> R =: ((cos , -@sin) ,: sin , cos)
I am curious why the definition isn't
(cos , -@sin) ,: (sin , cos)
It seems prettier (more balanced, more symmetric).
On Fri, Dec 18, 2020 at 3:56 AM Ben Gorte wrote:
> Perhaps I'm missing the point, but I would say:
>
> R =: ((cos , -@sin) ,: sin , co
Thanks to all who presented their ideas on how to address this issue with
J. As usual, I learn a lot from the users of this Forum!
F.
On Fri, Dec 18, 2020 at 8:58 AM Francesco Pedulla' wrote:
> Dear all,
> I need to represent the 2D rotation matrix 'R'
>
> R = |cos(t) -sin(t)|
>|sin(t)
Your solution is simple but does not easily generalize to more complex
cases (infact, a requirement I did not express...).
Thanks,
Francesco
On Fri, Dec 18, 2020 at 6:18 PM Raul Miller wrote:
> I would be tempted to use something like:
>R=: (2 1,:1 2) o. _1 1&*
>
> For example:
>R"0(0 3
Hi Ben,
in fact you got the point. I thought you needed a gerund to build a
structure of functions. Your solution is very clean and I like it a lot.
Thanks!
Francesco
On Fri, Dec 18, 2020 at 12:56 PM Ben Gorte wrote:
> Perhaps I'm missing the point, but I would say:
>
> R =: ((cos , -@sin) ,: s
n=.'1+(2*3)+(4*(5+6))'
'()'i.n
2 2 0 2 2 2 1 2 0 2 2 0 2 2 2 1 1
Now THAT is cool! I forgot one could use dyadic i. as "first occurrence".
That's a much simpler way to generate the amend indices.
.
Skip Cave
Cave Consulting LLC
On Fri, Dec 18, 2020 at 12:21 PM 'Michael Day' via Programming <
|. ('()'i.n)} ')','(',: n
or tacit
([:|. ('()'&i.)`(')','(',:])}) n
On Fri Dec 18, 2020 at 10:35 PM CET, Skip Cave wrote:
> Using curlyrt: (amend in place)
>
> n=.'1+(2*3)+(4*(5+6))'
>
>
> |.(+/1 2*'()'=/n)}n,(17#')'),:(17#'(')
>
> ((6+5)*4)+(3*2)+1
>
>
>
> Skip Cave
> Cave Consulting LLC
>
>
> O
Hi,
" amend in place " , that's very cool
Thanks for demonstrating a practical use case.
Jimmy
PS you can use #.(base 2 decode) to replace +/1 2*
On Fri, Dec 18, 2020 at 4:36 PM Skip Cave wrote:
> Using curlyrt: (amend in place)
>
> n=.'1+(2*3)+(4*(5+6))'
>
>
> |.(+/1 2*'()'=/n)}n,(17#'
Using curlyrt: (amend in place)
n=.'1+(2*3)+(4*(5+6))'
|.(+/1 2*'()'=/n)}n,(17#')'),:(17#'(')
((6+5)*4)+(3*2)+1
Skip Cave
Cave Consulting LLC
On Fri, Dec 18, 2020 at 12:21 PM 'Michael Day' via Programming <
programm...@jsoftware.com> wrote:
> Unlike in the K forum, the J-wires have been
Slightly different approach by converting over and back from integers and doing
some arithmetic
|.@:(+ (+/ @: (1 _1 * 40 41 =/ ]))) &. (3&u:) '1 + (2 * 3) + (4 * (5 + 6))'
((6 + 5) * 4) + (3 * 2) + 1
Cheers, bob
> On Dec 18, 2020, at 12:12, 'Michael Day' via Programming
> wrote:
>
> Yes
Right! 0 1 =/ '()' i. y is just '()' =/ y...
Yes, day 15/2 is unfortunately similarly slow for me, where I update a
length 3e7 array to track times. An equivalent scheme version manages
in <600ms. The gap in performance is odd...
Joseph
On 12/18/20, 'Michael Day' via Programming wrote:
> Yes, I
Yes, I ended up indexing brackets with I. '()' =/ y , assuming that
the left & right brackets
were balanced!
Part 2 was long-winded, recursive, but evaluated quickly, luckily.
My slowest routine for any of these posers is for part 2 of Day 15; best
I managed was
~ 40 seconds.
Cheers,
M
For part 1 I used pretty much the same approach, eval by way of
reversing (|.&.;:):
+/ {{".'('c}')'o}y['o c'=. I. 0 1=/'()'i.y=.|.&.;:y}}&> in
Mine differs mostly in avoiding library definitions, amending parens
by index. For part 2 I used ;: to calculate a depth vector for the
tokens and then
Thanks, Brian - I thought I needed to put in a temporary intermediate
value to avoid
reversing the first change, hence the three pairs rather than two.
But the enigma remains, doesn't it: what stops ']' becoming ')' ? Do
rplc and/or
stringreplace only work on a maximum of two pairs of before
What about the following?
