一些隐式约束 [英] Some implicit constraint

问题描述

当我执行这段代码(如下所示)时，它总是设置隐式约束.

When i execute this code(shown below) , it always sets implicit kind of constraint.

正如你在下面看到的，它总是说 D1 = D2 但没有这种明确的约束，也没有任何模式匹配来强制这样做.

As you can see below , it always says that D1 = D2 but there is no such explicit constraints nor any pattern matching which forces this.

或者换句话说，D1 和 D2 之间有一些引用，这样每当 D1 被初始化时，D2被自动初始化.我看不出这是怎么回事.有人能解释一下吗，我试着用调试器弄明白了，但没有帮助.

Or in otherwords there is some reference between D1 and D2 such that whenever D1 gets initialized, D2 gets automatically initialized. I can't see how this is happening. Can someone explain me, i tried ot figure it out with debugger but it did not help.

这是一个谜题GERALD + DONALD = ROBERT"，最初三个列表包含这些变量.

It's a puzzle "GERALD + DONALD = ROBERT", initially three lists contains these variables.

如果有人想测试它，我会在下面添加代码:

I add code below if anyone wants to test it:

sum(N1,N2,N):-
    sum1(N1,N2,N,0,0,[0,1,2,3,4,5,6,7,8,9],_).

sum1([],[],[],0,0,Digits,Digits).    
sum1([D1|N1],[D2|N2],[D|N],C1,C,Digs1,Digs2):-
    sum1(N1,N2,N,C1,C2,Digs1,Digs2),
    digitSum(D1,D2,C2,D,C,Digs2,Digs).

digitSum(D1,D2,C1,D,C,Digs1,Digs):-
    del(D1,Digs1,Digs2),
    del(D2,Digs2,Digs3),
    del(D,Digs3,Digs),
    S is D1 + D2 + C1,
    D is S mod 10,
    C is D div 10.

del(A,L,L):-
    nonvar(A),!.
del(A,[A|L],L).
del(A,[B|L],[B|L1]):-
    del(A,L,L1).

查询:

?- sum( [D,O,N,A,L,D], [G,E,R,A,L,D], [R,O,B,E,R,T] ).

推荐答案

当它说 D1 = D2 那么你应该看到 D1 和 D2 列表中有相同的变量，List 是一个函子，一个变量在整个函子中可见.

When it says D1 = D2 then you should see that both D1 and D2 have same variable from the list and List is a functor and one variable is visible in whole functor.

你应该看到，当递归结束时，你有来自 GERALD 和 DONALD 的 D's(最后一个元素)，因为这个 D在整个函子中可见，D1 和 D2 都指向同一个变量.

You should see that when the recursion ends, then you have have D's(last element) from GERALD and DONALD , since this D is visible in whole functor, D1 and D2 both refer to same variable.

这篇关于一些隐式约束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IT屋！