我试图编写谓词divide(L,Len,Slist)
,当Slist可以与从List L分配的长度为Len的列表统一时,这将是正确的。例如
divide([1,2,3,4,5,6,7],3,Slist).
应该给出这样的答案
Slist=[1,2,3];
Slist=[2,3,4];
Slist=[3,4,5];
Slist=[4,5,6];
Slist=[5,6,7];
但是我找不到比length(X,Len), sublist(L,X).
更好的方法,但它确实工作得太慢了。
应该如何划分谓词?
答案 0 :(得分:2)
或者你可以在this great answer中使用@false提到的DCG:
seq([]) --> [].
seq([E|Es]) --> [E], seq(Es).
divide(List, Length, Result) :-
length(Result, Length),
phrase((seq(_), seq(Result), seq(_)), List).
答案 1 :(得分:1)
子列表/ 2似乎没有按预期工作:
?- [library(dialect/sicstus/lists)].
% library(dialect/sicstus/lists) compiled into sicstus_lists 0,00 sec, 14 clauses
true.
?- L=[1,2,3,4,5,6], length(T, 3),sublist(T,L).
L = [1, 2, 3, 4, 5, 6],
T = [1, 2, 3] ;
L = [1, 2, 3, 4, 5, 6],
T = [1, 2, 4] ;
L = [1, 2, 3, 4, 5, 6],
T = [1, 2, 5] ;
....
您可以使用append / 3代替:
?- L=[1,2,3,4,5,6], length(T, 3), append(_, Q, L), append(T, _, Q).
L = [1, 2, 3, 4, 5, 6],
T = [1, 2, 3],
Q = [1, 2, 3, 4, 5, 6] ;
L = [1, 2, 3, 4, 5, 6],
T = [2, 3, 4],
Q = [2, 3, 4, 5, 6] ;
L = [1, 2, 3, 4, 5, 6],
T = [3, 4, 5],
Q = [3, 4, 5, 6] ;
L = [1, 2, 3, 4, 5, 6],
T = Q, Q = [4, 5, 6] ;
false.
我认为它不是很快,只是必不可少......