Java Python CSS 322 – Security and Cryptography
Simplified AES
Example
1 Simplified AES Example
Lets assume the inputs for the encryption are:
• 16-bit Plaintext, P: 1101 0111 0010 1000
• 16-bit Key, K: 0100 1010 1111 0101
1.1 Key Generation
The first step is to generate the sub-keys. This is called Key Generation or Key Expansion:
The input key, K, is split into 2 words, w0 and w1:
w0 = 0100 1010
w1 = 1111 0101
The first sub-key, Key0, is in fact just the input key: Key0 = w0w1 = K
The other sub-keys are generated as follows:
w2 = w0 XOR 10000000 XOR SubNib(RotNib(w1))
(Note: RotNib() is “rotate the nibbles”, which is equivalent to swapping the nibbles)
= 0100 1010 XOR 10000000 XOR SubNib(0101 1111)
(Note: SubNib() is “apply S-Box substitution on nibbles using encryption S-Box”)
= 1100 1010 XOR SubNib(0101 1111)
= 1100 1010 XOR 0001 0111
= 1101 1101
w3
= w2 XOR w1
= 1101 1101 XOR 1111 0101
= 0010 1000
w4
= w2 XOR 0011 0000 XOR SubNib(RotNib(w3))
= 1101 1101 XOR 0011 0000 XOR SubNib(1000 0010)
= 1110 1101 XOR 0110 1010
= 1000 0111
w5
= w4 XOR w3
= 1000 0111 XOR 0010 1000
= 1010 1111
Now the sub-keys are:
Key0 = w0w1
= 0100 1010 1111 0101
Key1
= w2w3
= 1101 1101 0010 1000
Key2
= w4w5
= 1000 0111 1010 1111
1.2 Encryption
Now let’s do the encryption. There is an initial operation (Add Round Key), followed by the main Round, followed by the final Round. (Note, the main difference in the real DES is that the main Round is repeated many times).
Remember, the output of each operation is used as the input to the next operation, always operating on 16-bits. The 16-bits can be viewed as a state matrix of nibbles.
1.2.1 Add Round 0 Key
Plaintext XOR Key1
= 1101 0111 0010 1000 XOR
0100 1010 1111 0101
= 1001 1101 1101 1101
1.2.2 Round 1
CSS 322 – Security and CryptographyMatlab Nibble Substitution (S-boxes). Each nibble in the input is used in the Encryption S-Box to generate an output nibble.
Input = 1001 1101 1101 1101
utput = 0010 1110 1110 1110
Shift Row. Swap 2nd nibble and 4th nibble (note, in this example, its not so easy to see since 2nd and 4th nibbles are the same!)
= 0010 1110 1110 1110
Mix Columns. Apply the matrix multiplication with the constant matrix, Me, using GF(24 ). For GF(24 ), the addition operation is simply an XOR, and for the multiplication operation you can use a lookup table.
Me = 1 4
4 1
S = 0010 1110 = S00’ S01’
1110 1110 S10’ S11’
S’ = Me x S
S00’
= 0010 XOR (4 x 1110)
= 0010 XOR (4 x E)
= 0010 XOR D
= 0010 XOR 1101
= 1111
S10’
= (4 x 0010) XOR 1110
= 1000 XOR 1110
= 0110
S01’
= 1110 XOR (4 x 1110)
= 1110 XOR (4 x E)
= 1110 XOR 1101
= 0011
S11’
= (4 x 1110) XOR 1110
= 1101 XOR 1110
= 0011
utput = S00’ S10’ S01’ S11’
= 1111 0110 0011 0011
Add Round 1 Key.
= 1111 0110 0011 0011 XOR
1101 1101 0010 1000
= 0010 1011 0001 1011
1.2.3 Final Round
Nibble Substitution (S-boxes)
= 1010 0011 0100 0011
Shift Row (2nd and 4th)
= 1010 0011 0100 0011
Add Round 2 Key
1010 0011 0100 0011 XOR
1000 0111 1010 1111
= 0010 0100 1110 1100
Now we have the final ciphertext.
Ciphertext = 0010 0100 1110 1100
1.3 Decryption
Now lets decrypt. Note that we use the same keys generated during the encryption (that is, the decryptor would generate the round sub-keys using the input key K, using the encryption S-Box)
