TL;DR

On with Advent of Code puzzle 01 from 2021: some tricky comparisons.

The 2021 edition of Advent of Code started with a trick. In hindsight, it was a sort of manifesto: there will be a lot of brute forcing to do, but you might find a clever solution every now and then.

And soâ€¦ I was gullible enough to fall in the trap, and implemented it the full way:

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

sub MAIN ($filename =$?FILE.subst(/\.raku$/, '.sample')) { my$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 ($filename) {$filename.IO.basename.IO.lines.Array;
} ## end sub get_inputs ($filename = undef) sub solve ($inputs) {
return (part1($inputs), part2($inputs));
}

sub count-increases (@inputs) {
my $count = (1 .. @inputs.end) .map({@inputs[$_] > @inputs[$_ - 1] ?? 1 !! 0 }) .sum; } sub part1 ($inputs) { return count-increases($inputs) } sub part2 ($inputs) {
return count-increases(
(1 ..^ $inputs.end).map({$inputs[($_-1)..($_+1)].sum})
);
}


The trap is in part 2. Iâ€™ve been tricked into calculating the sliding window sum, but it was not needed. Consider four consecutive values, yielding two values to be compared:

$..., x_n, x_{n+1}, x_{n+2}, x_{n+3}, ...$

The two sums would be:

$S_n = x_n + x_{n+1} + x_{n+2} \\ S_{n+1} = x_{n+1} + x_{n+2} + x_{n+3}$

Thereâ€™s a lot of similarity between the twoâ€¦ because they share two items out of three. So the comparison can be simplified like this:

$S_n \gtrless S_{n+1} \\ x_n + x_{n+1} + x_{n+2} \gtrless x_{n+1} + x_{n+2} + x_{n+3} \\ x_n \gtrless x_{n+3}$

Soâ€¦ no need to do sums, itâ€™s sufficient to skip two items for doing the comparison!

Stay safe folks!

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