Computer Science Canada :: View topic - missionaries and cannibals

% Main control block and printing

find :-
path([3,3,left],[0,0,right],[[3,3,left]],_).
output([]) :- nl, nl.
output([[A,B,String]|T]) :-
output(T),
write(B), write(' ~~ '), write(A), write(': '), write(String), nl.

% Base case

path([A,B,C],[A,B,C],_,MoveList):-
nl,nl,output(MoveList).

% Recursive call to solve the problem

path([A,B,C],[D,E,F],Traversed,Moves) :-
move([A,B,C],[I,J,K],Out),
legal([I,J,K]), % Don't use this move unless it's safe.
not(member([I,J,K],Traversed)),
path([I,J,K],[D,E,F],[[I,J,K]|Traversed],[ [[I,J,K],[A,B,C],Out] | Moves ]).

% Move commands and descriptions of the move

move([A,B,left],[C,B,right],'One missionary crosses the river') :-
A > 0, C is A - 1.
move([A,B,left],[C,B,right],'Two missionaries cross the river') :-
A > 1, C is A - 2.
move([A,B,left],[C,D,right],'One missionary and One cannibal cross the river') :-
A > 0, B > 0, C is A - 1, D is B - 1.
move([A,B,left],[A,D,right],'One cannibal crosses the river') :-
B > 0, D is B - 1.
move([A,B,left],[A,D,right],'Two cannibals cross the river') :-
B > 1, D is B - 2.
move([A,B,right],[C,B,left],'One missionary returns from the other side') :-
A < 3, C is A + 1.
move([A,B,right],[C,B,left],'Two missionaries return from the other side') :-
A < 2, C is A + 2.
move([A,B,right],[C,D,left],'One missionary and One cannibal return from the other side') :-
A < 3, B < 3, C is A + 1, D is B + 1.
move([A,B,right],[A,D,left],'One cannibal returns from the other side') :-
B < 3, D is B + 1.
move([A,B,right],[A,D,left],'Two cannibals return from the other side') :-
B < 2, D is B + 2.

% Legal move definition where B is missionaries and A is cannibals:

legal([B,A,_]) :-
(A =< B ; B = 0),
C is 3-A, D is 3-B,
(C =< D; D = 0).

Author:	Tubs [ Tue Feb 12, 2008 12:23 pm ]
Post subject:	missionaries and cannibals
Quote: In the missionaries and cannibals problem, three missionaries and three cannibals must cross a river using a boat which can carry at most two people, under the constraint that, for both banks, if there are missionaries present on the bank, they cannot be outnumbered by cannibals (if they were, the cannibals would eat the missionaries.) The boat cannot cross the river by itself with no people on board. I have so far developed a program in Prolog to solve this problem, but the question requires that the program solve for all possible solutions. My first idea was to record the first solution and check that the next solution isn't identical to the first, but would this not just incur an infinate loop when the program restarts? I am a total noob at Prolog and the prof is sub-par, does anyone have any suggestions? I can provide code if necessary.