From my research online, I am confused about the difference between a 'monoalphabetic cipher' and 'monoalphabetic substitution cipher'. However, I have got my answer from the websites Kifanga and Cornell.edu for reference:

The following is my answer:

A monoalphabetic substitution cipher is a cipher in which each occurrence of a plaintext symbol is replaced by a corresponding ciphertext symbol to generate ciphertext. The key for such a cipher is a table of the correspondence or a function from which the correspondence is computed.

Monalphabetic substitution ciphers are easy to break using a decryption method called letter frequency analysis. This is done by studying the text in the language of the cipher, and the frequency of each letter can be determined. For example, in the English language, the most frequent letter is E followed by T. By substituting the most frequent letter in the ciphertext with the letter E the second most frequent with the letter T and so on I will end up with the original plaintext.

Are my definition and description accurate? If not, what is an accurate description and why does it then provide a poor level of security?

  • $\begingroup$ Surprisingly (to me), I couldn't find any particularly good existing question that this would be a duplicate of, at least not with a cursory search. Most of my own answers on related questions just link to Wikipedia for a definition. $\endgroup$ Dec 28 '18 at 6:16
  • $\begingroup$ My understanding is that a "monoalphabetic cipher" and "monoalphabetic substitution cipher" are exactly the same thing (synonyms). Wikipedia: monoalphabetic cipher $\endgroup$
    – David Cary
    Oct 22 '19 at 15:00

A monoalphabetic substitution cipher uses a fixed permutation of an alphabet $A$ namely $$\pi:A\rightarrow A,$$ to encrypt a plaintext $P=(P_1,\ldots P_n)\in A^n$ of length $n$ into the ciphertext $C=(C_1,\ldots,C_n) \in A^n$ via $$C_i=\pi(P_i),\quad i=1,\ldots,n.$$

For a natural language, the letter frequencies are non-uniform. Thus, given a long enough ciphertext, the original text can be easily recovered.

The length of plaintext required for unique recovery of the ciphertext is called the unicity distance, see here defined by Shannon, and is related to the redundancy in English, which is a result of the non uniform letter distribution. Most models of English give a unicity distance of around 30 letters.

Edit: To see how easy it is to attack monoalphabetic substitution, consider the letter frequencies of English, say as shown here. The four most common letters (E,T,A,O) have a total probability of around $0.38$. Thus if we have a 30 letter ciphertext (in a ciphertext only attack), the number of occurences of the letters in the set $$\mathrm{U}=\{\sigma(E),\sigma(T),\sigma(A),\sigma(O)\}$$ is highly likely to be the four most common ciphertext letters, and on average about $0.38 \times 30 \approx 11$ letters will belong to this set. There are only $4!=24$ permutations of this set and trying all will likely expose some English like fragments, helping in the cryptanalysis.

The probability that less than, say $t=8$, ciphertext letters belong to $\mathrm{U}$ can also easily be evaluated using the binomial distribution $\mathrm{Bin}(40,0.38).$

This attack can be further strengthened by considering common digrams (two consecutive letters), trigrams etc. and forming hypotheses on those, not only is the letter $\sf{T}$ very common, the digram $\sf{TH}$ is very common in English, etc.

One good countermeasure to protect this encryption to some extend is to use a Homophonic substitution, i.e., a new alphabet of larger cardinality and splitting the letters of English more or less uniformly, where each common letter is mapped to more than one new ciphertext symbol pseudorandomly. Simon Singh's website describes this rather well here. However by no means does this countermeasure provide modern cryptographic strength against computer based attacks.

  • $\begingroup$ Excellent. There is a distinction between fractionation (straddling checkerboard) and homophonic substitution, of course. And homophonic substitution starts to look like a one-time pad if values are chosen randomly. ie. 1st A = V, 2nd A= H, etc. $\endgroup$
    – Patriot
    Aug 17 '19 at 11:00

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