Day 13 Notes

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| Part 1 |
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$ elixir day13part1.exs
4808

Thoughts:

Pretty easy.
Filter out the x items, divide the current time by the bus ID, and the remainder is how many
minutes to wait. Take the minimum.


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| Part 2 |
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$ elixir day13part2.exs
*runs forever*

$ elixir day13part2-cheating.exs
741745043105674

Thoughts:

Another one that's impossible to brute force. Currently stuck.
Have created a correct algorithm that produces the correct answer for the examples.
However, it will never complete for the real input.

There must be some other relationship that I'm missing. Maybe something to do with common factors?

Attempted to improve by starting with the multiple of all the ids and searching downwards, but
that doesn't complete either (I'm not interested in brute-force, so won't let it run for a long
time).

Wondering if the fact that three of the examples use the same IDs and differ only in offset
is important.

Going to come back to this later with a clear mind. Not sure I'll get it without a hint.
Seems like I have a knowledge gap.

** Update 16 Dec - 3 days later **

I did lots of reading into the Chinese Remainder Theorem, but wasn't able to work out
how to apply that here. I'm giving in and asking for help. Voltone on Elixir Forum had
the answer, so I adapted it to my program and it outputted the right answer - so I'm
claiming the star :P, although I'm not going to claim I solved this, so I added that
version to day13part2-cheating.exs with credit. Will definitely come back to this to
understand it properly.


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| Overall Thoughts |
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Doesn't really feel like a "coding" problem, because I implemented the algorithm but there's
obviously another trick. But will judge that once I know the answer.

Working out things from first principles is hard! But I think the thought process is more
valuable than just getting the right answer. So far I've been a bit competitive due to the
private leaderboards, but I guess that ends now.