mistake in Z definition found,
Zl =: ] [ _3 Z: [
makes this work,
((4;~4{.]) ((1>:@{::]),&<~[,~ _3{.0{::])`(_2 Z:1:)@.(4=0#@~.@{::])
((100 Zl ]) F..) (4}.])) ex
but still
((4;~4{.]) ((1>:@{::]),&<~[,~ _3{.0{::])`(_2 Z:1:)@.(4=0#@~.@{::])
((100 Zl ]) F.) (4}.])) ex
|domain error
On Thursday, December 8, 2022 at 02:15:39 p.m. EST, 'Pascal Jasmin'
via Programming <programm...@jsoftware.com> wrote:
The short circuit you are trying to avoid is to skip the full
partitioning part
4 #@~.\4
one approach is to separate into the simplest possible partition,
then hope that i. special code finds the short circuit.
4 + (1 i.~ 4 = #@~."1) 4 ]\ ex
your power approach is completely reasonable.
A fold approach is complicated by seeking the index rather than a
"computed result"
There might be too much manipulated computation involved, but the
general approach would be:
x parameter to fold is (a: ,< 0) NB. initial y.
if head cell of y is shorter than 4, then append x to last 3 chars.
(better to preprocess y such that x (to F.) has first 4 chars, and y
omits first 4.)
if head is length 4, then apply test ((1 >:@{:: ] ) ,&<~ [,~
_3{.0{::])`(_2 Z: 1:)@.(4 = #@~. 0 {:: ])
((4;~4{.]) ((1>:@{::]),&<~[,~ _3{.0{::])`(_2 Z:1:)@.(4=0#@~.@{::]) (]
F..) (4}.])) ex
┌────┬─┐
│jpqm│7│
└────┴─┘
I don't understand why this fails: (F. instead of F..)
Z =: ] [ Z:
((4;~4{.]) ((1>:@{::]),&<~[,~ _3{.0{::])`(_2
Z:1:)@.(4=0#@~.@{::]) ((100 Z~ _3:) F.) (4}.])) ex
|domain error
or
((4;~4{.]) ((1>:@{::]),&<~[,~ _3{.0{::])`(_2 Z:1:)@.(4=0#@~.@{::])
(] F.) (4}.])) ex
|domain error
also it is not documented whether F. or F: are forward or reverse.
On Thursday, December 8, 2022 at 11:28:12 a.m. EST, 'Michael Day' via
Programming <programm...@jsoftware.com> wrote:
Having remarked recently (re Advent of Code day 6):
Very easy today, in J at least, and probably in APL & K/Q.
(Though it
suggests one should learn the new fold features, as it's
inefficient to
examine all the data for something that might occur early. No
problem
with 4kb, but ....)
I've just had a look at the Jwiki entries about Fold, and wonder how it
would work
for day 6.
The problem is so easy for J-ers that I don't think this will spoil it
for forum users:
ex
mjqjpqmgbljsphdztnvjfqwrcgsmlb
4 + 4 i.~ 4 #@~.\ ex
7
QED
#data NB. the size of my data set
4096
ts'4 + 4 i.~ 4 #@~.\ data' NB. time & space
0.0008555 67232
I didn't bother making a function for this!
One way to stop early, rather than process the whole array, is this:
ts'4 + (# - #@:((}.`])@.(4&([ = #@:~.@:{.))^:_)) data'
0.0007299 4736
(I know it's ugly, and non-optimal!)
Similar time but with ca 14x space-saving.
But there must be some sort of Fold to progressively examine 4-windows,
or 14-windows in part 2, until the first appropriate window is found,
when
it stops. That should save time as well as space. Not necessary
here, of
course, but if there was a real application with many megabytes of
data...
Cheers,
Mike
On 06/12/2022 19:46, Brian Schott wrote:
I have successfully solved Day 5 using the following looping verb
`tomove`
but cannot craft a Fold version, and would like help.
*[FYI I have included an attachment that seems to load and execute
properly, but I could NOT get the email versions to load correctly,
presumably because of some funny characters.]*
tomove=: dyad define
while. #y do.
x=.x move {. y
y =. }.y
end.
x
)
NB. The verb `move` for part 1 is very simple:
move =: dyad define
'n f t'=. _1 1 1*y - 0 1 1
take =. |.n{.>f{x
left =. <n}.>f{x
put =. <take,~>t{x
(left,put) (f,t)}x
)
NB. the inputs are produced as follows.
top =: 0 :0
[D]
[N] [C]
[Z] [M] [P]
1 2 3
)
state=.deb each ;/ |:|.}:1 5 9&{;._2 top
bottom =: 0 :0
move 1 from 2 to 1
move 3 from 1 to 3
move 2 from 2 to 1
move 1 from 1 to 2
)
alpha =. a. {~97+i.26
moves =. ".-.&alpha ;._2 bottom
NB. finally the result is produced by the next phrase
{:every state tomove moves
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