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On a practice question for my intro cryptography exam, it asks the following:

Assuming that keys are chosen with equal likelihood, the shift cipher provides:

   A) computational security

   B) perfect secrecy

   C) semantic security

   D) none of the above

I chose "none of the above", though I wasn't too sure. But I was certain it was not "perfect secrecy". However, when I review the exam solution key, it says "perfect secrecy" is, in fact, the answer.

Given that the message space of a shift cipher is huge, and the key space's cardinality is only 26, how can it possibly provide perfect secrecy under any circumstances?

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4 Answers4

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The Caesar cipher (aka Shift cipher) has, as you said, a key space of size 26. To achieve perfect secrecy, it thus can have at most 26 plaintexts and ciphertexts. With a message space of one character (and every key only used once), it would fit the definition of perfect secrecy.

For the usual use with messages longer than one character, or multiple messages with the same key, none of the given options fit (other than the "None" option).

Paŭlo Ebermann
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For perfect secrecy:

$$number\_ of\_keys >= number\_of\_cipher >= number\_of\_plaintext$$

According to Shannon's perfect secrecy theorem:

let,

$$number\_ of\_keys = number\_of\_cipher = number\_of\_plaintext$$

then we have perfect secrecy if and only if:

  1. each key is used with same probability, and
  2. for each (plain,cipher) pair there is unique key.

So, with these rules, shift cipher has perfect secrecy.

Mike Edward Moras
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Nishant Gandhi
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Shift cipher or ceasar cipher attains perfect secrecy only in the special case with the assumption that $26$ keys are used in equal probability in the shift cipher, and to encrypt each symbol we use a different key which is choosen equiprobably (i.e. perfectly random) from the key space.

It is easy to check all keys given a plaintext when the key is fixed for a sequence of plaintext. but if we use different key for each symbol then it is not possible to check all keys.

Suppose that you have a sequence of plaintext symbols of length $100$. If each key for each symbol is chosen equiprobably then you land up with $26^{100}$ possible keys. That is not possible to exhaustively search.

Raoul722
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Ankit Anurag
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Shift cipher does not satisfy the perfect secrecy property if message length ≥ 2.

Directly quote from https://www.ics.uci.edu/~stasio/fall04/lect1.pdf

Proof:

Take $m_1 = “AB”, m_2 = “AZ”, c = “BC”$

Then $$ \exists k ∈ K, s.t.Enc(k, m_1) = c$$ Namely k = 1.

However, for all $k ∈ K$ we have $Enc(k, m2) \neq c$, and hence $$Prob[Enc(K, m_1) = c] = 1/26 $$ while $$Prob[Enc(K, m2) = c] = 0$$ So the perfect secrecy requirement is violated, which requires above two probabilities to be equal.

Weishi Z
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