diff --git a/src/Year_2022/Day01.py b/src/Year_2022/Day01.py
new file mode 100644
index 0000000..65e2975
--- /dev/null
+++ b/src/Year_2022/Day01.py
@@ -0,0 +1,98 @@
+# --- Day 1: Calorie Counting ---
+
+# Santa's reindeer typically eat regular reindeer food, but they need a lot of
+# magical energy to deliver presents on Christmas. For that, their favorite
+# snack is a special type of star fruit that only grows deep in the jungle. The
+# Elves have brought you on their annual expedition to the grove where the
+# fruit grows.
+
+# To supply enough magical energy, the expedition needs to retrieve a minimum
+# of fifty stars by December 25th. Although the Elves assure you that the grove
+# has plenty of fruit, you decide to grab any fruit you see along the way, just
+# in case.
+
+# 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!
+
+# The jungle must be too overgrown and difficult to navigate in vehicles or
+# access from the air; the Elves' expedition traditionally goes on foot. As
+# your boats approach land, the Elves begin taking inventory of their supplies.
+# One important consideration is food - in particular, the number of Calories
+# each Elf is carrying (your puzzle input).
+
+# The Elves take turns writing down the number of Calories contained by the
+# various meals, snacks, rations, etc. that they've brought with them, one item
+# per line. Each Elf separates their own inventory from the previous Elf's
+# inventory (if any) by a blank line.
+
+# For example, suppose the Elves finish writing their items' Calories and end
+# up with the following list:
+
+# 1000
+# 2000
+# 3000
+
+# 4000
+
+# 5000
+# 6000
+
+# 7000
+# 8000
+# 9000
+
+# 10000
+
+# This list represents the Calories of the food carried by five Elves:
+
+#     The first Elf is carrying food with 1000, 2000, and 3000 Calories, a
+# total of 6000 Calories.
+#     The second Elf is carrying one food item with 4000 Calories.
+#     The third Elf is carrying food with 5000 and 6000 Calories, a total of
+# 11000 Calories.
+#     The fourth Elf is carrying food with 7000, 8000, and 9000 Calories, a
+# total of 24000 Calories.
+#     The fifth Elf is carrying one food item with 10000 Calories.
+
+# In case the Elves get hungry and need extra snacks, they need to know which
+# Elf to ask: they'd like to know how many Calories are being carried by the
+# Elf carrying the most Calories. In the example above, this is 24000 (carried
+# by the fourth Elf).
+
+# Find the Elf carrying the most Calories. How many total Calories is that Elf
+# carrying?
+
+from collections import Counter
+
+with open("P1.txt") as f:
+    cal_list = [line for line in f.read().strip().split("\n\n")]
+
+elfs = Counter()
+for idx, elf in enumerate(cal_list):
+    elfs[idx] = sum([int(food) for food in elf.split("\n")])
+
+print(elfs.most_common(1))
+
+# --- Part Two ---
+
+# By the time you calculate the answer to the Elves' question, they've already
+# realized that the Elf carrying the most Calories of food might eventually run
+# out of snacks.
+
+# To avoid this unacceptable situation, the Elves would instead like to know
+# the total Calories carried by the top three Elves carrying the most Calories.
+# That way, even if one of those Elves runs out of snacks, they still have two
+# backups.
+
+# In the example above, the top three Elves are the fourth Elf (with 24000
+# Calories), then the third Elf (with 11000 Calories), then the fifth Elf (with
+# 10000 Calories). The sum of the Calories carried by these three elves is
+# 45000.
+
+# Find the top three Elves carrying the most Calories. How many Calories are
+# those Elves carrying in total?
+
+top_three = elfs.most_common(3)
+
+print(sum(cals for elf, cals in top_three))
diff --git a/src/Year_2022/Day02.py b/src/Year_2022/Day02.py
new file mode 100644
index 0000000..c24d0f0
--- /dev/null
+++ b/src/Year_2022/Day02.py
@@ -0,0 +1,110 @@
+# --- Day 2: Rock Paper Scissors ---
+
+# The Elves begin to set up camp on the beach. To decide whose tent gets to be
+# closest to the snack storage, a giant Rock Paper Scissors tournament is
+# already in progress.
+
+# Rock Paper Scissors is a game between two players. Each game contains many
+# rounds; in each round, the players each simultaneously choose one of Rock,
+# Paper, or Scissors using a hand shape. Then, a winner for that round is
+# selected: Rock defeats Scissors, Scissors defeats Paper, and Paper defeats
+# Rock. If both players choose the same shape, the round instead ends in a
+# draw.
+
+# Appreciative of your help yesterday, one Elf gives you an encrypted strategy
+# guide (your puzzle input) that they say will be sure to help you win. "The
+# first column is what your opponent is going to play: A for Rock, B for Paper,
+# and C for Scissors. The second column--" Suddenly, the Elf is called away to
+# help with someone's tent.
+
+# The second column, you reason, must be what you should play in response: X
+# for Rock, Y for Paper, and Z for Scissors. Winning every time would be
+# suspicious, so the responses must have been carefully chosen.
+
+# The winner of the whole tournament is the player with the highest score. Your
+# total score is the sum of your scores for each round. The score for a single
+# round is the score for the shape you selected (1 for Rock, 2 for Paper, and 3
+# for Scissors) plus the score for the outcome of the round (0 if you lost, 3
+# if the round was a draw, and 6 if you won).
+
+# Since you can't be sure if the Elf is trying to help you or trick you, you
+# should calculate the score you would get if you were to follow the strategy
+# guide.
+
+# For example, suppose you were given the following strategy guide:
+
+# A Y
+# B X
+# C Z
+
+# This strategy guide predicts and recommends the following:
+
+#     In the first round, your opponent will choose Rock (A), and you should
+# choose Paper (Y). This ends in a win for you with a score of 8 (2 because you
+# chose Paper + 6 because you won).
+#     In the second round, your opponent will choose Paper (B), and you should
+# choose Rock (X). This ends in a loss for you with a score of 1 (1 + 0).
+#     The third round is a draw with both players choosing Scissors, giving you
+# a score of 3 + 3 = 6.
+
+# In this example, if you were to follow the strategy guide, you would get a
+# total score of 15 (8 + 1 + 6).
+
+# What would your total score be if everything goes exactly according to your
+# strategy guide?
+
+game_score_dic = {
+    "A": {"X": 3 + 1, "Y": 6 + 2, "Z": 0 + 3},
+    "B": {"X": 0 + 1, "Y": 3 + 2, "Z": 6 + 3},
+    "C": {"X": 6 + 1, "Y": 0 + 2, "Z": 3 + 3},
+}
+
+with open("P2.txt") as f:
+    strategy_list = [line for line in f.read().strip().split("\n")]
+
+score_dic = 0
+for game in strategy_list:
+    p1, p2 = game.split()
+    score_dic += game_score_dic[p1][p2]
+
+
+print(score_dic)
+
+
+# --- Part Two ---
+
+# The Elf finishes helping with the tent and sneaks back over to you. "Anyway,
+# the second column says how the round needs to end: X means you need to lose,
+# Y means you need to end the round in a draw, and Z means you need to win.
+# Good luck!"
+
+# The total score is still calculated in the same way, but now you need to
+# figure out what shape to choose so the round ends as indicated. The example
+# above now goes like this:
+
+#     In the first round, your opponent will choose Rock (A), and you need the
+# round to end in a draw (Y), so you also choose Rock. This gives you a score
+# of 1 + 3 = 4.
+#     In the second round, your opponent will choose Paper (B), and you choose
+# Rock so you lose (X) with a score of 1 + 0 = 1.
+#     In the third round, you will defeat your opponent's Scissors with Rock
+# for a score of 1 + 6 = 7.
+
+# Now that you're correctly decrypting the ultra top secret strategy guide, you
+# would get a total score of 12.
+
+# Following the Elf's instructions for the second column, what would your total
+# score be if everything goes exactly according to your strategy guide?
+
+game_score_2nd_dic = {
+    "A": {"X": 0 + 3, "Y": 3 + 1, "Z": 6 + 2},
+    "B": {"X": 0 + 1, "Y": 3 + 2, "Z": 6 + 3},
+    "C": {"X": 0 + 2, "Y": 3 + 3, "Z": 6 + 1},
+}
+
+score_2nd = 0
+for game in strategy_list:
+    p1, p2 = game.split()
+    score_2nd += game_score_2nd_dic[p1][p2]
+
+print(score_2nd)
diff --git a/src/Year_2022/Day03.py b/src/Year_2022/Day03.py
new file mode 100644
index 0000000..00d9c90
--- /dev/null
+++ b/src/Year_2022/Day03.py
@@ -0,0 +1,128 @@
+# --- Day 3: Rucksack Reorganization ---
+
+# One Elf has the important job of loading all of the rucksacks with supplies
+# for the jungle journey. Unfortunately, that Elf didn't quite follow the
+# packing instructions, and so a few items now need to be rearranged.
+
+# Each rucksack has two large compartments. All items of a given type are meant
+# to go into exactly one of the two compartments. The Elf that did the packing
+# failed to follow this rule for exactly one item type per rucksack.
+
+# The Elves have made a list of all of the items currently in each rucksack
+# (your puzzle input), but they need your help finding the errors. Every item
+# type is identified by a single lowercase or uppercase letter (that is, a and
+# A refer to different types of items).
+
+# The list of items for each rucksack is given as characters all on a single
+# line. A given rucksack always has the same number of items in each of its two
+# compartments, so the first half of the characters represent items in the
+# first compartment, while the second half of the characters represent items in
+# the second compartment.
+
+# For example, suppose you have the following list of contents from six
+# rucksacks:
+
+# vJrwpWtwJgWrhcsFMMfFFhFp
+# jqHRNqRjqzjGDLGLrsFMfFZSrLrFZsSL
+# PmmdzqPrVvPwwTWBwg
+# wMqvLMZHhHMvwLHjbvcjnnSBnvTQFn
+# ttgJtRGJQctTZtZT
+# CrZsJsPPZsGzwwsLwLmpwMDw
+
+#    The first rucksack contains the items vJrwpWtwJgWrhcsFMMfFFhFp, which
+# means its first compartment contains the items vJrwpWtwJgWr, while the second
+# compartment contains the items hcsFMMfFFhFp. The only item type that appears
+# in both compartments is lowercase p.
+#    The second rucksack's compartments contain jqHRNqRjqzjGDLGL and
+# rsFMfFZSrLrFZsSL. The only item type that appears in both compartments is
+# uppercase L.
+#    The third rucksack's compartments contain PmmdzqPrV and vPwwTWBwg; the
+# only common item type is uppercase P.
+#    The fourth rucksack's compartments only share item type v.
+#    The fifth rucksack's compartments only share item type t.
+#    The sixth rucksack's compartments only share item type s.
+
+# To help prioritize item rearrangement, every item type can be converted to a
+# priority:
+
+#   Lowercase item types a through z have priorities 1 through 26.
+#   Uppercase item types A through Z have priorities 27 through 52.
+
+# In the above example, the priority of the item type that appears in both
+# compartments of each rucksack is 16 (p), 38 (L), 42 (P), 22 (v), 20 (t), and
+# 19 (s); the sum of these is 157.
+
+# Find the item type that appears in both compartments of each rucksack. What
+# is the sum of the priorities of those item types?
+
+with open("P3.txt") as f:
+    items_list = [line for line in f.read().strip().split()]
+
+priorities = 0
+for items in items_list:
+    first, second = items[: len(items) // 2], items[len(items) // 2 :]
+    char = list(set(first) & set(second))[0]
+    if char.isupper():
+        priorities += ord(char) - 38
+    else:
+        priorities += ord(char) - 96
+
+print(priorities)
+
+# --- Part Two ---
+
+# As you finish identifying the misplaced items, the Elves come to you with
+# another issue.
+
+# For safety, the Elves are divided into groups of three. Every Elf carries a
+# badge that identifies their group. For efficiency, within each group of three
+# Elves, the badge is the only item type carried by all three Elves. That is,
+# if a group's badge is item type B, then all three Elves will have item type B
+# somewhere in their rucksack, and at most two of the Elves will be carrying
+# any other item type.
+
+# The problem is that someone forgot to put this year's updated authenticity
+# sticker on the badges. All of the badges need to be pulled out of the
+# rucksacks so the new authenticity stickers can be attached.
+
+# Additionally, nobody wrote down which item type corresponds to each group's
+# badges. The only way to tell which item type is the right one is by finding
+# the one item type that is common between all three Elves in each group.
+
+# Every set of three lines in your list corresponds to a single group, but each
+# group can have a different badge item type. So, in the above example, the
+# first group's rucksacks are the first three lines:
+
+# vJrwpWtwJgWrhcsFMMfFFhFp
+# jqHRNqRjqzjGDLGLrsFMfFZSrLrFZsSL
+# PmmdzqPrVvPwwTWBwg
+
+# And the second group's rucksacks are the next three lines:
+
+# wMqvLMZHhHMvwLHjbvcjnnSBnvTQFn
+# ttgJtRGJQctTZtZT
+# CrZsJsPPZsGzwwsLwLmpwMDw
+
+# In the first group, the only item type that appears in all three rucksacks is
+# lowercase r; this must be their badges. In the second group, their badge item
+# type must be Z.
+
+# Priorities for these items must still be found to organize the sticker
+# attachment efforts: here, they are 18 (r) for the first group and 52 (Z) for
+# the second group. The sum of these is 70.
+
+# Find the item type that corresponds to the badges of each three-Elf group.
+# What is the sum of the priorities of those item types?
+
+from more_itertools import grouper
+
+new_priorities = 0
+for triplet in grouper(items_list, 3):
+    first, second, third = triplet
+    char = list(set(first) & set(second) & set(third))[0]
+    if char.isupper():
+        new_priorities += ord(char) - 38
+    else:
+        new_priorities += ord(char) - 96
+
+print(new_priorities)
diff --git a/src/Year_2022/Day04.py b/src/Year_2022/Day04.py
new file mode 100644
index 0000000..ecce3a8
--- /dev/null
+++ b/src/Year_2022/Day04.py
@@ -0,0 +1,104 @@
+# --- Day 4: Camp Cleanup ---
+
+# Space needs to be cleared before the last supplies can be unloaded from the
+# ships, and so several Elves have been assigned the job of cleaning up
+# sections of the camp. Every section has a unique ID number, and each Elf is
+# assigned a range of section IDs.
+
+# However, as some of the Elves compare their section assignments with each
+# other, they've noticed that many of the assignments overlap. To try to
+# quickly find overlaps and reduce duplicated effort, the Elves pair up and
+# make a big list of the section assignments for each pair (your puzzle input).
+
+# For example, consider the following list of section assignment pairs:
+
+# 2-4,6-8
+# 2-3,4-5
+# 5-7,7-9
+# 2-8,3-7
+# 6-6,4-6
+# 2-6,4-8
+
+# For the first few pairs, this list means:
+
+#     Within the first pair of Elves, the first Elf was assigned sections 2-4
+# (sections 2, 3, and 4), while the second Elf was assigned sections 6-8
+# (sections 6, 7, 8).
+#     The Elves in the second pair were each assigned two sections.
+#     The Elves in the third pair were each assigned three sections: one got
+# sections 5, 6, and 7, while the other also got 7, plus 8 and 9.
+
+# This example list uses single-digit section IDs to make it easier to draw;
+# your actual list might contain larger numbers. Visually, these pairs of
+# section assignments look like this:
+
+# .234.....  2-4
+# .....678.  6-8
+
+# .23......  2-3
+# ...45....  4-5
+
+# ....567..  5-7
+# ......789  7-9
+
+# .2345678.  2-8
+# ..34567..  3-7
+
+# .....6...  6-6
+# ...456...  4-6
+
+# .23456...  2-6
+# ...45678.  4-8
+
+# Some of the pairs have noticed that one of their assignments fully contains
+# the other. For example, 2-8 fully contains 3-7, and 6-6 is fully contained by
+# 4-6. In pairs where one assignment fully contains the other, one Elf in the
+# pair would be exclusively cleaning sections their partner will already be
+# cleaning, so these seem like the most in need of reconsideration. In this
+# example, there are 2 such pairs.
+
+# In how many assignment pairs does one range fully contain the other?
