# Advent of Code 2021 - Day 22

By Eric Burden | January 1, 2022

It’s that time of year again! Just like last year, I’ll be posting my solutions to the Advent of Code puzzles. This year, I’ll be solving the puzzles in Julia. I’ll post my solutions and code to GitHub as well. I had a blast with AoC last year, first in R, then as a set of exercises for learning Rust, so I’m really excited to see what this year’s puzzles will bring. If you haven’t given AoC a try, I encourage you to do so along with me!

# Day 22 - Reactor Reboot

Find the problem description HERE.

## The Input - Cubic Impressions

The input for today’s puzzle isn’t super complicated, consisting of lines that start with the word ‘on’ or ‘off’, followed by a string in the format of “x=..,y=..,z=..”, where the values are positive or negative numbers that represent the bounds of a cube. So, with a function that can parse one line, we can parse all the lines.


# Data Structures --------------------------------------------------------------

# We'll represent each of the cubes of lights as a set of x, y, and z-ranges
# that encompass all the points of the cube
struct Cube
x::UnitRange{Int64}
y::UnitRange{Int64}
z::UnitRange{Int64}
end

# Constructor that ensures that the first number in the x, y, or z-range is
# the smaller number.
function Cube(x1, x2, y1, y2, z1, z2)
if x1 > x2; (x1, x2) = (x2, x1); end
if y1 > y2; (y1, y2) = (y2, y1); end
if z1 > z2; (z1, z2) = (z2, z1); end
Cube(x1:x2, y1:y2, z1:z2)
end

# Constructor that takes a line from the input and produces a Cube
function Cube(s::AbstractString)
re = r".*x=(-?\d+)\.\.(-?\d+),y=(-?\d+)\.\.(-?\d+),z=(-?\d+)\.\.(-?\d+)"
bounds = parse.(Int, match(re, s).captures)
return Cube(bounds...)
end

# An Instruction encompasses a Cube and whether the instruction is to turn
# 'on' all the lights in that cube (state == True) or to turn the lights 'off'
struct Instruction
state::Bool
cube::Cube
end

# Constructor that takes a line from the input and returns an Instruction
function Instruction(s::AbstractString)
re = r"^(on|off) (.*)\$"
(statestr, cubestr) = match(re, s).captures
state = statestr == "on"
cube = Cube(cubestr)
return Instruction(state, cube)
end

# Ingest the Data -------------------------------------------------------------

# Parse the input file into a list of Instructions
function ingest(path)
return open(path) do f
[Instruction(l) for l in readlines(f)]
end
end


We’ll keep the parsing simple by representing each cube as a set of ranges bounded by the values in the input. Based on how large some of the values in the input are, we won’t be able to keep track of every point.

## Part One - Minicube

Time to reboot the submarine! As we all know, most every technological problem is solved with a reboot. In fact, if we can figure this one out, we should be good for Days 23-25 as well! And, as it turns out, we were overly pessimistic when parsing the input. We actually only care about the portions of the instruction cubes in the range from -50 -> 50, so we can just keep track of which lights are on in a 3D Array. This will be easier than we thought!

# Helper Functions -------------------------------------------------------------

# It is often convenient to get the upper and lower bounds of the
# x, y, and z-ranges that comprise a Cube
function bounds(cube::Cube)
(x1, x2) = (cube.x.start, cube.x.stop)
(y1, y2) = (cube.y.start, cube.y.stop)
(z1, z2) = (cube.z.start, cube.z.stop)
return (x1, x2, y1, y2, z1, z2)
end

# Check whether a Cube overlaps the region being considered for the
# initialization sequence.
function overlap(lights::BitArray, cube::Cube)
(x1, x2, y1, y2, z1, z2) = bounds(cube)
(lightx, lighty, lightz) = size(lights)
return (x1 <= lightx && x2 >= 1
&& y1 <= lighty && y2 >= 1
&& z1 <= lightz && z2 >= 1)
end

# Shift a cube so that all its dimensions are positive and have at least a
# a value of 1, so that the coordinates map cleanly to an Array.
function displace(cube::Cube, offset::Int64)
cubebounds = map(c -> c + offset, bounds(cube))
return Cube(cubebounds...)
end

