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

*From*: Tom St Denis <tom@xxxxxxx>*Date*: Tue, 8 Mar 2011 06:36:48 -0800 (PST)

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?: http://telecom.tbi.net/ascii7o.html

So DES parity key rule means the starting key string must come from this

table?

It like an IBM typewriter!

As required if the whole table is used whenever you need a byte for the key

string

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

keys?

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.

Tom

.

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