Some of the other unimplemented tasks did not seem difficult; the Catalan numbers/Pascal's triangle in particular got my attention and I implemented it: http://rosettacode.org/wiki/Catalan_numbers/Pascal%27s_triangle#J
On Thu, Apr 17, 2014 at 12:22 PM, Dan Bron <[email protected]> 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 > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
