Include Paulo Moura's Logtalk OO LP system
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Logtalk/examples/searching/SCRIPT
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Logtalk/examples/searching/SCRIPT
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=================================================================
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Logtalk - Object oriented extension to Prolog
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Release 2.8.4
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Copyright (c) 1998-2001 Paulo Moura. All Rights Reserved.
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=================================================================
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% farmer, cabbage, goat and wolf problem
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| ?- farmer::initial_state(Initial), depth_first(10)::solve(farmer, Initial, Path), farmer::print_path(Path).
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cgwf.<__>..........____
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c_w_..........<__>.f_g_
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c_wf.<__>..........__g_
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__w_..........<__>.fcg_
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_gwf.<__>.........._c__
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_g__..........<__>.fc_w
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_g_f.<__>.........._c_w
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____..........<__>.fcgw
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Path = [(north,north,north,north),(north,south,north,south),(north,south,north,north),(south,south,north,south),(south,north,north,north),(south,north,south,south),(south,north,south,north),(south,south,south,south)],
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Initial = (north,north,north,north) ?
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yes
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% missionaires and cannibals problem, solved using a hill-climbing strategy
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| ?- miss_cann::initial_state(Initial), hill_climbing(16)::solve(miss_cann, Initial, Path, Cost), miss_cann::print_path(Path).
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MMMCCC.<__>..........
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MMCC..........<__>.MC
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MMMCC.<__>..........C
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MMM..........<__>.CCC
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MMMC.<__>..........CC
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MC..........<__>.MMCC
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MMCC.<__>..........MC
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CC..........<__>.MMMC
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CCC.<__>..........MMM
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C..........<__>.MMMCC
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CC.<__>..........MMMC
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..........<__>.MMMCCC
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Cost = 15,
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Path = [((3,3),esq,0,0),((2,2),dir,1,1),((3,2),esq,0,1),((3,0),dir,0,3),((3,1),esq,0,2),((1,1),dir,2,2),((2,2),esq,1,1),((0,2),dir,3,1),((0,3),esq,3,0),((0,1),dir,3,2),((0,2),esq,3,1),((0,0),dir,3,3)],
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Initial = ((3,3),esq,0,0)
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yes
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% same problem as above with the addition of a monitor to measure hill-climbing performance
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| ?- performance::init, miss_cann::initial_state(Initial), hill_climbing(16)::solve(miss_cann, Initial, Path, Cost), miss_cann::print_path(Path), performance::report.
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MMMCCC.<__>..........
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MMCC..........<__>.MC
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MMMCC.<__>..........C
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MMM..........<__>.CCC
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MMMC.<__>..........CC
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MC..........<__>.MMCC
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MMCC.<__>..........MC
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CC..........<__>.MMMC
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CCC.<__>..........MMM
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C..........<__>.MMMCC
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CC.<__>..........MMMC
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..........<__>.MMMCCC
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solution length: 12
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number of state transitions: 27
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ratio solution length / state transitions: 0.4444444444444444
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minimum branching degree: 2
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average branching degree: 2.5555555555555554
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maximum branching degree: 3
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time: 0.067999999999756255
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Cost = 15,
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Path = [((3,3),esq,0,0),((2,2),dir,1,1),((3,2),esq,0,1),((3,0),dir,0,3),((3,1),esq,0,2),((1,1),dir,2,2),((2,2),esq,1,1),((0,2),dir,3,1),((0,3),esq,3,0),((0,1),dir,3,2),((0,2),esq,3,1),((0,0),dir,3,3)],
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Initial = ((3,3),esq,0,0) ?
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yes
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% water jugs problem solved using a breadth and a depth first strategy, with performance monitors
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% it's interesting to compare the results
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| ?- performance::init, water_jug::initial_state(Initial), breadth_first(6)::solve(water_jug, Initial, Path), water_jug::print_path(Path), performance::report.
