Jip J. Dekker
f2a1c4e389
git-subtree-dir: software/mza git-subtree-split: f970a59b177c13ca3dd8aaef8cc6681d83b7e813
121 lines
4.8 KiB
MiniZinc
121 lines
4.8 KiB
MiniZinc
% RUNS ON mzn20_fd
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% RUNS ON mzn-fzn_fd
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% RUNS ON mzn20_fd_linear
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% RUNS ON mzn20_mip
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%-----------------------------------------------------------------------------%
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% The wolf, goat, cabbage problem
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%
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% A farmer has to take a wolf, goat and cabbage across a bridge
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% He can only take one thing at a time
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% The wolf and goat can't be left together alone (without the farmer)
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% The goat and cabbage can't be left alone together
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%-----------------------------------------------------------------------------%
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% horizon is the maximum number of steps that might be required in the plan
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int: horizon = 20;
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%-----------------------------------------------------------------------------%
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% -1..1 represent the three locations:
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% -1 is on the left bank of the river
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% 0 is on the bridge
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% 1 is on the right bank
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% A boolean, for each object, time and location,
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% holds if the object is at the location at that time
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array [1..horizon,-1..1] of var bool: wolf;
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array [1..horizon,-1..1] of var bool: goat;
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array [1..horizon,-1..1] of var bool: cabbage;
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array [1..horizon,-1..1] of var bool: farmer;
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%-----------------------------------------------------------------------------%
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% Things can only move from bank to bridge, or from bridge to bank, in one step
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constraint forall(t in 2..horizon,loc1 in -1..1,loc2 in -1..1)
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(wolf[t-1,loc1] /\ wolf[t,loc2] -> loc1-loc2<2 /\ loc2-loc1<2);
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constraint forall(t in 2..horizon,loc1 in -1..1,loc2 in -1..1)
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(goat[t-1,loc1] /\ goat[t,loc2] -> loc1-loc2<2 /\ loc2-loc1<2);
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constraint forall(t in 2..horizon,loc1 in -1..1,loc2 in -1..1)
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(cabbage[t-1,loc1] /\ cabbage[t,loc2] -> loc1-loc2<2 /\ loc2-loc1<2);
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constraint forall(t in 2..horizon,loc1 in -1..1,loc2 in -1..1)
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(farmer[t-1,loc1] /\ farmer[t,loc2] -> loc1-loc2<2 /\ loc2-loc1<2);
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% Can't leave wolf and goat, or goat and cabbage, together
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constraint (forall(t in 1..horizon,loc in -1..1)
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( (wolf[t,loc] /\ goat[t,loc] -> farmer[t,loc])
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/\
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(goat[t,loc] /\ cabbage[t,loc] -> farmer[t,loc])
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) );
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% The wolf is somewhere at each time point, and is only in one place
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constraint forall(t in 1..horizon)
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(exists(loc in -1..1)
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(wolf[t,loc] /\ forall(loc2 in -1..1 where loc2!=loc)
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(not wolf[t,loc2])
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)
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);
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% Similarly for the goat, cabbage and farmer...
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constraint forall(t in 1..horizon)
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(exists(loc in -1..1)
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(goat[t,loc] /\ forall(loc2 in -1..1 where loc2!=loc)
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(not goat[t,loc2])
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)
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);
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constraint forall(t in 1..horizon)
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(exists(loc in -1..1)
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(cabbage[t,loc] /\ forall(loc2 in -1..1 where loc2!=loc)
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(not cabbage[t,loc2])
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)
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);
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constraint forall(t in 1..horizon)
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(exists(loc in -1..1)
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(farmer[t,loc] /\ forall(loc2 in -1..1 where loc2!=loc)
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(not farmer[t,loc2])
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)
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);
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% The wolf can ony be on the bridge if:
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% (a) The farmer is on the bridge, and neither the goat nor the cabbage is
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% (b) The farmer was previously on the same bank as the wolf
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% (c) The farmer goes subsequently to the same bank as the wolf
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constraint forall(t in 2..horizon-1)
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((wolf[t,0] -> farmer[t,0] /\ not goat[t,0] /\ not cabbage[t,0]
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/\ (wolf[t-1,1] <-> farmer[t-1,1]) /\ not farmer[t-1,0]
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/\ (wolf[t+1,1] <-> farmer[t+1,1]) /\ not farmer[t+1,0])
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/\
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% Similarly for the cabbage
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(cabbage[t,0] -> farmer[t,0] /\ not goat[t,0] /\ not wolf[t,0]
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/\ (cabbage[t-1,1] <-> farmer[t-1,1]) /\ not farmer[t-1,0]
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/\ (cabbage[t+1,1] <-> farmer[t+1,1]) /\ not farmer[t+1,0])
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/\
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% and for the goat
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(goat[t,0] -> farmer[t,0] /\ not wolf[t,0] /\ not cabbage[t,0]
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/\ (goat[t-1,1] <-> farmer[t-1,1]) /\ not farmer[t-1,0]
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/\ (goat[t+1,1] <-> farmer[t+1,1]) /\ not farmer[t+1,0])
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);
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% The animals all start on the right bank and finish on the left bank
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constraint
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(wolf[1,1] /\ goat[1,1] /\ cabbage[1,1]
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/\
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wolf[horizon,-1] /\ goat[horizon,-1] /\ cabbage[horizon,-1]
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);
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solve ::bool_search(array1d(1..3*horizon,wolf)++array1d(1..3*horizon,goat)++array1d(1..3*horizon,cabbage)++array1d(1..3*horizon,farmer), input_order, indomain_max, complete) satisfy;
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%-----------------------------------------------------------------------------%
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output [
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"wolf : ", show(wolf), "\n",
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"goat : ", show(goat), "\n",
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"cabbage : ", show(cabbage), "\n",
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"farmer : ", show(farmer), "\n"
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];
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%-----------------------------------------------------------------------------%
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%-----------------------------------------------------------------------------%
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