Editing Talk:1286: Encryptic

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Maybe the last clues could be intended to be a hash collision? With 64-bit blocks that seems unlikely, but maybe it's a trick?
 
Maybe the last clues could be intended to be a hash collision? With 64-bit blocks that seems unlikely, but maybe it's a trick?
 
[[User:Quantum7|Quantum7]] ([[User talk:Quantum7|talk]]) 21:29, 6 November 2013 (UTC)
 
[[User:Quantum7|Quantum7]] ([[User talk:Quantum7|talk]]) 21:29, 6 November 2013 (UTC)
:Finding the probability of a collision amounts to the birthday problem. Assuming the hash function gives all 2^64 hash values with equal probability and there are 153 million unique message blocks (probably right within an order of magnitude), we have:
 
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: <pre>Pr[collision] = 1 - exp(-153000000^2/(2*2^64)) = 0.000634</pre>
 
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: So the probability of a hash collision from different passwords is still quite low, even with such a large number of passwords. So it's worth assuming that all the identical hash blocks are from the same message, and keep looking for the poke-mash password.
 
: BTW, getting a 50% chance of a collision requires about <tt>sqrt(-2^65*ln(.5))</tt> = 5 billion unique passwords.
 
:[[User:Quantum7|Quantum7]] ([[User talk:Quantum7|talk]]) 21:46, 6 November 2013 (UTC)
 
  
 
'''Explanation for the last 3 - Keyboard Mash'''
 
'''Explanation for the last 3 - Keyboard Mash'''

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