end, I have the impressions that it all boils down to the
rank of u .v, and how . works in general.
Adrien Mathieu
On 16/05/2022 21:50, Raul Miller wrote:
First off, this is a good question.
That said, it's matrix divide because we are "dividing by" a matrix.
But maybe it&#x
Is it a
mistake on NuVoc?
Adrien Mathieu
--
For information about J forums see http://www.jsoftware.com/forums.htm
Hello,
If you know how to use a terminal, you can try Termux. If you don't, you
can try to learn the very basics, which are enough for what you are
trying to do.
Regards
On 11/4/21 17:57, Thomas Arneson wrote:
I mistakenly changed the name of a file called from startup_android.ijs. Now j
s
ately the square
root of a, there is a solution.
For instance, starting with b = 151 (taken at random), Newton's method
will quickly converge to 151.1753, which is indeed an approximation of a
solution.
Adrien Mathieu
On 19/10/2021 11:08, Elijah Stone wrote:
On Tue, 19 Oct 2021, Elijah St
take a
look at nix (which packages J). I am not sure whether nix packages JQt,
but if it doesn't you can always to on their github repository and open
an issue in which you ask for it to be packaged (or even package it
yourself!), and it should be done within next version (usually).
Cheers,
fically on nouns, u is called m and v n, although in practice it
makes no difference.
Adrien Mathieu
Le dim. 10 oct. 2021 à 22:34, P Padilcdx a écrit :
> Thank you for the quick reply. Got the adverb part, thank you. But I’m
> still missing something fundamental. If u=< and C=“, V=uC i
ot;)y is <"0 y
(according to the above definition).
--
Adrien Mathieu
On 10/10/2021 21:43, P Padilcdx wrote:
J noob so pardon the noob question. As the subject indicates, I’m confused as to how
or why <“0 y turns into 0(<“)y when interpreted as a hook. Looked at the Primer
and LJ and th
I turned this question into a puzzle page, if you have more ideas you
can post them there.
--
Adrien Mathieu
On 06/10/2021 19:31, Raul Miller wrote:
Generally speaking, when sorting performance is actually the issue, we
use J's sorting primitive.
Implementations of sorting using
exity is not better than if they were not).
This does not mean it could not work, but it's not completely satisfactory.
Adrien Mathieu
On 05/10/2021 20:37, Adrien Mathieu wrote:
Thank you for this reply, this is pretty much what I was looking for.
I'll probably still have questions,
Thank you for this reply, this is pretty much what I was looking for.
I'll probably still have questions, but I need to assimilate this code
first (so small and yet so information-heavy).
Adrien Mathieu
On 05/10/2021 19:56, Marshall Lochbaum wrote:
While I don't think arrays will
ow do
you solve a problem when you encounter one in J!
Adrien Mathieu
On 05/10/2021 20:14, Hauke Rehr wrote:
Forget about this one, you don’t want to compute a matrix
when sorting. And this wouldn’t actually leverage rank for
getting the “looping” done.
Shame on me.
Still I think it’s possible
he merging problem, whereas
it is possible for my example.
Adrien Mathieu
On 05/10/2021 19:13, Hauke Rehr wrote:
I didn’t express that very well, sorry for that.
Rank helps when you want data parallelism.
Apply to /this/ data, after chopping it into chunks.
Not to different data computed from th
't even know if the two are
completely equivalent (ie. if you can get a solution with the same
complexity using both for every problem).
Adrien Mathieu
On 05/10/2021 19:00, Hauke Rehr wrote:
I don’t quite understand.
Rank always applies to data (space). Recursion to program flow (ti
on of mergesort. So the loop is
transformed into recursion.
What I wanted to know if there is a way in which loop is translated into
rank (if you get what I mean).
Adrien Mathieu
On 05/10/2021 18:53, Gilles Kirouac wrote:
Does this page help?
https://code.jsoftware.com/wiki/NYCJUG/2012-05-08
See
Hello,
I am a beginner, and I would like to know if there is a way to write
mergesort without using loops, or an ugly translation of loops using the
^: conjunction.
In particular, I have the impression that the merge part is hard to achieve.
Thanks in advance,
Adrien Mathieu
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