+
+with open("P4.txt") as f:
+    section_pairs = [line for line in f.read().strip().split()]
+
+assignment_pairs = 0
+for elfs in section_pairs:
+    e1, e2 = [[int(x) for x in pairs.split("-")] for pairs in elfs.split(",")]
+    e1, e2 = {x for x in range(e1[0], e1[1] + 1)}, {
+        x for x in range(e2[0], e2[1] + 1)
+    }
+    if len(e1) == len(e1 | e2) or len(e2) == len(e1 | e2):
+        assignment_pairs += 1
+
+print(assignment_pairs)
+
+
+# --- Part Two ---
+
+# It seems like there is still quite a bit of duplicate work planned. Instead,
+# the Elves would like to know the number of pairs that overlap at all.
+
+# In the above example, the first two pairs (2-4,6-8 and 2-3,4-5) don't
+# overlap, while the remaining four pairs (5-7,7-9, 2-8,3-7, 6-6,4-6, and
+# 2-6,4-8) do overlap:
+
+#     5-7,7-9 overlaps in a single section, 7.
+#     2-8,3-7 overlaps all of the sections 3 through 7.
+#     6-6,4-6 overlaps in a single section, 6.
+#     2-6,4-8 overlaps in sections 4, 5, and 6.
+
+# So, in this example, the number of overlapping assignment pairs is 4.
+
+# In how many assignment pairs do the ranges overlap?
+
+assignment_pairs_overlaps = len(section_pairs)
+for elfs in section_pairs:
+    e1, e2 = [[int(x) for x in pairs.split("-")] for pairs in elfs.split(",")]
+    e1, e2 = {x for x in range(e1[0], e1[1] + 1)}, {
+        x for x in range(e2[0], e2[1] + 1)
+    }
+    if len(e1 & e2) == 0:
+        assignment_pairs_overlaps -= 1
+
+print(assignment_pairs_overlaps)
diff --git a/src/Year_2022/Day05.py b/src/Year_2022/Day05.py
new file mode 100644
index 0000000..79c1693
--- /dev/null
+++ b/src/Year_2022/Day05.py
@@ -0,0 +1,194 @@
+# --- Day 5: Supply Stacks ---
+
+# The expedition can depart as soon as the final supplies have been unloaded
+# from the ships. Supplies are stored in stacks of marked crates, but because
+# the needed supplies are buried under many other crates, the crates need to be
+# rearranged.
+
+# The ship has a giant cargo crane capable of moving crates between stacks. To
+# ensure none of the crates get crushed or fall over, the crane operator will
+# rearrange them in a series of carefully-planned steps. After the crates are
+# rearranged, the desired crates will be at the top of each stack.
+
+# The Elves don't want to interrupt the crane operator during this delicate
+# procedure, but they forgot to ask her which crate will end up where, and they
+# want to be ready to unload them as soon as possible so they can embark.
+
+# They do, however, have a drawing of the starting stacks of crates and the
+# rearrangement procedure (your puzzle input). For example:
+
+#     [D]
+# [N] [C]
+# [Z] [M] [P]
+#  1   2   3
+
+# move 1 from 2 to 1
+# move 3 from 1 to 3
+# move 2 from 2 to 1
+# move 1 from 1 to 2
+
+# In this example, there are three stacks of crates. Stack 1 contains two
+# crates: crate Z is on the bottom, and crate N is on top. Stack 2 contains
+# three crates; from bottom to top, they are crates M, C, and D. Finally, stack
+# 3 contains a single crate, P.
+
+# Then, the rearrangement procedure is given. In each step of the procedure, a
+# quantity of crates is moved from one stack to a different stack. In the first
+# step of the above rearrangement procedure, one crate is moved from stack 2 to
+# stack 1, resulting in this configuration:
+
+# [D]
+# [N] [C]
+# [Z] [M] [P]
+#  1   2   3
+
+# In the second step, three crates are moved from stack 1 to stack 3. Crates
+# are moved one at a time, so the first crate to be moved (D) ends up below the
+# second and third crates:
+
+#         [Z]
+#         [N]
+#     [C] [D]
+#     [M] [P]
+#  1   2   3
+
+# Then, both crates are moved from stack 2 to stack 1. Again, because crates
+# are moved one at a time, crate C ends up below crate M:
+
+#         [Z]
+#         [N]
+# [M]     [D]
+# [C]     [P]
+#  1   2   3
+
+# Finally, one crate is moved from stack 1 to stack 2:
+
+#         [Z]
+#         [N]
+#         [D]
+# [C] [M] [P]
+#  1   2   3
+
+# The Elves just need to know which crate will end up on top of each stack; in
+# this example, the top crates are C in stack 1, M in stack 2, and Z in stack
+# 3, so you should combine these together and give the Elves the message CMZ.
+
+# After the rearrangement procedure completes, what crate ends up on top of
+# each stack?
+
+from collections import defaultdict
+
+with open("P5.txt") as f:
+    stack_crates, instructions = [
+        line for line in f.read().strip().split("\n\n")
+    ]
+
+
+def parse_crates(crates):
+    stacks = defaultdict(str)
+    for row in crates.splitlines():
+        for idx, char in enumerate(row):
+            if not char.isalpha():
+                continue
+            stacks[idx // 4 + 1] += char
+    return stacks
+
+
+def parse_instruction(inst):
+    procedure = inst.split(" ")
+    units, from_crate, to_crate = [int(num) for num in procedure[1::2]]
+    return units, from_crate, to_crate
+
+
+def rearrange(stack, u, f, t):
+    move = stack[f][:u][::-1]
+    stack[t] = move + stack[t]
+    stack[f] = stack[f][u:]
+
+
+stacks = parse_crates(stack_crates)
+
+# needed for Part 2
+stacks_ = stacks.copy()
+
+for inst in instructions.splitlines():
+    u, f, t = parse_instruction(inst)
+    rearrange(stacks, u, f, t)
+
+print("".join([v[0] for k, v in sorted(stacks.items())]))
+
+# --- Part Two ---
+
+# As you watch the crane operator expertly rearrange the crates, you notice the
+# process isn't following your prediction.
+
+# Some mud was covering the writing on the side of the crane, and you quickly
+# wipe it away. The crane isn't a CrateMover 9000 - it's a CrateMover 9001.
+
+# The CrateMover 9001 is notable for many new and exciting features: air
+# conditioning, leather seats, an extra cup holder, and the ability to pick up
+# and move multiple crates at once.
+
+# Again considering the example above, the crates begin in the same
+# configuration:
+
+#     [D]
+# [N] [C]
+# [Z] [M] [P]
+#  1   2   3
+
+# Moving a single crate from stack 2 to stack 1 behaves the same as before:
+
+# [D]
+# [N] [C]
+# [Z] [M] [P]
+#  1   2   3
+
+# However, the action of moving three crates from stack 1 to stack 3 means that
+# those three moved crates stay in the same order, resulting in this new
+# configuration:
+
+#         [D]
+#         [N]
+#     [C] [Z]
+#     [M] [P]
+#  1   2   3
+
+# Next, as both crates are moved from stack 2 to stack 1, they retain their
+# order as well:
+
+#         [D]
+#         [N]
+# [C]     [Z]
+# [M]     [P]
+#  1   2   3
+
+# Finally, a single crate is still moved from stack 1 to stack 2, but now it's
+# crate C that gets moved:
+
+#         [D]
+#         [N]
+#         [Z]
+# [M] [C] [P]
+#  1   2   3
+
+# In this example, the CrateMover 9001 has put the crates in a totally
+# different order: MCD.
+
+# Before the rearrangement process finishes, update your simulation so that the
+# Elves know where they should stand to be ready to unload the final supplies.
+# After the rearrangement procedure completes, what crate ends up on top of
+# each stack?
+
+
+def rearrange_9001(stack, u, f, t):
+    move = stack[f][:u]
+    stack[t] = move + stack[t]
+    stack[f] = stack[f][u:]
+
+
+for inst in instructions.splitlines():
+    u, f, t = parse_instruction(inst)
+    rearrange_9001(stacks_, u, f, t)
+
+print("".join([v[0] for k, v in sorted(stacks_.items())]))
diff --git a/src/Year_2022/Day06.py b/src/Year_2022/Day06.py
new file mode 100644
index 0000000..6dca334
--- /dev/null
+++ b/src/Year_2022/Day06.py
@@ -0,0 +1,89 @@
+# --- Day 6: Tuning Trouble ---
+
+# The preparations are finally complete; you and the Elves leave camp on foot
+# and begin to make your way toward the star fruit grove.
+
+# As you move through the dense undergrowth, one of the Elves gives you a
+# handheld device. He says that it has many fancy features, but the most
+# important one to set up right now is the communication system.
+
+# However, because he's heard you have significant experience dealing with
+# signal-based systems, he convinced the other Elves that it would be okay to
+# give you their one malfunctioning device - surely you'll have no problem
+# fixing it.
+
+# As if inspired by comedic timing, the device emits a few colorful sparks.
+
+# To be able to communicate with the Elves, the device needs to lock on to
+# their signal. The signal is a series of seemingly-random characters that the
+# device receives one at a time.
+
+# To fix the communication system, you need to add a subroutine to the device
+# that detects a start-of-packet marker in the datastream. In the protocol
+# being used by the Elves, the start of a packet is indicated by a sequence of
+# four characters that are all different.
+
+# The device will send your subroutine a datastream buffer (your puzzle input);
+# your subroutine needs to identify the first position where the four most
+# recently received characters were all different. Specifically, it needs to
+# report the number of characters from the beginning of the buffer to the end
+# of the first such four-character marker.
+
+# For example, suppose you receive the following datastream buffer:
+
+# mjqjpqmgbljsphdztnvjfqwrcgsmlb
+
+# After the first three characters (mjq) have been received, there haven't been
+# enough characters received yet to find the marker. The first time a marker
+# could occur is after the fourth character is received, making the most recent
+# four characters mjqj. Because j is repeated, this isn't a marker.
+
+# The first time a marker appears is after the seventh character arrives. Once
+# it does, the last four characters received are jpqm, which are all different.
+# In this case, your subroutine should report the value 7, because the first
+# start-of-packet marker is complete after 7 characters have been processed.
+
+# Here are a few more examples:
+
+#     bvwbjplbgvbhsrlpgdmjqwftvncz: first marker after character 5
+#     nppdvjthqldpwncqszvftbrmjlhg: first marker after character 6
+#     nznrnfrfntjfmvfwmzdfjlvtqnbhcprsg: first marker after character 10
+#     zcfzfwzzqfrljwzlrfnpqdbhtmscgvjw: first marker after character 11
+
+# How many characters need to be processed before the first start-of-packet
+# marker is detected?
+
+from more_itertools import sliding_window
+
+with open("P6.txt") as f:
+    data_stream = [line for line in f.read().strip().split()][0]
+
+for idx, key in enumerate(sliding_window(data_stream, 4)):
+    if len(set(key)) == 4:
+        print(key, idx + 4)
+        break
+
+# --- Part Two ---
+
+# Your device's communication system is correctly detecting packets, but still
+# isn't working. It looks like it also needs to look for messages.
+
+# A start-of-message marker is just like a start-of-packet marker, except it
+# consists of 14 distinct characters rather than 4.
+
+# Here are the first positions of start-of-message markers for all of the above
+# examples:
+
+#     mjqjpqmgbljsphdztnvjfqwrcgsmlb: first marker after character 19
+#     bvwbjplbgvbhsrlpgdmjqwftvncz: first marker after character 23
+#     nppdvjthqldpwncqszvftbrmjlhg: first marker after character 23
+#     nznrnfrfntjfmvfwmzdfjlvtqnbhcprsg: first marker after character 29
+#     zcfzfwzzqfrljwzlrfnpqdbhtmscgvjw: first marker after character 26
+
+# How many characters need to be processed before the first start-of-message
+# marker is detected?
+
+for idx, key in enumerate(sliding_window(data_stream, 14)):
+    if len(set(key)) == 14:
+        print(key, idx + 14)
+        break
diff --git a/src/Year_2022/Day07.py b/src/Year_2022/Day07.py
new file mode 100644
index 0000000..ebb3cda
--- /dev/null
+++ b/src/Year_2022/Day07.py
@@ -0,0 +1,184 @@
+# --- Day 7: No Space Left On Device ---
+
+# You can hear birds chirping and raindrops hitting leaves as the expedition
+# proceeds. Occasionally, you can even hear much louder sounds in the distance;
+# how big do the animals get out here, anyway?
+
+# The device the Elves gave you has problems with more than just its
+# communication system. You try to run a system update:
+
+# $ system-update --please --pretty-please-with-sugar-on-top
+# Error: No space left on device
+
+# Perhaps you can delete some files to make space for the update?
+
+# You browse around the filesystem to assess the situation and save the
+# resulting terminal output (your puzzle input). For example:
+
+# $ cd /
+# $ ls
+# dir a
+# 14848514 b.txt
+# 8504156 c.dat
+# dir d
+# $ cd a
+# $ ls
+# dir e
+# 29116 f
+# 2557 g
+# 62596 h.lst
+# $ cd e
+# $ ls
+# 584 i
+# $ cd ..
+# $ cd ..
+# $ cd d
+# $ ls
+# 4060174 j
+# 8033020 d.log
+# 5626152 d.ext
+# 7214296 k
+
+# The filesystem consists of a tree of files (plain data) and directories
+# (which can contain other directories or files). The outermost directory is
+# called /. You can navigate around the filesystem, moving into or out of
+# directories and listing the contents of the directory you're currently in.
+
+# Within the terminal output, lines that begin with $ are commands you
+# executed, very much like some modern computers:
+
+#     cd means change directory. This changes which directory is the current
+# directory, but the specific result depends on the argument:
+#         cd x moves in one level: it looks in the current directory for the
+# directory named x and makes it the current directory.
+#         cd .. moves out one level: it finds the directory that contains the
+# current directory, then makes that directory the current directory.
+#         cd / switches the current directory to the outermost directory, /.
+#     ls means list. It prints out all of the files and directories
+# immediately contained by the current directory:
+#         123 abc means that the current directory contains a file named abc
+# with size 123.
+#         dir xyz means that the current directory contains a directory named
+# xyz.
+
+# Given the commands and output in the example above, you can determine that
+# the filesystem looks visually like this:
+
+# - / (dir)
+#   - a (dir)
+#     - e (dir)
+#       - i (file, size=584)
+#     - f (file, size=29116)
+#     - g (file, size=2557)
+#     - h.lst (file, size=62596)
+#   - b.txt (file, size=14848514)
+#   - c.dat (file, size=8504156)
+#   - d (dir)
+#     - j (file, size=4060174)
+#     - d.log (file, size=8033020)
+#     - d.ext (file, size=5626152)
+#     - k (file, size=7214296)
+
+# Here, there are four directories: / (the outermost directory), a and d (which
+# are in /), and e (which is in a). These directories also contain files of
+# various sizes.
+
+# Since the disk is full, your first step should probably be to find
+# directories that are good candidates for deletion. To do this, you need to
+# determine the total size of each directory. The total size of a directory is
+# the sum of the sizes of the files it contains, directly or indirectly.
+# (Directories themselves do not count as having any intrinsic size.)
+
+# The total sizes of the directories above can be found as follows:
+
+#     The total size of directory e is 584 because it contains a single file i
+# of size 584 and no other directories.
+#     The directory a has total size 94853 because it contains files f (size
+# 29116), g (size 2557), and h.lst (size 62596), plus file i indirectly (a
+# contains e which contains i).
+#     Directory d has total size 24933642.
+#     As the outermost directory, / contains every file. Its total size is
+# 48381165, the sum of the size of every file.
+
+# To begin, find all of the directories with a total size of at most 100000,
+# then calculate the sum of their total sizes. In the example above, these
+# directories are a and e; the sum of their total sizes is 95437 (94853 + 584).
+# (As in this example, this process can count files more than once!)
+
+# Find all of the directories with a total size of at most 100000. What is the
+# sum of the total sizes of those directories?
+
+with open("P7.txt") as f:
+    filesystem = [line for line in f.read().splitlines()]
+
+
+tree = []
+total = 0
+for line in filesystem:
+    if "$ cd" in line:
+        if ".." in line:
+            dir_size = tree.pop()
+            tree[-1] += dir_size
+            if dir_size < 100_000:
+                total += dir_size
+        else:
+            tree.append(0)
+    elif line[0].isdigit():
+        tree[-1] += int(line.split()[0])
+
+print(total)
+
+# --- Part Two ---
+
+# Now, you're ready to choose a directory to delete.
