Advent of Code --- Day 21: Fractal Art --- You find a program trying to generate some art. It uses a strange process that involves repeatedly enhancing the detail of an image through a set of rules. The image consists of a two-dimensional square grid of pixels that are either on (#) or off (.). The program always begins with this pattern: .#. ..# ### Because the pattern is both 3 pixels wide and 3 pixels tall, it is said to have a size of 3. Then, the program repeats the following process: • If the size is evenly divisible by 2, break the pixels up into 2x2 squares, and convert each 2x2 square into a 3x3 square by following the corresponding enhancement rule. • Otherwise, the size is evenly divisible by 3; break the pixels up into 3x3 squares, and convert each 3x3 square into a 4x4 square by following the corresponding enhancement rule. Because each square of pixels is replaced by a larger one, the image gains pixels and so its size increases. The artist's book of enhancement rules is nearby (your puzzle input); however, it seems to be missing rules. The artist explains that sometimes, one must rotate or flip the input pattern to find a match. (Never rotate or flip the output pattern, though.) Each pattern is written concisely: rows are listed as single units, ordered top-down, and separated by slashes. For example, the following rules correspond to the adjacent patterns: ../.# = .. .# .#. .#./..#/### = ..# ### #..# #..#/..../#..#/.##. = .... #..# .##. When searching for a rule to use, rotate and flip the pattern as necessary. For example, all of the following patterns match the same rule: .#. .#. #.. ### ..# #.. #.# ..# ### ### ##. .#. Suppose the book contained the following two rules: ../.# => ##./#../... .#./..#/### => #..#/..../..../#..# As before, the program begins with this pattern: .#. ..# ### The size of the grid (3) is not divisible by 2, but it is divisible by 3. It divides evenly into a single square; the square matches the second rule, which produces: #..# .... .... #..# The size of this enhanced grid (4) is evenly divisible by 2, so that rule is used. It divides evenly into four squares: #.|.# ..|.. --+-- ..|.. #.|.# Each of these squares matches the same rule (../.# => ##./#../...), three of which require some flipping and rotation to line up with the rule. The output for the rule is the same in all four cases: ##.|##. #..|#.. ...|... ---+--- ##.|##. #..|#.. ...|... Finally, the squares are joined into a new grid: ##.##. #..#.. ...... ##.##. #..#.. ...... Thus, after 2 iterations, the grid contains 12 pixels that are on. How many pixels stay on after 5 iterations? Your puzzle answer was 164. --- Part Two --- How many pixels stay on after 18 iterations? Your puzzle answer was 2355110. Both parts of this puzzle are complete! They provide two gold stars: ** At this point, all that is left is for you to admire your advent calendar. If you still want to see it, you can get your puzzle input. References Visible links . http://adventofcode.com/ . http://adventofcode.com/2017/about . http://adventofcode.com/2017/support . http://adventofcode.com/2017/events . http://adventofcode.com/2017/settings . http://adventofcode.com/2017/auth/logout . http://adventofcode.com/2017 . http://adventofcode.com/2017 . http://adventofcode.com/2017/leaderboard . http://adventofcode.com/2017/stats . http://adventofcode.com/2017/sponsors . http://adventofcode.com/2017/sponsors . http://adventofcode.com/2017 . http://adventofcode.com/2017/day/21/input