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13

n is the exponent. So when n is doubled from 64 to 128 it doesn't mean that you have to try twice as many values. It means that you have to try $2^{64}$ times the amount you were already trying (as $2^{128} = 2^{2\times64} = 2 ^{64+64} = 2^{64}\times2^{64}$). It is required to only search half of the key space on average (if average is the correct term ...


5

I know that some of them are pretty hard to crack, but since they are so commonly known is it even practical to consider using something like that as an encryption method considering the algorithms for encryption and decryption are commonly known (from a security perspective)? In fact, this is exactly what we want. Schneier's law Anyone, from the ...


4

I should begin by noting that this seems like an unusual assignment. I'm not sure why someone would explicitly have a goal of combining block ciphers and stream ciphers. First, let's summarize the difference between block and stream ciphers, since this may be useful for future readers. Block ciphers are so called because they operate over fixed lengths of ...


1

Here are some hints on how it's done on Mega: The password provided is passed through a KDF to derive a key, that is used to en-/decrypt the master key (later provided by the server through an API call). To bring it down to the crucial bits: The KDF applies $2^{16}$ rounds of AES-128 with it. The details can be found in the function prepare_key() of the ...


1

Your design seems to be a byte-wise generalization of Jennings' multiplexed generator rather than the alternating step generator. S. M. Jennings, “Multiplexed Sequences: Some Properties of the Minimum Polynomial,” Lecture Notes in Computer Science, vol. 149, 1983. I believe designs like hers [may even be byte based for efficiency] have been used in ...


1

The triple DES (3DES) block cipher works by essentially running the block through DES three times. Triple DES is also known as "DES EDE" (encrypt-decrypt-encrypt) and under the name given by the standard document: "TDEA". The TDEA algorithm is described in FIPS NIST Special Publication 800-67 Revision 1 where paragraph 3.2 describes the TDEA Keying Options. ...


1

At least it checks out: $$(3 + 2X^2 - 3X^4 + X^6)(-2 + 4X + 2X^2 + 4X^3 - 4X^4 + 2X^5 - 2X^6) = -6 + X + 6X^2 + X^3 - X^4 + 6X^5 - 6X^6 - 4X^2 + 8X^3 + 4X^4 + 8X^5 - 8X^6 + 4 - 4X + 6X^4 - X^5 - 6X^6 - 1 + X - 6X^2 + 6X^3 - 2X^6 + 4 + 2X + 4X^2 - 4X^3 + 2X^4 - 2X^5 = 1$$ multiplying out all the terms, using $X^7 = 1, X^8 = X, X^9 = X^2$ etc. and the fact ...



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