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>https://everybody.codes/events ETA 29 hours Discuss what programming language you will use NO AI ALLOWED

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𓀐 @++worDpL: I will check this out. I never heard of it. I hope it's better than aoc+

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𓀐 @-B6SeWoU: Why start at such an awkward time?+

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𓀐 @cb_kghWy: 3 x 3 4 x 4 5 x 5 ... ### #### ##### #.# #..# #...# ### #..# #...# #### #...# ##### >The first outpost had 3 enchanted pine cones, so it had to collect 5 more to build a 3 x 3 frame. >The second outpost had 11 pine cones and needed 1 more to complete a 4 x 4 frame. this makes no sense to me. can someone explain why 3 requires a 5x5 and has a remainder of 5?+

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𓀐 @5rL?7gqA: I forgot how to code without AI.+ 1

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𓀐 @Q+Cay2RH: i dont know how to code even with ai. i wrote this program yesterday from a tutorial, and i couldn't even print an integer without using printf (not allowed according to the tutorial, to build up skills), just the write function. i feel like a noskills vibecoder ;_; #include <unistd.h> #include <stdio.h> int ft_strlen(char *str){ int i = 0; while (str[i] != '\0'){ i++; } return i; } int main(void){ int i = ft_strlen("the size of the string is: \n"); write(1, &i , sizeof(i)); // <<--- right here return 0; } [\code]+

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𓀐 @fav6WZ=H: Imagine wasting your lifespan on this. Hours you will never get back.+

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𓀐 @vELkipb~: I will thank you :)+

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𓀐 @H8iB+mdv: Which language should I use if I'm optimizing for maximum (You)'s? Serious answers only+

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𓀐 @gkhgtZiz: >inb4 seething poorfags and tinkertrannies I'm looking forward to doing this year's programming puzzles together bros. At least AI hasn't taken this away from us yet.+

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𓀐 @fpksMYh+: >parts have different inputs+

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𓀐 @gFVxt7~Q: >bigboy every integer from 0 to 1,000,000 inclusive Part 1: 666166500 Part 2: 1500024 Part 3: 1334836000 48.703ms+

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𓀐 @cqNbz~Mu: What's the smart way to solve it? The logic I think is necessary is: Every alloy ID should have a list of alloy IDs that are weaker than it. For every pair of alloys N and M, where N is stronger than M, update list N to include alloy M, update any list K that includes N to also include M, update any list J that includes K to also include M, repeat as necessary. Is there a simpler way to do it?+ 1

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𓀐 @uaoCQYyo: mine solution was basically: >make a dict where the key is the allow and the value is a list of alloys known to be weaker than it >make a list of all alloys >find the alloy that isn't in any other alloy's "weaker than" list >that's the strongest >now iterate through the list again, ignoring the strongest alloy, to find the other alloy not in anyone's list >that's the second strongest alloy >keep iterating til you've assigned them all probably a faster way of doing it, but it's pretty simple conceptually+

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𓀐 @wYi_UW8f: Just the constant intrusive homoerotic thoughts in my head+

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𓀐 @Wr-pTKrx: Do you guys use auto complete? My internet went out recently. So I was trying to code for fun and realized I'm a retard without an LSP. So now I've just turned autocomplete off completely so I can just memorize everything. Or thats my plan.+

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𓀐 @WAgVyWZC: >part 3 Uh oh+

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𓀐 @Lb?T+Y9W: i don't know what this is i've been sick for a week and i don't know how to code but i want to do stuff because i have depression and i need to be distracted python i guess?+

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𓀐 @oMBVLDw_: I thought of a kino solution to part 3 that Im too lazy to write myself. Imagine ISO2-xyzw ijkl = n. We do i*x^3 + j*y^2... = 137, express i as a function of j k l, so we can find if the room to minimize it is sufficient. If it is we search for a way to minimize it and if we find it vote invalid if not (in either of the two), we consider i fixed, so we express everything in terms of j instead and so on, till the second to last var, at which point we can validate the vote+

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𓀐 @fCX*rFE6: man I feel like no one is playing this. aoc gets thousands of people. for this thing I'm in the top 100 even if I start late a couple hours late and take an hour to finish. programming is dead boys pack er up+ 1

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𓀐 @gfdX8sj6: It's weird how little participation we get considering there's basically no meaningful quality difference between AoC and EC. Last year I remember the /g/derboard with 30-40 users making up like a quarter of all participants toward the end of the event. Need more fa/g/gots to join us for their puzzle fix now that AoC is in zombie mode with only 12 puzzles/year.+

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𓀐 @Q_qG7p3b: Eric seems to really resent doing it for free. Seriously, stop whining and just put the puzzles in the bag.+

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𓀐 @mqrv5?7F: >part 3 I don't feel like doing it+

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𓀐 @nx56dn=j: This one felt like work+

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𓀐 @wCNJ?-s5: >if part == 1 >if part == 2 I don't even care that you imported solution but this way of solving these is so lame. You're supposed to solve each part in its own function while building up helper functions in a way that makes each subsequent part easier. If you build the wrong abstraction and you can't use it for a subsequent part, you get penalized.+

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