The most known example cipher reaching perfect secrecy is One-time Pad, which employs modulus addition for encryption and decryption.

Is there any other well-known cipher no less practical than OTP that reachs perfect secrecy?

  • 2
    $\begingroup$ Hiya :-) Technically, modular maths isn't a requirement. The transformation has only to be bijective, which is how OTPs were implemented decades ago. $\endgroup$
    – Paul Uszak
    Sep 8 at 12:08

2 Answers 2


There is a famous result about perfect secrecy, usually called Shannon's theorem:

If an encryption scheme has perfect secrecy, then $|\mathcal{K}| \ge |\mathcal{M}|$, where $\mathcal{K}$ is the set of possible keys and $\mathcal{M}$ is the set of possible plaintexts.

One-time pad achieves equality $|\mathcal{K}| = |\mathcal{M}|$, which is the fewest number of keys possible according to Shannon's theorem. If you seek a scheme that is "no less practical" than OTP, then I suppose that you too require $|\mathcal{K}| = |\mathcal{M}|$.

OTP also has a deterministic encryption algorithm, so any scheme that is "no less practical" should also have deterministic encryption. (Also, randomized encryption would needlessly increase the ciphertext size.)

With a deterministic encryption scheme, we can imagine writing out the entire truth table of $\textsf{Enc}(K,M)$, with rows indexed by $K \in \mathcal K$ and columns indexed by $M = \mathcal M$. As $|\mathcal M|=|\mathcal K|$, the table is a square. For decryption to be possible, every row must have distinct entries -- this corresponds to the property that $\textsf{Enc}(K,\cdot)$ is a bijection for every $K$. Every column must also have distinct entries: if the column corresponding to $M$ does not have distinct entries, then it is missing some $C$, meaning that no key encrypts $M$ to $C$. But then observing the ciphertext $C$ means the plaintext could not have been $M$, and this violates perfect secrecy. (We know $C$ appears somewhere in the table as encryption of some plaintext, otherwise the condition on rows doesn't hold.)

The truth table of $\textsf{Enc}$ is therefore a Latin square, making it the truth table of a quasigroup operation. Let's call the quasigroup $(\mathcal G, \boxplus)$. Then $\mathcal M$ and $\mathcal K$ are both isomorphic to $\mathcal G$ and the encryption algorithm is $\textsf{Enc}(K,M) = K \boxplus M$.

So the question of alternative perfectly secret encryption schemes reduces to the question of identifying a suitable quasigroup.


There's computational perfect secrecy and information-theoretic perfect secrecy, and OTP belongs to the latter.

If it's well-known information-theoretic perfect secrecy that you seek and is no less practical, then none.


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