Day 3 - Advent of Code 2024
Mix.install([:kino, :benchee])Links
Prompt
— Day 3: Mull It Over —
“Our computers are having issues, so I have no idea if we have any Chief Historians in stock! You’re welcome to check the warehouse, though,” says the mildly flustered shopkeeper at the North Pole Toboggan Rental Shop. The Historians head out to take a look.
The shopkeeper turns to you. “Any chance you can see why our computers are having issues again?”
The computer appears to be trying to run a program, but its memory (your puzzle input) is corrupted. All of the instructions have been jumbled up!
It seems like the goal of the program is just to multiply some numbers. It does that with instructions like mul(X,Y), where X and Y are each 1-3 digit numbers. For instance, mul(44,46) multiplies 44 by 46 to get a result of 2024. Similarly, mul(123,4) would multiply 123 by 4.
However, because the program’s memory has been corrupted, there are also many invalid characters that should be ignored, even if they look like part of a mul instruction. Sequences like mul(4*, mul(6,9!, ?(12,34), or mul ( 2 , 4 ) do nothing.
For example, consider the following section of corrupted memory:
xmul(2,4)%&mul[3,7]!@^donotmul(5,5)+mul(32,64]then(mul(11,8)``mul(8,5))
Only the four highlighted sections are real mul instructions. Adding up the result of each instruction produces 161 (2*4 + 5*5 + 11*8 + 8*5).
Scan the corrupted memory for uncorrupted mul instructions. What do you get if you add up all of the results of the multiplications?
To begin, get your puzzle input.
— Part Two —
As you scan through the corrupted memory, you notice that some of the conditional statements are also still intact. If you handle some of the uncorrupted conditional statements in the program, you might be able to get an even more accurate result.
There are two new instructions you’ll need to handle:
-
The
do()instruction enables future mul instructions. -
The
don't()instruction disables future mul instructions.
Only the most recent do() or don't() instruction applies. At the beginning of the program, mul instructions are enabled.
For example:
xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5))
This corrupted memory is similar to the example from before, but this time the mul(5,5) and mul(11,8) instructions are disabled because there is a don’t() instruction before them. The other mul instructions function normally, including the one at the end that gets re-enabled by a do() instruction.
This time, the sum of the results is 48 (2*4 + 8*5).
Handle the new instructions; what do you get if you add up all of the results of just the enabled multiplications?
Although it hasn’t changed, you can still get your puzzle input.
Input
input = Kino.Input.textarea("Please paste your input file:")input = input |> Kino.Input.read()"why()}''(!how()$~mul(420,484) ]}}mul(218,461),]/select()mul(93,56)';$-;*#$mul(162,415)mul(556,374)when()~when()<[select()^<(@mul(561,946);mul(97,699)select()+%when()~who():mul(387,15)>mul(927,207)~>~when()*who()'do()mul(454,740)when()%from(),~@%]from()mul(54,688)mul(338,694)what()select()~!< <;+/mul(373,536)&+mul(505,931)why()[mul(457,381) >/ mul(800,67)why()~what()[mul(807,815)&$who()mul(667,529)&how(843,372)what()mul(636,823)?*who()mul(960,331)where()?mul(648,668)how()/who()why()!'mul(649,819)[~:how()&what()]{mul(857,238)%-mul(603,559)mul(511,89):mul(888,328)*how()$/}]mul(177,966)who(777,724)why();:;mul(211,756)]:}+mul(297,394)>^how();$[mul(603,264)mul(794,883)when()why()~&select()from())mul(446,859) & ;:>?*what()/mul(388,763);},?when()do()where())<->what()mul(974,397)+why()select()#@mul(137,814)@when()why()when(697,786)mul(897,431)^}}&mul(74,810)?<&~] '&$]mul(49,565)(>where())]/mul(926,812)mul(842,573) how()-mul(126,526) @mul(818,934)?select()}^what()who()from()}mul(240,118)!from()^@?what()mul(77,983)+$what()mul(736,950)$&%?why()select();mul(213,409)where()*&*what():>~when()when()mul(87,245)$^ what()]&#select()mul(558,637)why()!mul(695,929[;*)+??mul(896,494)when())who()select()^>+how(641,113)mul(374,932);^?:where()>mul(780,265)^#;+when()}}mul(218,272)@$#mul(892,55)[what()?~{'when()}mul(527,984) {>?-'+-:-don't()select()#;/?]''