My Advent of Code solutions in Python.
kevinyap.ca/2019/12/going-fast-in-advent-of-code/
advent-of-code
python
Update 2022/25
+17
-32
+17
-32
2022/day25.py
+17
-32
2022/day25.py
···
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import fileinput
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from collections import Counter
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from itertools import zip_longest
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SNAFU_DIGITS = {
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'2': 2,
···
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REVERSE_DIGITS = {v: k for k, v in SNAFU_DIGITS.items()}
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def decimal_to_base(n, b):
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if n == 0:
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return [0]
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digits = []
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while n:
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digits.append(int(n % b))
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n //= b
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return digits[::-1]
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# Read problem input and solve problem.
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nums = [line.strip() for line in fileinput.input()]
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remainder = Counter()
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part_1 = ''
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for i, place in enumerate(zip_longest(*(reversed(n) for n in nums), fillvalue='0')):
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count = sum(SNAFU_DIGITS[c] for c in place) + remainder[i]
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while not -2 <= count <= 2:
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if count < 0:
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remainder[i+1] -= 1
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count += 5
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else:
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remainder[i+1] += 1
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count -= 5
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def snafu_to_decimal(s):
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n = 0
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for i, c in enumerate(reversed(s)):
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n += SNAFU_DIGITS[c] * (5 ** i)
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return n
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part_1 = REVERSE_DIGITS[count] + part_1
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def decimal_to_snafu(n):
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quinary_digits = decimal_to_base(n, 5)
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snafu_digits = []
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# Iterate in reverse order through the base-5 digits; if a number larger
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# than 2 is found, add 1 to the next "place" and take the remainder.
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for i, n in enumerate(reversed(quinary_digits)):
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if n >= 3:
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quinary_digits[-(i+2)] += 1
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n -= 5
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snafu_digits = [REVERSE_DIGITS[n]] + snafu_digits
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return ''.join(snafu_digits)
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# Read problem input and solve problem.
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total = sum(snafu_to_decimal(line.strip()) for line in fileinput.input())
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print("Part 1:", decimal_to_snafu(total))
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print("Part 1:", part_1)