Advent of Code

Feeds

01:40

Adám has removed Adám from the list of this room's owners.

totallyhuman

03:47

oh

Redwolf Programs

04:40

20m

tjjfvi

10m

Redwolf Programs

04:55

under 5m

emanresu A

πm

Redwolf Programs

yum yum

tjjfvi

2.5m

The Fifth Marshal

2m

Redwolf Programs

Maybe there's an oil spill in the ASCII art

The Fifth Marshal

04:58

1.5m

Razetime

maybe it's a squid tentacle

tjjfvi

or bubbles

from the vents

Redwolf Programs

Those have been white so far though

The Fifth Marshal

1m

Redwolf Programs

30s, glhf

hyper-neutrino

05:01

11/1

Unrelated String

52

let's go

nice!

lyxal

I have a nice 6 byter in vyxal that of course times out for the actual input

Unrelated String

epic

emanresu A

Dammit, made a mistake

Now I have to wait 40s

Redwolf Programs

67/139

Not bad

Still on the leaderboard, moved down one place

Unrelated String

05:05

i think the sole reason i climbed between points was i only hesistated briefly to copy paste what i already had into the triangular number formula

emanresu A

1011/295

@UnrelatedString Same

I probably would've got top 100 if I hadn't done Math.min(<array>)

Unrelated String

...does that do something fucky in js

rip

tjjfvi

So I just wildly guessed that the value wouldn't be more than the length of the array * 2

emanresu A

You need to do Math.min(...array), Math.min(array) returns NaN

tjjfvi

Which turned out to be fine

Unrelated String

05:07

lmao what

emanresu A

@tjjfvi I just guessed 2000

Unrelated String

i just used the max value in the input

hyper-neutrino

anyway

that was a massive difficulty drop from the weekend... lol

Redwolf Programs

@tjjfvi I guessed 1000 and it was fine lol

Unrelated String

since that's kind of a logical maximum

yeah that was actually insanely easy

hyper-neutrino

05:08

i could've probably ranked higher if i just onelined it

Unrelated String

lol

same actually

Redwolf Programs

My original way to calculate the triangle number thingy was with the sum of a range, but it was too slow

hyper-neutrino

coulda just gone for the min(sum(abs(x - i) for i in k) for x in range(min(k), max(k) + 1))

but i was not confident and thought it'd end up too long and messy

but that's actually a really nice looking one-liner

Redwolf Programs

@RedwolfPrograms So I had to use a recursive function since I didn't remember the formula

Unrelated String

it really is

hyper-neutrino

05:08

and then for p2 it's trivial to just make a g = lambda x: x * (x + 1) // 2 and throw a g(...) around something in that one-liner

oh well, i really can't be disappointed with my score lol

Redwolf Programs

I'm so happy I've managed to stay on the leaderboard

Unrelated String

i sort of stayed away from it worrying that p2 would be a nontrivial extension

Redwolf Programs

I was pretty certain I'd throw today, things've been going suspiciously well for me lol

The Fifth Marshal

Is there some non-O(n^2) way of doing part 2?

Unrelated String

@hyper-neutrino i almost did that but then i realized it's faster to literally just write abs(n-x)*(abs(n-x)+1)//2

Redwolf Programs

05:09

@TheFifthMarshal You can calculate triangular numbers in constant time

It's just a quadratic

The Fifth Marshal

What do triangular numbers have to do with this?

emanresu A

Part 2

Unrelated String

...

Redwolf Programs

That's what the movement cost is

The nth triangular number

hyper-neutrino

@UnrelatedString oh true

it's

The Fifth Marshal

05:10

Oh, I independently realized it was 1 + 2 + 3 + 4 + ..., looked up the sum of that sequence on Wikipedia, and didn't realize it was a triangular number.

hyper-neutrino

still O(n ^ 2)

to check every value and then check every value for every convergence point

right?

anyway is anyone else 500ing on personal stats page

Redwolf Programs

Oh, idk. I assumed from the way it was phrased that p1 wasn't also O(n^2) lol

The Fifth Marshal

I am

Redwolf Programs

I didn't earlier, lemme check

Yes, I am

hyper-neutrino

kekw

The Fifth Marshal

05:11

Part 1 can be done in O(n), because the median is always the best convergence point

Unrelated String

yeah i think the naive approach we probably all took is O([number of crabs] * [furthest crab position])

wait does that actually work

neat

Redwolf Programs

If I had written a faster way to calculate triangular numbers the first time around, I would've top-100'd on part 2, but I went with the sum of a range >:|

emanresu A

My code (reconstruction):

v=[...]; Math.min(...Array(2000).fill(0).map((_,x) => v.map(f => (a = Math.abs(f - v)) * (a+1) / 2)).reduce((a,b) => a+b)))

The Fifth Marshal

And I spent a lot of time on Part 2 undoing that optimization, and then wondering whether I needed a different one (because it felt stupid for Part 2 to be less efficient that Part 1)

hyper-neutrino

lol top is already 300 pts ahead sadge

emanresu A

05:13

Ugly butit woorks

hyper-neutrino

oh well, if i can hold my position i already met my goal i set, and maybe i can still climb a few places if i'm lucky

Razetime

part 2 somehow worked with a cartesian product

emanresu A

@hyper-neutrino You're ahead of Doorknob (at least on private)

hyper-neutrino

huh. interesting

emanresu A

@Razetime ಠ_ಠ how?

