Well, I decided to just go ahead and use AES128. Concluded that, as with all
computing stuff, it is too easy to get carried away by the "let's go for the
largest number" mentality whenever confronted with a choice and don't really
know any of the details. : )
With that out of the way, and just out of curiosity, could anyone clear up
the decision of having 2925 bytes? I can tell that divided by 65 it means
that each key will be 45 bytes long; but what is the rationale here? My
understanding of these issues is very dim, but how does this number of 45
bytes relate with the use of AES128? Similarly, why would someone using
AES256 pick 90-byte keys?
Thanks for educating me,
Marv
From: Bradley Worley <geekysuavo@xxxxxxxxx>
To: Marvin Lyndon <marvin.lyndon@xxxxxxxxxxx>
CC: linux-crypto@xxxxxxxxxxxx
Subject: Re: different instructions for use with aes256?
Date: Tue, 25 Oct 2005 15:37:49 -0500
that code seems to work just fine, since almost all examples on the
net use either 2925 bytes from /dev/random or 2880 bytes. however, i
guess if you really want to be picky you can up it to 5850 bytes
(double 2925), since you are technically doubling your key sizes.
(it's a wild guess, really.)
~ brad.
On 10/25/05, Marvin Lyndon <marvin.lyndon@xxxxxxxxxxx> wrote:
> Hi all,
>
> I have successfully followed all the steps in the loop-AES readme file.
> Since I couldn't find any reference to this, I would like to know
whether
> the step in which one creates the 65 random keys
>
> head -c 2925 /dev/random | uuencode -m - | head -n 66 | tail -n 65 \
> | gpg --symmetric -a >/a/usbstick/keyfile.gpg
>
> needs any modification for use in AES256 mode. Or is it enough to
replace
> all occurrences of AES128 with AES256 as one follows the README file?
>
> Thanks for any help
>
> Marv
>
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