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5

This begs the question, why would you in any real-world circumstance wish to reduce the difficulty for an attacker to break your cryptosystem? To answer your question practically, the only reasonable way I can think of to accomplish this is to simply reduce the entropy in the key. At 100%, all 128 bits of the key are used. At 50%, 64 bits of the key are ...


5

First, it's not said that AES is unbreakable, merely that none of the currently known attacks reduce the computational cost to a point where it's feasible. The current best attack on AES-128 takes 2^126.1 operations, if we had a computer (or cluster) several million times more efficient than any current computer and could operate at the thermodynamic ...


5

The most likely rationale to change the AES design is political. It's a NIST standard, designed in Western Europe. It's a bad idea! How much scrutiny has it received? Almost none. How much will it receive? Almost none. Bad idea.


3

I ("SEJPM" as of now) have contacted the authors asked them the same questions as in my question. I'm posting this as community wiki, as it's not my answer to this question but rather theirs. Now the responses follow: First off, the authors are working on a design rationale in english for their new cipher. As soon as it's published, it will be linked ...


3

You cannot encrypt 720 bits plaintext using just AES-128. AES is a 128 bit block cipher. Such a block cipher has an input of 128 bits of plaintext and an output of 128 bits ciphertext; and that's it. You need some kind of construction to make block ciphers encrypt larger or smaller plaintext. Such constructions are known as (block cipher) modes of ...


2

No, it won't "leak" information, as long as you're using a modern symmetric algorithm that's resistant to known-plaintext attacks. However, depending on the encryption mode used (and whether there's integrity checking or not), there can be other security implications, such as the data in the known spot being substituted.


1

How the cipher key is scheduled if it is less than 128 bit (for AES 128), or less than 192 bit (for AES 192) or less than 256 (for AES 256)? That is undefined; the AES specification does not address that possibility. The AES 128 algorithm assumes that you give it 128 bits of key (and tells you exactly what to do with that key); it says nothing about ...


1

To do EEA on a finite-field, you can't perform the operations using the operations in the ring of the integers (and they're not precisely the same operations in the field either, as you're working with bit vectors, not field elements). In particular: When you do "addition", you need to perform the addition as done in even characteristic fields (that is, ...


1

As I have duplicated your question by mistake and non of us have had an answer, I request help to the authors to know why those parameters where selected like that. The affine transformation is a vector space operation $(\mathbb{F}_{2})^8$, and the simplicity comes from the fact that, from the bunch of possible transformations the one used can be also ...


1

Finally I've went to the source and I've mailed the Rijndael's authors. They have answered very fast and very nice. I've understood the other way around. The affine transformation is over the vector space $((GF(2))^8$ and what they've say as simplicity was that, between all the possible affine transformations they select one that can also be described as ...



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