# Apply an instruction (turning lights on or off) to the lighted region. Just
# sets positions in lignts bounded by the coordinates of the instruction's
# Cube to either 1 or 0, depending on the type of instruction. Displaces
# the Cube before making this change.
function apply!(lights::BitArray, instruction::Instruction, offset = 51)
(lightx, lighty, lightz) = size(lights)
cube = displace(instruction.cube, offset)
overlap(lights, cube) || return
xrange = intersect(1:lightx, cube.x)
yrange = intersect(1:lighty, cube.y)
zrange = intersect(1:lightz, cube.z)
lights[xrange, yrange, zrange] .= instruction.state
end

# Solve Part One ---------------------------------------------------------------

# Simple, straightforward; represent the target region as a 3D BitArray and
# turn on/off the lights according to the instructions, in order. We use
# displace() to adjust the coordinates of each Cube so that its bounds fall
# within the bounds of lignts.
function part1(input)
lights = falses(101, 101, 101)
foreach(cube -> apply!(lights, cube), input)
return count(lights)
end


On second thought, yesterday’s puzzle has me a bit paranoid. Is it possible that this part was too easy…

## Part Two - Rubik’s Wonderland

sigh Of course it was too good to be true. I did try to create a 3D Array large enough to accommodate the entire range, but it wouldn’t fit into my laptop’s memory. I also tried processing the instructions in chunks, which would probably work, but it was going to take forever (approximately). So, it was back to the drawing board for some shudder geometry. I ended up drawing a diagram for myself to help work out the ‘shattering’ rules (see the code below). # Helper Functions -------------------------------------------------------------

# Check whether two cubes overlap
function overlap(a::Cube, b::Cube)
(ax1, ax2, ay1, ay2, az1, az2) = bounds(a)
(bx1, bx2, by1, by2, bz1, bz2) = bounds(b)
return (   ax1 <= bx2 && ax2 >= bx1
&& ay1 <= by2 && ay2 >= by1
&& az1 <= bz2 && az2 >= bz1)
end

# Check whether the Cubes from two Instructions overlap
overlap(a::Instruction, b::Instruction) = overlap(a.cube, b.cube)

# Mostly just used for sanity checking while developing the shatter() function
# below, but it's interesting enought to keep around. Returns a Cube
# representing the portions of a and b that overlap.
function Base.intersect(a::Cube, b::Cube)
overlap(a, b) || return Cube(0:0, 0:0, 0:0)
(ax1, ax2, ay1, ay2, az1, az2) = bounds(a)
(bx1, bx2, by1, by2, bz1, bz2) = bounds(b)
x = max(ax1, bx1):min(ax2, bx2)
y = max(ay1, by1):min(ay2, by2)
z = max(az1, bz1):min(az2, bz2)
return Cube(x, y, z)
end

# Volume of a cube. Tricky, right? Yeah, well, *I* mistakenly corrected for
# Julia's 1-indexing by adding one to each of these lengths and spent _way_
# too long debugging the wrong thing before I realized what I'd done...
function volume(cube::Cube)
xlen = length(cube.x)
ylen = length(cube.y)
zlen = length(cube.z)
return xlen * ylen * zlen
end

# The next set of functions are related to functionality to break apart one
# Cube around another Cube. This depends on creating new Cubes for each
# "zone" around the smaller Cube, 8 zones that align with the corners of the
# smaller Cube, 6 zones that align with the faces of the smaller Cube, and
# 12 zones that align with the edges of the smaller Cube. This way, if an
# 'off' Instruction has a cube that overlaps a Cube of lignts already on,
# we can 'cut out' the Cube that gets turned off from the Cube that's
# already on, replacing the remaining portions of the 'on' Cube with smaller
# Cubes and removing the overlapping section.

# For "face" and "edge" zones, for each dimension, there are four possible
# ranges that can be taken based on the relative coordinates of the target
# and template. Arguments are the min/max coordinates of target and
# template (from shatter()) in a single dimension. I find it easier to
# imagine these values as line endpoints on a number line, if one line
# stretches from D1>D2 and the other from d1>d2.
function middlerange(D1, D2, d1, d2)
D1 <= d1 && D2 >= d2 && return d1:d2
D1 > d1  && D2 >= d2 && return D1:d2
D1 <= d1 && D2 < d2  && return d1:D2
D1 > d1  && D2 < d2  && return D1:D2
return nothing
end

