Comply with mypy conventions

src/Year_2020/P1.py

@@ -0,0 +1,88 @@
+# --- Day 1: Report Repair ---
+
+# After saving Christmas five years in a row, you've decided to take a vacation
+# at a nice resort on a tropical island. Surely, Christmas will go on without you.
+
+# The tropical island has its own currency and is entirely cash-only. The gold
+# coins used there have a little picture of a starfish; the locals just call
+# them stars. None of the currency exchanges seem to have heard of them, but
+# somehow, you'll need to find fifty of these coins by the time you arrive so
+# you can pay the deposit on your room.
+
+# To save your vacation, you need to get all fifty stars by December 25th.
+
+# Collect stars by solving puzzles. Two puzzles will be made available on each
+# day in the Advent calendar; the second puzzle is unlocked when you complete
+# the first. Each puzzle grants one star. Good luck!
+
+# Before you leave, the Elves in accounting just need you to fix your expense
+# report (your puzzle input); apparently, something isn't quite adding up.
+
+# Specifically, they need you to find the two entries that sum to 2020 and then
+# multiply those two numbers together.
+
+# For example, suppose your expense report contained the following:
+
+# 1721
+# 979
+# 366
+# 299
+# 675
+# 1456
+
+# In this list, the two entries that sum to 2020 are 1721 and 299. Multiplying
+# them together produces 1721 * 299 = 514579, so the correct answer is 514579.
+
+# Of course, your expense report is much larger. Find the two entries that sum
+# to 2020; what do you get if you multiply them together?
+
+with open("files/P1.txt", "r") as f:
+    numbers = [int(number) for number in f.read().strip().split("\n")]
+
+
+def part_1() -> None:
+    for n in numbers:
+        for i in range(len(numbers)):
+            if n + numbers[i] == 2020:
+                print(f"{n} times {numbers[i]} is {n*numbers[i]}")
+                break
+        else:
+            continue
+        break
+
+
+# --- Part Two ---
+
+# The Elves in accounting are thankful for your help; one of them even offers
+# you a starfish coin they had left over from a past vacation. They offer you
+# a second one if you can find three numbers in your expense report that meet
+# the same criteria.
+
+# Using the above example again, the three entries that sum to 2020 are 979,
+# 366, and 675. Multiplying them together produces the answer, 241861950.
+
+# In your expense report, what is the product of the three entries that sum to
+# 2020?
+
+
+def part_2() -> None:
+    for n in numbers:
+        for i in range(len(numbers)):
+            for j in range(len(numbers)):
+                if n + numbers[i] + numbers[j] == 2020:
+                    print(
+                        f"{n} times {numbers[i]} times {numbers[j]} is "
+                        f"{n*numbers[i]*numbers[j]}"
+                    )
+                    break
+            else:
+                continue
+            break
+        else:
+            continue
+        break
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P10.py

@@ -0,0 +1,227 @@
+from functools import lru_cache
+
+import numpy as np
+
+# --- Day 10: Adapter Array ---
+
+# Patched into the aircraft's data port, you discover weather forecasts of a
+# massive tropical storm. Before you can figure out whether it will impact your
+# vacation plans, however, your device suddenly turns off!
+
+# Its battery is dead.
+
+# You'll need to plug it in. There's only one problem: the charging outlet near
+# your seat produces the wrong number of jolts. Always prepared, you make a
+# list of all of the joltage adapters in your bag.
+
+# Each of your joltage adapters is rated for a specific output joltage (your
+# puzzle input). Any given adapter can take an input 1, 2, or 3 jolts lower
+# than its rating and still produce its rated output joltage.
+
+# In addition, your device has a built-in joltage adapter rated for 3 jolts
+# higher than the highest-rated adapter in your bag. (If your adapter list were
+# 3, 9, and 6, your device's built-in adapter would be rated for 12 jolts.)
+
+# Treat the charging outlet near your seat as having an effective joltage
+# rating of 0.
+
+# Since you have some time to kill, you might as well test all of your
+# adapters. Wouldn't want to get to your resort and realize you can't even
+# charge your device!
+
+# If you use every adapter in your bag at once, what is the distribution of
+# joltage differences between the charging outlet, the adapters, and your
+# device?
+
+# For example, suppose that in your bag, you have adapters with the following
+# joltage ratings:
+
+# 16
+# 10
+# 15
+# 5
+# 1
+# 11
+# 7
+# 19
+# 6
+# 12
+# 4
+
+# With these adapters, your device's built-in joltage adapter would be rated
+# for 19 + 3 = 22 jolts, 3 higher than the highest-rated adapter.
+
+# Because adapters can only connect to a source 1-3 jolts lower than its
+# rating, in order to use every adapter, you'd need to choose them like this:
+
+#     The charging outlet has an effective rating of 0 jolts, so the only
+# adapters that could connect to it directly would need to have a joltage
+# rating of 1, 2, or 3 jolts. Of these, only one you have is an adapter rated 1
+# jolt (difference of 1).
+#     From your 1-jolt rated adapter, the only choice is your 4-jolt rated
+# adapter (difference of 3).
+#     From the 4-jolt rated adapter, the adapters rated 5, 6, or 7 are valid
+# choices. However, in order to not skip any adapters, you have to pick the
+# adapter rated 5 jolts (difference of 1).
+#     Similarly, the next choices would need to be the adapter rated 6 and then
+# the adapter rated 7 (with difference of 1 and 1).
+#     The only adapter that works with the 7-jolt rated adapter is the one
+# rated 10 jolts (difference of 3).
+#     From 10, the choices are 11 or 12; choose 11 (difference of 1) and then
+# 12 (difference of 1).
+#     After 12, only valid adapter has a rating of 15 (difference of 3), then
+# 16 (difference of 1), then 19 (difference of 3).
+#     Finally, your device's built-in adapter is always 3 higher than the
+# highest adapter, so its rating is 22 jolts (always a difference of 3).
+
+# In this example, when using every adapter, there are 7 differences of 1 jolt
+# and 5 differences of 3 jolts.
+
+# Here is a larger example:
+
+# 28
+# 33
+# 18
+# 42
+# 31
+# 14
+# 46
+# 20
+# 48
+# 47
+# 24
+# 23
+# 49
+# 45
+# 19
+# 38
+# 39
+# 11
+# 1
+# 32
+# 25
+# 35
+# 8
+# 17
+# 7
+# 9
+# 4
+# 2
+# 34
+# 10
+# 3
+
+# In this larger example, in a chain that uses all of the adapters, there are
+# 22 differences of 1 jolt and 10 differences of 3 jolts.
+
+# Find a chain that uses all of your adapters to connect the charging outlet to
+# your device's built-in adapter and count the joltage differences between the
+# charging outlet, the adapters, and your device. What is the number of 1-jolt
+# differences multiplied by the number of 3-jolt differences?
+
+with open("files/P10.txt", "r") as f:
+    inputs = [int(num) for num in f.read().strip().split("\n")]
+
+# Add zero (charging outlet) and max+3 (device own built-in adapter)
+joltages = sorted(inputs + [0, max(inputs) + 3])
+
+
+def part_1() -> None:
+    joltages_diffs = np.diff(joltages)
+
+    print(
+        f"The difference is {np.sum(joltages_diffs == 1) * np.sum(joltages_diffs == 3)}"
+    )
+
+
+# --- Part Two ---
+
+# To completely determine whether you have enough adapters, you'll need to
+# figure out how many different ways they can be arranged. Every arrangement
+# needs to connect the charging outlet to your device. The previous rules about
+# when adapters can successfully connect still apply.
+
+# The first example above (the one that starts with 16, 10, 15) supports the
+# following arrangements:
+
+# (0), 1, 4, 5, 6, 7, 10, 11, 12, 15, 16, 19, (22)
+# (0), 1, 4, 5, 6, 7, 10, 12, 15, 16, 19, (22)
+# (0), 1, 4, 5, 7, 10, 11, 12, 15, 16, 19, (22)
+# (0), 1, 4, 5, 7, 10, 12, 15, 16, 19, (22)
+# (0), 1, 4, 6, 7, 10, 11, 12, 15, 16, 19, (22)
+# (0), 1, 4, 6, 7, 10, 12, 15, 16, 19, (22)
+# (0), 1, 4, 7, 10, 11, 12, 15, 16, 19, (22)
+# (0), 1, 4, 7, 10, 12, 15, 16, 19, (22)
+
+# (The charging outlet and your device's built-in adapter are shown in
+# parentheses.) Given the adapters from the first example, the total number of
+# arrangements that connect the charging outlet to your device is 8.
+
+# The second example above (the one that starts with 28, 33, 18) has many
+# arrangements. Here are a few:
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 46, 47, 48, 49, (52)
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 46, 47, 49, (52)
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 46, 48, 49, (52)
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 46, 49, (52)
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 47, 48, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 46, 48, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 46, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 47, 48, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 47, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 48, 49, (52)
+
+# In total, this set of adapters can connect the charging outlet to your device
+# in 19208 distinct arrangements.
+
+# You glance back down at your bag and try to remember why you brought so many
+# adapters; there must be more than a trillion valid ways to arrange them!
+# Surely, there must be an efficient way to count the arrangements.
+
+# What is the total number of distinct ways you can arrange the adapters to
+# connect the charging outlet to your device?
+
+
+@lru_cache(maxsize=256)
+def recursion(node: int) -> int:
+    if node == len(joltages) - 1:
+        return 1
+    # consider last three nodes (or the end of the list, whichever is first),
+    # and for any that are valid jumps (value is no more than three more),
+    # find the number of paths from that node to the end.
+    return sum(
+        [
+            recursion(current_node)
+            for current_node in range(node + 1, min(node + 4, len(joltages)))
+            if joltages[current_node] - joltages[node] <= 3
+        ]
+    )
+
+
+def part_2() -> None:
+    total_num = recursion(0)
+    print(f"The total number of distint ways is {total_num}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P11.py

@@ -0,0 +1,367 @@
+# --- Day 11: Seating System ---
+
+# Your plane lands with plenty of time to spare. The final leg of your journey
+# is a ferry that goes directly to the tropical island where you can finally
+# start your vacation. As you reach the waiting area to board the ferry, you
+# realize you're so early, nobody else has even arrived yet!
+
+# By modeling the process people use to choose (or abandon) their seat in the
+# waiting area, you're pretty sure you can predict the best place to sit. You
+# make a quick map of the seat layout (your puzzle input).
+
+# The seat layout fits neatly on a grid. Each position is either floor (.), an
+# empty seat (L), or an occupied seat (#). For example, the initial seat layout
+# might look like this:
+
+# L.LL.LL.LL
+# LLLLLLL.LL
+# L.L.L..L..
+# LLLL.LL.LL
+# L.LL.LL.LL
+# L.LLLLL.LL
+# ..L.L.....
+# LLLLLLLLLL
+# L.LLLLLL.L
+# L.LLLLL.LL
+
+# Now, you just need to model the people who will be arriving shortly.
+# Fortunately, people are entirely predictable and always follow a simple set
+# of rules. All decisions are based on the number of occupied seats adjacent to
+# a given seat (one of the eight positions immediately up, down, left, right,
+# or diagonal from the seat). The following rules are applied to every seat
+# simultaneously:
+
+#     If a seat is empty (L) and there are no occupied seats adjacent to it,
+# the seat becomes occupied.
+#     If a seat is occupied (#) and four or more seats adjacent to it are also
+# occupied, the seat becomes empty.
+#     Otherwise, the seat's state does not change.
+
+# Floor (.) never changes; seats don't move, and nobody sits on the floor.
+
+# After one round of these rules, every seat in the example layout becomes
+# occupied:
+
+# #.##.##.##
+# #######.##
+# #.#.#..#..
+# ####.##.##
+# #.##.##.##
+# #.#####.##
+# ..#.#.....
+# ##########
+# #.######.#
+# #.#####.##
+
+# After a second round, the seats with four or more occupied adjacent seats
+# become empty again:
+
+# #.LL.L#.##
+# #LLLLLL.L#
+# L.L.L..L..
+# #LLL.LL.L#
+# #.LL.LL.LL
+# #.LLLL#.##
+# ..L.L.....
+# #LLLLLLLL#
+# #.LLLLLL.L
+# #.#LLLL.##
+
+# This process continues for three more rounds:
+
+# #.##.L#.##
+# #L###LL.L#
+# L.#.#..#..
+# #L##.##.L#
+# #.##.LL.LL
+# #.###L#.##
+# ..#.#.....
+# #L######L#
+# #.LL###L.L
+# #.#L###.##
+
+# #.#L.L#.##
+# #LLL#LL.L#
+# L.L.L..#..
+# #LLL.##.L#
+# #.LL.LL.LL
+# #.LL#L#.##
+# ..L.L.....
+# #L#LLLL#L#
+# #.LLLLLL.L
+# #.#L#L#.##
+
+# #.#L.L#.##
+# #LLL#LL.L#
+# L.#.L..#..
+# #L##.##.L#
+# #.#L.LL.LL
+# #.#L#L#.##
+# ..L.L.....
+# #L#L##L#L#
+# #.LLLLLL.L
+# #.#L#L#.##
+
+# At this point, something interesting happens: the chaos stabilizes and
+# further applications of these rules cause no seats to change state! Once
+# people stop moving around, you count 37 occupied seats.
+
+# Simulate your seating area by applying the seating rules repeatedly until no
+# seats change state. How many seats end up occupied?
+
+with open("files/P11.txt", "r") as f:
+    seats = [line for line in f.read().strip().split("\n")]
+
+
+def seats_around(seats: list[str], r: int, c: int) -> int:
+    total = 0
+    around_me = [
+        (-1, -1),
+        (-1, 0),
+        (-1, 1),
+        (0, -1),
+        (0, 1),
+        (1, -1),
+        (1, 0),
+        (1, 1),
+    ]
+    for row, col in around_me:
+        _row = r + row
+        _col = c + col
+        if (
+            _row >= 0
+            and _row < len(seats)
+            and _col >= 0
+            and _col < len(seats[c])
+        ):
+            total += seats[_row][_col] == "#"
+    return total
+
+
+def free_seats(v: list[str]) -> int:
+    total = 0
+    for row in v:
+        total += row.count("#")
+    return total
+
+
+def part_1(seats: list[str]) -> None:
+    R = len(seats)
+    C = len(seats[0])
+    while True:
+        has_changed = False
+        next_iter = []
+        for r in range(R):
+            new_row = ""
+            for c in range(C):
+                seat = seats[r][c]
+                if seat != ".":
+                    occupants = seats_around(seats, r, c)
+                    if seat == "L" and occupants == 0:
+                        seat = "#"
+                        has_changed = True
+                    elif seat == "#" and occupants >= 4:
+                        seat = "L"
+                        has_changed = True
+                new_row += seat
+            next_iter.append(new_row)
+        if not has_changed:
+            break
+        seats = next_iter
+
+    print(f"There are {free_seats(seats)} seats occupied")
+
+
+# --- Part Two ---
+
+# As soon as people start to arrive, you realize your mistake. People don't
+# just care about adjacent seats - they care about the first seat they can see
+# in each of those eight directions!
+
+# Now, instead of considering just the eight immediately adjacent seats,
+# consider the first seat in each of those eight directions. For example, the
+# empty seat below would see eight occupied seats:
+
+# .......#.
+# ...#.....
+# .#.......
+# .........
+# ..#L....#
+# ....#....
+# .........
+# #........
+# ...#.....
+
+# The leftmost empty seat below would only see one empty seat, but cannot see
+# any of the occupied ones:
+
+# .............
+# .L.L.#.#.#.#.
+# .............
+
+# The empty seat below would see no occupied seats:
+
+# .##.##.
+# #.#.#.#
+# ##...##
+# ...L...
+# ##...##
+# #.#.#.#
+# .##.##.
+
+# Also, people seem to be more tolerant than you expected: it now takes five or
+# more visible occupied seats for an occupied seat to become empty (rather than
+# four or more from the previous rules). The other rules still apply: empty
+# seats that see no occupied seats become occupied, seats matching no rule
+# don't change, and floor never changes.
+
+# Given the same starting layout as above, these new rules cause the seating
+# area to shift around as follows:
+
+# L.LL.LL.LL
+# LLLLLLL.LL
+# L.L.L..L..
+# LLLL.LL.LL
+# L.LL.LL.LL
+# L.LLLLL.LL
+# ..L.L.....
+# LLLLLLLLLL
+# L.LLLLLL.L
+# L.LLLLL.LL
+
+# #.##.##.##
+# #######.##
+# #.#.#..#..
+# ####.##.##
+# #.##.##.##
+# #.#####.##
+# ..#.#.....
+# ##########
+# #.######.#
+# #.#####.##
+
+# #.LL.LL.L#
+# #LLLLLL.LL
+# L.L.L..L..
+# LLLL.LL.LL
+# L.LL.LL.LL
+# L.LLLLL.LL
+# ..L.L.....
+# LLLLLLLLL#
+# #.LLLLLL.L
+# #.LLLLL.L#
+
+# #.L#.##.L#
+# #L#####.LL
+# L.#.#..#..
+# ##L#.##.##
+# #.##.#L.##
+# #.#####.#L
+# ..#.#.....
+# LLL####LL#
+# #.L#####.L
+# #.L####.L#
+
+# #.L#.L#.L#
+# #LLLLLL.LL
+# L.L.L..#..
+# ##LL.LL.L#
+# L.LL.LL.L#
+# #.LLLLL.LL
+# ..L.L.....
+# LLLLLLLLL#
+# #.LLLLL#.L
+# #.L#LL#.L#
+
+# #.L#.L#.L#
+# #LLLLLL.LL
+# L.L.L..#..
+# ##L#.#L.L#
+# L.L#.#L.L#
+# #.L####.LL
+# ..#.#.....
+# LLL###LLL#
+# #.LLLLL#.L
+# #.L#LL#.L#
+
+# #.L#.L#.L#
+# #LLLLLL.LL
+# L.L.L..#..
+# ##L#.#L.L#
+# L.L#.LL.L#
+# #.LLLL#.LL
+# ..#.L.....
+# LLL###LLL#
+# #.LLLLL#.L
+# #.L#LL#.L#
+
+# Again, at this point, people stop shifting around and the seating area
+# reaches equilibrium. Once this occurs, you count 26 occupied seats.
+
+# Given the new visibility method and the rule change for occupied seats
+# becoming empty, once equilibrium is reached, how many seats end up occupied?
+
+
+def seats_around_mod(seats: list[str], r: int, c: int) -> int:
+    total = 0
+    around_me = [
+        (-1, -1),
+        (-1, 0),
+        (-1, 1),
+        (0, -1),
+        (0, 1),
+        (1, -1),
+        (1, 0),
+        (1, 1),
+    ]
+    for row, col in around_me:
+        _row = r + row
+        _col = c + col
+        while (
+            _row >= 0
+            and _row < len(seats)
+            and _col < len(seats[c])
+            and seats[_row][_col] == "."
+        ):
+            _row += row
+            _col += col
+        if (
+            _row >= 0
+            and _row < len(seats)
+            and _col >= 0
+            and _col < len(seats[c])
+        ):
+            total += seats[_row][_col] == "#"
+    return total
+
+
+def part_2(seats: list[str]) -> None:
+    R = len(seats)
+    C = len(seats[0])
+    while True:
+        has_changed = False
+        next_iter = []
+        for r in range(R):
+            new_row = ""
+            for c in range(C):
+                seat = seats[r][c]
+                if seat != ".":
+                    occupants = seats_around_mod(seats, r, c)
+                    if seat == "L" and occupants == 0:
+                        seat = "#"
+                        has_changed = True
+                    elif seat == "#" and occupants >= 5:
+                        seat = "L"
+                        has_changed = True
+                new_row += seat
+            next_iter.append(new_row)
+        if not has_changed:
+            break
+        seats = next_iter
+
+    print(f"There are {free_seats(seats)} seats occupied")
+
+
+if __name__ == "__main__":
+    part_1(seats)
+    part_2(seats)

