AoC 2022/18 - Cooling surface


On with Advent of Code puzzle 18 from 2022: cubes and area calculation.

This puzzle happened to come up on a Sunday, when I had a lovely walk with my wife, talking about the puzzle and possible solutions. Very inspirational!

Reading the inputs is easy, because there’s no difficult parsing to do:

sub get-inputs ($filename) {
   [ ${ [ .comb(/ \-? \d+ /) «+» 1 ] }) ]

Each cube is represented by its coordinates in the grid. We can think of these coordinates as belonging to any point of the cube, as long as it’s the same point for all cubes. For sake of choosing one, let’s say it’s the corner pointing to the negatives for all dimensions.

Why add 1 to all coordinates, you might ask! Good question, I don’t know!

I was actually biten by the negative numbers, so the lonely \-? is there in evidence as a reminder.

Part 1 is about calculating the total surface of an aggregate of cubes, in a rigid tridimensional grid. Two cubes sharing a face are considered “merged” together and the common face is not part of the surface area we’re after. In other terms, two cubes sharing a face have the total surface area of 10 times the surface of each face (5 faces from each cube).

My surface area calculation function is the following:

sub surface ($inputs) {
   my $overall =;
   my @by-dimension = (0..2).map({ });
   state @faces = [1, 0, 0], [0, 1, 0], [0, 0, 1];
   for @$inputs -> $cube {
      die 'duplicate' if $overall{$cube.join(',')};
      for 0 .. 2 -> $dim {
         for 0 .. 1 -> $offset {
            my $coords = $cube «+» (@faces[$dim] «*» $offset);
            my $key = $coords.join(',');
            my $sh = @by-dimension[$dim];
            if $sh{$key} { $sh.unset($key) }
            else         { $sh.set($key)   }
   return [+] @by-dimension».elems;

Using a SetHash is easy, but any regular hash would do, e.g. in Perl. Anyway, we have it in Raku and using it adds to the readability. so why not?

The $overall variable is there only because I’m a paranoid and I wanted to double check that there were no duplicate cubes in the inputs, so we can disregard it.

We keep track of all faces, separating them by direction, in @by-dimension. Each cube tries to add its six faces; if a face is already there, then it means that the two participating cubes share that face and we can get rid of it ($sh.unset($key)), otherwise we add it to the lot ($sh.set($key)). As there are no duplicate cubes, each face can appear at most twice, so this will work.

After iterating through all cubes, we’re left with the singly-added faces only, so we just have to count them (by sum).

return [+] @by-dimension».elems;

In hindsight, I could have kept one single cauldron of faces, using their direction to generate their associated $key and avoiding the hypersum at the end. Anyway… it’s cool and does not harm.

Part 2 poses an interesting additional constraint: calculate only the surface that remains on the outside of the whole thing. In other terms, if there are “air bubbles” inside the whole thing, we have to disregard the area of the internal bubble surface.

I didn’t want to chase all these bubbles, so I thought it better to take a cast of the whole thing and use it for the area calculation.

The idea is to build the following aggregate of cubes:

  • find the bounding box of the whole thing, by finding the minimum and maximum value through all coordinates.
  • Consider a bigger bounding box, adding one voxel layer on every dimension. This creates a box that totally wraps our thing in all directions.
  • Find all voxels that are outside of the thing. This can be done with a global visit of all adjacent voxels, starting from any corner of the embedding box (which is outside, by construction), stopping at any cube belonging to the thing.

At this point, we’re left with a sort of plaster cast of the thing, with an exterior surface that is the same surface as the full box (so it’s very easy to calculate), and an interior surface that is what we are after.

As our surface() function above calculates the sum of the two, we can calculate our result by subtracting the area of the box surface from it.

All of this is implemented in the part2 function:

sub part2 ($inputs) {
   my $is-full = $inputs».join(',').Set;
   my $transposed = (0..2).map({$inputs»[$_]}).Array;

   # We immerse our droplet in a "box", totally surrounding it with
   # at least one layer in each direction
   my $mins = $transposed».min «-» 1;
   my $maxs = $transposed».max «+» 1;

   # This box has an exterior area that can be easily calculated.
   # If it has dimensions A, B, and C, then the area will be
   # 2 * (A*B + A*C + B*C). Note that A*B = (A*B*C)/C
   my $deltas = ($maxs «-» $mins) «+» 1;
   my $wrap = ([*] @$deltas);
   my $exterior = (($wrap «/» $deltas) «*» 2).sum;

   # Now we put our droplet in the box and fill it as much as we can,
   # creating a "cast" whose internal area is exactly what we are after.
   # It's just a plain visit through the graph of reachable voxels inside
   # the box, starting from one of the corners that we know *for sure* to
   # be on the outside.
   my %external;
   my @queue = [[$mins.Slip],$mins.join(',')],;
   state @offsets = [1, 0, 0], [0, 1, 0], [0, 0, 1];
   while @queue {
      my ($p, $pkey) = @queue.shift.Slip;
      next if %external{$pkey}:exists;
      %external{$pkey} = $p;
      for 0 .. 2 -> $dim {
         for -1, 1 -> $sign {
            my $q = $p «+» (@offsets[$dim] «*» $sign);
            next unless $mins[$dim] <= $q[$dim] <= $maxs[$dim];
            my $qkey = $q.join(',');
            next if $is-full{$qkey};
            @queue.push: [$q, $qkey];

   # Our "surface" function above will calculate the interal area *plus*
   # the external area that we already calculated above as $exterior
   return surface([%external.values]) - $exterior;

Full solution.

Stay safe!

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