bjorng
Advent of Code 2024 - Day 23
This was surprisingly easy.
After a quick attempt to use digraph, I implemented by own straightforward algorithm to find the groups. To my surprise, I didn’t need to do any optimizations to solve both parts. The combined runtime for solving both parts was 2.5 seconds.
I then added an optimization to keep track of the size of largest set seen so far and quickly discard any sets smaller than that number before checking for connectedness.
That reduced the runtime to 0.2 seconds.
https://github.com/bjorng/advent-of-code/blob/main/2024/day23/lib/day23.ex
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liamcmitchell
Another one that took a long time to reason about and even longer to get working around xmas distractions.
My part 2 runs in 100ms on my 2013 MBP so I’ve prob done something different.
Part 1 example (5.2ms): 7
Part 1 input (22.8ms): 1352
Part 2 example (0.9ms): "co,de,ka,ta"
Part 2 input (106.6ms): "dm,do,fr,gf,gh,gy,iq,jb,kt,on,rg,xf,ze"
For each node I sort all linked nodes by the number of times they are linked to each other. Then reduce the sorted list into the largest set by adding each node if it links to all existing nodes in the set.
https://github.com/liamcmitchell/advent-of-code/blob/6227b6dd61e22568ad110332b85d8ec03561ebad/2024/23/1.exs#L45-L68
antoine-duchenet
Pretty similar solution here (at least for part 2), even without further optimizations (I use simple lists and maps) it solves part 2 sub 500ms:
defmodule Y2024.D23 do
use Day, input: "2024/23", part1: ~c"l", part2: ~c"l"
defp part1(input) do
sorted_edges =
input
|> parse_input()
|> Enum.sort_by(&elem(&1, 1))
|> Enum.sort_by(&elem(&1, 0))
sorted_edges
|> Enum.chunk_by(&elem(&1, 0))
|> Enum.flat_map(fn set ->
set
|> pairs()
|> Enum.filter(&match?({{a1, _}, {a1, _}}, &1))
|> Enum.filter(fn {{_, a2}, {_, b2}} -> Enum.member?(sorted_edges, {a2, b2}) end)
|> Enum.map(fn {{a1, a2}, {a1, b2}} -> {a1, a2, b2} end)
end)
|> Enum.filter(fn
{"t" <> _, _, _} -> true
{_, "t" <> _, _} -> true
{_, _, "t" <> _} -> true
_ -> false
end)
|> Enum.count()
end
defp part2(input) do
links =
input
|> parse_input()
|> Enum.flat_map(fn {a, b} -> [{a, b}, {b, a}] end)
|> Enum.group_by(&elem(&1, 0), &elem(&1, 1))
links
|> Enum.reduce({[], 0}, fn {from, tos}, {_, size} = acc ->
tos
|> combinations()
|> Enum.sort_by(&Enum.count/1, :desc)
|> Enum.take_while(&(Enum.count(&1) >= size))
|> Enum.find(&full_mesh?(&1, links))
|> case do
nil -> acc
set -> {[from | set], Enum.count(set) + 1}
end
end)
|> elem(0)
|> Enum.sort()
|> Enum.join(",")
end
defp full_mesh?([], _), do: true
defp full_mesh?([h | tail], links) do
connected = Map.get(links, h)
tail
|> Enum.all?(&Enum.member?(connected, &1))
|> Kernel.and(full_mesh?(tail, links))
end
defp pairs([]), do: []
defp pairs([h | tail]), do: for(t <- tail, do: {h, t}) ++ pairs(tail)
defp combinations([]), do: [[]]
defp combinations([h | tail]) do
tails = combinations(tail)
for(t <- tails, do: [h | t]) ++ tails
end
defp parse_input(input), do: Enum.map(input, &parse_line/1)
defp parse_line(<<node1::bytes-size(2), "-", node2::bytes-size(2)>>) do
[node1, node2]
|> Enum.sort()
|> List.to_tuple()
end
end
@bjorng I see this in your code:
|> Enum.group_by(&elem(&1, 0))
|> Enum.map(fn {v, reachable} ->
{v, Enum.map(reachable, &elem(&1, 1))}
end)
|> Map.new
Is there any reason you do not use the third argument of Enum.group_by/3 to transform the values ?
rvnash
Another fairly easy day. Decided to do part1 and part2 by just getting all valid networks first, and filter them later. It was taking about 8 seconds. Then added a filter function to pick out only the valid combinations on the fly, and that helped.
https://github.com/rvnash/aoc2024/blob/main/lib/d23.ex#L63
Then, taking advantage of the knowledge I gained yesterday about Atom being the fastest type of key for a Map, converted the computer names to atoms and got 3X faster. Final results are ~110ms for both parts on my M1 Mac.
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