+
+# The total disk space available to the filesystem is 70000000. To run the
+# update, you need unused space of at least 30000000. You need to find a
+# directory you can delete that will free up enough space to run the update.
+
+# In the example above, the total size of the outermost directory (and thus the
+# total amount of used space) is 48381165; this means that the size of the
+# unused space must currently be 21618835, which isn't quite the 30000000
+# required by the update. Therefore, the update still requires a directory with
+# total size of at least 8381165 to be deleted before it can run.
+
+# To achieve this, you have the following options:
+
+#     Delete directory e, which would increase unused space by 584.
+#     Delete directory a, which would increase unused space by 94853.
+#     Delete directory d, which would increase unused space by 24933642.
+#     Delete directory /, which would increase unused space by 48381165.
+
+# Directories e and a are both too small; deleting them would not free up
+# enough space. However, directories d and / are both big enough! Between
+# these, choose the smallest: d, increasing unused space by 24933642.
+
+# Find the smallest directory that, if deleted, would free up enough space on
+# the filesystem to run the update. What is the total size of that directory?
+
+tree = []
+total = []
+for line in filesystem:
+    if "$ cd" in line:
+        if ".." in line:
+            dir_size = tree.pop()
+            tree[-1] += dir_size
+            total.append(dir_size)
+        else:
+            tree.append(0)
+    elif line[0].isdigit():
+        tree[-1] += int(line.split()[0])
+
+while tree:
+    total.append(tree.pop())
+
+    if tree:
+        tree[-1] += total[-1]
+
+total.sort()
+
+used_space = 70_000_000 - total[-1]
+for size_dir in total:
+    if used_space + size_dir >= 30_000_000:
+        print(size_dir)
+        break
diff --git a/src/Year_2022/Day08.py b/src/Year_2022/Day08.py
new file mode 100644
index 0000000..e0aa3c7
--- /dev/null
+++ b/src/Year_2022/Day08.py
@@ -0,0 +1,175 @@
+# --- Day 8: Treetop Tree House ---
+
+# The expedition comes across a peculiar patch of tall trees all planted
+# carefully in a grid. The Elves explain that a previous expedition planted
+# these trees as a reforestation effort. Now, they're curious if this would be
+# a good location for a tree house.
+
+# First, determine whether there is enough tree cover here to keep a tree house
+# hidden. To do this, you need to count the number of trees that are visible
+# from outside the grid when looking directly along a row or column.
+
+# The Elves have already launched a quadcopter to generate a map with the
+# height of each tree (your puzzle input). For example:
+
+# 30373
+# 25512
+# 65332
+# 33549
+# 35390
+
+# Each tree is represented as a single digit whose value is its height, where 0
+# is the shortest and 9 is the tallest.
+
+# A tree is visible if all of the other trees between it and an edge of the
+# grid are shorter than it. Only consider trees in the same row or column; that
+# is, only look up, down, left, or right from any given tree.
+
+# All of the trees around the edge of the grid are visible - since they are
+# already on the edge, there are no trees to block the view. In this example,
+# that only leaves the interior nine trees to consider:
+
+#     The top-left 5 is visible from the left and top. (It isn't visible from
+# the right or bottom since other trees of height 5 are in the way.)
+#     The top-middle 5 is visible from the top and right.
+#     The top-right 1 is not visible from any direction; for it to be visible,
+# there would need to only be trees of height 0 between it and an edge.
+#     The left-middle 5 is visible, but only from the right.
+#     The center 3 is not visible from any direction; for it to be visible,
+# there would need to be only trees of at most height 2 between it and an edge.
+#     The right-middle 3 is visible from the right.
+#     In the bottom row, the middle 5 is visible, but the 3 and 4 are not.
+
+# With 16 trees visible on the edge and another 5 visible in the interior, a
+# total of 21 trees are visible in this arrangement.
+
+# Consider your map; how many trees are visible from outside the grid?
+
+import numpy as np
+
+with open("/home/xfeluser/AoC_2022/P8.txt") as f:
+    grid = [int(num) for line in f.read().strip().split() for num in line]
+
+grid_arr = np.array(grid).reshape(int(len(grid) ** 0.5), int(len(grid) ** 0.5))
+
+perimeter = grid_arr.shape[0] * 2 + (grid_arr.shape[0] - 2) * 2
+
+
+def is_visible(arr, x, y):
+    point = arr[x, y]
+    top = arr[:x, y]
+    bottom = arr[x + 1 :, y]
+    left = arr[x, :y]
+    right = arr[x, y + 1 :]
+
+    return np.any(
+        [1 for pos in [top, bottom, left, right] if np.all(point - pos > 0)]
+    )
+
+
+total = 0
+for idx, element in enumerate(np.nditer(grid_arr[1:-1, 1:-1])):
+    x = 1 + (idx // grid_arr[1:-1, 1:-1].shape[0])
+    y = 1 + (idx % grid_arr[1:-1, 1:-1].shape[0])
+    if is_visible(grid_arr, x, y):
+        total += 1
+
+print(perimeter + total)
+
+
+# --- Part Two ---
+
+# Content with the amount of tree cover available, the Elves just need to know
+# the best spot to build their tree house: they would like to be able to see a
+# lot of trees.
+
+# To measure the viewing distance from a given tree, look up, down, left, and
+# right from that tree; stop if you reach an edge or at the first tree that is
+# the same height or taller than the tree under consideration. (If a tree is
+# right on the edge, at least one of its viewing distances will be zero.)
+
+# The Elves don't care about distant trees taller than those found by the rules
+# above; the proposed tree house has large eaves to keep it dry, so they
+# wouldn't be able to see higher than the tree house anyway.
+
+# In the example above, consider the middle 5 in the second row:
+
+# 30373
+# 25512
+# 65332
+# 33549
+# 35390
+
+#     Looking up, its view is not blocked; it can see 1 tree (of height 3).
+#     Looking left, its view is blocked immediately; it can see only 1 tree
+# (of height 5, right next to it).
+#     Looking right, its view is not blocked; it can see 2 trees.
+#     Looking down, its view is blocked eventually; it can see 2 trees (one of
+# height 3, then the tree of height 5 that blocks its view).
+
+# A tree's scenic score is found by multiplying together its viewing distance
+# in each of the four directions. For this tree, this is 4 (found by
+# multiplying 1 * 1 * 2 * 2).
+
+# However, you can do even better: consider the tree of height 5 in the middle
+# of the fourth row:
+
+# 30373
+# 25512
+# 65332
+# 33549
+# 35390
+
+#     Looking up, its view is blocked at 2 trees (by another tree with a height
+# of 5).
+#     Looking left, its view is not blocked; it can see 2 trees.
+#     Looking down, its view is also not blocked; it can see 1 tree.
+#     Looking right, its view is blocked at 2 trees (by a massive tree of
+# height 9).
+
+# This tree's scenic score is 8 (2 * 2 * 1 * 2); this is the ideal spot for the
+# tree house.
+
+# Consider each tree on your map. What is the highest scenic score possible for
+# any tree?
+
+from math import prod
+
+
+def visibility(arr, x, y):
+    point = arr[x, y]
+    top = arr[:x, y]
+    bottom = arr[x + 1 :, y]
+    left = arr[x, :y]
+    right = arr[x, y + 1 :]
+
+    return prod(
+        (
+            length_path(point, direction)
+            for direction in [top[::-1], bottom, left[::-1], right]
+        )
+    )
+
+
+def length_path(p, direction):
+    path_length = 0
+    for pos in direction:
+        if p - pos >= 0:
+            path_length += 1
+            if p - pos == 0:
+                break
+        else:
+            path_length += 1
+            break
+
+    return path_length
+
+
+total = 0
+maximum_path = 0
+for idx, element in enumerate(np.nditer(grid_arr[1:-1, 1:-1])):
+    x = 1 + (idx // grid_arr[1:-1, 1:-1].shape[0])
+    y = 1 + (idx % grid_arr[1:-1, 1:-1].shape[0])
+    maximum_path = max(visibility(grid_arr, x, y), maximum_path)
+
+print(maximum_path)
diff --git a/src/Year_2022/Day09.py b/src/Year_2022/Day09.py
new file mode 100644
index 0000000..1521c56
--- /dev/null
+++ b/src/Year_2022/Day09.py
@@ -0,0 +1,770 @@
+# --- Day 9: Rope Bridge ---
+
+# This rope bridge creaks as you walk along it. You aren't sure how old it is,
+# or whether it can even support your weight.
+
+# It seems to support the Elves just fine, though. The bridge spans a gorge
+# which was carved out by the massive river far below you.
+
+# You step carefully; as you do, the ropes stretch and twist. You decide to
+# distract yourself by modeling rope physics; maybe you can even figure out
+# where not to step.
+
+# Consider a rope with a knot at each end; these knots mark the head and the
+# tail of the rope. If the head moves far enough away from the tail, the tail
+# is pulled toward the head.
+
+# Due to nebulous reasoning involving Planck lengths, you should be able to
+# model the positions of the knots on a two-dimensional grid. Then, by
+# following a hypothetical series of motions (your puzzle input) for the head,
+# you can determine how the tail will move.
+
+# Due to the aforementioned Planck lengths, the rope must be quite short; in
+# fact, the head (H) and tail (T) must always be touching (diagonally adjacent
+# and even overlapping both count as touching):
+
+# ....
+# .TH.
+# ....
+
+# ....
+# .H..
+# ..T.
+# ....
+
+# ...
+# .H. (H covers T)
+# ...
+
+# If the head is ever two steps directly up, down, left, or right from the
+# tail, the tail must also move one step in that direction so it remains close
+# enough:
+
+# .....    .....    .....
+# .TH.. -> .T.H. -> ..TH.
+# .....    .....    .....
+
+# ...    ...    ...
+# .T.    .T.    ...
+# .H. -> ... -> .T.
+# ...    .H.    .H.
+# ...    ...    ...
+
+# Otherwise, if the head and tail aren't touching and aren't in the same row or
+# column, the tail always moves one step diagonally to keep up:
+
+# .....    .....    .....
+# .....    ..H..    ..H..
+# ..H.. -> ..... -> ..T..
+# .T...    .T...    .....
+# .....    .....    .....
+
+# .....    .....    .....
+# .....    .....    .....
+# ..H.. -> ...H. -> ..TH.
+# .T...    .T...    .....
+# .....    .....    .....
+
+# You just need to work out where the tail goes as the head follows a series of
+# motions. Assume the head and the tail both start at the same position,
+# overlapping.
+
+# For example:
+
+# R 4
+# U 4
+# L 3
+# D 1
+# R 4
+# D 1
+# L 5
+# R 2
+
+# This series of motions moves the head right four steps, then up four steps,
+# then left three steps, then down one step, and so on. After each step, you'll
+# need to update the position of the tail if the step means the head is no
+# longer adjacent to the tail. Visually, these motions occur as follows (s
+# marks the starting position as a reference point):
+
+# == Initial State ==
+
+# ......
+# ......
+# ......
+# ......
+# H.....  (H covers T, s)
+
+# == R 4 ==
+
+# ......
+# ......
+# ......
+# ......
+# TH....  (T covers s)
+
+# ......
+# ......
+# ......
+# ......
+# sTH...
+
+# ......
+# ......
+# ......
+# ......
+# s.TH..
+
+# ......
+# ......
+# ......
+# ......
+# s..TH.
+
+# == U 4 ==
+
+# ......
+# ......
+# ......
+# ....H.
+# s..T..
+
+# ......
+# ......
+# ....H.
+# ....T.
+# s.....
+
+# ......
+# ....H.
+# ....T.
+# ......
+# s.....
+
+# ....H.
+# ....T.
+# ......
+# ......
+# s.....
+
+# == L 3 ==
+
+# ...H..
+# ....T.
+# ......
+# ......
+# s.....
+
+# ..HT..
+# ......
+# ......
+# ......
+# s.....
+
+# .HT...
+# ......
+# ......
+# ......
+# s.....
+
+# == D 1 ==
+
+# ..T...
+# .H....
+# ......
+# ......
+# s.....
+
+# == R 4 ==
+
+# ..T...
+# ..H...
+# ......
+# ......
+# s.....
+
+# ..T...
+# ...H..
+# ......
+# ......
+# s.....
+
+# ......
+# ...TH.
+# ......
+# ......
+# s.....
+
+# ......
+# ....TH
+# ......
+# ......
+# s.....
+
+# == D 1 ==
+
+# ......
+# ....T.
+# .....H
+# ......
+# s.....
+
+# == L 5 ==
+
+# ......
+# ....T.
+# ....H.
+# ......
+# s.....
+
+# ......
+# ....T.
+# ...H..
+# ......
+# s.....
+
+# ......
+# ......
+# ..HT..
+# ......
+# s.....
+
+# ......
+# ......
+# .HT...
+# ......
+# s.....
+
+# ......
+# ......
+# HT....
+# ......
+# s.....
+
+# == R 2 ==
+
+# ......
+# ......
+# .H....  (H covers T)
+# ......
+# s.....
+
+# ......
+# ......
+# .TH...
+# ......
+# s.....
+
+# After simulating the rope, you can count up all of the positions the tail
+# visited at least once. In this diagram, s again marks the starting position
+# (which the tail also visited) and # marks other positions the tail visited:
+
+# ..##..
+# ...##.
+# .####.
+# ....#.
+# s###..
+
+# So, there are 13 positions the tail visited at least once.
+
+# Simulate your complete hypothetical series of motions. How many positions
+# does the tail of the rope visit at least once?
+
+with open("/home/xfeluser/AoC_2022/P9.txt") as f:
+    grid_positions = [line for line in f.read().strip().split("\n")]
+
+head = tail = (0, 0)
+visited = {(0, 0)}
+
+for motion in grid_positions:
+    direction, steps = motion.split()
+    for _ in range(int(steps)):
+        head = (
+            head[0] + (direction == "R") - (direction == "L"),
+            head[1] + (direction == "D") - (direction == "U"),
+        )
+
+        if max(abs(tail[0] - head[0]), abs(tail[1] - head[1])) == 2:
+            tail = (
+                tail[0] + (tail[0] < head[0]) - (tail[0] > head[0]),
+                tail[1] + (tail[1] < head[1]) - (tail[1] > head[1]),
+            )
+
+        visited.add(tail)
+
+print(len(visited))
+
+
+# --- Part Two ---
+
+# A rope snaps! Suddenly, the river is getting a lot closer than you remember.
+# The bridge is still there, but some of the ropes that broke are now whipping
+# toward you as you fall through the air!
+
+# The ropes are moving too quickly to grab; you only have a few seconds to
+# choose how to arch your body to avoid being hit. Fortunately, your simulation
+# can be extended to support longer ropes.
+
+# Rather than two knots, you now must simulate a rope consisting of ten knots.
+# One knot is still the head of the rope and moves according to the series of
+# motions. Each knot further down the rope follows the knot in front of it
+# using the same rules as before.
+
+# Using the same series of motions as the above example, but with the knots
+# marked H, 1, 2, ..., 9, the motions now occur as follows:
+
+# == Initial State ==
+
+# ......
+# ......
+# ......
+# ......
+# H.....  (H covers 1, 2, 3, 4, 5, 6, 7, 8, 9, s)
+
+# == R 4 ==
+
+# ......
+# ......
+# ......
+# ......
+# 1H....  (1 covers 2, 3, 4, 5, 6, 7, 8, 9, s)
+
+# ......
+# ......
+# ......
+# ......
+# 21H...  (2 covers 3, 4, 5, 6, 7, 8, 9, s)
+
+# ......
+# ......
+# ......
+# ......
+# 321H..  (3 covers 4, 5, 6, 7, 8, 9, s)
+
+# ......
+# ......
+# ......
+# ......
+# 4321H.  (4 covers 5, 6, 7, 8, 9, s)
+
+# == U 4 ==
+
+# ......
+# ......
+# ......
+# ....H.
+# 4321..  (4 covers 5, 6, 7, 8, 9, s)
+
+# ......
+# ......
+# ....H.
+# .4321.
+# 5.....  (5 covers 6, 7, 8, 9, s)
+
+# ......
+# ....H.
+# ....1.
