Question

我有这个代码来评估数字列表并返回与模式[G,A],[A,B],[B,C],[C,D],[D,E],[E,F],[F,G]匹配的列表。但是，我希望数字只是唯一的数字。 EG：（[0,2],[2,4],[4,19],[19,3],[3,5],[5,7],[7,0]）。不同的方法根据输入的数字返回true或false。但是，代码仍返回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]).

这是它返回的部分的一个例子：

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]] ;

这些都不是有效的，因为它们包含重复的数字。以下是输出中进一步向下的两个有效输出：

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.

这是我最初的电话：

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).

Answer 1

different/1是CLP（FD）库的一部分。只有在您致电label/1之前执行了different/1，您的all_different/1才有效。

因此，您可以在当前实施中使用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).（首选），也可以按如下方式重新安排代码：

<小时/> 原始代码中

:- 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).

的目的有点不清楚。更通用的解决方案可能如下所示：

?- 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.

解决方案的数量很大，但这是一个小的缩写示例：

$.get(
        'accNoRealMovs1.jsp',
        {mode:"0"},         
        function(responseText){

        // copy html in to tmpValue and then segregate to respective divs.
          $("body").append("<div id='tempValue'>"+responseText+"</div>").done(function(){

//1st div
$("#divAccMovementNRNew").html($("#tempValue").find("#divAccMovementNR").html());
$("#tempValue #divAccMovementNR").remove();

// 2nd div
$("#divAccMovementNR2New").html($("#tempValue").find("#divAccMovementNR2").html());
$("#tempValue #divAccMovementNR2").remove();



});  

});

<div id="divAccMovementNRNew"></div>
<div id="divAccMovementNR2New"></div>

为什么prolog在收到False后没有回溯？

1 个答案: