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我正試圖解決swi-prolog中的2個水罐問題:給定2個罐的容量分別爲4加侖和3加侖,我想要找到在水罐中獲得2加侖的步驟容量4和0在另一個。序列中的2個水罐
我在C++中使用bfs和dfs編寫了這個問題的程序:http://kartikkukreja.wordpress.com/2013/10/11/water-jug-problem/。現在,我試圖在序言中解決這個問題。我對語言完全陌生,我的代碼不會終止。
這裏是我到目前爲止的代碼:
% state(0, 0) is the initial state
% state(2, 0) is the goal state
% Jugs 1 and 2 have capacities 4 and 3 respectively
% P is a path to the goal state
% C is the list of visited states
solution(P) :-
path(0, 0, [state(0, 0)], P).
path(2, 0, [state(2, 0)|_], _).
path(0, 2, C, P) :-
not(member(state(2, 0), C)),
path(2, 0, [state(2, 0)|C], R),
P = ['Pour 2 gallons from 3-Gallon jug to 4-gallon.'|R].
path(X, Y, C, P) :-
X < 4,
not(member(state(4, Y), C)),
path(4, Y, [state(4, Y)|C], R),
P = ['Fill the 4-Gallon Jug.'|R].
path(X, Y, C, P) :-
Y < 3,
not(member(state(X, 3), C)),
path(X, 3, [state(X, 3)|C], R),
P = ['Fill the 3-Gallon Jug.'|R].
path(X, Y, C, P) :-
X > 0,
not(member(state(0, Y), C)),
path(0, Y, [state(0, Y)|C], R),
P = ['Empty the 4-Gallon jug on ground.'|R].
path(X, Y, C, P) :-
Y > 0,
not(member(state(X, 0), C)),
path(X, 0, [state(X, 0)|C], R),
P = ['Empty the 3-Gallon jug on ground.'|R].
path(X, Y, C, P) :-
X + Y >= 4,
X < 4,
Y > 0,
NEW_Y = Y - (4 - X),
not(member(state(4, NEW_Y), C)),
path(4, NEW_Y, [state(4, NEW_Y)|C], R),
P = ['Pour water from 3-Gallon jug to 4-gallon until it is full.'|R].
path(X, Y, C, P) :-
X + Y >=3,
X > 0,
Y < 3,
NEW_X = X - (3 - Y),
not(member(state(NEW_X, 3), C)),
path(NEW_X, 3, [state(NEW_X, 3)|C], R),
P = ['Pour water from 4-Gallon jug to 3-gallon until it is full.'|R].
path(X, Y, C, P) :-
X + Y =< 4,
Y > 0,
NEW_X = X + Y,
not(member(state(NEW_X, 0), C)),
path(NEW_X, 0, [state(NEW_X, 0)|C], R),
P = ['Pour all the water from 3-Gallon jug to 4-gallon.'|R].
path(X, Y, C, P) :-
X + Y =< 3,
X > 0,
NEW_Y = X + Y,
not(member(state(0, NEW_Y), C)),
path(0, NEW_Y, [state(0, NEW_Y)|C], R),
P = ['Pour all the water from 4-Gallon jug to 3-gallon.'|R].
任何幫助表示讚賞。