up to problem 15

src/Year_2015/P13.py

@@ -0,0 +1,123 @@
+# --- Day 13: Knights of the Dinner Table ---
+
+# In years past, the holiday feast with your family hasn't gone so well. Not
+# everyone gets along! This year, you resolve, will be different. You're going
+# to find the optimal seating arrangement and avoid all those awkward
+# conversations.
+
+# You start by writing up a list of everyone invited and the amount their
+# happiness would increase or decrease if they were to find themselves sitting
+# next to each other person. You have a circular table that will be just big
+# enough to fit everyone comfortably, and so each person will have exactly two
+# neighbors.
+
+# For example, suppose you have only four attendees planned, and you calculate
+# their potential happiness as follows:
+
+# Alice would gain 54 happiness units by sitting next to Bob.
+# Alice would lose 79 happiness units by sitting next to Carol.
+# Alice would lose 2 happiness units by sitting next to David.
+# Bob would gain 83 happiness units by sitting next to Alice.
+# Bob would lose 7 happiness units by sitting next to Carol.
+# Bob would lose 63 happiness units by sitting next to David.
+# Carol would lose 62 happiness units by sitting next to Alice.
+# Carol would gain 60 happiness units by sitting next to Bob.
+# Carol would gain 55 happiness units by sitting next to David.
+# David would gain 46 happiness units by sitting next to Alice.
+# David would lose 7 happiness units by sitting next to Bob.
+# David would gain 41 happiness units by sitting next to Carol.
+
+# Then, if you seat Alice next to David, Alice would lose 2 happiness units
+# (because David talks so much), but David would gain 46 happiness units
+# (because Alice is such a good listener), for a total change of 44.
+
+# If you continue around the table, you could then seat Bob next to Alice (Bob
+# gains 83, Alice gains 54). Finally, seat Carol, who sits next to Bob (Carol
+# gains 60, Bob loses 7) and David (Carol gains 55, David gains 41). The
+# arrangement looks like this:
+
+#      +41 +46
+# +55   David    -2
+# Carol       Alice
+# +60    Bob    +54
+#      -7  +83
+
+# After trying every other seating arrangement in this hypothetical scenario,
+# you find that this one is the most optimal, with a total change in happiness
+# of 330.
+
+# What is the total change in happiness for the optimal seating arrangement of
+# the actual guest list?
+
+import re
+from collections import defaultdict
+from itertools import permutations
+from typing import DefaultDict
+
+with open("files/P13.txt") as f:
+    sittings = [line for line in f.read().strip().split("\n")]
+
+
+arrengements: DefaultDict = defaultdict(dict)
+for happiness in sittings:
+    pattern = r"(?P<p1>\S+) would (?P<op>lose|gain) (?P<val>\d+) happiness units by sitting next to (?P<p2>\S+)."
+    matches = re.match(pattern, happiness)
+    p1, op, val, p2 = re.match(pattern, happiness).group(
+        "p1", "op", "val", "p2"
+    )
+    arrengements[p1][p2] = -int(val) if op == "lose" else int(val)
+
+
+def part_1() -> None:
+    max_happiness = 0
+    for ordering in permutations(arrengements.keys()):
+        happiness = sum(
+            arrengements[a][b] + arrengements[b][a]
+            for a, b in zip(ordering, ordering[1:])
+        )
+        happiness += (
+            arrengements[ordering[0]][ordering[-1]]
+            + arrengements[ordering[-1]][ordering[0]]
+        )
+        max_happiness = max(max_happiness, happiness)
+    print(f"The total change in happiness is {max_happiness}")
+
+
+# --- Part Two ---
+
+# In all the commotion, you realize that you forgot to seat yourself. At this
+# point, you're pretty apathetic toward the whole thing, and your happiness
+# wouldn't really go up or down regardless of who you sit next to. You assume
+# everyone else would be just as ambivalent about sitting next to you, too.
+
+# So, add yourself to the list, and give all happiness relationships that
+# involve you a score of 0.
+
+# What is the total change in happiness for the optimal seating arrangement
+# that actually includes yourself?
+
+
+def part_2() -> None:
+    for somebody in list(arrengements.keys()):
+        arrengements["me"][somebody] = 0
+        arrengements[somebody]["me"] = 0
+
+    max_happiness = 0
+    for ordering in permutations(arrengements.keys()):
+        happiness = sum(
+            arrengements[a][b] + arrengements[b][a]
+            for a, b in zip(ordering, ordering[1:])
+        )
+        happiness += (
+            arrengements[ordering[0]][ordering[-1]]
+            + arrengements[ordering[-1]][ordering[0]]
+        )
+        max_happiness = max(max_happiness, happiness)
+    print(
+        f"The total change in happiness is {max_happiness}, including myself"
+    )
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2015/files/P13.txt

