Here\'s an example of a MonoAlphabetic Substitution Cipher. Here\'s a letter sub
ID: 3877588 • Letter: H
Question
Here's an example of a MonoAlphabetic Substitution Cipher.
Here's a letter substitution table for such a cipher:
ABCDEFGHIJKLMNOPQRSTUVWXYZ
OEXPVYQNAZGCDKURHJLBIFMSWT
where the top row represents the plaintext letters, and the bottom row represents the corresponding ciphertext letters.
Using this table, we can encipher some plaintext into its corresponding ciphertext:
A bowl of Moose Tracks ice cream
O eumc uy Duulv Bjoxgl axv xjvod.
The top line is the plaintext, and the bottom line is the ciphertext. You can figure out how the translation is done by working through each letter of the plaintext, and matching it with the corresponding letter in the substitution table.
So, given some ciphertext, how would you determine the letter substitution table? Again, you really have two choices: Use brute force.
In this case, we try every possibility, until we find a reasonable looking plaintext.
Question 3: Given the approach described above, for a MonoAlphabetic Substitution cipher, how many possibilities are there for character mappings? Is this a feasible task?
Analyze the letter frequency of the ciphertext, and try to deduce the character mapping.
Explanation / Answer
How to determine the letter substitution table?
One approach is to use the Brute Force technique. In brute force method, we try all possible combinations and then find out the most suitable answer. To break a MonoAlphabetic Substitution cipher with this approach, there are 26! possibilities of character mappings. 26! = 403291461126605635584000000 possible character mappings. Even if a computer finds out all possible combinations at some time, it is quite a headache for a human to find out the reasonable looking text from this many combinations. Hence, it is not feasible.
Another approach is to analyze letter frequency of the ciphertext and compare it with plaintext letter frequency and deduce it. Every language has a letter frequency graph, English also has where each letter appears in a sentence with a probability of some percentage already known. As we are talking about monoalphabetic cipher, it cannot hide the letter frequency. We can calculate the letter frequencies of the letters present in the ciphertext and then compare these frequencies with the ones in the plaintext. This way, we can get a ciphertext-letter-plaintext-letter mapping.