Anders Englöf Ytterström
31bb5b7006
Lost 60 minutes due to misinterpreting this in p2: > *whenever* you generate a hash The code initially only did the 2016 stretching for the triplet hash, not the quintet hash. By doing it to both, pt 2 is solved. Not sure the lru cache actually speeds anything up. Many on the subreddit used the approach to generate the quintet first and look backwards 1000 times for a matching quintet (since quintets are more rare than triplets), this will most likely speed things up. Also, this solution do not store the found keys. Many other solutions do, I believe this is some presumptions.
75 lines
1.5 KiB
Python
75 lines
1.5 KiB
Python
import functools
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import re
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from hashlib import md5
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from output import answer # , matrix, D, DD, ADJ, ints, mhd, mdbg, vdbg
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n = 14
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title = "One-Time Pad"
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@answer(1, "64th key is at index {}")
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def part_1(presolved):
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return presolved[0]
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@answer(2, "64th key is at index {} using key stretching")
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def part_2(presolved):
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return presolved[1]
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def solve(s):
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p1 = run(s)
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p2 = run(s, p2=True)
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return p1, p2
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def run(s, p2=False):
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r3 = re.compile(r"(\w)\1{2}")
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c = 0
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i = 0
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H = md5(s.encode())
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while True:
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i += 1
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if p2:
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a = stretch(f"{s}{i}")
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else:
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Ha = H.copy()
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Ha.update(str(i).encode())
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a = Ha.hexdigest()
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if m := re.search(r3, a):
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x = m.group(1)
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r5 = re.compile(r"(" + x + r")\1{4}")
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for j in range(1000):
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if p2:
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b = stretch(f"{s}{str(i + 1 + j)}")
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else:
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Hb = H.copy()
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Hb.update(str(i + 1 + j).encode())
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b = Hb.hexdigest()
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if re.search(r5, b):
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c += 1
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break
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if c == 64:
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break
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return i
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@functools.lru_cache(maxsize=None)
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def stretch(s):
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for _ in range(2017):
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s = md5(s.encode()).hexdigest()
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return s
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if __name__ == "__main__":
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with open("./input/14.txt", "r") as f:
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inp = f.read().strip()
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inp = solve(inp)
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a = part_1(inp)
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b = part_2(inp)
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assert a == 18626
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assert b == 20092
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