为什么序言在收到 False 后不回溯?

Why does prolog not backtrack after receiving a False?

我有这段代码可以评估数字列表和 returns 匹配模式 [G,A],[A,B],[B,C],[C,D],[D,E],[E,F],[F,G] 的列表。但是,我希望这些数字只是唯一的数字。例如:([0,2],[2,4],[4,19],[19,3],[3,5],[5,7],[7,0])。根据输入的数字,不同的方法 returns true 或 false。然而,该代码仍然 returns 具有多个相似数字的数字。

r5(L,R):-
    R = [[G,A],[A,B],[B,C],[C,D],[D,E],[E,F],[F,G]],

    [A,B,C,D,E,F,G] ins 0 .. 27,

    different([A,B,C,D,E,F,G]),

    member([G,A],L),
    member([A,B],L),
    member([B,C],L),
    member([C,D],L),
    member([D,E],L),
    member([E,F],L),
    member([F,G],L),

    label([A,B,C,D,E,F,G]).

这是其中一部分的示例 returns:

R = [[0, 2], [2, 0], [0, 2], [2, 5], [5, 9], [9, 2], [2, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 5], [5, 23], [23, 17], [17, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 5], [5, 24], [24, 19], [19, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 5], [5, 26], [26, 12], [12, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 9], [9, 5], [5, 2], [2, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 9], [9, 15], [15, 2], [2, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 9], [9, 15], [15, 17], [17, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 9], [9, 15], [15, 19], [19, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 15], [15, 9], [9, 2], [2, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 21], [21, 1], [1, 12], [12, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 21], [21, 1], [1, 20], [20, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 21], [21, 11], [11, 12], [12, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 21], [21, 11], [11, 17], [17, 0]] ;
R = [[0, 2], [2, 0], [0, 2], [2, 21], [21, 24], [24, 19], [19, 0]] ;
R = [[0, 2], [2, 0], [0, 12], [12, 1], [1, 18], [18, 12], [12, 0]] ;
其中

None 是有效的,因为它们包含重复的数字。以下是输出中更下方的两个有效输出:

R = [[0, 2], [2, 9], [9, 4], [4, 8], [8, 23], [23, 17], [17, 0]] ;
R = [[0, 2], [2, 9], [9, 4], [4, 8], [8, 24], [24, 19], [19, 0]] ;

这是不同的方法:

different(X) :-
    sort(X, Sorted),
    length(X, OriginalLength),
    length(Sorted, SortedLength),
    OriginalLength == SortedLength.

这是我原来的电话:

r5_3([[0, 2],[2,0],[0, 12],[12,0],[0, 17],[17,0],[0, 19],[19,0],[0, 20],[20,0],[1, 4],[4,1],[1, 12],[12,1],[1, 18],[18,1],[1, 20],[20,1],[1, 21],[21,1],[2, 5],[5,2],[9, 2],[2,9],[2, 15],[15,2],[2, 21],[21,2],[8, 3],[3,8],[10, 3],[3,10],[16, 3],[3,16],[3, 22],[22,3],[25, 3],[3,25],[8, 4],[4,8],[9, 4],[4,9],[4, 23],[23,4],[26, 4],[4,26],[9, 5],[5,9],[5, 23],[23,5],[24, 5],[5,24],[26, 5],[5,26],[14, 6],[6,14],[17, 6],[6,17],[18, 6],[6,18],[24, 6],[6,24],[25, 6],[6,25],[18, 7],[7,18],[19, 7],[7,19],[22, 7],[7,22],[23, 7],[7,23],[26, 7],[7,26],[8, 14],[14,8],[8, 23],[23,8],[8, 24],[24,8],[9, 15],[15,9],[9, 13],[13,9],[16, 10],[10,16],[10, 20],[20,10],[10, 13],[13,10],[10, 27],[27,10],[11, 12],[12,11],[17, 11],[11,17],[11, 21],[21,11],[25, 11],[11,25],[11, 27],[27,11],[18, 12],[12,18],[26, 12],[12,26],[14, 15],[15,14],[16, 14],[14,16],[26, 14],[14,26],[17, 15],[15,17],[19, 15],[15,19],[16, 20],[20,16],[16, 22],[22,16],[17, 23],[23,17],[18, 27],[27,18],[24, 19],[19,24],[19, 27],[27,19],[25, 20],[20,25],[24, 21],[21,24],[13, 21],[21,13],[25, 22],[22,25],[13, 22],[22,13],[27, 13],[13,27]],R).

all_different/1 是 CLP(FD) 库的一部分。只有在调用 different/1 之前执行 label/1,您的 different/1 才会有效。

因此,您可以在当前实施中使用 all_different/1(首选),或者您可以 re-arrange 您的代码如下:

r5(L,R):-
    R = [[G,A],[A,B],[B,C],[C,D],[D,E],[E,F],[F,G]],

    [A,B,C,D,E,F,G] ins 0 .. 27,

    % all_different([A,B,C,D,E,F,G]), % preferred in place of 'different/1' below

    label([A,B,C,D,E,F,G]),        

    different([A,B,C,D,E,F,G]),

    member([G,A],L),
    member([A,B],L),
    member([B,C],L),
    member([C,D],L),
    member([D,E],L),
    member([E,F],L),
    member([F,G],L).


原始代码中 L 的用途有点不清楚。更通用的解决方案可能如下所示:

:- use_module(library(clpfd)).

% Define consecutive intervals consisting of Length intervals
%  with elements from 0 to MaxNumber
consecutive_intervals(MaxNumber, Length, Intervals):-
    length(Elements, Length),             % Establish the number of elements
    Elements ins 0 .. Max,                % Establish the range of each element
    all_different(Elements),              % Each element is different
    list_intervals(Elements, Intervals),  % Define consecutive intervals
    label(Elements).

% list_intervals(List, Intervals)
%    Intervals is a complete list of consecutive intervals with elements from List
list_intervals([X1,X2|Xs], [[X1,X2]|T]) :-
    list_intervals([X2|Xs], X1, T).
list_intervals([X2], X1, [[X2,X1]]).
list_intervals([X2,X3|Xs], X1, [[X2,X3]|T]) :-
    list_intervals([X3|Xs], X1, T).

解决方案的数量很多,但这里有一个简短的小例子:

?- consecutive_intervals(4, 3, R).
R = [[0, 1], [1, 2], [2, 0]] ;
R = [[0, 1], [1, 3], [3, 0]] ;
R = [[0, 1], [1, 4], [4, 0]] ;
R = [[0, 2], [2, 1], [1, 0]] ;
R = [[0, 2], [2, 3], [3, 0]] ;
R = [[0, 2], [2, 4], [4, 0]] ;
...
R = [[4, 3], [3, 0], [0, 4]] ;
R = [[4, 3], [3, 1], [1, 4]] ;
R = [[4, 3], [3, 2], [2, 4]] ;
false.