https://crypto.stackexchange.com/questions/70445/what-is-the-origin-of-the-phrase-dont-roll-your-own-crypto

Looks like Gary McGraw in this article was the first one to use the phrase “Don’t roll your own crypto” (at least that can be find in written form).

http://web.archive.org/web/20030629085904/http://www-106.ibm.com/developerworks/library/s-everything.html

Do you know where “Don’t roll your own crypto” comes from?

]]>Took a quick glance at the doc. Good work and a step or two in the right direction. There is a small typo at the end…

The scenarios we presented in this document are some of the examples where we **thing** the Pointer Authentication instructions would be useful. By providing a way for quickly verifying the integrity of…

I * thing* it should be “we

Also, this should be called Pointer Authorization, not Pointer Authentication, imho. If I have the time, I’ll take a second look at it because I caught something else that I forgot (multi-tasking is a b*tch.)

]]>https://www.qualcomm.com/media/documents/files/whitepaper-pointer-authentication-on-armv8-3.pdf

I hope QARMA doesn’t hit Schneier’s Law.

]]>Proof: trivial.

]]>Thanks for your attention, Johnny.

The ciphering algorithm consists, basicly, in the following steps:

############### Begining section ##############################

0. In the protoype presented, we use an alphabet of only 32 characteres.

1. There is a “vector” of 80 prime numbers, begining with 101. In that manner, ending zeroes are avoided and, furthermore, we got huge Gödel numbers

2. The key is supposed to have 80 characters

3. The Gödel-number (NG) of the key is calculated as usual:

p1^c1 * p2^c2*…

where pi are the prime numbers and ci are the ASCII values of the characters in the key

- The alphabet is altered, after creating NG, using pairs of its digits (modulo 32) as random numbers

################### Ciphering section ########################## - Take two digits of the NG, modulo 32, giving a number n
- Pick a character form the plane text, and find its position j in the modified alphabet
- Calculate cj as n XOR j
- Obtain the ciphered character at the cj position in the modified alphabet
- Check if there are only 240 digits remaining in NG

Case yes:

a) Form a new key, taking three digits from NG, (module 256), 80 times

b) Calculate a new NG

c) Modify the alphabet one more time

d) Go to 5

Case no: Go to 5

To decipher follow exactly the same steps

Possible improvements:

a) Extend the alfabet to 256

b) Create a different vector of prime numbers, depending on the user key.

c) Many others

(Agustin Sanchez / Creative Commons Attribution)

]]>Sounds interesting, but the contents are all in Spanish. Me hablamos no espanol.

So if some of the explanation could be translated, that would be great.

]]>