+# .432..
+# 5.....  (5 covers 6, 7, 8, 9, s)
+
+# ....H.
+# ....1.
+# ..432.
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# == L 3 ==
+
+# ...H..
+# ....1.
+# ..432.
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# ..H1..
+# ...2..
+# ..43..
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# .H1...
+# ...2..
+# ..43..
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# == D 1 ==
+
+# ..1...
+# .H.2..
+# ..43..
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# == R 4 ==
+
+# ..1...
+# ..H2..
+# ..43..
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# ..1...
+# ...H..  (H covers 2)
+# ..43..
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# ......
+# ...1H.  (1 covers 2)
+# ..43..
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# ......
+# ...21H
+# ..43..
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# == D 1 ==
+
+# ......
+# ...21.
+# ..43.H
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# == L 5 ==
+
+# ......
+# ...21.
+# ..43H.
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# ......
+# ...21.
+# ..4H..  (H covers 3)
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# ......
+# ...2..
+# ..H1..  (H covers 4; 1 covers 3)
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# ......
+# ...2..
+# .H13..  (1 covers 4)
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# ......
+# ......
+# H123..  (2 covers 4)
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# == R 2 ==
+
+# ......
+# ......
+# .H23..  (H covers 1; 2 covers 4)
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# ......
+# ......
+# .1H3..  (H covers 2, 4)
+# .5....
+# 6.....  (6 covers 7, 8, 9, s)
+
+# Now, you need to keep track of the positions the new tail, 9, visits. In this
+# example, the tail never moves, and so it only visits 1 position. However, be
+# careful: more types of motion are possible than before, so you might want to
+# visually compare your simulated rope to the one above.
+
+# Here's a larger example:
+
+# R 5
+# U 8
+# L 8
+# D 3
+# R 17
+# D 10
+# L 25
+# U 20
+
+# These motions occur as follows (individual steps are not shown):
+
+# == Initial State ==
+
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ...........H..............  (H covers 1, 2, 3, 4, 5, 6, 7, 8, 9, s)
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+
+# == R 5 ==
+
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ...........54321H.........  (5 covers 6, 7, 8, 9, s)
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+
+# == U 8 ==
+
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ................H.........
+# ................1.........
+# ................2.........
+# ................3.........
+# ...............54.........
+# ..............6...........
+# .............7............
+# ............8.............
+# ...........9..............  (9 covers s)
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+
+# == L 8 ==
+
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ........H1234.............
+# ............5.............
+# ............6.............
+# ............7.............
+# ............8.............
+# ............9.............
+# ..........................
+# ..........................
+# ...........s..............
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+
+# == D 3 ==
+
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# .........2345.............
+# ........1...6.............
+# ........H...7.............
+# ............8.............
+# ............9.............
+# ..........................
+# ..........................
+# ...........s..............
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+
+# == R 17 ==
+
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ................987654321H
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ...........s..............
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+
+# == D 10 ==
+
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ...........s.........98765
+# .........................4
+# .........................3
+# .........................2
+# .........................1
+# .........................H
+
+# == L 25 ==
+
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ...........s..............
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# H123456789................
+
+# == U 20 ==
+
+# H.........................
+# 1.........................
+# 2.........................
+# 3.........................
+# 4.........................
+# 5.........................
+# 6.........................
+# 7.........................
+# 8.........................
+# 9.........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ...........s..............
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+
+# Now, the tail (9) visits 36 positions (including s) at least once:
+
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# ..........................
+# #.........................
+# #.............###.........
+# #............#...#........
+# .#..........#.....#.......
+# ..#..........#.....#......
+# ...#........#.......#.....
+# ....#......s.........#....
+# .....#..............#.....
+# ......#............#......
+# .......#..........#.......
+# ........#........#........
+# .........########.........
+
+# Simulate your complete series of motions on a larger rope with ten knots.
+# How many positions does the tail of the rope visit at least once?
+
+snake = [(0, 0)] * 10
+visited = {(0, 0)}
+
+for motion in grid_positions:
+    direction, steps = motion.split()
+    for _ in range(int(steps)):
+        snake[0] = (
+            snake[0][0] + (direction == "R") - (direction == "L"),
+            snake[0][1] + (direction == "D") - (direction == "U"),
+        )
+
+        for idx in range(1, 10):
+            head = snake[idx - 1]
+            tail = snake[idx]
+
+            if max(abs(tail[0] - head[0]), abs(tail[1] - head[1])) == 2:
+                tail = (
+                    tail[0] + (tail[0] < head[0]) - (tail[0] > head[0]),
+                    tail[1] + (tail[1] < head[1]) - (tail[1] > head[1]),
+                )
+
+            snake[idx - 1] = head
+            snake[idx] = tail
+
+        visited.add(snake[-1])
+
+print(len(visited))
diff --git a/src/Year_2022/Day10.py b/src/Year_2022/Day10.py
new file mode 100644
index 0000000..4e1fd15
--- /dev/null
+++ b/src/Year_2022/Day10.py
@@ -0,0 +1,397 @@
+# --- Day 10: Cathode-Ray Tube ---
+
+# You avoid the ropes, plunge into the river, and swim to shore.
+
+# The Elves yell something about meeting back up with them upriver, but the
+# river is too loud to tell exactly what they're saying. They finish crossing
+# the bridge and disappear from view.
+
+# Situations like this must be why the Elves prioritized getting the
+# communication system on your handheld device working. You pull it out of your
+# pack, but the amount of water slowly draining from a big crack in its screen
+# tells you it probably won't be of much immediate use.
+
+# Unless, that is, you can design a replacement for the device's video system!
+# It seems to be some kind of cathode-ray tube screen and simple CPU that are
+# both driven by a precise clock circuit. The clock circuit ticks at a constant
+# rate; each tick is called a cycle.
+
+# Start by figuring out the signal being sent by the CPU. The CPU has a single
+# register, X, which starts with the value 1. It supports only two
+# instructions:
+
+#     addx V takes two cycles to complete. After two cycles, the X register is
+# increased by the value V. (V can be negative.)
+#     noop takes one cycle to complete. It has no other effect.
+
+# The CPU uses these instructions in a program (your puzzle input) to, somehow,
+# tell the screen what to draw.
+
+# Consider the following small program:
+
+# noop
+# addx 3
+# addx -5
+
+# Execution of this program proceeds as follows:
+
+#     At the start of the first cycle, the noop instruction begins execution.
+# During the first cycle, X is 1. After the first cycle, the noop instruction
+# finishes execution, doing nothing.
+#     At the start of the second cycle, the addx 3 instruction begins
+# execution. During the second cycle, X is still 1.
+#     During the third cycle, X is still 1. After the third cycle, the addx 3
+# instruction finishes execution, setting X to 4.
+#     At the start of the fourth cycle, the addx -5 instruction begins
+# execution. During the fourth cycle, X is still 4.
+#     During the fifth cycle, X is still 4. After the fifth cycle, the addx -5
+# instruction finishes execution, setting X to -1.
+
+# Maybe you can learn something by looking at the value of the X register
+# throughout execution. For now, consider the signal strength (the cycle number
+# multiplied by the value of the X register) during the 20th cycle and every 40
+# cycles after that (that is, during the 20th, 60th, 100th, 140th, 180th, and
+# 220th cycles).
+
+# For example, consider this larger program:
+
+# addx 15
+# addx -11
+# addx 6
+# addx -3
+# addx 5
+# addx -1
+# addx -8
+# addx 13
+# addx 4
+# noop
+# addx -1
+# addx 5
+# addx -1
+# addx 5
+# addx -1
+# addx 5
+# addx -1
+# addx 5
+# addx -1
+# addx -35
+# addx 1
+# addx 24
+# addx -19
+# addx 1
+# addx 16
+# addx -11
+# noop
+# noop
+# addx 21
+# addx -15
+# noop
+# noop
+# addx -3
+# addx 9
+# addx 1
+# addx -3
+# addx 8
+# addx 1
+# addx 5
+# noop
+# noop
+# noop
+# noop
+# noop
+# addx -36
+# noop
+# addx 1
+# addx 7
+# noop
+# noop
+# noop
+# addx 2
+# addx 6
+# noop
+# noop
+# noop
+# noop
+# noop
+# addx 1
+# noop
+# noop
+# addx 7
+# addx 1
+# noop
+# addx -13
+# addx 13
+# addx 7
+# noop
+# addx 1
+# addx -33
+# noop
+# noop
+# noop
+# addx 2
+# noop
+# noop
+# noop
+# addx 8
+# noop
+# addx -1
+# addx 2
+# addx 1
+# noop
+# addx 17
+# addx -9
+# addx 1
+# addx 1
+# addx -3
+# addx 11
+# noop
+# noop
+# addx 1
+# noop
+# addx 1
+# noop
+# noop
+# addx -13
+# addx -19
+# addx 1
+# addx 3
+# addx 26
+# addx -30
+# addx 12
+# addx -1
+# addx 3
+# addx 1
+# noop
+# noop
+# noop
+# addx -9
+# addx 18
+# addx 1
+# addx 2
+# noop
+# noop
+# addx 9
+# noop
+# noop
+# noop
+# addx -1
+# addx 2
+# addx -37
+# addx 1
+# addx 3
+# noop
+# addx 15
+# addx -21
+# addx 22
+# addx -6
+# addx 1
+# noop
+# addx 2
+# addx 1
+# noop
+# addx -10
+# noop
+# noop
+# addx 20
+# addx 1
+# addx 2
+# addx 2
+# addx -6
+# addx -11
+# noop
+# noop
+# noop
+
+# The interesting signal strengths can be determined as follows:
+
+#     During the 20th cycle, register X has the value 21, so the signal
+# strength is 20 * 21 = 420. (The 20th cycle occurs in the middle of the second
+# addx -1, so the value of register X is the starting value, 1, plus all of the
+# other addx values up to that point:
+# 1 + 15 - 11 + 6 - 3 + 5 - 1 - 8 + 13 + 4 = 21.)
+#     During the 60th cycle, register X has the value 19, so the signal
+# strength is 60 * 19 = 1140.
+#     During the 100th cycle, register X has the value 18, so the signal
+# strength is 100 * 18 = 1800.
+#     During the 140th cycle, register X has the value 21, so the signal
+# strength is 140 * 21 = 2940.
+#     During the 180th cycle, register X has the value 16, so the signal
+# strength is 180 * 16 = 2880.
+#     During the 220th cycle, register X has the value 18, so the signal
+# strength is 220 * 18 = 3960.
+
+# The sum of these signal strengths is 13140.
+
+# Find the signal strength during the 20th, 60th, 100th, 140th, 180th, and
+# 220th cycles. What is the sum of these six signal strengths?
+
+with open("/home/xfeluser/AoC_2022/P10.txt") as f:
+    program = [line for line in f.read().strip().split("\n")]
+
+
+X = 1
+execution = [1, 1]
+
+for instruction in program:
+    if "noop" in instruction:
+        execution.append(X)
+    else:
+        execution.append(X)
+        X += int(instruction.split()[1])
+        execution.append(X)
+
+cycles = [20, 60, 100, 140, 180, 220]
+print(sum(cycle * execution[cycle] for cycle in cycles))
+
+
+# --- Part Two ---
+
+# It seems like the X register controls the horizontal position of a sprite.
+# Specifically, the sprite is 3 pixels wide, and the X register sets the
+# horizontal position of the middle of that sprite. (In this system, there is
+# no such thing as "vertical position": if the sprite's horizontal position
+# puts its pixels where the CRT is currently drawing, then those pixels will be
+# drawn.)
+
+# You count the pixels on the CRT: 40 wide and 6 high. This CRT screen draws
+# the top row of pixels left-to-right, then the row below that, and so on. The
+# left-most pixel in each row is in position 0, and the right-most pixel in
+# each row is in position 39.
+
+# Like the CPU, the CRT is tied closely to the clock circuit: the CRT draws a
+# single pixel during each cycle. Representing each pixel of the screen as a #,
+# here are the cycles during which the first and last pixel in each row are
+# drawn:
+
+# Cycle   1 -> ######################################## <- Cycle  40
+# Cycle  41 -> ######################################## <- Cycle  80
+# Cycle  81 -> ######################################## <- Cycle 120
+# Cycle 121 -> ######################################## <- Cycle 160
+# Cycle 161 -> ######################################## <- Cycle 200
+# Cycle 201 -> ######################################## <- Cycle 240
+
+# So, by carefully timing the CPU instructions and the CRT drawing operations,
+# you should be able to determine whether the sprite is visible the instant
+# each pixel is drawn. If the sprite is positioned such that one of its three
+# pixels is the pixel currently being drawn, the screen produces a lit pixel
+# (#); otherwise, the screen leaves the pixel dark (.).
+
+# The first few pixels from the larger example above are drawn as follows:
+
+# Sprite position: ###.....................................
+
+# Start cycle   1: begin executing addx 15
+# During cycle  1: CRT draws pixel in position 0
+# Current CRT row: #
+
+# During cycle  2: CRT draws pixel in position 1
+# Current CRT row: ##
+# End of cycle  2: finish executing addx 15 (Register X is now 16)
+# Sprite position: ...............###......................
+
+# Start cycle   3: begin executing addx -11
+# During cycle  3: CRT draws pixel in position 2
+# Current CRT row: ##.
+
+# During cycle  4: CRT draws pixel in position 3
+# Current CRT row: ##..
+# End of cycle  4: finish executing addx -11 (Register X is now 5)
+# Sprite position: ....###.................................
+
+# Start cycle   5: begin executing addx 6
+# During cycle  5: CRT draws pixel in position 4
+# Current CRT row: ##..#
+
+# During cycle  6: CRT draws pixel in position 5
+# Current CRT row: ##..##
+# End of cycle  6: finish executing addx 6 (Register X is now 11)
+# Sprite position: ..........###...........................
+
+# Start cycle   7: begin executing addx -3
+# During cycle  7: CRT draws pixel in position 6
+# Current CRT row: ##..##.
+
+# During cycle  8: CRT draws pixel in position 7
+# Current CRT row: ##..##..
+# End of cycle  8: finish executing addx -3 (Register X is now 8)
+# Sprite position: .......###..............................
+
+# Start cycle   9: begin executing addx 5
+# During cycle  9: CRT draws pixel in position 8
+# Current CRT row: ##..##..#
+
+# During cycle 10: CRT draws pixel in position 9
+# Current CRT row: ##..##..##
+# End of cycle 10: finish executing addx 5 (Register X is now 13)
+# Sprite position: ............###.........................
+
+# Start cycle  11: begin executing addx -1
+# During cycle 11: CRT draws pixel in position 10
+# Current CRT row: ##..##..##.
+
+# During cycle 12: CRT draws pixel in position 11
+# Current CRT row: ##..##..##..
+# End of cycle 12: finish executing addx -1 (Register X is now 12)
+# Sprite position: ...........###..........................
+
+# Start cycle  13: begin executing addx -8
+# During cycle 13: CRT draws pixel in position 12
+# Current CRT row: ##..##..##..#
+
+# During cycle 14: CRT draws pixel in position 13
+# Current CRT row: ##..##..##..##
+# End of cycle 14: finish executing addx -8 (Register X is now 4)
+# Sprite position: ...###..................................
+
+# Start cycle  15: begin executing addx 13
+# During cycle 15: CRT draws pixel in position 14
+# Current CRT row: ##..##..##..##.
+
+# During cycle 16: CRT draws pixel in position 15
+# Current CRT row: ##..##..##..##..
+# End of cycle 16: finish executing addx 13 (Register X is now 17)
+# Sprite position: ................###.....................
+
+# Start cycle  17: begin executing addx 4
+# During cycle 17: CRT draws pixel in position 16
+# Current CRT row: ##..##..##..##..#
+
+# During cycle 18: CRT draws pixel in position 17
+# Current CRT row: ##..##..##..##..##
+# End of cycle 18: finish executing addx 4 (Register X is now 21)
+# Sprite position: ....................###.................
+
+# Start cycle  19: begin executing noop
+# During cycle 19: CRT draws pixel in position 18
+# Current CRT row: ##..##..##..##..##.
+# End of cycle 19: finish executing noop
+
+# Start cycle  20: begin executing addx -1
+# During cycle 20: CRT draws pixel in position 19
+# Current CRT row: ##..##..##..##..##..
