我有一个Prolog程序,用于查找可被三或五整除的列表中所有数字的平方和。但是它会返回多个结果,我不知道为什么。
% --divisibility tests--
div_test(N):-
% divisible by three?
0 is N mod 3.
div_test(N):-
% divisible by five?
0 is N mod 5.
% sum of an empty list is zero (base case)
square_sum([], Sum):-
Sum is 0.
% --recursive cases
square_sum([Head | Tail], Sum) :-
div_test(Head),
square_sum(Tail, TempSum),
Sum is Head*Head + TempSum.
square_sum([Head | Tail], Sum) :-
square_sum(Tail, TempSum),
Sum is TempSum.
给出以下输入:
?-square_sum([1,2,3,4,5],Sum).
我得到以下输出:
Sum = 34 ;
Sum = 9 ;
Sum = 25 ;
Sum = 0.
34是我应该得到的唯一输出
答案 0 :(得分:2)
% --recursive cases
square_sum([Head | Tail], Sum) :-
div_test(Head),
square_sum(Tail, TempSum),
Sum is Head*Head + TempSum.
square_sum([Head | Tail], Sum) :-
square_sum(Tail, TempSum),
Sum is TempSum.
在第二条规则中,您不测试div_test是否成功。 您可以在第一个规则中在div_test(Head)之后添加剪切(!),或在第二个规则中添加 \ + div_test(Head)。
你使用SWI-Prolog,所以你可以使用http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/lambda.pl找到的模块lambda.pl并写
:- use_module(library(lambda)).
div_test(N):-
% divisible by three?
0 is N mod 3.
div_test(N):-
% divisible by five?
0 is N mod 5.
square_sum(Lst, Sum) :-
foldl(\X^Y^Z^((div_test(X); div_test(X))
-> Z is Y + X*X
; Z = Y), Lst, 0, Sum).