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1

One time pad is definitely both easy to do and has perfect secrecy, but key management is a pain and can compromise security. Basically a Vigenère cipher with a key as long as the the message should be secure, because different keys can create ALL possible messages with equal probabilities. Again, it's a one time pad, so no KPA, CPA, or CCA security. ...

2

This can be broken. The exact nature of the attack will depend what modulus you use for the Hill cipher: are you working modulo a prime number, or working modulo 26? Working modulo a prime $p$ A simple attack, with no fancy mathematics needed. One simple attack is to start by requesting the encryption of the 26 messages AAAA, BBBB, CCCC, DDDD, ..., ZZZZ. ...

2

Whilst your altered vector value will influence the results, this extension of your script demonstrates that typically obtaining a single quote/reset pair is enough to reveal the secret version: def find_versions(quote,reset): """ Brute force search for version keys that produce reset from quote """ versions = [] for version in range(256): ...

4

The intend appears to be that vector is secret and the main key; and version (that, we are told, is secret) is an extension of that key (or variant selector). 1) I could say that ResetCode is a MAC of quote with key vector and version. 2) Never met this particular one. Anyone with common sense should laugh at it as snake oil if it pretends to be ...

5

Password strength is typically measured in bits of entropy, or in layman's terms, the amount of "true randomness" in the system. This is measured by the process of how the password is generated rather than by the number of bits in the output. It's a simple extension of Kerckhoff's principle: assume your attacker knows your process, and the only information ...

0

Quantum computers are not yet at the stage where they can be deployed to brute-forcing public RSA moduli. There is no evidence of a quantum computer using more than 7 qubits. The company D-Wave has made several bold claims, but offered little evidence. source: http://www.technologyreview.com/view/426586/worlds-largest-quantum-computation-uses-84-qubits/

-6

us govt has implemented 1024 qb computers, currently lockheed martin, nasa and the dpt or energy employ 512 qb computers. they use them for factoring things like all of the calculations of space travel, encryption and number crunching. as far as modern quantum computing goes and encryption DONT go by a private company such as dwave, they just wana sell ...

3

Yes, because there are no known weaknesses in (full) Salsa20. http://www.researchgate.net/publication/226808260_The_Salsa20_Family_of_Stream_Ciphers

2

A hash function is not supposed to provide this kind of security: if you have access to some message m and H(m), you can compute h(m) yourself for a given hash function h and check if they match. So, if you really have hash outputs, you can test the usual suspects MD4, MD5, SHA-1, SHA-2, RIPEMD and a few others depending on the hash length. If it's a ...

5

Yes, you are remembering correctly. Yes, this is a reasonable method to find the key length. The reason why this works is because, typically, the plaintext is not uniformly random. For instance, rather than a random bit-string, the plaintext might be some English text, encoded in ASCII. If $X,Y$ represent two random English letters, encoded in ASCII, ...

0

In addition to the special case analytical attacks for small public exponents, I wouldn't use a low value of e due to Partial Key Exposure. See "Exposing an RSA Private Key Given a Small Fraction of its Bits.": Our results show that RSA, and particularly low public exponent RSA, are vulnerable to partial key exposure. Edit: added quote

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