生成列表-几何级数 [英] Generate list - geometric progression

问题描述

我想使用带有4个参数的谓词生成几何级数列表-生成级数的列表，该列表的长度，起始元素以及级数的乘数.到目前为止，我所做的只是只有一个3参数谓词来生成几何级数而不会停止:

I'd like to generate a geometric progression list using a predicate with 4 parameters - the list where the progression will be generated, the length of this list, the start element, and the multiplier of the progression. What I've done so far is having only a 3-parameter predicate to generate the geometric progression without stopping :

gengeom([X],X,_).
gengeom([H|Tail],H,Q):-X is H*Q,gengeom(Tail,X,Q).

此查询为我提供了开始元素1和乘数2的所有进度:

And this query gives me all progressions with start element 1 and multiplier 2 :

?-gengeom(L,1,2),write(L),nl,fail.

谁能帮我写出我真正想拥有的4参数谓词(在列表的长度变为某个数字后，它会停止生成更多数字)?

Can anyone help me write the 4-parameter predicate I'd really like to have (that stops generating any more numbers after the length of the list has become a certain number) ?

推荐答案

仅添加一个倒数参数将起作用，并且将保留代码的漂亮生成属性:

just adding a countdown parameter will work, and will preserve the nice generative property of your code:

gengeom([X],X,_,_).
gengeom([H|Tail],H,Q,N) :- N>1, M is N-1, X is H*Q, gengeom(Tail,X,Q,M).

?- gengeom(L,1,2,3).
L = [1] ;
L = [1, 2] ;
L = [1, 2, 4] ;
false.

当然，您可以使用findlog/3(Prolog的列表生成器")使结构更紧凑:

of course, you could get somewhat more compact using findall/3, the Prolog 'list generator':

gengeom(L,H,Q,N) :-
    findall(V, (between(H,N,M), V is Q**(M-1)), L).

但是此代码段(类似于@ joel76'帖子)将仅构建最终"列表...

but this snippet (similar to @joel76' post) will build just the 'final' list...

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