( '()',')(' ) stringreplace|. '1 + (2 * 3) + (4 * (5 + 6))'
((6 + 5) * 4) + (3 * 2) + 1
On Fri, Dec 18, 2020 at 1:21 PM 'Michael Day' via Programming <
programm...@jsoftware.com> wrote:
> Unlike in the K forum, the J-wires have been virtually silent on this
> year
For the first part you can use charsub -
https://www.jsoftware.com/help/user/lib_strings.htm#charsub
'())('charsub |.'1*2+(3*4)'
(4*3)+2*1
On Fri, Dec 18, 2020 at 19:21 'Michael Day' via Programming <
programm...@jsoftware.com> wrote:
> Unlike in the K forum, the J-wires have been virtually
Unlike in the K forum, the J-wires have been virtually silent on this
year's Advent of Code...
Anyway, today's problem, number 18, might interest J & APLers as part
1 is ALMOST
plug-it-in to J/APL; there's just one twist, so to speak, namely we
need to reverse an
arithmetic expression,
I would be tempted to use something like:
R=: (2 1,:1 2) o. _1 1&*
For example:
R"0(0 30p1 45p1 60p1%180)
1 0
0 1
0.866025 _0.5
0.5 0.866025
0.707107 _0.707107
0.707107 0.707107
0.5 _0.866025
0.866025 0.5
FYI,
--
Raul
On Fri, Dec 18
Hi,
bitten by shape (again) :
$&.>x
┌┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┐
││1│1│1│1│1│1│1│1│1│1│1│1│1│1│1│1│1│1│1│
└┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┘
Thanks,
Jimmy
On Fri, Dec 18, 2020 at 11:38 AM Hauke Rehr wrote:
> yes, I suspect there must be shape issues
>
>] x =: ;/
yes, I suspect there must be shape issues
] x =: ;/ 18 8 13 15 16 9 14 5 2 1 3 0 4 6 12 10 7 17 11 19
┌──┬─┬──┬──┬──┬─┬──┬─┬─┬─┬─┬─┬─┬─┬──┬──┬─┬──┬──┬──┐
│18│8│13│15│16│9│14│5│2│1│3│0│4│6│12│10│7│17│11│19│
└──┴─┴──┴──┴──┴─┴──┴─┴─┴─┴─┴─┴─┴─┴──┴──┴─┴──┴──┴──┘
/: x
11 9 8 10 12 7 13 16 1 5 15 1
I would like to know what $&.>x looks like.
On Fri, Dec 18, 2020 at 8:30 AM Jimmy Gauvin wrote:
> Hi all,
>
> I am failing at understanding the difference between sorting a boxed
> numeric vector and the unboxed vector :
>
>x
> ┌──┬─┬──┬──┬──┬─┬──┬─┬─┬─┬─┬─┬─┬─┬──┬──┬─┬──┬──┬──┐
> │18│8
Hi all,
I am failing at understanding the difference between sorting a boxed
numeric vector and the unboxed vector :
x
┌──┬─┬──┬──┬──┬─┬──┬─┬─┬─┬─┬─┬─┬─┬──┬──┬─┬──┬──┬──┐
│18│8│13│15│16│9│14│5│2│1│3│0│4│6│12│10│7│17│11│19│
└──┴─┴──┴──┴──┴─┴──┴─┴─┴─┴─┴─┴─┴─┴──┴──┴─┴──┴──┴──┘
3!:0 each x NB.
Perhaps we have both been missing the point.
I hastily said
“you won’t be able to ‘matrix multiply gerund matrices’”
In J, you can actually achieve that.
And if that’s what OP wanted, we indeed missed the point.
A solution would then keep R as-is in order to let the user
make good use of Js funct
Perhaps I'm missing the point, but I would say:
R =: ((cos , -@sin) ,: sin , cos)
R 1r6p1
0.866025 _0.5
0.5 0.866025
Ben
On Fri, 18 Dec 2020 at 18:59, Francesco Pedulla' wrote:
> Dear all,
> I need to represent the 2D rotation matrix 'R'
>
> R = |cos(t) -sin(t)|
>|sin(t) cos(t)|
Hello
I played some years ago with 2D rotation as part of general linear mappnigs
with complex numbers.
A small demo below.
Esa
NB. ID rotdemo.ijs v0.0 28.06.17
NB. NO rotate some points
load 'graph'
tographcoord=:[:_0.98&+1.96&*
randompts=. 3 : 0
(y,2) $ tographcoord ?(+:y) #0
)
rot=: 4
Sorry, I made a mistake: the order of the entries is wrong.
And what your code does is compute (R pi)%4 rather than
R(pi%4) so I guess my solution (with swapped sin entries)
is what you wanted.
Am 18.12.20 um 09:09 schrieb Hauke Rehr:
> First of all, I wonder which result you expect.
> Is the resu
First of all, I wonder which result you expect.
Is the result of “R mf pi%4” what you want?
Personally, I woldn’t mind if R’s shape reflects
what we call a matrix as long as its result does
(after all, you won’t be able to “matrix multiply
gerund matrices”)
So my solution would look like this:
R
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