0
$\begingroup$

Question

I am extremely new to cryptography and AES and so I am not confident that I did this question orrectly. Please can you check my answers and let me know if anything jumps out as incorrect, Thank you!

Here is the Question:

Consider the key expansion procedure for AES encryption.

The given four subkeys are: w4 = a0f afe17, w5 = 88542cb1, w6 = 23a33939 and w7 = 2a6c7605 using hexadecimal notation.

(a) Complete the following procedure to generate the next subkey w8.

i. Generate the temporary subkey wt = w______.

ii. Rotate (round-end) the binary sequence wt to the left for 8 positions and obtain wt =______.

iii. Substitute wt byte by byte using Table 1 and obtain wt =_______ .

iv. Generate the round constant r8 =________ for w8.

v. wt = wt ⊕ r8 =_________ .

vi. w8 = wt ⊕ w4 = ________.

My Answer

i) 

wt = w4

   = a0fafe17

ii)

Eight positions in binary = two positions in Hex:

wt = fafe17a0

iii)

wt = adbbf0e0

iv)

r8 = [ rc8 00 00 00]

rc8 = 2*rc7

    = 2*2a

    = 54

r8 = [ 54 00 00 00 ]

v)

wt = wt ⊕ r8

   = f9bbf0e0

vi)

w8 = wt ⊕ w4

   = 59410ef7

S Table

$\endgroup$
  • 1
    $\begingroup$ well, i can tell right off the bat your round constant is wrong, I think it is supposed to be 0x02, and 0x54 is not valid for ANY round $\endgroup$ – Richie Frame Jun 4 at 22:16
  • $\begingroup$ @RichieFrame Thank you! So it should be: r8 = [ 02 00 00 00 ] $\endgroup$ – John Jun 5 at 8:02
  • $\begingroup$ also sbox(0xFA) = 0x2D $\endgroup$ – Richie Frame Jun 5 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.