+
+# During cycle 21: CRT draws pixel in position 20
+# Current CRT row: ##..##..##..##..##..#
+# End of cycle 21: finish executing addx -1 (Register X is now 20)
+# Sprite position: ...................###..................
+
+# Allowing the program to run to completion causes the CRT to produce the
+# following image:
+
+# ##..##..##..##..##..##..##..##..##..##..
+# ###...###...###...###...###...###...###.
+# ####....####....####....####....####....
+# #####.....#####.....#####.....#####.....
+# ######......######......######......####
+# #######.......#######.......#######.....
+
+# Render the image given by your program. What eight capital letters appear on
+# your CRT?
+
+crt = []
+for cycle in range(241):
+    if not cycle % 40:
+        crt.append("\n")
+    crt.append("█" if cycle % 40 - execution[cycle + 1] in {-1, 0, 1} else " ")
+
+print("".join(crt))
diff --git a/src/Year_2022/Day11.py b/src/Year_2022/Day11.py
new file mode 100644
index 0000000..f1087ac
--- /dev/null
+++ b/src/Year_2022/Day11.py
@@ -0,0 +1,439 @@
+# --- Day 11: Monkey in the Middle ---
+
+# As you finally start making your way upriver, you realize your pack is much
+# lighter than you remember. Just then, one of the items from your pack goes
+# flying overhead. Monkeys are playing Keep Away with your missing things!
+
+# To get your stuff back, you need to be able to predict where the monkeys will
+# throw your items. After some careful observation, you realize the monkeys
+# operate based on how worried you are about each item.
+
+# You take some notes (your puzzle input) on the items each monkey currently
+# has, how worried you are about those items, and how the monkey makes
+# decisions based on your worry level. For example:
+
+# Monkey 0:
+#   Starting items: 79, 98
+#   Operation: new = old * 19
+#   Test: divisible by 23
+#     If true: throw to monkey 2
+#     If false: throw to monkey 3
+
+# Monkey 1:
+#   Starting items: 54, 65, 75, 74
+#   Operation: new = old + 6
+#   Test: divisible by 19
+#     If true: throw to monkey 2
+#     If false: throw to monkey 0
+
+# Monkey 2:
+#   Starting items: 79, 60, 97
+#   Operation: new = old * old
+#   Test: divisible by 13
+#     If true: throw to monkey 1
+#     If false: throw to monkey 3
+
+# Monkey 3:
+#   Starting items: 74
+#   Operation: new = old + 3
+#   Test: divisible by 17
+#     If true: throw to monkey 0
+#     If false: throw to monkey 1
+
+# Each monkey has several attributes:
+
+#     Starting items lists your worry level for each item the monkey is
+# currently holding in the order they will be inspected.
+#     Operation shows how your worry level changes as that monkey inspects an
+# item. (An operation like new = old * 5 means that your worry level after the
+# monkey inspected the item is five times whatever your worry level was before
+# inspection.)
+#     Test shows how the monkey uses your worry level to decide where to throw
+# an item next.
+#         If true shows what happens with an item if the Test was true.
+#         If false shows what happens with an item if the Test was false.
+
+# After each monkey inspects an item but before it tests your worry level, your
+# relief that the monkey's inspection didn't damage the item causes your worry
+# level to be divided by three and rounded down to the nearest integer.
+
+# The monkeys take turns inspecting and throwing items. On a single monkey's
+# turn, it inspects and throws all of the items it is holding one at a time and
+# in the order listed. Monkey 0 goes first, then monkey 1, and so on until each
+# monkey has had one turn. The process of each monkey taking a single turn is
+# called a round.
+
+# When a monkey throws an item to another monkey, the item goes on the end of
+# the recipient monkey's list. A monkey that starts a round with no items could
+# end up inspecting and throwing many items by the time its turn comes around.
+# If a monkey is holding no items at the start of its turn, its turn ends.
+
+# In the above example, the first round proceeds as follows:
+
+# Monkey 0:
+#   Monkey inspects an item with a worry level of 79.
+#     Worry level is multiplied by 19 to 1501.
+#     Monkey gets bored with item. Worry level is divided by 3 to 500.
+#     Current worry level is not divisible by 23.
+#     Item with worry level 500 is thrown to monkey 3.
+#   Monkey inspects an item with a worry level of 98.
+#     Worry level is multiplied by 19 to 1862.
+#     Monkey gets bored with item. Worry level is divided by 3 to 620.
+#     Current worry level is not divisible by 23.
+#     Item with worry level 620 is thrown to monkey 3.
+# Monkey 1:
+#   Monkey inspects an item with a worry level of 54.
+#     Worry level increases by 6 to 60.
+#     Monkey gets bored with item. Worry level is divided by 3 to 20.
+#     Current worry level is not divisible by 19.
+#     Item with worry level 20 is thrown to monkey 0.
+#   Monkey inspects an item with a worry level of 65.
+#     Worry level increases by 6 to 71.
+#     Monkey gets bored with item. Worry level is divided by 3 to 23.
+#     Current worry level is not divisible by 19.
+#     Item with worry level 23 is thrown to monkey 0.
+#   Monkey inspects an item with a worry level of 75.
+#     Worry level increases by 6 to 81.
+#     Monkey gets bored with item. Worry level is divided by 3 to 27.
+#     Current worry level is not divisible by 19.
+#     Item with worry level 27 is thrown to monkey 0.
+#   Monkey inspects an item with a worry level of 74.
+#     Worry level increases by 6 to 80.
+#     Monkey gets bored with item. Worry level is divided by 3 to 26.
+#     Current worry level is not divisible by 19.
+#     Item with worry level 26 is thrown to monkey 0.
+# Monkey 2:
+#   Monkey inspects an item with a worry level of 79.
+#     Worry level is multiplied by itself to 6241.
+#     Monkey gets bored with item. Worry level is divided by 3 to 2080.
+#     Current worry level is divisible by 13.
+#     Item with worry level 2080 is thrown to monkey 1.
+#   Monkey inspects an item with a worry level of 60.
+#     Worry level is multiplied by itself to 3600.
+#     Monkey gets bored with item. Worry level is divided by 3 to 1200.
+#     Current worry level is not divisible by 13.
+#     Item with worry level 1200 is thrown to monkey 3.
+#   Monkey inspects an item with a worry level of 97.
+#     Worry level is multiplied by itself to 9409.
+#     Monkey gets bored with item. Worry level is divided by 3 to 3136.
+#     Current worry level is not divisible by 13.
+#     Item with worry level 3136 is thrown to monkey 3.
+# Monkey 3:
+#   Monkey inspects an item with a worry level of 74.
+#     Worry level increases by 3 to 77.
+#     Monkey gets bored with item. Worry level is divided by 3 to 25.
+#     Current worry level is not divisible by 17.
+#     Item with worry level 25 is thrown to monkey 1.
+#   Monkey inspects an item with a worry level of 500.
+#     Worry level increases by 3 to 503.
+#     Monkey gets bored with item. Worry level is divided by 3 to 167.
+#     Current worry level is not divisible by 17.
+#     Item with worry level 167 is thrown to monkey 1.
+#   Monkey inspects an item with a worry level of 620.
+#     Worry level increases by 3 to 623.
+#     Monkey gets bored with item. Worry level is divided by 3 to 207.
+#     Current worry level is not divisible by 17.
+#     Item with worry level 207 is thrown to monkey 1.
+#   Monkey inspects an item with a worry level of 1200.
+#     Worry level increases by 3 to 1203.
+#     Monkey gets bored with item. Worry level is divided by 3 to 401.
+#     Current worry level is not divisible by 17.
+#     Item with worry level 401 is thrown to monkey 1.
+#   Monkey inspects an item with a worry level of 3136.
+#     Worry level increases by 3 to 3139.
+#     Monkey gets bored with item. Worry level is divided by 3 to 1046.
+#     Current worry level is not divisible by 17.
+#     Item with worry level 1046 is thrown to monkey 1.
+
+# After round 1, the monkeys are holding items with these worry levels:
+
+# Monkey 0: 20, 23, 27, 26
+# Monkey 1: 2080, 25, 167, 207, 401, 1046
+# Monkey 2:
+# Monkey 3:
+
+# Monkeys 2 and 3 aren't holding any items at the end of the round; they both
+# inspected items during the round and threw them all before the round ended.
+
+# This process continues for a few more rounds:
+
+# After round 2, the monkeys are holding items with these worry levels:
+# Monkey 0: 695, 10, 71, 135, 350
+# Monkey 1: 43, 49, 58, 55, 362
+# Monkey 2:
+# Monkey 3:
+
+# After round 3, the monkeys are holding items with these worry levels:
+# Monkey 0: 16, 18, 21, 20, 122
+# Monkey 1: 1468, 22, 150, 286, 739
+# Monkey 2:
+# Monkey 3:
+
+# After round 4, the monkeys are holding items with these worry levels:
+# Monkey 0: 491, 9, 52, 97, 248, 34
+# Monkey 1: 39, 45, 43, 258
+# Monkey 2:
+# Monkey 3:
+
+# After round 5, the monkeys are holding items with these worry levels:
+# Monkey 0: 15, 17, 16, 88, 1037
+# Monkey 1: 20, 110, 205, 524, 72
+# Monkey 2:
+# Monkey 3:
+
+# After round 6, the monkeys are holding items with these worry levels:
+# Monkey 0: 8, 70, 176, 26, 34
+# Monkey 1: 481, 32, 36, 186, 2190
+# Monkey 2:
+# Monkey 3:
+
+# After round 7, the monkeys are holding items with these worry levels:
+# Monkey 0: 162, 12, 14, 64, 732, 17
+# Monkey 1: 148, 372, 55, 72
+# Monkey 2:
+# Monkey 3:
+
+# After round 8, the monkeys are holding items with these worry levels:
+# Monkey 0: 51, 126, 20, 26, 136
+# Monkey 1: 343, 26, 30, 1546, 36
+# Monkey 2:
+# Monkey 3:
+
+# After round 9, the monkeys are holding items with these worry levels:
+# Monkey 0: 116, 10, 12, 517, 14
+# Monkey 1: 108, 267, 43, 55, 288
+# Monkey 2:
+# Monkey 3:
+
+# After round 10, the monkeys are holding items with these worry levels:
+# Monkey 0: 91, 16, 20, 98
+# Monkey 1: 481, 245, 22, 26, 1092, 30
+# Monkey 2:
+# Monkey 3:
+
+# ...
+
+# After round 15, the monkeys are holding items with these worry levels:
+# Monkey 0: 83, 44, 8, 184, 9, 20, 26, 102
+# Monkey 1: 110, 36
+# Monkey 2:
+# Monkey 3:
+
+# ...
+
+# After round 20, the monkeys are holding items with these worry levels:
+# Monkey 0: 10, 12, 14, 26, 34
+# Monkey 1: 245, 93, 53, 199, 115
+# Monkey 2:
+# Monkey 3:
+
+# Chasing all of the monkeys at once is impossible; you're going to have to
+# focus on the two most active monkeys if you want any hope of getting your
+# stuff back. Count the total number of times each monkey inspects items over
+# 20 rounds:
+
+# Monkey 0 inspected items 101 times.
+# Monkey 1 inspected items 95 times.
+# Monkey 2 inspected items 7 times.
+# Monkey 3 inspected items 105 times.
+
+# In this example, the two most active monkeys inspected items 101 and 105
+# times. The level of monkey business in this situation can be found by
+# multiplying these together: 10605.
+
+# Figure out which monkeys to chase by counting how many items they inspect
+# over 20 rounds. What is the level of monkey business after 20 rounds of
+# stuff-slinging simian shenanigans?
+
+from dataclasses import dataclass
+from math import prod
+
+with open("/home/xfeluser/AoC_2022/P11.txt") as f:
+    monkeys_list = [
+        [line for line in monkey.split("\n")]
+        for monkey in f.read().split("\n\n")
+    ]
+
+
+def parse_monkeys(lst):
+    monkeys = []
+
+    for monkey in lst:
+        items = [int(i) for i in monkey[1][18:].split(",")]
+        op = monkey[2][23:].split()
+        divisor = int(monkey[3][21:])
+        # True is the 0th item, False is the 1st!
+        dest = [int(monkey[4][29]), int(monkey[5][30])]
+        monkeys.append(Monkey(items, op, divisor, dest))
+
+    return monkeys
+
+
+@dataclass
+class Monkey:
+    items: list[int]
+    op: list[str]
+    divisor: int
+    dest: list[int]
+    act: int = 0
+
+
+def inspection(monkey):
+    for item in monkey.items:
+        op, value = monkey.op
+        if value == "old":
+            value = item
+        else:
+            value = int(value)
+        if op == "*":
+            item = (item * value) // 3
+        elif op == "+":
+            item = (item + value) // 3
+
+        # False if it divisible!
+        is_bored = bool(item % monkey.divisor)
+        monkey_receiving = monkey.dest[is_bored]
+        throw_to_monkey = monkeys[monkey_receiving]
+        throw_to_monkey.items.append(item)
+        monkey.act += 1
+
+    monkey.items = []
+
+
+monkeys = parse_monkeys(monkeys_list)
+rounds = 20
+for _ in range(rounds):
+    for monkey in monkeys:
+        inspection(monkey)
+
+print(prod(sorted([m.act for m in monkeys])[-2:]))
+
+
+# --- Part Two ---
+
+# You're worried you might not ever get your items back. So worried, in fact,
+# that your relief that a monkey's inspection didn't damage an item no longer
+# causes your worry level to be divided by three.
+
+# Unfortunately, that relief was all that was keeping your worry levels from
+# reaching ridiculous levels. You'll need to find another way to keep your
+# worry levels manageable.
+
+# At this rate, you might be putting up with these monkeys for a very long time
+# - possibly 10000 rounds!
+
+# With these new rules, you can still figure out the monkey business after
+# 10000 rounds. Using the same example above:
+
+# == After round 1 ==
+# Monkey 0 inspected items 2 times.
+# Monkey 1 inspected items 4 times.
+# Monkey 2 inspected items 3 times.
+# Monkey 3 inspected items 6 times.
+
+# == After round 20 ==
+# Monkey 0 inspected items 99 times.
+# Monkey 1 inspected items 97 times.
+# Monkey 2 inspected items 8 times.
+# Monkey 3 inspected items 103 times.
+
+# == After round 1000 ==
+# Monkey 0 inspected items 5204 times.
+# Monkey 1 inspected items 4792 times.
+# Monkey 2 inspected items 199 times.
+# Monkey 3 inspected items 5192 times.
+
+# == After round 2000 ==
+# Monkey 0 inspected items 10419 times.
+# Monkey 1 inspected items 9577 times.
+# Monkey 2 inspected items 392 times.
+# Monkey 3 inspected items 10391 times.
+
+# == After round 3000 ==
+# Monkey 0 inspected items 15638 times.
+# Monkey 1 inspected items 14358 times.
+# Monkey 2 inspected items 587 times.
+# Monkey 3 inspected items 15593 times.
+
+# == After round 4000 ==
+# Monkey 0 inspected items 20858 times.
+# Monkey 1 inspected items 19138 times.
+# Monkey 2 inspected items 780 times.
+# Monkey 3 inspected items 20797 times.
+
+# == After round 5000 ==
+# Monkey 0 inspected items 26075 times.
+# Monkey 1 inspected items 23921 times.
+# Monkey 2 inspected items 974 times.
+# Monkey 3 inspected items 26000 times.
+
+# == After round 6000 ==
+# Monkey 0 inspected items 31294 times.
+# Monkey 1 inspected items 28702 times.
+# Monkey 2 inspected items 1165 times.
+# Monkey 3 inspected items 31204 times.
+
+# == After round 7000 ==
+# Monkey 0 inspected items 36508 times.
+# Monkey 1 inspected items 33488 times.
+# Monkey 2 inspected items 1360 times.
+# Monkey 3 inspected items 36400 times.
+
+# == After round 8000 ==
+# Monkey 0 inspected items 41728 times.
+# Monkey 1 inspected items 38268 times.
+# Monkey 2 inspected items 1553 times.
+# Monkey 3 inspected items 41606 times.
+
+# == After round 9000 ==
+# Monkey 0 inspected items 46945 times.
+# Monkey 1 inspected items 43051 times.
+# Monkey 2 inspected items 1746 times.
+# Monkey 3 inspected items 46807 times.
