[Challenge] implementing the ambiguous operator in D

[Andrei Alexandrescu]

On 09/05/2010 07:59 AM, Philippe Sigaud wrote:
>
>
> On Sun, Sep 5, 2010 at 12:00, Simen kjaeraas <simen.kjaras at gmail.com
> <mailto:simen.kjaras at gmail.com>> wrote:
>
>     Philippe Sigaud <philippe.sigaud at gmail.com
>     <mailto:philippe.sigaud at gmail.com>> wrote:
>
>         I guess the D syntax would be
>         auto x = amb([1,2,3]);
>         auto y =amb([4,5,6]);
>         x*y == 8; // forces desambiguation => the ambs explore the
>         possibilities.
>         assert(x == amb([2]));
>         assert(y == amb([4]));
>
>         There is only one value left, no more ambiguity.
>
>
>     But what happens in the case where there is more than one value left?
>
>
> If I understand correctly the linked Ruby and Haskell versions, the
> search stops as soon as you find a possibility. But then, I'm no pro on
> this.
>
>
>     That's one of the reasons I feel that doing the combination, then
>     filtering it, is more correct. There is also the fact that the above
>     syntax does not let you call arbitrary functions in the requirements,
>     something filter does.
>
>
> The combining and filtering is interesting (in this case, it'd be like
> doing a  list comprehension).
> But the real challenge in the SO question is the 'going back in time'
> part, which I have trouble to understand : how can you modify x and y
> through a multiplication and a comparison?
>
> I did a range comprehension maybe one year ago, which would be partially
> equivalent to the problem:
>
> auto solution = comp!("[a,b]", "a*b == 8")([1,2,3], [4,5,6]);
>
> solution is a range, in this case a one element range, containing only
> [2,4].

Here we need to have a crossProduct range, as we discussed in the thread 
"Combining infinite ranges". Then it's easy to combine it with filter:

auto solutions = filter!"a*b == 8"(crossProduct([1,2,3], [4,5,6]));


Andrei