Affine Cipher:

$Enc(x) = (ax + b) \mod m$

$Dec(x) = a^{-1}(x-b) \mod m$

For a brute-force key search, we need to do $a \cdot b$ encryptions in the worst case.

For a meet-in-the-middle attack, how many encryptions/decryptions do we need to do in the worse case?

Meet-in-the-middle attack definition: Affine cipher can be broken up into two separate encryptions and decryptions.

$Enc(x) = (ax)\mod m = c_1$

$Enc(c_1) = (c_1 + b) \mod m = c_2$

$Dec(c_2) = (x-b) \mod m = c_1$

$Dec(c_1) = a^{-1}(c_1) \mod m = x$

Now you can start by encryption x over all possible values of $a$ and decrypt $c_2$ over all possible values of $b$ then compare the two tables and find a match over $c_1$

Which now I realize has answered my question that it uses $a + b$ encryptions.

Side question: How is $a^{-1}$ calculated? $m - a = a^{-1}$ or maybe $m - a \mod m = a^{-1}$


1 Answer 1


I don't know what you mean by a 'meet-in-the-middle' attack. The obvious way to attack this cipher is a known plaintext attack, that is, encrypt any two distinct plaintexts $x_1, x_2$ (and hence two encryptions), getting their encryptions $y_1, y_2$ :

$y_1 = (a x_1 + b) \mod m$

$y_2 = (a x_2 + b) \mod m$

and solve for $a, b$ as:

$a = (y_1 - y_2) (x_1 - x_2) ^ {-1} \mod m$

$b = (x_1y_2 - x_2y_1) (x_1 - x_2) ^ {-1} \mod m$

Now, as for your side question of what $a^{-1}$ means, well, that is the modular inverse of $a$; if we claim that $c = a^{-1}$, this is equivalent to claiming that $a \cdot c = 1 \mod m$

The modular inverse of $a$ can be found efficiently (given a, m) by the Extended Euclidean Method


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.