+
+# == After round 10000 ==
+# Monkey 0 inspected items 52166 times.
+# Monkey 1 inspected items 47830 times.
+# Monkey 2 inspected items 1938 times.
+# Monkey 3 inspected items 52013 times.
+
+# After 10000 rounds, the two most active monkeys inspected items 52166 and
+# 52013 times. Multiplying these together, the level of monkey business in this
+# situation is now 2713310158.
+
+# Worry levels are no longer divided by three after each item is inspected;
+# you'll need to find another way to keep your worry levels manageable.
+# Starting again from the initial state in your puzzle input, what is the level
+# of monkey business after 10000 rounds?
+
+
+def inspection_long(monkey):
+    for item in monkey.items:
+        op, value = monkey.op
+        if value == "old":
+            value = item
+        else:
+            value = int(value)
+        magic_divisor = prod([m.divisor for m in monkeys])
+        if op == "*":
+            item = (item * value) % magic_divisor
+        elif op == "+":
+            item = (item + value) % magic_divisor
+
+        # False if it divisible!
+        is_bored = bool(item % monkey.divisor)
+        monkey_receiving = monkey.dest[is_bored]
+        throw_to_monkey = monkeys[monkey_receiving]
+        throw_to_monkey.items.append(item)
+        monkey.act += 1
+
+    monkey.items = []
+
+
+monkeys = parse_monkeys(monkeys_list)
+rounds = 10_000
+for _ in range(rounds):
+    for monkey in monkeys:
+        inspection_long(monkey)
+
+print(prod(sorted([m.act for m in monkeys])[-2:]))
diff --git a/src/Year_2022/Day12.py b/src/Year_2022/Day12.py
new file mode 100644
index 0000000..67feca6
--- /dev/null
+++ b/src/Year_2022/Day12.py
@@ -0,0 +1,140 @@
+# --- Day 12: Hill Climbing Algorithm ---
+
+# You try contacting the Elves using your handheld device, but the river you're
+# following must be too low to get a decent signal.
+
+# You ask the device for a heightmap of the surrounding area (your puzzle
+# input). The heightmap shows the local area from above broken into a grid;
+# the elevation of each square of the grid is given by a single lowercase
+# letter, where a is the lowest elevation, b is the next-lowest, and so on up
+# to the highest elevation, z.
+
+# Also included on the heightmap are marks for your current position (S) and
+# the location that should get the best signal (E). Your current position (S)
+# has elevation a, and the location that should get the best signal (E) has
+# elevation z.
+
+# You'd like to reach E, but to save energy, you should do it in as few steps
+# as possible. During each step, you can move exactly one square up, down,
+# left, or right. To avoid needing to get out your climbing gear, the elevation
+# of the destination square can be at most one higher than the elevation of
+# your current square; that is, if your current elevation is m, you could step
+# to elevation n, but not to elevation o. (This also means that the elevation
+# of the destination square can be much lower than the elevation of your
+# current square.)
+
+# For example:
+
+# Sabqponm
+# abcryxxl
+# accszExk
+# acctuvwj
+# abdefghi
+
+# Here, you start in the top-left corner; your goal is near the middle. You
+# could start by moving down or right, but eventually you'll need to head
+# toward the e at the bottom. From there, you can spiral around to the goal:
+
+# v..v<<<<
+# >v.vv<<^
+# .>vv>E^^
+# ..v>>>^^
+# ..>>>>>^
+
+# In the above diagram, the symbols indicate whether the path exits each
+# square moving up (^), down (v), left (<), or right (>). The location that
+# should get the best signal is still E, and . marks unvisited squares.
+
+# This path reaches the goal in 31 steps, the fewest possible.
+
+# What is the fewest steps required to move from your current position to the
+# location that should get the best signal?
+
+with open("/home/xfeluser/AoC_2022/P12.txt") as f:
+    diagram = {
+        (x, y): ord(e)
+        for y, line in enumerate(f.read().strip().split("\n"))
+        for x, e in enumerate(line.strip())
+    }
+
+start = [k for k in diagram if diagram[k] == ord("S")][0]
+end = [k for k in diagram if diagram[k] == ord("E")][0]
+# Use expected value for starting/ending point
+diagram[start] = ord("a")
+diagram[end] = ord("z")
+
+queue = [(start, 0)]
+pos_to_steps = {}
+
+while queue:
+    cur, steps = queue.pop()
+
+    if cur not in pos_to_steps or steps < pos_to_steps[cur]:
+        pos_to_steps[cur] = steps
+
+        for offset in ((1, 0), (-1, 0), (0, 1), (0, -1)):
+            new_pos = cur[0] + offset[0], cur[1] + offset[1]
+
+            if new_pos in diagram and diagram[new_pos] - diagram[cur] <= 1:
+                queue.append((new_pos, steps + 1))
+
+print(pos_to_steps[end])
+
+
+# --- Part Two ---
+
+# As you walk up the hill, you suspect that the Elves will want to turn this
+# into a hiking trail. The beginning isn't very scenic, though; perhaps you can
+# find a better starting point.
+
+# To maximize exercise while hiking, the trail should start as low as possible:
+# elevation a. The goal is still the square marked E. However, the trail should
+# still be direct, taking the fewest steps to reach its goal. So, you'll need
+# to find the shortest path from any square at elevation a to the square marked
+# E.
+
+# Again consider the example from above:
+
+# Sabqponm
+# abcryxxl
+# accszExk
+# acctuvwj
+# abdefghi
+
+# Now, there are six choices for starting position (five marked a, plus the
+# square marked S that counts as being at elevation a). If you start at the
+# bottom-left square, you can reach the goal most quickly:
+
+# ...v<<<<
+# ...vv<<^
+# ...v>E^^
+# .>v>>>^^
+# >^>>>>>^
+
+# This path reaches the goal in only 29 steps, the fewest possible.
+
+# What is the fewest steps required to move starting from any square with
+# elevation a to the location that should get the best signal?
+
+starts = []
+pos_to_steps = {}
+
+for start in [k for k in diagram if diagram[k] == ord("a")]:
+    queue = [(start, 0)]
+
+    while queue:
+        cur, steps = queue.pop()
+
+        if cur not in pos_to_steps or steps < pos_to_steps[cur]:
+            pos_to_steps[cur] = steps
+
+            for offset in ((1, 0), (-1, 0), (0, 1), (0, -1)):
+                new_pos = cur[0] + offset[0], cur[1] + offset[1]
+
+                if new_pos in diagram and diagram[new_pos] - diagram[cur] <= 1:
+                    queue.append((new_pos, steps + 1))
+
+    if end in pos_to_steps:
+        starts.append(pos_to_steps[end])
+
+print(min(starts))
diff --git a/src/Year_2022/Day13.py b/src/Year_2022/Day13.py
new file mode 100644
index 0000000..f651adb
--- /dev/null
+++ b/src/Year_2022/Day13.py
@@ -0,0 +1,218 @@
+# --- Day 13: Distress Signal ---
+
+# You climb the hill and again try contacting the Elves. However, you instead
+# receive a signal you weren't expecting: a distress signal.
+
+# Your handheld device must still not be working properly; the packets from the
+# distress signal got decoded out of order. You'll need to re-order the list of
+# received packets (your puzzle input) to decode the message.
+
+# Your list consists of pairs of packets; pairs are separated by a blank line.
+# You need to identify how many pairs of packets are in the right order.
+
+# For example:
+
+# [1,1,3,1,1]
+# [1,1,5,1,1]
+
+# [[1],[2,3,4]]
+# [[1],4]
+
+# [9]
+# [[8,7,6]]
+
+# [[4,4],4,4]
+# [[4,4],4,4,4]
+
+# [7,7,7,7]
+# [7,7,7]
+
+# []
+# [3]
+
+# [[[]]]
+# [[]]
+
+# [1,[2,[3,[4,[5,6,7]]]],8,9]
+# [1,[2,[3,[4,[5,6,0]]]],8,9]
+
+# Packet data consists of lists and integers. Each list starts with [, ends
+# with ], and contains zero or more comma-separated values (either integers or
+# other lists). Each packet is always a list and appears on its own line.
+
+# When comparing two values, the first value is called left and the second
+# value is called right. Then:
+
+#     If both values are integers, the lower integer should come first. If the
+# left integer is lower than the right integer, the inputs are in the right
+# order. If the left integer is higher than the right integer, the inputs are
+# not in the right order. Otherwise, the inputs are the same integer; continue
+# checking the next part of the input.
+#     If both values are lists, compare the first value of each list, then the
+# second value, and so on. If the left list runs out of items first, the inputs
+# are in the right order. If the right list runs out of items first, the inputs
+# are not in the right order. If the lists are the same length and no
+# comparison makes a decision about the order, continue checking the next part
+# of the input.
+#     If exactly one value is an integer, convert the integer to a list which
+# contains that integer as its only value, then retry the comparison. For
+# example, if comparing [0,0,0] and 2, convert the right value to [2] (a list
+# containing 2); the result is then found by instead comparing [0,0,0] and [2].
+
+# Using these rules, you can determine which of the pairs in the example are in
+# the right order:
+
+# == Pair 1 ==
+# - Compare [1,1,3,1,1] vs [1,1,5,1,1]
+#   - Compare 1 vs 1
+#   - Compare 1 vs 1
+#   - Compare 3 vs 5
+#     - Left side is smaller, so inputs are in the right order
+
+# == Pair 2 ==
+# - Compare [[1],[2,3,4]] vs [[1],4]
+#   - Compare [1] vs [1]
+#     - Compare 1 vs 1
+#   - Compare [2,3,4] vs 4
+#     - Mixed types; convert right to [4] and retry comparison
+#     - Compare [2,3,4] vs [4]
+#       - Compare 2 vs 4
+#         - Left side is smaller, so inputs are in the right order
+
+# == Pair 3 ==
+# - Compare [9] vs [[8,7,6]]
+#   - Compare 9 vs [8,7,6]
+#     - Mixed types; convert left to [9] and retry comparison
+#     - Compare [9] vs [8,7,6]
+#       - Compare 9 vs 8
+#         - Right side is smaller, so inputs are not in the right order
+
+# == Pair 4 ==
+# - Compare [[4,4],4,4] vs [[4,4],4,4,4]
+#   - Compare [4,4] vs [4,4]
+#     - Compare 4 vs 4
+#     - Compare 4 vs 4
+#   - Compare 4 vs 4
+#   - Compare 4 vs 4
+#   - Left side ran out of items, so inputs are in the right order
+
+# == Pair 5 ==
+# - Compare [7,7,7,7] vs [7,7,7]
+#   - Compare 7 vs 7
+#   - Compare 7 vs 7
+#   - Compare 7 vs 7
+#   - Right side ran out of items, so inputs are not in the right order
+
+# == Pair 6 ==
+# - Compare [] vs [3]
+#   - Left side ran out of items, so inputs are in the right order
+
+# == Pair 7 ==
+# - Compare [[[]]] vs [[]]
+#   - Compare [[]] vs []
+#     - Right side ran out of items, so inputs are not in the right order
+
+# == Pair 8 ==
+# - Compare [1,[2,[3,[4,[5,6,7]]]],8,9] vs [1,[2,[3,[4,[5,6,0]]]],8,9]
+#   - Compare 1 vs 1
+#   - Compare [2,[3,[4,[5,6,7]]]] vs [2,[3,[4,[5,6,0]]]]
+#     - Compare 2 vs 2
+#     - Compare [3,[4,[5,6,7]]] vs [3,[4,[5,6,0]]]
+#       - Compare 3 vs 3
+#       - Compare [4,[5,6,7]] vs [4,[5,6,0]]
+#         - Compare 4 vs 4
+#         - Compare [5,6,7] vs [5,6,0]
+#           - Compare 5 vs 5
+#           - Compare 6 vs 6
+#           - Compare 7 vs 0
+#             - Right side is smaller, so inputs are not in the right order
+
+# What are the indices of the pairs that are already in the right order? (The
+# first pair has index 1, the second pair has index 2, and so on.) In the above
+# example, the pairs in the right order are 1, 2, 4, and 6; the sum of these
+# indices is 13.
+
+# Determine which pairs of packets are already in the right order. What is the
+# sum of the indices of those pairs?
+
+from ast import literal_eval
+
+with open("/home/xfeluser/AoC_2022/P13.txt") as f:
+    packets = [literal_eval(line) for line in f.read().strip().split()]
+
+
+def is_ordered(left, right):
+    if type(left) is list or type(right) is list:
+        if type(left) is not list:
+            left = [left]
+        if type(right) is not list:
+            right = [right]
+        for rec_left, rec_right in zip(left, right):
+            rec = is_ordered(rec_left, rec_right)
+            if rec != 0:
+                return rec
+        return len(left) - len(right)
+    else:
+        return left - right
+
+
+count = 0
+for idx, (left, right) in enumerate(zip(packets[::2], packets[1::2]), start=1):
+    if is_ordered(left, right) >= 0:
+        continue
+    count += idx
+
+print(count)
+
+
+# --- Part Two ---
+
+# Now, you just need to put all of the packets in the right order. Disregard
+# the blank lines in your list of received packets.
+
+# The distress signal protocol also requires that you include two additional
+# divider packets:
+
+# [[2]]
+# [[6]]
+
+# Using the same rules as before, organize all packets - the ones in your list
+# of received packets as well as the two divider packets - into the correct
+# order.
+
+# For the example above, the result of putting the packets in the correct order
+# is:
+
+# []
+# [[]]
+# [[[]]]
+# [1,1,3,1,1]
+# [1,1,5,1,1]
+# [[1],[2,3,4]]
+# [1,[2,[3,[4,[5,6,0]]]],8,9]
+# [1,[2,[3,[4,[5,6,7]]]],8,9]
+# [[1],4]
+# [[2]]
+# [3]
+# [[4,4],4,4]
+# [[4,4],4,4,4]
+# [[6]]
+# [7,7,7]
+# [7,7,7,7]
+# [[8,7,6]]
+# [9]
+
+# Afterward, locate the divider packets. To find the decoder key for this
+# distress signal, you need to determine the indices of the two divider packets
+# and multiply them together. (The first packet is at index 1, the second
+# packet is at index 2, and so on.) In this example, the divider packets are
+# 10th and 14th, and so the decoder key is 140.
+
+# Organize all of the packets into the correct order. What is the decoder key
+# for the distress signal?
+
+from functools import cmp_to_key
+
+packets += [[[2]], [[6]]]
+packets_sorted = sorted(packets, key=cmp_to_key(is_ordered))
+print((packets_sorted.index([[2]]) + 1) * (packets_sorted.index([[6]]) + 1))
diff --git a/src/Year_2022/Day14.py b/src/Year_2022/Day14.py
new file mode 100644
index 0000000..d15ecb9
--- /dev/null
+++ b/src/Year_2022/Day14.py
@@ -0,0 +1,254 @@
+# --- Day 14: Regolith Reservoir ---
+
+# The distress signal leads you to a giant waterfall! Actually, hang on - the
+# signal seems like it's coming from the waterfall itself, and that doesn't
+# make any sense. However, you do notice a little path that leads behind the
+# waterfall.
+
+# Correction: the distress signal leads you behind a giant waterfall! There
+# seems to be a large cave system here, and the signal definitely leads further
+# inside.
+
+# As you begin to make your way deeper underground, you feel the ground rumble
+# for a moment. Sand begins pouring into the cave! If you don't quickly figure
+# out where the sand is going, you could quickly become trapped!
+
+# Fortunately, your familiarity with analyzing the path of falling material
+# will come in handy here. You scan a two-dimensional vertical slice of the
+# cave above you (your puzzle input) and discover that it is mostly air with
+# structures made of rock.
+
+# Your scan traces the path of each solid rock structure and reports the x,y
+# coordinates that form the shape of the path, where x represents distance to
+# the right and y represents distance down. Each path appears as a single line
+# of text in your scan. After the first point of each path, each point
+# indicates the end of a straight horizontal or vertical line to be drawn from
+# the previous point. For example:
+
+# 498,4 -> 498,6 -> 496,6
+# 503,4 -> 502,4 -> 502,9 -> 494,9
+
+# This scan means that there are two paths of rock; the first path consists of
+# two straight lines, and the second path consists of three straight lines.
+# (Specifically, the first path consists of a line of rock from 498,4 through
+# 498,6 and another line of rock from 498,6 through 496,6.)
