Re: A cipher operates with more possible plaintexts than possible keys. What does this mean?

On Mar 8, 9:06 am, "ping pong" <mosescua...@xxxxxxxx> wrote:
Initial Des key string:
I looked up the table of 8 bit codes with odd number of ones.
I think this is  full table?:

So DES parity key rule means the starting key string must come from this
It like an IBM typewriter!

As required if the whole table is used whenever you need a byte for the key
then there are indeed: (2^7). To choose from.

So that looks good tyo me?

Now To the real Question:
1. Is the DES plaintext also to be restricted by this parity rule related to
So this table is also to be used for the plaintext.

No. The plaintext is 64-bits and they're all used.

2. If the DES plaintext is not also restricted by this parity rule:
a. The is looks as if it operates with 2^64 possible plaintexts and 2^56
possible keys.
set of possible keys = 0.00390625 x set of possible plaintexts?

Correct, so what?

b. Exactly what does it mean to operate a cipher such that
the number of possible plaintexts and possible keys are unequal in this way
and by this amount?

It means you can't use it as a hash construction.

3. Somehow I heard that ciphers were supposed to be:
- "balanced" mode?

Um, no. AES supports keys larger than the block size. The only
balanced I can think of is "balanced Feistel network" but that's a
whole other issue.

     - or have more possible keys than possible plaintexts?

See AES.