this post was submitted on 05 Dec 2023

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Advent Of Code

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🎁 - 2023 DAY 5 SOLUTIONS -🎁 (programming.dev)

submitted 11 months ago* (last edited 11 months ago) by [email protected] to c/[email protected]

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Day 5: If You Give a Seed a Fertilizer

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[–] [email protected] 7 points 11 months ago* (last edited 11 months ago) (2 children)

Ooof. Part 1 was easy enough, but for part two I initially went with the naive solution of trying every single seed which took more than a minute (I never really measured). Although that got me the right answer, to me that was just unacceptable.

I proceeded to try and combine all mappings into one but gave up after spending way too much time on it.

Then I had the idea that the lowest number in the end must lie at the beginning of a range somewhere. Either the start of a seed range in the beginning or the start of a range in one of the mappings. Any in-between numbers must end up with a higher result. So I considered the start points of all ranges, went through the mappings in reverse to find out if that point is actually within a seed range, and only tested those starting points.

Finally I had only 183 points to test which ran much faster (0.9ms).

[–] [email protected] 3 points 11 months ago

Then I had the idea that the lowest number in the end must lie at the beginning of a range somewhere. Either the start of a seed range in the beginning or the start of a range in one of the mappings.

This really helped me out. I was stuck on either trying every single seed, or working backwards and trying every single location from 0 to infinity, and couldn't wrap my head around how to filter down the list to be manageable. Your comment made it all make sense.

In the end, was able to work backwards with the 172 lowest locations in each range to get potential seeds, and from there was able to get a short list of 89 valid seeds (including the original seed values) to then figure out which one has the shortest location.

Thanks!

[–] [email protected] 1 points 11 months ago* (last edited 11 months ago) (1 children)

I'm a little confused about this one. The mappings are total, that is any number that is not defined explicitly gets mapped to itself. So it's easy to create an example where the lowest number does not get mentioned within a range:

seeds: 0 3

seed-to-soil map:
10 0 2

soil-to-fertilizer map:
100 200 5

fertilizer-to-water map:
100 200 5

water-to-light map:
100 200 5

light-to-temperature map:
100 200 5

temperature-to-humidity map:
100 200 5

humidity-to-location map:
100 200 5

Here, we have seeds 0, 1 and 2. seed 0 gets mapped to location 10, seed 1 gets mapped to location 11 and seed 2 gets mapped to location 2. That means location 2 would be the answer, but it's not a start of any range. I guess this just doesn't happen in any of the inputs?

EDIT: actually it's double weird. If you implemented a backwards search, that is you create reverse mappings and then try out all locations (which is what I and many others did), the result of the above example is location 0, whereas if you create a forwards brute force of all seeds, the result is 2. For the reverse approach to work in all cases, the mappings would have to be bijective.

[–] [email protected] 1 points 11 months ago

Indeed, my solution fails on this input (returns 10, which is the location to seed 0), but it can be easily solved by also adding the ends of each range as well.

Maybe the input was quite forgiving. Thinking about it more, reversing the mapping can get quite involved, because it is neither surjective nor injective, so the inverse can actually have any number of results.

In your example there is no input that maps to 0, but there are two inputs that map to 11 (1 and 11). If the seed-to-soil map also included 10 20 2, 21 would also map to 11.