@@ -0,0 +1,56 @@
+Alice would gain 54 happiness units by sitting next to Bob.
+Alice would lose 81 happiness units by sitting next to Carol.
+Alice would lose 42 happiness units by sitting next to David.
+Alice would gain 89 happiness units by sitting next to Eric.
+Alice would lose 89 happiness units by sitting next to Frank.
+Alice would gain 97 happiness units by sitting next to George.
+Alice would lose 94 happiness units by sitting next to Mallory.
+Bob would gain 3 happiness units by sitting next to Alice.
+Bob would lose 70 happiness units by sitting next to Carol.
+Bob would lose 31 happiness units by sitting next to David.
+Bob would gain 72 happiness units by sitting next to Eric.
+Bob would lose 25 happiness units by sitting next to Frank.
+Bob would lose 95 happiness units by sitting next to George.
+Bob would gain 11 happiness units by sitting next to Mallory.
+Carol would lose 83 happiness units by sitting next to Alice.
+Carol would gain 8 happiness units by sitting next to Bob.
+Carol would gain 35 happiness units by sitting next to David.
+Carol would gain 10 happiness units by sitting next to Eric.
+Carol would gain 61 happiness units by sitting next to Frank.
+Carol would gain 10 happiness units by sitting next to George.
+Carol would gain 29 happiness units by sitting next to Mallory.
+David would gain 67 happiness units by sitting next to Alice.
+David would gain 25 happiness units by sitting next to Bob.
+David would gain 48 happiness units by sitting next to Carol.
+David would lose 65 happiness units by sitting next to Eric.
+David would gain 8 happiness units by sitting next to Frank.
+David would gain 84 happiness units by sitting next to George.
+David would gain 9 happiness units by sitting next to Mallory.
+Eric would lose 51 happiness units by sitting next to Alice.
+Eric would lose 39 happiness units by sitting next to Bob.
+Eric would gain 84 happiness units by sitting next to Carol.
+Eric would lose 98 happiness units by sitting next to David.
+Eric would lose 20 happiness units by sitting next to Frank.
+Eric would lose 6 happiness units by sitting next to George.
+Eric would gain 60 happiness units by sitting next to Mallory.
+Frank would gain 51 happiness units by sitting next to Alice.
+Frank would gain 79 happiness units by sitting next to Bob.
+Frank would gain 88 happiness units by sitting next to Carol.
+Frank would gain 33 happiness units by sitting next to David.
+Frank would gain 43 happiness units by sitting next to Eric.
+Frank would gain 77 happiness units by sitting next to George.
+Frank would lose 3 happiness units by sitting next to Mallory.
+George would lose 14 happiness units by sitting next to Alice.
+George would lose 12 happiness units by sitting next to Bob.
+George would lose 52 happiness units by sitting next to Carol.
+George would gain 14 happiness units by sitting next to David.
+George would lose 62 happiness units by sitting next to Eric.
+George would lose 18 happiness units by sitting next to Frank.
+George would lose 17 happiness units by sitting next to Mallory.
+Mallory would lose 36 happiness units by sitting next to Alice.
+Mallory would gain 76 happiness units by sitting next to Bob.
+Mallory would lose 34 happiness units by sitting next to Carol.
+Mallory would gain 37 happiness units by sitting next to David.
+Mallory would gain 40 happiness units by sitting next to Eric.
+Mallory would gain 18 happiness units by sitting next to Frank.
+Mallory would gain 7 happiness units by sitting next to George.

