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Consider a variable one-time pad, that is, $\mathcal{M}:=\{0,1\}^{\leq \ell}$ is the set of plain text. Now, this scheme is not perfectly secret, since you can take two plain text of different size, say $|m_1| = 1, |m_2| = 2$ and considering a cipher text $c$ of length 1, the next happens: $$Pr(E(k, m_1) = c) = \frac{1}{2},\ Pr(E(k, m_2) = c) = 0.$$

Thus, how can I make a construction of this variable one-time pad such that it's perfectly secret? Is it even possible?

I tried to make sub-one-time pads, i.e., $\ell$ one-time pads, but it doesn't work when you have two messages of the same length (same as above), so my other idea was to extend all messages to be length $\ell$ by adding zeros to the right. The problem though, is that if you consider $\ell = 4$, how can you decrypt the messages 1, 10, 100, 1000?

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  • $\begingroup$ your plaintexts are the same size, no? $\endgroup$
    – kodlu
    Mar 28, 2022 at 23:51
  • $\begingroup$ @kodlu No, they are at most size $\ell$. $\endgroup$
    – Lug322d
    Mar 29, 2022 at 0:40
  • $\begingroup$ Hi Lug and welcome :-) Review the question please as it's confusing. What exactly are you asking? We love one time pads here though... $\endgroup$
    – Paul Uszak
    Mar 29, 2022 at 1:43
  • $\begingroup$ Thanks @PaulUszak! In simple words, I'm trying to form a variable length one-time pad that it's perfectly secret. $\endgroup$
    – Lug322d
    Mar 29, 2022 at 2:41
  • $\begingroup$ Okay. Yes OTP is informational secure. But you're talking formulae. Is there a device to produce the key material? And I don't understand the "variable" bit. Do you mean that |key| = |plaintext|? And what's $|m_1| = 1, |m_2| = 1$? One bit? $\endgroup$
    – Paul Uszak
    Mar 29, 2022 at 3:00

1 Answer 1

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The (binary) one-time pad is indeed proven to be perfectly secure in an information-theoretic sense, assuming the following: the message length exactly $n$ and a shared source of uniform randomness.

It is often overlooked that this security definition is not appropriate for general contexts where variable-length data is sent. For instance, an application where are yes and no are the only message sent would be insecure when applying the one-time pad naïvely.

Solution 1: Message padding The easiest way would be to apply a length hiding mechanism, a padding scheme that pads messages to the same length and then encrypts the padded message. Namely. for messages of length $l$, messages can be padded to length $k = l + 2$ (long can work as well). The padding of $m$ is $pad(m) = m \|10^{k - |m| - 1}$. This can be done since the problem statement does not restrict the length of the pads.

Solution 2: Encoding onto a group. Since there are $k = 2^l$ messages, another idea would be to encode (bijectively) messages into a group-like structure with the same cardinality; from there, one can apply the OTP over the group. Decryption requires decoding. The simplest example would be $(\mathbb{Z}/k\mathbb{Z}, + )$

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  • $\begingroup$ So I was reading again this solution and now I got one question for each solution. For solution 1, how do you decrypt the message to obtain $m$ again? The receiver can't know which bits they need to remove since the padding was also encrypted. And for solution 2, I think the number of possible messages is $\sum_{i=0}^\ell 2^i$ since it's variable. $\endgroup$
    – Lug322d
    Mar 29, 2022 at 19:33
  • $\begingroup$ @Lug322d, regarding 1) It is usually assumed that the legit receiver has access to the pads. So decryption works by applying the pad, and the undo the padding; obviously, another hidden point is no one did not modify that ciphertext. Regarding 2) That's a good point, I guess it also depends on what you consider as message. Do you consider "01" and "001" as different messages? $\endgroup$ Mar 29, 2022 at 20:15
  • $\begingroup$ I see, that makes sense. And yes, 01 can be distinct to 001. Still you could make that group though, but it will not be a finite field which, if it were, it will simplify a lot of things. $\endgroup$
    – Lug322d
    Mar 29, 2022 at 21:41
  • $\begingroup$ The second method would still work even if you differentiate 01 and 001. You just work in a larger group $\endgroup$ Mar 29, 2022 at 21:47

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