1993-11-30 - Re: Cryptosplit 2.0

Header Data

From: Brad Huntting <huntting@glarp.com>
To: m5@vail.tivoli.com (Mike McNally)
Message Hash: 0851fd9e6531d09923ce6f3470de4ab1a130bd85f5b9b15561a6f4fb69b2a638
Message ID: <199311301731.AA06857@misc.glarp.com>
Reply To: <9311301342.AA25298@vail.tivoli.com>
UTC Datetime: 1993-11-30 17:32:12 UTC
Raw Date: Tue, 30 Nov 93 09:32:12 PST

Raw message

From: Brad Huntting <huntting@glarp.com>
Date: Tue, 30 Nov 93 09:32:12 PST
To: m5@vail.tivoli.com (Mike McNally)
Subject: Re: Cryptosplit 2.0
In-Reply-To: <9311301342.AA25298@vail.tivoli.com>
Message-ID: <199311301731.AA06857@misc.glarp.com>
MIME-Version: 1.0
Content-Type: text/plain



Brad Huntting writes:
> If I remember coorectly it's KerberosV uses an MD5 hash of /dev/mem.
> 
> Still, probably not 128 bits worth of entropy.

Mike responds:
> Gee, that seems pretty amazing.  On a typical workstation, there's a
> heck of a lot going on; in the megabytes of data in /dev/mem I'd think
> it quite unlikely that there's a practical way to predict or recreate
> a configuration.

Well, Assuming I was just creating a key and not doing something
else at the same time, it would be pretty easy to predict what
processes were running.  It has about 6000 pages of physical memory.
From knowing what processes are running, you could probably narrow
down the pages they would have in memory to at most half again that
number.  This gives you mabey 9000 choose 6000 posiblilitys for
what process memory looks like.  That gives about exp(3000*ln(7000))
or 2^37k posiblilities...  I suppose that is more than 128 bits.

So if you cant predict which pages will land where in memory (which
may be a false assumption), this is probably a good method for
getting a random number on a unix box.


brad





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