>4096bit mutating encryption
what the fuck does that even mean
4096bit mutating encryption
John Foster
Wyatt Hernandez
> 4096 bit
self-explanatory. encryption uses 4096 bit key.
> mutating
probably means the exact way it encrypts shifts each time, so you can't just bruteforce it (and even if you could it's 4096 bit so good luck.)
Nathaniel Bell
X-Men encryption. It is powerful, but its past is troubled and it is still has dumb vulnerabilities that negates its benefits.
Jacob Gutierrez
>this, unironically.
Christopher Adams
not op, and dumb question, but how do they specify a bit length result for a key in an encryption algo?
Colton Anderson
Chase Sullivan
OP here, really? Why not just call it X-NENcryption or something like that?
Luis Thompson
>not op, and dumb question, but how do they specify a bit length result for a key in an encryption algo?
What do you mean exactly? Very simplistically, a (block) cipher *really* shuffles things around using permutations. The number of those permutations is huge, lets say you want to encrypt a single 128-bit block, you have (2^128)! (factorial) permutations to pick from.
The key simply selects which permutation to use out of all the possible ones. The way it selects the permutation given a key, depends on the cipher used.
Is your question how they choose what keysize to use?
Well for starters, the key size should be big enough to give you a lot of possible permutations, so it's gonna be impractical to try each and every one to decrpypt the text in brute force fashion.
But it shouldn't be huge, because if it's far, far too big, apart from computationally expensive, then the possible permutations it can choose would be the whole set of permutations for the plaintext. And that's a bad thing, because if your keysize allows your key to choose from almost *all* the permutations, there's a very high chance some keys would choose especially bad permutations, like the identity one (plaintext=ciphertext), or just a single pair of bit permuted while everything else remains the same etc. .
Using a reasonable key, say 256bit for a 128bit block, it means the key can choose from 2^256 permutations. But all the permutations are (2^128)! which is massively bigger than 2^256, so the chance your key will choose a "bad" one is very small.
Asher Lewis
It means they're paranoid
Kayden Anderson
How can encryption be 4096-bit if most processors are either 32 or 64-bit?
Jayden Parker
32/64 bit s are the size of the instructions that run on the CPU
Aaron Carter
>4096bit
kek, probably RSA. Should be using ECC in current year.
>mutating
probably a weird way of saying it's double ratchet
Kevin Sanders
>researchers have now demonstrated that they can break even RSA 4096 bit encryption with little more than a few emails and a ... smartphone sitting on the desk.
>...
>So it's not going to be something that the NSA tries to randomly use on 300 million of us.
Yeah, sure thing buddy.
Jeremiah Morris
>it encrypts shifts each time
?
Joshua Thompson
I work in cryptography & this is incredibly wrong. Why would you even bother typing this out?
fuck Jow Forums man
Justin Scott
>I work in cryptography
[x] Doubt
>this is incredibly wrong.
Explain what is wrong. You wouldn't make a baseless claim and handwave your way out, would you?
Asher Perry
A 384 bit ECC key is almost 2^128 times more secure than a 4096 bit RSA key.
Blake Thompson
?
Blake Collins
ok buddy. retard.
Andrew Turner
>Explain what is wrong.
RSA isn't a block cipher, the bit length in RSA refers to the primes not the key, a 128-bit block cipher permutes sets of cardinality 2^128 not (2^128)!, no one has proven the existence of idempotent block cipher keys, idempotent keys are less likely to be chosen for larger key sizes, 128-bit keys aren't brute-forceable for any algo in use, aes256 is barely more computationally intensive than aes128 for a pre-quantum security level increase of 2^128x, etc.
Christian Perez
Fully homomorphic schema.
Ryan Cox
>RSA isn't a block cipher
Yeah lemme stop you right there. Why did you even think I was talking about RSA to begin with?
RSA is asymmetric and so of course not all the possible combinations of a specified keysize form a valid key. For instance, very crudely, you only need to check those with the additional property that divide your big_integer and are less than sqrt(big_integer).
Lastly, RSA *is* a block cipher. It's just used as a single block cipher for small data (like encrypting the symmetric key in hybrid schemes) and not as your typical EBC/CBC cipher for the whole text.
And it's not a hugely complicated reason why, just that EBC is deprecated for security, and CBC makes somes assumptions about randomness that cannot be *guaranteed* if we try to chain RSA blocks; and if it turns out those assumptions don't hold, then the whole thing becomes a fancy EBC mode cipher which again is not secure. (Still, it's not proven definitely that it's impossible to find a good way to chain RSA blocks).
Lastly it's computationally expensive and has overhead.
>a 128-bit block cipher permutes sets of cardinality 2^128 not (2^128)!
But we're saying the same thing. The cardinality of the set is 2^128, so all the possible permutations of that set have cardinality (2^128)!.
What I was saying is that you can view the key, simplistically, as the thing that "chooses" which permutation you use for your block.
>28-bit keys aren't brute-forceable, aes256 is barely more computationally intensive than aes128
Yeah agreed with that.
The big thing about quantum is not just factorisation of integers, but there are ways to securely exchange a private key (for symmetrical encryption), through a public channel, without actually "revealing" the key. That is, you can detect "eavesdropping" with high probability, so you stop at that bit and start again from scratch, until the exchange completes successfully.
This way quantum can depreciate the whole public key cryptography scheme.