这是使用普通整数算术和回溯的直接递归实现:
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.