Day 8: Two-Factor Authentication (1st & 2nd)

Author: rab

README.md

@@ -13,6 +13,7 @@ My Personal [Leaderboard] stats:
 <pre>
       --Part 1---   --Part 2---
 Day   Rank  Score   Rank  Score
+  8   2224      0   2177      0
   7    210      0    330      0  This one really gave me trouble. 3rd answer for * and 2nd for **
   6    294      0    271      0
   5   2658      0   2387      0

day08.rb

@@ -0,0 +1,145 @@
+# --- Day 8: Two-Factor Authentication ---
+#
+# You come across a door implementing what you can only assume is an implementation of two-factor
+# authentication after a long game of requirements telephone.
+#
+# To get past the door, you first swipe a keycard (no problem; there was one on a nearby
+# desk). Then, it displays a code on a little screen, and you type that code on a keypad. Then,
+# presumably, the door unlocks.
+#
+# Unfortunately, the screen has been smashed. After a few minutes, you've taken everything apart
+# and figured out how it works. Now you just have to work out what the screen would have
+# displayed.
+#
+# The magnetic strip on the card you swiped encodes a series of instructions for the screen; these
+# instructions are your puzzle input. The screen is 50 pixels wide and 6 pixels tall, all of which
+# start off, and is capable of three somewhat peculiar operations:
+#
+#
+# rect AxB turns on all of the pixels in a rectangle at the top-left of the screen which is A wide
+# and B tall.
+#
+# rotate row y=A by B shifts all of the pixels in row A (0 is the top row) right by B
+# pixels. Pixels that would fall off the right end appear at the left end of the row.
+#
+# rotate column x=A by B shifts all of the pixels in column A (0 is the left column) down by B
+# pixels. Pixels that would fall off the bottom appear at the top of the column.
+#
+# For example, here is a simple sequence on a smaller screen:
+#
+# rect 3x2 creates a small rectangle in the top-left corner:
+# ###....
+# ###....
+# .......
+#
+# rotate column x=1 by 1 rotates the second column down by one pixel:
+# #.#....
+# ###....
+# .#.....
+#
+# rotate row y=0 by 4 rotates the top row right by four pixels:
+# ....#.#
+# ###....
+# .#.....
+#
+# rotate column x=1 by 1 again rotates the second column down by one pixel, causing the bottom pixel
+# to wrap back to the top:
+# .#..#.#
+# #.#....
+# .#.....
+#
+# As you can see, this display technology is extremely powerful, and will soon dominate the
+# tiny-code-displaying-screen market.  That's what the advertisement on the back of the display
+# tries to convince you, anyway.
+#
+# There seems to be an intermediate check of the voltage used by the display: after you swipe your
+# card, if the screen did work, how many pixels should be lit?
+#
+# Your puzzle answer was 116.
+#
+# --- Part Two ---
+#
+# You notice that the screen is only capable of displaying capital letters; in the font it uses,
+# each letter is 5 pixels wide and 6 tall.
+#
+# After you swipe your card, what code is the screen trying to display?
+#
+# Your puzzle answer was UPOJFLBCEZ.
+
+require_relative 'input'
+
+day = __FILE__[/\d+/].to_i(10)
+input = Input.for_day(day)
+
+# puts "solving day #{day} from input\n#{input.inspect}"
+
+class Display
+  def initialize(cols=50, rows=6)
+    @display = Array.new(rows) { Array.new(cols, '.') }
+  end
+
+  # rect AxB
+  # turns on all of the pixels in a rectangle at the top-left
+  # of the screen which is A wide and B tall.
+  def rect(wide, tall)
+    tall.times do |row|
+      wide.times do |col|
+        @display[row][col] = '#'
+      end
+    end
+    self
+  end
+
+  # rotate row y=A by B
+  # shifts all of the pixels in row A (0 is the top row) right by B pixels.
+  # Pixels that would fall off the right end appear at the left end of the row.
+  def row(row, by)
+    by.times { @display[row].unshift @display[row].pop }
+    self
+  end
+
+  # rotate column x=A by B
+  # shifts all of the pixels in column A (0 is the left column) down by B pixels.
+  # Pixels that would fall off the bottom appear at the top of the column.
+  def column(col, by)
+    tmp = @display.map {|row| row[col] }.join
+    by.times { tmp[0,0]= tmp[-1]; tmp.chop! }
+    tmp.each_char.with_index {|c,i| @display[i][col] = c }
+    self
+  end
+
+  def to_s
+    @display.map {|row| row.join }.join("\n")
+  end
+
+  def count
+    @display.map {|row| row.count('#') }.reduce(:+)
+  end
+end
+
+# display = Display.new(7,3)
+# input = <<-EOL
+# rect 3x2
+# rotate column x=1 by 1
+# rotate row y=0 by 4
+# rotate column x=1 by 1
+# EOL
+
+display = Display.new(50,6)
+
+input.each_line do |line|
+  case line
+  when /^rect (\d+)x(\d+)$/
+    display.rect($1.to_i,$2.to_i)
+  when /^rotate column x=(\d+) by (\d+)$/
+    display.column($1.to_i,$2.to_i)
+  when /^rotate row y=(\d+) by (\d+)$/
+    display.row($1.to_i,$2.to_i)
+  else
+    fail "Bad Command? #{line}"
+  end
+
+  # puts display, nil
+end
+
+puts display, display.count