I am somewhat new to Cryptography and I'm struggling with this proof I came across in my Intro to Cryptography course. I understand the proof of security for a one-time pad:

Pr[M = m | C = c] = Pr[M = m| C = c] * Pr[M = m] / Pr[C = c] = (1/2^n) * Pr[M = m] / (1/2^n) = Pr[M = m]

I was wondering what happens if m > n? Pr[M = m|C = c] != Pr[C = c] but I want to know why and exactly what values these probabilities acquire.


  • $\begingroup$ It looks like you have $m$ to be defined as a possible message (i.e. a value), and $n$ to be defined as (roughly) $|\mathcal{C}| = 2^n$, where $\mathcal{C}$ is the space of all possible ciphertexts (so something akin to the "dimension" of the space). Is this interpretation correct? If not, can you clarify what you mean by $n$ and $m$? $\endgroup$ – Mark Sep 20 '20 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.