Advent-of-code VS advent-of-code

Wow, almost identical to mine: https://github.com/MichalMarsalek/Advent-of-code/blob/master/2022/Nim/day6.nim

Beautiful Nim! My solution is actually very similar.

Or you can just add only one letter from each pair and hence avoid the double counting.

[LANGUAGE: Python 3] 153/75 Raw solution

Not sure what you mean by that, I just whipped up a parity-based part 2 solution and it doesn't do any doubling (unlike my original solution).

One could instead count |F7 (that's what I do in my refactored solution), but counting all the bends would miscount the vertical segments (FJ would end up canceling itself out).

Edit: Cleaned up code. I initially was going to code up a generalized solution but realized that it's complicated enough that I shouldn't stay up to do it now, it's already past 1AM here. Perhaps on the weekend?

Hm, thinking about it I think I actually have something coded up in my lib.math module which would apply here. Take a look at chinese_remainder_incongruence or offset_chinese_remainder_incongruence (which delegates to the former but accepts a slightly different set of expressions that make more sense in my head.) Obviously converting the cycles to all the incongruencies is absurd, but the first step in my algorithm is converting to candidate valid congruencies which does match this problem well.

That sounds like what I suggested here, actually. I don't have anything in my library with quite the right API yet, but I already have most of what you describe coded out. (It looks like I whipped it up for 2017 Day 13.)

[LANGUAGE: Python 3] Embarrassing/Embarrassing Ugly raw solution code

[LANGUAGE: Python 3] 66/101 Raw solution code

Python 3 21/12

Because I hated that duplication so much I already cleaned up the code before posting.