# Break a target cube into up to 27 segments, based on the location of
# template. I find it easiest to image that every corner of template that
# lies inside target emits a line in each direction, carving up target.
# Returns all the pieces _except_ for the piece that intersects with the
# template.
function shatter(target::Cube, template::Cube)
# I'm using capital letters for the target, and lowercase letters for
# the template Cube.
(X1, X2, Y1, Y2, Z1, Z2) = bounds(target)
(x1, x2, y1, y2, z1, z2) = bounds(template)

pieces = Vector{Cube}()

# These checks are used to determine if any of the points of the target
# cube stray into one of the 26 cubic 'zones' adjacent to the template
# cube. This includes 12 "edge" zones, 8 "corner" zones, and six "face"
# zones.
xoffsets = (X1 < x1, X1 <= x2 || x1 <= X2, x2 < X2)
yoffsets = (Y1 < y1, Y1 <= y2 || y1 <= Y2, y2 < Y2)
zoffsets = (Z1 < z1, Z1 <= z2 || z1 <= Z2, z2 < Z2)

# The ranges to use to construct the pieces. For each axis: the range when
# the target extends past the template in the negative (xyz)-direction,
# when the target occupies the same (xyz)-range as the template, and
# when the target extends past the template in the positive direction
xranges = (X1:x1-1, middlerange(X1, X2, x1, x2), x2+1:X2)
yranges = (Y1:y1-1, middlerange(Y1, Y2, y1, y2), y2+1:Y2)
zranges = (Z1:z1-1, middlerange(Z1, Z2, z1, z2), z2+1:Z2)

# Generate the pieces
for (a, b, c) in Iterators.product(1:3, 1:3, 1:3)
(a, b, c) == (2, 2, 2) && continue # Skip the template range
if xoffsets[a] && yoffsets[b] && zoffsets[c]
push!(pieces, Cube(xranges[a], yranges[b], zranges[c]))
end
end

# Sanity check, ensures that the volume of the pieces returned + the volume
# removed is equal to the volume of the target Cube. Not needed to run
# the solution, but it was handy for debugging.
# @assert sum(volume.(pieces)) + volume(intersect(template, target)) == volume(target)

return pieces
end

# Convenience function to shatter() the cubes held by two Instructions.
# Returns a list of Instructions containing each of the pieces.
function shatter(target::Instruction, template::Instruction)
newcubes = shatter(target.cube, template.cube)
return Instruction.(target.state, newcubes)
end

# Given an iterable containing Instructions, add up the volume of all the
# Instructions that turn lignts on.
function countlights(instructionset)
lightson = 0
for instruction in instructionset
!instruction.state && continue
lightson += volume(instruction.cube)
end
return lightson
end

# Solve Part Two ---------------------------------------------------------------

# Add each instruction to a set of final instructions. As each instruction is
# added, have it shatter() any Instruction it overlaps with before adding
# it to the set of final instructions, thereby replacing the overlapping
# region with itself. This means that instructions to turn lights off will
# "overwrite" regions that turn lights on, and instructions that turn lights on
# dont' get double-counted.
function part2(input)
finalinstructions = Set{Instruction}()
for instrin in input
for instrout in finalinstructions
overlap(instrin, instrout) || continue # No overlaps? No problem!

newinstrs = shatter(instrout, instrin)
delete!(finalinstructions, instrout)

# shatter() will return an empty list if target fits completely
# inside template, so there's no parts of target to add back
# after shattering in that case.
isempty(newinstrs) && continue
push!(finalinstructions, newinstrs...)
end

# Only need to store the 'on' Instructions.
instrin.state && push!(finalinstructions, instrin)
end

return countlights(finalinstructions)
end


Somehow, it’s always math, isn’t it? I’m pretty satisfied with this approach, though. There’s possibly a way to do this without needing to explicitly turn off cubes, if we worked from the end of the list backwards and only turned on the ones that were going to stay lit, but that seems complicated and I’m happy enough with this solution.

## Wrap Up

Oh, AoC, when will I learn that I should never relax when part one seems pretty easy compared to previous days? This was a somewhat time-consuming puzzle, but I’m really happy with the way shatter() worked out. That moment when I realized I should be able to sum the volumes of all the pieces to get the original volume? Should have been obvious? Yes. Was it still a great feeling? Absolutely! Today leaned heavily on Julia’s destructuring functionality, and provided some great practice there, so I’m going to call today a win-win!

If you found a different solution (or spotted a mistake in one of mine), please drop me a line!