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5

This is not a mathematical proof. A notable place it fails to be a proof is here: Pay attention to which cipher text I use, look up to match the message with the cipher-number below. $$cipher1⊕cipher3=character1⊕character3⊕IV1⊕IV2$$ (Note that the cipher BOTH use the SAME KEY, but they remain secure because of the two different IV) This line is ...

4

The system described in the quoted article depends on the security of AES for random keys (not only on the theoretical unbreakability of the OTP) in at least at least two things: the encryption of large files, as apparent in the quotes of the question; the initial establishment of the OTP, as shown by this other quote The first step is always optical, ...

3

Simply put: No. First recall that this is a mis-use of the term "One Time Pad" So lets call it a vigenere cipher instead. You can determine this is insecure with a simple algebraic combination: $\text{attack} = cipher_1 + cipher_2 + cipher_3 + cipher_4 \\ \text{Simplify: } \\ \text{attack} = character_1 + key_1 + IV_1 + character_2 + key_2 + IV_1 + ... 2 First, the system as I understand it does not use a one-time-pad at all, but a steam cipher. The keystream is generated via a CSPRNG, and is not truly random, so it cannot be a one-time-pad. They are misrepresenting it. But, to answer your question, let's assume the system did in fact use a one-time-pad, and used it in the manner described. In this ... 2 Actually, it's not a hare-brained idea at all; you certainly can do integrity checking using a one-time pad. However, I believe that you'll need to use the one-time pad bits a bit faster than you'd expect, to achieve a forgery probability of at most$2^{-32}$, I believe you'll need at least 64 pad bits per packet (assuming informational theoretical security ... 1 The right-hand side of your equation is just the average probability that a random message encrypted with a random key gives the ciphertext$c$. (Exercise: Why? Hint:${\rm avg}(x_1, x_2, \dotsc, x_n) = (x_1 + x_2 + \dotsb + x_n) / n$.) However, if$k$,$m$and$c$are all fixed (and you haven't indicated that they aren't), then$P[E_k(m)=c]$is trivially ... 1 Here's my interpretation of what is happening here, though it's hard to be sure without more context. A$c \in C$has been chosen, and we are asking "What is the probability$E_k(m)=c$is true if$m \in M$and$k \in K$are chosen uniformly at random?" To find this probability, you divide the total number of outcomes of your choices, which is$|M||K|$, by ... 1 I'm not fully sure if the equation is correct, as I think$P[E_{k_i}(m_i)=c_i]=1$, as the encryption always produces a cipher text (which is what you formulated I think, as there's no definition of$c_i$). If you change the$c_i$to$c$however you'd calculate the average probability that any message-key-pair encrypts to$c\$. I didn't take part in such a ...

1

First, you wouldn't call it a one-time-pad any longer if you reused the key. Second the security of the OTP can only be proven if the key is as long as the message. (which wouldn't be the case if you'd reuse a key) Third, OTP usually use XOR operation to combine key and message. If the key is never reused, you're safe, but if an attacker can mount a known ...

1

To answer the question of whether they would need some special technique to OTP encrypt a message longer than their key: No, to realize their claim of more than 3000 text messages they can use standard OTP. You are correct that with 0.5MB OTP key you can encrypt at most 0.5MB of data. 0.5MB is 500.000Bytes meaning as many characters can be encrypted ...

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