Question

- Perform Add round Key operation for the below

input =

A1	10	24	57
1F	29	41	DE
6B	BD	E3	6A
A9	25	18	12

key =

A1	13	24	57
1F	29	41	DE
6B	BD	E3	6A
A9	25	18	63

Answer 1

Add round key operation is quite simple, we just need to XOR each byte in the given state to the corresponding byte in the given KEY.

i.e. let assume our given matrix is in the form

state: a0 a1 a2 a3                                 key = A0 A1 A2 A3

         b0 b1 b2 b3                                           B0 B1 B2 B3

         c0 c1 c2 c3                                           C0 C1 C2 C3

         d0 d1 d2 d3                                           D0 D1 D2 D3

and after applying round key operation we will get the resultant matrix of the same form where each byte is the xor of state and key.

Truth table of XOR

A   B   (A XOR B)

0    0        0

0    1        1

1    0        1

1    1        0

so the first byte of resultant matrix (a0) = a0 xor A0, and so on,

let resultant matrix is : A B C D

                                    E F G H

                                    I J K L

                                   M N O P

now we have to calculate each byte of resultant matrix

A = a0 xor A0

   = A1 xor A1

   = 1010 0001 xor 1010 0001 (in binary)

   = 0000 0000 (in binary)

   = 0

B = a1 xor A1

   = 10 xor 13

   = 0001 0000 xor 0001 0011

   = 0000 0011

   = 3

C = a2 xor A2

   = 24 xor 24

   = 0010 0100 xor 0010 0100

   = 0000 0000

   = 0

D = a3 xor A3

   = 57 xor 57

   = 0101 0111 xor 0101 0111

   = 0000 0000

   = 0

E = b0 xor B0

   = 1F xor 1F

   = 0001 1111 xor 0001 1111

   = 0000 0000

   = 0

F = b1 xor B1

   = 29 xor 29

   = 0010 1001 xor 0010 1001

   = 0000 0000

    = 0

G = b2 xor B2

    = 41 xor 41

= 0100 0001 xor 0100 0001

    = 0000 0000

    = 0

H = b3 xor B3

   = DE xor DE

   = 1101 1110 xor 1101 1110

   = 0000 0000

   = 0

I = c0 xor C0

= 6B xor 6B

= 0110 1011 xor 0110 1011

= 0000 0000

= 0

J = c1 xor C1

= BD xor BD

= 1011 1101 xor 1011 1101

= 0000 0000

= 0

K = c2 xor c2

   = E3 xor E3

   = 1110 0011 xor 1110 0011

   = 0000 0000

   = 0

L = c3 xor C3

   = 6A xor 6A

   = 0110 1010 xor 0110 1010

   = 0000 0000

   = 0

M = d0 xor D0

    = A9 xor A9

    = 1010 1001 xor 1010 1001

    = 0000 0000

    = 0

N = d1 xor D1

    = 25 xor 25

    = 0010 0101 xor 0010 0101

    = 0000 0000

    = 0

O = d2 xor D2

    = 18 xor 18

    = 0001 1000 xor 0001 1000

    = 0000 0000

    = 0

P = d3 xor D3

    = 12 xor 63

    = 0001 0010 xor 0110 0011

    = 0111 0001

    = 71

so our solution matrix will be:

resultant matrix: 0    3    0    0

                           0    0    0    0

                           0    0    0    0

                           0    0    0    71