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4-gallon jug: 0
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3-gallon jug: 0
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4-gallon jug: 0
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3-gallon jug: 3
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4-gallon jug: 3
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3-gallon jug: 0
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4-gallon jug: 3
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3-gallon jug: 3
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4-gallon jug: 4
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3-gallon jug: 2
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4-gallon jug: 0
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3-gallon jug: 2
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solution length: 6
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number of state transitions: 105
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ratio solution length / state transitions: 0.05714285714285714
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minimum branching degree: 2
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average branching degree: 3.6315789473684212
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maximum branching degree: 4
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time: 0.20000000000027285
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Path = [(0,0),(0,3),(3,0),(3,3),(4,2),(0,2)],
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Initial = (0,0) ?
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yes
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| ?- performance::init, water_jug::initial_state(Initial), depth_first(10)::solve(water_jug, Initial, Path), water_jug::print_path(Path), performance::report.
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4-gallon jug: 0
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3-gallon jug: 0
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4-gallon jug: 4
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3-gallon jug: 0
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4-gallon jug: 4
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3-gallon jug: 3
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4-gallon jug: 0
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3-gallon jug: 3
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4-gallon jug: 3
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3-gallon jug: 0
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4-gallon jug: 3
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3-gallon jug: 3
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4-gallon jug: 4
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3-gallon jug: 2
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4-gallon jug: 0
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3-gallon jug: 2
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solution length: 8
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number of state transitions: 12
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ratio solution length / state transitions: 0.6666666666666666
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minimum branching degree: 1
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average branching degree: 2.0
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maximum branching degree: 3
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time: 0.021999999999934516
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Path = [(0,0),(4,0),(4,3),(0,3),(3,0),(3,3),(4,2),(0,2)],
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Initial = (0,0) ?
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yes
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% eight puzzle solved using a hill-climbing strategy
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| ?- performance::init, eight_puzzle::initial_state(five_steps, Initial), hill_climbing(25)::solve(eight_puzzle, Initial, Path, Cost), eight_puzzle::print_path(Path), performance::report.
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283
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164
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7 5
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283
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1 4
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765
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2 3
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184
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765
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23
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184
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765
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123
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84
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765
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123
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8 4
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765
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solution length: 6
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number of state transitions: 15
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ratio solution length / state transitions: 0.4
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minimum branching degree: 2
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average branching degree: 3.1333333333333333
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maximum branching degree: 4
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time: 0.050000000000181899
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Cost = 5,
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Path = [[2/1,1/2,1/3,3/3,3/2,3/1,2/2,1/1,2/3],[2/2,1/2,1/3,3/3,3/2,3/1,2/1,1/1,2/3],[2/3,1/2,1/3,3/3,3/2,3/1,2/1,1/1,2/2],[1/3,1/2,2/3,3/3,3/2,3/1,2/1,1/1,2/2],[1/2,1/3,2/3,3/3,3/2,3/1,2/1,1/1,2/2],[2/2,1/3,2/3,3/3,3/2,3/1,2/1,1/1,1/2]],
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Initial = [2/1,1/2,1/3,3/3,3/2,3/1,2/2,1/1,2/3] ?
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yes
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% eight puzzle solved using a best-first strategy
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| ?- performance::init, eight_puzzle::initial_state(five_steps, Initial), best_first(25)::solve(eight_puzzle, Initial, Path, Cost), eight_puzzle::print_path(Path), performance::report.
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283
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164
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7 5
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283
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1 4
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765
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2 3
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184
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765
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23
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184
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765
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123
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84
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765
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123
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8 4
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765
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solution length: 6
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number of state transitions: 15
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ratio solution length / state transitions: 0.4
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minimum branching degree: 2
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average branching degree: 3.1333333333333333
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maximum branching degree: 4
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time: 0.046000000000276486
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Cost = 5,
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Path = [[2/1,1/2,1/3,3/3,3/2,3/1,2/2,1/1,2/3],[2/2,1/2,1/3,3/3,3/2,3/1,2/1,1/1,2/3],[2/3,1/2,1/3,3/3,3/2,3/1,2/1,1/1,2/2],[1/3,1/2,2/3,3/3,3/2,3/1,2/1,1/1,2/2],[1/2,1/3,2/3,3/3,3/2,3/1,2/1,1/1,2/2],[2/2,1/3,2/3,3/3,3/2,3/1,2/1,1/1,1/2]],
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Initial = [2/1,1/2,1/3,3/3,3/2,3/1,2/2,1/1,2/3] ?
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yes
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% turn off performance monitor
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| ?- performance::stop.
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