I'm not sure if this counts as well. It is still brute force though:
(example used in several PL's)
p=:2 3 2 -.@-: 3 5 7&|
>:^:p^: (_) 0
23
On 17-04-14 18:22, Dan Bron wrote:
I sporadically scan the list of open tasks for J on RosettaCode:
http://rosettacode.org/wiki/Reports:Tasks_not_implemented_in_J
Most of these tasks lack a J solution for a good reason (e.g. stream
processing, low-level IO, heavy-duty GUI, multithreading, or just too many
tuits). But some fit nicely into J's domain, and simply need someone to
post a solution.
Chinese remainder thereom is an example of the latter:
http://rosettacode.org/wiki/Chinese_remainder_theorem
The algorithm looks pretty straightforward to me, and indeed I could
brute-force it by writing an explicit verb which matches, loop-for-loop,
the Python solution (for example).
But instead, I'd like to show off J's native grace and elegance;
unfortunately, it's been a long while since my undergraduate math days,
and I don't think I have the algebraic chops to tackle this problem
elegantly. Though I do have enough of a native instinct left to suspect
that there's room for a very pretty solution taking advantage of some of
J's more distinctive primitives (in particular +. #: %. seem relevant).
So, what's the most elegant way you could implement the Chinese remainder
theorem in J?
-Dan
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