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Of those you listed, AES is the best to study. Not only is it the standard that is used everywhere, it has a huge literature of people explaining it and analyzing it, far larger than any of the others on your list. Also, compared to the others on your list it is easier to understand why AES strongly resists certain major classes of attack (like linear and ...


4

First, it is important to learn the basics behind all symmetric ciphers. You can get this from Handbook of Applied Cryptography, see Chapter 7, especially 7.1, 7.2, 7.3. If you understand those three sections, you will be off on the right foot. From there, I would suggest just diving right into AES. It isn't that terribly difficult (yes, there are easier ...


2

You say I have never studied a cipher before In that case I would recommend the following: Sign up for the Stanford online class on Cryptography on Coursera. This is a great introduction to Cryptography and this will conver block ciphers. Get a library card with your local public library and ask them to get some textbooks on Cryptography for you. ...


2

None of Twofish, Serpent and AES are currently known as broken, so as far as security is concerned, you can use any of them. AES has a slight advantage because it's very widely used, so if it gets broken you're more likely to hear about it and get relevant software updates quickly. The Snowden postings haven't changed much as far as cryptography usage is ...


1

Camellia perhaps? NESSIE (EU) and CRYPTREC (Japan) both endorse it. Your requirements are brief and mysterious so it's difficult to give further suggestions.


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Some points towards an answer: Why HMAC-SHA3? HMAC and its security proofs have been devised for Merkle-Damgård hashes, and SHA3 is not one. HMAC-SHA256 would be fine (Updated per comment: the Keccak submission does endorse its use with HMAC, using a block size parameters of 576 (resp. 832, 1088, 1152) bits for the hash with output of 512 (resp 384, 256, ...


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The way the iterations work is that it roughly increases your security (in bits) by $\log_2(iterations)$. So you would still need $\frac{\log{2}}{\log{97}}\cdot (256 - \log_2(10000)) \approx 37$ characters in your password to have 256-bits of security. Think of it this way, if you have $2^{256}$ possible keys, that is an astronomically large number. Much ...



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