# 2023 AoC Day 12 – Hot Springs

This is a solution to Advent of Code 2023 day 12, written in Raku.

### Part One

For each row, count all of the different arrangements of operational and broken springs that meet the given criteria. What is the sum of those counts?

I started out with a brute-force approach that generates all patterns and then counts those that match the expected groupings. I was aiming for an easy to debug solution. This approach is quite memory hungry and slow.

my $total = 0; for '12-input.txt'.IO.lines { my ($springs, $groups) = .split(' '); my @groups =$groups.split(',')>>.Int;

my @patterns = [0],;

for $springs.comb ->$c {
given $c { when '.' { for @patterns -> @p { @p.push(0) unless @p[*-1] == 0; } } when '#' { for @patterns -> @p { @p[*-1] += 1; } } when '?' { my @clones = @patterns.map(-> @p { @p.clone; }); for @patterns -> @p { @p.push(0) unless @p[*-1] == 0; } for @clones -> @c { @c[*-1] += 1; } @patterns.append(@clones); } } } for @patterns -> @p { @p.pop if @p[*-1] == 0 } my$arrangements = +@patterns.grep(-> @p { @p ~~ @groups });
$total +=$arrangements;

# say "{$springs} -> {@groups} = {$arrangements}";
}
say "Sum of counts is {$total}"; say "Took " ~ (now - ENTER now).base(10,2) ~ " seconds"; Sum of counts is 7506 Took 49.83 seconds  ### Part Two As you look out at the field of springs, you feel like there are way more springs than the condition records list. When you examine the records, you discover that they were actually folded up this whole time! To unfold the records, on each row, replace the list of spring conditions with five copies of itself (separated by ?) and replace the list of contiguous groups of damaged springs with five copies of itself (separated by ,). Unfold your condition records; what is the new sum of possible arrangement counts? The unfolded dataset in part 2 needs a different approach. I tried running the solution from part 1 just to see how it fared and I was set to run out of 64GB memory before completing the first few rows. I chose to switch to a recursive solution that would be easy to memoise, i.e. cache the results of recursive calls. I also applied some Raku optimisation patterns but these were probably unnecessary given how quickly it ran. class Record { has @.springs; has @.groups; has int$.nsprings;
has int $.ngroups; has %.cache; submethod TWEAK {$!nsprings = +@!springs;
$!ngroups = +@!groups; } method count-arrangements(int \spos = 0, int \gpos = 0, int \count = 0) { my$key = "{spos}:{gpos}:{count}";
return %!cache{$key} if %!cache{$key}:exists;

my $n = 0; if spos ==$!nsprings {
if gpos == $!ngroups && count == 0 {$n = 1;
} elsif (gpos + 1) == $!ngroups && count == @!groups[gpos] {$n = 1;
}
} else {

if gpos == $!ngroups && count > 0 { # blown all groups } elsif gpos <$!ngroups && count > @!groups[gpos] {
# blown this group
} else {
my $char = @!springs[spos]; if$char eq '.' {
if count > 0 && count == @!groups[gpos] {
$n = self.count-arrangements(spos + 1, gpos + 1); } elsif count == 0 {$n = self.count-arrangements(spos + 1, gpos);
}
} elsif $char eq '#' { if count > 0 && count == @!groups[gpos] { # blown this group } else {$n = self.count-arrangements(spos + 1, gpos, count + 1);
}
} else {
# Try a '#'
$n = self.count-arrangements(spos + 1, gpos, count + 1); # Try a '.' if count > 0 && count == @!groups[gpos] {$n += self.count-arrangements(spos + 1, gpos + 1);
} elsif count == 0 {
$n += self.count-arrangements(spos + 1, gpos); } } } } %!cache{$key} = $n; return$n;
}
}

my @totals = '12-input.txt'.IO.lines.map(
-> $line { my$start = now;
my ($springs,$groups) = $line.split(' '); # Unfold for part 2$springs = ($springs xx 5).join('?');$groups = ($groups xx 5).join(','); my @springs =$springs.comb;
my @groups = $groups.split(',')>>.Int; my$row = Record.new(:@springs, :@groups);
my $total =$row.count-arrangements();

# say "{$line} - {$total} took " ~ (now - $start).base(10,2) ~ " seconds" ;$total
});

say "Sum of counts is {[+] @totals}";
say "Took " ~ (now - ENTER now).base(10,2) ~ " seconds";
Sum of counts is 548241300348335
Took 2.60 seconds