src/Year_2020/P12.py

@@ -0,0 +1,185 @@
+import math
+
+# --- Day 12: Rain Risk ---
+
+# Your ferry made decent progress toward the island, but the storm came in
+# faster than anyone expected. The ferry needs to take evasive actions!
+
+# Unfortunately, the ship's navigation computer seems to be malfunctioning;
+# rather than giving a route directly to safety, it produced extremely
+# circuitous instructions. When the captain uses the PA system to ask if anyone
+# can help, you quickly volunteer.
+
+# The navigation instructions (your puzzle input) consists of a sequence of
+# single-character actions paired with integer input values. After staring at
+# them for a few minutes, you work out what they probably mean:
+
+#     Action N means to move north by the given value.
+#     Action S means to move south by the given value.
+#     Action E means to move east by the given value.
+#     Action W means to move west by the given value.
+#     Action L means to turn left the given number of degrees.
+#     Action R means to turn right the given number of degrees.
+#     Action F means to move forward by the given value in the direction the ship is currently facing.
+
+# The ship starts by facing east. Only the L and R actions change the direction
+# the ship is facing. (That is, if the ship is facing east and the next
+# instruction is N10, the ship would move north 10 units, but would still move
+# east if the following action were F.)
+
+# For example:
+
+# F10
+# N3
+# F7
+# R90
+# F11
+
+# These instructions would be handled as follows:
+
+#     F10 would move the ship 10 units east (because the ship starts by facing
+# east) to east 10, north 0.
+#     N3 would move the ship 3 units north to east 10, north 3.
+#     F7 would move the ship another 7 units east (because the ship is still
+# facing east) to east 17, north 3.
+#     R90 would cause the ship to turn right by 90 degrees and face south; it
+# remains at east 17, north 3.
+#     F11 would move the ship 11 units south to east 17, south 8.
+
+# At the end of these instructions, the ship's Manhattan distance (sum of the
+# absolute values of its east/west position and its north/south position) from
+# its starting position is 17 + 8 = 25.
+
+# Figure out where the navigation instructions lead. What is the Manhattan
+# distance between that location and the ship's starting position?
+
+with open("files/P12.txt", "r") as f:
+    instructions = [line for line in f.read().strip().split("\n")]
+
+
+def part_1() -> None:
+    # dirs are E, S, W, and N
+    dirs = [[1, 0], [0, -1], [-1, 0], [0, 1]]
+    current_pos = [0, 0]
+    current_dir = 0
+
+    for instr in instructions:
+        _dir, _num = instr[:1], instr[1:]
+        if _dir == "E":
+            current_pos[0] += int(_num)
+        elif _dir == "W":
+            current_pos[0] -= int(_num)
+        elif _dir == "N":
+            current_pos[1] += int(_num)
+        elif _dir == "S":
+            current_pos[1] -= int(_num)
+        elif _dir == "F":
+            current_pos[0] += int(_num) * dirs[current_dir][0]
+            current_pos[1] += int(_num) * dirs[current_dir][1]
+        else:
+            if _dir == "R":
+                # rotations are clockwise
+                current_dir += int(_num) // 90
+            else:
+                current_dir -= int(_num) // 90
+            # avoid rotation overflow, just in case
+            current_dir %= 4
+
+    manhattan_distance = abs(current_pos[0]) + abs(current_pos[1])
+    print(f"The Manhattan distance is {manhattan_distance}")
+
+
+# --- Part Two ---
+
+# Before you can give the destination to the captain, you realize that the
+# actual action meanings were printed on the back of the instructions the whole
+# time.
+
+# Almost all of the actions indicate how to move a waypoint which is relative
+# to the ship's position:
+
+#     Action N means to move the waypoint north by the given value.
+#     Action S means to move the waypoint south by the given value.
+#     Action E means to move the waypoint east by the given value.
+#     Action W means to move the waypoint west by the given value.
+#     Action L means to rotate the waypoint around the ship left
+# (counter-clockwise) the given number of degrees.
+#     Action R means to rotate the waypoint around the ship right (clockwise)
+# the given number of degrees.
+#     Action F means to move forward to the waypoint a number of times equal to
+# the given value.
+
+# The waypoint starts 10 units east and 1 unit north relative to the ship. The
+# waypoint is relative to the ship; that is, if the ship moves, the waypoint
+# moves with it.
+
+# For example, using the same instructions as above:
+
+#     F10 moves the ship to the waypoint 10 times (a total of 100 units east
+# and 10 units north), leaving the ship at east 100, north 10. The waypoint
+# stays 10 units east and 1 unit north of the ship.
+#     N3 moves the waypoint 3 units north to 10 units east and 4 units north of
+# the ship. The ship remains at east 100, north 10.
+#     F7 moves the ship to the waypoint 7 times (a total of 70 units east and
+# 28 units north), leaving the ship at east 170, north 38. The waypoint stays
+# 10 units east and 4 units north of the ship.
+#     R90 rotates the waypoint around the ship clockwise 90 degrees, moving it
+# to 4 units east and 10 units south of the ship. The ship remains at east 170,
+# north 38.
+#     F11 moves the ship to the waypoint 11 times (a total of 44 units east and
+# 110 units south), leaving the ship at east 214, south 72. The waypoint stays
+# 4 units east and 10 units south of the ship.
+
+# After these operations, the ship's Manhattan distance from its starting
+# position is 214 + 72 = 286.
+
+# Figure out where the navigation instructions actually lead. What is the
+# Manhattan distance between that location and the ship's starting position?
+
+
+def part_2() -> None:
+
+    ship_pos = [0, 0]
+    waypoint_pos = [10, 1]
+    waypoint_dir = 0
+
+    for instr in instructions:
+        _dir, _num = instr[:1], instr[1:]
+        if _dir == "E":
+            waypoint_pos[0] += int(_num)
+        elif _dir == "W":
+            waypoint_pos[0] -= int(_num)
+        elif _dir == "N":
+            waypoint_pos[1] += int(_num)
+        elif _dir == "S":
+            waypoint_pos[1] -= int(_num)
+        elif _dir == "F":
+            ship_pos[0] += int(_num) * waypoint_pos[0]
+            ship_pos[1] += int(_num) * waypoint_pos[1]
+        else:
+            if _dir == "R":
+                # rotations are clockwise
+                _num = str(360 - int(_num))
+            waypoint_dir += int(_num)
+            # avoid rotation overflow
+            waypoint_dir %= 360
+            # convert to radians to make life easier
+            radians = math.radians(int(_num))
+            waypoint_pos = [
+                round(
+                    (waypoint_pos[0] * math.cos(radians))
+                    - (waypoint_pos[1] * math.sin(radians))
+                ),
+                round(
+                    (waypoint_pos[0] * math.sin(radians))
+                    + (waypoint_pos[1] * math.cos(radians))
+                ),
+            ]
+
+    manhattan_distance = abs(ship_pos[0]) + abs(ship_pos[1])
+    print(f"The Manhattan distance is {manhattan_distance}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P13.py

@@ -0,0 +1,209 @@
+# --- Day 13: Shuttle Search ---
+
+# Your ferry can make it safely to a nearby port, but it won't get much
+# further. When you call to book another ship, you discover that no ships
+# embark from that port to your vacation island. You'll need to get from the
+# port to the nearest airport.
+
+# Fortunately, a shuttle bus service is available to bring you from the sea
+# port to the airport! Each bus has an ID number that also indicates how often
+# the bus leaves for the airport.
+
+# Bus schedules are defined based on a timestamp that measures the number of
+# minutes since some fixed reference point in the past. At timestamp 0, every
+# bus simultaneously departed from the sea port. After that, each bus travels
+# to the airport, then various other locations, and finally returns to the sea
+# port to repeat its journey forever.
+
+# The time this loop takes a particular bus is also its ID number: the bus with
+# ID 5 departs from the sea port at timestamps 0, 5, 10, 15, and so on. The bus
+# with ID 11 departs at 0, 11, 22, 33, and so on. If you are there when the bus
+# departs, you can ride that bus to the airport!
+
+# Your notes (your puzzle input) consist of two lines. The first line is your
+# estimate of the earliest timestamp you could depart on a bus. The second line
+# lists the bus IDs that are in service according to the shuttle company;
+# entries that show x must be out of service, so you decide to ignore them.
+
+# To save time once you arrive, your goal is to figure out the earliest bus you
+# can take to the airport. (There will be exactly one such bus.)
+
+# For example, suppose you have the following notes:
+
+# 939
+# 7,13,x,x,59,x,31,19
+
+# Here, the earliest timestamp you could depart is 939, and the bus IDs in
+# service are 7, 13, 59, 31, and 19. Near timestamp 939, these bus IDs depart
+# at the times marked D:
+
+# time   bus 7   bus 13  bus 59  bus 31  bus 19
+# 929      .       .       .       .       .
+# 930      .       .       .       D       .
+# 931      D       .       .       .       D
+# 932      .       .       .       .       .
+# 933      .       .       .       .       .
+# 934      .       .       .       .       .
+# 935      .       .       .       .       .
+# 936      .       D       .       .       .
+# 937      .       .       .       .       .
+# 938      D       .       .       .       .
+# 939      .       .       .       .       .
+# 940      .       .       .       .       .
+# 941      .       .       .       .       .
+# 942      .       .       .       .       .
+# 943      .       .       .       .       .
+# 944      .       .       D       .       .
+# 945      D       .       .       .       .
+# 946      .       .       .       .       .
+# 947      .       .       .       .       .
+# 948      .       .       .       .       .
+# 949      .       D       .       .       .
+
+# The earliest bus you could take is bus ID 59. It doesn't depart until
+# timestamp 944, so you would need to wait 944 - 939 = 5 minutes before it
+# departs. Multiplying the bus ID by the number of minutes you'd need to wait
+# gives 295.
+
+# What is the ID of the earliest bus you can take to the airport multiplied by
+# the number of minutes you'll need to wait for that bus?
+
+from copy import copy
+from math import gcd
+
+with open("files/P13.txt", "r") as f:
+    start = int(f.readline())
+    buses_x = [bus for bus in f.readline().strip().split(",")]
+    buses = [int(bus) for bus in buses_x if bus != "x"]
+
+
+def part_1() -> None:
+    wait = copy(start)
+    found = False
+    while not found:
+        for bus in buses:
+            if wait % bus == 0:
+                print(f"Solution to part 1 is {str((wait - start) * bus)}")
+                found = True
+                break
+            wait += 1
+
+
+# --- Part Two ---
+
+# The shuttle company is running a contest: one gold coin for anyone that can
+# find the earliest timestamp such that the first bus ID departs at that time
+# and each subsequent listed bus ID departs at that subsequent minute. (The
+# first line in your input is no longer relevant.)
+
+# For example, suppose you have the same list of bus IDs as above:
+
+# 7,13,x,x,59,x,31,19
+
+# An x in the schedule means there are no constraints on what bus IDs must
+# depart at that time.
+
+# This means you are looking for the earliest timestamp (called t) such that:
+
+#     Bus ID 7 departs at timestamp t.
+#     Bus ID 13 departs one minute after timestamp t.
+#     There are no requirements or restrictions on departures at two or three
+# minutes after timestamp t.
+#     Bus ID 59 departs four minutes after timestamp t.
+#     There are no requirements or restrictions on departures at five minutes
+# after timestamp t.
+#     Bus ID 31 departs six minutes after timestamp t.
+#     Bus ID 19 departs seven minutes after timestamp t.
+
+# The only bus departures that matter are the listed bus IDs at their specific
+# offsets from t. Those bus IDs can depart at other times, and other bus IDs
+# can depart at those times. For example, in the list above, because bus ID 19
+# must depart seven minutes after the timestamp at which bus ID 7 departs, bus
+# ID 7 will always also be departing with bus ID 19 at seven minutes after
+# timestamp t.
+
+# In this example, the earliest timestamp at which this occurs is 1068781:
+
+# time     bus 7   bus 13  bus 59  bus 31  bus 19
+# 1068773    .       .       .       .       .
+# 1068774    D       .       .       .       .
+# 1068775    .       .       .       .       .
+# 1068776    .       .       .       .       .
+# 1068777    .       .       .       .       .
+# 1068778    .       .       .       .       .
+# 1068779    .       .       .       .       .
+# 1068780    .       .       .       .       .
+# 1068781    D       .       .       .       .
+# 1068782    .       D       .       .       .
+# 1068783    .       .       .       .       .
+# 1068784    .       .       .       .       .
+# 1068785    .       .       D       .       .
+# 1068786    .       .       .       .       .
+# 1068787    .       .       .       D       .
+# 1068788    D       .       .       .       D
+# 1068789    .       .       .       .       .
+# 1068790    .       .       .       .       .
+# 1068791    .       .       .       .       .
+# 1068792    .       .       .       .       .
+# 1068793    .       .       .       .       .
+# 1068794    .       .       .       .       .
+# 1068795    D       D       .       .       .
+# 1068796    .       .       .       .       .
+# 1068797    .       .       .       .       .
+
+# In the above example, bus ID 7 departs at timestamp 1068788 (seven minutes
+# after t). This is fine; the only requirement on that minute is that bus ID 19
+# departs then, and it does.
+
+# Here are some other examples:
+
+#     The earliest timestamp that matches the list 17,x,13,19 is 3417.
+#     67,7,59,61 first occurs at timestamp 754018.
+#     67,x,7,59,61 first occurs at timestamp 779210.
+#     67,7,x,59,61 first occurs at timestamp 1261476.
+#     1789,37,47,1889 first occurs at timestamp 1202161486.
+
+# However, with so many bus IDs in your list, surely the actual earliest
+# timestamp will be larger than 100000000000000!
+
+# What is the earliest timestamp such that all of the listed bus IDs depart at
+# offsets matching their positions in the list?
+
+
+# adapted from https://stackoverflow.com/questions/32728526/lcm-using-recursive
+def lcm(my_list: list[list[int]]) -> int:
+    lcm = my_list[0][0]
+    for pair in my_list[1:]:
+        lcm = lcm * pair[0] // gcd(lcm, pair[0])
+    return lcm
+
+
+def part_2() -> None:
+    my_buses = []
+    for idx, bus in enumerate(buses_x):
+        if bus != "x":
+            my_buses.append([int(bus), idx])
+
+    value = 0
+    add_to = 1
+    for i in range(1, len(my_buses)):
+        good = False
+        while not good:
+            good = True
+            for j in range(i + 1):
+                if (value + my_buses[j][1]) % my_buses[j][0] != 0:
+                    good = False
+                    break
+            if good is True:
+                # find the offset between buses
+                add_to = lcm(my_buses[:j])
+                break
+            else:
+                value += add_to
+
+    print(f"Solution to part 2 is {value}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P14.py

@@ -0,0 +1,199 @@
+# --- Day 14: Docking Data ---
+
+# As your ferry approaches the sea port, the captain asks for your help again.
+# The computer system that runs this port isn't compatible with the docking
+# program on the ferry, so the docking parameters aren't being correctly
+# initialized in the docking program's memory.
+
+# After a brief inspection, you discover that the sea port's computer system
+# uses a strange bitmask system in its initialization program. Although you
+# don't have the correct decoder chip handy, you can emulate it in software!
+
+# The initialization program (your puzzle input) can either update the bitmask
+# or write a value to memory. Values and memory addresses are both 36-bit
+# unsigned integers. For example, ignoring bitmasks for a moment, a line like
+# mem[8] = 11 would write the value 11 to memory address 8.
+
+# The bitmask is always given as a string of 36 bits, written with the most
+# significant bit (representing 2^35) on the left and the least significant bit
+# (2^0, that is, the 1s bit) on the right. The current bitmask is applied to
+# values immediately before they are written to memory: a 0 or 1 overwrites the
+# corresponding bit in the value, while an X leaves the bit in the value
+# unchanged.
+
+# For example, consider the following program:
+
+# mask = XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
+# mem[8] = 11
+# mem[7] = 101
+# mem[8] = 0
+
+# This program starts by specifying a bitmask (mask = ....). The mask it
+# specifies will overwrite two bits in every written value: the 2s bit is
+# overwritten with 0, and the 64s bit is overwritten with 1.
+
+# The program then attempts to write the value 11 to memory address 8. By
+# expanding everything out to individual bits, the mask is applied as follows:
+
+# value:  000000000000000000000000000000001011  (decimal 11)
+# mask:   XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
+# result: 000000000000000000000000000001001001  (decimal 73)
+
+# So, because of the mask, the value 73 is written to memory address 8 instead.
+# Then, the program tries to write 101 to address 7:
+
+# value:  000000000000000000000000000001100101  (decimal 101)
+# mask:   XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
+# result: 000000000000000000000000000001100101  (decimal 101)
+
+# This time, the mask has no effect, as the bits it overwrote were already the
+# values the mask tried to set. Finally, the program tries to write 0 to
+# address 8:
+
+# value:  000000000000000000000000000000000000  (decimal 0)
+# mask:   XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
+# result: 000000000000000000000000000001000000  (decimal 64)
+
+# 64 is written to address 8 instead, overwriting the value that was there
+# previously.
+
+# To initialize your ferry's docking program, you need the sum of all values
+# left in memory after the initialization program completes. (The entire 36-bit
+# address space begins initialized to the value 0 at every address.) In the
+# above example, only two values in memory are not zero - 101 (at address 7)
+# and 64 (at address 8) - producing a sum of 165.
+
+# Execute the initialization program. What is the sum of all values left in
+# memory after it completes? (Do not truncate the sum to 36 bits.)
+
+with open("files/P14.txt", "r") as f:
+    instructions = [line for line in f.read().strip().split("\n")]
+
+
+def part_1() -> None:
+    memory = {}
+    for instruction in instructions:
+        if instruction.startswith("mask"):
+            mask = instruction.split(" = ")[1]
+            mask_or = int(mask.replace("X", "0"), 2)
+            mask_and = int(mask.replace("X", "1"), 2)
+        elif instruction.startswith("mem"):
+            idx = int(instruction.split("[")[1].split("]")[0])
+            val = int(instruction.split(" = ")[1])
+            memory[idx] = (val & mask_and) | mask_or
+
+    values_in_memory = sum([val for val in memory.values()])
+    print(f"The sum of all values in memory is {values_in_memory}")
+
+
+# --- Part Two ---
+
+# For some reason, the sea port's computer system still can't communicate with
+# your ferry's docking program. It must be using version 2 of the decoder chip!
+
+# A version 2 decoder chip doesn't modify the values being written at all.
+# Instead, it acts as a memory address decoder. Immediately before a value is
+# written to memory, each bit in the bitmask modifies the corresponding bit of
+# the destination memory address in the following way:
+
+#     If the bitmask bit is 0, the corresponding memory address bit is
+# unchanged.
+#     If the bitmask bit is 1, the corresponding memory address bit is
+# overwritten with 1.
+#     If the bitmask bit is X, the corresponding memory address bit is
+# floating.
+
+# A floating bit is not connected to anything and instead fluctuates
+# unpredictably. In practice, this means the floating bits will take on all
+# possible values, potentially causing many memory addresses to be written all
+# at once!
+
+# For example, consider the following program:
+
+# mask = 000000000000000000000000000000X1001X
+# mem[42] = 100
+# mask = 00000000000000000000000000000000X0XX
+# mem[26] = 1
+
+# When this program goes to write to memory address 42, it first applies the
+# bitmask:
+
+# address: 000000000000000000000000000000101010  (decimal 42)
+# mask:    000000000000000000000000000000X1001X
+# result:  000000000000000000000000000000X1101X
+
+# After applying the mask, four bits are overwritten, three of which are
+# different, and two of which are floating. Floating bits take on every
+# possible combination of values; with two floating bits, four actual memory
+# addresses are written:
+
+# 000000000000000000000000000000011010  (decimal 26)
+# 000000000000000000000000000000011011  (decimal 27)
+# 000000000000000000000000000000111010  (decimal 58)
+# 000000000000000000000000000000111011  (decimal 59)
+
+# Next, the program is about to write to memory address 26 with a different
+# bitmask:
+
+# address: 000000000000000000000000000000011010  (decimal 26)
+# mask:    00000000000000000000000000000000X0XX
+# result:  00000000000000000000000000000001X0XX
+
+# This results in an address with three floating bits, causing writes to eight
+# memory addresses:
+
+# 000000000000000000000000000000010000  (decimal 16)
+# 000000000000000000000000000000010001  (decimal 17)
+# 000000000000000000000000000000010010  (decimal 18)
+# 000000000000000000000000000000010011  (decimal 19)
+# 000000000000000000000000000000011000  (decimal 24)
+# 000000000000000000000000000000011001  (decimal 25)
+# 000000000000000000000000000000011010  (decimal 26)
+# 000000000000000000000000000000011011  (decimal 27)
+
+# The entire 36-bit address space still begins initialized to the value 0 at
+# every address, and you still need the sum of all values left in memory at the
+# end of the program. In this example, the sum is 208.
+
+# Execute the initialization program using an emulator for a version 2 decoder
+# chip. What is the sum of all values left in memory after it completes?
+
+
+def get_indexes(mask: str) -> list[int]:
+    try:
+        firstx = mask.index("X")
+        return get_indexes(
+            mask[:firstx] + "0" + mask[firstx + 1 :]
+        ) + get_indexes(mask[:firstx] + "1" + mask[firstx + 1 :])
+    except ValueError:
+        return [int(mask, 2)]
+
+
+def part_2() -> None:
+    memory = {}
+    for instruction in instructions:
+        if instruction.startswith("mask"):
+            mask = instruction.split(" ")[2]
+        elif instruction.startswith("mem"):
+            idx = int(instruction.split("[")[1].split("]")[0])
+            val = int(instruction.split(" ")[2])
+            idx_mask = ""
+            # pad with 0s (up to 36) the index
+            for m, i in zip(mask, f"{idx:036b}"):
+                if m == "0":
+                    # do not change the bit value
+                    idx_mask += i
+                elif m == "1" or m == "X":
+                    idx_mask += m
+
+            idxs = get_indexes(idx_mask)
+            for idx in idxs:
+                memory[idx] = val
+
+    values_in_memory = sum([val for val in memory.values()])
+    print(f"The sum of all values in memory is {values_in_memory}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P15.py

