递归Prolog谓词? [英] recursive Prolog predicate?

问题描述

我目前正在做一个项目，我想在 Prolog 中实现辅助谓词

i am currently working on a project and i want to implement helper predicate in Prolog

break_down(N, L)

其工作原理如下

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

对任何正整数 N 依此类推.

and so on for any positive integer N .

我已经尝试并实现了一个代码，它只生成第一个结果，我无法得到其余的结果，这是我的代码

i have tried and implemented a code which generates only the first result and i cannot get the rest of the results , and this is my code

break_down(1,[1]).
break_down(N,L):-
   N>0,
   N1 is N-1,
   break_down(N1,L1),
   append(L1,[1],L).

只生成第一个输出结果:

which generates only the first output result :

 L = [1, 1, 1, 1] ;

任何建议如何编辑我的代码以获得其余的?

any suggestion how to edit my code to get the rest ?

推荐答案

这是一个使用普通整数算法和回溯的直接递归实现:

Here's a straight-forward recursive implementation using plain integer arithmetic and backtracking:

break_down(N,L) :-
    break_ref_down(N,1,L).       % reference item is initially 1

break_ref_down(0,_,[]).
break_ref_down(N,Z0,[Z|Zs]) :-
    between(Z0,N,Z),             % multiple choices
    N0 is N-Z,
    break_ref_down(N0,Z,Zs).     % pass on current item as reference

示例查询:

?- break_down(8,Zs).
  Zs = [1,1,1,1,1,1,1,1]
; Zs = [1,1,1,1,1,1,2]
; Zs = [1,1,1,1,1,3]
; Zs = [1,1,1,1,2,2]
; Zs = [1,1,1,1,4]
; Zs = [1,1,1,2,3]
; Zs = [1,1,1,5]
; Zs = [1,1,2,2,2]
; Zs = [1,1,2,4]
; Zs = [1,1,3,3]
; Zs = [1,1,6]
; Zs = [1,2,2,3]
; Zs = [1,2,5]
; Zs = [1,3,4]
; Zs = [1,7]
; Zs = [2,2,2,2]
; Zs = [2,2,4]
; Zs = [2,3,3]
; Zs = [2,6]
; Zs = [3,5]
; Zs = [4,4]
; Zs = [8]
; false.

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