Solution to Problem 14 in Python

src/Year_2016/P14.py

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+# --- Day 14: One-Time Pad ---
+
+# In order to communicate securely with Santa while you're on this mission,
+# you've been using a one-time pad that you generate using a pre-agreed
+# algorithm. Unfortunately, you've run out of keys in your one-time pad, and so
+# you need to generate some more.
+
+# To generate keys, you first get a stream of random data by taking the MD5 of
+# a pre-arranged salt (your puzzle input) and an increasing integer index
+# (starting with 0, and represented in decimal); the resulting MD5 hash should
+# be represented as a string of lowercase hexadecimal digits.
+
+# However, not all of these MD5 hashes are keys, and you need 64 new keys for
+# your one-time pad. A hash is a key only if:
+
+#     It contains three of the same character in a row, like 777. Only consider
+# the first such triplet in a hash.
+#     One of the next 1000 hashes in the stream contains that same character
+# five times in a row, like 77777.
+
+# Considering future hashes for five-of-a-kind sequences does not cause those
+# hashes to be skipped; instead, regardless of whether the current hash is a
+# key, always resume testing for keys starting with the very next hash.
+
+# For example, if the pre-arranged salt is abc:
+
+#     The first index which produces a triple is 18, because the MD5 hash of
+# abc18 contains ...cc38887a5.... However, index 18 does not count as a key for
+# your one-time pad, because none of the next thousand hashes (index 19 through
+# index 1018) contain 88888.
+#     The next index which produces a triple is 39; the hash of abc39 contains
+# eee. It is also the first key: one of the next thousand hashes (the one at
+# index 816) contains eeeee.
+#     None of the next six triples are keys, but the one after that, at index
+# 92, is: it contains 999 and index 200 contains 99999.
+#     Eventually, index 22728 meets all of the criteria to generate the 64th
+# key.
+
+# So, using our example salt of abc, index 22728 produces the 64th key.
+
+# Given the actual salt in your puzzle input, what index produces your 64th
+# one-time pad key?
+
+
+import re
+from hashlib import md5
+
+input = "ahsbgdzn"
+
+
+def generate_hash(num):
+    md5hash = md5(bytes(input + str(num), "ascii")).hexdigest()
+    quintuples = re.findall(r"(\w)\1{4}", md5hash)
+
+    return md5hash, quintuples
+
+
+def part_1():
+    hashes = [generate_hash(num) for num in range(1000)]
+    index = 0
+    keys = 0
+    while True:
+        hash, _ = hashes.pop(0)
+        hashes.append(generate_hash(index + 1000))
+
+        match = re.search(r"(\w)\1{2}", hash)
+        if match:
+            letter = re.search(r"(\w)\1{2}", hash).group(1)
+            if any(letter in item[1] for item in hashes):
+                keys += 1
+                if keys == 64:
+                    print(f"{index} produces the 64th one-time pad key")
+                    return
+        index += 1
+
+
+# --- Part Two ---
+
+# Of course, in order to make this process even more secure, you've also
+# implemented key stretching.
+
+# Key stretching forces attackers to spend more time generating hashes.
+# Unfortunately, it forces everyone else to spend more time, too.
+
+# To implement key stretching, whenever you generate a hash, before you use it,
+# you first find the MD5 hash of that hash, then the MD5 hash of that hash, and
+# so on, a total of 2016 additional hashings. Always use lowercase hexadecimal
+# representations of hashes.
+
+# For example, to find the stretched hash for index 0 and salt abc:
+
+#     Find the MD5 hash of abc0: 577571be4de9dcce85a041ba0410f29f.
+#     Then, find the MD5 hash of that hash: eec80a0c92dc8a0777c619d9bb51e910.
+#     Then, find the MD5 hash of that hash: 16062ce768787384c81fe17a7a60c7e3.
+#     ...repeat many times...
+#     Then, find the MD5 hash of that hash: a107ff634856bb300138cac6568c0f24.
+
+# So, the stretched hash for index 0 in this situation is a107ff.... In the end,
+# you find the original hash (one use of MD5), then find the
+# hash-of-the-previous-hash 2016 times, for a total of 2017 uses of MD5.
+
+# The rest of the process remains the same, but now the keys are entirely
+# different. Again for salt abc:
+
+#     The first triple (222, at index 5) has no matching 22222 in the next
+# thousand hashes.
+#     The second triple (eee, at index 10) hash a matching eeeee at index 89,
+# and so it is the first key.
+#     Eventually, index 22551 produces the 64th key (triple fff with matching
+# fffff at index 22859.
+
+# Given the actual salt in your puzzle input and using 2016 extra MD5 calls of
+# key stretching, what index now produces your 64th one-time pad key?
+
+
+def key_stretching(num):
+    md5hash = input + str(num)
+
+    for _ in range(2017):
+        md5hash = md5(bytes(md5hash, "ascii")).hexdigest()
+    quintuples = re.findall(r"(\w)\1{4}", md5hash)
+
+    return md5hash, quintuples
+
+
+def part_2():
+    hashes = [key_stretching(num) for num in range(1000)]
+    index = 0
+    keys = 0
+    while True:
+        hash, _ = hashes.pop(0)
+        hashes.append(key_stretching(index + 1000))
+
+        match = re.search(r"(\w)\1{2}", hash)
+        if match:
+            letter = re.search(r"(\w)\1{2}", hash).group(1)
+            if any(letter in item[1] for item in hashes):
+                keys += 1
+                if keys == 64:
+                    print(f"{index} produces the 64th one-time pad key")
+                    return
+        index += 1
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()