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02:30
my day 12 solution is not optimized, but still finishes in reasonable time
 
4 hours later…
06:41
0
Q: Screenshot of the Week Contest #54: Happy Holidays 2021!

Robotnikand welcome to the 54th edition of the Screenshot of the Week! To start with, congratulations to the winner of the previous contest! arghtype's evening ride in days-gone, won with a handy 16 votes: To submit a screenshot, simply post it as an answer to this question, but please take note of the...

2
 
4 hours later…
10:31
7
Q: What game is this? (First-person character holding an orange mass labeled "Paleberry")

Guest15335 I found this picture on a discord server and I was wondering what it was.

11:06
anyone recognizes the watch here? It is from a specific game?
It doesn't appear to be from an existing game
@Unionhawk it's a magic eye picture.
just squint and the letters will jump out at you
#### #### #    ####   ##  ##  ###  ####
#    #    #    #       # #  # #  # #
###  ###  #    ###     # #    #  # ###
#    #    #    #       # # ## ###  #
#    #    #    #    #  # #  # # #  #
#### #    #### #     ##   ### #  # #
today's really feels like a problem for regex...
11:43
Part 2 is obviously an efficiency problem, but I can't work out what the abstraction I'm supposed to use is
12:24
@Ronan same
If I can't just change 10 to 40, is it really efficient?
12:49
Well the example has the polymer length in the billions, so you can't just write it out
13:09
weekly reminder
Still not resolved apparently, if you live in EU do NOT buy repentance dlc from the WEB version of the shop.
It is still missing the disclaimer that warns customer that the dlc does only work with the digital version of the base game
@Ronan I'm guessing it's like a triange number or something, i.e. there's a well known formular for figuring out the counts of each element.
Perhaps, I just managed so solve it another way though
13:27
noice. I will have a look when I get home
14:00
@Ronan Today's? I haven't finished part 1 yet, but I'm assuming we're going to have to do something like 1000 iterations for part 2.
Yeah
I suspect you will have to find repeating sections.
Which means my part 1 solution as I have it envisioned will require significant rework for part 2.
my part 1 solution certainly does
 
1 hour later…
15:19
Gah
Self, stop looking at the Bridge before you've solved the day's puzzles
Sorry, I was trying to keep it vague
Oh no, you're good
I don't expect this to be spoiler-free
It's just a habit to pop into the Bridge and I need to remember to wait lol
15:53
Ah crap, my part 1 works on the example but not on the real data
Oh
16:20
Ugh, yeah, I'm going to have to completely rework this for part 2
16:47
Yo any of you guys have a great tryout GraphQL db?
Have to learn it to apply on a job and i can't seem to find somewhere that has a set of data I could query for a tryout site
You mean a sample api you can query?
17:06
oooh, right, before I forget.
@SaintWacko I think there is a rerun of past Expeditions on NMS.
You should still have a few days to try getting the Normandy from Expedition 2
17:24
When you're proud of part 1, but get to part 2 and realize there's some algorithm you clearly were supposed to know how to use
18:03
@Sterno Yuuuuup
Ugh
So I rewrote my part 2 and it worked in .025s
But the number was off by one
It's not actually that complicated, but I definitely wasn't sure it would work before I tried it
AoC said it was too high, so I just decreased it by 1 and it worked, but...
I'd still like my code to return the right number
Huh weird
I mean, I've got an idea why
I'm tracking the count of each polymer pair, and then at the end I add up each polymer based on the pairs and divide it by two to account for the fact that the pairs overlap
But that doesn't properly account for the fact that either end of the string doesn't overlap
And I'm not sure how to figure out which polymer is the one that's going to be off by one
Oh wait, yes I do
Duh
Oh. No, apparently I do not
Okay part 2 done
@SaintWacko I have a suggestion, but I don't want to spoil it for you.
but I tried doing something similar, I think.
18:15
@Wipqozn No, go ahead. At this point I've answered the problem correctly
@SaintWacko I'd count the characters as you go, instead of calculating it all at the end. so I had a dictionary which tracked the pairs, and another which tracked individual characters.
@Wipqozn Oh huh
Then you don't need to worry about any overlapping.
And as you add a new character you just increment its count
By the number of pairs it's being inserted into, yup.
18:17
Let me try that
although I still have some parts from an earlier solution kicking around haha.
okay now I'm actually using the function I made earlier haha.
The bug I ran into was I was forgetting to zero out old pairs when I inserted a character.
Oh that's a completely different solution to mine
Wups, now it's too high by 3 lol
Oh
Hadn't set the initial count of the polymers in the template
That works!
Thanks
That's a much better solution
More efficient, too
@SaintWacko I originally thought of trying to count it at the end, but then I thought that accounting for the overlapping would be a huge pain in the ass, so I just counted it as I went. Plus it is more efficient.
@Ronan Is ts typescript?
18:23
Yeah
Just ignore anything right after the colons and it's plain js
Am I the only one that always makes a dedicated program for each part?
I always like to keep my original part 1 solution around, even when part 2 is just a 1 line addition. Not sure why.
Sometimes I do that
18:41
I keep it all version-controlled, but generally the part 1 program gets upgraded to part 2 in place
@Wipqozn No, I always do that
There's my part 2
And, by extension, my whole repo
@Wipqozn Yeah, I knew I was going to have a problem with the overlapping, but I figured I'd solve it when I got to it. Then I couldn't figure out how to solve it lol
@Wipqozn I always do Part1 and Part2 functions even if Part2 is often copied in part from Part1
19:02
@Unionhawk Yeah, I always start my part 2 by copying part 1. And usually it's just a minor modification. Not so much today...
@SaintWacko Same. First attempt at part 2, out of memory exception after like 5 minutes. Second attempt, 9 milliseconds
@SaintWacko If you did it anything remotely like how I did it, you probably forgot to add in the last character of the string
@Wipqozn I had a dictionary which tracked pairs. Final count could be calculated by just looking at the first character of each pair (because the 2nd character would be covered by another pair). But then you have to make sure you add in the very last character of the whole thing.
(since the last char of the whole thing isn't counted in the first char of any pairs)
19:25
@Sterno Ohh, that's also smart
I thought that was my problem at first, that I was missing the last character
I hadn't put enough thought into it and I just added all the characters together and divided by 2
19:54
Well, this discussion means I'll be able to do part 1 the fast way and not have to rework any of it for part 2.
 
2 hours later…
21:28
@MBraedley Sorry, we shouldn't talk about it here :(
No, I was going to have to figure it out somehow and probably look for hints.
@Wipqozn I have a single program for both parts that will output both answers.
as that's what I've seen on the reddit

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