# Tag Info

3

For a scheme to be information-theoretically secure, you need that $$\Pr[M=m\mid C=c]=\Pr[M=m\mid C=c^\prime]$$ for all $c,c^\prime$ (that is, any ciphertext has the same probability $M=m$, so the ciphertext doesn't change the probability $M=m$). Let's suppose we have a $c$ and a $c^\prime$. Both of them have the same number of ones and zeroes, because both ...

4

If you perform the distribution digitally (using networks) then you have a problem. Unless you use another one time pad you lose the perfect confidentiality as the distribution itself won't deliver perfect security. But using another one time pad is pointless: you would lose exactly as many key bits as you are distributing, while you are only protecting the ...

0

Are M and C correlated? No. The distribution that the values the M and C may take on are independent of each other. The condition probability distribution that M has is unchanged no matter what the observed C value is. This is true whether you are using the statistical meaning of correlation, or whether you are looking more specifically at linear ...

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 ...

-2

This is NOT secure. I have found the weakness with mathematics. So this is our two messages: Sending the first message: $character_1 ⊕ key_1 ⊕ IV_1 = cipher_1$ $character_2 ⊕ key_2 ⊕ IV_1 = cipher_2$ Sending the second message: $character_3 ⊕ key_1 ⊕ IV_2 = cipher_3$ $character_4 ⊕ key_2 ⊕ IV_2 = cipher_4$ Now see what happens if we XOR all of ...

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 ...

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 ...

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 + ...

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 ...

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