I haven't seen anyone spell it out here, but the answer to your enigma
is the same as the reason that you didn't need to perform a sequence
of replacements:
The code is designed such that all replacements happen against the
original string.
So you would have to stringreplace again to complete the
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,
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