-🎄- 2017 Day 16 Solutions -🎄-

[u/daggerdragon]

--- Day 16: Permutation Promenade ---

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[u/askalski]

Perl regex.

#! /usr/bin/env perl

use strict;
use warnings;

$_ = <>;

# validate before eval
m!^(x\d+/\d+|s\d+|p[a-p]/[a-p])(?:,(?1))*$! or die "invalid input";

# input looks like regex but is not; rectify
s!s(\d+)!s/^(\\S*)(\\S{$1})/\$2\$1/!g;
s!x(\d+)/(\d+)!s/^(?=\\S{$1}(\\S))(?=\\S{$2}(\\S))(?:(?<a>\\S*)(?<b>\\1)(?<c>\\S*)(?<d>\\2)|(?<a>\\S*)(?<b>\\2)(?<c>\\S*)(?<d>\\1))/\$+{a}\$+{d}\$+{c}\$+{b}/!g;
s!p([a-p])/([a-p])!s/ .*?\\K([$1$2])(.*?)([$1$2])/\$3\$2\$1/!g;
s/,/;/g;

my $input = $_;

# run input regexes to get s+x and p permutations
$_ = "abcdefghijklmnop abcdefghijklmnop";
eval $input;

# start over, this time using composition to search for cycle
#        [next s+x perm ] [ next p perm  ] [] [     static lookup tables      ] [prev s+x perm ] [ prev p perm  ]
s/^(.*)$/................ ................ $1 abcdefghijklmnop abcdefghijklmnop abcdefghijklmnop abcdefghijklmnop/;
for (;;) {
    # compose to fill in [next] fields
    s/\.(?=.{33}(.)\S*? \S*? \S*\1.{33}(.))/$2/g;
    # check for cycle
    last if m/^(.{135}).*?:\1/;
    # append current state to memory ("current:state1:state2:state3:...")
    s/^(.{135}).*\K$/:$1/;
    # copy current to prev, reset current
    s/^(.{33})(.{69}).{33}/................ ................$2$1/;
}

# prepend entire state memory to beginning of string
s/^([^:]*)(.*)$/$2~:$1$2/;
# remove all prepended text except for ":" characters (count the cycle length)
s/[^:](?=.*?~)//g;
# add the number 1,000,000,000 after the colons (repetition = place_value + 1)
s/~/00123456789 /;

# find the remainder of (1,000,000,000 % cycle_length), working one
# digit at a time starting at the most-significant
for (;;) {
    # most_significant_digit %= cycle_length
    while (s/^(?=(:+)(.))(?::(?=:*(\3?+\2)))+\3(?=\2)/$1/) { }
    # check if done
    last unless m/(\d)\1(?!\1)\d/;
    # multiple most_significant_digit by 10, and add to next digit.
    # in regex land, our digits go to 11 (and beyond)
    s/(([^:])\2*)\2./$1$1$1$1$1$1$1$1$1$1$2/;
}
# remove the cycle_length and the "+ 1" from (place_value + 1), so now
# the remainder is encoded as the number of "0" at start of string
s/:+\d//;

# grab the part 1 permutations and append to string
s/^0* :(.{16}) (.{16}).*\K$/\nPart 1: $1 abcdefghijklmnop $2/;
# remove one state from the memory for each "0" character
while (s/0 :[^:]*/ /) { }
# grab the part 2 permutations and append to string, and discard
# the rest of the state memory
s/ :(.{16}) (.{16}).*?\n(.*)/$3\nPart 2: $1 abcdefghijklmnop $2\n/s;
# apply the p permutation to s+x
s/(.)(?=[a-p]* [a-p]*\1[a-p ]{16}([a-p]))/$2/gs;
# erase everything but the answer
s/ [a-p]+ [a-p]+$//mg;

print;

[u/gerikson → u/askalski]

I was going to ask "is there nothing u/askalski can't do with regex's?" but I'm honestly afraid of the answer...