Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

just a follow up question with this question..

How does the key schedule of Rijndael looks for keysizes other than 128 bit?

A 256-bit key can still encrypt a block of plaintext in 128 bit?

but in this tutorial, to encrypt 128-bit plaintext, It need 128-bit of secret key also.. enter image description here

can someone provide me a clear explanation about using of 192, 256 bit key to encrypt a 128-bit block of plaintext?

The only thing that I can understand is that If you will encrypt a 128-bit of plaintext, it need to use 128 bit of secret key, to encrypt a 192 bit of plaintext, it need to use 192 bit of secret key, to encrypt a 256 bit of plaintext, it need to use 256 of secret key.. that is the only thing that I know based on my research..

share|improve this question
Is there still something open about the key schedule after your previous question? – Paŭlo Ebermann May 16 '12 at 23:57
up vote 6 down vote accepted

One very important thing you are missing is the idea behind AddRoundKey and the Key Schedule.

Each round, there is a new round-key. The round key is derived from the main key. So, say you have a 128-bit block (AES) and a 256-bit key. The internal state is 128-bits and the round key is 128-bits. The specification states that for this configuration, there will be 14 rounds. Since the round-key changes for each round, it is easy to see that all 256-bits of the main key will be used over the 14 rounds.

To see exactly how all 256-bits of the main key are used, you'll have to dive into the details of the key schedule.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.