), not classA(R,P), not classB(R,P). :- not 6 { assigned(P,R) : paper(P) } 9, reviewer(R). assignedB(P,R) :- classB(R,P), assigned(P,R). :- 3 { assignedB(P,R) : paper(P) }, reviewer(R). #minimize { 1,P,R : assignedB(P,R), paper(P), reviewer(R) }. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 106 / 541

  73. Case studies Reviewer Assignment Reviewer Assignment by Ilkka Niemel¨ a paper(p1). reviewer(r1). classA(r1,p1). classB(r1,p2). coi(r1,p3). paper(p2). reviewer(r2). classA(r1,p3). classB(r1,p4). coi(r1,p6). [...] { assigned(P,R) : reviewer(R) } = 3 :- paper(P). :- assigned(P,R), coi(R,P). :- assigned(P,R), not classA(R,P), not classB(R,P). :- not 6 { assigned(P,R) : paper(P) } 9, reviewer(R). assignedB(P,R) :- classB(R,P), assigned(P,R). :- 3 { assignedB(P,R) : paper(P) }, reviewer(R). #minimize { 1,P,R : assignedB(P,R), paper(P), reviewer(R) }. Torsten Schaub (KRR@UP) Answer Set Solving in Practice October 20, 2018 106 / 541

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