在不使用maplist的情况下对列表中的所有成员(包括子列表Prolog)进行平方 [英] square all members in the list including sublist Prolog without using maplist

问题描述

我对Prolog并不陌生，他试图实现一种深平方谓词，该谓词对列表以及子列表中的所有数字求平方.我写了一些有效的代码，但没有给我期望的输出.

I am new to Prolog and trying to implement a sort of a deep square predicate, which squares all the numbers in the list and also in the sublists. I wrote some sort of a working code but its not giving me the output i expect.

代码:

dsquare([],S).
dsquare([H|T],[R|S]):- number(H), dsquare(T,S), R is H*H, !.
dsquare([H|T],S):- isList(H), dsquare(H,S).
dsquare([H|T],[R|S]) :- dsquare(T,S), R = H, !.

当前输出:

2?- dsquare([[2],4,a],X).

X = [4| _VDHV] ;

X = [[2], 16, a| _VDNM] ;

fail.

预期输出:

X = [[4]，16，a]

X = [[4], 16, a]

我也想知道为什么我在输出中得到那些'_VDHV'和'_VDNM'. 任何帮助将不胜感激.

Also i wanted to know why am i getting those '_VDHV' and '_VDNM' in my output. Any help would be much appreciated.

好的，所以我将代码更新为:

ok so I updated my code as:

dsquare([],[]).
dsquare([H|T],[R|S]):- number(H), R is H*H, dsquare(T,S).
dsquare([H|T],[R|S]):- isList(H), dsquare(H,R), dsquare(T,S).
dsquare([H|T],[R|S]) :- R=H, dsquare(T,S).

但是我得到的输出是:

13?- dsquare([a,3,[[2]],b,4],X).

X = [a, 9, [[4]], b, 16] ;

X = [a, 9, [[4]], b, 4] ;

X = [a, 9, [[2]], b, 16] ;

X = [a, 9, [[2]], b, 4] ;

X = [a, 9, [[2]], b, 16] ;

X = [a, 9, [[2]], b, 4] ;

X = [a, 9, [[2]], b, 16] ;

X = [a, 9, [[2]], b, 4] ;

X = [a, 3, [[4]], b, 16] ;

X = [a, 3, [[4]], b, 4] ;

X = [a, 3, [[2]], b, 16] ;

X = [a, 3, [[2]], b, 4] ;

X = [a, 3, [[2]], b, 16] ;

X = [a, 3, [[2]], b, 4] ;

X = [a, 3, [[2]], b, 16] ;

X = [a, 3, [[2]], b, 4] ;

fail.

我不知道它如何获得这么多结果.

I have no clue how it gets so many results.

编辑 最后可行的解决方案是

Edit finally the working solution is

dsquare([],[]).
dsquare([H|T],[R|S]) :- number(H), !, R is H*H, dsquare(T,S).
dsquare([H|T],[R|S]) :- isList(H), !, dsquare(H,R), dsquare(T,S).
dsquare([H|T],[H|S]) :- dsquare(T,S).

推荐答案

您的序言应该警告您有关第一规则和第三规则中的单个".

your Prolog should warn you about a 'singleton' in your first and third rules.

尝试

dsquare([],[]).
...
dsquare([H|T],[S|R]):- isList(H), dsquare(H,S), dsquare(T,R).

在没有动力原因的情况下，OT切勿裁切.

OT don't place cuts without a motivated reason.

edit 您会获得更多结果，因为最后一条规则在回溯时被触发了.现在可能是将剪切片段放置在需要的位置 的时候了(即在代码进入受条件保护的分支之后):

edit you get more results beacuse the last rule get fired on backtracking. Now could be the time to place the cuts where needed (i.e. after the code entered a branch guarded by a condition):

dsquare([],[]).
dsquare([H|T],[R|S]) :- number(H), !, R is H*H, dsquare(T,S).
dsquare([H|T],[R|S]) :- isList(H), !, dsquare(H,R), dsquare(T,S).
dsquare([H|T],[R|S]) :- R=H, dsquare(T,S).

或考虑考虑重复代码的重构:

or consider a refactoring that accounts for repeated code:

dsquare([],[]).
dsquare([H|T],[R|S]) :-
  (  number(H)
  -> R is H*H
  ;  isList(H)
  -> dsquare(H,R)
  ;  R=H
  ),
  dsquare(T,S).

edit 上面的定义(我用'if/then/else'测试过)看起来不错:

edit the above definition (I tested that with 'if/then/else') seems fine:

1 ?- dsquare([[2],4,a],X).
X = [[4], 16, a].

2 ?- dsquare([a,[3],[[[5]]],[2],a],X).
X = [a, [9], [[[25]]], [4], a].

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