adventofcode/2016/day14/problem

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Advent of Code
--- Day 14: One-Time Pad ---
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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.
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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.
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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:
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 It contains three of the same character in a row, like 777. Only
consider the first such triplet in a hash.
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 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.
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For example, if the pre-arranged salt is abc:
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 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.
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So, using our example salt of abc, index 22728 produces the 64th key.
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Given the actual salt in your puzzle input, what index produces your 64th
one-time pad key?
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Your puzzle input was cuanljph.
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Your puzzle answer was 23769.
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--- Part Two ---
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Of course, in order to make this process even more secure, you've also
implemented key stretching.
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Key stretching forces attackers to spend more time generating hashes.
Unfortunately, it forces everyone else to spend more time, too.
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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.
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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.
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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.
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 The second triple (eee, at index 10) hash a matching eeeee at index 89,
and so it is the first key.
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 Eventually, index 22551 produces the 64th key (triple fff with matching
fffff at index 22859.
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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?
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Your puzzle answer was 20606.
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Your puzzle input was cuanljph.
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. https://en.wikipedia.org/wiki/One-time_pad
. https://en.wikipedia.org/wiki/Security_through_obscurity
. https://en.wikipedia.org/wiki/MD5
. https://en.wikipedia.org/wiki/Salt_(cryptography)
. https://en.wikipedia.org/wiki/MD5#Security
. https://en.wikipedia.org/wiki/Key_stretching
. http://adventofcode.com/2016