179 lines
5.9 KiB
Python
179 lines
5.9 KiB
Python
# --- Day 11: Chronal Charge ---
|
|
|
|
# You watch the Elves and their sleigh fade into the distance as they head
|
|
# toward the North Pole.
|
|
|
|
# Actually, you're the one fading. The falling sensation returns.
|
|
|
|
# The low fuel warning light is illuminated on your wrist-mounted device.
|
|
# Tapping it once causes it to project a hologram of the situation: a 300x300
|
|
# grid of fuel cells and their current power levels, some negative. You're not
|
|
# sure what negative power means in the context of time travel, but it can't be
|
|
# good.
|
|
|
|
# Each fuel cell has a coordinate ranging from 1 to 300 in both the X
|
|
# (horizontal) and Y (vertical) direction. In X,Y notation, the top-left cell
|
|
# is 1,1, and the top-right cell is 300,1.
|
|
|
|
# The interface lets you select any 3x3 square of fuel cells. To increase your
|
|
# chances of getting to your destination, you decide to choose the 3x3 square
|
|
# with the largest total power.
|
|
|
|
# The power level in a given fuel cell can be found through the following
|
|
# process:
|
|
|
|
# Find the fuel cell's rack ID, which is its X coordinate plus 10.
|
|
# Begin with a power level of the rack ID times the Y coordinate.
|
|
# Increase the power level by the value of the grid serial number (your
|
|
# puzzle input).
|
|
# Set the power level to itself multiplied by the rack ID.
|
|
# Keep only the hundreds digit of the power level (so 12345 becomes 3;
|
|
# numbers with no hundreds digit become 0).
|
|
# Subtract 5 from the power level.
|
|
|
|
# For example, to find the power level of the fuel cell at 3,5 in a grid with
|
|
# serial number 8:
|
|
|
|
# The rack ID is 3 + 10 = 13.
|
|
# The power level starts at 13 * 5 = 65.
|
|
# Adding the serial number produces 65 + 8 = 73.
|
|
# Multiplying by the rack ID produces 73 * 13 = 949.
|
|
# The hundreds digit of 949 is 9.
|
|
# Subtracting 5 produces 9 - 5 = 4.
|
|
|
|
# So, the power level of this fuel cell is 4.
|
|
|
|
# Here are some more example power levels:
|
|
|
|
# Fuel cell at 122,79, grid serial number 57: power level -5.
|
|
# Fuel cell at 217,196, grid serial number 39: power level 0.
|
|
# Fuel cell at 101,153, grid serial number 71: power level 4.
|
|
|
|
# Your goal is to find the 3x3 square which has the largest total power. The
|
|
# square must be entirely within the 300x300 grid. Identify this square using
|
|
# the X,Y coordinate of its top-left fuel cell. For example:
|
|
|
|
# For grid serial number 18, the largest total 3x3 square has a top-left corner
|
|
# of 33,45 (with a total power of 29); these fuel cells appear in the middle of
|
|
# this 5x5 region:
|
|
|
|
# -2 -4 4 4 4
|
|
# -4 4 4 4 -5
|
|
# 4 3 3 4 -4
|
|
# 1 1 2 4 -3
|
|
# -1 0 2 -5 -2
|
|
|
|
# For grid serial number 42, the largest 3x3 square's top-left is 21,61 (with a
|
|
# total power of 30); they are in the middle of this region:
|
|
|
|
# -3 4 2 2 2
|
|
# -4 4 3 3 4
|
|
# -5 3 3 4 -4
|
|
# 4 3 3 4 -3
|
|
# 3 3 3 -5 -1
|
|
|
|
# What is the X,Y coordinate of the top-left fuel cell of the 3x3 square with
|
|
# the largest total power?
|
|
|
|
from collections import defaultdict
|
|
|
|
with open("files/P11.txt") as f:
|
|
grid_serial_number = int(f.read().strip().split()[0])
|
|
|
|
|
|
def get_value(x: int, y: int) -> int:
|
|
rack_ID = x + 10
|
|
power_level = rack_ID * y
|
|
power_level += grid_serial_number
|
|
power_level *= rack_ID
|
|
power_level = (power_level // 100) % 10
|
|
return power_level - 5
|
|
|
|
|
|
partial_sums_p1: dict[tuple[int, int], int] = defaultdict(int)
|
|
|
|
|
|
def calculate_ps(x, y):
|
|
return (
|
|
get_value(x + 1, y + 1)
|
|
+ partial_sums_p1[x, y - 1]
|
|
+ partial_sums_p1[x - 1, y]
|
|
- partial_sums_p1[x - 1, y - 1]
|
|
)
|
|
|
|
|
|
def get_partial_sums(arr_size: int) -> dict[tuple[int, int], int]:
|
|
for y in range(arr_size):
|
|
for x in range(arr_size):
|
|
partial_sums_p1[(x, y)] = calculate_ps(x, y)
|
|
return partial_sums_p1
|
|
|
|
|
|
def part_1() -> None:
|
|
arr_size, square_size = 300, 3
|
|
grid_sums = {}
|
|
|
|
partial_sums = get_partial_sums(arr_size)
|
|
|
|
for size in range(1, square_size + 1):
|
|
for j in range(size - 1, arr_size):
|
|
for i in range(size - 1, arr_size):
|
|
gp = (
|
|
partial_sums[(i, j)]
|
|
+ partial_sums[(i - size, j - size)]
|
|
- partial_sums[(i - size, j)]
|
|
- partial_sums[(i, j - size)]
|
|
)
|
|
grid_sums[gp] = (i - size + 2, j - size + 2, size)
|
|
|
|
x3, y3, _ = grid_sums[max(grid_sums)]
|
|
print(f"Largest total power is located at {x3},{y3}")
|
|
|
|
|
|
# --- Part Two ---
|
|
|
|
# You discover a dial on the side of the device; it seems to let you select a
|
|
# square of any size, not just 3x3. Sizes from 1x1 to 300x300 are supported.
|
|
|
|
# Realizing this, you now must find the square of any size with the largest
|
|
# total power. Identify this square by including its size as a third parameter
|
|
# after the top-left coordinate: a 9x9 square with a top-left corner of 3,5 is
|
|
# identified as 3,5,9.
|
|
|
|
# For example:
|
|
|
|
# For grid serial number 18, the largest total square (with a total power
|
|
# of 113) is 16x16 and has a top-left corner of 90,269, so its identifier is
|
|
# 90,269,16.
|
|
# For grid serial number 42, the largest total square (with a total power
|
|
# of 119) is 12x12 and has a top-left corner of 232,251, so its identifier is
|
|
# 232,251,12.
|
|
|
|
# What is the X,Y,size identifier of the square with the largest total power?
|
|
|
|
|
|
def part_2() -> None:
|
|
arr_size = 300
|
|
grid_sums = {}
|
|
|
|
partial_sums = get_partial_sums(arr_size)
|
|
|
|
for size in range(1, arr_size):
|
|
for j in range(size - 1, arr_size):
|
|
for i in range(size - 1, arr_size):
|
|
gp = (
|
|
partial_sums[(i, j)]
|
|
+ partial_sums[(i - size, j - size)]
|
|
- partial_sums[(i - size, j)]
|
|
- partial_sums[(i, j - size)]
|
|
)
|
|
grid_sums[gp] = (i - size + 2, j - size + 2, size)
|
|
|
|
x_max, y_max, size_max = grid_sums[max(grid_sums)]
|
|
print(f"Largest total power identifier is {x_max},{y_max},{size_max}")
|
|
|
|
|
|
if __name__ == "__main__":
|
|
part_1()
|
|
part_2()
|