@@ -0,0 +1,107 @@
+# --- Day 15: Rambunctious Recitation ---
+
+# You catch the airport shuttle and try to book a new flight to your vacation
+# island. Due to the storm, all direct flights have been cancelled, but a route
+# is available to get around the storm. You take it.
+
+# While you wait for your flight, you decide to check in with the Elves back at
+# the North Pole. They're playing a memory game and are ever so excited to
+# explain the rules!
+
+# In this game, the players take turns saying numbers. They begin by taking
+# turns reading from a list of starting numbers (your puzzle input). Then, each
+# turn consists of considering the most recently spoken number:
+
+#     If that was the first time the number has been spoken, the current player
+# says 0.
+#     Otherwise, the number had been spoken before; the current player
+# announces how many turns apart the number is from when it was previously
+# spoken.
+
+# So, after the starting numbers, each turn results in that player speaking
+# aloud either 0 (if the last number is new) or an age (if the last number is a
+# repeat).
+
+# For example, suppose the starting numbers are 0,3,6:
+
+#     Turn 1: The 1st number spoken is a starting number, 0.
+#     Turn 2: The 2nd number spoken is a starting number, 3.
+#     Turn 3: The 3rd number spoken is a starting number, 6.
+#     Turn 4: Now, consider the last number spoken, 6. Since that was the first
+# time the number had been spoken, the 4th number spoken is 0.
+#     Turn 5: Next, again consider the last number spoken, 0. Since it had been
+# spoken before, the next number to speak is the difference between the turn
+# number when it was last spoken (the previous turn, 4) and the turn number of
+# the time it was most recently spoken before then (turn 1). Thus, the 5th
+# number spoken is 4 - 1, 3.
+#     Turn 6: The last number spoken, 3 had also been spoken before, most
+# recently on turns 5 and 2. So, the 6th number spoken is 5 - 2, 3.
+#     Turn 7: Since 3 was just spoken twice in a row, and the last two turns
+# are 1 turn apart, the 7th number spoken is 1.
+#     Turn 8: Since 1 is new, the 8th number spoken is 0.
+#     Turn 9: 0 was last spoken on turns 8 and 4, so the 9th number spoken is
+# the difference between them, 4.
+#     Turn 10: 4 is new, so the 10th number spoken is 0.
+
+# (The game ends when the Elves get sick of playing or dinner is ready,
+# whichever comes first.)
+
+# Their question for you is: what will be the 2020th number spoken? In the
+# example above, the 2020th number spoken will be 436.
+
+# Here are a few more examples:
+
+#     Given the starting numbers 1,3,2, the 2020th number spoken is 1.
+#     Given the starting numbers 2,1,3, the 2020th number spoken is 10.
+#     Given the starting numbers 1,2,3, the 2020th number spoken is 27.
+#     Given the starting numbers 2,3,1, the 2020th number spoken is 78.
+#     Given the starting numbers 3,2,1, the 2020th number spoken is 438.
+#     Given the starting numbers 3,1,2, the 2020th number spoken is 1836.
+
+# Given your starting numbers, what will be the 2020th number spoken?
+
+from collections import defaultdict
+
+with open("files/P15.txt", "r") as f:
+    sequence = [int(val) for val in f.read().strip().split(",")]
+
+
+numbers = defaultdict(list)
+for idx, number in enumerate(sequence):
+    numbers[number].append(idx)
+
+
+def part_1_and_2(num: int) -> None:
+    last = 0  # list(numbers.keys())[-1].copy()
+    for i in range(len(sequence), num):
+        if len(numbers[last]) < 2:
+            numbers[0].append(i)
+            last = 0
+        else:
+            last = numbers[last][-1] - numbers[last][-2]
+            numbers[last].append(i)
+        # because of indexing starts at 0
+        if i == 2019:
+            print(f"The {i+1}th number spoken is {last}")
+
+    print(f"The {num}th number spoken is {last}")
+
+
+# --- Part Two ---
+
+# Impressed, the Elves issue you a challenge: determine the 30000000th number
+# spoken. For example, given the same starting numbers as above:
+
+#     Given 0,3,6, the 30000000th number spoken is 175594.
+#     Given 1,3,2, the 30000000th number spoken is 2578.
+#     Given 2,1,3, the 30000000th number spoken is 3544142.
+#     Given 1,2,3, the 30000000th number spoken is 261214.
+#     Given 2,3,1, the 30000000th number spoken is 6895259.
+#     Given 3,2,1, the 30000000th number spoken is 18.
+#     Given 3,1,2, the 30000000th number spoken is 362.
+
+# Given your starting numbers, what will be the 30000000th number spoken?
+
+
+if __name__ == "__main__":
+    part_1_and_2(30_000_000)

src/Year_2020/P16.py

@@ -0,0 +1,245 @@
+# --- Day 16: Ticket Translation ---
+
+# As you're walking to yet another connecting flight, you realize that one of
+# the legs of your re-routed trip coming up is on a high-speed train. However,
+# the train ticket you were given is in a language you don't understand. You
+# should probably figure out what it says before you get to the train station
+# after the next flight.
+
+# Unfortunately, you can't actually read the words on the ticket. You can,
+# however, read the numbers, and so you figure out the fields these tickets
+# must have and the valid ranges for values in those fields.
+
+# You collect the rules for ticket fields, the numbers on your ticket, and the
+# numbers on other nearby tickets for the same train service (via the airport
+# security cameras) together into a single document you can reference (your
+# puzzle input).
+
+# The rules for ticket fields specify a list of fields that exist somewhere on
+# the ticket and the valid ranges of values for each field. For example, a rule
+# like class: 1-3 or 5-7 means that one of the fields in every ticket is named
+# class and can be any value in the ranges 1-3 or 5-7 (inclusive, such that 3
+# and 5 are both valid in this field, but 4 is not).
+
+# Each ticket is represented by a single line of comma-separated values. The
+# values are the numbers on the ticket in the order they appear; every ticket
+# has the same format. For example, consider this ticket:
+
+# .--------------------------------------------------------.
+# | ????: 101    ?????: 102   ??????????: 103     ???: 104 |
+# |                                                        |
+# | ??: 301  ??: 302             ???????: 303      ??????? |
+# | ??: 401  ??: 402           ???? ????: 403    ????????? |
+# '--------------------------------------------------------'
+
+# Here, ? represents text in a language you don't understand. This ticket might
+# be represented as 101,102,103,104,301,302,303,401,402,403; of course, the
+# actual train tickets you're looking at are much more complicated. In any
+# case, you've extracted just the numbers in such a way that the first number
+# is always the same specific field, the second number is always a different
+# specific field, and so on - you just don't know what each position actually
+# means!
+
+# Start by determining which tickets are completely invalid; these are tickets
+# that contain values which aren't valid for any field. Ignore your ticket for
+# now.
+
+# For example, suppose you have the following notes:
+
+# class: 1-3 or 5-7
+# row: 6-11 or 33-44
+# seat: 13-40 or 45-46
+
+# your ticket:
+# 7,1,14
+
+# nearby tickets:
+# 7,3,47
+# 40,4,50
+# 55,2,20
+# 38,6,12
+
+# It doesn't matter which position corresponds to which field; you can identify
+# invalid nearby tickets by considering only whether tickets contain values
+# that are not valid for any field. In this example, the values on the first
+# nearby ticket are all valid for at least one field. This is not true of the
+# other three nearby tickets: the values 4, 55, and 12 are are not valid for
+# any field. Adding together all of the invalid values produces your ticket
+# scanning error rate: 4 + 55 + 12 = 71.
+
+# Consider the validity of the nearby tickets you scanned. What is your ticket
+# scanning error rate?
+
+from typing import Set
+
+with open("files/P16.txt", "r") as f:
+    rules_raw, my_ticket, nearby_tickets = [
+        f.split("\n") for f in f.read().strip().split("\n\n")
+    ]
+
+
+def part_1() -> None:
+    rules = []
+    for rule in rules_raw:
+        name = rule.split(": ")[0]
+        valid_numbers = set()
+        for range_numbers in rule.split(": ")[1].split(" or "):
+            numbers = [int(number) for number in range_numbers.split("-")]
+            first, last = range(*numbers).start, range(*numbers).stop
+            list_numbers = list(range(first, last + 1))
+            for number in list_numbers:
+                valid_numbers.add(number)
+        rules.append([name, valid_numbers])
+
+    # Create a single set with all rules
+    all_rules = set()
+    for rule in rules:
+        all_rules |= rule[1]
+
+    all_tickets = [list(map(int, o.split(","))) for o in nearby_tickets[1:]]
+
+    ticket_scanning_error_rate = sum(
+        [
+            ticket
+            for ticket_numbers in all_tickets
+            for ticket in ticket_numbers
+            if ticket not in all_rules
+        ]
+    )
+
+    print(f"The ticket scanning error rate is {ticket_scanning_error_rate}")
+
+
+# --- Part Two ---
+
+# Now that you've identified which tickets contain invalid values, discard
+# those tickets entirely. Use the remaining valid tickets to determine which
+# field is which.
+
+# Using the valid ranges for each field, determine what order the fields appear
+# on the tickets. The order is consistent between all tickets: if seat is the
+# third field, it is the third field on every ticket, including your ticket.
+
+# For example, suppose you have the following notes:
+
+# class: 0-1 or 4-19
+# row: 0-5 or 8-19
+# seat: 0-13 or 16-19
+
+# your ticket:
+# 11,12,13
+
+# nearby tickets:
+# 3,9,18
+# 15,1,5
+# 5,14,9
+
+# Based on the nearby tickets in the above example, the first position must be
+# row, the second position must be class, and the third position must be seat;
+# you can conclude that in your ticket, class is 12, row is 11, and seat is 13.
+
+# Once you work out which field is which, look for the six fields on your
+# ticket that start with the word departure. What do you get if you multiply
+# those six values together?
+
+
+def from_part_1():
+    # from part_1
+    rules = []
+    for rule in rules_raw:
+        name = rule.split(": ")[0]
+        valid_numbers = set()
+        for range_numbers in rule.split(": ")[1].split(" or "):
+            numbers = [int(number) for number in range_numbers.split("-")]
+            first, last = range(*numbers).start, range(*numbers).stop
+            list_numbers = list(range(first, last + 1))
+            for number in list_numbers:
+                valid_numbers.add(number)
+        rules.append([name, valid_numbers])
+
+    all_rules = set()
+    for rule in rules:
+        all_rules |= rule[1]
+
+    return rules, all_rules
+
+
+def remove_invalid(all_rules):
+    valid_tickets = []
+    for line in nearby_tickets[1:]:
+        valid = True
+        for ticket in [int(t) for t in line.split(",")]:
+            if ticket not in all_rules:
+                valid = False
+                break
+        if valid:
+            valid_tickets.append(line)
+    return valid_tickets
+
+
+def check_validity(rules, ticket_values):
+    # check validity of sets
+    possibles = []
+    for c in rules:
+        possibles.append([c[0], []])
+        for idx, ticket in enumerate(ticket_values):
+            valid = True
+            for value in ticket:
+                if value not in c[1]:
+                    valid = False
+                    break
+            if valid:
+                possibles[-1][1].append(idx)
+    return possibles
+
+
+def get_ordered_fields(sorted_possibles):
+    # get each field and its respective order
+    ordered_fields = []
+    for idx, (field, values) in enumerate(sorted_possibles):
+        # first field
+        if len(values) == 1:
+            ordered_fields.append([field, values[0]])
+        else:
+            value = [
+                v for v in values if v not in sorted_possibles[idx - 1][1]
+            ][0]
+            ordered_fields.append([field, value])
+    return ordered_fields
+
+
+def part_2() -> None:
+    rules, all_rules = from_part_1()
+
+    valid_tickets = remove_invalid(all_rules)
+
+    # collect 20 sets of all respective values
+    ticket_values = [set() for x in range(len(rules))]
+    for line in valid_tickets:
+        index = 0
+        for element in line.split(","):
+            ticket_values[index].add(int(element))
+            index += 1
+
+    possibles = check_validity(rules, ticket_values)
+
+    # sort possibles by the length of values each set holds
+    sorted_possibles = sorted(possibles, key=lambda l: (len(l[1]), l))
+
+    sorted_fields = get_ordered_fields(sorted_possibles)
+
+    # parse "my ticket"
+    myticket = [
+        int(v) for elements in my_ticket[1:] for v in elements.split(",")
+    ]
+
+    res = 1
+    for (field, idx) in sorted_fields:
+        if field.startswith("departure"):
+            res *= myticket[idx]
+    print(f"The final result is {res}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P17.py

@@ -0,0 +1,527 @@
+# --- Day 17: Conway Cubes ---
+
+# As your flight slowly drifts through the sky, the Elves at the Mythical
+# Information Bureau at the North Pole contact you. They'd like some help
+# debugging a malfunctioning experimental energy source aboard one of their
+# super-secret imaging satellites.
+
+# The experimental energy source is based on cutting-edge technology: a set of
+# Conway Cubes contained in a pocket dimension! When you hear it's having
+# problems, you can't help but agree to take a look.
+
+# The pocket dimension contains an infinite 3-dimensional grid. At every
+# integer 3-dimensional coordinate (x,y,z), there exists a single cube which is
+# either active or inactive.
+
+# In the initial state of the pocket dimension, almost all cubes start
+# inactive. The only exception to this is a small flat region of cubes (your
+# puzzle input); the cubes in this region start in the specified active (#) or
+# inactive (.) state.
+
+# The energy source then proceeds to boot up by executing six cycles.
+
+# Each cube only ever considers its neighbors: any of the 26 other cubes where
+# any of their coordinates differ by at most 1. For example, given the cube at
+# x=1,y=2,z=3, its neighbors include the cube at x=2,y=2,z=2, the cube at
+# x=0,y=2,z=3, and so on.
+
+# During a cycle, all cubes simultaneously change their state according to the
+# following rules:
+
+#     If a cube is active and exactly 2 or 3 of its neighbors are also active,
+# the cube remains active. Otherwise, the cube becomes inactive.
+#     If a cube is inactive but exactly 3 of its neighbors are active, the cube
+# becomes active. Otherwise, the cube remains inactive.
+
+# The engineers responsible for this experimental energy source would like you
+# to simulate the pocket dimension and determine what the configuration of
+# cubes should be at the end of the six-cycle boot process.
+
+# For example, consider the following initial state:
+
+# .#.
+# ..#
+# ###
+
+# Even though the pocket dimension is 3-dimensional, this initial state
+# represents a small 2-dimensional slice of it. (In particular, this initial
+# state defines a 3x3x1 region of the 3-dimensional space.)
+
+# Simulating a few cycles from this initial state produces the following
+# configurations, where the result of each cycle is shown layer-by-layer at
+# each given z coordinate (and the frame of view follows the active cells in
+# each cycle):
+
+# Before any cycles:
+
+# z=0
+# .#.
+# ..#
+# ###
+
+
+# After 1 cycle:
+
+# z=-1
+# #..
+# ..#
+# .#.
+
+# z=0
+# #.#
+# .##
+# .#.
+
+# z=1
+# #..
+# ..#
+# .#.
+
+
+# After 2 cycles:
+
+# z=-2
+# .....
+# .....
+# ..#..
+# .....
+# .....
+
+# z=-1
+# ..#..
+# .#..#
+# ....#
+# .#...
+# .....
+
+# z=0
+# ##...
+# ##...
+# #....
+# ....#
+# .###.
+
+# z=1
+# ..#..
+# .#..#
+# ....#
+# .#...
+# .....
+
+# z=2
+# .....
+# .....
+# ..#..
+# .....
+# .....
+
+
+# After 3 cycles:
+
+# z=-2
+# .......
+# .......
+# ..##...
+# ..###..
+# .......
+# .......
+# .......
+
+# z=-1
+# ..#....
+# ...#...
+# #......
+# .....##
+# .#...#.
+# ..#.#..
+# ...#...
+
+# z=0
+# ...#...
+# .......
+# #......
+# .......
+# .....##
+# .##.#..
+# ...#...
+
+# z=1
+# ..#....
+# ...#...
+# #......
+# .....##
+# .#...#.
+# ..#.#..
+# ...#...
+
+# z=2
+# .......
+# .......
+# ..##...
+# ..###..
+# .......
+# .......
+# .......
+
+# After the full six-cycle boot process completes, 112 cubes are left in the
+# active state.
+
+# Starting with your given initial configuration, simulate six cycles. How many
+# cubes are left in the active state after the sixth cycle?
+
+import itertools as iter
+
+with open("files/P17.txt", "r") as f:
+    init_state = [line for line in f.read().strip().split("\n")]
+
+
+def get_active_points(dim: int):
+    ap = set()
+    for y, z in enumerate(init_state):
+        for x, c in enumerate(z):
+            if c == "#":
+                ap.add(tuple([x, y] + [0] * (dim - 2)))
+    return ap
+
+
+def part_1() -> None:
+    active_points = get_active_points(dim=3)
+    for it in range(6):
+        new_active_points = set()
+        # check x,y,z point
+        for x in range(-10 - it, it + 10):
+            for y in range(-10 - it, it + 10):
+                for z in range(-2 - it, it + 2):
+                    point = (x, y, z)
+
+                    # for the current point, check the neighbors
+                    num_actives = 0
+                    for delta in iter.product(range(-1, 2), repeat=3):
+                        if delta != (0,) * 3:
+                            if (
+                                tuple([a + b for a, b in zip(point, delta)])
+                                in active_points
+                            ):
+                                num_actives += 1
+
+                    # apply rules
+                    if point in active_points and (
+                        num_actives == 2 or num_actives == 3
+                    ):
+                        new_active_points.add(point)
+                    if point not in active_points and num_actives == 3:
+                        new_active_points.add(point)
+
+        # next iteration
+        active_points = new_active_points
+    print(f"After 6 cycles in 3D, we have {len(active_points)} active cubes")
+
+
+# --- Part Two ---
+
+# For some reason, your simulated results don't match what the experimental
+# energy source engineers expected. Apparently, the pocket dimension actually
+# has four spatial dimensions, not three.
+
+# The pocket dimension contains an infinite 4-dimensional grid. At every
+# integer 4-dimensional coordinate (x,y,z,w), there exists a single cube
+# (really, a hypercube) which is still either active or inactive.
+
+# Each cube only ever considers its neighbors: any of the 80 other cubes where
+# any of their coordinates differ by at most 1. For example, given the cube at
+# x=1,y=2,z=3,w=4, its neighbors include the cube at x=2,y=2,z=3,w=3, the cube
+# at x=0,y=2,z=3,w=4, and so on.
+
+# The initial state of the pocket dimension still consists of a small flat
+# region of cubes. Furthermore, the same rules for cycle updating still apply:
+# during each cycle, consider the number of active neighbors of each cube.
+
+# For example, consider the same initial state as in the example above. Even
+# though the pocket dimension is 4-dimensional, this initial state represents a
+# small 2-dimensional slice of it. (In particular, this initial state defines a
+# 3x3x1x1 region of the 4-dimensional space.)
+
+# Simulating a few cycles from this initial state produces the following
+# configurations, where the result of each cycle is shown layer-by-layer at
+# each given z and w coordinate:
+
+# Before any cycles:
+
+# z=0, w=0
+# .#.
+# ..#
+# ###
+
+
+# After 1 cycle:
+
+# z=-1, w=-1
+# #..
+# ..#
+# .#.
+
+# z=0, w=-1
+# #..
+# ..#
+# .#.
+
+# z=1, w=-1
+# #..
+# ..#
+# .#.
+
+# z=-1, w=0
+# #..
+# ..#
+# .#.
+
+# z=0, w=0
+# #.#
+# .##
+# .#.
+
+# z=1, w=0
+# #..
+# ..#
+# .#.
+
+# z=-1, w=1
+# #..
+# ..#
+# .#.
+
+# z=0, w=1
+# #..
+# ..#
+# .#.
+
+# z=1, w=1
+# #..
+# ..#
+# .#.
+
+
+# After 2 cycles:
+
+# z=-2, w=-2
+# .....
+# .....
+# ..#..
+# .....
+# .....
+
+# z=-1, w=-2
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=0, w=-2
+# ###..
+# ##.##
+# #...#
+# .#..#
+# .###.
+
+# z=1, w=-2
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=2, w=-2
+# .....
+# .....
+# ..#..
+# .....
+# .....
+
+# z=-2, w=-1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=-1, w=-1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=0, w=-1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=1, w=-1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=2, w=-1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=-2, w=0
+# ###..
+# ##.##
+# #...#
+# .#..#
+# .###.
+
+# z=-1, w=0
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=0, w=0
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=1, w=0
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=2, w=0
+# ###..
+# ##.##
+# #...#
+# .#..#
+# .###.
+
+# z=-2, w=1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=-1, w=1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=0, w=1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=1, w=1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=2, w=1
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=-2, w=2
+# .....
+# .....
+# ..#..
+# .....
+# .....
+
+# z=-1, w=2
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=0, w=2
+# ###..
+# ##.##
+# #...#
+# .#..#
+# .###.
+
+# z=1, w=2
+# .....
+# .....
+# .....
+# .....
+# .....
+
+# z=2, w=2
+# .....
+# .....
+# ..#..
+# .....
+# .....
+
+# After the full six-cycle boot process completes, 848 cubes are left in the
+# active state.
+
+# Starting with your given initial configuration, simulate six cycles in a
+# 4-dimensional space. How many cubes are left in the active state after the
+# sixth cycle?
+
+
+def part_2() -> None:
+    active_points = get_active_points(dim=4)
+    for it in range(6):
+        new_active_points = set()
+        # check x,y,z point
+        for x in range(-10 - it, it + 10):
+            for y in range(-10 - it, it + 10):
+                for z in range(-2 - it, it + 2):
+                    for w in range(-2 - it, it + 2):
+                        point = (x, y, z, w)
+
+                        # for the current point, check the neighbors
+                        num_actives = 0
+                        for delta in iter.product(range(-1, 2), repeat=4):
+                            if delta != (0,) * 4:
+                                if (
+                                    tuple(
+                                        [a + b for a, b in zip(point, delta)]
+                                    )
+                                    in active_points
+                                ):
+                                    num_actives += 1
+
+                        # apply rules
+                        if point in active_points and (
+                            num_actives == 2 or num_actives == 3
+                        ):
+                            new_active_points.add(point)
+                        if point not in active_points and num_actives == 3:
+                            new_active_points.add(point)
+
+        # next iteration
+        active_points = new_active_points
+    print(f"After 6 cycles in 4D, we have {len(active_points)} active cubes")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P18.py

