recursive Prolog predicate?

Question

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

break_down(N, L)

which works as follows

?- 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.

and so on for any positive integer N .

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).

which generates only the first output result :

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

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

Based upon the problem statement, you need to be adding some numbers up somewhere, but your solution doesn't add any numbers. It only appends. — lurker, Mar 26 '15 at 11:48

repeat · Accepted Answer · 2015-09-02T09:15:44.860

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

Sample query:

?- 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.

repeat · Answer 2 · 2015-09-02T09:14:16.573

Here's an implementation based on clpfd.

:- use_module(library(clpfd)).

As the predicate break_downFD/2 is non-recursive, the code is both readable and simple:

break_downFD(N,Zs) :-
    length(Max,N),        % multiple choices
    append(_,Zs,Max),
    Zs ins 1..N,
    sum(Zs,#=,N),
    chain(Zs,#=<),        % enforce sequence is non-descending
    labeling([],Zs).      % multiple choices, possibly

Sample query using SWI-Prolog:

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

recursive Prolog predicate?

2 Answers2

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