尾部整数的尾部恢复列表

发布于 2025-01-31 07:58:23 字数 705 浏览 2 评论 0原文

例如

?- divisors(14,What).
What = [2, 7] ?

除数，

?- divisors(14,[2, 8]).
no

除非我试图验证[]不是列表，否则不幸的是成功：

?- divisors(14,[]).
true

可以在调试器中看到为什么它错误地成功了，但是我看不到如何解决这个问题：

divisors_aux(1,_,L,L).
divisors_aux(D,N,List,Aux) :- D>1,0 is N mod D,D1 is D-1,
                              divisors_aux(D1,N,List,[D|Aux]).
divisors_aux(D,N,List,Aux) :- D>1,D1 is D-1,divisors_aux(D1,N,List,Aux).

divisors(N,List) :- N>1,nonvar(N),D is N-1,divisors_aux(D,N,List,[]).

我知道有循环方法（foreach，foreach，foreach，foreach，foreach，等）但是，我试图根据我到目前为止所学到的知识（Covington等人的前100页）来使这种“蛮力”方法起作用

。谢谢。

原文

I am trying to make a simple tail-recursive loop to collect the divisors of an integer, divisors(N,List) and its working when List is not instantiated, for example,

?- divisors(14,What).
What = [2, 7] ?

And I can verify that the wrong list is not the list of divisors,

?- divisors(14,[2, 8]).
no

Unless I try to verify that [] is not the list, this unfortunately succeeds:

?- divisors(14,[]).
true

can see in the debugger why it is erroneously succeeding, but I don't see how to fix this:

divisors_aux(1,_,L,L).
divisors_aux(D,N,List,Aux) :- D>1,0 is N mod D,D1 is D-1,
                              divisors_aux(D1,N,List,[D|Aux]).
divisors_aux(D,N,List,Aux) :- D>1,D1 is D-1,divisors_aux(D1,N,List,Aux).

divisors(N,List) :- N>1,nonvar(N),D is N-1,divisors_aux(D,N,List,[]).

I know there are looping approaches (foreach, etc.) but I am trying to make this "brute force" approach work based on what I have learned so far (roughly first 100 pages of Covington et. al.)

Thanks.

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清秋悲枫 2025-02-07 07:58:23

您的问题是，当divisors_aux的第二子句/3失败（或者成功并要求其他解决方案）时，它将始终回到第三子句。

数字是N。

您可能需要使用\+（0是n mod d）来捍卫第三子句，以便当当前否则构造：

divisors_aux(1,_,L,L).
divisors_aux(D,N,List,Aux) :- D>1,
                              (0 is N mod D -> Aux1=[D|Aux] ; Aux1=Aux),
                              D1 is D-1,
                              divisors_aux(D1,N,List,Aux1).

divisors(N,List) :- N>1,nonvar(N),D is N-1,divisors_aux(D,N,List,[]).

Your problem is that when the second clause of divisors_aux/3 fails (or if it succeeds and you ask for another solution) it will always backtrack to the third clause.

You may want to guard the third clause with \+(0 is N mod D) so that it fails when the current number is a divisor of N.

Or alternatively join both clauses with an if-then-else construct:

divisors_aux(1,_,L,L).
divisors_aux(D,N,List,Aux) :- D>1,
                              (0 is N mod D -> Aux1=[D|Aux] ; Aux1=Aux),
                              D1 is D-1,
                              divisors_aux(D1,N,List,Aux1).

divisors(N,List) :- N>1,nonvar(N),D is N-1,divisors_aux(D,N,List,[]).

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