/cAES

A really bad implementation of the AES block cipher in C. Written by someone that doesn't know C😕

Primary LanguageC

cAES

A really bad implementation of the AES block cipher in C. Written by someone that doesn't know C😕

🔷 What is AES?

Advanced Encryption Standard (AES) is a symmetric block cipher. Here is a high-level overview of how it works (spcifically in the implementation I do here, other ways exist):

  1. Key expansion - 11 keys are generated from the original secret key and the syncornization. I havn't implemented this yet so I can't go into too much details (becausue I don't know too much details...).

  2. Initial key addition - the bits of the message are XORed with the bits of the first generated round key.

  3. Loop 9 times:

    1. Substitude bytes - All the bytes of the message are substituted according to the s-box lookup table. The byte b would be replaced by s according to the formula:
    s = b ^ (b <<< 1) ^ (b <<< 2) ^ (b <<< 3) ^ (b <<< 4) ^ 0x63

    Where ^ is the bit-wise XOR operation, and <<< is the left circular bit-shift.

    The inverse operation to this is given by

    b = (s <<< 1) ^ (s <<< 3) ^ (s <<< 6) ^ 0x05
    1. Shift rows - The bytes of msg are shaped into a 4 by 4 grid. At this stage, each row is shifted left (circuraly) by one byte more than the row above it. So the first row is un-changed, the second row is shifted by one byte to the left, the third by 2 bytes, and the forth by 3 bytes.
    2. Mix columns - Each column of the message grid is considered as a column vector. We multiply that column vector by some matrix (in the Rijndael's Galois field) to get a new column to put in-place of the old one. The matrix is (I wish there was latex support on github...)
    2 3 1 1
    1 2 3 1
    1 1 2 3
    3 1 1 2

    What does it mean to be under the Rijndael's Galois field you ask? Well, I won't cover it in depth here (check the official specification chapter 4) but here is a brief explenation.

    Think of a byte comprised of 8 bits, let's say 10010101 for example. Now think of each digit as the coefficient of a polynomial, the polynomial we get is (again, I wish I had latex math rendering here, but OK):

    P(x) = 1x^7 + 0x^6 + 0x^5 + 1x^4 +0x^3 +1x^2 + 0x + 1

    See how the bits of the byte come in the polynomial? Now this is our byte. If we want to add two bytes, we do so just like we would two polynomials (element-wise), but with one small twist. Instead of regular addition, we use the XOR operation, so 1 + 1 is not 2, it's 0 (you can think of it like addition modulo 2).

    Multiplication is the interesting part, because we need to multiply all the elements by one another (we'll look at an example later). You may also ask yourself, what happens with stuff like x^7 times x^7, it should be x^14 but we only have 8 bits in our byte, so what would we do? The answer is that we do all the multiplication modulo (x^4 - 1), which I won't go into now (again, check the specifications! chapter 4).

    Anyways, just remember that when you want to do something like 5*3 remember that that's not 15! that should be 00000101 times 00000011 which is equivilent to the mulitplication of the polynomials:

    5 = 00000101 = x^2 + 1

    and

    3 = 00000011 = x + 1

    which if you carry out you would get

    x^3 + x^2 = 00000110 = 6

    So 5 times 3 is equal to 6 (weird, I know).

    By the way, the inverse of the mix columns matrix is
    14 11 13 9
    9 14 11 13
    13 9 14 11
    31 13 9 14

    If you were wondering...

    1. Add key - The message is XORed with another round key (different key each iteration, generated from the key expansion step).

  4. Final round (making 10 in total). The same as before but without the mix columns step.

    1. Substitude bytes
    2. Shift rows
    3. Add key