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Message-ID: <d737b430-2343-b5a7-d47a-0598b1cd225f@johannes-bauer.com>
Date: Fri, 19 Oct 2018 20:17:54 +0200
From: zugtprgfwprz@...rnkuller.de
To: oss-security@...ts.openwall.com
Subject: Re: Travis CI MITM RCE
Hey Jakub,
On 18.10.2018 17:10, Jakub Wilk wrote:
> Nitpicking, but for an ideal n-bit hash function, on avergage you need
> 2ⁿ (not 2ⁿ⁻¹) evalutations of the function to find the preimage.
Huh, wow! I would also have had the gut-feeling of 2^(n-1) and had to
code a little program to verify the facts:
import hashlib
ps = [ ]
for q in range(20000):
for p in range(10000):
z = q | (p << 32)
z = int.to_bytes(z, length = 8, byteorder = "little")
if hashlib.md5(z).digest()[0] == 0:
break
ps.append(p)
print(sum(ps) / len(ps))
And indeed, you're absolutely correct. Thanks for the comment!
Have a great weekend,
Joe
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