Affine Cipher

Encryption uses a classic alphabet, and two integers, called coefficients or keys A and B.

Example: Encrypt DCODE with the alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ and keys A=5, B=3.

For each letter of the alphabet is associated to the value of its position in the alphabet (start at 0).

Example: A=0, B=1,..., Z=25. It is possible to use A=1, ..., Y=25, Z=0 with the alphabet ZABCDEFGHIJKLMNOPQRSTUVWXY (not recommended).

For each letter of value \( x \) of the plain text, is associated a value \( y \), resulting of the affine function \( y = A \times x + B \mod 26 \) (with \( 26 \) the alphabet size). For each value \( y \), corresponds a letter with the same position in the alphabet, it is the ciphered letter.

Example:

The Affine cipher text is the replacement of the letters by the new ones.

Example: DCODE is crypted SNVSX

Affine decryption requires to know the two keys A and B and the alphabet.

Example: Decrypt the ciphered message SNVSX with keys A=5 and B=3

For each letter of the alphabet corresponds the value of its position in the alphabet.

Example: The alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ, starting at 0 gives A=0, B=1, ..., Z=25. There is also A=1, ..., Y=25 and Z=0 with the alphabet ZABCDEFGHIJKLMNOPQRSTUVWXY.

For each letter of value \( y \) of the message, corresponds a value \( x \), result of the inverse function \( x = A' \times (y-B) \mod 26 \) (with \( 26 \) the alphabet size)

The value \( A' \) is an integer such as \( A \times A' = 1 \mod 26 \) (with \( 26 \) the alphabet size). To find \( A' \), calculate a modular inverse.

Example: A coefficient \( A' \) for \( A = 5 \) with an alphabet size of \( 26 \) is \( 21 \) because \( 5 \times 21 = 105 \equiv 1 \mod 26 \).

Example: For S (\(y=18\)), \( x = A' \times (18-B) = 21*(18-3) \equiv 315 \mod 26 = 3 \)

For each value \( x \), corresponds a letter with the same position in the alphabet: the coded letter.

Example: For S (\(x=3\)) corresponds the letter at position 3: D.

The plain text is the replacement of all characters with calculated new letters.

Example: The original plain text is DCODE.

A message encrypted by Affine has a coincidence index close to the plain text language's one.

To crack Affine, it is possible to bruteforce all A and B coefficients. Use the Brute-force attack option.

If the alphabet is 26 characters long, then A coefficient has only 12 possible values, and B has 26 values, so there are only 312 test to try.

Calculate the modular inverse of A, modulo the length of the alphabet.

B' has the same value as B.

The value of A' depends on A but also on the alphabet's length, if it is a classic one, it is 26 characters long. The values of A' are then:

The Bezout's theorem indicates that A' only exists if A and 26 (alphabet length) are coprime. This limits A values to 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23 and 25 (if the alphabet is 26 characters long)

Yes, but an automatic decryption process becomes impossible, a single ciphered letter will have multiple plain letters possible.

Yes, but it exists a positive corresponding value, a value of A = -1 is equals to a value of A = 25 (because 25 = -1 mod 26).

No, B can take any value.

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