src/Year_2021/P12.py

@@ -0,0 +1,221 @@
+# --- Day 12: Passage Pathing ---
+
+# With your submarine's subterranean subsystems subsisting suboptimally, the
+# only way you're getting out of this cave anytime soon is by finding a path
+# yourself. Not just a path - the only way to know if you've found the best
+# path is to find all of them.
+
+# Fortunately, the sensors are still mostly working, and so you build a rough
+# map of the remaining caves (your puzzle input). For example:
+
+# start-A
+# start-b
+# A-c
+# A-b
+# b-d
+# A-end
+# b-end
+
+# This is a list of how all of the caves are connected. You start in the cave
+# named start, and your destination is the cave named end. An entry like b-d
+# means that cave b is connected to cave d - that is, you can move between
+# them.
+
+# So, the above cave system looks roughly like this:
+
+#     start
+#     /   \
+# c--A-----b--d
+#     \   /
+#      end
+
+# Your goal is to find the number of distinct paths that start at start, end at
+# end, and don't visit small caves more than once. There are two types of
+# caves: big caves (written in uppercase, like A) and small caves (written in
+# lowercase, like b). It would be a waste of time to visit any small cave more
+# than once, but big caves are large enough that it might be worth visiting
+# them multiple times. So, all paths you find should visit small caves at most
+# once, and can visit big caves any number of times.
+
+# Given these rules, there are 10 paths through this example cave system:
+
+# start,A,b,A,c,A,end
+# start,A,b,A,end
+# start,A,b,end
+# start,A,c,A,b,A,end
+# start,A,c,A,b,end
+# start,A,c,A,end
+# start,A,end
+# start,b,A,c,A,end
+# start,b,A,end
+# start,b,end
+
+# (Each line in the above list corresponds to a single path; the caves visited
+# by that path are listed in the order they are visited and separated by
+# commas.)
+
+# Note that in this cave system, cave d is never visited by any path: to do so,
+# cave b would need to be visited twice (once on the way to cave d and a second
+# time when returning from cave d), and since cave b is small, this is not
+# allowed.
+
+# Here is a slightly larger example:
+
+# dc-end
+# HN-start
+# start-kj
+# dc-start
+# dc-HN
+# LN-dc
+# HN-end
+# kj-sa
+# kj-HN
+# kj-dc
+
+# The 19 paths through it are as follows:
+
+# start,HN,dc,HN,end
+# start,HN,dc,HN,kj,HN,end
+# start,HN,dc,end
+# start,HN,dc,kj,HN,end
+# start,HN,end
+# start,HN,kj,HN,dc,HN,end
+# start,HN,kj,HN,dc,end
+# start,HN,kj,HN,end
+# start,HN,kj,dc,HN,end
+# start,HN,kj,dc,end
+# start,dc,HN,end
+# start,dc,HN,kj,HN,end
+# start,dc,end
+# start,dc,kj,HN,end
+# start,kj,HN,dc,HN,end
+# start,kj,HN,dc,end
+# start,kj,HN,end
+# start,kj,dc,HN,end
+# start,kj,dc,end
+
+# Finally, this even larger example has 226 paths through it:
+
+# fs-end
+# he-DX
+# fs-he
+# start-DX
+# pj-DX
+# end-zg
+# zg-sl
+# zg-pj
+# pj-he
+# RW-he
+# fs-DX
+# pj-RW
+# zg-RW
+# start-pj
+# he-WI
+# zg-he
+# pj-fs
+# start-RW
+
+# How many paths through this cave system are there that visit small caves at
+# most once?
+
+from collections import defaultdict
+from typing import DefaultDict
+
+caves_map = defaultdict(list)
+with open("files/P12.txt") as f:
+    for row in f:
+        start, end = row.strip().split("-")
+        caves_map[start].append(end)
+        caves_map[end].append(start)
+
+
+def dfs(
+    node: str,
+    graph: DefaultDict[str, list[str]],
+    visited: set[str],
+    already_visited: bool,
+    counter: int = 0,
+):
+    if node == "end":
+        return 1
+    for neighbour in graph[node]:
+        if neighbour.isupper():
+            counter += dfs(neighbour, graph, visited, already_visited)
+        else:
+            if neighbour not in visited:
+                counter += dfs(
+                    neighbour, graph, visited | {neighbour}, already_visited
+                )
+            elif already_visited and neighbour not in {"start", "end"}:
+                counter += dfs(neighbour, graph, visited, False)
+    return counter
+
+
+def part_1() -> None:
+    ans = dfs("start", caves_map, {"start"}, already_visited=False)
+    print(f"There are {ans} paths visiting small caves")
+
+
+# --- Part Two ---
+
+# After reviewing the available paths, you realize you might have time to visit
+# a single small cave twice. Specifically, big caves can be visited any number
+# of times, a single small cave can be visited at most twice, and the remaining
+# small caves can be visited at most once. However, the caves named start and
+# end can only be visited exactly once each: once you leave the start cave, you
+# may not return to it, and once you reach the end cave, the path must end
+# immediately.
+
+# Now, the 36 possible paths through the first example above are:
+
+# start,A,b,A,b,A,c,A,end
+# start,A,b,A,b,A,end
+# start,A,b,A,b,end
+# start,A,b,A,c,A,b,A,end
+# start,A,b,A,c,A,b,end
+# start,A,b,A,c,A,c,A,end
+# start,A,b,A,c,A,end
+# start,A,b,A,end
+# start,A,b,d,b,A,c,A,end
+# start,A,b,d,b,A,end
+# start,A,b,d,b,end
+# start,A,b,end
+# start,A,c,A,b,A,b,A,end
+# start,A,c,A,b,A,b,end
+# start,A,c,A,b,A,c,A,end
+# start,A,c,A,b,A,end
+# start,A,c,A,b,d,b,A,end
+# start,A,c,A,b,d,b,end
+# start,A,c,A,b,end
+# start,A,c,A,c,A,b,A,end
+# start,A,c,A,c,A,b,end
+# start,A,c,A,c,A,end
+# start,A,c,A,end
+# start,A,end
+# start,b,A,b,A,c,A,end
+# start,b,A,b,A,end
+# start,b,A,b,end
+# start,b,A,c,A,b,A,end
+# start,b,A,c,A,b,end
+# start,b,A,c,A,c,A,end
+# start,b,A,c,A,end
+# start,b,A,end
+# start,b,d,b,A,c,A,end
+# start,b,d,b,A,end
+# start,b,d,b,end
+# start,b,end
+
+# The slightly larger example above now has 103 paths through it, and the even
+# larger example now has 3509 paths through it.
+
+# Given these new rules, how many paths through this cave system are there?
+
+
+def part_2() -> None:
+    ans = dfs("start", caves_map, {"start"}, already_visited=True)
+    print(f"There are {ans} paths in total")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2021/P13.py

@@ -0,0 +1,184 @@
+# --- Day 13: Transparent Origami ---
+
+# You reach another volcanically active part of the cave. It would be nice if
+# you could do some kind of thermal imaging so you could tell ahead of time
+# which caves are too hot to safely enter.
+
+# Fortunately, the submarine seems to be equipped with a thermal camera! When
+# you activate it, you are greeted with:
+
+# Congratulations on your purchase! To activate this infrared thermal imaging
+# camera system, please enter the code found on page 1 of the manual.
+
+# Apparently, the Elves have never used this feature. To your surprise, you
+# manage to find the manual; as you go to open it, page 1 falls out. It's a
+# large sheet of transparent paper! The transparent paper is marked with random
+# dots and includes instructions on how to fold it up (your puzzle input).
+# For example:
+
+# 6,10
+# 0,14
+# 9,10
+# 0,3
+# 10,4
+# 4,11
+# 6,0
+# 6,12
+# 4,1
+# 0,13
+# 10,12
+# 3,4
+# 3,0
+# 8,4
+# 1,10
+# 2,14
+# 8,10
+# 9,0
+
+# fold along y=7
+# fold along x=5
+
+# The first section is a list of dots on the transparent paper. 0,0 represents
+# the top-left coordinate. The first value, x, increases to the right. The
+# second value, y, increases downward. So, the coordinate 3,0 is to the right
+# of 0,0, and the coordinate 0,7 is below 0,0. The coordinates in this example
+# form the following pattern, where # is a dot on the paper and . is an empty,
+# unmarked position:
+
+# ...#..#..#.
+# ....#......
+# ...........
+# #..........
+# ...#....#.#
+# ...........
+# ...........
+# ...........
+# ...........
+# ...........
+# .#....#.##.
+# ....#......
+# ......#...#
+# #..........
+# #.#........
+
+# Then, there is a list of fold instructions. Each instruction indicates a line
+# on the transparent paper and wants you to fold the paper up (for horizontal
+# y=... lines) or left (for vertical x=... lines). In this example, the first
+# fold instruction is fold along y=7, which designates the line formed by all
+# of the positions where y is 7 (marked here with -):
+
+# ...#..#..#.
+# ....#......
+# ...........
+# #..........
+# ...#....#.#
+# ...........
+# ...........
+# -----------
+# ...........
+# ...........
+# .#....#.##.
+# ....#......
+# ......#...#
+# #..........
+# #.#........
+
+# Because this is a horizontal line, fold the bottom half up. Some of the dots
+# might end up overlapping after the fold is complete, but dots will never
+# appear exactly on a fold line. The result of doing this fold looks like this:
+
+# #.##..#..#.
+# #...#......
+# ......#...#
+# #...#......
+# .#.#..#.###
+# ...........
+# ...........
+
+# Now, only 17 dots are visible.
+
+# Notice, for example, the two dots in the bottom left corner before the
+# transparent paper is folded; after the fold is complete, those dots appear in
+# the top left corner (at 0,0 and 0,1). Because the paper is transparent, the
+# dot just below them in the result (at 0,3) remains visible, as it can be seen
+# through the transparent paper.
+
+# Also notice that some dots can end up overlapping; in this case, the dots
+# merge together and become a single dot.
+
+# The second fold instruction is fold along x=5, which indicates this line:
+
+# #.##.|#..#.
+# #...#|.....
+# .....|#...#
+# #...#|.....
+# .#.#.|#.###
+# .....|.....
+# .....|.....
+
+# Because this is a vertical line, fold left:
+
+# #####
+# #...#
+# #...#
+# #...#
+# #####
+# .....
+# .....
+
+# The instructions made a square!
+
+# The transparent paper is pretty big, so for now, focus on just completing the
+# first fold. After the first fold in the example above, 17 dots are visible -
+# dots that end up overlapping after the fold is completed count as a single
+# dot.
+
+# How many dots are visible after completing just the first fold instruction on
+# your transparent paper?
+
+from copy import copy
+
+with open("files/P13.txt", "r") as f:
+    dots, folds = [line for line in f.read().strip().split("\n\n")]
+
+_dots = {tuple(map(int, row.split(","))) for row in dots.split("\n")}
+_folds = [(s[11] == "y", int(s[13:])) for s in folds.split("\n")]
+
+
+def fold(dim, value, dots):
+    folded = {dot for dot in dots if dot[dim] < value}
+    for x, y in dots - folded:
+        folded.add((x, 2 * value - y) if dim else (2 * value - x, y))
+    return folded
+
+
+def part_1():
+    ans = len(fold(*_folds[0], _dots))
+    print(f"There are {ans} dots visible after 1 fold")
+
+
+# --- Part Two ---
+
+# Finish folding the transparent paper according to the instructions. The
+# manual says the code is always eight capital letters.
+
+# What code do you use to activate the infrared thermal imaging camera system?
+
+
+def part_2():
+    dots = copy(_dots)
+    for dim, value in _folds:
+        dots = fold(dim, value, dots)
+
+    maxX, maxY = map(max, zip(*dots))
+
+    print("The code is: \n")
+    for y in range(maxY + 1):
+        for x in range(maxX + 1):
+            print("X" if (x, y) in dots else " ", end="")
+        print()
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2021/files/P12.txt

@@ -0,0 +1,22 @@
+end-MY
+MY-xc
+ho-NF
+start-ho
+NF-xc
+NF-yf
+end-yf
+xc-TP
+MY-qo
+yf-TP
+dc-NF
+dc-xc
+start-dc
+yf-MY
+MY-ho
+EM-uh
+xc-yf
+ho-dc
+uh-NF
+yf-ho
+end-uh
+start-NF

src/Year_2021/files/P13.txt

@@ -0,0 +1,987 @@
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