Day 5 Notes
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| Part 1 |
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$ elixir day5part1.exs
906
Thoughts:
Could do something clever with binaries or the Bitwise module?
The number of rows is even (128), but we're zero-indexed (0-127), so how to deal with calculating half?
- in binary, can just shift one bit. But again, we need to halve a range (e.g. 32 to 63) so not sure
how to do that.
Instead of storing the range, I decided to store the current "front" seat, and how many rows back to count.
Then we can work with 128 which divides by two without worrying about rounding problems.
When there are only two seats left, just return the correct one based on the flag.
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| Part 2 |
+--------+
$ elixir day5part2.exs
519
Thoughts:
Not really sure what the missing front/back seats was about. Decided just to just compare the sorted list
with itself shifted by one (by zipping them), and finding the first ID without a next ID, and assume
the next ID is my seat.
That turned out to be the right answer.
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| Overall Thoughts |
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Looking forward to seeing how others implemented the seat ID parsing.
Did a lot of refactoring here but the end result, especially for part 2, is pretty clean I think.
Even if I decided to refactor the parser, it can just be swapped in for part 2. Zipping was a nice idea.
I quite enjoyed this one, maybe because it's a Saturday and I wasn't burnt out from work :-)
** Update **
After reading the Elixir Forum thread: https://elixirforum.com/t/advent-of-code-2020-day-5/35997/
I really had to have a go at just parsing the boarding pass directly into an integer. I said I wouldn't
commit code to my solutions after reading forums, so I'll add it here instead :-) It converts each
character to a bit on the fly, and interprets the result as a 10-bit integer. 🦾
def char_to_bit(c) when c in 'FL', do: <<0::1>>
def char_to_bit(c) when c in 'BR', do: <<1::1>>
def seat_id_bitstring(boarding_pass \\ "FBFBBFFRLR") do
<<id::integer-10>> = for <<char <- boarding_pass>>, into: <<>>, do: char_to_bit(char)
id
end