@@ -0,0 +1,188 @@
+# --- Day 18: Operation Order ---
+
+# As you look out the window and notice a heavily-forested continent slowly
+# appear over the horizon, you are interrupted by the child sitting next to
+# you. They're curious if you could help them with their math homework.
+
+# Unfortunately, it seems like this "math" follows different rules than you
+# remember.
+
+# The homework (your puzzle input) consists of a series of expressions that
+# consist of addition (+), multiplication (*), and parentheses ((...)). Just
+# like normal math, parentheses indicate that the expression inside must be
+# evaluated before it can be used by the surrounding expression. Addition still
+# finds the sum of the numbers on both sides of the operator, and
+# multiplication still finds the product.
+
+# However, the rules of operator precedence have changed. Rather than
+# evaluating multiplication before addition, the operators have the same
+# precedence, and are evaluated left-to-right regardless of the order in which
+# they appear.
+
+# For example, the steps to evaluate the expression 1 + 2 * 3 + 4 * 5 + 6 are
+# as follows:
+
+# 1 + 2 * 3 + 4 * 5 + 6
+#   3   * 3 + 4 * 5 + 6
+#       9   + 4 * 5 + 6
+#          13   * 5 + 6
+#              65   + 6
+#                  71
+
+# Parentheses can override this order; for example, here is what happens if
+# parentheses are added to form 1 + (2 * 3) + (4 * (5 + 6)):
+
+# 1 + (2 * 3) + (4 * (5 + 6))
+# 1 +    6    + (4 * (5 + 6))
+#      7      + (4 * (5 + 6))
+#      7      + (4 *   11   )
+#      7      +     44
+#             51
+
+# Here are a few more examples:
+
+#     2 * 3 + (4 * 5) becomes 26.
+#     5 + (8 * 3 + 9 + 3 * 4 * 3) becomes 437.
+#     5 * 9 * (7 * 3 * 3 + 9 * 3 + (8 + 6 * 4)) becomes 12240.
+#     ((2 + 4 * 9) * (6 + 9 * 8 + 6) + 6) + 2 + 4 * 2 becomes 13632.
+
+# Before you can help with the homework, you need to understand it yourself.
+# Evaluate the expression on each line of the homework; what is the sum of the
+# resulting values?
+
+with open("files/P18.txt", "r") as f:
+    homework = [line for line in f.read().strip().split("\n")]
+
+
+def eval(n1, n2, op) -> int:
+    if op == "+":
+        return n1 + n2
+    elif op == "*":
+        return n1 * n2
+
+
+def simplify(expr) -> str:
+    while len(expr) > 1:
+        n1 = int(expr.pop())
+        op = expr.pop()
+        n2 = int(expr.pop())
+        result = eval(n1, n2, op)
+        expr.append(str(result))
+
+    return expr[0]
+
+
+def part_1() -> None:
+    total = 0
+    for line in homework:
+        line = line.replace("(", "( ")
+        line = line.replace(")", " )")
+        expr = line.split()
+
+        stack = []
+        for char in expr:
+            if char.isdigit() or char in ["+", "*", "("]:
+                stack.append(char)
+            elif char == ")":
+                ordered_stack = []
+                while stack[len(stack) - 1] != "(":
+                    ordered_stack.append(stack.pop())
+                stack.pop()
+
+                result = simplify(ordered_stack)
+                stack.append(result)
+        stack.reverse()
+        result = simplify(stack)
+        total += int(result)
+
+    print(f"The result of evaluating my homework is {total}")
+
+
+# --- Part Two ---
+
+# You manage to answer the child's questions and they finish part 1 of their
+# homework, but get stuck when they reach the next section: advanced math.
+
+# Now, addition and multiplication have different precedence levels, but
+# they're not the ones you're familiar with. Instead, addition is evaluated
+# before multiplication.
+
+# For example, the steps to evaluate the expression 1 + 2 * 3 + 4 * 5 + 6 are
+# now as follows:
+
+# 1 + 2 * 3 + 4 * 5 + 6
+#   3   * 3 + 4 * 5 + 6
+#   3   *   7   * 5 + 6
+#   3   *   7   *  11
+#      21       *  11
+#          231
+
+# Here are the other examples from above:
+
+#     1 + (2 * 3) + (4 * (5 + 6)) still becomes 51.
+#     2 * 3 + (4 * 5) becomes 46.
+#     5 + (8 * 3 + 9 + 3 * 4 * 3) becomes 1445.
+#     5 * 9 * (7 * 3 * 3 + 9 * 3 + (8 + 6 * 4)) becomes 669060.
+#     ((2 + 4 * 9) * (6 + 9 * 8 + 6) + 6) + 2 + 4 * 2 becomes 23340.
+
+# What do you get if you add up the results of evaluating the homework problems
+# using these new rules?
+
+
+def simplify_rules(expr) -> str:
+    while len(expr) > 1:
+        ordered_stack, hold = [], []
+        done = False
+
+        while not done and len(expr) > 0:
+            char = expr.pop()
+            if char == "+":
+                done = True
+            hold.append(char)
+
+        if not done and len(expr) == 0:
+            result = simplify(hold)
+            expr.append(result)
+        else:
+            ordered_stack.append(expr.pop())
+            for _ in range(2):
+                ordered_stack.append(hold.pop())
+
+            result = simplify(ordered_stack)
+            hold.append(result)
+
+            while len(hold) > 0:
+                expr.append(hold.pop())
+
+    return expr[0]
+
+
+def part_2() -> None:
+    total = 0
+    for line in homework:
+        line = line.replace("(", "( ")
+        line = line.replace(")", " )")
+        expr = line.split()
+
+        stack = []
+        for char in expr:
+            if char.isdigit() or char in ["+", "*", "("]:
+                stack.append(char)
+            elif char == ")":
+                ordered_stack = []
+                while stack[len(stack) - 1] != "(":
+                    ordered_stack.append(stack.pop())
+                stack.pop()
+
+                result = simplify_rules(ordered_stack)
+                stack.append(result)
+        stack.reverse()
+        result = simplify_rules(stack)
+        total += int(result)
+
+    print(f"The result of evaluating my homework is {total}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P19.py

@@ -0,0 +1,113 @@
+# --- Day 19: Monster Messages ---
+
+# You land in an airport surrounded by dense forest. As you walk to your high
+# speed train, the Elves at the Mythical Information Bureau contact you again.
+# They think their satellite has collected an image of a sea monster!
+# Unfortunately, the connection to the satellite is having problems, and many
+# of the messages sent back from the satellite have been corrupted.
+
+# They sent you a list of the rules valid messages should obey and a list of
+# received messages they've collected so far (your puzzle input).
+
+# The rules for valid messages (the top part of your puzzle input) are numbered
+# and build upon each other. For example:
+
+# 0: 1 2
+# 1: "a"
+# 2: 1 3 | 3 1
+# 3: "b"
+
+# Some rules, like 3: "b", simply match a single character (in this case, b).
+
+# The remaining rules list the sub-rules that must be followed; for example,
+# the rule 0: 1 2 means that to match rule 0, the text being checked must match
+# rule 1, and the text after the part that matched rule 1 must then match rule
+# 2.
+
+# Some of the rules have multiple lists of sub-rules separated by a pipe (|).
+# This means that at least one list of sub-rules must match. (The ones that
+# match might be different each time the rule is encountered.) For example, the
+# rule 2: 1 3 | 3 1 means that to match rule 2, the text being checked must
+# match rule 1 followed by rule 3 or it must match rule 3 followed by rule 1.
+
+# Fortunately, there are no loops in the rules, so the list of possible matches
+# will be finite. Since rule 1 matches a and rule 3 matches b, rule 2 matches
+# either ab or ba. Therefore, rule 0 matches aab or aba.
+
+# Here's a more interesting example:
+
+# 0: 4 1 5
+# 1: 2 3 | 3 2
+# 2: 4 4 | 5 5
+# 3: 4 5 | 5 4
+# 4: "a"
+# 5: "b"
+
+# Here, because rule 4 matches a and rule 5 matches b, rule 2 matches two
+# letters that are the same (aa or bb), and rule 3 matches two letters that are
+# different (ab or ba).
+
+# Since rule 1 matches rules 2 and 3 once each in either order, it must match
+# two pairs of letters, one pair with matching letters and one pair with
+# different letters. This leaves eight possibilities: aaab, aaba, bbab, bbba,
+# abaa, abbb, baaa, or babb.
+
+# Rule 0, therefore, matches a (rule 4), then any of the eight options from
+# rule 1, then b (rule 5): aaaabb, aaabab, abbabb, abbbab, aabaab, aabbbb,
+# abaaab, or ababbb.
+
+# The received messages (the bottom part of your puzzle input) need to be
+# checked against the rules so you can determine which are valid and which are
+# corrupted. Including the rules and the messages together, this might look
+# like:
+
+# 0: 4 1 5
+# 1: 2 3 | 3 2
+# 2: 4 4 | 5 5
+# 3: 4 5 | 5 4
+# 4: "a"
+# 5: "b"
+
+# ababbb
+# bababa
+# abbbab
+# aaabbb
+# aaaabbb
+
+# Your goal is to determine the number of messages that completely match rule
+# 0. In the above example, ababbb and abbbab match, but bababa, aaabbb, and
+# aaaabbb do not, producing the answer 2. The whole message must match all of
+# rule 0; there can't be extra unmatched characters in the message. (For
+# example, aaaabbb might appear to match rule 0 above, but it has an extra
+# unmatched b on the end.)
+
+# How many messages completely match rule 0?
+
+import re
+
+with open("files/P19.txt", "r") as f:
+    rules, messages = [f.split("\n") for f in f.read().strip().split("\n\n")]
+
+
+def dfs(tree, node, depth):
+    rule = ""
+    for next in tree[node].split():
+        if next == "|":
+            rule += next
+        elif next.isdigit():
+            rule += dfs(tree, next, depth + 1)
+        else:
+            return next[1]
+    return "(" + rule + ")"
+
+
+def part_1():
+    rules_dict = dict([rule.split(": ") for rule in rules])
+    rule_0 = re.compile(dfs(rules_dict, "0", 0))
+    total = sum([1 for message in messages if rule_0.fullmatch(message)])
+
+    print(f"There are {total} messages completely matching rule 0")
+
+
+if __name__ == "__main__":
+    part_1()

src/Year_2020/P2.py

@@ -0,0 +1,97 @@
+# --- Day 2: Password Philosophy ---
+
+# Your flight departs in a few days from the coastal airport; the easiest way
+# down to the coast from here is via toboggan.
+
+# The shopkeeper at the North Pole Toboggan Rental Shop is having a bad day.
+# "Something's wrong with our computers; we can't log in!" You ask if you can
+# take a look.
+
+# Their password database seems to be a little corrupted: some of the passwords
+# wouldn't have been allowed by the Official Toboggan Corporate Policy that was
+# in effect when they were chosen.
+
+# To try to debug the problem, they have created a list (your puzzle input) of
+# passwords (according to the corrupted database) and the corporate policy when
+# that password was set.
+
+# For example, suppose you have the following list:
+
+# 1-3 a: abcde
+# 1-3 b: cdefg
+# 2-9 c: ccccccccc
+
+# Each line gives the password policy and then the password. The password
+# policy indicates the lowest and highest number of times a given letter must
+# appear for the password to be valid. For example, 1-3 a means that the
+# password must contain a at least 1 time and at most 3 times.
+
+# In the above example, 2 passwords are valid. The middle password, cdefg, is
+# not; it contains no instances of b, but needs at least 1. The first and third
+# passwords are valid: they contain one a or nine c, both within the limits of
+# their respective policies.
+
+# How many passwords are valid according to their policies?
+
+
+with open("files/P2.txt", "r") as f:
+    passwds = [line.split() for line in f.read().strip().split("\n")]
+
+
+def part_1() -> None:
+    correct_passwds = 0
+    for passwd in passwds:
+        min_, max_ = passwd[0].split("-")
+
+        if passwd[2].count(passwd[1][:-1]) >= int(min_) and passwd[2].count(
+            passwd[1][:-1]
+        ) <= int(max_):
+            correct_passwds += 1
+
+    print(f"There are {correct_passwds} valid passwords in the database")
+
+
+# --- Part Two ---
+
+# While it appears you validated the passwords correctly, they don't seem to be
+# what the Official Toboggan Corporate Authentication System is expecting.
+
+# The shopkeeper suddenly realizes that he just accidentally explained the
+# password policy rules from his old job at the sled rental place down the
+# street! The Official Toboggan Corporate Policy actually works a little
+# differently.
+
+# Each policy actually describes two positions in the password, where 1 means
+# the first character, 2 means the second character, and so on. (Be careful;
+# Toboggan Corporate Policies have no concept of "index zero"!) Exactly one of
+# these positions must contain the given letter. Other occurrences of the
+# letter are irrelevant for the purposes of policy enforcement.
+
+# Given the same example list from above:
+
+#     1-3 a: abcde is valid: position 1 contains a and position 3 does not.
+#     1-3 b: cdefg is invalid: neither position 1 nor position 3 contains b.
+#     2-9 c: ccccccccc is invalid: both position 2 and position 9 contain c.
+
+# How many passwords are valid according to the new interpretation of the
+# policies?
+
+
+def part_2() -> None:
+    correct_passwds = 0
+    for passwd in passwds:
+        pos1, pos2 = passwd[0].split("-")
+        checks = 0
+        if passwd[2][int(pos1) - 1] == passwd[1][:-1]:
+            checks += 1
+        if passwd[2][int(pos2) - 1] == passwd[1][:-1]:
+            checks += 1
+        if checks == 1:
+            correct_passwds += 1
+
+    print(f"There are {correct_passwds} valid passwords in the database")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P3.py

@@ -0,0 +1,171 @@
+# --- Day 3: Toboggan Trajectory ---
+
+# With the toboggan login problems resolved, you set off toward the airport.
+# While travel by toboggan might be easy, it's certainly not safe: there's very
+# minimal steering and the area is covered in trees. You'll need to see which
+# angles will take you near the fewest trees.
+
+# Due to the local geology, trees in this area only grow on exact integer
+# coordinates in a grid. You make a map (your puzzle input) of the open squares
+# (.) and trees (#) you can see. For example:
+
+# ..##.......
+# #...#...#..
+# .#....#..#.
+# ..#.#...#.#
+# .#...##..#.
+# ..#.##.....
+# .#.#.#....#
+# .#........#
+# #.##...#...
+# #...##....#
+# .#..#...#.#
+
+# These aren't the only trees, though; due to something you read about once
+# involving arboreal genetics and biome stability, the same pattern repeats to
+# the right many times:
+
+# ..##.........##.........##.........##.........##.........##.......  --->
+# #...#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..
+# .#....#..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#.
+# ..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#
+# .#...##..#..#...##..#..#...##..#..#...##..#..#...##..#..#...##..#.
+# ..#.##.......#.##.......#.##.......#.##.......#.##.......#.##.....  --->
+# .#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#
+# .#........#.#........#.#........#.#........#.#........#.#........#
+# #.##...#...#.##...#...#.##...#...#.##...#...#.##...#...#.##...#...
+# #...##....##...##....##...##....##...##....##...##....##...##....#
+# .#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#  --->
+
+# You start on the open square (.) in the top-left corner and need to reach the
+# bottom (below the bottom-most row on your map).
+
+# The toboggan can only follow a few specific slopes (you opted for a cheaper
+# model that prefers rational numbers); start by counting all the trees you
+# would encounter for the slope right 3, down 1:
+
+# From your starting position at the top-left, check the position that is right
+# 3 and down 1. Then, check the position that is right 3 and down 1 from there,
+# and so on until you go past the bottom of the map.
+
+# The locations you'd check in the above example are marked here with O where
+# there was an open square and X where there was a tree:
+
+# ..##.........##.........##.........##.........##.........##.......  --->
+# #..O#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#..
+# .#....X..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#.
+# ..#.#...#O#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#
+# .#...##..#..X...##..#..#...##..#..#...##..#..#...##..#..#...##..#.
+# ..#.##.......#.X#.......#.##.......#.##.......#.##.......#.##.....  --->
+# .#.#.#....#.#.#.#.O..#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#
+# .#........#.#........X.#........#.#........#.#........#.#........#
+# #.##...#...#.##...#...#.X#...#...#.##...#...#.##...#...#.##...#...
+# #...##....##...##....##...#X....##...##....##...##....##...##....#
+# .#..#...#.#.#..#...#.#.#..#...X.#.#..#...#.#.#..#...#.#.#..#...#.#  --->
+
+# In this example, traversing the map using this slope would cause you to
+# encounter 7 trees.
+
+# Starting at the top-left corner of your map and following a slope of right 3
+# and down 1, how many trees would you encounter?
+
+
+with open("files/P3.txt", "r") as f:
+    forest = [line.split() for line in f.read().strip().split("\n")]
+
+
+def part_1() -> None:
+    trees_found, pos = 0, 0
+    for idx, line in enumerate(forest):
+        extended_line = ",".join(line * 32).replace(",", "")
+        if extended_line[pos] == "#":
+            trees_found += 1
+        pos += 3
+
+    print(f"We found {trees_found} trees in our way")
+
+
+# --- Part Two ---
+
+# Time to check the rest of the slopes - you need to minimize the probability
+# of a sudden arboreal stop, after all.
+
+# Determine the number of trees you would encounter if, for each of the
+# following slopes, you start at the top-left corner and traverse the map all
+# the way to the bottom:
+
+#     Right 1, down 1.
+#     Right 3, down 1. (This is the slope you already checked.)
+#     Right 5, down 1.
+#     Right 7, down 1.
+#     Right 1, down 2.
+
+# In the above example, these slopes would find 2, 7, 3, 4, and 2 tree(s)
+# respectively; multiplied together, these produce the answer 336.
+
+# What do you get if you multiply together the number of trees encountered on
+# each of the listed slopes?
+
+
+def part_2() -> None:
+    pos = 0
+    trees_first = 0
+    # first slope: right 1, down 1
+    for idx, line in enumerate(forest):
+        extended_line = ",".join(line * 11).replace(",", "")
+        if extended_line[pos] == "#":
+            trees_first += 1
+        pos += 1
+
+    # reset pos
+    pos = 0
+    trees_second = 0
+    # second slope: right 3, down 1
+    for idx, line in enumerate(forest):
+        extended_line = ",".join(line * 32).replace(",", "")
+        if extended_line[pos] == "#":
+            trees_second += 1
+        pos += 3
+
+    # reset pos
+    pos = 0
+    trees_third = 0
+    # third slope: right 5, down 1
+    for idx, line in enumerate(forest):
+        extended_line = ",".join(line * 55).replace(",", "")
+        if extended_line[pos] == "#":
+            trees_third += 1
+        pos += 5
+
+    # reset pos
+    pos = 0
+    trees_fourth = 0
+    # fourth slope: right 7, down 1
+    for idx, line in enumerate(forest):
+        extended_line = ",".join(line * 77).replace(",", "")
+        if extended_line[pos] == "#":
+            trees_fourth += 1
+        pos += 7
+
+    # reset pos
+    pos = 0
+    trees_fifth = 0
+    # fifth slope: right 1, down 2
+    for idx, line in enumerate(forest):
+        if idx % 2 == 0:
+            # break
+            extended_line = ",".join(line * 6).replace(",", "")
+            if extended_line[pos] == "#":
+                trees_fifth += 1
+            pos += 1
+
+    total_trees = (
+        trees_first * trees_second * trees_third * trees_fourth * trees_fifth
+    )
+
+    print(f"We found {total_trees} trees in our way")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P4.py

