I am having a very difficult time trying to understand what exactly this question is asking.

How many valid English plaintexts are there for the ciphertext message CJU using a length-3, one-time pad of cyclic shifts, (i, j, k)?

If the encrypted question message is CJU and the one time pad is i, j, k would there not be exactly 1 solution? We are given the one time pad i, j, k so we can derive the plaintext simply by some alagebra

(x + 9) % 26 = 3

(y + 10) % 26 = 10

(z + 11) % 26 = 21

Where x, y, and z are the plaintext letters T, Z, J?

I feel like this is a trick question that I am not understanding

  • $\begingroup$ I think you are misreading the statement, $(i, j, k)$ isn't supposed to be the key $(9, 10, 11)$, but some key. $\endgroup$
    – Aristu
    Commented Oct 16, 2016 at 3:36
  • $\begingroup$ how does i, j, k help me narrow down my resulting set? $\endgroup$ Commented Oct 16, 2016 at 3:39

1 Answer 1


$(i,j,k)$ is not a specific OPT. It's three variable names that you can use to talk about the key. The question is just asking, "Given all possible three-character OTPs, how many valid English plaintexts are there for the ciphertext CJU?"

In other words, how many different values are there for the triple $(i,j,k)$ such that "CJU" decrypts to an actual English word.

  • $\begingroup$ ahhhh, so $(i, j, k)$ is just a generic. So the solution is every possible english 3 letter word since we do not know the pad $\endgroup$ Commented Oct 16, 2016 at 18:26
  • $\begingroup$ Where i !=j and j != k and i != k $\endgroup$ Commented Oct 16, 2016 at 18:33
  • $\begingroup$ @Programatic There's no reason to assume that the successive characters of the pad are different. Indeed, if you did include the rule that consecutive characters of the pad must be different, that would leak a small amount of information. $\endgroup$ Commented Oct 16, 2016 at 18:58

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