Unrelated String

05:14

what

hyper-neutrino

right i guess i'm punished less for fucking up yesterday on private than public since tckmn didn't choke i think

lyxal

y'all already done while I have to figure out how to optimise my solution in a language that does not do speed well

Razetime

@emanresuA {⊑𝕩/˜(⊢=⌊´)𝕩}•Show +´˘(1+↕⌈´inp)({2÷˜𝕩×𝕩+1}|∘-)⌜inp this is my code, go fish lmao

hyper-neutrino

it's a damn shame i can't check the private leaderboards

Unrelated String

yep

Razetime

05:15

all the leaderboards giving a 500

emanresu A

@hyper-neutrino Wdym?

I have it open right now, I can chek

Unrelated String

500

emanresu A

I've got it cached

if anyone wants to see

may be out of date

Unrelated String

it's back now

hyper-neutrino

tbh the most important thing is i am improving (or at least i feel like i am) so even if my score doesn't end up where i want this year i can always keep trying again and keep getting better

which is a better result than the internet points anyway xD

Unrelated String

05:17

that's the spirit

hyper-neutrino

always on that sigma male grindset

Unrelated String

...i feel like i'm definitely doing worse than last year but i can blame that on finals

totally gonna pop off in a week when i actually have free time

Razetime

@emanresuA runs instantly

emanresu A

I'm doing ∞ times better than last year

Unrelated String

true

Razetime

05:19

maybe the interpreter is just amazing or i did some optimization without knowing it

totallyhuman

these are getting suspiciously easy

maybe it's a setup for a bombshell of a wednesday

hyper-neutrino

uh oh

Unrelated String

lmao

wonder what day the total completion after a day will be less than the part-1-only completion on day 3

Redwolf Programs

@hyper-neutrino Top 10 in the world is seriously impressive even if you weren't steadily improving lol

Unrelated String

^^^

emanresu A

05:20

^^

totallyhuman

^

Razetime

btnlq

v

emanresu A

^

(Infinite loop go brr)

hyper-neutrino

@RedwolfPrograms true. i do have an issue with always comparing up so i'm never really satisfied, but thanks for the reminder :P

lyxal

05:27

for reference, my p1 in vyxal: v-ȧv∑g

emanresu A

We have 1-byters for v∑ and -ȧ

lyxal

@emanresuA the one byter for -ȧ doesn't vectorise properly with numbers and lists

emanresu A

Does vε work?

lyxal

53 secs ago, by lyxal

@emanresuA the one byter for -ȧ doesn't vectorise properly with numbers and lists

I just said that it doesn't

emanresu A

Oh. That's... annoying, is there somewhere else we can put it?

lyxal

05:30

i'll think about it after i finish p2

totallyhuman

wait

the cheapest position for part2 is just the mean

or at least on my input

but that must be coincidental

'cause there's no reason that wouldn't be true for part1

Razetime

> Part 1 can be done in O(n), because the median is always the best convergence point

totallyhuman

right

but can anybody confirm on their input if the mean is the cheapest position

@totallyhuman wait no i'm dumb

hyper-neutrino

median

totallyhuman

for part 2

median is true for part 1

hyper-neutrino

05:45

oh wait you were talking about p2

yeah i'm sure that's a lucky coincidence lol

highly doubt that'd work in general tho i cba to prove it :P

Razetime

for part 1 i mistakenly assumed the input had the answer in it

somehow that worked

hyper-neutrino

lol

totallyhuman

lyxal

@totallyhuman nope

wouldn't have worked

totallyhuman

hey look at that it's true for part 2

:P

lyxal

05:49

p2 in vyxal: Gʀƛ?-ȧ:›*½∑;g

totallyhuman

in other news private leaderboards are up and personal stats are disabled

Razetime

06:07

meanwhile 2020 day 7 was this: adventofcode.com/2020/day/7

Unrelated String

i forgot that one was only day 7

that was something

Bubbler

Oh right, day 7s in 2015, 2017, 2018, 2020 all required handling graph structure in some way

2016 d7 was simple string and 2019 was intcode (obviously)

totallyhuman

what if the last couple easy ones are a setup into a bigger challenge that involves them but you can't brute force it any more

that's it i'm becoming an aoc conspiracy theorist

Razetime

@UnrelatedString string manip nightmare

i was literally dying in apl

Unrelated String

oh i can imagine

Bubbler

06:18

Well, when most of the interesting things are revealed on the very day (esp. in aoc reddit) there's very little point of building up across days

Unrelated String

i think i recall the parsing was relatively easy compared to the rest of the logic if not so much the parsing for other problems in my case using python

Bubbler

parsing was frick and doing the rest was also frick

Unrelated String

might have been tempted to use regex but i think i just used a bunch of splits or something

btnlq

O(n) solution for part 2: Try it online!

Unrelated String

nice

Bubbler

06:27

(multiline messages do not support markdown)

hyper-neutrino

oh, cool!

Razetime

so someone wrote an intcode soltuion for today's problem: reddit.com/r/adventofcode/comments/rar7ty/2021_day_7_solutions/…

3

hyper-neutrino

15:40

reddit.com/r/adventofcode/comments/rawxad/…

cc @totallyhuman someone wrote a paper on why the mean works for p2

or specifically, the answer will be extremely close to the mean, so if you take the mean, you only need to check two values

or something like that. maybe i'm reading it wrong

totallyhuman

16:56

ooh

but that also means that very smart people found no closed solution for it

SaintWacko

19:07

Man, I really need to be at least looking at these at 11

This was another super easy one

Took me 13 minutes

Granted, it's a brute force solution. Although I could speed it up some by starting at a more sensible point

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