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Adapted my primality test algorithm to do a check for if a number is a "Perfect Number" (the sum of all its dividends equals the number itself). This one was fun and turned out to be super fast as well. It was written in Haskell.
@freemo Another cool feature I found is how to quickly find prime numbers for RSA.
@LWFlouisa Yea we have a lot of really cool probabilistic methods for finding large primes. Did you discover a new novel way or just read about an existing one?
@freemo It has to do with how math is taught in high school, as in my mine they tended to spout facts, rather than emphasize why those facts were the case.
@LWFlouisa I agree they do a very poor job at teaching math. Common core in america is a step in the right direction at least but still has its flaws
@freemo So something that has a positive charge will always be followed by a negative charge. 9 is always positive, and 3 and 6 are always negatively charged.
@freemo I think I have a preview somewhere, I'll look for it.
Heere we go: https://www.instagram.com/p/Bky-LSSgtY8/?taken-by=sarahrweaver
It's still theoretical at this point. But has worked in my experiments.
@freemo But the theory goes that using this method, you can skip primes that aren't perfect.
@LWFlouisa All primes are perfect to me :)
@freemo What a 59 * 53 example!
@freemo Yea I prefer the term compatible primes: 59 and 53; 197 and 193, and so on work better than say 3 * 197.
@freemo The compatibility allows you to generate mathematically related co-primes more easily where phin * phin + 1 mod phin is equal to 1.
@freemo My guess is this is why people are wanting to move toward other methods besides being hard to compute.
@LWFlouisa Well hard to compute isnt guaranteed to always be hard to compute. Most cryptography today will be useful against a Quantum Computer.
@freemo Especially symmetric systems.
Its one reason I focus on information splitting/dispersal over key exchange.
In the asymmetric solitaire modification, randomness is useful in finding a chain of small primes that can be used for key exchange.
Such as in using a different public key to shift each different portion of a symmetric key.
@freemo With traditional asymmetric, you'd still prone to chosen plaintext, although I'd probably need to ask Bruce Schneier what that means exactly. His book wasn't real clear about this.
@freemo In this context, because of the nature of the vortex as a viable one way function: p (or 6) and q (or 3) must be chosen such that 8 and 5 are easy not find a quantum computer to find, so much play the fact that primes are easy to find against itself.
@freemo To make the vortex a viable one way function I mean.
@LWFlouisa Anything based on primes at all will fail against a QC.
@freemo Yea it will be interesting to see how QC will effect cryptography.
Not sure if it's as strong against steganography. Maybe?
@LWFlouisa There are QC resistant cryptographic algorithms that can run on conventional computers. None of them however rely on the difficulty to verify prime numbers however.
@freemo Code Based lAttice is an example, I think. Not sure.
@freemo I guess that's why key exchange is so hard though. One of my ideas was hiding an encrypted key in plain sight.
@LWFlouisa Hard how? We have "Perfect forward Security" which allows for key exchange with some rather nice security guarantees.
@freemo Oh security wise very easy. I'm thinking more in keeping the encrypted key from being discovered.
@freemo Namely, keeping the public key public, while keeping the ciphertext hidden in plain sight.
@LWFlouisa A symetric encryption across your private key accomplishes just that
@freemo I may have to look into that.
@freemo Mainly i want the hiding in plaint sight incorporated into the encryption program itself.
@freemo Situational, but if one doesn't have time to embed and image and encrypt a key separately, such as in high risk time constraints.
@freemo Ex. After I encrypt plaintext 8, I take the ciphertext and embed it into an image using steghide.
@freemo One thing I'm plugging into this algorithm: when you multiple 59 and 53 you get 3127, with a phi of 3016. Normally candidates are based on adding phi + 1. In mine the candidates are found by phi * 2 +1 or -1 depending on the point on a vortex chart.