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Would it be useful for companies who need to keep their data safe? Not exactly. The One-Time-Pad is extremely inconvenient. If your client has to encrypt a piece of plaintext that's 4GB large, then they will not only have to generate 4GB of random data, they also will have to share that pad with the receivers of that message, making it a total of 8GB of ...

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You are creating a key stream using a hash function. This is often called a stream mode of operation (although it doesn't seem to be a well defined term). This is a known method of creating a key stream. An OTP requires the key stream to be completely random. This is because it would otherwise be possible to brute force the key. If you can brute force the ...

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I just came across this question and was surprised that no one referenced the paper: A Natural Language Approach to Automated Cryptanalysis of Two-time Pads by Mason et al. at ACM CCS 2006. This shows how to solve this problem in an automated and intelligent way.

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By two-time pad I assume you mean using a one time pad key to encrypt two messages. Lets say $K$ is the key and $m_1, m_2$ are your messages. Then from the ciphertexts $c_1 = m_1 \oplus K$ and $c_2 = m_2 \oplus K$ an adversary could trivially learn information such as $x = c_1 \oplus c_2 = m_1 \oplus m_2$. Whether or not this information is "exploitable" may ...

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There is no security difference; there are a handful of practical ones: With xor, you can have the same code to do encryption and decryption With xor, you don't have to pick a 'word size'; a larger CPU can handle 4 or 8 bytes at a time, while a microcontroller can handle 1 byte at a time, without changing the ciphertext With xor, you don't have to worry ...

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Important note: One-time pad does not "not give any information about the content of the message". This is a fallacy. It still does reveal information. For example, simply knowing that there exists an exchanged message does tell us some information. If the assumptions of the one-time pad are satisfied, what we will have is an encryption method that ...

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Preliminary on notation: in the question, it seems advisable to change $C=T_{k}M$  to $C=T_{k}(M)$ , and change $M=T^{-1}_{k}C$  to $M={T_k}^{-1}(C)$ it is necessary to change $T_{k}T^{-1}_{k}=I$  to ${T_k}^{-1}\circ T_k=I$ , meaning that the function obtained by applying $T_k$ then its inverse ${T_k}^{-1}$ is identity, with $\forall M, ... 1 I think it covers both symmetric and asymmetric systems. Given a transformation$T_k$, if computing$T^{-1}_k$is easy then this is a symmetric cryptosystem. On the other hand, if this is difficult than it is an asymmetric one. Note that in this case, we are ''hard-coding'' the key$k$inside the transformation. Thus we are just given the transformation ... 1 Your definition only covers symmetric encryption since the same "index"$k$is used in$T_k$and$T^{-1}_k$(i.e., encryption and decryption use the same key) If you want to give a generic definition that covers both types of encryption, you could say that the transformations are$T_k$and$T^{-1}_{k'}$and that in the case of symmetric encryption$k'=k\$.

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what I was originally thinking of was a way to compress random data. this compression data could be encrypted using one-time-pad and then used as a key. the approach of that is impractical. I have new approach to infinite one-time-pads. in every message include a reference to a randomly picked public file(s). encrypt said files to produce higher entropy ...

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