Oops, I did my name recognition fail where I stopped paying attention
after the first letter. Sorry about that.

Meanwhile, (and perhaps distantly related?) I would not worry about
performing a unique after every intersection because if the underlying
representation already deals in unique elements, intersection will not
violate that constraint.

Thanks,

-- 
Raul


On Tue, Jun 6, 2017 at 6:11 AM, 'Mike Day' via Programming
<[email protected]> wrote:
> My point being,  if Michael Rice is exploring how to solve peg solitaire,
> it's often useful to work on small(er) problems, before perhaps getting
> stuck in a long loop,  or seeing the memory climb and one's pc seize up
> and need rebooting.
>
> But he'll know all this stuff anyway!
>
> BTW, re Raul's useful thoughts on union matters, same thread, earlier post.
> I've just noticed that the intersection I posted,  ix = [-.-. ,  is,
> perhaps, "lazy";
> better might be ~.@([-.-.),  but neither is appropriate for bit-vector
> comparisons!
>
> Cheers,
> Mike
>
>
> On 06/06/2017 10:28, Raul Miller wrote:
>>
>> On Tue, Jun 6, 2017 at 4:09 AM, 'Mike Day' via Programming
>> <[email protected]> wrote:
>>>
>>> if I were developing a solver for solitaire,  I'd include a variable as a
>>> parameter for
>>>
>>> the size of problem,  eg the number of rows,  1 2 3 etc,  or the
>>> ravel-size,
>>> eg 1 3 6 etc.
>>
>> You had not specified that previously, but note that implementing this
>> is a simple change.
>>
>> For example:
>>
>> flip=: [ ~: i.@#@[ e. ]
>>
>> solitaire=:3 :0
>>    5 solitaire y
>> :
>>    path=. (y flip~ 1#~ 0 0.5 0.5 p. x) search i.0 3
>>    if. #path do. path else. 'No solution' end.
>> )
>>
>> That said, the 'No solution' cases require a time-consuming exhaustive
>> search, and in the few tests I did with sizes other than 5, I was
>> hitting no solution cases. It's time consuming because each
>> permutation of moves that leads to a final board state gets tested.
>>
>> Still, I hope this helps,
>>
>
>
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