65 lines
2.0 KiB
Plaintext
65 lines
2.0 KiB
Plaintext
Day 13 Notes
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+--------+
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| Part 1 |
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+--------+
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$ elixir day13part1.exs
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4808
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Thoughts:
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Pretty easy.
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Filter out the x items, divide the current time by the bus ID, and the remainder is how many
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minutes to wait. Take the minimum.
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+--------+
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| Part 2 |
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+--------+
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$ elixir day13part2.exs
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*runs forever*
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$ elixir day13part2-cheating.exs
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741745043105674
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Thoughts:
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Another one that's impossible to brute force. Currently stuck.
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Have created a correct algorithm that produces the correct answer for the examples.
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However, it will never complete for the real input.
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There must be some other relationship that I'm missing. Maybe something to do with common factors?
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Attempted to improve by starting with the multiple of all the ids and searching downwards, but
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that doesn't complete either (I'm not interested in brute-force, so won't let it run for a long
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time).
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Wondering if the fact that three of the examples use the same IDs and differ only in offset
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is important.
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Going to come back to this later with a clear mind. Not sure I'll get it without a hint.
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Seems like I have a knowledge gap.
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** Update 16 Dec - 3 days later **
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I did lots of reading into the Chinese Remainder Theorem, but wasn't able to work out
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how to apply that here. I'm giving in and asking for help. Voltone on Elixir Forum had
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the answer, so I adapted it to my program and it outputted the right answer - so I'm
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claiming the star :P, although I'm not going to claim I solved this, so I added that
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version to day13part2-cheating.exs with credit. Will definitely come back to this to
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understand it properly.
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+------------------+
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| Overall Thoughts |
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+------------------+
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Doesn't really feel like a "coding" problem, because I implemented the algorithm but there's
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obviously another trick. But will judge that once I know the answer.
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Working out things from first principles is hard! But I think the thought process is more
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valuable than just getting the right answer. So far I've been a bit competitive due to the
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private leaderboards, but I guess that ends now.
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