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5. Given the plaintext {000102030405060708090A0B0C0D0E0F) and the key {01010101010101010101010101010101}: a. Show the origina

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Answer #1

As Given :

The plaintext is {000102030405060708090A0B0C0D0E0F} and

The key is {01010101010101010101010101010101}

a. Show the original contents of State, displayed as a 4x4 matrix.

Answer :

\begin{bmatrix} &00 &04 &08 &0C\\ &01 &05 &09 &0D \\ &02 &06 &0A &0E\\ &03 &07 &0B &0F \end{bmatrix}

b. Show the value of State after initial AddRoundKey.

Answer :

Key = \begin{bmatrix} &01 &01 &01 &01\\ &01 &01 &01 &01 \\ &01 &01 &01 &01\\ &01 &01 &01 &01 \end{bmatrix}

Now, State \oplus Key is :

\begin{bmatrix} &00 &04 &08 &0C\\ &01 &05 &09 &0D \\ &02 &06 &0A &0E\\ &03 &07 &0B &0F \end{bmatrix}   \oplus   \begin{bmatrix} &01 &01 &01 &01\\ &01 &01 &01 &01 \\ &01 &01 &01 &01\\ &01 &01 &01 &01 \end{bmatrix} = \begin{bmatrix} &01 &05 &09 &0D \\ &00 &04 &08 &0C\\ &03 &07 &0B &0F\\ &02 &06 &0A &0E \end{bmatrix}

c. Show the value of State after SubBytes.

Answer :

\begin{bmatrix} &01 &05 &09 &0D \\ &00 &04 &08 &0C\\ &03 &07 &0B &0F\\ &02 &06 &0A &0E \end{bmatrix}  \rightarrow\begin{bmatrix} &7C &6B &01 &D7\\ &63 &F2 &30 &FE \\ &7B &C5 &2B &76\\ &77 &6F &67 &AB \end{bmatrix}

d. Show the value of State after ShiftRows

Answer :

\begin{bmatrix} &7C &6B &01 &D7\\ &63 &F2 &30 &FE \\ &7B &C5 &2B &76\\ &77 &6F &67 &AB \end{bmatrix}  \rightarrow\begin{bmatrix} &7C &6B &01 &D7\\ &F2 &30 &FE &63 \\ &2B &76 &7B &C5\\ &AB &77 &6F &67 \end{bmatrix}

e. Show the value of State after MixColumns

Answer :

\begin{bmatrix} &02 &03 &01 &01\\ &01 &02 &03 &01 \\ &01 &01 &02 &03\\ &03 &01 &01 &02 \end{bmatrix}\begin{bmatrix} &7C &6B &01 &D7\\ &F2 &30 &FE &63 \\ &2B &76 &7B &C5\\ &AB &77 &6F &67 \end{bmatrix} =  \begin{bmatrix} &74 &E7 &0F &A2\\ &55 &E6 &04 &22 \\ &3E &2E &B8 &8C\\ &F6 &15 &58 &0B \end{bmatrix}


answered by: ANURANJAN SARSAM
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Answer #3

As Given :

The plaintext is {000102030405060708090A0B0C0D0E0F} and

The key is {01010101010101010101010101010101}

a. Show the original contents of State, displayed as a 4x4 matrix.

Answer :

\begin{bmatrix} &00 &04 &08 &0C\\ &01 &05 &09 &0D \\ &02 &06 &0A &0E\\ &03 &07 &0B &0F \end{bmatrix}

b. Show the value of State after initial AddRoundKey.

Answer :

Key = \begin{bmatrix} &01 &01 &01 &01\\ &01 &01 &01 &01 \\ &01 &01 &01 &01\\ &01 &01 &01 &01 \end{bmatrix}

Now, State \oplus Key is :

\begin{bmatrix} &00 &04 &08 &0C\\ &01 &05 &09 &0D \\ &02 &06 &0A &0E\\ &03 &07 &0B &0F \end{bmatrix}   \oplus   \begin{bmatrix} &01 &01 &01 &01\\ &01 &01 &01 &01 \\ &01 &01 &01 &01\\ &01 &01 &01 &01 \end{bmatrix} = \begin{bmatrix} &01 &05 &09 &0D \\ &00 &04 &08 &0C\\ &03 &07 &0B &0F\\ &02 &06 &0A &0E \end{bmatrix}

c. Show the value of State after SubBytes.

Answer :

\begin{bmatrix} &01 &05 &09 &0D \\ &00 &04 &08 &0C\\ &03 &07 &0B &0F\\ &02 &06 &0A &0E \end{bmatrix}  \rightarrow\begin{bmatrix} &7C &6B &01 &D7\\ &63 &F2 &30 &FE \\ &7B &C5 &2B &76\\ &77 &6F &67 &AB \end{bmatrix}

d. Show the value of State after ShiftRows

Answer :

\begin{bmatrix} &7C &6B &01 &D7\\ &63 &F2 &30 &FE \\ &7B &C5 &2B &76\\ &77 &6F &67 &AB \end{bmatrix}  \rightarrow\begin{bmatrix} &7C &6B &01 &D7\\ &F2 &30 &FE &63 \\ &2B &76 &7B &C5\\ &AB &77 &6F &67 \end{bmatrix}

e. Show the value of State after MixColumns

Answer :

\begin{bmatrix} &02 &03 &01 &01\\ &01 &02 &03 &01 \\ &01 &01 &02 &03\\ &03 &01 &01 &02 \end{bmatrix}\begin{bmatrix} &7C &6B &01 &D7\\ &F2 &30 &FE &63 \\ &2B &76 &7B &C5\\ &AB &77 &6F &67 \end{bmatrix} =  \begin{bmatrix} &74 &E7 &0F &A2\\ &55 &E6 &04 &22 \\ &3E &2E &B8 &8C\\ &F6 &15 &58 &0B \end{bmatrix}

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