timespacex'|. 281474976710656&|@*/\.<.1+2*i.-2^12'
0.00096 331904
   (281474976710656&|@*/\x:1+2*i.2^5) -: |. 281474976710656&|@*/\.<.1+2*i.-2^5
1

I hope this helps,

-- 
Raul


On Tue, May 16, 2017 at 10:31 AM, 'Mike Day' via Programming
<[email protected]> wrote:
> Thanks again,  Raul.
>
> Well - I'd actually meant ~2^24 odd numbers,  1... _1+2^25 - trust me
> to make a mistake in trying to simplify the presentation!   Let's call it
> a typo.
>
> I don't think my post was a spoiler,  as the problem needs quite a bit
> of insight elsewhere;  this array helps in getting the answer.
>
> You're right, of course, for that expression, ~50 seconds on this laptop,
> which produces one extended scalar result,  but unfortunately I need the
> whole vector of cumulative products, 1 3 15 105...
>
> These time and space tests are for lower size arrays:
>     1 ts'281474976710656&|@*/\x:1+2*i.2^10'
>
> 1.90882 628864
>
>    1 ts'datatype 281474976710656&|@*/\x:1+2*i.2^11'
>
> 7.91167 1.25146e6
>
>    1 ts'q =: 281474976710656&|@*/\x:1+2*i.2^12'
>
> 30.8606 2.49638e6
>
>     7!:5<'q'   NB. space required for ~2^12 extended elements.
>
> 589824
>
>     2 10 >.@^.589824  NB. rounded up log 2 & log 10 space.
>
> 20 6
>
>
> So 2^24 elements are likely to need around 2^40, 10^12 bytes.
> This laptop only has about 16GB.
>
>
> So,  I think the cludgy "cumoddprod" or its like remains the
> answer for a few more Moore cycles!
>
> Thanks,
>
>
> Mike
>
> On 16/05/2017 13:44, Raul Miller wrote:
>>
>> Forgive me for asking, but isn't 33554431 = _1+2^25 ?
>>
>> Anyway, this seems to work for me:
>>
>>     281474976710656&|@*/x:1+2*i.2^23
>>
>> I'll leave out the answer, out of respect for project euler (though
>> maybe the rules allow that now?).
>>
>>     timespacex'281474976710656&|@*/x:1+2*i.2^23'
>> 12.457 1.67937e9
>>
>> ... and needs about 13 seconds and about 2 gig ram.
>>
>> That said, I suppose it would be nice if x m&|@* y worked reliably for
>> fixed precision integers x, m and y
>>
>> Thanks,
>>
>
>
>
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