TL;DR

On with Advent of Code puzzle 15 from 2022: there’s always to learn from the Megathread.

This day’s solution was a bad exercise in brute force for me, so I’m not commenting my solution. As a matter of fact, I deleted my solution to the second part actually, and opted for reimplementing it in a more rational way (although not completely correct).

The full solution is available, of course.

The gist of how part 2 should be addressed is to avoid going fully brutal and only figure out intersections of the outer “shells” (actually, perimeters) for each sensor’s reach.

This can also be combined with a simple change of coordinates that turns the diamond-shaped ranges into plain squares, which are extremely easy to search for intersections.

So here’s my part 2:

sub part2 ($inputs, $midsize) {
   my $size = $midsize * 2;
   my @t-rex = $inputs.map: ->($x, $y, $bx, $by) {
      my @r = (1 + ($x - $bx, $y - $by)».abs.sum) «*» (-1, 1);
      (($x + $y) «+» @r, ($x - $y) «+» @r).flat.Array;
   };

   for @t-rex -> $h {
      my ($hx-min, $hx-max, $hy-min, $hy-max) = |$h;
      for @t-rex -> $v {
         next if $h === $v;
         my ($vx-min, $vx-max, $vy-min, $vy-max) = |$v;
         PAIR:
         for ($hy-min, $hy-max) X ($vx-min, $vx-max) -> ($hy, $vx) {
            next unless $vy-min <= $hy <= $vy-max
                     && $hx-min <= $vx <= $hx-max;
            next PAIR if ($vx + $hy) % 2;
            for @t-rex -> ($cx-min, $cx-max, $cy-min, $cy-max) {
               next PAIR if $cx-min < $vx < $cx-max
                         && $cy-min < $hy < $cy-max;
            }
            my $x = ($vx + $hy) div 2;
            next PAIR unless 0 <= $x <= $size;
            my $y = ($vx - $hy) div 2;
            next PAIR unless 0 <= $y <= $size;
            return $x * $size + $y;
         }
      }
   }

   return 'part2';
}

The $midsize is the 10 or 2000000 value that should be otherwise hardcoded.

After calculating the intersections, we go back in the “normal” coordinates system and check if the intersection is within the bounds. When we find one such position… we’re done.

How is this not completely correct? Well… there are four corner cases, located at the very four corners of the search area, where the missing beacon might be located, and this solution would not find it.

Easy to address and left as an exercise for the reader, have fun!