Question

我想在两个列表（相同的长度）中计算相等且位于相同位置的元素数量。例如：假设我们有列表A = [3,6,7,9]和B = [2,6,4,9]，我想在屏幕上打印消息，“找到2头公牛”。

到目前为止，我已经做到了这一点：

bulls([],[]).
bulls([Ha|Ta],[Hb|Tb]) :-
    Ha = Hb,
    writeln('bull found'),
    bulls(Ta,Tb);
    bulls(Ta,Tb).

每次在两个列表中存在于同一位置的元素时，都会打印出“bull found”消息。在我看来，我想做出这样的事情：

bulls([],[],_).
bulls([Ha|Ta],[Hb|Tb],Counter) :-
    Ha = Hb,
    NewCounter is Counter + 1,
    bulls(Ta,Tb,NewCounter);
    bulls(Ta,Tb,NewCounter).

bulls(List1,List2):- bulls(List1,List2,0).

bulls从另一个规则中调用，该规则将列表传递给它。如何制作它以便将“公牛”的值打印到屏幕上。有什么帮助吗？

修改所以在 Suki的帖子之后，我让这个测试程序测试了2个列表：

bulls([],[],X), write(X), write('bulls found'),fail.
bulls([Ha|Ta],[Hb|Tb],Counter) :-
    Ha = Hb,
    NewCounter is Counter + 1,
    bulls(Ta,Tb,NewCounter);
    bulls(Ta,Tb,NewCounter).

check(List1,List2):- 
    bulls(List1,List2,0).


start:-
    A=[1,1,1,1],
    B=[2,1,2,1],
    writeln(A),writeln(B),
    check(A,B).

我得到了这个结果

1 ?- start.
[1,1,1,1]
[2,1,2,1]
ERROR: bulls/3: Arguments are not sufficiently instantiated

我做错了什么？

Answer 1

关于您编辑的节目：

第一个条款不是条款，这是一个目标！它应该是这样的：

bulls([],[],X) :- write(X), write(' bulls found').

你应该放弃fail，顺便说一句。

在第二个条款中，您需要使用“if-then-else”，并在“else”-branch中使用Counter代替NewCounter：

bulls([Ha|Ta],[Hb|Tb],Counter) :-
  (
    Ha == Hb
  ->
    NewCounter is Counter + 1,
    bulls(Ta,Tb,NewCounter)
  ;
    bulls(Ta,Tb,Counter)
  ).

Answer 2

假设你的prolog实现中存在谓词findall，length，nth0：
（以下内容来自swi-prolog）

A = [3,6,7,9], 
B = [2,6,4,9], 
findall(Y, ((nth0(X, A, Y), nth0(X, B, Y))), Y),
length(Y, LenY),
write(LenY), write(' Bulls Found').

[trace]  ?- A = [3,6,7,9], 
B = [2,6,4,9], 
findall(Y, ((nth0(X, A, Y), nth0(X, B, Y))), Y),
length(Y, LenY),
write(LenY), write(' Bulls Found').
   Call: (7) _G2814=[3, 6, 7, 9] ? creep
   Exit: (7) [3, 6, 7, 9]=[3, 6, 7, 9] ? creep
   Call: (7) _G2829=[2, 6, 4, 9] ? creep
   Exit: (7) [2, 6, 4, 9]=[2, 6, 4, 9] ? creep
^  Call: (7) findall(_G2834, (nth0(_G2832, [3, 6, 7, 9], _G2834), nth0(_G2832, [2, 6, 4, 9], _G2834)), _G2834) ? creep
   Call: (13) lists:nth0(_G2832, [3, 6, 7, 9], _G2834) ? creep
   Exit: (13) lists:nth0(0, [3, 6, 7, 9], 3) ? creep
   Call: (13) lists:nth0(0, [2, 6, 4, 9], 3) ? creep
   Fail: (13) lists:nth0(0, [2, 6, 4, 9], 3) ? creep
   Redo: (13) lists:nth0(_G2832, [3, 6, 7, 9], _G2834) ? creep
   Exit: (13) lists:nth0(1, [3, 6, 7, 9], 6) ? creep
   Call: (13) lists:nth0(1, [2, 6, 4, 9], 6) ? creep
   Exit: (13) lists:nth0(1, [2, 6, 4, 9], 6) ? creep
   Redo: (13) lists:nth0(_G2832, [3, 6, 7, 9], _G2834) ? creep
   Exit: (13) lists:nth0(2, [3, 6, 7, 9], 7) ? creep
   Call: (13) lists:nth0(2, [2, 6, 4, 9], 7) ? creep
   Fail: (13) lists:nth0(2, [2, 6, 4, 9], 7) ? creep
   Redo: (13) lists:nth0(_G2832, [3, 6, 7, 9], _G2834) ? creep
   Exit: (13) lists:nth0(3, [3, 6, 7, 9], 9) ? creep
   Call: (13) lists:nth0(3, [2, 6, 4, 9], 9) ? creep
   Exit: (13) lists:nth0(3, [2, 6, 4, 9], 9) ? creep
^  Exit: (7) findall([6, 9], user: (nth0(_G2832, [3, 6, 7, 9], [6, 9]), nth0(_G2832, [2, 6, 4, 9], [6, 9])), [6, 9]) ? creep
   Call: (7) length([6, 9], _G2848) ? creep
   Exit: (7) length([6, 9], 2) ? creep
   Call: (7) write(2) ? creep
2
   Exit: (7) write(2) ? creep
   Call: (7) write(' Bulls Found') ? creep
 Bulls Found
   Exit: (7) write(' Bulls Found') ? creep
A = [3, 6, 7, 9],
B = [2, 6, 4, 9],
Y = [6, 9],
LenY = 2.

专柜，Prolog

2 个答案: