Merge branch 'master' of ssh://git.elnota.space:10222/daviddoji/Advent_of_code

This commit is contained in:
David Doblas Jiménez 2022-04-04 17:13:09 +02:00
commit e52821eee1
7 changed files with 880 additions and 0 deletions

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src/Year_2015/P8.py Normal file
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# --- Day 8: Matchsticks ---
# Space on the sleigh is limited this year, and so Santa will be bringing his
# list as a digital copy. He needs to know how much space it will take up when
# stored.
# It is common in many programming languages to provide a way to escape special
# characters in strings. For example, C, JavaScript, Perl, Python, and even PHP
# handle special characters in very similar ways.
# However, it is important to realize the difference between the number of
# characters in the code representation of the string literal and the number of
# characters in the in-memory string itself.
# For example:
# "" is 2 characters of code (the two double quotes), but the string
# contains zero characters.
# "abc" is 5 characters of code, but 3 characters in the string data.
# "aaa\"aaa" is 10 characters of code, but the string itself contains six
# "a" characters and a single, escaped quote character, for a total of 7 characters in the string data.
# "\x27" is 6 characters of code, but the string itself contains just one
# - an apostrophe ('), escaped using hexadecimal notation.
# Santa's list is a file that contains many double-quoted string literals, one
# on each line. The only escape sequences used are \\ (which represents a
# single backslash), \" (which represents a lone double-quote character), and
# \x plus two hexadecimal characters (which represents a single character with
# that ASCII code).
# Disregarding the whitespace in the file, what is the number of characters of
# code for string literals minus the number of characters in memory for the
# values of the strings in total for the entire file?
# For example, given the four strings above, the total number of characters of
# string code (2 + 5 + 10 + 6 = 23) minus the total number of characters in
# memory for string values (0 + 3 + 7 + 1 = 11) is 23 - 11 = 12.
with open("files/P8.txt") as f:
lines = [line for line in f.read().strip().split()]
# TIL: eval()
def part_1() -> None:
res = sum(len(line) - len(eval(line)) for line in lines)
print(f"There are {res} characters.")
# --- Part Two ---
# Now, let's go the other way. In addition to finding the number of characters
# of code, you should now encode each code representation as a new string and
# find the number of characters of the new encoded representation, including
# the surrounding double quotes.
# For example:
# "" encodes to "\"\"", an increase from 2 characters to 6.
# "abc" encodes to "\"abc\"", an increase from 5 characters to 9.
# "aaa\"aaa" encodes to "\"aaa\\\"aaa\"", an increase from 10 characters to
# 16.
# "\x27" encodes to "\"\\x27\"", an increase from 6 characters to 11.
# Your task is to find the total number of characters to represent the newly
# encoded strings minus the number of characters of code in each original
# string literal. For example, for the strings above, the total encoded length
# (6 + 9 + 16 + 11 = 42) minus the characters in the original code
# representation (23, just like in the first part of this puzzle) is
# 42 - 23 = 19.
def part_2() -> None:
# 2 for the ""
# just count numbers of \\ and new "
res = sum(2 + line.count("\\") + line.count('"') for line in lines)
print(f"There are {res} characters.")
if __name__ == "__main__":
part_1()
part_2()

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src/Year_2015/files/P8.txt Normal file
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"qxfcsmh"
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"wtoodejqmrrgslhvnk\"pi\"ldnogpth"
"njro\x68qgbx\xe4af\"\\suan"

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# --- Day 8: Two-Factor Authentication ---
# You come across a door implementing what you can only assume is an
# implementation of two-factor authentication after a long game of
# requirements telephone.
# To get past the door, you first swipe a keycard (no problem; there was one on
# a nearby desk). Then, it displays a code on a little screen, and you type
# that code on a keypad. Then, presumably, the door unlocks.
# Unfortunately, the screen has been smashed. After a few minutes, you've taken
# everything apart and figured out how it works. Now you just have to work out
# what the screen would have displayed.
# The magnetic strip on the card you swiped encodes a series of instructions
# for the screen; these instructions are your puzzle input. The screen is 50
# pixels wide and 6 pixels tall, all of which start off, and is capable of
# three somewhat peculiar operations:
# rect AxB turns on all of the pixels in a rectangle at the top-left of the
# screen which is A wide and B tall.
# rotate row y=A by B shifts all of the pixels in row A (0 is the top row)
# right by B pixels. Pixels that would fall off the right end appear at the
# left end of the row.
# rotate column x=A by B shifts all of the pixels in column A (0 is the
# left column) down by B pixels. Pixels that would fall off the bottom appear
# at the top of the column.
# For example, here is a simple sequence on a smaller screen:
# rect 3x2 creates a small rectangle in the top-left corner:
# ###....
# ###....
# .......
# rotate column x=1 by 1 rotates the second column down by one pixel:
# #.#....
# ###....
# .#.....
# rotate row y=0 by 4 rotates the top row right by four pixels:
# ....#.#
# ###....
# .#.....
# rotate column x=1 by 1 again rotates the second column down by one pixel,
# causing the bottom pixel to wrap back to the top:
# .#..#.#
# #.#....
# .#.....
# As you can see, this display technology is extremely powerful, and will soon
# dominate the tiny-code-displaying-screen market. That's what the
# advertisement on the back of the display tries to convince you, anyway.
# There seems to be an intermediate check of the voltage used by the display:
# after you swipe your card, if the screen did work, how many pixels should be
# lit?
import numpy as np
import numpy.typing as npt
with open("files/P8.txt") as f:
instructions = [line for line in f.read().strip().split("\n")]
def rect(instruction: str, grid: npt.NDArray[int]) -> npt.NDArray[int]:
_, row = instruction.split("x")
_, col = _.split()
grid[: int(row), : int(col)] = 1
def rotate_column(
instruction: str, grid: npt.NDArray[int]
) -> npt.NDArray[int]:
_, cols = instruction.split("=")
x, _, steps = cols.split()
new_col = np.roll(grid[:, int(x) : int(x) + 1], int(steps), axis=0)
grid[:, int(x) : int(x) + 1] = new_col
def rotate_row(instruction: str, grid: npt.NDArray[int]) -> npt.NDArray[int]:
_, rows = instruction.split("=")
y, _, steps = rows.split()
new_row = np.roll(grid[int(y) : int(y) + 1, :], int(steps), axis=1)
grid[int(y) : int(y) + 1, :] = new_row
# dimensions given by the problem
init_grid = np.zeros((6, 50), dtype=int)
def part_1() -> None:
for instruction in instructions:
if instruction.startswith("rect"):
rect(instruction, init_grid)
elif instruction.startswith("rotate row"):
rotate_row(instruction, init_grid)
elif instruction.startswith("rotate column"):
rotate_column(instruction, init_grid)
lit = np.sum(init_grid)
print(f"There should be {lit} pixels lit")
# --- Part Two ---
# You notice that the screen is only capable of displaying capital letters; in
# the font it uses, each letter is 5 pixels wide and 6 tall.
# After you swipe your card, what code is the screen trying to display?
def display(grid: npt.NDArray[int]) -> None:
print(
"\n".join("".join("X" if one else " " for one in row) for row in grid)
)
def part_2() -> None:
print("The code is the following:\n")
display(init_grid)
if __name__ == "__main__":
part_1()
part_2()

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rect 1x1
rotate row y=0 by 20
rect 1x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 3
rect 2x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 3
rect 2x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 4
rect 2x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 3
rect 2x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 5
rect 1x1
rotate row y=0 by 2
rect 1x1
rotate row y=0 by 6
rect 5x1
rotate row y=0 by 2
rect 1x3
rotate row y=2 by 8
rotate row y=0 by 8
rotate column x=0 by 1
rect 7x1
rotate row y=2 by 24
rotate row y=0 by 20
rotate column x=5 by 1
rotate column x=4 by 2
rotate column x=2 by 2
rotate column x=0 by 1
rect 7x1
rotate column x=34 by 2
rotate column x=22 by 1
rotate column x=15 by 1
rotate row y=2 by 18
rotate row y=0 by 12
rotate column x=8 by 2
rotate column x=7 by 1
rotate column x=5 by 2
rotate column x=2 by 1
rotate column x=0 by 1
rect 9x1
rotate row y=3 by 28
rotate row y=1 by 28
rotate row y=0 by 20
rotate column x=18 by 1
rotate column x=15 by 1
rotate column x=14 by 1
rotate column x=13 by 1
rotate column x=12 by 2
rotate column x=10 by 3
rotate column x=8 by 1
rotate column x=7 by 2
rotate column x=6 by 1
rotate column x=5 by 1
rotate column x=3 by 1
rotate column x=2 by 2
rotate column x=0 by 1
rect 19x1
rotate column x=34 by 2
rotate column x=24 by 1
rotate column x=23 by 1
rotate column x=14 by 1
rotate column x=9 by 2
rotate column x=4 by 2
rotate row y=3 by 5
rotate row y=2 by 3
rotate row y=1 by 7
rotate row y=0 by 5
rotate column x=0 by 2
rect 3x2
rotate column x=16 by 2
rotate row y=3 by 27
rotate row y=2 by 5
rotate row y=0 by 20
rotate column x=8 by 2
rotate column x=7 by 1
rotate column x=5 by 1
rotate column x=3 by 3
rotate column x=2 by 1
rotate column x=1 by 2
rotate column x=0 by 1
rect 9x1
rotate row y=4 by 42
rotate row y=3 by 40
rotate row y=1 by 30
rotate row y=0 by 40
rotate column x=37 by 2
rotate column x=36 by 3
rotate column x=35 by 1
rotate column x=33 by 1
rotate column x=32 by 1
rotate column x=31 by 3
rotate column x=30 by 1
rotate column x=28 by 1
rotate column x=27 by 1
rotate column x=25 by 1
rotate column x=23 by 3
rotate column x=22 by 1
rotate column x=21 by 1
rotate column x=20 by 1
rotate column x=18 by 1
rotate column x=17 by 1
rotate column x=16 by 3
rotate column x=15 by 1
rotate column x=13 by 1
rotate column x=12 by 1
rotate column x=11 by 2
rotate column x=10 by 1
rotate column x=8 by 1
rotate column x=7 by 2
rotate column x=5 by 1
rotate column x=3 by 3
rotate column x=2 by 1
rotate column x=1 by 1
rotate column x=0 by 1
rect 39x1
rotate column x=44 by 2
rotate column x=42 by 2
rotate column x=35 by 5
rotate column x=34 by 2
rotate column x=32 by 2
rotate column x=29 by 2
rotate column x=25 by 5
rotate column x=24 by 2
rotate column x=19 by 2
rotate column x=15 by 4
rotate column x=14 by 2
rotate column x=12 by 3
rotate column x=9 by 2
rotate column x=5 by 5
rotate column x=4 by 2
rotate row y=5 by 5
rotate row y=4 by 38
rotate row y=3 by 10
rotate row y=2 by 46
rotate row y=1 by 10
rotate column x=48 by 4
rotate column x=47 by 3
rotate column x=46 by 3
rotate column x=45 by 1
rotate column x=43 by 1
rotate column x=37 by 5
rotate column x=36 by 5
rotate column x=35 by 4
rotate column x=33 by 1
rotate column x=32 by 5
rotate column x=31 by 5
rotate column x=28 by 5
rotate column x=27 by 5
rotate column x=26 by 3
rotate column x=25 by 4
rotate column x=23 by 1
rotate column x=17 by 5
rotate column x=16 by 5
rotate column x=13 by 1
rotate column x=12 by 5
rotate column x=11 by 5
rotate column x=3 by 1
rotate column x=0 by 1

View File

@ -88,5 +88,78 @@ def part_1() -> None:
print(f"The name of the bottom program is {root_program}") print(f"The name of the bottom program is {root_program}")
# --- Part Two ---
# The programs explain the situation: they can't get down. Rather, they could
# get down, if they weren't expending all of their energy trying to keep the
# tower balanced. Apparently, one program has the wrong weight, and until it's
# fixed, they're stuck here.
# For any program holding a disc, each program standing on that disc forms a
# sub-tower. Each of those sub-towers are supposed to be the same weight, or
# the disc itself isn't balanced. The weight of a tower is the sum of the
# weights of the programs in that tower.
# In the example above, this means that for ugml's disc to be balanced, gyxo,
# ebii, and jptl must all have the same weight, and they do: 61.
# However, for tknk to be balanced, each of the programs standing on its disc
# and all programs above it must each match. This means that the following sums
# must all be the same:
# ugml + (gyxo + ebii + jptl) = 68 + (61 + 61 + 61) = 251
# padx + (pbga + havc + qoyq) = 45 + (66 + 66 + 66) = 243
# fwft + (ktlj + cntj + xhth) = 72 + (57 + 57 + 57) = 243
# As you can see, tknk's disc is unbalanced: ugml's stack is heavier than the
# other two. Even though the nodes above ugml are balanced, ugml itself is too
# heavy: it needs to be 8 units lighter for its stack to weigh 243 and keep the
# towers balanced. If this change were made, its weight would be 60.
# Given that exactly one program is the wrong weight, what would its weight
# need to be to balance the entire tower?
def child_values(node, node_map, node_values):
weights, unbalanced = [], []
children = node_map[node]
for child in children:
if child in node_map.keys():
child_weight, child_balance = child_values(
child, node_map, node_values
)
value = sum(child_weight) + node_values[child]
unbalanced.append(child_balance)
else:
value = node_values[child]
weights.append(value)
if len(set(weights)) != 1:
unbalanced.append((node_map[node], weights))
return weights, unbalanced
def part_2() -> None:
node_map, node_values = {}, {}
for line in towers:
if "->" in line:
lhs, rhs = line.split(" -> ")
parent, value = lhs.split()
node_values[parent] = int(value[1:-1])
_childs = rhs.split(", ")
node_map[parent] = [child for child in _childs]
else:
child, value = line.split()
node_values[child] = int(value[1:-1])
_, unbalance = child_values("rqwgj", node_map, node_values)
heavier_child_idx = unbalance[0][0][5][1].index(max(unbalance[0][0][5][1]))
heavier_child = unbalance[0][0][5][0][heavier_child_idx]
unbalanced_child = node_values[heavier_child]
print(f"To balance the tower, the weight must be {unbalanced_child - 8}")
if __name__ == "__main__": if __name__ == "__main__":
part_1() part_1()
part_2()

117
src/Year_2021/P7.py Normal file
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@ -0,0 +1,117 @@
# --- Day 7: The Treachery of Whales ---
# A giant whale has decided your submarine is its next meal, and it's much
# faster than you are. There's nowhere to run!
# Suddenly, a swarm of crabs (each in its own tiny submarine - it's too deep
# for them otherwise) zooms in to rescue you! They seem to be preparing to
# blast a hole in the ocean floor; sensors indicate a massive underground cave
# system just beyond where they're aiming!
# The crab submarines all need to be aligned before they'll have enough power
# to blast a large enough hole for your submarine to get through. However, it
# doesn't look like they'll be aligned before the whale catches you! Maybe you
# can help?
# There's one major catch - crab submarines can only move horizontally.
# You quickly make a list of the horizontal position of each crab (your puzzle
# input). Crab submarines have limited fuel, so you need to find a way to make
# all of their horizontal positions match while requiring them to spend as
# little fuel as possible.
# For example, consider the following horizontal positions:
# 16,1,2,0,4,2,7,1,2,14
# This means there's a crab with horizontal position 16, a crab with horizontal
# position 1, and so on.
# Each change of 1 step in horizontal position of a single crab costs 1 fuel.
# You could choose any horizontal position to align them all on, but the one
# that costs the least fuel is horizontal position 2:
# Move from 16 to 2: 14 fuel
# Move from 1 to 2: 1 fuel
# Move from 2 to 2: 0 fuel
# Move from 0 to 2: 2 fuel
# Move from 4 to 2: 2 fuel
# Move from 2 to 2: 0 fuel
# Move from 7 to 2: 5 fuel
# Move from 1 to 2: 1 fuel
# Move from 2 to 2: 0 fuel
# Move from 14 to 2: 12 fuel
# This costs a total of 37 fuel. This is the cheapest possible outcome; more
# expensive outcomes include aligning at position 1 (41 fuel), position 3 (39
# fuel), or position 10 (71 fuel).
# Determine the horizontal position that the crabs can align to using the least
# fuel possible. How much fuel must they spend to align to that position?
with open("files/P7.txt") as f:
positions = [int(num) for num in f.read().strip().split(",")]
def part_1() -> None:
fuel = []
for pos in positions:
sum = 0
for elem in positions:
sum += abs(elem - pos)
fuel.append(sum)
print(f"It is needed {min(fuel)} units of fuel")
# --- Part Two ---
# The crabs don't seem interested in your proposed solution. Perhaps you
# misunderstand crab engineering?
# As it turns out, crab submarine engines don't burn fuel at a constant rate.
# Instead, each change of 1 step in horizontal position costs 1 more unit of
# fuel than the last: the first step costs 1, the second step costs 2, the
# third step costs 3, and so on.
# As each crab moves, moving further becomes more expensive. This changes the
# best horizontal position to align them all on; in the example above, this
# becomes 5:
# Move from 16 to 5: 66 fuel
# Move from 1 to 5: 10 fuel
# Move from 2 to 5: 6 fuel
# Move from 0 to 5: 15 fuel
# Move from 4 to 5: 1 fuel
# Move from 2 to 5: 6 fuel
# Move from 7 to 5: 3 fuel
# Move from 1 to 5: 10 fuel
# Move from 2 to 5: 6 fuel
# Move from 14 to 5: 45 fuel
# This costs a total of 168 fuel. This is the new cheapest possible outcome;
# the old alignment position (2) now costs 206 fuel instead.
# Determine the horizontal position that the crabs can align to using the least
# fuel possible so they can make you an escape route! How much fuel must they
# spend to align to that position?
# TIL :D
def triangular_cost(steps: int) -> int:
return steps * (steps + 1) // 2
def part_2() -> None:
fuel = []
for pos in positions:
sum = 0
for elem in positions:
sum += triangular_cost(abs(elem - pos))
fuel.append(sum)
print(f"It is needed {min(fuel)} units of fuel")
if __name__ == "__main__":
part_1()
part_2()

View File

@ -0,0 +1 @@
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