Re: What's the probability that 2 files of n bytes have the same hash using SHA1?
- From: unruh <unruh@xxxxxxxxxxxxxxxxxxxxxxx>
- Date: Tue, 15 Dec 2009 16:28:12 GMT
On 2009-12-15, PeterMacGonagan <rodrigue.roland@xxxxxxxxx> wrote:
Ok, thank you for all your answers.
I read about the birthday paradox and I think I undestand it. As we
can read, we can approximate the number of attempts to generate a
collision using brute force by:
Q(H)\approx \sqrt{\frac{\pi}{2}H}.
For a 160-bit hash, there are 2^160 possibilities (approximately
1,4615e+48 different outputs). So, it'll take approximately 1.515e+24
attempts to generate a collision (2^80 = 1.2e+24).
Well, depends on what you mean by "collision". It it means "find two
files with the same hash" you are right. If it means ( find a second
file with the same hash as this given file) then this is wrong. It would
take 2^159 on average.
.
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