111 lines
4.0 KiB
Python
111 lines
4.0 KiB
Python
# --- Day 5: How About a Nice Game of Chess? ---
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# You are faced with a security door designed by Easter Bunny engineers that
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# seem to have acquired most of their security knowledge by watching hacking
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# movies.
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# The eight-character password for the door is generated one character at a
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# time by finding the MD5 hash of some Door ID (your puzzle input) and an
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# increasing integer index (starting with 0).
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# A hash indicates the next character in the password if its hexadecimal
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# representation starts with five zeroes. If it does, the sixth character in
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# the hash is the next character of the password.
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# For example, if the Door ID is abc:
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# The first index which produces a hash that starts with five zeroes is
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# 3231929, which we find by hashing abc3231929; the sixth character of the
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# hash, and thus the first character of the password, is 1.
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# 5017308 produces the next interesting hash, which starts with
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# 000008f82..., so the second character of the password is 8.
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# The third time a hash starts with five zeroes is for abc5278568,
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# discovering the character f.
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# In this example, after continuing this search a total of eight times, the
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# password is 18f47a30.
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# Given the actual Door ID, what is the password?
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# Your puzzle input is abbhdwsy.
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from collections import defaultdict
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from hashlib import md5
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from typing import DefaultDict
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_input = "abbhdwsy"
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def part_1() -> None:
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passwd = []
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end = "9" * len(_input)
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for num in range(int(end)):
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testing = _input + str(num)
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result = md5(testing.encode()).hexdigest()
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if result.startswith("0" * 5):
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passwd.append(result[5])
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if len(passwd) == 8:
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print(f"The password is {''.join(passwd)}")
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break
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# --- Part Two ---
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# As the door slides open, you are presented with a second door that uses a
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# slightly more inspired security mechanism. Clearly unimpressed by the last
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# version (in what movie is the password decrypted in order?!), the Easter
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# Bunny engineers have worked out a better solution.
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# Instead of simply filling in the password from left to right, the hash now
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# also indicates the position within the password to fill. You still look for
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# hashes that begin with five zeroes; however, now, the sixth character
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# represents the position (0-7), and the seventh character is the character to
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# put in that position.
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# A hash result of 000001f means that f is the second character in the
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# password. Use only the first result for each position, and ignore invalid
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# positions.
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# For example, if the Door ID is abc:
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# The first interesting hash is from abc3231929, which produces 0000015...;
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# so, 5 goes in position 1: _5______.
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# In the previous method, 5017308 produced an interesting hash; however, it
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# is ignored, because it specifies an invalid position (8).
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# The second interesting hash is at index 5357525, which produces
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# 000004e...; so, e goes in position 4: _5__e___.
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# You almost choke on your popcorn as the final character falls into place,
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# producing the password 05ace8e3.
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# Given the actual Door ID and this new method, what is the password? Be extra
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# proud of your solution if it uses a cinematic "decrypting" animation.
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def part_2() -> None:
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passwd = defaultdict(list)
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end = "9" * len(_input)
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for num in range(int(end)):
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testing = _input + str(num)
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result = md5(testing.encode()).hexdigest()
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if (
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result.startswith("0" * 5)
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# only numbers are allow
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and str(result[5]).isdecimal()
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# the passwd is 8 letters long
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and int(result[5]) < 8
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):
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passwd[str(result[5])].append(result[6])
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if len(passwd.keys()) == 8:
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ordered_passwd = dict(sorted(passwd.items()))
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# take the first result for each key
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my_lst = [elem[0] for elem in ordered_passwd.values()]
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print(f"The password is {''.join(my_lst)}")
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break
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if __name__ == "__main__":
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part_1()
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part_2()
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