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我是一个完全的编程初学者，必须使用 GNU Prolog 在 Prolog 中创建和解决一个谜语，类似于爱因斯坦谜语，尽管不太复杂。我一直在尝试为下表中包含的项目创建一个谜语。

到目前为止，我的代码看起来像这样，但我真的不完全理解我做错了什么或就在这里。我可以在 GNU Prolog 中编译代码，但它无法解开谜团：

middle(M,[_,M,_]).
right(A,B,[[_|A]|B]).
left(A,B,[A|[B|_]]).
run:-
   X = [_,_,_],
   middle([_,brown,_],X),   /* the brown guinea pig lives in the middle of the cage */
   member([brown,carrots,_],X), /* the brown guinea pig loves to eat carrots */
   member([_,salad,giggles],X), /* the salad eating guinea pig giggles */
   right([_,salad,_],[brown,_,_],X),    /* the salad eating guinea pig sits to the right of the brown guinea pig */
   left([black,_,_],[_,_,squeaks],X),   /* the black guinea pig sleeps to the left of the squeaking guinea pig */
   member([black,_,grumbles],X),    /* the black guinea pig grumbles */
   member([grey,_,giggles],X),  /* the grey guinea pig giggles*/
   write(X),nl, /* write out all fur colors */
   write('the '),write(N),write(' guinea pig loves to eat cucumbers'),nl. /* answer to the question */

我将不胜感激任何帮助，因为我对这些事情很不熟悉，但必须为我正在上课的课程找出解决方案。任何提示都会有很大帮助。谢谢！

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1 回答 1

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只是要小心，并保持统一。把每件东西放在自己的地方。而且，您已经破坏了一些需要修复的列表代码。所以，给你。

middle(M, [_,M,_]).
right(A,B,X) :- left(B,A,X).
left(A,B,X) :- append(_, [A,B | _], X).

run :-
   X = [_,_,_],
   middle([_       ,brown,_      ],X),   /* the brown guinea pig - middle of the cage */
   member([_       ,brown,carrots],X),   /* the brown guinea pig loves to eat carrots */
   member([giggles ,_    ,salad  ],X),   /* the salad-eating guinea pig giggles */
   right( [_       ,_    ,salad  ],      /* the salad-eating guinea pig sits */
          [_       ,brown,_      ],X),   /*    to the right of the brown guinea pig */
   left(  [_       ,black,_      ],      /* the black guinea pig sleeps to the left */
          [squeaks ,_    ,_      ],X),   /*    of the squeaking guinea pig */
   member([grumbles,black,_      ],X),   /* the black guinea pig grumbles */
   member([giggles ,grey ,_      ],X),   /* the grey guinea pig giggles */
   member([_       ,EC ,cucumbers],X),   /* a guinea pig that loves to eat cucumbers */

   X = [[_,A,_],[_,B,_],[_,C,_]], write([A,B,C]), nl, /* write out all fur colors */
   write('the '), write(EC),
   write(' guinea pig loves to eat cucumbers'), nl. /* the answer to the question */

但是，有些人可能会争辩说，人类程序员在这里灌输了太多他的理解（即有些受骗），他在模型中制作了三个位置来表示他知道豚鼠在这个宇宙中具有的三个属性。

不过，这不是必需的。以下是我们如何让“计算机”使用某种“可扩展记录”自行解决所有问题 ：

attr(A, [N-V | R]):- memberchk( N-X, A), X = V, attr(A, R).
attr(_, []).

color(A, B):- attr( A, [color-B]).

pigs( Pigs):-
  length( Pigs,N), 
  N rem 2 =:= 1, Middle is N div 2,    /* there _is_ a middle - list length is odd */
  nth0( Middle,Pigs,P1), attr( P1, [color-brown]), 
  member( P2, Pigs),     attr( P2, [color-brown, eats-carrots]),
  member( P3, Pigs),     attr( P3, [eats-salad, sound-giggles]),
  right( P4,P4b,Pigs),   attr( P4, [eats-salad]),   attr( P4b, [color-brown]),
  left(  P5,P5b,Pigs),   attr( P5, [color-black]),  attr( P5b, [sound-squeaks]),
  member( P6, Pigs),     attr( P6, [color-black, sound-grumbles]),
  member( P7, Pigs),     attr( P7, [color-grey,  sound-giggles]),
  member( P8, Pigs),     attr( P8, [eats-cucumbers, color-EatsCucumbers]),
  length( Furs, N),      maplist( color, Pigs, Furs),
  writeln( Furs),        writeln( EatsCucumbers),  nl, !.

测试：

14 ?- time(( pigs(_P), maplist(writeln,_P), ! )).
[black,brown,grey]
black

[color-black, sound-grumbles, eats-cucumbers|_G1484]
[color-brown, eats-carrots,   sound-squeaks |_G1424]
[eats-salad,  sound-giggles,  color-grey    |_G1463]
/* % 287 inferences, 0.000 CPU in 0.030 seconds (0% CPU, Infinite Lips) */
true.
于 2013-12-05T19:03:18.370 回答