Advent of Code 2024 day 3 in Lua and Regex

Published: 01-16-25

Lua ?

Ok I know I said I would do the advent of codes in Zig but day 3 felt tailor made for Lua. I did a few days last year in Lua and it helped tremendously with Nvim plugins and configuration, and showed me the power of Lua regex.
Zig doesn’t yet have a regex library and the ones that are 3rd party seem somewhat unmaintained. So I decided to reach for good old Lua and it was suprisingly easy and the code is surprisingly performant.

Part one

essentially we just need to find and perform all the ‘multiplication’ operations in our input
- they look like mul(num1, num2), and the numbers can be at most 3 digits long. There are a bunch of ‘partial matches’ and a simple regex will solve it easily

Lua Regex

the Lua regex syntax has some differences to what I would consider ‘normal’ regex but works surprisingly well and I’d bet it is part of what made Lua a great choice for Nvim.
for this problem we want to match mul(num, num), with each number being at most 3 digits long
match ‘mul(’
- mul%( -> the % before the ( denotes that this is a char to match not part of a regex
matching for a number, at most 3 digits long
- (%d%d?%d?) -> here the () denote a regex expression, %d represents a digit, and %d? represents an optional digit.
matching the ending parenthesis ’)’
- %) -> match for the closing parenthesis char.

Part 1 solution

here we can use a simple gmatch() to match for the occurrences of digits within a mult()
- we convert these strings to numbers, multiply them and add the product to our sum

-- Part 1
local sum = 0

-- open the file
local file = io.open("input.txt", "r")
if file then
  -- iterate across file lines
  for line in file:lines() do
    for x, y in line:gmatch("mul%((%d%d?%d?),(%d%d?%d?)%)") do
      sum = sum + (tonumber(x) * tonumber(y))
    end
  end
  file:close()
  print(sum)
else
  print("Cannot open file.")
end

and part 1 takes roughly 750 micro seconds to run! I’m always surprised by Lua’s performance and simplicity. Just 13 lines of code (without blank lines)!
- I am using LuaJit, so performance is better than ‘regular’ Lua.

Part 2

part two is a little more complex.
there are also do() and don't() strings in our input and we only sum the multiplications between do()s
- do achieve this we track the indexes of do() and don't() to make sure our current position is within a do block.
- when doing a regex with find() we can declare parameters for the location of the start and end of the pattern found.
  - this is the mechanism that allows us to find where we have “do”s and “don’t”s.

local sum = 0
local enabled = true -- whether we are in a 'do()' block

local file = io.open("input.txt", "r")
if file then
  for line in file:lines() do
    local pos = 1 -- current position (Lua is 1 indexed)
    local n = #line -- length of current line

    -- while still in the same line
    while pos <= n do
      -- indexes of "do" block
      local doStart, doEnd = line:find("do%(%)", pos)
      if doStart == pos then
        enabled = true -- hit "do"
        pos = doEnd + 1 -- go to end of "do"
      else
        -- indexes of "don't()" block
        local dontStart, dontEnd = line:find("don't%(%)", pos)
        if dontStart == pos then
          enabled = false -- hit "don't"
          pos = dontEnd + 1 -- go to end of "don't"
        else
          -- find "mult()"
          local mulStart, mulEnd, x, y = line:find("mul%((%d%d?%d?),(%d%d?%d?)%)", pos)
          if mulStart == pos then
            -- if enabled add product of x and y to the sum
            if enabled then
              sum = sum + (tonumber(x) * tonumber(y))
            end
            pos = mulEnd + 1
          else
            -- move forward one position
            pos = pos + 1
          end
        end
      end
    end
  end
  file:close() -- Always close the file when you're done
  print(sum)   -- This should print 161 for the given example
else
  print("Cannot open file.")
end