Re: Decimal encryption

[Greg Rose]

"Hal Finney" wrote:
So, you don't have a 133-bit block cipher lying around? No worries, I'll
sell you one ;-). Actually that is easy too. Take a trustworthy 128-bit
block cipher like AES. To encrypt, do:

1. Encrypt the first 128 bits (ECB mode)
2. Encrypt the last 128 bits (also ECB mode).

I didn't understand this at first, but I finally saw that the point is to
do the encryptions in-place; step 1 replaces the first 128 bits of the
data with the encryption, and similarly for step 2. This is equivalent
to doing CBC mode with a fixed IV of 0, and ciphertext stealing for the
final partial block of 5 bits.

Yes, I guess it is... hadn't thought of it that way. But yes, I confirm 
that I meant to do the encryptions in place.

To decrypt, do decryptions in the reverse order, obviously. It's easy to
see that this is a secure permutation if AES itself is, depending on
your definition of secure; if you add a third step, to re-encrypt the
first 128 bits, it is truly secure. (Without the third step, tweaking a
bit in the first 5 bits will often leave the last 5 unchanged on
decryption, which is clearly a distinguishing attack; the third
encryption makes it an all-or-nothing transform.)

I am not familiar with the security proof here, do you have a reference?
Or is it an exercise for the student?

It's a degenerate case of Rivest's All-or-nothing transform (which 
applies to larger, multi-block blocks, if you know what I mean :-) ). I 
believe he gave a security proof, some 6ish years ago. But I could be 
confabulating.

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