GPG Implementation of Symmetric Operations,
and To-Self Encryption
Sergi Blanch i Torné
sbt at megacceso.com
Sun Jun 4 23:22:24 CEST 2006
Hey! Too much things to answer... I will only respond some that I can know the
answer, sorry.
A Diumenge 04 Juny 2006 22:36, utternoncesense at gmail.com va escriure:
(...)
> Thirdly, GPG is based upon a hybrid system entirely. The data of any
> file is ALWAYS encrypted symmetrically, and a symmetric key is made
> for each encryption use. The symmetric key used is then encrypted
> with the public key of the recipient and the whole thing is bundled
> together.
Think, for example that you want to send 100MB of information to 10 people.
With in a pure asymmetric system yo will encrypt it one by one and then send
at less 1GB (possible more). Using a hybric scheme (not less secure) you will
send 100MB symmetrically encrypted and a little more of 10k where you have
the symmetric key encrypted one by one.
> RSA & ElGamal use keys around 1024-2048 usually. EC uses 160-224 bit
> keys, but is based on mostly different math (it may be equivalent at
> some level, but I'm neither aware nor able to understand anythig
> beyond yes or no on that topic). AES uses 256 bit. It's not allowed
> to go over 256 bit. This is because it's an entirely different area
> of cryptography? Block Ciphers as opposed to integer factorization,
> discrete logs, or curvature? And comparing key lengths between the
> three areas (IF/DS, EC, Block) without any normalization is like
> comparing the engine in a semi to one in a sedan without considering
> the weights of the vehicles - They both enable the vehicle to go 80
> (encrypted to some rigor) but the semi needs a much larger one because
> the truck weighs more (easier to test factors than undo block
> ciphers). Right?
To stablish the equivalence of security between different algorithm, also from
different nature (like compare symmetric/asymmetric) we could use a formula
that relates how many basic operations are needed to broke it, with how many
basic operations the computer could do per second. Then you have one very
much optimistic time.
If some one find a new atack to one cryptosystem, this equivalences will
change.
(...)
Sorry for the partial answer. Some one else could answer you better than I.
/Sergi.
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