@@ -0,0 +1,228 @@
+# --- Day 4: Passport Processing ---
+
+# You arrive at the airport only to realize that you grabbed your North Pole
+# Credentials instead of your passport. While these documents are extremely
+# similar, North Pole Credentials aren't issued by a country and therefore
+# aren't actually valid documentation for travel in most of the world.
+
+# It seems like you're not the only one having problems, though; a very long
+# line has formed for the automatic passport scanners, and the delay could
+# upset your travel itinerary.
+
+# Due to some questionable network security, you realize you might be able to
+# solve both of these problems at the same time.
+
+# The automatic passport scanners are slow because they're having trouble
+# detecting which passports have all required fields. The expected fields are
+# as follows:
+
+#     byr (Birth Year)
+#     iyr (Issue Year)
+#     eyr (Expiration Year)
+#     hgt (Height)
+#     hcl (Hair Color)
+#     ecl (Eye Color)
+#     pid (Passport ID)
+#     cid (Country ID)
+
+# Passport data is validated in batch files (your puzzle input). Each passport
+# is represented as a sequence of key:value pairs separated by spaces or
+# newlines. Passports are separated by blank lines.
+
+# Here is an example batch file containing four passports:
+
+# ecl:gry pid:860033327 eyr:2020 hcl:#fffffd
+# byr:1937 iyr:2017 cid:147 hgt:183cm
+
+# iyr:2013 ecl:amb cid:350 eyr:2023 pid:028048884
+# hcl:#cfa07d byr:1929
+
+# hcl:#ae17e1 iyr:2013
+# eyr:2024
+# ecl:brn pid:760753108 byr:1931
+# hgt:179cm
+
+# hcl:#cfa07d eyr:2025 pid:166559648
+# iyr:2011 ecl:brn hgt:59in
+
+# The first passport is valid - all eight fields are present. The second
+# passport is invalid - it is missing hgt (the Height field).
+
+# The third passport is interesting; the only missing field is cid, so it looks
+# like data from North Pole Credentials, not a passport at all! Surely, nobody
+# would mind if you made the system temporarily ignore missing cid fields.
+# Treat this "passport" as valid.
+
+# The fourth passport is missing two fields, cid and byr. Missing cid is fine,
+# but missing any other field is not, so this passport is invalid.
+
+# According to the above rules, your improved system would report 2 valid
+# passports.
+
+# Count the number of valid passports - those that have all required fields.
+# Treat cid as optional. In your batch file, how many passports are valid?
+
+
+with open("files/P4.txt", "r") as f:
+    passports = [line.split() for line in f.read().strip().split("\n\n")]
+
+
+def part_1() -> None:
+    valid_passports = 0
+    for passport in passports:
+        if len(passport) > 7:
+            valid_passports += 1
+        if len(passport) == 7:
+            valid_fields = 0
+            for field in passport:
+                if "cid" not in field:
+                    valid_fields += 1
+            if valid_fields == 7:
+                valid_passports += 1
+
+    print(f"There are {valid_passports} valid passports")
+
+
+# --- Part Two ---
+
+# The line is moving more quickly now, but you overhear airport security
+# talking about how passports with invalid data are getting through. Better add
+# some data validation, quick!
+
+# You can continue to ignore the cid field, but each other field has strict
+# rules about what values are valid for automatic validation:
+
+#     byr (Birth Year) - four digits; at least 1920 and at most 2002.
+#     iyr (Issue Year) - four digits; at least 2010 and at most 2020.
+#     eyr (Expiration Year) - four digits; at least 2020 and at most 2030.
+#     hgt (Height) - a number followed by either cm or in:
+#         If cm, the number must be at least 150 and at most 193.
+#         If in, the number must be at least 59 and at most 76.
+#     hcl (Hair Color) - a # followed by exactly six characters 0-9 or a-f.
+#     ecl (Eye Color) - exactly one of: amb blu brn gry grn hzl oth.
+#     pid (Passport ID) - a nine-digit number, including leading zeroes.
+#     cid (Country ID) - ignored, missing or not.
+
+# Your job is to count the passports where all required fields are both present
+# and valid according to the above rules. Here are some example values:
+
+# byr valid:   2002
+# byr invalid: 2003
+
+# hgt valid:   60in
+# hgt valid:   190cm
+# hgt invalid: 190in
+# hgt invalid: 190
+
+# hcl valid:   #123abc
+# hcl invalid: #123abz
+# hcl invalid: 123abc
+
+# ecl valid:   brn
+# ecl invalid: wat
+
+# pid valid:   000000001
+# pid invalid: 0123456789
+
+# Here are some invalid passports:
+
+# eyr:1972 cid:100
+# hcl:#18171d ecl:amb hgt:170 pid:186cm iyr:2018 byr:1926
+
+# iyr:2019
+# hcl:#602927 eyr:1967 hgt:170cm
+# ecl:grn pid:012533040 byr:1946
+
+# hcl:dab227 iyr:2012
+# ecl:brn hgt:182cm pid:021572410 eyr:2020 byr:1992 cid:277
+
+# hgt:59cm ecl:zzz
+# eyr:2038 hcl:74454a iyr:2023
+# pid:3556412378 byr:2007
+
+# Here are some valid passports:
+
+# pid:087499704 hgt:74in ecl:grn iyr:2012 eyr:2030 byr:1980
+# hcl:#623a2f
+
+# eyr:2029 ecl:blu cid:129 byr:1989
+# iyr:2014 pid:896056539 hcl:#a97842 hgt:165cm
+
+# hcl:#888785
+# hgt:164cm byr:2001 iyr:2015 cid:88
+# pid:545766238 ecl:hzl
+# eyr:2022
+
+# iyr:2010 hgt:158cm hcl:#b6652a ecl:blu byr:1944 eyr:2021 pid:093154719
+
+# Count the number of valid passports - those that have all required fields and
+# valid values. Continue to treat cid as optional. In your batch file, how many
+# passports are valid?
+
+
+def is_valid(passport: dict[str, str]) -> bool:
+    keys = ["byr", "iyr", "eyr", "hgt", "hcl", "ecl", "pid"]
+    for key in keys:
+        if key not in passport:
+            return False
+    if not any([passport[key] != 4 for key in ["byr", "iyr", "eyr"]]):
+        return False
+    if passport["byr"] < "1920" or passport["byr"] > "2002":
+        return False
+    if passport["iyr"] < "2010" or passport["iyr"] > "2020":
+        return False
+    if passport["eyr"] < "2020" or passport["eyr"] > "2030":
+        return False
+    if not any(["cm" in passport["hgt"], "in" in passport["hgt"]]):
+        return False
+    _hgt = int(passport["hgt"][:-2])
+    if "cm" in passport["hgt"]:
+        if _hgt < 150 or _hgt > 193:
+            return False
+    if "in" in passport["hgt"]:
+        if _hgt < 59 or _hgt > 76:
+            return False
+    if "#" not in passport["hcl"]:
+        return False
+    _hcl = passport["hcl"].split("#")[-1]
+    if len(_hcl) != 6:
+        return False
+    try:
+        int(_hcl, 16)
+    except ValueError:
+        return False
+    if not any(
+        [
+            passport["ecl"]
+            in [
+                "amb",
+                "blu",
+                "brn",
+                "gry",
+                "grn",
+                "hzl",
+                "oth",
+            ]
+        ]
+    ):
+        return False
+    if len(passport["pid"]) != 9:
+        return False
+    return True
+
+
+def part_2() -> None:
+    valid_passports = 0
+    for passport in passports:
+        _items: list[list[str]] = [
+            item.split(":") for item in passport if item
+        ]
+        passport_dict: dict[str, str] = dict(_items)
+        if is_valid(passport_dict):
+            valid_passports += 1
+    print(f"There are {valid_passports} valid passports")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P5.py

@@ -0,0 +1,128 @@
+# --- Day 5: Binary Boarding ---
+
+# You board your plane only to discover a new problem: you dropped your
+# boarding pass! You aren't sure which seat is yours, and all of the flight
+# attendants are busy with the flood of people that suddenly made it through
+# passport control.
+
+# You write a quick program to use your phone's camera to scan all of the
+# nearby boarding passes (your puzzle input); perhaps you can find your seat
+# through process of elimination.
+
+# Instead of zones or groups, this airline uses binary space partitioning to
+# seat people. A seat might be specified like FBFBBFFRLR, where F means
+# "front", B means "back", L means "left", and R means "right".
+
+# The first 7 characters will either be F or B; these specify exactly one of
+# the 128 rows on the plane (numbered 0 through 127). Each letter tells you
+# which half of a region the given seat is in. Start with the whole list of
+# rows; the first letter indicates whether the seat is in the front (0 through
+# 63) or the back (64 through 127). The next letter indicates which half of
+# that region the seat is in, and so on until you're left with exactly one row.
+
+# For example, consider just the first seven characters of FBFBBFFRLR:
+
+#     Start by considering the whole range, rows 0 through 127.
+#     F means to take the lower half, keeping rows 0 through 63.
+#     B means to take the upper half, keeping rows 32 through 63.
+#     F means to take the lower half, keeping rows 32 through 47.
+#     B means to take the upper half, keeping rows 40 through 47.
+#     B keeps rows 44 through 47.
+#     F keeps rows 44 through 45.
+#     The final F keeps the lower of the two, row 44.
+
+# The last three characters will be either L or R; these specify exactly one of
+# the 8 columns of seats on the plane (numbered 0 through 7). The same process
+# as above proceeds again, this time with only three steps. L means to keep the
+# lower half, while R means to keep the upper half.
+
+# For example, consider just the last 3 characters of FBFBBFFRLR:
+
+#     Start by considering the whole range, columns 0 through 7.
+#     R means to take the upper half, keeping columns 4 through 7.
+#     L means to take the lower half, keeping columns 4 through 5.
+#     The final R keeps the upper of the two, column 5.
+
+# So, decoding FBFBBFFRLR reveals that it is the seat at row 44, column 5.
+
+# Every seat also has a unique seat ID: multiply the row by 8, then add the
+# column. In this example, the seat has ID 44 * 8 + 5 = 357.
+
+# Here are some other boarding passes:
+
+#     BFFFBBFRRR: row 70, column 7, seat ID 567.
+#     FFFBBBFRRR: row 14, column 7, seat ID 119.
+#     BBFFBBFRLL: row 102, column 4, seat ID 820.
+
+# As a sanity check, look through your list of boarding passes. What is the
+# highest seat ID on a boarding pass?
+
+with open("files/P5.txt", "r") as f:
+    boarding_passes = [code for code in f.read().strip().split("\n")]
+
+
+def binary_search(
+    arr: str, upper_limit: int, var_lower: str, var_upper: str
+) -> int:
+    range_l = 0
+    range_h = upper_limit
+    range_m = 0
+    for char in arr:
+        range_m = (range_h + range_l) // 2
+        if char == var_lower:
+            range_h = range_m
+        elif char == var_upper:
+            range_l = range_m + 1
+    return int(range_l)
+
+
+def part_1() -> None:
+    seat_ID: int = 0
+    for code in boarding_passes:
+        row, col = code[:7], code[-3:]
+
+        _row = binary_search(row, 127, "F", "B")
+        _col = binary_search(col, 7, "L", "R")
+        _seat_ID: int = _row * 8 + _col
+        seat_ID = max(seat_ID, _seat_ID)
+
+    print(f"The highest seat ID on a boarding pass is {seat_ID}")
+
+
+# --- Part Two ---
+
+# Ding! The "fasten seat belt" signs have turned on. Time to find your seat.
+
+# It's a completely full flight, so your seat should be the only missing
+# boarding pass in your list. However, there's a catch: some of the seats at
+# the very front and back of the plane don't exist on this aircraft, so they'll
+# be missing from your list as well.
+
+# Your seat wasn't at the very front or back, though; the seats with IDs +1 and
+# -1 from yours will be in your list.
+
+# What is the ID of your seat?
+
+
+def part_2() -> None:
+    seat_IDs: list[int] = []
+    missing_seats: list[int] = []
+    for code in boarding_passes:
+        row, col = code[:7], code[-3:]
+
+        _row = binary_search(row, 127, "F", "B")
+        _col = binary_search(col, 7, "L", "R")
+        _seat_ID: int = _row * 8 + _col
+        seat_IDs.append(_seat_ID)
+
+    for ID in range(len(boarding_passes)):
+        if ID in seat_IDs:
+            pass
+        else:
+            missing_seats.append(ID)
+    print(f"The ID of my seat is {missing_seats[-1]}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P6.py

@@ -0,0 +1,138 @@
+# --- Day 6: Custom Customs ---
+
+# As your flight approaches the regional airport where you'll switch to a much
+# larger plane, customs declaration forms are distributed to the passengers.
+
+# The form asks a series of 26 yes-or-no questions marked a through z. All you
+# need to do is identify the questions for which anyone in your group answers
+# "yes". Since your group is just you, this doesn't take very long.
+
+# However, the person sitting next to you seems to be experiencing a language
+# barrier and asks if you can help. For each of the people in their group, you
+# write down the questions for which they answer "yes", one per line. For
+# example:
+
+# abcx
+# abcy
+# abcz
+
+# In this group, there are 6 questions to which anyone answered "yes": a, b, c,
+# x, y, and z. (Duplicate answers to the same question don't count extra; each
+# question counts at most once.)
+
+# Another group asks for your help, then another, and eventually you've
+# collected answers from every group on the plane (your puzzle input). Each
+# group's answers are separated by a blank line, and within each group, each
+# person's answers are on a single line. For example:
+
+# abc
+
+# a
+# b
+# c
+
+# ab
+# ac
+
+# a
+# a
+# a
+# a
+
+# b
+
+# This list represents answers from five groups:
+
+#     The first group contains one person who answered "yes" to 3 questions: a,
+# b, and c.
+#     The second group contains three people; combined, they answered "yes" to
+# 3 questions: a, b, and c.
+#     The third group contains two people; combined, they answered "yes" to 3
+# questions: a, b, and c.
+#     The fourth group contains four people; combined, they answered "yes" to
+# only 1 question, a.
+#     The last group contains one person who answered "yes" to only 1 question,
+# b.
+
+# In this example, the sum of these counts is 3 + 3 + 3 + 1 + 1 = 11.
+
+# For each group, count the number of questions to which anyone answered "yes".
+# What is the sum of those counts?
+
+with open("files/P6.txt", "r") as f:
+    answers_groups = [line.split() for line in f.read().strip().split("\n\n")]
+
+
+def part_1() -> None:
+    counts: int = 0
+    for answers_group in answers_groups:
+        yes = len(
+            set(
+                [
+                    answer
+                    for individual_answers in answers_group
+                    for answer in individual_answers
+                ]
+            )
+        )
+        counts += yes
+
+    print(f"The sum of counts is {counts}")
+
+
+# --- Part Two ---
+
+# As you finish the last group's customs declaration, you notice that you
+# misread one word in the instructions:
+
+# You don't need to identify the questions to which anyone answered "yes"; you
+# need to identify the questions to which everyone answered "yes"!
+
+# Using the same example as above:
+
+# abc
+
+# a
+# b
+# c
+
+# ab
+# ac
+
+# a
+# a
+# a
+# a
+
+# b
+
+# This list represents answers from five groups:
+
+#     In the first group, everyone (all 1 person) answered "yes" to 3
+# questions: a, b, and c.
+#     In the second group, there is no question to which everyone answered
+# "yes".
+#     In the third group, everyone answered yes to only 1 question, a. Since
+# some people did not answer "yes" to b or c, they don't count.
+#     In the fourth group, everyone answered yes to only 1 question, a.
+#     In the fifth group, everyone (all 1 person) answered "yes" to 1 question,
+# b.
+
+# In this example, the sum of these counts is 3 + 0 + 1 + 1 + 1 = 6.
+
+# For each group, count the number of questions to which everyone answered
+# "yes". What is the sum of those counts?
+
+
+def part_2() -> None:
+    counts: int = 0
+    for answers_group in answers_groups:
+        yes = len(set.intersection(*map(set, answers_group)))
+        counts += yes
+
+    print(f"The sum of counts is {counts}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P7.py

@@ -0,0 +1,168 @@
+from typing import Dict
+
+# --- Day 7: Handy Haversacks ---
+
+# You land at the regional airport in time for your next flight. In fact, it
+# looks like you'll even have time to grab some food: all flights are currently
+# delayed due to issues in luggage processing.
+
+# Due to recent aviation regulations, many rules (your puzzle input) are being
+# enforced about bags and their contents; bags must be color-coded and must
+# contain specific quantities of other color-coded bags. Apparently, nobody
+# responsible for these regulations considered how long they would take to
+# enforce!
+
+# For example, consider the following rules:
+
+# light red bags contain 1 bright white bag, 2 muted yellow bags.
+# dark orange bags contain 3 bright white bags, 4 muted yellow bags.
+# bright white bags contain 1 shiny gold bag.
+# muted yellow bags contain 2 shiny gold bags, 9 faded blue bags.
+# shiny gold bags contain 1 dark olive bag, 2 vibrant plum bags.
+# dark olive bags contain 3 faded blue bags, 4 dotted black bags.
+# vibrant plum bags contain 5 faded blue bags, 6 dotted black bags.
+# faded blue bags contain no other bags.
+# dotted black bags contain no other bags.
+
+# These rules specify the required contents for 9 bag types. In this example,
+# every faded blue bag is empty, every vibrant plum bag contains 11 bags (5
+# faded blue and 6 dotted black), and so on.
+
+# You have a shiny gold bag. If you wanted to carry it in at least one other
+# bag, how many different bag colors would be valid for the outermost bag? (In
+# other words: how many colors can, eventually, contain at least one shiny gold
+# bag?)
+
+# In the above rules, the following options would be available to you:
+
+#     A bright white bag, which can hold your shiny gold bag directly.
+#     A muted yellow bag, which can hold your shiny gold bag directly, plus
+# some other bags.
+#     A dark orange bag, which can hold bright white and muted yellow bags,
+# either of which could then hold your shiny gold bag.
+#     A light red bag, which can hold bright white and muted yellow bags,
+# either of which could then hold your shiny gold bag.
+
+# So, in this example, the number of bag colors that can eventually contain at
+# least one shiny gold bag is 4.
+
+# How many bag colors can eventually contain at least one shiny gold bag? (The
+# list of rules is quite long; make sure you get all of it.)
+
+with open("files/P7.txt", "r") as f:
+    lines = [code for code in f.read().strip().split("\n")]
+
+# to make a dictionary of bags
+bag_types = []
+all_bags: Dict[str, str] = {}
+for line in lines:
+    main_bag = line.split(" bags contain ")[0]
+    # get a string with the content of each bag
+    contains = line.split(" bags contain ")[1:][0]
+    # house keeping
+    each_contain = contains.split(",")
+    each_contain = [cnt.lstrip() for cnt in each_contain]
+    # get a list of strings with the content of each bag
+    each_contain = [" ".join(cont.split(" ")[:-1]) for cont in each_contain]
+    # get a dictionary with bags as keys and number of them as values
+    each_contain = {
+        " ".join(cont.split(" ")[1:]): cont.split(" ")[0]
+        for cont in each_contain
+    }
+
+    if main_bag not in bag_types:
+        bag_types.append(main_bag)
+    if all_bags.get(main_bag):
+        each_contain.update(all_bags[main_bag])
+    all_bags[main_bag] = each_contain
+
+
+def check_bag(bags: Dict[str, str], my_bag: str, current_bag: str) -> int:
+    if current_bag == my_bag:
+        return 1
+    if bags.get(current_bag) is None:
+        return 0
+    else:
+        counts = []
+        for k, v in bags[current_bag].items():
+            counts.append(check_bag(bags, my_bag, k))
+        return max(counts)
+
+
+def part_1() -> None:
+    found_bags = 0
+    my_bag = "shiny gold"
+    for k, v in all_bags.items():
+        if k != my_bag:
+            found_bags += check_bag(all_bags, my_bag, k)
+    print(f"{found_bags} bags can contain a {my_bag} bag.")
+
+
+# --- Part Two ---
+
+# It's getting pretty expensive to fly these days - not because of ticket
+# prices, but because of the ridiculous number of bags you need to buy!
+
+# Consider again your shiny gold bag and the rules from the above example:
+
+#     faded blue bags contain 0 other bags.
+#     dotted black bags contain 0 other bags.
+#     vibrant plum bags contain 11 other bags: 5 faded blue bags and 6 dotted
+# black bags.
+#     dark olive bags contain 7 other bags: 3 faded blue bags and 4 dotted
+# black bags.
+
+# So, a single shiny gold bag must contain 1 dark olive bag (and the 7 bags
+# within it) plus 2 vibrant plum bags (and the 11 bags within each of those):
+# 1 + 1*7 + 2 + 2*11 = 32 bags!
+
+# Of course, the actual rules have a small chance of going several levels
+# deeper than this example; be sure to count all of the bags, even if the
+# nesting becomes topologically impractical!
+
+# Here's another example:
+
+# shiny gold bags contain 2 dark red bags.
+# dark red bags contain 2 dark orange bags.
+# dark orange bags contain 2 dark yellow bags.
+# dark yellow bags contain 2 dark green bags.
+# dark green bags contain 2 dark blue bags.
+# dark blue bags contain 2 dark violet bags.
+# dark violet bags contain no other bags.
+
+# In this example, a single shiny gold bag must contain 126 other bags.
+
+# How many individual bags are required inside your single shiny gold bag?
+
+
+bags_contains: Dict[str, str] = {}
+for k, v in all_bags.items():
+    bags_contains[k] = []
+    try:
+        for kk, vv in v.items():
+            bags_contains[k] += [kk] * int(vv)
+    except ValueError:
+        pass
+
+
+def count_bags(current_bag: str) -> int:
+    if current_bag == " " or bags_contains.get(current_bag) is None:
+        return 0
+
+    cnt = len(bags_contains[current_bag])
+    nbags = []
+    for k in bags_contains[current_bag]:
+        nbags.append(count_bags(k))
+    return sum(nbags) + cnt
+
+
+def part_2() -> None:
+    my_bag = "shiny gold"
+    nbags = count_bags(my_bag)
+
+    print(f"My shiny gold bag can contain {nbags} bags")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2020/P8.py

