Tag Info

Hot answers tagged

14

In complete honesty: if you have to ask this question, it's overwhelmingly unlikely that you have actually succeeded in breaking the security of AES. At best, you may have discovered a well-known attack against misuse of particular block cipher modes; for instance, plaintext recovery with a chosen-ciphertext attack against ECB, or blind manipulation of the ...


8

Assuming you really had broken AES or another frequently used algorithm that is thought to be secure, the first step would be to prove it. Write the code for the attack. Verify that it works on randomly generated data of the kind it requires. If it can break some challenge (e.g. these), do it. Post the results to the challenger or show the results ...


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 ...


3

Assuming for the moment that your claim is correct, I would suggest caution in revealing the details of your findings. After having your results validated by one or two people with the skills to do so (and whom you trust to keep things confidential), then some sort of general announcement (without specifics) would be best, to give people time (say three ...


3

They are referring to the formula for B├ęzout's identity which calculates the greatest common divisor, extended to a finite field, where 1 is the multiplicative neutral element. From the relation $b(x)a(x)+m(x)c(x)=1$, $$b^{-1}(x)=a(x)\mod m(x)$$ I cant figure out what $a(x)$ and $c(x)$ are? In this case, $b(x)$ is the element for which the inverse is ...


3

I'm not seeing any particularity between the two field that they build, but exists? First off, one basic truth about finite fields is that there is at most one finite field of a given size. Given that both $x^7 + x + 1$ and $x^7 + x^4 + x^3 + x^2 + 1$ are both irreducible, they both generate the same finite field. Where they differ is the ...


2

What you are looking to do can be done, but if you expect it to just work, you are mistaken. Performing the calculation of the inverse here is the easy part, the hard part is the choice of affine transform polynomial and vector, as incorrect choice will lead to an insecure s-box. Multiplicative Inverse The most simple method of calculating the inverse is ...


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

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

The key schedule uses constants that differ between the key sizes. For arbitrary sized keys you would have to define an algorithm for deriving them. Each key size also uses a different number of rounds, for which you would have to do the same. Also, what's the point? 256-bit keys are enough for all eternity. Using a longer keylength variant would likely ...



Only top voted, non community-wiki answers of a minimum length are eligible