问题描述
我有一个序言谓词:
Add( [A|B] , Answer ) :-
...
~ Add everything in the list to come up with answer
...
我现在想实现 AddUnique
,当我给它两次变量时,它将为列表 except 中的所有内容返回唯一值.
I would now like to implement AddUnique
that would return unique values for everything in the list except when I give it the variable twice.
这里有一些逻辑上等价的东西:
Here are somethings that are logically equivalent:
?- AddUnique([A, B, C], 10).
等价于:?- Add([A, B, C], 10), A !=B, B != C, A != C.
还有:
?- AddUnique([A, B, B], 10).
等价于:?- Add([A, B, B], 10), A !=B.
还有:
?- AddUnique([A, B, B], 10).
NOT 等价于:?- Add([A, B, B], 10), A != B, B!=B.
如果 ?- AddUnique([A,B,C,D], 4).
给定它应该返回 false,因为它不能带有加到四的唯一正整数.
If ?- AddUnique([A,B,C,D], 4).
is given it should return false since it cannot come with unique positive integers that add to four.
如果 ?- AddUnique([A,A,A,A], 4).
给出它应该返回 A=1
.
If ?- AddUnique([A,A,A,A], 4).
is given it should return A=1
.
问题:如何在不执行类似操作的情况下将 A != B, B != C, A != C.
逻辑移动到谓词中>A != A?
Question: How can I move the A != B, B != C, A != C.
logic inside the predicate without doing something like this A != A
?
推荐答案
这是我想出的解决方案.它只会将输入分配为小于十的数字,但效果很好!
This is the solution that I came up with. It will only assign the input to be numbers less than ten but works great for that!
addUnique( A, Answer ) :-
used(A,[0,1,2,3,4,5,6,7,8,9],_),
add(A,Answer).
add( [A|B] , Answer ) :-
~ Add everything in the list to come up with answer ~.
% ================================
% Ensures that all variables are unique.
% ================================
% Base case: Assigned variables unique values
used([], Nin, Nin).
% Have already assigned a value to this variable
used([A|B], Nin, Nout) :-
integer(A),
helper(B,Nin,Nout).
% Have not assigned a value to this variable yet
% Assign it and remove it from the list.
used( [A|B] , Nin, Nout) :-
member(A,Nin),
delete(Nin,A,Temp),
helper(B,Temp,Nout).
这篇关于Prolog初学者:如何为谓词中的每个变量创建唯一值的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!