序言替换 [英] Prolog substitution

问题描述

如何用另一个包含要替换的变量的列表替换一个列表.例如

How can I replace a list with another list that contain the variable to be replaced. for example

rep([x, d, e, z, x, z, p], [x=z, z=x, d=c], R).
R = [z, c, e, x, z, x, p]

x 到 z 和 z 被替换后不会改变.

the x to z and z doesn't change after it has been replaced.

到目前为止我只做了一个没有列表的

so far I did only the one without the list

rep([], _, []).
rep(L1, H1=H2, L2) :-
   rep(L1, H1, H2, L2).

rep([],_,_,[]).
rep([H|T], X1, X2, [X2|L]) :-
   H=X1,
   rep(T,X1,X2,L),
   !.
rep([H|T],X1,X2,[H|L]) :-
   rep(T,X1,X2,L).

推荐答案

我发现您的代码相当混乱.一方面，您有 rep/3 和 rep/4，但它们都没有在您传递变量绑定列表的第二个位置有列表.H1=H2 不可能匹配一个列表，这是唯一一个检查第二个参数的 rep/3 子句.如果这是课堂作业，看起来你有点落后，我建议你花一些时间在之前的材料上.

I find your code rather confused. For one thing, you have rep/3 and rep/4, but none of them have a list in the second position where you're passing the list of variable bindings. H1=H2 cannot possibly match a list, and that's the only rep/3 clause that examines the second argument. If this is a class assignment, it looks like you're a little bit behind and I'd suggest you spend some time on the previous material.

解决方案比您想象的要简单:

The solution is simpler than you'd think:

rep([], _, []).
rep([X|Xs], Vars, [Y|Rest]) :-    member(X=Y, Vars), rep(Xs, Vars, Rest).
rep([X|Xs], Vars, [X|Rest]) :- \+ member(X=_, Vars), rep(Xs, Vars, Rest).

我们使用 member/2 在列表中查找变量绑定"(用引号引起来，因为它们是原子而不是真正的 Prolog 变量).如果它在列表中，则 Y 是替代品，否则我们继续使用 X.您会看到这具有预期的效果:

We're using member/2 to find a "variable binding" in the list (in quotes because these are atoms and not true Prolog variables). If it's in the list, Y is the replacement, otherwise we keep using X. And you see this has the desired effect:

?- rep([x, d, e, z, x, z, p], [x=z, z=x, d=c], R).
R = [z, c, e, x, z, x, p] ;
false.

直接使用或"可以提高效率(并为我们节省一个选择点):

This could be made somewhat more efficient using "or" directly (and save us a choice point):

rep([], _, []).
rep([X|Xs], Vars, [Y|Ys]) :- 
  (member(X=Y, Vars), ! ; X=Y), 
  rep(Xs, Vars, Ys).

见:

?- rep([x, d, e, z, x, z, p], [x=z, z=x, d=c], R).
R = [z, c, e, x, z, x, p].

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