TL;DR

On with Advent of Code puzzle 14 from 2021: taming exponential growth in the evolution mechanism.

This is another take to a two-fold puzzle in which the first part is well within the bounds of doing that with simple brute force, but the second one is proibitively forbidden from doing so.

This time I didn’t have to think too much about this fact, just by looking at the results in the puzzle example:

In the above example, the most common element is B (occurring 2192039569602 times) and the least common element is H (occurring 3849876073 times); subtracting these produces 2188189693529.

I surely don’t have all that memory!

The key here is that we only need counting stuff, but we are not required to keep the whole polymer sequence. This means that something like this:

``````ABABABABABABABABABAB
``````

can be represented as `AB` occurring 10 times and `BA` occurring 9.

Each pair will then evolve independently, but all equal pairs will evolve in the same manner. So our generic pair `AB` will evolve like this (assuming we have to insert a `C`):

• the number of `AC` is incremented by 10 units;
• the number of `CB` is incremented by 10 units.

In the same spirit, if our generic pair `BA` is expanded with a `D` in between:

• the number of `BD` is incremented by 9 units;
• the number of `DA` is incremented by 9 units.

We calculate the evolution in a separate hash, then assume the new hash is the starting point of a future iteration.

This can then be used in both parts, with different number of iterations.

``````#!/usr/bin/env raku
use v6;

sub MAIN (\$filename = \$?FILE.subst(/\.raku\$/, '.tmp')) {
my \$inputs = get-inputs(\$filename);
my (\$part1, \$part2) = solve(\$inputs);

my \$highlight = "\e[1;97;45m";
my \$reset     = "\e[0m";
put "part1 \$highlight\$part1\$reset";
put "part2 \$highlight\$part2\$reset";
}

sub get-inputs (\$filename) {
my @words = \$filename.IO.lines.comb: / \w+ /;
my \$start = @words.shift;
my %new-letter-for = @words;
return {
start => \$start,
nlfor => %new-letter-for,
};
}

sub solve (\$inputs) {
my \$nlfor = \$inputs<nlfor>;
my @s = \$inputs<start>.comb: / \w /;
my \$bag = (@s Z @s[1 .. *])».join('').Bag.Hash;
my %count = @s.Bag.Hash;
my \$part1;
for 1 .. 40 -> \$step {
my %new;
for \$bag.kv -> \$key, \$factor {
my @items;
if \$nlfor{\$key}:exists {
my \$m = \$nlfor{\$key};
%count{\$m} += \$factor;
my (\$l, \$r) = \$key.comb: / \w /;
@items.push: \$l ~ \$m, \$m ~ \$r;
}
else { @items.push: \$key };
for @items -> \$item {
%new{\$item} //= 0;
%new{\$item} += \$factor;
}
}
\$bag = %new;
\$part1 = %count.values.max - %count.values.min if \$step == 10;
}
return \$part1, %count.values.max - %count.values.min;
}
``````

The `%new` is the collector for the result of a new iteration, which is eventually kept for future ones in `\$bag`.

The intermediate result at step 10 is collected along the way, so the `solve` function does a single iteration from 1 to 40.

I guess it’s everything, please ask if you have questions and otherwise… stay safe!

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