Key Expansion / Key Schedule
Last updated
Last updated
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 )
bits for AES-128
bits for AES-192
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 :
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 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.
Letβs 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.
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, XORβing with the round constant amounts to XORβing with just its leftmost byte.
The multiplication applied here is the same as in Mix Column operation when multiplying by 2.
As explained before, AES-128 need a total key lenght of 1408 bits ( ).
As each word has a size of 32 bits, ( )44 words are needed.
To serve as the round key for the round, must be a multiple of 4.
These will obviously serve as the round key for the round. For example :
is the beginning of the 4-word group and is obtained by using :
The first word of the new 4-word group is obtained by XORβing the first word of the last group ( ) with the result of a function g() applied to the last word of the previous 4-word group
The function consists of the following 3 steps :
The round constant for the round is noted .
The only non-zero byte in the round constants, , obeys the following recursion: