Advent Of Code 2021, Day 9: Smoke Basin

by Abigail Challenge

We are given a heightmap as follows:

2199943210
3987894921
9856789892
8767896789
9899965678

Each number (all single digits) corresponds to a height: 0 is the lowest, and 9 is the heightest.

Part One

Part One asks us the find the low points: points which are lower than any of its orthogonally adjacent locations. The risk level of a point is its heigth plus one. We need to report the sum of the risk levels of the low points.

For the given example, the answer is 15.

Part Two

Part Two defines basins:

A basin is all locations that eventually flow downward to a single low point. Therefore, every low point has a basin, although some basins are very small. Locations of height 9 do not count as being in any basin, and all other locations will always be part of exactly one basin.

That's a bit of a tricky definition. But it does mean that the input is such that every basin has exactly one low point, and every basin is either bounded by the edge of the input, or by 9s.

The size of a basin is the number of points in a basin. For Part Two, we are asked to find the three largests basins, and report the product of their sizes.

For the example input, we get 1134 as answer, as the three largest basins have sizes 9, 9, and 14, and 1134 is the product of those three numbers.

Solution

Perl

We first read in the map. For that, we use a 2-dimensional array. At the end of each row, we add a 9, and we add an additional row with all 9s. The advantage of that is, we can look at all the neighbours of a point by adding/substracting 1 from its indices without worrying about boundaries — indexing by -1 gives the last element of an array.

my @floor = map {[(/[0-9]/g), 9]} <>;
push @floor => [(9) x @{$floor }]; my$X =   @floor      - 1;
my $Y = @{$floor } - 1;

We now create a method basin_size which takes three arguments: $x,$y (the coordinates of a low point), and $floor, the height map. Not only will it count the size of the basin, it also sets the height of any point of the basin to 9. This not only prevents us from counting a point in the basin more than once, it also has a very tiny speed up later on. We're using a basic breadth-first search. sub basin_size ($x, $y,$floor) {
my $size = 0; my @todo = ([$x, $y]); while (my$point = shift @todo) {
my ($px,$py) = @$point; next if $$floor [px] [py] == 9;$$floor [$px] [$py] = 9;$size ++;
push @todo =>  [$px - 1,$py],     [$px + 1,$py],
[$px,$py - 1], [$px,$py + 1];
}
$size; } We can now scan the map, and if we have a low point, calculate the size of the basin. We keep the basin sizes in an array: my @basins; foreach my$x (0 .. $X - 1) { foreach my$y (0 .. $Y - 1) { if ($floor [$x] [$y] < $floor [$x - 1] [$y] &&$floor [$x] [$y] < $floor [$x + 1] [$y] &&$floor [$x] [$y] < $floor [$x]     [$y - 1] &&$floor [$x] [$y] < $floor [$x]     [$y + 1]) {$sum += $floor [$x] [$y] + 1; push @basins => basin_size$x, $y, \@floor; } } } Given the basin sizes, it's easy to calculate the product of the three largests: @basins = sort {$a <=> $b} @basins; say "Solution 1: ",$sum;
say "Solution 2: ", $basins [-1] *$basins [-2] * \$basins [-3];

Find the full program on GitHub.