I see.
The only clause that adds a 1 to the solution is in line 26. It takes the
first number (in the example, n=2), takes the first n vars (line 34) and
force their sum to be n, making all of them equal to 1. So there is no way
to have a solution starting with (1 0)


On Thu, Mar 6, 2014 at 2:13 PM, David Nolen <[email protected]> wrote:

> On Thu, Mar 6, 2014 at 12:10 PM, Mauro Lopes <[email protected]> wrote:
>
>> (1 0 0 1) and (0 1 0 1) are wrong for me.
>>
>
> But which portion of your program should disallow this? If I can identify
> that it will be much simpler for me to determine the flaw in core.logic's
> FD functionality. I've encountered bugs like this in the past - they almost
> always boil down to some subtle problem around logic var aliasing. We have
> a bunch of tests to check for this but it looks like you've come across a
> case I've missed.
>
> Thanks,
> David
>
>
>> The idea is that, say, (f 8 [2 3]) should give me all ways of putting two
>> sequences of ones (with sizes 2 and 3, in this order), separated by some
>> zeros, in an 8-length row.
>>
>> It is like solving a one-row nonogram. Then I just use it for all rows
>> and columns to make a full nonogram solver:
>>
>> http://github.com/maurolopes/nonogramsolver
>>
>> As I said, my program seems to be working now. I'm only giving you
>> context so it is easier to understand my code and debug core.logic.
>>
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