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You are using a Vernam-encryption (simple XOR), as for the one-time pad. The general principle for Vernam is that it is perfectly secure as long as you never reuse the same key for more than one message, and gets utterly broken as soon as it is reused even once (this is the "two-time pad"). The key here is the hashed password, the message the key. If one ...

Using the password itself (or anything similar predictable) instead of an independent random value as the salt denies the whole benefit of salt: Same passwords (passphrases) give now the same key, instead a different one. So, if two users happen to choose the same favorite image as their password, they get the same key, and thus an attacker can use this ...

The reason that salts are used is that people tend to choose the same passwords, and not at all randomly. Many used passwords out there are short real words, to make it easy to remember, but this also enables for an attack. As you may know, passwords are generally not stored in cleartext, but rather hashed. If you are unsure of the purpose of a ...

Both scrypt and pbkdf2 have variable length outputs, and each bit of the output is effectively independent on every other bit. So, one obvious way would be just to ask for enough output for both keys. For example, if the two keys are each 128 bits, then ask scrypt (or pbkdf2) for 256 bits of output; use the first 128 bits as the first key, and the second ...

Short answer: Use bcrypt. Long answer: First, "hashing a password with the password" is an undefined statement. Regardless, you are well into "inventing your own crypto" territory. Assuming you mean something along the lines of SHA-256(password + password), this is a phenomenally bad password digesting scheme. Being unsalted, your approach is vulnerable to ...

Yes, scrypt achieves this. Scrypt has a variable-length output, so just generate as much output as you need. For instance, you can ask it for 256 bits of output, then use the first 128 bits for one key and the second 128 bits for the other key. While PBKDF2 also has a variable-length output, I do not recommend that you use it in the same way. It has a ...

Don't bother with changing the actual cipher algorithm. Read about Kerckhoffs's principle: you should only change things like the key and the IV, not the actual algorithm. In order to test your avalanche, flip one bit in your key. That should change about half the bits in your output. For cipher design, Applied Cryptography has already been suggested. ...

One option would be to generate a random key, split it using Shamir's secret sharing, then encrypt each of the split parts individually under a key derived from each user's password. So for example: key = read from os.urandom() d1,d2,...d5 = split(key=key,n=5,k=3) e1 = encrypt(d1, KDF(PW1)), e2 = encrypt(d2, KDF(PW2))... key can then be derived from all ...

Here's a possible scenario: 1) Your password is put through a slow KDF such as Scrypt. The output of Scrypt can be configured to take a long time to calculate, and as such, can mitigate the risk of brute-forcing passwords. See here. 2) The output length of Scrypt is also configurable. So assume that half of the output becomes the encryption key for ...

I think what you are looking for is a Password-Based Key Derivation Function (PBKDF). You can take a moderately strong password, like 12-14 random letters and numbers (no dictionary words though!), and throw it into the PBKDF function together with some other parameters, e.g. salt, number of iterations and the desired key length. After that you have a ...

Search for passwords on IT Security and you will find tons of advice on how to store passwords, and how not to. Your scheme is not a good method for hashing passwords: it is a fast hash, it lacks any salt, and it unnecessarily limits the password length. People have studied this at great length: before trying to re-invent the wheel, I suggest you read up ...

What you are looking for is a Pseudo Random Function that should be indistinguishable from uniform, even if the key material that is passed to it is not. One potential problem with your scheme is that the AES key schedule is not particularly good at extracting the entropy from keys that are not selected (pseudo-)randomly, such as passwords and pass-phrases. ...

It is a random number that is needed to access the encrypted data, along with the password. If an attacker does not know the password, and is trying to guess it with a brute-force attack, then every password he tries has to be tried with each salt value. So, for a one-bit salt (0 or 1), this makes the encryption twice as hard to break in this way. A two ...