@@ -0,0 +1,181 @@
+from typing import List, Tuple
+
+# --- Day 8: Handheld Halting ---
+
+# Your flight to the major airline hub reaches cruising altitude without
+# incident. While you consider checking the in-flight menu for one of those
+# drinks that come with a little umbrella, you are interrupted by the kid
+# sitting next to you.
+
+# Their handheld game console won't turn on! They ask if you can take a look.
+
+# You narrow the problem down to a strange infinite loop in the boot code
+# (your puzzle input) of the device. You should be able to fix it, but first
+# you need to be able to run the code in isolation.
+
+# The boot code is represented as a text file with one instruction per line of
+# text. Each instruction consists of an operation (acc, jmp, or nop) and an
+# argument (a signed number like +4 or -20).
+
+#     acc increases or decreases a single global value called the accumulator
+# by the value given in the argument. For example, acc +7 would increase the
+# accumulator by 7. The accumulator starts at 0. After an acc instruction, the
+# instruction immediately below it is executed next.
+#     jmp jumps to a new instruction relative to itself. The next instruction
+# to execute is found using the argument as an offset from the jmp instruction;
+# for example, jmp +2 would skip the next instruction, jmp +1 would continue to
+# the instruction immediately below it, and jmp -20 would cause the instruction
+# 20 lines above to be executed next.
+#     nop stands for No OPeration - it does nothing. The instruction
+# immediately below it is executed next.
+
+# For example, consider the following program:
+
+# nop +0
+# acc +1
+# jmp +4
+# acc +3
+# jmp -3
+# acc -99
+# acc +1
+# jmp -4
+# acc +6
+
+# These instructions are visited in this order:
+
+# nop +0  | 1
+# acc +1  | 2, 8(!)
+# jmp +4  | 3
+# acc +3  | 6
+# jmp -3  | 7
+# acc -99 |
+# acc +1  | 4
+# jmp -4  | 5
+# acc +6  |
+
+# First, the nop +0 does nothing. Then, the accumulator is increased from 0 to
+# 1 (acc +1) and jmp +4 sets the next instruction to the other acc +1 near the
+# bottom. After it increases the accumulator from 1 to 2, jmp -4 executes,
+# setting the next instruction to the only acc +3. It sets the accumulator to
+# 5, and jmp -3 causes the program to continue back at the first acc +1.
+
+# This is an infinite loop: with this sequence of jumps, the program will run
+# forever. The moment the program tries to run any instruction a second time,
+# you know it will never terminate.
+
+# Immediately before the program would run an instruction a second time, the
+# value in the accumulator is 5.
+
+# Run your copy of the boot code. Immediately before any instruction is
+# executed a second time, what value is in the accumulator?
+
+with open("files/P8.txt", "r") as f:
+    lines = [code for code in f.read().strip().split("\n")]
+
+
+def part_1(lines: List[str], debug: bool = False) -> Tuple[int, bool]:
+    accu = 0
+    executed = []
+    nline = 0
+    while True:
+        try:
+            instr, acc = lines[nline].split(" ")
+            if nline in executed:
+                # we are in a for loop :(
+                break
+            else:
+                executed.append(nline)
+                if instr == "acc":
+                    accu += int(acc)
+                    nline += 1
+                    continue
+                elif instr == "jmp":
+                    nline = nline + int(acc)
+                    continue
+                elif instr == "nop":
+                    nline += 1
+                    continue
+        except IndexError:
+            if debug:
+                print("Code has run successfully")
+            return accu, True
+    if debug:
+        print(f"The value of the accumulator was {accu} before exiting")
+    return accu, False
+
+
+# --- Part Two ---
+
+# After some careful analysis, you believe that exactly one instruction is
+# corrupted.
+
+# Somewhere in the program, either a jmp is supposed to be a nop, or a nop is
+# supposed to be a jmp. (No acc instructions were harmed in the corruption of
+# this boot code.)
+
+# The program is supposed to terminate by attempting to execute an instruction
+# immediately after the last instruction in the file. By changing exactly one
+# jmp or nop, you can repair the boot code and make it terminate correctly.
+
+# For example, consider the same program from above:
+
+# nop +0
+# acc +1
+# jmp +4
+# acc +3
+# jmp -3
+# acc -99
+# acc +1
+# jmp -4
+# acc +6
+
+# If you change the first instruction from nop +0 to jmp +0, it would create a
+# single-instruction infinite loop, never leaving that instruction. If you
+# change almost any of the jmp instructions, the program will still eventually
+# find another jmp instruction and loop forever.
+
+# However, if you change the second-to-last instruction (from jmp -4 to
+# nop -4), the program terminates! The instructions are visited in this order:
+
+# nop +0  | 1
+# acc +1  | 2
+# jmp +4  | 3
+# acc +3  |
+# jmp -3  |
+# acc -99 |
+# acc +1  | 4
+# nop -4  | 5
+# acc +6  | 6
+
+# After the last instruction (acc +6), the program terminates by attempting to
+# run the instruction below the last instruction in the file. With this change,
+# after the program terminates, the accumulator contains the value 8 (acc +1,
+# acc +1, acc +6).
+
+# Fix the program so that it terminates normally by changing exactly one jmp
+# (to nop) or nop (to jmp). What is the value of the accumulator after the
+# program terminates?
+
+
+def part_2() -> None:
+    for nline, _ in enumerate(lines):
+        inst, acc = lines[nline].split(" ")
+        # change instructions
+        if inst == "nop":
+            new_inst = "jmp"
+        elif inst == "jmp":
+            new_inst = "nop"
+
+        # make a copy of the original code
+        new_lines = lines.copy()
+        # change line according to the instructions in the original code
+        new_lines[nline] = " ".join((new_inst, acc))
+        accu, has_ended = part_1(new_lines)
+        if has_ended:
+            print(f"The value of the accumulator was {accu} after terminates")
+            break
+
+
+if __name__ == "__main__":
+    part_1(lines, debug=True)
+    part_2()

src/Year_2020/P9.py

@@ -0,0 +1,149 @@
+import sys
+from itertools import combinations
+
+# --- Day 9: Encoding Error ---
+
+# With your neighbor happily enjoying their video game, you turn your attention
+# to an open data port on the little screen in the seat in front of you.
+
+# Though the port is non-standard, you manage to connect it to your computer
+# through the clever use of several paperclips. Upon connection, the port
+# outputs a series of numbers (your puzzle input).
+
+# The data appears to be encrypted with the eXchange-Masking Addition System
+# (XMAS) which, conveniently for you, is an old cypher with an important
+# weakness.
+
+# XMAS starts by transmitting a preamble of 25 numbers. After that, each number
+# you receive should be the sum of any two of the 25 immediately previous
+# numbers. The two numbers will have different values, and there might be more
+# than one such pair.
+
+# For example, suppose your preamble consists of the numbers 1 through 25 in a
+# random order. To be valid, the next number must be the sum of two of those
+# numbers:
+
+#     26 would be a valid next number, as it could be 1 plus 25 (or many other
+# pairs, like 2 and 24).
+#     49 would be a valid next number, as it is the sum of 24 and 25.
+#     100 would not be valid; no two of the previous 25 numbers sum to 100.
+#     50 would also not be valid; although 25 appears in the previous 25
+# numbers, the two numbers in the pair must be different.
+
+# Suppose the 26th number is 45, and the first number (no longer an option, as
+# it is more than 25 numbers ago) was 20. Now, for the next number to be valid,
+# there needs to be some pair of numbers among 1-19, 21-25, or 45 that add up
+# to it:
+
+#     26 would still be a valid next number, as 1 and 25 are still within the
+# previous 25 numbers.
+#     65 would not be valid, as no two of the available numbers sum to it.
+#     64 and 66 would both be valid, as they are the result of 19+45 and 21+45
+# respectively.
+
+# Here is a larger example which only considers the previous 5 numbers (and has
+# a preamble of length 5):
+
+# 35
+# 20
+# 15
+# 25
+# 47
+# 40
+# 62
+# 55
+# 65
+# 95
+# 102
+# 117
+# 150
+# 182
+# 127
+# 219
+# 299
+# 277
+# 309
+# 576
+
+# In this example, after the 5-number preamble, almost every number is the sum
+# of two of the previous 5 numbers; the only number that does not follow this
+# rule is 127.
+
+# The first step of attacking the weakness in the XMAS data is to find the
+# first number in the list (after the preamble) which is not the sum of two of
+# the 25 numbers before it. What is the first number that does not have this
+# property?
+
+
+with open("files/P9.txt", "r") as f:
+    numbers = [int(num) for num in f.read().strip().split("\n")]
+
+
+def part_1() -> int:
+    window = 25
+    for i in range(window, len(numbers)):
+        if all(
+            [
+                x + y != numbers[i]
+                for x, y in combinations(numbers[i - window : i], 2)
+            ]
+        ):
+            break
+
+    print(f"The first number without the property is {numbers[i]}")
+    return numbers[i]
+
+
+# --- Part Two ---
+
+# The final step in breaking the XMAS encryption relies on the invalid number
+# you just found: you must find a contiguous set of at least two numbers in
+# your list which sum to the invalid number from step 1.
+
+# Again consider the above example:
+
+# 35
+# 20
+# 15
+# 25
+# 47
+# 40
+# 62
+# 55
+# 65
+# 95
+# 102
+# 117
+# 150
+# 182
+# 127
+# 219
+# 299
+# 277
+# 309
+# 576
+
+# In this list, adding up all of the numbers from 15 through 40 produces the
+# invalid number from step 1, 127. (Of course, the contiguous set of numbers in
+# your actual list might be much longer.)
+
+# To find the encryption weakness, add together the smallest and largest number
+# in this contiguous range; in this example, these are 15 and 47, producing 62.
+
+# What is the encryption weakness in your XMAS-encrypted list of numbers?
+
+
+def part_2() -> None:
+    target = invalid_number
+    for i in range(len(numbers)):
+        for j in range(i + 1, len(numbers)):
+            if sum(numbers[i:j]) == target:
+                print(
+                    f"The encryption weakness is {min(numbers[i:j]) + max(numbers[i:j])}"
+                )
+                return
+
+
+if __name__ == "__main__":
+    invalid_number = part_1()
+    part_2()

src/Year_2020/__init__.py

@@ -0,0 +1 @@
+__version__ = "0.1.0"

src/Year_2020/files/P1.txt

@@ -0,0 +1,200 @@
+1865
+1179
+1328
+1390
+322
+1999
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+1808
+1380
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+1748
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+1708
+700
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+1537
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+1289
+1905
+1307
+1845
+1855
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+1554
+1570
+1931
+1780
+1929
+1980
+1978
+1998
+1371
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+1817
+1722
+1545
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+1640
+1414
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+1744
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+1990
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+1919
+1431
+2002
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+1317
+1282
+1349
+1839
+1575
+1730
+1849
+1959
+1971
+2009
+1641
+1402
+1924
+1757
+1605
+1693
+1762
+283
+1525
+1964
+1715
+1602

src/Year_2020/files/P10.txt

@@ -0,0 +1,102 @@
+99
+128
+154
+160
+61
+107
+75
+38
+15
+11
+129
+94
+157
+84
+121
+14
+119
+48
+30
+10
+55
+108
+74
+104
+91
+45
+134
+109
+164
+66
+146
+44
+116
+89
+79
+32
+149
+1
+136
+58
+96
+7
+60
+23
+31
+3
+65
+110
+90
+37
+43
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+52
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+51
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+36
+4
+163
+85
+42
+59
+67
+64
+86
+49
+2
+145
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+22
+24
+33
+137
+16
+27
+70
+133
+130
+20
+21
+83
+143
+100
+41
+76
+17

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src/Year_2020/files/P12.txt

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src/Year_2020/files/P13.txt

@@ -0,0 +1,2 @@
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src/Year_2020/files/P14.txt

@@ -0,0 +1,592 @@
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+mem[44008] = 265991679
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+mem[9648] = 23284946
+mask = 100010000010XX0XX01000000X011100X1X0
+mem[30270] = 421
+mem[52379] = 86815089
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src/Year_2020/files/P15.txt

@@ -0,0 +1 @@
+6,3,15,13,1,0

src/Year_2020/files/P16.txt

@@ -0,0 +1,265 @@
+departure location: 28-184 or 203-952
+departure station: 43-261 or 283-958
+departure platform: 43-549 or 564-970
+departure track: 30-724 or 732-970
+departure date: 37-650 or 657-973
+departure time: 28-911 or 922-965
+arrival location: 41-855 or 863-970
+arrival station: 26-304 or 324-970
+arrival platform: 45-896 or 903-963
+arrival track: 34-458 or 466-962
+class: 43-337 or 363-954
+duration: 33-239 or 260-973
+price: 34-600 or 606-961
+route: 25-686 or 711-973
+row: 36-101 or 124-963
+seat: 25-794 or 806-949
+train: 38-139 or 164-952
+type: 37-619 or 627-956
+wagon: 35-62 or 75-963
+zone: 40-479 or 490-960
+
+your ticket:
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+
+nearby tickets:
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src/Year_2020/files/P17.txt

@@ -0,0 +1,8 @@
+...#...#
+#######.
+....###.
+.#..#...
+#.#.....
+.##.....
+#.####..
+#....##.

src/Year_2020/files/P18.txt

@@ -0,0 +1,377 @@
+6 * 6 * (9 + 4) + (8 * 7)
+3 * 7 + 3 * (3 + 6 * 7 + 9 + 4) * 9
+2 + 5 + (8 * 7) * 4 + 4 + 7
+6 + 3 + 4
+2 * 8 * 5 * ((6 + 9 + 3 * 6) * 4 * 6 + 3) * 7
+7 + 3 + 4 * 6 + ((8 + 6 * 5 * 6 * 2 * 4) * 9 * 9 * 3)
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src/Year_2020/files/P19.txt

@@ -0,0 +1,450 @@
+104: 23 105 | 105 23
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src/Year_2020/files/P7.txt

