Advent Of Code

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Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2024

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🏃‍♀️ - 2024 DAY 20 SOLUTIONS -🏃‍♀️ (programming.dev)

submitted 5 days ago by [email protected] to c/[email protected]

16 comments fedilink hide all child comments

Day 20: Race Condition

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[–] [email protected] 3 points 5 days ago (2 children)

Hey - I've a question about this. Why is it correct? (Or is it?)

If you have two maps for positions in the maze that give (distance to end) and (distance from start), then you can select for points p1, p2 such that

d(p1, p2) + distance-to-end(p1) + distance-to-start(p2) <= best - 100

however, your version seems to assume that distance-to-end(p) = best - distance-to-start(p) - surely this isn't always the case?

[–] [email protected] 4 points 5 days ago (1 children)

There is exactly one path without cheating, so yes, the distance to one end is always the total distance minus the distance to the other end.

[–] [email protected] 2 points 5 days ago (1 children)

Gotcha, thanks. I just re-read the problem statement and looked at the input and my input has the strongest possible version of that constraint: the path is unbranching and has start and end at the extremes. Thank-you!

[–] [email protected] 2 points 5 days ago

I missed that line too:

Because there is only a single path from the start to the end

So I also did my pathfinding for every variation in the first part, but realised something must be wrong with my approach when I saw part 2.

[–] [email protected] 3 points 5 days ago (1 children)

(I ask because everyone's solution seems to make the same assumption - that is, that you're finding a shortcut onto the same path, as opposed to onto a different path.)

[–] [email protected] 1 points 5 days ago

Some others have answered already, but yes, there was a well-hidden line in the problem description about the map having only a single path from start to end..