Prolog 更改事实顺序会产生不同的结果

Question

Prolog 的新手。我有一个无法解释的小故障，但如果我添加其他字典(X)事实，该程序似乎可以工作。该程序采用一个字符串列表，其中的字母已被编码并生成一个解码列表。每个字母代表单词列表中的一个不同字母。所以 go([abccd,edfgh,ade,ifb,kdl],X) 返回 X = ['HELLO', 'WORLD', 'HOW', 'ARE', 'YOU']。问题是如果字典('HOW')事实出现在字典('YOU')事实之前，那么程序返回 X = ['HELLO', 'WORLD', 'HOW', 'ARE', 'HOW'] .这是故障代码:

/*word bank*/
dictionary('HELLO').
dictionary('WORLD').
dictionary('HOW').
dictionary('ARE').
dictionary('YOU').

/*This spits out a single list where
the lengths of words in the dictionary
are matched to each word in the encoded 
message, so [abccd,edfgh,ade,ifb,kdl]
matches [HELLO,WORLD,HOW,ARE,HOW] or
any combination*/ 
sameLength([X|XTail],[Y|YTail]) :-
   dictionary(Y),
   name(X,L1),name(Y,L2),
   length(L1,Z),length(L2,Z),
   sameLength(XTail,YTail).
sameLength([],[]).

/*Turns a list of lists into
a single list*/
oneWord([X|XTail],Y) :-
   name(X,L),
   append(L,Z,Y),
   oneWord(XTail,Z).
oneWord([],[]).

/*This replaces the letters that are in
the dictionary, with the letters in the
message.  If at any point a letter has
been replaced because it is UPPERCASE,
and that letter is being replaced by
something else then fail, other wise,
the letter has to be lowercase*/
replaceLetters(List,[X|XTail],[Y|YTail],Result) :-
   (X<91,X=Y);(X>96),
   replaceP(X,Y,List,Result1),
   replaceLetters(Result1,XTail,YTail,Result).
replaceLetters(Result,[],[],Result).

/*the call to action*/
go(X,Y) :-
   sameLength(X,Y),
   oneWord(X,A),
   oneWord(Y,B),
   replaceLetters(A,A,B,C),
   B=C,
   !.

/*replace thanks to @repeat*/
replaceP(_, _, [], []).
replaceP(O, R, [O|T], [R|T2]) :- replaceP(O, R, T, T2).
replaceP(O, R, [H|T], [H|T2]) :- dif(H,O), replaceP(O, R, T, T2).

我想补充一点，Prolog 很酷。谢谢你的帮助。

Answer 1

我的解决方案让我明白问题是从哪里来的

go(X,Y) :-
    go(X,Y,[]).

go([],[],_Dict).
go([W|Ws],[T|Ts],Dict) :-
    assoc_codes(W,T,Dict,DictUpd),
    go(Ws,Ts,DictUpd).

assoc_codes(W,T,Dict,DictUpd) :-
    atom_codes(W, Cs),
    dictionary(T),
    atom_codes(T, Tcs),
    upd_dict(Cs, Tcs, Dict, DictUpd).

upd_dict([], [], DictUpd, DictUpd). % done, all went well
upd_dict([C|Cs], [D|Ds], Dict, DictUpd) :-
        memberchk(C-T, Dict)    % if C already 'assigned', must match D
    ->  T = D,
        upd_dict(Cs, Ds, Dict, DictUpd)
    ;   \+ ( memberchk(X-D, Dict), X \= C ),
        upd_dict(Cs, Ds, [C-D|Dict], DictUpd).

注意 else 分支:当 C 还没有被赋值时，D 也是如此。如果我们将它注释掉(即

...
;   % \+ ( memberchk(X-D, Dict), X \= C ),
...

) 我们有你报告的问题

?- go([abccd,edfgh,ade,ifb,kdl],X).
X = ['HELLO', 'WORLD', 'HOW', 'ARE', 'HOW'] ;
X = ['HELLO', 'WORLD', 'HOW', 'ARE', 'YOU'] ;
false.

它来自这样一个事实，即 'L' 必须同时分配给 'c' 和 'g'。
取消注释测试后，我们有

?- go([abccd,edfgh,ade,ifb,kdl],X).
false.

?- go([abccd,edfch,ade,ifb,kdl],X).
X = ['HELLO', 'WORLD', 'HOW', 'ARE', 'YOU'] ;
false.