The APL Orchard

@Adám what solution? (am I allowed to ask that)

is it in APLcart or something?

Adám

4:53 PM

—

?

Adám

I can't speak about it :-)

fair enough

That means it's in the vault.

xpqz

5:34 PM

@RikedyP did you manage to get to the video now?

haha, my answer for phase1:10 gets timeout. I guess someone else had came up with a more clever one.

5:45 PM

I fixed mine to pass the test but it is still long. See if I could get some time improve it later

How long is long?

Fun APL program that solves a problem similar to aoc2021/4p1: {⊃∊⍺∘{⍸∨/5=(+/,+⌿)⍤2∨⌿⍺=⍤3 0⊢⍵}¨,\⍵}

given a set of bingo cards and the vector of numbers announced, find the first bingo card to win

looks inefficient

can you give an example?

cg.esolangs.gay/15

except my code deliberately ignores diagonals

and yeah, it maybe is inefficient, i didn't really care about efficiency

i think it's elegant though

I think {⊃⍋⌊/(⌈⌿⍤2⌊⌈/)⍵⍳⍺} works

5:58 PM

> the first number called will always be 0, as will be the centre space of all of the cards. this emulates the "free space" from the real game.
> there will always be a winner, and the game will always end when the last number is called.

i feel like one could take advantage of these somehow.

i didn't

(⊃⍤⍋⌈⌿⍤2⌊/⍤⌊⌈/)⍳⍨ as a train maybe

github.com/rak1507/Advent-Of-Code-APL/blob/main/2021/…

i initially wanted to model my solution using something that isn't too APLy because i had to use something else in the end

fair enough

⊃∊b∘{{⍸↑∨/∨/¨¨5=¨(+/¨⍵) (+⌿¨⍵)}↑∨/⍺∘=¨⍵}¨,\v was my first solution assuming b is a vector of boards and v are the guesses

it's rather scalar and box heavy, but the above explains it.

      B←?10000 5 5⍴100
      N←?⍨100
      ]runtime -c 'B ((⊃⍤⍋⌈⌿⍤2⌊/⍤⌊⌈/)⍳⍨) N' 'B {⊃∊⍺∘{⍸∨/5=(+/,+⌿)⍤2∨⌿⍺=⍤3 0⊢⍵}¨,\⍵} N'

  B ((⊃⍤⍋⌈⌿⍤2⌊/⍤⌊⌈/)⍳⍨) N                  → 1.0E¯3 |       0%
  B {⊃∊⍺∘{⍸∨/5=(+/,+⌿)⍤2∨⌿⍺=⍤3 0⊢⍵}¨,\⍵} N → 1.6E0  | +164800% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

6:04 PM

      B←?10000 5 5⍴100
      N←?⍨100
      ]runtime -c 'B {(⊃⍤⍋⌈⌿⍤2⌊/⍤⌊⌈/)⍵⍳⍺} N' 'B {⊃∊⍺∘{⍸∨/5=(+/,+⌿)⍤2∨⌿⍺=⍤3 0⊢⍵}¨,\⍵} N'

  B {(⊃⍤⍋⌈⌿⍤2⌊/⍤⌊⌈/)⍵⍳⍺} N                 → 1.0E¯3 |       0%
  B {⊃∊⍺∘{⍸∨/5=(+/,+⌿)⍤2∨⌿⍺=⍤3 0⊢⍵}¨,\⍵} N → 1.0E0  | +103450% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

i get (barely) better timings

are you on 18.2?

nope!

ah interesting

i feel like my solution is worse by a factor of n than yours when it comes to computational complexity

as the percentage is pretty much the same for 100 boards

I think so because ,\ is O(n^2)

6:07 PM

:(

apl pls

but this isn't really needed though i think

well ,\ can't really be anything other than O(n^2)

there's probably a way to ¨ over N and update state that would still be O(n) slightly modifying your solution

dzaima

even if ,\ were O(n), you'd still be doing O(n^2) operations with the result's items

> the game will always end when the last number is called.

am i interpreting this correctly

does this mean that the game always finishes when the last number is called

so this would theoretically mean i don't have to check prefixes of omega, just omega.

@PaulMansour I got 38 chars, and later 32 using a different method

assuming this is the case my submission becomes much faster than it used to be

6:11 PM

I realized there are something about @ I didn't know

i get a feeling that first 8 problems in phase 1 were really simple and i made satisfactory solutions to all of them within 30 minutes

while the other 2 problems are actually really hard to solve in a nice way to me

phase1:9 is a new one.

@KamilaSzewczyk did you find the @ hint in problem 10 helpful?

p2:3 feels like a question that'd pop up in an Excel/MS Access problem from the HS finals here

@LdBeth i have not solved problem 10 in an elegant way, but i can let you know once I do.

my solution to p10 is 20 chars and uses @ but I wouldn't say it's elegant

⋄'CI11'⎕FMT 123456789 couldn't get this to work (sorry if this is a spoiler, I don't think it is bc I don't think ⎕FMT can do that on strings)

TryAPL

@rak1507 123,456,789

6:26 PM

that's smart

but yeah i think it's a dead end

yeah

> Note that the number of digits in the character representation might exceed the number of digits that can be represented as a 32-bit integer.

mainly because of this

@KamilaSzewczyk I have 23 conv dfn, pretty ugly

@LdBeth Wrong ping

yep, I was hoping there was some way ⎕FMT could do that but on strings, but it doesn't seem so

The reason for the requirement of long input was I'm sure to specifically exclude ⎕FMT

6:29 PM

⎕FMT is clunky anyway, afaik there's no way to say 'give me as many digits as I need'

In that regard, yes.

7:32 PM

@rak1507 because of the use of ∊?

8:07 PM

got a 36 char for phase1:9

I guess that's the minimal possible unless someone can come up a clever encoding trick

@LdBeth 32 here.

@LdBeth the answer is yes, i did

I came up with a 27 byte solution.

For 10 a 25 byte is possible.

hope you guys have fun on phase2