Kerberos vulnerability

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A Kerberos V4 session key is chosen by calling random() repeatedly.
THe PRNG is seeded with srandom(time.tv_usec ^ time.tv_sec ^ p ^ n++),
where p is a static integer set to getpid() ^ gethostid() on the first
call and n is a static counter.

Is there any entropy here???  Most, if not all, Kerberos servers run one
time synchronization protocol or another, which reduces the entropy to a
few bits at most.

DEADBEAT <na5877@anon.penet.fi>


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> -----BEGIN PGP SIGNED MESSAGE-----
> 
> A Kerberos V4 session key is chosen by calling random() repeatedly.
> THe PRNG is seeded with srandom(time.tv_usec ^ time.tv_sec ^ p ^ n++),
> where p is a static integer set to getpid() ^ gethostid() on the first
> call and n is a static counter.
> 
> Is there any entropy here???  Most, if not all, Kerberos servers run one
> time synchronization protocol or another, which reduces the entropy to a
> few bits at most.
> 
> DEADBEAT <na5877@anon.penet.fi>

usec grainlessness typically doesn't approach anything like a usec on most
OS implimentations either.


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