93 lines
2.9 KiB
Python
93 lines
2.9 KiB
Python
import logging
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import os
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import sys
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VERBOSE = os.getenv("AOC_VERBOSE") == "True"
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logging.basicConfig(level=logging.INFO if VERBOSE else logging.WARNING)
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def reachable(
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map: list[str], tiles: set[tuple[int, int]], steps: int
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) -> set[tuple[int, int]]:
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n_rows, n_cols = len(map), len(map[0])
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for _ in range(steps):
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tiles = {
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(i + di, j + dj)
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for i, j in tiles
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for di, dj in ((-1, 0), (+1, 0), (0, -1), (0, +1))
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if map[(i + di) % n_rows][(j + dj) % n_cols] != "#"
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}
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return tiles
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map = sys.stdin.read().splitlines()
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start = next(
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(i, j) for i in range(len(map)) for j in range(len(map[i])) if map[i][j] == "S"
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)
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# part 1
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answer_1 = len(reachable(map, {start}, 6 if len(map) < 20 else 64))
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print(f"answer 1 is {answer_1}")
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# part 2
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# the initial map is a square and contains an empty rhombus whose diameter is the size
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# of the map, and has only empty cells around the middle row and column
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#
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# after ~n/2 steps, the first map is filled with a rhombus, after that we get a bigger
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# rhombus every n steps
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#
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# we are going to find the number of cells reached for the initial rhombus, n steps
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# after and n * 2 steps after, and then interpolate the value to get a 2nd order
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# polynomial
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#
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cycle = len(map)
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rhombus = (len(map) - 3) // 2 + 1
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values: list[int] = []
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values.append(len(tiles := reachable(map, {start}, rhombus)))
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values.append(len(tiles := reachable(map, tiles, cycle)))
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values.append(len(tiles := reachable(map, tiles, cycle)))
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if logging.root.getEffectiveLevel() == logging.INFO:
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n_rows, n_cols = len(map), len(map[0])
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rows = [
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[
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map[i % n_rows][j % n_cols] if (i, j) not in tiles else "O"
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for j in range(-2 * cycle, 3 * cycle + 1)
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]
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for i in range(-2 * cycle, 3 * cycle + 1)
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]
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for i in range(len(rows)):
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for j in range(len(rows[i])):
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if (i // cycle) % 2 == (j // cycle) % 2:
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rows[i][j] = f"\033[91m{rows[i][j]}\033[0m"
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print("\n".join("".join(row) for row in rows))
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logging.info(f"values to fit: {values}")
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# the value we are interested in (26501365) can be written as R + K * C where R is the
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# step at which we find the first rhombus, and K the repeat step, so instead of fitting
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# for X values (R, R + K, R + 2 K), we are going to fit for (0, 1, 2), giving us much
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# simpler equation for the a, b and c coefficient
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#
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# we get:
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# - (a * 0² + b * 0 + c) = y1 => c = y1
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# - (a * 1² + b * 1 + c) = y2 => a + b = y2 - y1
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# => b = y2 - y1 - a
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# - (a * 2² + b * 2 + c) = y3 => 4a + 2b = y3 - y1
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# => 4a + 2(y2 - y1 - a) = y3 - y1
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# => a = (y1 + y3) / 2 - y2
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#
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y1, y2, y3 = values
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a, b, c = (y1 + y3) // 2 - y2, 2 * y2 - (3 * y1 + y3) // 2, y1
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n = (26501365 - rhombus) // cycle
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answer_2 = a * n * n + b * n + c
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print(f"answer 2 is {answer_2}")
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