require_relative 'util'

# --- Day 25: Let It Snow ---

# Merry Christmas! Santa is booting up his weather machine; looks like
# you might get a white Christmas after all.

# The weather machine beeps! On the console of the machine is a copy
# protection message asking you to enter a code from the instruction
# manual. Apparently, it refuses to run unless you give it that
# code. No problem; you'll just look up the code in the--

# "Ho ho ho", Santa ponders aloud. "I can't seem to find the manual."

# You look up the support number for the manufacturer and give them a
# call. Good thing, too - that 49th star wasn't going to earn itself.

# "Oh, that machine is quite old!", they tell you. "That model went
# out of support six minutes ago, and we just finished shredding all
# of the manuals. I bet we can find you the code generation algorithm,
# though."

# After putting you on hold for twenty minutes (your call is very
# important to them, it reminded you repeatedly), they finally find an
# engineer that remembers how the code system works.

# The codes are printed on an infinite sheet of paper, starting in the
# top-left corner. The codes are filled in by diagonals: starting with
# the first row with an empty first box, the codes are filled in
# diagonally up and to the right. This process repeats until the
# infinite paper is covered. So, the first few codes are filled in in
# this order:

#        | 1   2   3   4   5   6
#     ---+---+---+---+---+---+---+
#      1 |  1   3   6  10  15  21
#      2 |  2   5   9  14  20
#      3 |  4   8  13  19
#      4 |  7  12  18
#      5 | 11  17
#      6 | 16

# For example, the 12th code would be written to row 4, column 2; the
# 15th code would be written to row 1, column 5.

# The voice on the other end of the phone continues with how the codes
# are actually generated. The first code is 20151125. After that, each
# code is generated by taking the previous one, multiplying it by
# 252533, and then keeping the remainder from dividing that value by
# 33554393.

# So, to find the second code (which ends up in row 2, column 1),
# start with the previous value, 20151125. Multiply it by 252533 to
# get 5088824049625. Then, divide that by 33554393, which leaves a
# remainder of 31916031. That remainder is the second code.

# "Oh!", says the voice. "It looks like we missed a scrap from one of
# the manuals. Let me read it to you." You write down his numbers:

#        |    1         2         3         4         5         6
#     ---+---------+---------+---------+---------+---------+---------+
#      1 | 20151125  18749137  17289845  30943339  10071777  33511524
#      2 | 31916031  21629792  16929656   7726640  15514188   4041754
#      3 | 16080970   8057251   1601130   7981243  11661866  16474243
#      4 | 24592653  32451966  21345942   9380097  10600672  31527494
#      5 |    77061  17552253  28094349   6899651   9250759  31663883
#      6 | 33071741   6796745  25397450  24659492   1534922  27995004

# "Now remember", the voice continues, "that's not even all of the
# first few numbers; for example, you're missing the one at 7,1 that
# would come before 6,2. But, it should be enough to let your-- oh,
# it's time for lunch! Bye!" The call disconnects.

# Santa looks nervous. Your puzzle input contains the message on the
# machine's console. What code do you give the machine?

input = File.open('25.txt') { |f| f.read.scan(/\d+/).map(&:to_i) }

# [1, 1], [2, 1], [1, 2], [3, 1], [2, 2], [1, 3], [4, 1], [3, 2], [2,
# 3], [4, 1]

def cantor
  Enumerator.new do |y|
    sum = 2
    i = 1
    loop do
      y << [sum - i, i]
      i += 1
      if i == sum
        sum += 1
        i = 1
      end
    end
  end
end

def easy(input)
  code = 20_151_125
  cantor.each do |row_col|
    return code if row_col == input
    code *= 252_533
    code %= 33_554_393
  end
end

assert(easy([1, 1]) == 20_151_125)
assert(easy([2, 1]) == 31_916_031)
assert(easy([3, 1]) == 16_080_970)
assert(easy([1, 3]) == 17_289_845)
assert(easy([6, 6]) == 27_995_004)
puts "easy(input): #{easy(input)}"