Prolog办公室拼图

Question

我试图为学校做一个Prolog作业，基本上是想找出谁有什么办公室。问题是：

亨特、劳拉、吉姆、萨利和杰克在同一幢大楼里工作，有五个相邻的办公室。亨特不在第五间办公室工作，劳拉不在第一间办公室工作。吉姆不在第一间或最后一间办公室工作，也不在杰克或劳拉旁边的办公室。莎莉在比劳拉高的办公室工作。谁在哪个办公室工作？

所以我有下面的代码，但是我收到了：

uncaught exception: error(type_error(fd_evaluable,abs/1),(#\=)/2)

这是密码。

puzzle(P) :-
        findWhosOffice(P,Z),
        labeling([],Z).

findWhosOffice(P, Z) :-
        P = [hunter-Hunter, jack-Jack, jim-Jim, laura,Laura,sally,Sally],
        Z = [Hunter, Laura, Jim, Sally, Jack],

        Hunter #\= 5,
        Laura #\= 1,
        Jim #\= 1,
        Jim #\= 5,
        abs(Jim-Jack) #\=1,
        abs(Jim-Laura) #\=1,
        Sally #> Laura.

Answer 1

问题是您还没有为Z中的变量定义域。也有一个限制是缺少的，以使它成为一个有效和唯一的解决方案。

Answer 2

下面是我的解决方案，我看到您没有使用all_different()：

:-use_module(library(clpfd)).

puzzle(Employee,Employee_Office_Room):-
    Employee=[Hunter, Laura, Jim, Sally, Jack], Employee ins 1..5,
    Employee_Office_Room=[First,_Second,_Third,_Fourth,Fifth], Employee_Office_Room ins 1..5,
    all_different(Employee),
    all_different(Employee_Office_Room),
    
    Hunter#\=Fifth,
    Laura#\=First,
    Jim#\=First,
    Jim#\=Fifth,
    not_nextto(Jim,Laura),
    not_nextto(Jim,Jack),
    Sally#=Laura-1,
    labeling([],Employee),labeling([],Employee_Office_Room).
    
    not_nextto(A,B):-abs(A-B)#\=1.

?-puzzle(Employee,Employee_Office_Room).
Employee = [1, 4, 2, 3, 5],
Employee_Office_Room = [1, 2, 3, 4, 5]
Employee = [1, 4, 2, 3, 5],
Employee_Office_Room = [1, 2, 3, 5, 4]
Employee = [1, 4, 2, 3, 5],
Employee_Office_Room = [1, 2, 4, 3, 5]
Employee = [1, 4, 2, 3, 5],
Employee_Office_Room = [1, 2, 4, 5, 3]
Employee = [1, 4, 2, 3, 5],
Employee_Office_Room = [1, 2, 5, 3, 4]....