AoC day 7, part 2. In principle, this problem is eminently regexable: All steps a beam takes can be described by a regex, ^ corresponds to choice/alternative, and the number of ways the regex matches is the solution. In practice, this takes forever. I tried to speed things up with embedded code blocks and a cache, but it quickly gets overly complex and I suspect there may be bugs lurking in the interaction between embedded code blocks, closures, local, and backtracking.

In the end I gave up and rewrote the regex as direct code with manual backtracking. I guess the cache makes it an instance of "dynamic programming".

use v5.36;<br><br>my $grid = do { local $/; readline };<br><br>my $w = index($grid, "\n") + 1;<br><br>$grid .= 'E' x $w . "\n";<br><br>my $p = index $grid, 'S';<br>$p += $w;<br><br>my $count = 0;<br>my (@ alt, @cache);<br>while () {<br>    while (substr($grid, $p, 1) eq '.') {<br>        $p += $w;<br>    }<br>    if (substr($grid, $p, 1) eq '^') {<br>        if (defined(my $c = $cache[$p])) {<br>            $count += $c;<br>        } else {<br>            push @ alt,<br>                [$p, $count],<br>                $p + 1;<br>            $p--;<br>            next;<br>        }<br>    } else {<br>        $count++;<br>    }<br>    while (@ alt && ref $alt[-1]) {<br>        my ($ap, $ac) = @{pop @ alt};<br>        $cache[$ap] = $count - $ac;<br>    }<br>    $p = pop @ alt // last;<br>}<br><br>say $count;<br>

#AdventOfCode #perl