Solution to problem 15 in Python

src/Year_2016/P15.py

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+# --- Day 15: Timing is Everything ---
+
+# The halls open into an interior plaza containing a large kinetic sculpture.
+# The sculpture is in a sealed enclosure and seems to involve a set of identical
+# spherical capsules that are carried to the top and allowed to bounce through
+# the maze of spinning pieces.
+
+# Part of the sculpture is even interactive! When a button is pressed, a capsule
+# is dropped and tries to fall through slots in a set of rotating discs to
+# finally go through a little hole at the bottom and come out of the sculpture.
+# If any of the slots aren't aligned with the capsule as it passes, the capsule
+# bounces off the disc and soars away. You feel compelled to get one of those
+# capsules.
+
+# The discs pause their motion each second and come in different sizes; they
+# seem to each have a fixed number of positions at which they stop. You decide
+# to call the position with the slot 0, and count up for each position it
+# reaches next.
+
+# Furthermore, the discs are spaced out so that after you push the button, one
+# second elapses before the first disc is reached, and one second elapses as the
+# capsule passes from one disc to the one below it. So, if you push the button
+# at time=100, then the capsule reaches the top disc at time=101, the second
+# disc at time=102, the third disc at time=103, and so on.
+
+# The button will only drop a capsule at an integer time - no fractional seconds
+# allowed.
+
+# For example, at time=0, suppose you see the following arrangement:
+
+# Disc #1 has 5 positions; at time=0, it is at position 4.
+# Disc #2 has 2 positions; at time=0, it is at position 1.
+
+# If you press the button exactly at time=0, the capsule would start to fall;
+# it would reach the first disc at time=1. Since the first disc was at position
+# 4 at time=0, by time=1 it has ticked one position forward. As a five-position
+# disc, the next position is 0, and the capsule falls through the slot.
+
+# Then, at time=2, the capsule reaches the second disc. The second disc has
+# ticked forward two positions at this point: it started at position 1, then
+# continued to position 0, and finally ended up at position 1 again. Because
+# there's only a slot at position 0, the capsule bounces away.
+
+# If, however, you wait until time=5 to push the button, then when the capsule
+# reaches each disc, the first disc will have ticked forward 5+1 = 6 times (to
+# position 0), and the second disc will have ticked forward 5+2 = 7 times (also
+# to position 0). In this case, the capsule would fall through the discs and
+# come out of the machine.
+
+# However, your situation has more than two discs; you've noted their positions
+# in your puzzle input. What is the first time you can press the button to get
+# a capsule?
+
+import re
+
+with open("files/P15.txt") as f:
+    arrangements = [
+        tuple(map(int, re.findall("\d+", line))) for line in f.readlines()
+    ]
+
+
+def part_1():
+    tik = 0
+    while True:
+        for i, (_, positions, _, position) in enumerate(arrangements, start=1):
+            if not (position + tik + i) % positions:
+                continue
+            break
+        else:
+            print(tik)
+            break
+        tik += 1
+
+
+# --- Part Two ---
+
+# After getting the first capsule (it contained a star! what great fortune!),
+# the machine detects your success and begins to rearrange itself.
+
+# When it's done, the discs are back in their original configuration as if it
+# were time=0 again, but a new disc with 11 positions and starting at position 0
+# has appeared exactly one second below the previously-bottom disc.
+
+# With this new disc, and counting again starting from time=0 with the
+# configuration in your puzzle input, what is the first time you can press the
+# button to get another capsule?
+
+
+def part_2():
+    arrangements.append((1, 11, 0, 0))
+    tik = 0
+    while True:
+        for i, (_, positions, _, position) in enumerate(arrangements, start=1):
+            if not (position + tik + i) % positions:
+                continue
+            break
+        else:
+            print(tik)
+            break
+        tik += 1
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2016/files/P15.txt

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+Disc #1 has 7 positions; at time=0, it is at position 0.
+Disc #2 has 13 positions; at time=0, it is at position 0.
+Disc #3 has 3 positions; at time=0, it is at position 2.
+Disc #4 has 5 positions; at time=0, it is at position 2.
+Disc #5 has 17 positions; at time=0, it is at position 0.
+Disc #6 has 19 positions; at time=0, it is at position 7.