#Day 3 Gear Ratios
thorn sparrow
prisma dock
this is the stupidest fucking engine schematic ive ever seen
elves are terrible engineers
thorn sparrow
no surprise, they ruin christmas every year
tall lance
Oh another parsing exercise..
nova meteor
Wow. This problem looks hard enough to do in a normal language.
still trout
Oh honey new the AoC problem is up
I had sleep plans but I guess I can be up when it's almost morning
hooooly mother
More parsing indeed
nova meteor
If they keep on this trajectory, I might have to drop out now. I'm pretty sure I could do this, but these are taking me hours and I really need to do other things.
prisma dock
historically speaking, they definitely will
nova meteor
But they feel a lot harder this year. Unless I'm just remembering poorly.
still trout
It's my first time doing AoC but so far this is the only one that looks fairly difficult
worn ingot
i love string manipulation so much
thorn sparrow
both parts C++
I should make an mdspan LayoutPolicy to handle oob access for these
tall lance
I need a shower after writing this code. At least it's done.
haughty steppe
both parts in python
||
#!/usr/bin/env python3
import math
def getnum(s: str, i: int) -> int:
l, r = i, i
while (0 <= l - 1) and (s[l - 1].isdigit()):
l -= 1
while (r + 1 < W) and (s[r + 1].isdigit()):
r += 1
return int(s[l : r + 1])
p1, p2 = 0, 0
adj = [(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)]
G = open("input.txt").read().splitlines()
H, W = len(G), len(G[0])
for y in range(H):
for x in range(W):
sym = G[y][x]
if sym in ".0123456789":
continue
nums = set()
for i, j in adj:
i, j = i + y, j + x
if G[i][j].isdigit():
nums.add(getnum(G[i], j))
p1 += sum(nums)
p2 += math.prod(nums) if len(nums) == 2 and sym == "*" else 0
print(p1, p2)
||
^ won't work if there are multiple numbers of same value adjacent to a symbol .. e.g. 123$123
quartz badge
Probably over-engineered, but it helps my brain. Github: ||https://github.com/bigbass1997/aoc-2023/blob/main/day03/src/main.rs||
still trout
Only part 1 for now, in Mr. Krabs' language, would not recommend looking because it was made in only 30min and it's just horrible. will fix it after I sleep 😴
inner coyote
both parts, Kotlin
Abusing LINQ
||```cs
var nums = world.Where(n => n.typ == otype.num);
var sum2 = world.Where(c => c.typ == otype.sym && (char)c.contents == '*')
.Select(c => c.left_coord)
.Select(g => nums.Where(n => n.surrounding().Contains(g)).Select(n => n.contents).ToArray())
.Where(gn => gn.Length == 2)
.Select(gn => gn[0] * gn[1])
.Sum();
Console.WriteLine("Part 2: " + sum2);
Just looked at part 2, will have to rewrite my code anyway because it sucks LOL
tall lance
Just looked at part 2, will have to rewrite my code anyway because it sucks LOL
I think you can still do it with some adjustments. In general I always just parse everything to a manageable data structure first. It's more code, but most of the time it helps
still trout
I think you can still do it with some adjustments. In general I always just pars...
I was gonna do that, I had a pretty little struct I was proud to make, then decided I was too sleep deprived to bother. I'll probably bring it back after I wake up lmao
I guess I just wanted to make a dumb and naive solution to validate my first idea
knotty ermine
improved simplified solution
haughty steppe
@knotty ermine for part2 .filter shouldn't it be .. numbers.count() == 2 ?
nevermind, just checked my own input ... there are no symbols that have more than 2 adjacent numbers
knotty ermine
technically yes but yeah it never happens in the input
||I’m surprised there’s no gotcha in the input where a number is attached to two gears||
haughty steppe
my script would be ok with that, .. but wouldn't be able to handle this || 123*123|||
quartz badge
||I’m surprised there’s no gotcha in the input where a number is attached to two...
While I was reading the question, I feared whether or not ||we had to wrap around to the opposite edge of the data. But didn't seem to be the case in the end. The directions never explicitly mentioned such a scenario.||
haughty steppe
😨 that would be cruel
quartz badge
hehe cruelty wouldn't stop AoC from doing it.
haughty steppe
then we should also add numbers that span vertically 😂
quartz badge
ha yeah I feared that too! since again, the instructions didn't mention them one way or another.
tall lance
Something like that will definitely happen further down the month.. It always does
quartz badge
I'm just wondering when the first graph-theory problem will happen.
tall lance
We'll get some BFS/Dijkstra again probably too. At least now I have a nice crate for that 🙂
Remainder theorem..
This is what I mean by it gets easier the more years you do 😆 You start recognizing the problems
magic spindle
Rust day today
tall lance
Is that Rust with some custom macro wrapper?
brittle quiver
yes. the macro is here https://github.com/maia-s/aoc/blob/main/2023/src/lib.rs
thorny musk
The problem from the first day is still the hardest one yet tbh
shell edge
ugly haskell part 1 and 2
still trout
Finally woke up and finished part 2, too long to post here lol. Will improve later
He did day 4 easily though