# How many oracle calls to break an affine cipher in a chosen plaintext attack?

Assuming the ability to launch Chosen Plaintext Attacks (CPA), how many oracle calls does an attacker need to break the affine cipher? How does it work?

• What happens if the attacker asks the oracle to encrypt $x = 0$? – Thomas Feb 25 '13 at 9:13

The answer is 1 I think because if we apply a chosen plain text attack, then we need at least 2 encrypted alphabets and two plaintext alphabets corresponding to the related encrypted alphabets. Then we will have two variable and two equations.

By solving those you can get the additive key as well as the multiplicative key.

The answer is one. Choose the alphabet as the plaintext message and send it to the oracle.

Assuming I haven't fully answered your homework question yet, your job is to think about what the oracle's reply would look like, and how you might use it to help break other messages.