!mul(56,714)/why()>mul(819,178)*~[+}],mul(793,717)from(75,849)~-,+($/,mul(719,587)where()#:@;*$+#?mul(919,859)who()+what(), @;>mul(89,488)where()?from()#$mul(680,657)mul(619,642)?[+don't()$:]{select()$what()@mul(619,164)}!select()mul(528,754)mul(199,830)how()where()+from()/~$who()(mul(49,273(mul(32,974)do()*select()]mul(42,960)%@?:[##::mul(802,384)]{##where())what()[mul(610,271)how()#@>what()where()don't()'&)>mul%?mul(449,76)when()%[<{^mul(944,356)what()(*;<{$<~mul(476,918){who(){-mul(124,698);?who()who(610,775)what()'&mul(579,336)why()<{'mul(953,943):%mul(651,439)/who(870,472)from()select()( -] $mul(16,267)mul(865,452)!select()who(){)$%$mul(888,41)],when(737,918)[#@#'mul(655,759)}/$(?#do()when()how()mul(160,406)]$mul(392,433)([;!##^from()mul(56,722)what()*#%select()(>#mul(554,881)'mul(355,774)&/mul!~*)mul(32,426)where()[how()how()mul(160,991)(where()w" <> ...Solution
defmodule Day03 do
def part1(input) do
~r/mul\((\d{1,3})\,(\d{1,3})\)/
|> Regex.scan(input, capture: :all_but_first)
|> Enum.map(&do_mul/1)
|> Enum.sum()
end
def part2(input) do
~r/(mul\((\d{1,3})\,(\d{1,3})\)|do\(\)|don\'t\(\))/
|> Regex.scan(input, capture: :all_but_first)
|> handle(:enabled)
|> Enum.sum()
end
defp do_mul([a, b]), do: do_mul(a, b)
defp do_mul(a, b), do: String.to_integer(a) * String.to_integer(b)
defp handle([["don't()"] | tail], :enabled),
do: handle(tail, :disabled)
defp handle([["do()"] | tail], :disabled),
do: handle(tail, :enabled)
defp handle([[_, a, b] | tail], :enabled),
do: [do_mul(a, b) | handle(tail, :enabled)]
defp handle([_ | tail], status),
do: handle(tail, status)
defp handle([], _status),
do: []
end{:module, Day03, <<70, 79, 82, 49, 0, 0, 13, ...>>, {:handle, 2}}Scan the corrupted memory for uncorrupted mul instructions. What do you get if you add up all of the results of the multiplications?
Your puzzle answer was 174561379.
Day03.part1(input)174561379Handle the new instructions; what do you get if you add up all of the results of just the enabled multiplications?
Your puzzle answer was 106921067.
Day03.part2(input)106921067Both parts of this puzzle are complete! They provide two gold stars: **
At this point, you should return to your Advent calendar and try another puzzle.
If you still want to see it, you can get your puzzle input.
Tests
ExUnit.start(auto_run: false)
defmodule Day03Test do
use ExUnit.Case, async: false
setup_all do
[
input: "xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5))"
]
end
describe "part1/1" do
test "returns expected value", %{input: input} do
assert Day03.part1(input) == 161
end
end
describe "part2/1" do
test "returns expected value", %{input: input} do
assert Day03.part2(input) == 48
end
end
end
ExUnit.run()..
Finished in 0.00 seconds (0.00s async, 0.00s sync)
2 tests, 0 failures
Randomized with seed 720197%{total: 2, failures: 0, excluded: 0, skipped: 0}Benchmarking
defmodule Benchmarking do
# https://github.com/bencheeorg/benchee
def run(input) do
Benchee.run(
%{
"Part 1" => fn -> Day03.part1(input) end,
"Part 2" => fn -> Day03.part2(input) end
},
memory_time: 2,
reduction_time: 2
)
nil
end
end{:module, Benchmarking, <<70, 79, 82, 49, 0, 0, 7, ...>>, {:run, 1}}Benchmarking.run(input)Operating System: macOS
CPU Information: Apple M1
Number of Available Cores: 8
Available memory: 8 GB
Elixir 1.15.7
Erlang 26.1.1
Benchmark suite executing with the following configuration:
warmup: 2 s
time: 5 s
memory time: 2 s
reduction time: 2 s
parallel: 1
inputs: none specified
Estimated total run time: 22 s
Benchmarking Part 1 ...
Benchmarking Part 2 ...
Name ips average deviation median 99th %
Part 1 2.19 K 456.17 μs ±15.56% 441.81 μs 659.33 μs
Part 2 1.08 K 923.81 μs ±13.59% 916.87 μs 969.28 μs
Comparison:
Part 1 2.19 K
Part 2 1.08 K - 2.03x slower +467.64 μs
Memory usage statistics:
Name average deviation median 99th %
Part 1 302.74 KB ±0.00% 302.74 KB 302.74 KB
Part 2 433.62 KB ±0.00% 433.62 KB 433.62 KB
Comparison:
Part 1 302.74 KB
Part 2 433.62 KB - 1.43x memory usage +130.87 KB
Reduction count statistics:
Name Reduction count
Part 1 22.71 K
Part 2 27.51 K - 1.21x reduction count +4.80 K
**All measurements for reduction count were the same**nil