@@ -0,0 +1,594 @@
+muted tomato bags contain 1 bright brown bag, 1 dotted gold bag, 2 faded gray bags, 1 posh yellow bag.
+posh brown bags contain 1 dark lime bag, 5 mirrored crimson bags, 1 striped chartreuse bag.
+dotted violet bags contain 4 striped white bags.
+mirrored black bags contain 1 clear yellow bag.
+light aqua bags contain 3 clear silver bags, 2 shiny coral bags, 2 muted tomato bags, 2 drab turquoise bags.
+faded violet bags contain no other bags.
+shiny lime bags contain 3 muted brown bags, 4 dull gray bags.
+dark green bags contain 3 muted magenta bags.
+faded bronze bags contain 5 clear lime bags, 3 muted gold bags.
+clear lavender bags contain 4 dark beige bags, 3 mirrored crimson bags, 2 bright blue bags.
+striped bronze bags contain 5 muted olive bags, 4 clear chartreuse bags, 2 vibrant blue bags.
+dim bronze bags contain 1 clear indigo bag.
+dull crimson bags contain 3 dotted violet bags, 2 pale teal bags, 3 plaid bronze bags, 3 faded aqua bags.
+mirrored bronze bags contain 3 striped maroon bags, 4 shiny gold bags, 4 light indigo bags, 5 clear orange bags.
+vibrant white bags contain 4 light lime bags.
+dark blue bags contain 1 faded black bag, 5 mirrored plum bags, 4 muted cyan bags.
+drab bronze bags contain 5 mirrored plum bags, 2 bright turquoise bags, 1 clear gray bag, 3 faded chartreuse bags.
+mirrored lime bags contain 2 faded red bags.
+posh violet bags contain 3 dull chartreuse bags, 5 clear plum bags, 2 mirrored chartreuse bags, 5 dull magenta bags.
+dull bronze bags contain 2 wavy purple bags, 4 drab salmon bags, 5 dull chartreuse bags.
+dull black bags contain 4 shiny crimson bags, 5 dotted turquoise bags, 4 clear beige bags, 2 striped maroon bags.
+striped aqua bags contain 1 faded violet bag, 1 plaid teal bag.
+clear silver bags contain 1 faded green bag.
+dark yellow bags contain 4 shiny plum bags, 2 dull gray bags, 5 light lime bags, 4 shiny orange bags.
+dull coral bags contain 2 striped turquoise bags, 4 mirrored fuchsia bags, 5 faded yellow bags.
+shiny yellow bags contain 5 wavy maroon bags, 1 striped fuchsia bag.
+shiny black bags contain 2 muted tomato bags, 3 muted maroon bags, 4 shiny orange bags.
+posh aqua bags contain 3 wavy cyan bags, 5 posh brown bags, 4 dim plum bags.
+plaid maroon bags contain 5 faded aqua bags, 5 mirrored olive bags, 5 plaid gold bags, 2 vibrant gray bags.
+mirrored gold bags contain 3 shiny tomato bags, 3 shiny crimson bags.
+muted lavender bags contain 2 muted tan bags.
+vibrant magenta bags contain 1 dull purple bag, 2 vibrant coral bags, 5 faded silver bags.
+striped tomato bags contain 3 faded beige bags, 3 pale tan bags, 2 drab orange bags.
+dull turquoise bags contain 3 striped green bags, 2 shiny blue bags, 2 dim purple bags.
+shiny aqua bags contain 1 clear salmon bag.
+drab silver bags contain 2 shiny gold bags, 4 posh yellow bags.
+plaid gold bags contain 3 bright lavender bags, 5 plaid cyan bags, 1 bright brown bag, 5 faded turquoise bags.
+wavy lavender bags contain 2 plaid lavender bags, 3 plaid red bags, 4 posh cyan bags, 4 bright aqua bags.
+dark bronze bags contain 4 muted cyan bags, 4 clear cyan bags.
+plaid lime bags contain 1 posh teal bag.
+mirrored magenta bags contain 5 dotted turquoise bags.
+light turquoise bags contain 5 posh tan bags, 2 posh aqua bags.
+dark indigo bags contain 5 clear crimson bags, 3 posh lime bags.
+mirrored brown bags contain 4 posh white bags, 1 dim lime bag, 2 pale magenta bags, 1 mirrored chartreuse bag.
+vibrant lavender bags contain 5 posh plum bags, 1 posh aqua bag.
+dim crimson bags contain 3 vibrant beige bags, 4 clear indigo bags, 3 dim lime bags, 4 striped olive bags.
+vibrant violet bags contain 5 vibrant red bags, 1 striped white bag.
+dull silver bags contain 4 striped bronze bags, 1 plaid plum bag, 1 dim tomato bag.
+plaid brown bags contain 5 drab blue bags, 2 light turquoise bags, 5 dull lavender bags.
+dim aqua bags contain 4 plaid lavender bags, 5 muted maroon bags.
+faded plum bags contain 3 striped tan bags.
+drab black bags contain 5 clear bronze bags, 4 plaid red bags, 1 drab orange bag, 5 clear lavender bags.
+pale bronze bags contain 5 dim plum bags, 1 drab lavender bag.
+mirrored purple bags contain 5 mirrored crimson bags.
+mirrored red bags contain 5 striped magenta bags, 4 mirrored chartreuse bags, 5 pale cyan bags.
+posh lime bags contain 5 dull lavender bags, 1 faded turquoise bag, 2 striped coral bags, 3 striped green bags.
+vibrant black bags contain 2 drab salmon bags, 4 clear tomato bags, 2 faded turquoise bags.
+striped orange bags contain 2 plaid blue bags, 5 pale teal bags, 1 drab maroon bag.
+dotted tomato bags contain 1 dim white bag, 5 dotted magenta bags, 4 dull tomato bags, 1 dull gold bag.
+posh orange bags contain 3 posh gold bags, 4 light lime bags, 2 faded violet bags, 5 faded turquoise bags.
+muted gold bags contain 4 faded maroon bags, 5 dark turquoise bags, 4 bright turquoise bags, 3 plaid silver bags.
+striped yellow bags contain 4 drab fuchsia bags, 1 dim lime bag.
+dotted green bags contain 4 clear beige bags.
+faded fuchsia bags contain 1 dull blue bag, 4 bright coral bags.
+shiny lavender bags contain 3 mirrored beige bags, 1 dark coral bag, 1 muted turquoise bag, 3 light brown bags.
+bright maroon bags contain 5 faded lime bags, 3 dim brown bags, 3 vibrant yellow bags.
+dark maroon bags contain 2 drab red bags.
+dark white bags contain 3 wavy turquoise bags, 4 dotted red bags, 3 plaid fuchsia bags.
+dotted aqua bags contain 4 clear green bags, 3 drab magenta bags, 1 dull tomato bag, 3 striped coral bags.
+faded red bags contain 3 plaid purple bags, 3 dark lime bags.
+wavy yellow bags contain 1 bright yellow bag, 2 light chartreuse bags, 1 dull chartreuse bag.
+posh beige bags contain 5 shiny orange bags, 4 drab silver bags, 1 striped cyan bag.
+dull fuchsia bags contain 5 faded lavender bags, 1 dark cyan bag, 2 clear tomato bags, 3 mirrored chartreuse bags.
+muted silver bags contain 5 muted brown bags, 2 striped brown bags, 4 mirrored bronze bags, 5 faded plum bags.
+dim white bags contain 1 dull bronze bag, 3 dotted beige bags, 4 wavy tan bags, 4 vibrant white bags.
+faded green bags contain 1 clear indigo bag, 2 pale cyan bags, 1 wavy brown bag.
+pale blue bags contain 2 posh turquoise bags.
+muted green bags contain 1 dull bronze bag.
+dull indigo bags contain 4 shiny silver bags, 1 wavy red bag, 4 muted orange bags.
+muted plum bags contain 1 shiny lime bag, 4 plaid tomato bags, 2 faded indigo bags.
+striped magenta bags contain 1 mirrored chartreuse bag, 1 faded beige bag, 3 bright blue bags, 2 clear yellow bags.
+clear coral bags contain 4 dark turquoise bags, 2 posh magenta bags, 2 shiny aqua bags, 5 pale aqua bags.
+pale olive bags contain 4 shiny crimson bags, 3 wavy cyan bags, 1 vibrant beige bag, 3 dark beige bags.
+striped plum bags contain 2 light salmon bags.
+dim indigo bags contain 2 shiny lavender bags, 4 clear lavender bags, 1 bright tomato bag, 1 vibrant bronze bag.
+striped green bags contain 5 muted cyan bags.
+shiny salmon bags contain 1 wavy brown bag.
+wavy aqua bags contain 5 muted white bags, 1 mirrored gray bag, 5 dull lavender bags.
+posh gold bags contain 1 vibrant beige bag, 3 dull chartreuse bags, 4 mirrored brown bags.
+wavy blue bags contain 3 shiny gray bags, 4 posh turquoise bags.
+dark fuchsia bags contain 2 posh fuchsia bags, 4 bright indigo bags, 3 posh maroon bags, 2 posh aqua bags.
+shiny purple bags contain 5 clear maroon bags, 3 bright indigo bags.
+dull chartreuse bags contain no other bags.
+light beige bags contain 4 dim purple bags, 1 posh red bag, 4 clear aqua bags, 1 striped coral bag.
+wavy purple bags contain 3 bright turquoise bags.
+clear yellow bags contain 2 posh plum bags.
+clear violet bags contain 5 faded chartreuse bags, 5 clear green bags, 3 wavy brown bags.
+striped white bags contain 3 light white bags.
+clear green bags contain 4 muted bronze bags.
+drab crimson bags contain 1 posh tan bag.
+dark turquoise bags contain 2 shiny tomato bags, 5 wavy maroon bags, 5 dim aqua bags.
+drab magenta bags contain 1 pale gray bag, 4 striped olive bags, 2 posh brown bags.
+dotted gray bags contain 1 dull tomato bag, 1 posh yellow bag.
+plaid magenta bags contain 5 clear green bags, 2 dotted bronze bags, 5 drab black bags.
+striped indigo bags contain 4 mirrored magenta bags, 5 posh beige bags.
+bright fuchsia bags contain 4 dotted orange bags, 2 faded green bags, 1 bright turquoise bag.
+mirrored teal bags contain 1 mirrored chartreuse bag, 2 drab brown bags, 2 dull black bags.
+dotted purple bags contain 5 dotted turquoise bags, 5 dotted orange bags, 5 shiny gold bags.
+dim lavender bags contain 3 shiny coral bags, 5 mirrored crimson bags, 2 clear yellow bags, 2 muted orange bags.
+shiny gray bags contain 3 vibrant purple bags.
+mirrored white bags contain 1 mirrored brown bag, 3 striped purple bags.
+light cyan bags contain 4 bright maroon bags, 3 posh violet bags, 4 faded aqua bags, 5 dull tan bags.
+wavy gray bags contain 4 mirrored violet bags, 1 muted orange bag.
+striped silver bags contain 2 bright silver bags, 5 posh aqua bags.
+dull green bags contain 5 mirrored chartreuse bags, 2 dim beige bags.
+faded black bags contain 5 wavy turquoise bags, 1 muted tan bag, 1 dull chartreuse bag.
+posh crimson bags contain 2 muted tan bags, 4 striped beige bags, 1 mirrored lavender bag, 5 faded white bags.
+pale teal bags contain 4 posh aqua bags, 5 mirrored plum bags, 1 muted turquoise bag.
+muted tan bags contain 1 striped coral bag, 1 mirrored brown bag, 1 posh gold bag, 4 muted cyan bags.
+posh indigo bags contain 5 shiny tomato bags, 1 clear salmon bag, 3 dotted black bags, 1 striped maroon bag.
+clear gray bags contain 1 bright violet bag.
+striped olive bags contain no other bags.
+mirrored silver bags contain 4 shiny maroon bags, 4 muted coral bags.
+clear brown bags contain 2 dotted magenta bags, 3 clear aqua bags, 5 faded lime bags.
+striped lavender bags contain 4 dull turquoise bags.
+plaid teal bags contain 3 muted bronze bags, 2 drab orange bags.
+faded blue bags contain 3 bright teal bags, 2 posh brown bags.
+drab gold bags contain 3 dotted black bags, 5 striped yellow bags.
+plaid gray bags contain 2 drab magenta bags, 4 bright yellow bags, 5 pale beige bags.
+mirrored coral bags contain 1 shiny fuchsia bag, 5 vibrant gray bags, 1 wavy maroon bag.
+drab fuchsia bags contain 3 dull silver bags, 5 posh brown bags.
+wavy olive bags contain 2 plaid lavender bags.
+vibrant maroon bags contain 4 drab brown bags, 1 dark indigo bag, 1 clear gray bag.
+faded magenta bags contain 1 dim crimson bag.
+pale aqua bags contain 3 dotted turquoise bags.
+dotted orange bags contain 2 light coral bags.
+dim coral bags contain 4 pale gold bags.
+faded turquoise bags contain 5 posh gold bags, 2 bright turquoise bags, 1 drab magenta bag.
+striped tan bags contain 3 bright red bags.
+drab green bags contain 3 muted gold bags, 3 plaid gold bags.
+posh red bags contain 3 mirrored red bags, 2 striped fuchsia bags, 3 faded violet bags, 1 bright black bag.
+drab lime bags contain 2 dotted brown bags.
+plaid tomato bags contain 4 muted olive bags, 2 striped tan bags.
+striped blue bags contain 3 drab lavender bags.
+bright bronze bags contain 2 clear gold bags, 2 striped gold bags, 4 dim coral bags, 5 vibrant purple bags.
+muted white bags contain 4 light olive bags.
+pale violet bags contain 1 bright teal bag, 5 bright lavender bags, 2 faded green bags, 5 clear gold bags.
+striped violet bags contain 2 drab turquoise bags, 2 bright red bags, 3 vibrant blue bags.
+dull beige bags contain 2 vibrant beige bags.
+light lime bags contain 1 bright turquoise bag, 3 vibrant beige bags, 2 wavy bronze bags, 1 muted cyan bag.
+striped gray bags contain 5 vibrant brown bags, 3 wavy red bags, 3 plaid bronze bags, 4 wavy aqua bags.
+dark violet bags contain 5 dim indigo bags, 1 dim chartreuse bag, 5 dim purple bags, 1 striped magenta bag.
+dim turquoise bags contain 3 muted tan bags.
+pale green bags contain 2 pale olive bags, 5 wavy red bags, 5 pale lime bags.
+dark beige bags contain 1 light green bag, 4 drab beige bags, 2 dim cyan bags.
+drab maroon bags contain 3 light coral bags, 1 clear crimson bag.
+dark gold bags contain 1 dim black bag, 1 dotted gold bag, 3 dim gray bags, 5 wavy red bags.
+drab white bags contain 1 pale gray bag.
+mirrored olive bags contain 4 mirrored crimson bags, 3 faded beige bags.
+pale yellow bags contain 4 shiny maroon bags, 5 dim tomato bags, 4 shiny olive bags, 1 faded olive bag.
+striped lime bags contain 2 light lavender bags, 4 shiny aqua bags, 2 pale maroon bags.
+mirrored tomato bags contain 2 clear crimson bags, 2 pale beige bags, 5 striped purple bags, 4 clear tomato bags.
+dark black bags contain 1 mirrored orange bag, 4 shiny brown bags.
+plaid chartreuse bags contain 1 dim beige bag, 2 vibrant red bags, 1 bright lavender bag, 5 mirrored tomato bags.
+posh gray bags contain 5 dotted blue bags.
+mirrored crimson bags contain no other bags.
+pale silver bags contain 3 plaid cyan bags.
+shiny green bags contain 4 dotted violet bags, 3 drab olive bags.
+light brown bags contain 2 vibrant yellow bags, 3 dim tomato bags.
+muted yellow bags contain 1 drab magenta bag, 5 muted olive bags, 3 clear crimson bags, 2 mirrored brown bags.
+mirrored plum bags contain 4 muted tan bags, 1 posh brown bag, 3 faded silver bags, 1 dull gold bag.
+dark chartreuse bags contain 1 dim orange bag, 2 dull indigo bags, 1 dark beige bag.
+wavy brown bags contain 3 striped chartreuse bags.
+faded white bags contain 5 striped magenta bags, 2 muted turquoise bags, 2 bright cyan bags, 3 clear chartreuse bags.
+bright green bags contain 1 light salmon bag.
+bright salmon bags contain 3 posh tan bags, 2 drab orange bags.
+drab plum bags contain 2 posh turquoise bags, 4 dark salmon bags, 3 posh lime bags, 4 light coral bags.
+drab tomato bags contain 5 faded teal bags, 4 mirrored black bags, 4 mirrored lavender bags, 1 drab salmon bag.
+faded olive bags contain 4 light green bags, 4 clear bronze bags.
+plaid bronze bags contain 3 dim plum bags, 2 posh white bags, 1 dotted gray bag, 2 dull gray bags.
+light salmon bags contain 2 faded violet bags.
+bright red bags contain 1 posh plum bag, 4 clear crimson bags.
+faded maroon bags contain 3 striped olive bags, 5 wavy bronze bags, 3 bright lavender bags.
+mirrored orange bags contain 1 bright silver bag, 3 bright red bags, 1 drab brown bag.
+dotted indigo bags contain 1 dull black bag, 2 posh white bags.
+wavy teal bags contain 1 dotted tomato bag, 4 faded beige bags, 5 dull beige bags, 1 clear crimson bag.
+dim magenta bags contain 2 posh maroon bags, 1 dim yellow bag, 4 faded red bags, 5 pale olive bags.
+light red bags contain 3 dim silver bags, 5 drab maroon bags, 3 shiny red bags.
+plaid red bags contain 1 posh yellow bag, 5 muted turquoise bags, 1 bright red bag.
+pale magenta bags contain 4 dim plum bags, 5 dull tomato bags, 3 bright turquoise bags, 1 dark lime bag.
+posh yellow bags contain 5 posh brown bags.
+bright lavender bags contain 5 clear aqua bags, 1 posh yellow bag, 4 pale magenta bags.
+dim fuchsia bags contain 4 muted green bags, 3 mirrored beige bags, 5 dark brown bags, 1 dull crimson bag.
+vibrant salmon bags contain 4 clear lime bags, 5 plaid brown bags, 5 pale white bags.
+dim black bags contain 3 faded magenta bags, 5 dull gray bags, 2 vibrant blue bags, 2 vibrant bronze bags.
+dotted yellow bags contain 3 drab turquoise bags, 3 muted cyan bags, 3 clear gray bags.
+mirrored beige bags contain 1 drab salmon bag, 1 striped olive bag, 5 dotted gold bags, 2 faded beige bags.
+striped brown bags contain 5 bright maroon bags, 2 light gold bags.
+striped chartreuse bags contain no other bags.
+dark lime bags contain 3 dull chartreuse bags, 5 dim violet bags, 2 mirrored crimson bags.
+faded salmon bags contain 2 dull chartreuse bags, 4 posh brown bags, 3 mirrored olive bags, 5 light green bags.
+bright brown bags contain 2 vibrant bronze bags, 2 dim gray bags.
+dull tan bags contain 5 posh indigo bags.
+dim silver bags contain 1 vibrant tomato bag, 4 dim maroon bags.
+clear lime bags contain 4 clear gray bags, 1 clear tan bag, 4 posh brown bags, 1 dark indigo bag.
+shiny beige bags contain 5 plaid lavender bags, 4 drab red bags, 4 striped green bags, 2 vibrant tomato bags.
+dim tomato bags contain 3 clear yellow bags, 2 wavy gold bags, 3 pale cyan bags.
+dotted tan bags contain 3 pale white bags.
+dark crimson bags contain 1 dim coral bag, 4 muted brown bags, 2 clear chartreuse bags.
+plaid coral bags contain 1 dark yellow bag, 3 muted tomato bags.
+faded lavender bags contain 1 muted lavender bag, 5 mirrored red bags, 1 muted orange bag, 3 light coral bags.
+vibrant crimson bags contain 3 plaid plum bags, 2 bright olive bags.
+faded chartreuse bags contain 5 dim cyan bags.
+dark gray bags contain 5 clear plum bags, 5 faded aqua bags.
+shiny crimson bags contain 4 mirrored gray bags, 2 dull green bags.
+muted purple bags contain 5 dull tomato bags.
+light coral bags contain no other bags.
+pale beige bags contain 3 dark indigo bags, 5 bright turquoise bags, 3 vibrant lavender bags, 1 pale magenta bag.
+posh turquoise bags contain 4 faded green bags, 2 light coral bags, 1 light teal bag, 3 dim tomato bags.
+clear aqua bags contain 2 wavy cyan bags, 1 muted yellow bag.
+posh silver bags contain 4 plaid fuchsia bags, 2 dotted violet bags, 3 drab olive bags.
+dark silver bags contain 4 vibrant purple bags, 1 vibrant cyan bag.
+dull plum bags contain 3 dim red bags, 3 wavy turquoise bags, 2 mirrored beige bags.
+muted magenta bags contain 3 faded lime bags, 2 muted crimson bags, 1 mirrored gray bag.
+plaid indigo bags contain 4 wavy aqua bags, 1 plaid silver bag, 1 light blue bag, 3 wavy magenta bags.
+wavy violet bags contain 2 striped bronze bags.
+plaid yellow bags contain 3 dull plum bags, 1 striped blue bag, 5 faded aqua bags.
+dotted bronze bags contain 3 bright lavender bags, 1 wavy lime bag, 2 dotted cyan bags, 1 mirrored beige bag.
+striped coral bags contain 4 dark cyan bags, 4 wavy bronze bags, 2 dotted magenta bags.
+posh tan bags contain 4 mirrored brown bags, 3 dotted teal bags, 5 light lime bags.
+bright white bags contain 2 muted indigo bags, 2 wavy aqua bags.
+posh cyan bags contain 4 muted tomato bags, 1 posh tan bag, 4 mirrored lavender bags, 2 drab blue bags.
+wavy gold bags contain 1 dim beige bag, 5 dotted teal bags, 2 dotted turquoise bags.
+mirrored cyan bags contain 2 drab silver bags, 5 muted orange bags, 5 mirrored olive bags.
+dim orange bags contain 3 striped blue bags.
+muted indigo bags contain 4 clear aqua bags, 2 mirrored tan bags, 2 clear green bags, 2 posh brown bags.
+dull tomato bags contain 4 posh plum bags.
+plaid orange bags contain 1 striped coral bag.
+posh bronze bags contain 3 vibrant purple bags, 2 dim brown bags, 1 clear violet bag, 4 mirrored gold bags.
+dotted chartreuse bags contain 3 dull crimson bags.
+faded coral bags contain 1 wavy tomato bag.
+pale salmon bags contain 2 dull purple bags, 2 dark tomato bags.
+wavy crimson bags contain 4 bright fuchsia bags, 5 light lime bags, 1 posh tan bag.
+pale fuchsia bags contain 2 wavy black bags, 1 muted crimson bag, 2 dotted aqua bags.
+striped fuchsia bags contain 1 mirrored violet bag, 3 dull chartreuse bags, 1 shiny lavender bag.
+plaid cyan bags contain 2 vibrant lavender bags.
+dotted beige bags contain 4 dim beige bags, 2 vibrant lavender bags.
+mirrored turquoise bags contain 4 mirrored beige bags, 3 dotted turquoise bags, 1 striped maroon bag.
+faded indigo bags contain 3 dark lime bags, 5 mirrored salmon bags, 3 drab red bags.
+dotted silver bags contain 3 dark yellow bags, 4 posh crimson bags.
+light silver bags contain 2 wavy orange bags, 2 vibrant maroon bags, 4 wavy beige bags.
+vibrant cyan bags contain 1 shiny fuchsia bag, 3 posh plum bags.
+pale crimson bags contain 1 dotted gold bag, 4 mirrored brown bags, 1 mirrored olive bag.
+vibrant beige bags contain 4 dull white bags.
+faded tomato bags contain 4 mirrored salmon bags, 5 faded green bags.
+dull maroon bags contain 5 mirrored olive bags, 4 dim crimson bags.
+muted aqua bags contain 2 plaid olive bags, 1 shiny indigo bag.
+shiny teal bags contain 4 striped bronze bags, 3 dotted aqua bags.
+dotted cyan bags contain 3 clear yellow bags, 5 faded maroon bags, 3 dim gray bags.
+wavy beige bags contain 1 posh cyan bag, 4 wavy magenta bags, 5 dim green bags.
+bright silver bags contain 4 clear beige bags, 2 faded chartreuse bags, 4 clear aqua bags.
+striped purple bags contain 2 clear chartreuse bags, 4 dotted magenta bags.
+clear gold bags contain 2 vibrant lavender bags, 3 posh aqua bags, 4 light green bags, 3 pale cyan bags.
+dull cyan bags contain 5 dim tomato bags, 1 dim orange bag.
+light crimson bags contain 1 dull green bag.
+faded crimson bags contain 4 plaid coral bags, 3 dotted orange bags.
+pale gray bags contain 3 dark lime bags.
+faded gold bags contain 1 muted crimson bag, 1 plaid red bag, 5 wavy cyan bags.
+plaid plum bags contain 4 bright blue bags.
+wavy tomato bags contain 5 posh brown bags, 2 striped maroon bags, 4 faded indigo bags, 3 clear gold bags.
+shiny fuchsia bags contain 2 mirrored gray bags.
+wavy magenta bags contain 2 posh olive bags.
+dim purple bags contain 1 vibrant yellow bag.
+wavy black bags contain 1 pale green bag.
+wavy silver bags contain 3 dull yellow bags, 5 faded magenta bags, 4 mirrored black bags, 3 dull indigo bags.
+shiny cyan bags contain 2 shiny salmon bags, 2 pale olive bags, 1 shiny red bag.
+striped cyan bags contain 5 pale magenta bags, 3 faded silver bags.
+dim salmon bags contain 2 vibrant aqua bags, 1 drab red bag.
+bright chartreuse bags contain 1 shiny tomato bag, 2 wavy yellow bags, 2 wavy silver bags.
+drab coral bags contain 1 drab silver bag.
+pale white bags contain 2 faded red bags, 4 pale gray bags, 2 mirrored lavender bags.
+dim violet bags contain 3 dull white bags.
+muted red bags contain 3 shiny lime bags, 3 plaid beige bags, 2 shiny aqua bags.
+bright teal bags contain 3 dim tomato bags, 3 mirrored turquoise bags, 5 vibrant yellow bags.
+striped maroon bags contain 1 dim plum bag, 2 wavy cyan bags, 1 dull beige bag.
+vibrant green bags contain 5 dotted salmon bags, 4 mirrored olive bags.
+dim tan bags contain 2 shiny white bags, 3 posh lime bags, 1 dotted teal bag.
+muted olive bags contain 4 mirrored chartreuse bags.
+muted salmon bags contain 1 clear salmon bag.
+dim beige bags contain 2 pale gray bags, 2 mirrored olive bags, 3 clear yellow bags.
+light tomato bags contain 1 pale white bag, 2 dark teal bags, 5 shiny lime bags.
+striped beige bags contain 4 bright plum bags, 1 plaid olive bag, 2 pale crimson bags.
+light indigo bags contain 4 bright fuchsia bags, 1 dull gold bag, 5 dark beige bags, 3 mirrored teal bags.
+drab gray bags contain 3 dark coral bags, 5 mirrored teal bags, 5 dim purple bags, 4 mirrored violet bags.
+dark teal bags contain 2 muted orange bags.
