My Advent of Code solutions in Python.
kevinyap.ca/2019/12/going-fast-in-advent-of-code/
advent-of-code
python
Add solution for 2023/21
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+104
2023/day21.py
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2023/day21.py
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import fileinput
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from collections import defaultdict, deque, Counter
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from utils import Point
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PART_1_STEPS = 64
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PART_2_STEPS = 26501365
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def search(start, max_dist):
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horizon = deque([(start, 0)])
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cost_so_far = {}
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while horizon:
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curr, dist = horizon.pop()
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if dist > max_dist:
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continue
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if curr in cost_so_far:
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continue
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cost_so_far[curr] = dist
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for n in curr.neighbours_4():
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if GRAPH.get(n) != '.':
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continue
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horizon.appendleft((n, dist + 1))
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return sum(1 for c, v in cost_so_far.items() if v % 2 == ((max_dist % 2)))
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# Read problem input.
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GRAPH = {}
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START = None
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for y, line in enumerate(fileinput.input()):
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for x, c in enumerate(line.strip()):
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p = Point(x, y)
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if c == 'S':
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START = p
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GRAPH[p] = '.'
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else:
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GRAPH[p] = c
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# Solve part 1.
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print("Part 1:", search(START, 64))
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# Solve part 2.
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HEIGHT = y + 1
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WIDTH = x + 1
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assert HEIGHT == WIDTH
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assert START.x == START.y == (HEIGHT // 2) == (WIDTH // 2)
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RADIUS = START.x
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assert (PART_2_STEPS % WIDTH) == RADIUS
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# Define the key search-start points for blocks within the infinite diamond.
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EDGE_STARTS = [Point(x, y) for x, y in [(RADIUS, 0), (WIDTH - 1, RADIUS), (0, RADIUS), (RADIUS, HEIGHT - 1)]]
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CORNER_STARTS = [Point(x, y) for x, y in [(0, 0), (0, HEIGHT - 1), (WIDTH - 1, 0), (WIDTH - 1, HEIGHT - 1)]]
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# Start computing the final answer to part 2. We leverage the following properties of the problem input:
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# - There is a straight path from the central starting point to the outer edges of the block.
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# - The step length is equal to the block radius mod block length, meaning there is perfect
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# symmetry even on the outside edges of the diamond that gets formed.
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# First, add in the centre piece of the diamond.
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part_2 = search(START, PART_2_STEPS)
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# Compute how many times we need to add "full edge pieces" in with parity.
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num_edge_pieces = Counter()
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for i, x in enumerate(range(WIDTH, PART_2_STEPS - RADIUS, WIDTH), start=1):
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num_edge_pieces[i % 2] += 1
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edge_addition = 0
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for s in EDGE_STARTS:
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# Add "fully searched" edge blocks.
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edge_addition += num_edge_pieces[PART_2_STEPS % 2] * search(s, PART_2_STEPS)
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edge_addition += num_edge_pieces[(PART_2_STEPS + 1) % 2] * search(s, PART_2_STEPS + 1)
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# Accomodate the outer tips of the infinite diamond.
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edge_addition += search(s, WIDTH - 1)
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part_2 += edge_addition
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# Compute how many "corner pieces" are scattered on the edge of the diamond with parity.
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num_corner_pieces = Counter() # track parity
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for i, _ in enumerate(range(WIDTH, PART_2_STEPS - WIDTH * 1, WIDTH), start=0):
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num_corner_pieces[i % 2] += i
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corner_addition = 0
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for s in CORNER_STARTS:
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# Add "fully searched" corner blocks.
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corner_addition += search(s, PART_2_STEPS) * num_corner_pieces[1]
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corner_addition += search(s, PART_2_STEPS + 1) * num_corner_pieces[0]
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# Accomodate the outer edges of the infinite diamond.
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corner_addition += search(s, WIDTH + RADIUS - 1) * ((PART_2_STEPS // WIDTH) - 1)
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corner_addition += search(s, RADIUS - 1) * ((PART_2_STEPS // WIDTH))
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part_2 += corner_addition
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print("Part 2:", part_2)