this post was submitted on 06 Dec 2024
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Advent Of Code

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An unofficial home for the advent of code community on programming.dev!

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.

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I know that we can brute force it by placing an obstacle at every valid position in the path, but is there a more elegant / efficient solution?

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[–] Karmmah 3 points 2 weeks ago (1 children)

I'm also struggling with part 2. At the moment I am following the path and in each place I look if can connect with an earlier part of the path to the right by placing an obstacle, but without cutting off another earlier part of the path with that obstacle. I have indexes for each path location to see where in the path they first appear.

However at the moment I still have a bug counting too many possibilities. Any other ideas?

[–] [email protected] 2 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

I am using same solution, and getting right count with test input, but the actual input gives too many.

Annoying, and really hard to debug. I made renderer to visualize, but unable to find the bug.

One misunderstanding was that I counted same walls twice, because the result should not count same added wall twice if it has same x,y.

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

Yeah something like that happend for me too, and later I counted too little because it would not recognise some possible solutions. Finally I solved it by just walking along the path and at each location put an obstacle in front of it and then check if the changed map results in a path enters longer than 10000 steps. Not pretty but at least it lead to the right result with a runtime of ~3 seconds. I would have saved a lot of time if I had tried the "brute force" way before, so I guess lesson learned.

[–] [email protected] 2 points 2 weeks ago

You can track by just keeping list of all positions and movement direction, if two are same, it means that it is in a loop.