+dim plum bags contain 1 posh brown bag, 4 dotted orange bags, 3 dotted gold bags, 2 drab salmon bags.
+dim green bags contain 3 dark beige bags, 3 bright tan bags, 4 striped white bags, 5 faded white bags.
+drab teal bags contain 1 light gray bag.
+dark aqua bags contain 5 vibrant green bags, 3 shiny teal bags, 3 dull red bags.
+wavy lime bags contain 5 mirrored brown bags, 2 dark cyan bags, 4 dull green bags, 1 wavy cyan bag.
+muted violet bags contain 3 light plum bags.
+posh green bags contain 2 shiny fuchsia bags, 3 posh aqua bags, 5 dotted turquoise bags.
+clear blue bags contain 4 vibrant gray bags, 4 shiny plum bags.
+muted brown bags contain 1 dark lime bag, 1 mirrored olive bag, 1 faded beige bag.
+dotted black bags contain 5 pale beige bags, 1 dotted tomato bag, 2 wavy cyan bags, 3 wavy aqua bags.
+muted chartreuse bags contain 5 muted lavender bags, 5 clear aqua bags, 5 bright maroon bags, 4 plaid bronze bags.
+bright purple bags contain 4 bright fuchsia bags, 3 pale tomato bags, 5 mirrored tomato bags.
+drab orange bags contain 5 light crimson bags, 4 dark beige bags.
+clear white bags contain 2 shiny salmon bags.
+posh lavender bags contain 4 posh tan bags, 1 posh magenta bag.
+faded purple bags contain 2 dull maroon bags, 2 mirrored coral bags, 4 vibrant green bags.
+dotted olive bags contain 5 dull chartreuse bags, 2 wavy magenta bags, 1 pale tan bag, 3 dim turquoise bags.
+muted crimson bags contain 1 striped olive bag, 3 dull chartreuse bags, 1 light bronze bag.
+striped turquoise bags contain 2 light blue bags.
+bright indigo bags contain 3 dull green bags, 1 pale teal bag.
+drab purple bags contain 3 bright aqua bags, 4 dull tan bags, 2 mirrored indigo bags.
+wavy salmon bags contain 4 dark olive bags, 5 dull chartreuse bags, 3 vibrant purple bags, 1 pale lime bag.
+drab olive bags contain 5 dim crimson bags, 3 posh green bags, 3 clear brown bags, 1 posh teal bag.
+muted orange bags contain 5 shiny orange bags, 3 muted black bags, 5 bright blue bags, 1 mirrored olive bag.
+drab salmon bags contain 5 light coral bags.
+dark salmon bags contain 4 bright lavender bags, 5 clear gray bags.
+posh blue bags contain 1 drab coral bag, 3 faded green bags.
+shiny violet bags contain 1 posh yellow bag, 4 light turquoise bags, 4 drab silver bags, 2 muted teal bags.
+clear tomato bags contain 3 bright plum bags, 3 muted indigo bags, 2 muted maroon bags, 3 dotted cyan bags.
+dotted coral bags contain 4 dark tomato bags, 5 dotted magenta bags, 5 wavy gold bags.
+striped teal bags contain 1 plaid brown bag, 2 bright plum bags, 1 mirrored blue bag.
+pale maroon bags contain 4 posh gold bags, 5 striped magenta bags, 2 dim brown bags, 1 clear chartreuse bag.
+muted lime bags contain 5 dull lavender bags, 3 plaid gray bags, 3 shiny tomato bags, 4 drab tomato bags.
+mirrored green bags contain 2 striped purple bags.
+posh chartreuse bags contain 4 dull tan bags, 2 plaid coral bags, 1 plaid cyan bag, 4 clear bronze bags.
+shiny bronze bags contain 3 striped indigo bags, 4 mirrored white bags, 1 faded coral bag, 1 clear violet bag.
+dull olive bags contain 1 muted coral bag, 3 clear tomato bags, 3 bright tan bags.
+light maroon bags contain 5 mirrored tomato bags.
+light chartreuse bags contain 3 dark indigo bags, 2 mirrored coral bags.
+light bronze bags contain 1 bright yellow bag.
+dim chartreuse bags contain 3 plaid gray bags, 2 faded teal bags.
+dotted lavender bags contain 3 bright teal bags, 3 vibrant green bags.
+dark tan bags contain 5 posh black bags.
+muted blue bags contain 2 mirrored salmon bags, 4 striped olive bags.
+mirrored chartreuse bags contain 2 bright turquoise bags.
+clear cyan bags contain 5 dotted gold bags, 3 shiny fuchsia bags, 3 dim violet bags.
+light white bags contain 2 vibrant bronze bags.
+pale orange bags contain 3 plaid aqua bags.
+faded brown bags contain 2 shiny blue bags, 1 dark cyan bag.
+plaid salmon bags contain 5 wavy yellow bags, 3 faded aqua bags, 1 dull yellow bag.
+shiny blue bags contain 2 bright violet bags, 3 light lime bags, 5 clear blue bags.
+dull orange bags contain 4 faded turquoise bags, 2 dim lavender bags.
+dull gold bags contain 2 dotted gray bags.
+vibrant tan bags contain 2 posh turquoise bags, 2 muted orange bags, 4 plaid blue bags.
+shiny white bags contain 1 pale chartreuse bag, 4 dark tomato bags.
+light olive bags contain 3 striped bronze bags, 3 posh white bags.
+dark lavender bags contain 3 drab magenta bags, 3 plaid black bags, 2 dull red bags.
+shiny turquoise bags contain 1 vibrant black bag, 5 wavy fuchsia bags.
+clear fuchsia bags contain 2 dull aqua bags, 2 shiny coral bags.
+shiny coral bags contain 2 pale magenta bags, 4 muted bronze bags, 4 light coral bags.
+striped salmon bags contain 3 posh olive bags.
+shiny olive bags contain 5 pale chartreuse bags.
+dotted brown bags contain 1 mirrored brown bag.
+dull red bags contain 1 faded chartreuse bag, 1 dull chartreuse bag, 5 clear lavender bags, 5 wavy salmon bags.
+bright beige bags contain 1 posh crimson bag, 1 posh chartreuse bag.
+plaid crimson bags contain 5 wavy magenta bags, 4 muted lime bags, 1 muted maroon bag, 3 dim green bags.
+wavy maroon bags contain 1 wavy cyan bag, 1 bright turquoise bag, 4 shiny coral bags, 3 drab beige bags.
+vibrant fuchsia bags contain 2 dull tomato bags, 1 light coral bag.
+pale brown bags contain 5 striped gray bags, 1 light plum bag, 1 pale chartreuse bag, 2 shiny tomato bags.
+light plum bags contain 4 bright yellow bags, 2 dark tomato bags, 4 plaid coral bags.
+dull gray bags contain 1 mirrored olive bag.
+light yellow bags contain 5 bright magenta bags, 5 pale gray bags, 4 mirrored red bags.
+dull lime bags contain 3 dark indigo bags, 3 mirrored salmon bags, 5 striped white bags, 3 posh aqua bags.
+drab yellow bags contain 4 shiny white bags, 3 faded green bags.
+bright gray bags contain 5 drab olive bags.
+bright coral bags contain 3 mirrored brown bags, 5 dim bronze bags, 5 muted white bags, 1 dim crimson bag.
+dim maroon bags contain 1 vibrant red bag, 3 faded fuchsia bags, 3 vibrant blue bags.
+plaid turquoise bags contain 3 bright yellow bags, 5 dark purple bags.
+plaid violet bags contain 1 clear yellow bag, 2 posh violet bags, 3 mirrored bronze bags, 3 vibrant tan bags.
+dotted maroon bags contain 2 striped fuchsia bags, 3 striped white bags, 2 light blue bags, 5 plaid cyan bags.
+dull blue bags contain 1 wavy brown bag.
+drab tan bags contain 1 clear indigo bag, 4 faded violet bags, 3 pale cyan bags.
+posh maroon bags contain 5 striped aqua bags, 3 wavy coral bags, 3 mirrored olive bags.
+dim blue bags contain 4 dotted maroon bags, 5 striped black bags, 5 plaid gray bags.
+faded teal bags contain 3 clear indigo bags, 5 dark beige bags, 1 shiny cyan bag, 3 bright tomato bags.
+clear bronze bags contain 1 muted yellow bag.
+dark plum bags contain 3 faded white bags, 2 wavy brown bags, 5 pale white bags.
+wavy cyan bags contain 3 mirrored brown bags, 5 faded silver bags, 1 dim lime bag, 5 clear indigo bags.
+posh plum bags contain no other bags.
+wavy plum bags contain 4 dull indigo bags, 3 faded black bags, 4 bright turquoise bags, 5 bright red bags.
+dim red bags contain 2 light cyan bags.
+pale coral bags contain 2 wavy brown bags, 1 dull gray bag.
+posh teal bags contain 2 striped purple bags, 1 dotted gray bag.
+clear tan bags contain 5 striped maroon bags.
+wavy chartreuse bags contain 5 bright chartreuse bags.
+posh black bags contain 2 mirrored gray bags, 3 mirrored beige bags, 2 clear aqua bags, 2 bright plum bags.
+muted maroon bags contain 5 vibrant red bags, 1 plaid lavender bag.
+vibrant silver bags contain 1 mirrored salmon bag, 5 clear lime bags, 4 shiny violet bags.
+pale indigo bags contain 4 clear maroon bags, 4 wavy cyan bags.
+mirrored salmon bags contain 4 pale gray bags, 4 dim beige bags.
+pale lavender bags contain 1 wavy turquoise bag, 2 striped green bags, 4 light coral bags, 4 clear chartreuse bags.
+drab beige bags contain no other bags.
+wavy coral bags contain 3 muted cyan bags, 3 dull bronze bags.
+plaid black bags contain 2 posh orange bags, 4 shiny tan bags, 5 mirrored brown bags, 3 shiny plum bags.
+light gray bags contain 2 dim crimson bags, 1 faded salmon bag.
+shiny plum bags contain 2 dull tomato bags, 1 mirrored tan bag, 2 dull green bags, 1 dark lime bag.
+drab blue bags contain 5 mirrored chartreuse bags, 5 dull tomato bags, 2 mirrored beige bags, 4 posh plum bags.
+faded orange bags contain 5 striped yellow bags, 1 drab gray bag, 5 vibrant purple bags, 3 muted fuchsia bags.
+plaid purple bags contain 1 faded silver bag, 4 shiny gold bags, 4 dim cyan bags.
+pale tan bags contain 3 shiny gold bags, 2 shiny lavender bags, 3 pale cyan bags.
+bright crimson bags contain 5 clear cyan bags.
+dotted lime bags contain 2 posh violet bags, 5 striped beige bags, 4 bright teal bags.
+posh purple bags contain 5 mirrored silver bags, 1 dim tan bag, 1 pale white bag.
+vibrant red bags contain 3 pale chartreuse bags, 2 dim purple bags, 4 wavy maroon bags, 1 dull gray bag.
+light fuchsia bags contain 1 drab maroon bag, 2 faded violet bags.
+bright blue bags contain 4 drab salmon bags, 4 dotted gold bags.
+bright turquoise bags contain 2 dotted gold bags, 2 dim violet bags, 1 vibrant bronze bag.
+light orange bags contain 1 plaid olive bag.
+light gold bags contain 2 clear plum bags, 2 striped bronze bags, 4 dark yellow bags, 2 posh orange bags.
+dotted magenta bags contain 4 wavy cyan bags.
+muted teal bags contain 1 clear beige bag, 4 plaid brown bags, 2 posh plum bags.
+light black bags contain 3 drab bronze bags, 2 plaid silver bags, 3 mirrored coral bags.
+wavy turquoise bags contain 5 dotted orange bags.
+muted fuchsia bags contain 3 dull black bags, 1 striped coral bag, 1 bright magenta bag.
+posh white bags contain 5 bright blue bags, 2 pale gray bags, 2 dark lime bags, 4 dotted gold bags.
+striped black bags contain 5 drab maroon bags, 1 faded green bag, 2 mirrored silver bags, 2 shiny white bags.
+striped gold bags contain 4 dim gray bags.
+dull violet bags contain 3 light plum bags, 4 faded turquoise bags, 3 mirrored cyan bags.
+striped red bags contain 5 plaid orange bags, 1 clear plum bag, 2 vibrant lavender bags.
+bright cyan bags contain 1 dim brown bag, 4 mirrored turquoise bags.
+dim cyan bags contain 5 bright turquoise bags.
+shiny chartreuse bags contain 2 striped white bags.
+vibrant blue bags contain 5 drab tan bags, 4 dotted magenta bags.
+mirrored aqua bags contain 3 shiny gold bags, 3 drab brown bags, 3 faded black bags, 3 clear tan bags.
+dark cyan bags contain 4 pale gray bags, 1 dotted gold bag, 4 dotted orange bags, 4 pale magenta bags.
+wavy fuchsia bags contain 5 mirrored teal bags, 1 dotted gray bag.
+pale gold bags contain 3 posh lavender bags, 2 drab brown bags.
+shiny orange bags contain 2 dim gray bags.
+posh tomato bags contain 5 bright brown bags.
+clear orange bags contain 4 posh plum bags, 5 dull gray bags.
+pale purple bags contain 1 drab purple bag, 5 bright crimson bags.
+drab brown bags contain 1 bright red bag, 3 light turquoise bags, 1 posh plum bag, 2 dull chartreuse bags.
+shiny tan bags contain 2 dotted gold bags, 2 plaid orange bags, 3 clear cyan bags.
+faded cyan bags contain 2 wavy cyan bags, 2 posh teal bags.
+clear maroon bags contain 2 light coral bags.
+drab cyan bags contain 3 wavy bronze bags, 5 posh aqua bags, 4 dull crimson bags, 1 shiny gold bag.
+plaid white bags contain 4 wavy brown bags.
+bright tan bags contain 2 shiny silver bags.
+drab turquoise bags contain 2 faded beige bags, 4 vibrant lavender bags.
+faded yellow bags contain 2 dull black bags, 3 vibrant yellow bags, 4 plaid silver bags, 3 posh teal bags.
+bright lime bags contain 3 striped aqua bags, 4 dull tomato bags, 5 dim gray bags.
+plaid tan bags contain 2 plaid coral bags, 5 drab blue bags, 4 drab orange bags.
+clear purple bags contain 2 drab tomato bags, 3 mirrored bronze bags, 2 dark teal bags, 4 dull green bags.
+wavy red bags contain 4 faded tan bags, 4 wavy cyan bags.
+vibrant purple bags contain 5 mirrored salmon bags, 1 dark salmon bag.
+wavy tan bags contain 1 shiny olive bag, 2 faded maroon bags.
+vibrant gray bags contain 1 mirrored olive bag, 2 dotted turquoise bags, 4 striped chartreuse bags, 1 dull tomato bag.
+mirrored gray bags contain 4 bright blue bags, 1 striped olive bag.
+dull aqua bags contain 1 vibrant purple bag, 1 faded maroon bag, 3 muted yellow bags.
+clear red bags contain 1 posh violet bag.
+light purple bags contain 2 shiny olive bags, 2 striped yellow bags, 4 dotted turquoise bags, 2 shiny violet bags.
+dull brown bags contain 2 wavy red bags, 4 clear green bags.
+wavy bronze bags contain 2 dull tomato bags.
+muted beige bags contain 5 wavy aqua bags, 4 striped maroon bags, 2 drab tan bags.
+muted black bags contain 1 posh green bag, 1 clear green bag, 5 striped bronze bags.
+light green bags contain 4 dim lime bags, 1 shiny gold bag, 3 shiny crimson bags, 2 muted yellow bags.
+mirrored fuchsia bags contain 1 dull bronze bag.
+pale tomato bags contain 2 vibrant maroon bags, 2 muted orange bags, 1 wavy brown bag, 2 dull cyan bags.
+drab chartreuse bags contain 5 mirrored magenta bags.
+shiny indigo bags contain 4 pale teal bags, 3 dull silver bags, 5 mirrored olive bags.
+bright aqua bags contain 3 bright violet bags.
+shiny magenta bags contain 2 bright red bags.
+vibrant orange bags contain 3 mirrored black bags, 5 mirrored olive bags.
+mirrored indigo bags contain 4 vibrant coral bags, 1 mirrored crimson bag, 2 mirrored red bags.
+plaid blue bags contain 5 wavy aqua bags, 2 plaid fuchsia bags, 3 wavy purple bags, 1 dim lime bag.
+dotted plum bags contain 4 striped chartreuse bags, 4 dotted yellow bags, 4 light yellow bags, 3 striped red bags.
+light violet bags contain 5 shiny chartreuse bags, 1 wavy lime bag, 1 vibrant beige bag.
+dotted salmon bags contain 2 drab maroon bags, 5 muted black bags, 1 posh violet bag, 1 dull lime bag.
+vibrant gold bags contain 5 dull magenta bags, 1 clear aqua bag, 3 mirrored beige bags, 1 pale cyan bag.
+mirrored lavender bags contain 3 clear gold bags.
+dotted gold bags contain 1 drab beige bag, 1 mirrored crimson bag.
+bright magenta bags contain 1 striped maroon bag, 1 striped bronze bag, 5 mirrored beige bags, 1 dotted maroon bag.
+clear black bags contain 1 faded chartreuse bag, 5 light blue bags.
+bright tomato bags contain 3 mirrored tan bags, 2 mirrored olive bags, 3 muted bronze bags, 1 posh violet bag.
+mirrored violet bags contain 3 striped chartreuse bags, 3 dim lime bags, 5 pale chartreuse bags, 1 posh tan bag.
+faded gray bags contain 4 light crimson bags, 3 muted turquoise bags.
+dark red bags contain 5 muted green bags, 4 mirrored maroon bags.
+clear crimson bags contain 2 muted olive bags, 5 striped chartreuse bags.
+vibrant lime bags contain 1 dim crimson bag, 2 faded coral bags, 3 light crimson bags.
+vibrant bronze bags contain no other bags.
+dotted white bags contain 4 dim purple bags, 2 mirrored chartreuse bags.
+clear indigo bags contain 3 mirrored olive bags, 2 posh brown bags.
+pale lime bags contain 5 faded salmon bags, 5 striped tan bags, 4 light white bags, 3 pale chartreuse bags.
+vibrant coral bags contain 1 muted orange bag.
+clear teal bags contain 1 posh plum bag.
+wavy white bags contain 2 dim chartreuse bags, 1 dull tomato bag, 1 dull fuchsia bag, 4 faded silver bags.
+dull yellow bags contain 2 mirrored gray bags, 2 shiny olive bags, 4 clear beige bags, 2 dim crimson bags.
+posh fuchsia bags contain 3 dull teal bags.
+dotted teal bags contain 3 dull white bags, 5 pale gray bags.
+vibrant tomato bags contain 4 dull green bags.
+dull purple bags contain 3 shiny red bags, 1 dim crimson bag, 5 drab turquoise bags.
+vibrant aqua bags contain 1 bright cyan bag.
+clear plum bags contain 5 drab magenta bags, 2 mirrored salmon bags.
+dark tomato bags contain 1 dull beige bag, 1 faded violet bag.
+bright violet bags contain 3 dim lime bags, 4 muted cyan bags, 4 dull gold bags.
+dim teal bags contain 3 mirrored orange bags.
+dark magenta bags contain 2 light tan bags, 5 bright violet bags, 4 dim aqua bags, 5 bright teal bags.
+faded silver bags contain 2 dotted gold bags.
+light lavender bags contain 5 mirrored chartreuse bags.
+vibrant chartreuse bags contain 1 pale indigo bag, 2 drab tan bags, 2 striped bronze bags.
+dotted blue bags contain 3 faded gray bags, 3 muted coral bags, 4 drab red bags.
+drab lavender bags contain 1 dull bronze bag.
+wavy green bags contain 2 vibrant brown bags.
+bright gold bags contain 5 shiny blue bags.
+dull lavender bags contain 4 vibrant beige bags, 4 wavy purple bags.
+shiny tomato bags contain 5 plaid olive bags.
+faded aqua bags contain 5 mirrored olive bags.
+clear beige bags contain 4 shiny fuchsia bags, 5 wavy bronze bags.
+pale black bags contain 1 shiny lavender bag, 4 muted olive bags, 3 striped gold bags.
+faded beige bags contain 3 clear yellow bags, 1 bright turquoise bag.
+dull salmon bags contain 1 plaid tomato bag, 1 dull coral bag, 4 dotted coral bags.
+drab aqua bags contain 5 muted olive bags.
+plaid fuchsia bags contain 2 wavy lime bags, 3 mirrored gray bags, 3 clear crimson bags.
+pale turquoise bags contain 3 bright cyan bags, 3 posh blue bags, 3 dark olive bags.
+dotted red bags contain 3 pale black bags, 3 light blue bags, 4 clear gold bags, 4 pale olive bags.
+dark olive bags contain 5 shiny coral bags.
+dull white bags contain no other bags.
+shiny silver bags contain 3 posh plum bags, 4 posh brown bags, 4 dull chartreuse bags, 1 drab beige bag.
+muted gray bags contain 5 posh coral bags.
+light magenta bags contain 3 vibrant cyan bags, 1 striped tan bag, 5 dark maroon bags, 2 wavy cyan bags.
+dim brown bags contain 3 dull green bags, 2 dull beige bags, 2 wavy gold bags, 5 wavy bronze bags.
+vibrant brown bags contain 1 striped violet bag, 2 dull beige bags.
+pale chartreuse bags contain 5 shiny gold bags, 4 dotted gold bags.
+vibrant plum bags contain 4 mirrored gray bags, 1 bright silver bag, 3 bright cyan bags, 2 dull gray bags.
+muted bronze bags contain 2 mirrored beige bags.
+plaid green bags contain 3 striped tan bags, 4 drab maroon bags.
+dark orange bags contain 4 drab beige bags, 1 striped beige bag, 1 dim indigo bag.
+bright olive bags contain 4 bright turquoise bags.
+drab indigo bags contain 4 striped tan bags, 5 dotted gold bags, 4 shiny crimson bags, 2 dull chartreuse bags.
+light teal bags contain 5 dotted orange bags, 3 posh brown bags, 3 shiny gold bags.
+dotted crimson bags contain 5 posh turquoise bags, 1 clear salmon bag, 5 faded salmon bags.
+shiny red bags contain 4 bright blue bags, 2 posh lime bags, 5 shiny plum bags.
+dark brown bags contain 5 posh magenta bags.
+drab violet bags contain 5 posh indigo bags, 2 light chartreuse bags, 2 vibrant crimson bags.
+plaid beige bags contain 3 drab olive bags, 5 dim fuchsia bags, 5 wavy turquoise bags.
+bright plum bags contain 2 mirrored salmon bags, 4 faded tan bags.
+plaid lavender bags contain 2 striped coral bags, 1 light coral bag.
+shiny brown bags contain 2 wavy tan bags, 1 pale salmon bag, 2 clear bronze bags.
+clear turquoise bags contain 2 dotted beige bags.
+wavy orange bags contain 4 shiny green bags, 1 pale magenta bag, 4 dark plum bags, 4 mirrored olive bags.
+vibrant turquoise bags contain 3 dotted black bags, 4 wavy salmon bags.
+posh olive bags contain 1 dull black bag.
+pale red bags contain 5 plaid tomato bags.
+dim gold bags contain 5 wavy violet bags.
+dotted fuchsia bags contain 4 pale plum bags, 2 dim salmon bags, 2 clear salmon bags, 4 vibrant blue bags.
+wavy indigo bags contain 3 muted fuchsia bags, 1 drab maroon bag.
+clear chartreuse bags contain 3 mirrored olive bags, 1 posh yellow bag, 1 faded salmon bag, 5 drab salmon bags.
+drab red bags contain 2 faded lime bags, 3 bright fuchsia bags, 5 bright olive bags, 2 striped maroon bags.
+dotted turquoise bags contain 5 dull white bags, 3 bright olive bags, 4 clear crimson bags.
+muted coral bags contain 1 posh green bag, 4 dim violet bags, 2 faded silver bags, 3 dim plum bags.
+plaid olive bags contain 2 striped cyan bags, 4 drab red bags, 2 clear beige bags, 5 plaid coral bags.
+vibrant indigo bags contain 1 vibrant maroon bag.
+clear olive bags contain 3 vibrant purple bags, 3 plaid green bags.
+light blue bags contain 5 plaid lavender bags, 1 mirrored violet bag, 4 muted yellow bags.
+muted turquoise bags contain 3 plaid brown bags, 5 clear aqua bags, 5 drab silver bags.
+muted cyan bags contain 5 shiny fuchsia bags.
+plaid aqua bags contain 3 posh green bags, 2 dull crimson bags.
+dull teal bags contain 2 drab red bags.
+dim olive bags contain 3 dark plum bags.
+bright black bags contain 4 shiny salmon bags, 3 dull white bags, 4 clear brown bags.
+mirrored maroon bags contain 1 drab turquoise bag, 2 plaid red bags, 3 faded black bags.
+pale plum bags contain 1 dark salmon bag.
+posh magenta bags contain 4 bright tan bags.
+faded tan bags contain 1 pale magenta bag, 1 drab beige bag, 2 dull beige bags, 1 shiny silver bag.
+dark coral bags contain 4 shiny orange bags, 2 dull green bags, 2 dim crimson bags.
+plaid silver bags contain 2 dim black bags, 2 mirrored tan bags, 4 shiny plum bags, 4 wavy teal bags.
+dim gray bags contain 3 dull green bags, 3 dull white bags, 2 clear yellow bags.
+clear magenta bags contain 5 pale chartreuse bags, 5 light magenta bags, 4 mirrored gray bags.
+posh coral bags contain 1 muted plum bag, 4 dim turquoise bags.
+dim yellow bags contain 3 striped white bags.
+mirrored tan bags contain 1 dim cyan bag, 3 wavy purple bags.
+vibrant olive bags contain 5 striped fuchsia bags, 3 shiny silver bags.
+pale cyan bags contain 5 vibrant bronze bags, 2 clear crimson bags, 1 dim violet bag.
+vibrant teal bags contain 1 pale white bag, 5 clear gold bags, 2 striped white bags.
+shiny gold bags contain 1 shiny coral bag, 5 posh white bags, 3 wavy cyan bags.
+dim lime bags contain 1 bright turquoise bag.
+posh salmon bags contain 1 dotted aqua bag, 3 faded cyan bags.
+vibrant yellow bags contain 4 dim violet bags, 2 striped bronze bags, 3 clear crimson bags.
+dull magenta bags contain 5 clear green bags.
+bright orange bags contain 1 bright indigo bag, 3 clear tan bags, 5 dark brown bags.
+mirrored blue bags contain 3 dotted aqua bags, 4 dotted magenta bags.
+clear salmon bags contain 5 striped chartreuse bags, 5 drab brown bags.
+light tan bags contain 3 drab orange bags, 5 posh purple bags, 3 faded plum bags.
+dark purple bags contain 2 dim plum bags, 4 dark beige bags, 1 dull tomato bag, 4 dull gray bags.
+shiny maroon bags contain 3 dotted aqua bags.
+faded lime bags contain 2 light turquoise bags.
+mirrored yellow bags contain 4 muted orange bags, 4 shiny black bags, 1 shiny green bag.
+bright yellow bags contain 1 plaid bronze bag, 2 shiny gold bags.
+striped crimson bags contain 5 dotted black bags.

src/Year_2020/files/P8.txt

@@ -0,0 +1,611 @@
+jmp +265
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