# Tag Info

4

A stream cipher, by definition, acts on individual bits at a time. This effectively means that bits later in the plaintext cannot affect bits around the beginning of the ciphertext. It can, however, feed the ciphertext back into the state of the PRNG to affect all future bits. This can be seen in the CFB (Cipher Feedback) block cipher mode. It acts similarly ...

0

Are you implementing this for cryptographic purpose ? If so, you should use True Random Number Generators (TRNGs) to generate your master key.Your key should have following properties: Statistical independence = For a given generated sequence of values, a particular value should not be more likely to appear next. Uniform distribution = All numbers are ...

0

You have to be sure that blocks of data would have sufficient size such that equivalent key size remains secure. Also the key should only be used once. Once you have a secure PRG then you can employ a stream cipher like XOR encryption. This comes however with the disadvantage of having key size equal your data, and the key should be used once

9

XORing a master key (presumably a long term key) with data is a very dangerous idea. If any data key is leaked, then the master key may be easily calculated, thus leaking all keys. ($m$ for the master key, $d_x$ for all data keys) $$c_x = d_x \oplus m$$ then somehow $d_4$ is leaked $$m = d_4 \oplus c_4$$ $$d_x = c_x \oplus m$$ You'd be better off applying a ...

-1

If done right, XOR can be used to encrypt data. One way to do this, is to generate a pseudo-random key-stream using your master key and XOR the key-stream, which needs to be as long as the data, with the data. The key-stream can be generated using a block-cipher like AES. This is how AES-CTR (Counter) Mode works.

3

At the present time only exists xnor, which is the complement of xor. 0 0 = 1 0 1 = 0 1 0 = 0 1 1 = 1 = xnor The others truth tables ignores entirely one of the inputs and are not suitable for accumulation of entropy.

3

While the other answer is correct, there are multiple ways to disprove a theorem. In your case you want to disprove $$\Pr[\mathcal M=m|\mathcal C=c]=\Pr[\mathcal M=m]$$ holds for $$c=m\oplus k \oplus \operatorname{rev}(k)$$. The first one is to argue using the probabilities and show that the above equation doesn't hold for this scheme. The second (formally ...

3

You know that the definition of perfect secrecy is $P(M = m \mid C = c) = P(M = m)$ for all $m$ and $c$ and that each key is only used once. This mean that know about $c$ not affect $P(M=m)$ and if you know $c$ you can't access any information about $m$. But as yyyyyyy mentioned,since $k\oplus \operatorname{rev}(k)$ is symmetric, if $\operatorname{Enc}(k,m)$ ...

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