+
+# The sand is pouring into the cave from point 500,0.
+
+# Drawing rock as #, air as ., and the source of the sand as +, this becomes:
+
+
+#   4     5  5
+#   9     0  0
+#   4     0  3
+# 0 ......+...
+# 1 ..........
+# 2 ..........
+# 3 ..........
+# 4 ....#...##
+# 5 ....#...#.
+# 6 ..###...#.
+# 7 ........#.
+# 8 ........#.
+# 9 #########.
+
+# Sand is produced one unit at a time, and the next unit of sand is not
+# produced until the previous unit of sand comes to rest. A unit of sand is
+# large enough to fill one tile of air in your scan.
+
+# A unit of sand always falls down one step if possible. If the tile
+# immediately below is blocked (by rock or sand), the unit of sand attempts to
+# instead move diagonally one step down and to the left. If that tile is
+# blocked, the unit of sand attempts to instead move diagonally one step down
+# and to the right. Sand keeps moving as long as it is able to do so, at each
+# step trying to move down, then down-left, then down-right. If all three
+# possible destinations are blocked, the unit of sand comes to rest and no
+# longer moves, at which point the next unit of sand is created back at the
+# source.
+
+# So, drawing sand that has come to rest as o, the first unit of sand simply
+# falls straight down and then stops:
+
+# ......+...
+# ..........
+# ..........
+# ..........
+# ....#...##
+# ....#...#.
+# ..###...#.
+# ........#.
+# ......o.#.
+# #########.
+
+# The second unit of sand then falls straight down, lands on the first one, and
+# then comes to rest to its left:
+
+# ......+...
+# ..........
+# ..........
+# ..........
+# ....#...##
+# ....#...#.
+# ..###...#.
+# ........#.
+# .....oo.#.
+# #########.
+
+# After a total of five units of sand have come to rest, they form this
+# pattern:
+
+# ......+...
+# ..........
+# ..........
+# ..........
+# ....#...##
+# ....#...#.
+# ..###...#.
+# ......o.#.
+# ....oooo#.
+# #########.
+
+# After a total of 22 units of sand:
+
+# ......+...
+# ..........
+# ......o...
+# .....ooo..
+# ....#ooo##
+# ....#ooo#.
+# ..###ooo#.
+# ....oooo#.
+# ...ooooo#.
+# #########.
+
+# Finally, only two more units of sand can possibly come to rest:
+
+# ......+...
+# ..........
+# ......o...
+# .....ooo..
+# ....#ooo##
+# ...o#ooo#.
+# ..###ooo#.
+# ....oooo#.
+# .o.ooooo#.
+# #########.
+
+# Once all 24 units of sand shown above have come to rest, all further sand
+# flows out the bottom, falling into the endless void. Just for fun, the path
+# any new sand takes before falling forever is shown here with ~:
+
+# .......+...
+# .......~...
+# ......~o...
+# .....~ooo..
+# ....~#ooo##
+# ...~o#ooo#.
+# ..~###ooo#.
+# ..~..oooo#.
+# .~o.ooooo#.
+# ~#########.
+# ~..........
+# ~..........
+# ~..........
+
+# Using your scan, simulate the falling sand. How many units of sand come to
+# rest before sand starts flowing into the abyss below?
+
+with open("/home/xfeluser/AoC_2022/P14.txt") as f:
+    scan_traces = [line for line in f.read().strip().split("\n")]
+
+
+def cave_generation():
+    cave = set()
+
+    for line in scan_traces:
+        points = line.split(" -> ")
+        for point0, point1 in zip(points, points[1:]):
+            x0, y0 = [int(p) for p in point0.split(",")]
+            x1, y1 = [int(p) for p in point1.split(",")]
+            for x in range(min(x0, x1), max(x0, x1) + 1):
+                for y in range(min(y0, y1), max(y0, y1) + 1):
+                    cave.add((x, y))
+
+    return cave
+
+
+def sand_simulation(deepest):
+    sand, dirs = 0, [(0, 1), (-1, 1), (1, 1)]
+    while True:
+        x, y = 500, 0
+        while True:
+            if y >= deepest or (500, 0) in cave:
+                return sand
+            for dx, dy in dirs:
+                nx, ny = x + dx, y + dy
+                if (nx, ny) not in cave:
+                    break
+            else:
+                cave.add((x, y))
+                sand += 1
+                break
+            x, y = nx, ny
+
+
+cave = cave_generation()
+deepest = max(y for (x, y) in cave)
+print(sand_simulation(deepest))
+
+
+# --- Part Two ---
+
+# You realize you misread the scan. There isn't an endless void at the bottom
+# of the scan - there's floor, and you're standing on it!
+
+# You don't have time to scan the floor, so assume the floor is an infinite
+# horizontal line with a y coordinate equal to two plus the highest y
+# coordinate of any point in your scan.
+
+# In the example above, the highest y coordinate of any point is 9, and so the
+# floor is at y=11. (This is as if your scan contained one extra rock path
+# like -infinity,11 -> infinity,11.) With the added floor, the example above
+# now looks like this:
+
+#         ...........+........
+#         ....................
+#         ....................
+#         ....................
+#         .........#...##.....
+#         .........#...#......
+#         .......###...#......
+#         .............#......
+#         .............#......
+#         .....#########......
+#         ....................
+# <-- etc #################### etc -->
+
+# To find somewhere safe to stand, you'll need to simulate falling sand until a
+# unit of sand comes to rest at 500,0, blocking the source entirely and
+# stopping the flow of sand into the cave. In the example above, the situation
+# finally looks like this after 93 units of sand come to rest:
+
+# ............o............
+# ...........ooo...........
+# ..........ooooo..........
+# .........ooooooo.........
+# ........oo#ooo##o........
+# .......ooo#ooo#ooo.......
+# ......oo###ooo#oooo......
+# .....oooo.oooo#ooooo.....
+# ....oooooooooo#oooooo....
+# ...ooo#########ooooooo...
+# ..ooooo.......ooooooooo..
+# #########################
+
+# Using your scan, simulate the falling sand until the source of the sand
+# becomes blocked. How many units of sand come to rest?
+
+
+cave = cave_generation()
+even_deepest = deepest + 2
+# 1000 in both directions should be enough
+for x in range(-1000, 1000):
+    cave.add((x, even_deepest))
+print(sand_simulation(even_deepest))
diff --git a/src/Year_2022/Day15.py b/src/Year_2022/Day15.py
new file mode 100644
index 0000000..47d3df1
--- /dev/null
+++ b/src/Year_2022/Day15.py
@@ -0,0 +1,190 @@
+# --- Day 15: Beacon Exclusion Zone ---
+
+# You feel the ground rumble again as the distress signal leads you to a large
+# network of subterranean tunnels. You don't have time to search them all, but
+# you don't need to: your pack contains a set of deployable sensors that you
+# imagine were originally built to locate lost Elves.
+
+# The sensors aren't very powerful, but that's okay; your handheld device
+# indicates that you're close enough to the source of the distress signal to
+# use them. You pull the emergency sensor system out of your pack, hit the big
+# button on top, and the sensors zoom off down the tunnels.
+
+# Once a sensor finds a spot it thinks will give it a good reading, it attaches
+# itself to a hard surface and begins monitoring for the nearest signal source
+# beacon. Sensors and beacons always exist at integer coordinates. Each sensor
+# knows its own position and can determine the position of a beacon precisely;
+# however, sensors can only lock on to the one beacon closest to the sensor as
+# measured by the Manhattan distance. (There is never a tie where two beacons
+# are the same distance to a sensor.)
+
+# It doesn't take long for the sensors to report back their positions and
+# closest beacons (your puzzle input). For example:
+
+# Sensor at x=2, y=18: closest beacon is at x=-2, y=15
+# Sensor at x=9, y=16: closest beacon is at x=10, y=16
+# Sensor at x=13, y=2: closest beacon is at x=15, y=3
+# Sensor at x=12, y=14: closest beacon is at x=10, y=16
+# Sensor at x=10, y=20: closest beacon is at x=10, y=16
+# Sensor at x=14, y=17: closest beacon is at x=10, y=16
+# Sensor at x=8, y=7: closest beacon is at x=2, y=10
+# Sensor at x=2, y=0: closest beacon is at x=2, y=10
+# Sensor at x=0, y=11: closest beacon is at x=2, y=10
+# Sensor at x=20, y=14: closest beacon is at x=25, y=17
+# Sensor at x=17, y=20: closest beacon is at x=21, y=22
+# Sensor at x=16, y=7: closest beacon is at x=15, y=3
+# Sensor at x=14, y=3: closest beacon is at x=15, y=3
+# Sensor at x=20, y=1: closest beacon is at x=15, y=3
+
+# So, consider the sensor at 2,18; the closest beacon to it is at -2,15. For
+# the sensor at 9,16, the closest beacon to it is at 10,16.
+
+# Drawing sensors as S and beacons as B, the above arrangement of sensors and
+# beacons looks like this:
+
+#                1    1    2    2
+#      0    5    0    5    0    5
+#  0 ....S.......................
+#  1 ......................S.....
+#  2 ...............S............
+#  3 ................SB..........
+#  4 ............................
+#  5 ............................
+#  6 ............................
+#  7 ..........S.......S.........
+#  8 ............................
+#  9 ............................
+# 10 ....B.......................
+# 11 ..S.........................
+# 12 ............................
+# 13 ............................
+# 14 ..............S.......S.....
+# 15 B...........................
+# 16 ...........SB...............
+# 17 ................S..........B
+# 18 ....S.......................
+# 19 ............................
+# 20 ............S......S........
+# 21 ............................
+# 22 .......................B....
+
+# This isn't necessarily a comprehensive map of all beacons in the area,
+# though. Because each sensor only identifies its closest beacon, if a sensor
+# detects a beacon, you know there are no other beacons that close or closer to
+# that sensor. There could still be beacons that just happen to not be the
+# closest beacon to any sensor. Consider the sensor at 8,7:
+
+#                1    1    2    2
+#      0    5    0    5    0    5
+# -2 ..........#.................
+# -1 .........###................
+#  0 ....S...#####...............
+#  1 .......#######........S.....
+#  2 ......#########S............
+#  3 .....###########SB..........
+#  4 ....#############...........
+#  5 ...###############..........
+#  6 ..#################.........
+#  7 .#########S#######S#........
+#  8 ..#################.........
+#  9 ...###############..........
+# 10 ....B############...........
+# 11 ..S..###########............
+# 12 ......#########.............
+# 13 .......#######..............
+# 14 ........#####.S.......S.....
+# 15 B........###................
+# 16 ..........#SB...............
+# 17 ................S..........B
+# 18 ....S.......................
+# 19 ............................
+# 20 ............S......S........
+# 21 ............................
+# 22 .......................B....
+
+# This sensor's closest beacon is at 2,10, and so you know there are no beacons
+# that close or closer (in any positions marked #).
+
+# None of the detected beacons seem to be producing the distress signal, so
+# you'll need to work out where the distress beacon is by working out where it
+# isn't. For now, keep things simple by counting the positions where a beacon
+# cannot possibly be along just a single row.
+
+# So, suppose you have an arrangement of beacons and sensors like in the
+# example above and, just in the row where y=10, you'd like to count the number
+# of positions a beacon cannot possibly exist. The coverage from all sensors
+# near that row looks like this:
+
+#                  1    1    2    2
+#        0    5    0    5    0    5
+#  9 ...#########################...
+# 10 ..####B######################..
+# 11 .###S#############.###########.
+
+# In this example, in the row where y=10, there are 26 positions where a beacon
+# cannot be present.
+
+# Consult the report from the sensors you just deployed. In the row where
+# y=2000000, how many positions cannot contain a beacon?
+
+with open("/home/xfeluser/AoC_2022/P15.txt") as f:
+    data = [line for line in f.read().strip().split("\n")]
+
+
+def is_possible(x, y):
+    for sx, sy, d in sensors:
+        if abs(x - sx) + abs(y - sy) <= d and (x, y) not in beacons:
+            return False
+    return True
+
+
+sensors, beacons = set(), set()
+for line in data:
+    parts = line.split()
+    sx, sy = int(parts[2][2:-1]), int(parts[3][2:-1])
+    bx, by = int(parts[8][2:-1]), int(parts[9][2:])
+    d = abs(sx - bx) + abs(sy - by)
+    sensors.add((sx, sy, d))
+    beacons.add((bx, by))
+
+count = 0
+y = 2_000_000
+for x in range(
+    min(x - d for x, _, d in sensors), max(x + d for x, _, d in sensors)
+):
+    if not is_possible(x, y) and (x, y) not in beacons:
+        count += 1
+
+print(count)
+
+# --- Part Two ---
+
+# Your handheld device indicates that the distress signal is coming from a
+# beacon nearby. The distress beacon is not detected by any sensor, but the
+# distress beacon must have x and y coordinates each no lower than 0 and no
+# larger than 4000000.
+
+# To isolate the distress beacon's signal, you need to determine its tuning
+# frequency, which can be found by multiplying its x coordinate by 4000000 and
+# then adding its y coordinate.
+
+# In the example above, the search space is smaller: instead, the x and y
+# coordinates can each be at most 20. With this reduced search area, there is
+# only a single position that could have a beacon: x=14, y=11. The tuning
+# frequency for this distress beacon is 56000011.
+
+# Find the only possible position for the distress beacon. What is its tuning
+# frequency?
