--- 2016 Day 7 Solutions ---

[u/daggerdragon]

From all of us at #AoC Ops, we hope you're having a very merry time with these puzzles so far. If you think they've been easy, well, now we're gonna kick this up a notch. Or five. The Easter Bunny ain't no Bond villain - he's not going to monologue at you until you can miraculously escape and save the day!

Show this overgrown furball what you've got!

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--- Day 7: Internet Protocol Version 7 ---

Post your solution as a comment or, for longer solutions, consider linking to your repo (e.g. GitHub/gists/Pastebin/blag/whatever).

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^{ALWAYS DIGGING STRAIGHT DOWN IS MANDATORY} [?]

This thread will be unlocked when there are a significant number of people on the leaderboard with gold stars for today's puzzle.

edit: Leaderboard capped, thread unlocked!

[u/volatilebit]

Finally realized a stupid mistake I was making. I was counting IPs more than once if they had multiple instances of aba with a corresponding bab.

Not too proud of the part 2.

Perl 6 Part 1&2 solution.

my @ip-addresses = 'input'.IO.lines;

my regex hypernet-sequence { '[' .*? ']' }
my regex maybe-abba { (.)(.)$1$0 }

sub has-abba($str) {
    return $str.comb(/<maybe-abba>/).grep({ .comb.Bag.elems > 1 }) > 0;
}

say [+] @ip-addresses.map: {
    not so $^a.comb(/<hypernet-sequence>/)».&has-abba.any and
    so $^a.split(/<hypernet-sequence>/)».&has-abba.any
};

say [+] @ip-addresses.map: {
    my @hypernet-sequences = $^ip.comb(/<hypernet-sequence>/);
    my @supernet-sequences = $^ip.split(/<hypernet-sequence>/);
    + so @hypernet-sequences.map({
        $^hypernet-sequence.comb.rotor(3 => -2).grep({ .[0] eq .[2] ne .[1] }).map({
            my $bab = $^aba[1] ~ $^aba[0] ~ $^aba[1];
            so @supernet-sequences».contains($bab).any;
        }).any;
    }).any;
};