Day 10 Part 2
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My (attempted) solutions to [Advent of Code 2020](https://adventofcode.com/2020) in Elixir.
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<img width="975" alt="image" src="https://user-images.githubusercontent.com/498229/101746064-b0663000-3b0e-11eb-81bd-92597dba329e.png">
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<img width="975" alt="image" src="https://user-images.githubusercontent.com/498229/101786711-97c33d80-3b41-11eb-831a-d6a9b3670f30.png">
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## Strategy
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42
day10/README
42
day10/README
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@ -5,7 +5,7 @@ Day 10 Notes
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+--------+
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$ elixir day10part1.exs
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**solved - will add answer after part 2 is complete**
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2450
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Thoughts:
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@ -15,23 +15,47 @@ implementing exactly what the instructions say.
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In this case, we just need to sort the data, and compute the difference between all the pairs.
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+-------------------+
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| Part 2 [UNSOLVED] |
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+-------------------+
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+--------+
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| Part 2 |
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+--------+
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$ elixir day10part2.exs
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...
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32396521357312
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Thoughts:
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So far defeated on this one. I'm obviously missing a fundamental trick.
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Fudge me this took me a long time (hours, including a walk in the park...)
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I went down a massive rabbit hold calculating combinations, which didn't pay off.
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Will leave this for now, and maybe come back to it later.
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I spent far too long trying to find a mathematical answer based on combinations, and went down
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another rabbit hole identifying the "removable" items and then trying to calculate the answer
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based on that information.
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Turns out the trick I was missing is graph theory.
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Finally, after drawing out the example on paper, I realised that the number of combinations
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forms a directed graph, where the weight of each node inherits the weight of all its parents.
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For the example case (excuse my ascii drawing skills):
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0(1)-1(1)-4(1)-5(1)-|
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| ` | |
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|---6(2) |
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| | |
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|---7(4)-|
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10(4)-11(4)
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` |
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12(8)-18(8)-19(8)
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So just need to make an algorithm that builds this graph, by iterating the sorted list
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and checking the next 3 elements, keeping track of the weights as we go. It's actually simpler
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than that, because we don't need to keep the connections around, just the weightings for each node.
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Finally, the weight of the max joltage adapter is the answer.
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+------------------+
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| Overall Thoughts |
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+------------------+
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Not committing the input for now, in case somebody decides to @ me my answer or something.
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Part 1 was very easy, part 2 was fiendish. Glad this wasn't an interview question. Sometimes my
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brain just doesn't process certain types of problem. This was one of them. Once I made the
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connection to a weighted graph it was fine.
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@ -0,0 +1,31 @@
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defmodule Day10Part2 do
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def run do
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File.read!("input")
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|> String.split("\n", trim: true)
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|> Enum.map(&String.to_integer/1)
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|> count_arrangements()
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|> IO.puts()
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end
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def count_arrangements(list), do: count_arrangements(%{0 => 1}, [0 | Enum.sort(list)])
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def count_arrangements(arrs_to, [last]), do: arrs_to[last]
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def count_arrangements(arrs_to, [current | rest]) do
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rest
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|> Enum.take(3)
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|> count_reachable(current, arrs_to)
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|> count_arrangements(rest)
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end
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def count_reachable([], _a, arrs_to), do: arrs_to
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def count_reachable([b | rest], a, arrs_to) when b - a <= 3 do
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arrs_to = Map.update(arrs_to, b, arrs_to[a], &(&1 + arrs_to[a]))
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count_reachable(rest, a, arrs_to)
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end
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def count_reachable([_b | rest], a, arrs_to), do: count_reachable(rest, a, arrs_to)
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end
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Day10Part2.run()
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@ -0,0 +1,104 @@
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2
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1
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@ -1,8 +1,7 @@
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defmodule DayREPLACE_MEPart1 do
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def run do
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File.read!("input")
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|> String.trim()
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|> String.split("\n")
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|> String.split("\n", trim: true)
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end
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end
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