Cube Conundrum Part 2

chicken.org

@@ -177,7 +177,6 @@ these together produces =281=.
 #+RESULTS: day1-solution-2
 : 53866
 
-
 ** Puzzle Input
 
 Jump to [[#headline-10][day 2]].
@@ -1350,7 +1349,7 @@ basically, it is nothing more than a recursive ~and~ statement.
   And now, everything can be put together:
 
 #+NAME: day2-part1-solution
-    #+begin_src scheme :noweb no-export
+    #+begin_src scheme :noweb yes :var input=day2-input
   (import (chicken string)
           (chicken keyword)
           (chicken irregex))
@@ -1367,7 +1366,89 @@ basically, it is nothing more than a recursive ~and~ statement.
 
 ** Part Two
 
-To be done...
+*** Quest
+
+The Elf says they've stopped producing snow because they aren't getting any *water*! He isn't sure
+why the water stopped; however, he can show you how to get to the water source to check it out for
+yourself. It's just up ahead!
+
+As you continue your walk, the Elf poses a second question: in each game you played, what is the
+*fewest number of cubes of each color* that could have been in the bag to make the game possible?
+
+Again consider the example games from earlier:
+
+#+begin_example
+Game 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green
+Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue
+Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red
+Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red
+Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green
+#+end_example
+
+- In game 1, the game could have been played with as few as 4 red, 2 green, and 6 blue cubes. If any
+  color had even one fewer cube, the game would have been impossible.
+- Game 2 could have been played with a minimum of 1 red, 3 green, and 4 blue cubes.
+- Game 3 must have been played with at least 20 red, 13 green, and 6 blue cubes.
+- Game 4 required at least 14 red, 3 green, and 15 blue cubes.
+- Game 5 needed no fewer than 6 red, 3 green, and 2 blue cubes in the bag.
+
+The *power* of a set of cubes is equal to the numbers of red, green, and blue cubes multiplied
+together. The power of the minimum set of cubes in game 1 is =48=. In games 2-5 it was =12=, =1560=,
+=630=, and =36=, respectively. Adding up these five powers produces the sum *=2286=*.
+
+For each game, find the minimum set of cubes that must have been present. *What is the sum of the
+power of these sets?*
+
+*** Puzzle Solution
+
+Most code can be reused from the first part; only the success check and the main function have to be
+rewritten to accommodate the new requirements.
+
+The success check gets replaced by a function that multiplies the minima together.
+
+#+NAME: day2-part2-powercalc
+#+begin_src scheme
+  (define (powercalc draws)
+    (* (apply max (map (lambda (draw)
+                         (alist-ref #:red draw eqv? 1))
+                       draws))
+       (apply max (map (lambda (draw)
+                         (alist-ref #:green draw eqv? 1))
+                       draws))
+       (apply max (map (lambda (draw)
+                         (alist-ref #:blue draw eqv? 1))
+                       draws))))
+#+end_src
+
+And the main function gets modified to sum everything up.
+
+#+NAME: day2-part2-main-function
+#+begin_src scheme
+  (define (game-set-power input)
+    (let ((games (record-fold input)))
+      (foldl + 0
+             (map (lambda (game)
+                    (powercalc (cdr game)))
+                  games))))
+#+end_src
+
+The full thing put together:
+
+#+NAME: day2-part2-solution
+#+begin_src scheme :noweb yes :var input=day2-input
+  (import (chicken string)
+          (chicken keyword)
+          (chicken irregex))
+
+  <<day2-part1-draw-processing>>
+  <<day2-part1-draw-splitting>>
+  <<day2-part1-record-splitting>>
+  <<day2-part2-powercalc>>
+  <<day2-part2-main-function>>
+#+end_src
+
+#+RESULTS: day2-part1-solution
+: 78375
 
 ** Puzzle Input