this post was submitted on 04 Dec 2025
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Advent Of Code

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An unofficial home for the advent of code community on programming.dev! Other challenges are also welcome!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

Everybody Codes is another collection of programming puzzles with seasonal events.

EC 2025

AoC 2025

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console.log('Hello World')

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Day 4: Printing Department

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[–] Quant@programming.dev 2 points 2 weeks ago

Uiua

Quite simple this one. Part 2 still takes a few seconds because I'm essentially checking off each roll individually.

Run with example input

Code

$ ..@@.@@@@.
$ @@@.@.@.@@
$ @@@@@.@.@@
$ @.@@@@..@.
$ @@.@@@@.@@
$ .@@@@@@@.@
$ .@.@.@.@@@
$ @.@@@.@@@@
$ .@@@@@@@@.
$ @.@.@@@.@.
⊜∘⊸≠@\n
=@@

Rolls ← ⍣(⧻⊚˙⍤)∞⊸⊡1_1
Removable ← (
  ⬚0⧈Rolls[3_3 1_1 1_1]
  ⊚≤4
)

Remove ← ⍜⊡(˜↯0⧻)

PartOne ← ⧻Removable

PartTwo ← (
  ⊙0
  ⍥(
    ⊙(⊙+⟜⧻)⟜Removable
    ˜Remove
  )∞
  ◌
)
&pf "Part One: "
&p ⊸PartOne

&pf "Part Two: "
&p PartTwo

Old Part 2Before seeing mykl's solution this was my solution for part 2

PartTwoOld ← (
  0⊙0
  ⍢(⊙⊙◌
    ⊙⊸Removable
    +⊙⊸⧻
    ⊙⊸Remove
  | ¬≍◌)
  ⊙⋅◌
)

It's basically the same, just that I used a while-do-loop, making the check for the ending condition myself (which took me a bit to get right because I still find loops in Uiua a bit confusing).
Using the repeat-loop as above also gets rid of the dip's (⊙). I could've removed them here as well but I was already deep in the trouble of getting the loop to work correctly and I liked the little face at the beginning 0⊙0