family(family_member(oleg, barmin, birth_date(6, 1, 98), false), family_member(kate, barmin, birth_date(1, 10, 97), true),
children(family_member(vasy, barmin, birth_date(5, 12, 07), true))).
family(family_member(artem, kudinov, birth_date(6, 1, 57), true), family_member(ann, kudinov, birth_date(1, 10, 67), false),
children(family_member(julia, barmin, birth_date(5, 12, 88), false))).
family(family_member(kola, avramov, birth_date(6, 1, 57), false), family_member(nastya, avramov, birth_date(1, 10, 67), false),
children(family_member(masha, avramov, birth_date(5, 12, 88), false),family_member(fedya, avramov, birth_date(5, 12, 88), false))).
family(family_member(ivan, petrov, birth_date(6, 1, 57), true), family_member(daria, petrov, birth_date(1, 10, 67), false),
children(family_member(a00000000, petrov, birth_date(5, 12, 88), false),family_member(warihaerh, petrov, birth_date(5, 12, 88), false),
family_member(b00000000, petrov, birth_date(5, 12, 88), false))).
family(family_member(ivan, ivanov, birth_date(6, 1, 57), true), family_member(daria, ivanov, birth_date(1, 10, 67), false),
children(family_member(a00000000, ivanov, birth_date(5, 12, 88), false),family_member(warihaerh, ivanov, birth_date(5, 12, 88), false),
family_member(orihthth, ivanov, birth_date(5, 12, 88), false), family_member(shgsgh, ivanov, birth_date(5, 12, 88), false))).
familyWifeWorkFalse(Y) :-
Y = family_member(_, _, _, false),
family(_, family_member(_, _, _, false), _).
oneChildrenMan(X) :-
family(X, _, children(_)).
twoChildrenMan(X) :-
family(X, _, children(_, _)).
moreTwoChildrenMan(X) :-
+oneChildrenMan(X),
+twoChildrenMan(X),
family(X, _, _).
大家好!我需要帮助来完成我的prolog任务。我有几个家庭,第一个变量是husbend,第二个变量是妻子,第三个变量是孩子。我需要创建一个规则,它将输出有 3 个或更多孩子的家庭。我创建了输出一个和两个子项的规则,然后尝试在第三条规则中排除它们。但我得到了:
?- moreTwoChildrenMan(X).
false.
几天来,我一直在努力完成这项任务,但我没有得到任何结果。有没有知道如何做到这一点?
你的moreTwoChildrenMan/1
谓词的问题在于你错误地使用了Prolog的回溯。
目标否定算子+
强制Prolog的执行机制搜索前一个目标的替代解决方案,并且在moreTwoChildrenMan/1
的代码中oneChildrenMan(X)
之前没有目标。这就是谓词失败的原因。
但是,如果您只是将目标移动到目标之前family(X, _, _)
oneChildrenMan(X)
它会给您所需的结果,因为现在执行将回溯family(X, _, _)
的解决方案:
moreTwoChildrenMan(X) :-
family(X, _, _), % find some family member X
+ oneChildrenMan(X), % filter out those with one child
+ twoChildrenMan(X). % filter out those with two children