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Cryptography

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    $\begingroup$ While this is certainly amusingly written, it does miss one point: if the probability of being mauled by a runaway Gorilla is $2^{-60}$, then the probability of being mauled by two runaway Gorillas is not $0.5 \times 2^{-60}$, but $(2^{-60})^2 = 2^{-120}$. Hence, you can't really expect to be mauled by 250000 successive gorillas before you find a collision; however, you are still far more likely to be mauled by one than find a collision. $\endgroup$ Commented Nov 11, 2011 at 21:49
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    $\begingroup$ @poncho: $2^{-60}$ per day. So $2^{-120}$ is the probability of encountering two gorillas the same day. You can view it with a time frame: on average, you will meet a gorilla every $2^{60}$ days. You will get a SHA-256 collision every $2^{76}$ days (there was a mistake in my estimate, so 65000 gorillas, not 250000)(assuming you regenerate the $2^{90}$ 1MB blocks every day). So you really get $2^{16}$ gorillas for every collision -- but not in one go, as a massive gorilla army attack ! (that would be spooky) $\endgroup$ Commented Nov 11, 2011 at 22:00
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    $\begingroup$ Ah, I missed that point. On the other hand, I was going through the probability that you really will be attacked by a Gorilla that escaped from the zoo: a quick Google shows at least three people who have actually been attacked by Gorilla's escaping from a zoo in the last decade (none severely). That bounds the probability of such an event to about $3/(7000000000 \times 365 \times 10) \approx 2^{-43}$. Hence, finding a collision isn't that much more likely than being attacked by two separate Gorillas in the same day (!) $\endgroup$ Commented Nov 12, 2011 at 14:39
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    $\begingroup$ @Ricky: if we knew how to handcraft data blocks specifically to trigger a SHA-256 collision, with better success than with random blocks, then this would be advertised as a break on SHA-256. No such break is currently known on SHA-256. Current methods for attacking MD5 and SHA-1 appear unlikely to apply to SHA-256 (this has been tried). $\endgroup$ Commented Nov 13, 2011 at 14:48
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    $\begingroup$ Just remember that gorilla escapes are not necessarily independent events. :-) $\endgroup$ Commented Dec 22, 2012 at 20:03