Decimal encryption
Tim Dierks
tim at dierks.org
Wed Aug 27 14:45:02 EDT 2008
On Wed, Aug 27, 2008 at 11:05 AM, Philipp Gühring <pg at futureware.at> wrote:
> I am searching for symmetric encryption algorithms for decimal strings.
> Since the 132,877 bits is similar to 128 bit encryption (like eg. AES),
> I would like to use an algorithm with a somewhat comparable strength to AES.
> But the problem is that I have 132,877 bits, not 128 bits. And I can't
> cut it off or enhance it, since the result has to be a 40 digit decimal
> number again.
I believe the most straightforward thing to do is to build a balanced
4-round Feistel cipher [ http://en.wikipedia.org/wiki/Feistel_cipher ]
that uses AES as its mixing function, but which operates within a
field of 10^20; you can then encrypt a value within F_10^40 as a
single block operation (ECB mode), taking 4 AES operations and some
other math do to so.
In this usage, each 20-digit side of the cipher would be expressed as
a bit string with ~66 bits, zero-padded to make a 128-bit block. You
should also use the round number in the input; you can put it in the
top 2 bits of the block. This block would then be encrypted with AES,
resulting in a 128-bit output block. You would then reduce this
128-bit value modulo 10^20 to give you a 20-digit output value from
your f() function; that value can be added, modulo 10^20, into the
other 20-digit side of the network (or subtracted on decryption).
A couple of notes:
- I believe 4 rounds should be secure, but someone else on this list
should validate this.
- As simply described here, this is unbalanced, because 2^128 is not
an even multiple of 10^20, so some 20-digit output values of f() are
more likely than others. To avoid this problem, if the 128-bit result
of the AES encryption is less than 2^128 % 10^20
(63374607431768211456), reencrypt the 128-bit output block with AES
again and iterate. This will happen approximately one time in 5e18, so
it's not clear that it's a real vulnerability; it's certainly not a
performance problem.
Good luck; please feel free to ask if any of this isn't clear.
- Tim
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