# Key Expansion / Key Schedule

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Each round use its own round key into the "**Add Key**" step that is derived from the original encryption key.

So, the Key Schedule will allow creating a different key for each encryption round, each of these sub-keys being 128 bits

The entire key length is then 128 times n+1 ( where n is the amount of round, depending of the original key size )

$128 \cdot 11 = 1408$ bits for AES-128

$128 \cdot 13 = 1664$ bits for AES-192

$128 \cdot 15 = 1920$ bits for AES-256

To make the key expansion, the original key is divided into 32 bits blocks called `words`

Here is 4 words for AES-128, 6 for AES-192 and 8 for AES-256

The next words are calculated following these graph :

Algorithm steps

The following explanation will be based on AES-128. Some little ajustement ( such as the amount of generated words ) have to be done to make it applicable to the others key size.

The first four words group

The first four words are provided by the original key.

W0 =

`key[0:31]`

W1 =

`key[32:63]`

W2 =

`key[64:95]`

W3 =

`key[96:127]`

Here, we have the four words of the key used for the round 0 ( before the 10 loops )

`w4`

,`w5`

,`w6`

,`w7`

is the round key for round 1`w8`

,`w9`

,`w10`

,`w11`

the round key for round 2,and so on.

The others words groups

Let’s say that we have the four words of the round key for the i th round:

And we need to determine the words

Using the Figure 1, we can write :

Note that except for the first word in a new 4-word grouping, each word is an XOR of the previous word and the corresponding word in the previous 4-word grouping.

The first word of each groups

The g() function

Perform a one-byte left circular rotation on the argument 4-byte word.

Perform a byte substitution for each byte of the word using the same "S-box" in the SubBytes step of the encryption rounds

XOR the bytes obtained from the previous step with a round constant.

The round constant is a word whose three rightmost bytes are always zero.

Therefore, XOR’ing with the round constant amounts to XOR’ing with just its leftmost byte.

The multiplication applied here is the same as in Mix Column operation when multiplying by 2.

Python implementation

As explained before, AES-128 need a total key lenght of 1408 bits ( $11 \cdot 128$ ).

As each word has a size of 32 bits, ( $\frac{1408}{32} = 44$ )44 words are needed.

To serve as the round key for the $i^{th}$ round, $i$ must be a multiple of 4.

These will obviously serve as the round key for the $i/4^{th}$ round. For example :

$w_i \; w_{i+1} \; w_{i+2} \; w_{i+3}$

$w_{i+4} \; w_{i+5} \; w_{i+6} \; w_{i+7}$

$w_{i+5} = w_{i+4} \oplus w_{i+1} \\ w_{i+6} = w_{i+5} \oplus w_{i+2} \\ w_{i+7} = w_{i+6} \oplus w_{i+3} \\$

$w_{i+4}$ is the beginning of the 4-word group and is obtained by using :

$w_{i+4} = w_i \oplus g(w_{i+3})$

The first word of the new 4-word group is obtained by XOR’ing the first word of the last group ( $w_i$ ) with the result of a function g() applied to the last word of the previous 4-word group

The function $g()$consists of the following 3 steps :

The **round constant** for the $i^{th}$round is noted $Rcon[i]$.

$Rcon[i] = (RC[i], 0 , 0 ,0 )$

The only non-zero byte in the round constants, $RC[i]$, obeys the following recursion:

$RC[1] = 0x01 \\ RC[i] = 0x02 \cdot RC[i-1]$