+
+import sys
+
+for sx, sy, d in sensors:
+    for dx in range(d + 2):
+        dy = (d + 1) - dx
+        for mx, my in [(-1, 1), (1, -1), (-1, -1), (1, 1)]:
+            x, y = sx + (dx * mx), sy + (dy * my)
+            if not (0 <= x <= 4_000_000 and 0 <= y <= 4_000_000):
+                continue
+            if is_possible(x, y):
+                print(x * 4_000_000 + y)
+                sys.exit()
diff --git a/src/Year_2022/files/P1.txt b/src/Year_2022/files/P1.txt
new file mode 100644
index 0000000..96dba2d
--- /dev/null
+++ b/src/Year_2022/files/P1.txt
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diff --git a/src/Year_2022/files/P10.txt b/src/Year_2022/files/P10.txt
new file mode 100644
index 0000000..50f6b08
--- /dev/null
+++ b/src/Year_2022/files/P10.txt
@@ -0,0 +1,140 @@
+noop
+noop
+noop
+addx 5
+noop
+addx 1
+noop
+addx 4
+addx 25
+addx -20
+noop
+noop
+addx 5
+addx 3
+noop
+addx 2
+noop
+noop
+addx -1
+addx 6
+addx 1
+noop
+addx 4
+noop
+addx -37
+noop
+noop
+noop
+addx 3
+addx 32
+addx -25
+addx 2
+addx 3
+noop
+addx 2
+addx 3
+noop
+addx 2
+addx 2
+addx -24
+addx 25
+addx 5
+addx 2
+addx 8
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+addx 18
+addx 5
+addx -39
+addx 11
+addx -9
+addx 6
+addx -2
+addx 5
+addx 4
+addx -4
+addx 3
+addx 5
+addx 2
+noop
+addx -1
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+addx -21
+addx 22
+addx 3
+addx 1
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+noop
+noop
+addx -35
+noop
+noop
+noop
+noop
+addx 37
+addx -33
+noop
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+addx 2
+addx -1
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+addx 1
+addx 5
+addx 2
+addx -19
+addx 21
+addx 1
+addx 5
+addx -31
+addx 36
+noop
+addx 3
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+addx -38
+noop
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+addx 7
+addx 14
+addx -4
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+addx 5
+addx 2
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+addx -15
+addx 6
+addx 2
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+addx -27
+addx 25
+addx 5
+noop
+addx 7
+addx -2
+addx 5
+addx -40
+noop
+addx 7
+noop
+addx -1
+addx 2
+addx 5
+addx -1
+addx 1
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+noop
+addx -2
+noop
+addx 3
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diff --git a/src/Year_2022/files/P11.txt b/src/Year_2022/files/P11.txt
new file mode 100644
index 0000000..2af0021
--- /dev/null
+++ b/src/Year_2022/files/P11.txt
@@ -0,0 +1,55 @@
+Monkey 0:
+  Starting items: 63, 57
+  Operation: new = old * 11
+  Test: divisible by 7
+    If true: throw to monkey 6
+    If false: throw to monkey 2
+
+Monkey 1:
+  Starting items: 82, 66, 87, 78, 77, 92, 83
+  Operation: new = old + 1
+  Test: divisible by 11
+    If true: throw to monkey 5
+    If false: throw to monkey 0
+
+Monkey 2:
+  Starting items: 97, 53, 53, 85, 58, 54
+  Operation: new = old * 7
+  Test: divisible by 13
+    If true: throw to monkey 4
+    If false: throw to monkey 3
+
+Monkey 3:
+  Starting items: 50
+  Operation: new = old + 3
+  Test: divisible by 3
+    If true: throw to monkey 1
+    If false: throw to monkey 7
+
+Monkey 4:
+  Starting items: 64, 69, 52, 65, 73
+  Operation: new = old + 6
+  Test: divisible by 17
+    If true: throw to monkey 3
+    If false: throw to monkey 7
+
+Monkey 5:
+  Starting items: 57, 91, 65
+  Operation: new = old + 5
+  Test: divisible by 2
+    If true: throw to monkey 0
+    If false: throw to monkey 6
+
+Monkey 6:
+  Starting items: 67, 91, 84, 78, 60, 69, 99, 83
+  Operation: new = old * old
+  Test: divisible by 5
+    If true: throw to monkey 2
+    If false: throw to monkey 4
+
+Monkey 7:
+  Starting items: 58, 78, 69, 65
+  Operation: new = old + 7
+  Test: divisible by 19
+    If true: throw to monkey 5
+    If false: throw to monkey 1
diff --git a/src/Year_2022/files/P12.txt b/src/Year_2022/files/P12.txt
new file mode 100644
index 0000000..4994f18
--- /dev/null
+++ b/src/Year_2022/files/P12.txt
@@ -0,0 +1,41 @@
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diff --git a/src/Year_2022/files/P13.txt b/src/Year_2022/files/P13.txt
new file mode 100644
index 0000000..61f439b
--- /dev/null
+++ b/src/Year_2022/files/P13.txt
@@ -0,0 +1,449 @@
+[[[0,10]],[7,9,[[2,1],[7]],4,[[5,0,5,7,8],0,0,[10,3,3,2]]],[[1,[3,8,8,6,8],6,[1,3,7,9],[9,1,1,7]],[],[[1,0,4,10],8,5]],[]]
+[[10,3,[[4],0,8]],[7,6,[]]]
+
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+[[[[]],[[8],[],[5,1]],4,1]]
+
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+[[3,[8,[1,8,3],[0,10],7],[[8,6,2],[10,7,6,7],4,[1,9,4,6,9]],[[10,5,10],4,[4,1,2,2,8]],[4]],[],[6,6,[[7,6],[5,6,7,5,7]]],[[[0,1,5],0,0,6],[]],[]]
+
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+[[0,3],[0,9]]
+
+[[],[8,[],5],[]]
+[[9]]
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+
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+[[[[],3,[0,1,6,0]]],[1,[[6],10]],[[[],1,[2,8]]],[[[10,9,10,2],[1,2,10],[9,7,7,6,5],[5,9,2,4]],[[5,10]]]]
+
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+[[7,[3],[[3,0],[4,0,0,5,5],[10,7,2,10],5,[2,4]]],[[5,1,2,9],[[8,4],[3]],9],[]]
+
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+
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+
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+
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+[[1,[2,[],9,[2,9,3]]]]
+
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+
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+[[[],[]],[]]
+
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+[[8,[1],8,0],[9,[[10,1,9],[],0,[8,5,7,0,0]],0,[1,[6],6]],[[1],[1,[]],[8,4,[8,5,6,4,7],[9,2],8],2,[0,[9,2,7]]],[[9,[4,1,8,7,0],5],6,8,0,2],[[2,0],10,[[4,6,2,0]],[[]],[[]]]]
+
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+
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+
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diff --git a/src/Year_2022/files/P14.txt b/src/Year_2022/files/P14.txt
new file mode 100644
index 0000000..c5d837b
--- /dev/null
+++ b/src/Year_2022/files/P14.txt
@@ -0,0 +1,134 @@
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diff --git a/src/Year_2022/files/P15.txt b/src/Year_2022/files/P15.txt
new file mode 100644
index 0000000..e41a21c
--- /dev/null
+++ b/src/Year_2022/files/P15.txt
@@ -0,0 +1,35 @@
+Sensor at x=545406, y=2945484: closest beacon is at x=772918, y=2626448
+Sensor at x=80179, y=3385522: closest beacon is at x=772918, y=2626448
+Sensor at x=2381966, y=3154542: closest beacon is at x=2475123, y=3089709
+Sensor at x=2607868, y=1728571: closest beacon is at x=2715626, y=2000000
+Sensor at x=746476, y=2796469: closest beacon is at x=772918, y=2626448
+Sensor at x=911114, y=2487289: closest beacon is at x=772918, y=2626448
+Sensor at x=2806673, y=3051666: closest beacon is at x=2475123, y=3089709
+Sensor at x=1335361, y=3887240: closest beacon is at x=2505629, y=4282497
+Sensor at x=2432913, y=3069935: closest beacon is at x=2475123, y=3089709
+Sensor at x=1333433, y=35725: closest beacon is at x=1929144, y=529341
+Sensor at x=2289207, y=1556729: closest beacon is at x=2715626, y=2000000
+Sensor at x=2455525, y=3113066: closest beacon is at x=2475123, y=3089709
+Sensor at x=3546858, y=3085529: closest beacon is at x=3629407, y=2984857
+Sensor at x=3542939, y=2742086: closest beacon is at x=3629407, y=2984857
+Sensor at x=2010918, y=2389107: closest beacon is at x=2715626, y=2000000
+Sensor at x=3734968, y=3024964: closest beacon is at x=3629407, y=2984857
+Sensor at x=2219206, y=337159: closest beacon is at x=1929144, y=529341
+Sensor at x=1969253, y=890542: closest beacon is at x=1929144, y=529341
+Sensor at x=3522991, y=3257032: closest beacon is at x=3629407, y=2984857
+Sensor at x=2303155, y=3239124: closest beacon is at x=2475123, y=3089709
+Sensor at x=2574308, y=111701: closest beacon is at x=1929144, y=529341
+Sensor at x=14826, y=2490395: closest beacon is at x=772918, y=2626448
+Sensor at x=3050752, y=2366125: closest beacon is at x=2715626, y=2000000
+Sensor at x=3171811, y=2935106: closest beacon is at x=3629407, y=2984857
+Sensor at x=3909938, y=1033557: closest beacon is at x=3493189, y=-546524
+Sensor at x=1955751, y=452168: closest beacon is at x=1929144, y=529341
+Sensor at x=2159272, y=614653: closest beacon is at x=1929144, y=529341
+Sensor at x=3700981, y=2930103: closest beacon is at x=3629407, y=2984857
+Sensor at x=3236266, y=3676457: closest beacon is at x=3373823, y=4223689
+Sensor at x=3980003, y=3819278: closest beacon is at x=3373823, y=4223689
+Sensor at x=1914391, y=723058: closest beacon is at x=1929144, y=529341
+Sensor at x=474503, y=1200604: closest beacon is at x=-802154, y=776650
+Sensor at x=2650714, y=3674470: closest beacon is at x=2505629, y=4282497
+Sensor at x=1696740, y=586715: closest beacon is at x=1929144, y=529341
+Sensor at x=3818789, y=2961752: closest beacon is at x=3629407, y=2984857
diff --git a/src/Year_2022/files/P2.txt b/src/Year_2022/files/P2.txt
new file mode 100644
index 0000000..895876a
--- /dev/null
+++ b/src/Year_2022/files/P2.txt
@@ -0,0 +1,2500 @@
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diff --git a/src/Year_2022/files/P3.txt b/src/Year_2022/files/P3.txt
new file mode 100644
index 0000000..5282006
--- /dev/null
+++ b/src/Year_2022/files/P3.txt
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diff --git a/src/Year_2022/files/P4.txt b/src/Year_2022/files/P4.txt
new file mode 100644
index 0000000..29559d0
--- /dev/null
+++ b/src/Year_2022/files/P4.txt
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diff --git a/src/Year_2022/files/P5.txt b/src/Year_2022/files/P5.txt
new file mode 100644
index 0000000..39caf42
--- /dev/null
+++ b/src/Year_2022/files/P5.txt
@@ -0,0 +1,511 @@
+[N]             [R]             [C]
+[T] [J]         [S] [J]         [N]
+[B] [Z]     [H] [M] [Z]         [D]
+[S] [P]     [G] [L] [H] [Z]     [T]
+[Q] [D]     [F] [D] [V] [L] [S] [M]
+[H] [F] [V] [J] [C] [W] [P] [W] [L]
+[G] [S] [H] [Z] [Z] [T] [F] [V] [H]
+[R] [H] [Z] [M] [T] [M] [T] [Q] [W]
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+
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diff --git a/src/Year_2022/files/P6.txt b/src/Year_2022/files/P6.txt
new file mode 100644
index 0000000..13a015a
--- /dev/null
+++ b/src/Year_2022/files/P6.txt
@@ -0,0 +1 @@
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diff --git a/src/Year_2022/files/P7.txt b/src/Year_2022/files/P7.txt
new file mode 100644
index 0000000..c988c21
--- /dev/null
+++ b/src/Year_2022/files/P7.txt
@@ -0,0 +1,1013 @@
+$ cd /
+$ ls
+dir jmtrrrp
+dir jssnn
+dir lbrmb
+11968 pcccp
+$ cd jmtrrrp
+$ ls
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+dir fmgsql
+$ cd fmgsql
+$ ls
+dir dbnsfp
+dir vvp
+$ cd dbnsfp
+$ ls
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+$ cd ..
+$ cd vvp
+$ ls
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+$ cd ..
+$ cd ..
+$ cd ..
+$ cd jssnn
+$ ls
+dir bphfqs
+dir dbnsfp
+dir pcccp
+dir snr
+dir zjbvwsnv
+$ cd bphfqs
+$ ls
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+$ cd ..
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+$ ls
+dir hgvh
+dir jtqdcmsz
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+$ cd hgvh
+$ ls
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+$ cd qjnbg
+$ ls
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+$ cd bqzfpr
+$ ls
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+$ cd ..
+$ cd ..
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+$ ls
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+$ ls
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+$ ls
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+$ ls
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+$ ls
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+$ ls
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+$ cd lczflft
+$ ls
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+$ ls
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+$ ls
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+$ ls
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+$ ls
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+$ cd ..
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+$ ls
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+$ cd ..
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+$ cd ..
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+$ ls
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+$ ls
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+dir tzhslnv
+$ cd rbvbhnvs
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+$ ls
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+$ cd dhltrqqq
+$ ls
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+$ cd mfd
+$ ls
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+$ cd ..
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+$ ls
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+dir qpbqqj
+$ cd nhdrnthj
+$ ls
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+dir zjbvwsnv
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+$ cd ..
+$ cd ..
+$ cd qpbqqj
+$ ls
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+$ cd ttvp
+$ ls
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+$ cd ..
+$ cd ..
+$ cd ..
+$ cd smmb
+$ ls
+dir dbnsfp
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+dir mjpfnfns
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+dir pmnssvd
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+$ cd dbnsfp
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+$ cd ..
+$ cd mjpfnfns
+$ ls
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+$ cd ..
+$ cd nsdpfnhb
+$ ls
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+$ cd ..
+$ cd pmnssvd
+$ ls
+165816 dfvl.czb
+144561 gbn
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+$ cd ..
+$ cd ..
+$ cd ..
+$ cd fcqwgzp
+$ ls
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+$ cd ..
+$ cd jssnn
+$ ls
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+dir tzqfn
+dir zjbvwsnv
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+$ cd ccblfvl
+$ ls
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+dir hmrdghbr
+dir jssnn
+dir qzzb
+dir tmdlcv
+$ cd hmrdghbr
+$ ls
+dir jvgpwrbs
+$ cd jvgpwrbs
+$ ls
+dir wzdv
+$ cd wzdv
+$ ls
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+$ cd ..
+$ cd ..
+$ cd ..
+$ cd jssnn
+$ ls
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+$ cd ..
+$ cd qzzb
+$ ls
+dir mgpql
+dir src
+dir zvdgc
+$ cd mgpql
+$ ls
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+$ cd ..
+$ cd src
+$ ls
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+$ cd ..
+$ cd zvdgc
+$ ls
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+$ cd ..
+$ cd ..
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+$ ls
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+$ cd ..
+$ cd ..
+$ cd ..
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+$ ls
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+dir ptgn
+dir pzgpqp
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+$ cd ptgn
+$ ls
+dir jssnn
+dir qpbqqj
+dir rpd
+$ cd jssnn
+$ ls
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+$ cd ..
+$ cd qpbqqj
+$ ls
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+$ cd ..
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+$ ls
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+$ cd ..
+$ cd ..
+$ cd pzgpqp
+$ ls
+dir pcccp
+$ cd pcccp
+$ ls
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+$ cd ..
+$ cd ..
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+$ cd tzqfn
+$ ls
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+dir gdl
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+dir wndpdj
+$ cd cbpfvdp
+$ ls
+dir cvfr
+dir qpbqqj
+$ cd cvfr
+$ ls
+dir jfrnvts
+dir qpbqqj
+$ cd jfrnvts
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+$ cd vwdn
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+$ cd ..
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+$ cd qpbqqj
+$ ls
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+$ cd ..
+$ cd ..
+$ cd qpbqqj
+$ ls
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+$ cd jssnn
+$ ls
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+$ cd ..
+$ cd ..
+$ cd ..
+$ cd gdl
+$ ls
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+64007 mjzjgfg.jsb
+$ cd ..
+$ cd wndpdj
+$ ls
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+186962 fql.mww
+dir hlmhtfz
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+$ cd hlmhtfz
+$ ls
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+$ cd ..
+$ cd ..
+$ cd ..
+$ cd zjbvwsnv
+$ ls
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+87552 qtz.ccw
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+$ cd ..
+$ cd ..
+$ cd nwglfhpl
+$ ls
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+dir cwq
+dir dlgrsw
+267814 frhlttn.nmd
+dir hmprt
+dir qpbqqj
+dir wnfzznfh
+$ cd cwq
+$ ls
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+dir pcccp
+dir zjbvwsnv
+dir zzhjfmnr
+$ cd pcccp
+$ ls
+dir pcccp
+$ cd pcccp
+$ ls
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+$ cd ..
+$ cd ..
+$ cd zjbvwsnv
+$ ls
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+dir ngqbdqp
+$ cd ngqbdqp
+$ ls
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+$ cd ..
+$ cd ..
+$ cd zzhjfmnr
+$ ls
+dir dbnsfp
+$ cd dbnsfp
+$ ls
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+$ cd ..
+$ cd ..
+$ cd ..
+$ cd dlgrsw
+$ ls
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+dir zrntbl
+$ cd zrntbl
+$ ls
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+$ cd ..
+$ cd ..
+$ cd hmprt
+$ ls
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+$ cd ..
+$ cd qpbqqj
+$ ls
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+$ cd ..
+$ cd wnfzznfh
+$ ls
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+$ cd ..
+$ cd ..
+$ cd pcccp
+$ ls
+dir rdzntr
+dir rvccq
+$ cd rdzntr
+$ ls
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+$ cd ..
+$ cd rvccq
+$ ls
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+dir msgwsnjq
+dir pggz
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+$ cd msgwsnjq
+$ ls
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+$ cd ..
+$ cd pggz
+$ ls
+dir cnjqsqj
+$ cd cnjqsqj
+$ ls
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+$ cd ..
+$ cd ..
+$ cd ..
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+$ ls
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+$ cd ..
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+$ ls
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+$ cd ..
+$ cd zssln
+$ ls
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+$ cd ..
+$ cd ..
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+$ ls
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+$ cd ..
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+$ ls
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+$ cd ..
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+$ ls
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+dir gnrztn
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+dir pzjnrwt
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+$ ls
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+$ cd ..
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+$ ls
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+dir nbfdtwzr
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+$ cd ..
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+$ ls
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+$ cd ..
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+$ ls
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+$ cd ..
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+$ ls
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diff --git a/src/Year_2022/files/P8.txt b/src/Year_2022/files/P8.txt
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