题目是这样的, 用swi-prolog写一个小AI程序:
有两个空水壶和无限的水,水壶都没有刻度..一个能装7升,另一个能装4升,用深度优先搜索,量出准确的5升水.水壶从空状态(0,0)开始,到(5,X).
代码如下,但出现了一点小问题,不能运行.
solve(5,_).
solve(X,Y):- X < 7,
\+ member((7,Y),L),
concat(E,[(X,Y)],L),
write('Fill 7 litre jug from tap.\n'),
solve(7,Y).
solve(X,Y):- Y < 4,
\+ member((X,4),L),
concat(E,[(X,Y)],L),
write('Fill 4 litre jug from tap.\n'),
solve(X,4).
solve(X,Y):- X+Y >= 7,
Y > 0,
Z is Y - (7 - X),
\+ member((7,Z),L),
concat(E,[(X,Z)],L),
write('Fill 7 litre jug with 4 litre jug.\n'),
solve(7,Z).
solve(X,Y):- X+Y >= 4,
X > 0,
Z is X - (4 - Y),
\+ member((Z,4),L),
concat(E,[(Z,Y)],L),
write('Fill 4 litre jug with 7 litre jug.\n'),
solve(Z,4).
solve(X,Y):- X+Y < 4,
Y > 0,
Z is X + Y,
\+ member((Z,0),L),
concat(E,[(Z,Y)],L),
write('Empty 4 litre jug into 7 litre jug.\n'),
solve(Z,0).
solve(X,Y):- X+Y < 7,
X > 0,
Z is X+Y,
\+ member((0,Z),L),
concat(E,[(X,Z)],L),
write('Empty 7 litre jug into 4 litre jug.\n'),
solve(0,Z).
solve(X,Y):- X > 0,
\+ member((0,Y),L),
concat(E,[(X,Y)],L),
write('Empty 7 litre jug.\n'),
solve(0,Y).
solve(X,Y):- Y > 0,
\+ member((X,0),L),
concat(E,[(X,Y)],L),
write('Empty 4 litre jug.\n'),
solve(X,0).
member(X,[X|L]).
member(X,[Y|L]):-X \= Y, member(X,L).
concat([],L,L).
concat([X|A],B,[X|L]):- concat(A,B,L).
有两个空水壶和无限的水,水壶都没有刻度..一个能装7升,另一个能装4升,用深度优先搜索,量出准确的5升水.水壶从空状态(0,0)开始,到(5,X).
代码如下,但出现了一点小问题,不能运行.
solve(5,_).
solve(X,Y):- X < 7,
\+ member((7,Y),L),
concat(E,[(X,Y)],L),
write('Fill 7 litre jug from tap.\n'),
solve(7,Y).
solve(X,Y):- Y < 4,
\+ member((X,4),L),
concat(E,[(X,Y)],L),
write('Fill 4 litre jug from tap.\n'),
solve(X,4).
solve(X,Y):- X+Y >= 7,
Y > 0,
Z is Y - (7 - X),
\+ member((7,Z),L),
concat(E,[(X,Z)],L),
write('Fill 7 litre jug with 4 litre jug.\n'),
solve(7,Z).
solve(X,Y):- X+Y >= 4,
X > 0,
Z is X - (4 - Y),
\+ member((Z,4),L),
concat(E,[(Z,Y)],L),
write('Fill 4 litre jug with 7 litre jug.\n'),
solve(Z,4).
solve(X,Y):- X+Y < 4,
Y > 0,
Z is X + Y,
\+ member((Z,0),L),
concat(E,[(Z,Y)],L),
write('Empty 4 litre jug into 7 litre jug.\n'),
solve(Z,0).
solve(X,Y):- X+Y < 7,
X > 0,
Z is X+Y,
\+ member((0,Z),L),
concat(E,[(X,Z)],L),
write('Empty 7 litre jug into 4 litre jug.\n'),
solve(0,Z).
solve(X,Y):- X > 0,
\+ member((0,Y),L),
concat(E,[(X,Y)],L),
write('Empty 7 litre jug.\n'),
solve(0,Y).
solve(X,Y):- Y > 0,
\+ member((X,0),L),
concat(E,[(X,Y)],L),
write('Empty 4 litre jug.\n'),
solve(X,0).
member(X,[X|L]).
member(X,[Y|L]):-X \= Y, member(X,L).
concat([],L,L).
concat([X|A],B,[X|L]):- concat(A,B,L).