Answered Essay: The number is: 01000011 01001001 01010000 01001000 01000101 01010010 01010100 0

Hi, i am having trouble with the following AES question in my encryption class. Can you help me?

2. [40 pts] For the given plaintext, what would be the state (intermediate cipher) after the first round of AES? Show the value of Co. A ciphertext block right after the first key addition layer (01000011 01001001 01010000 01001000 01000101 01010010 01010100 01000101

The number is: 01000011 01001001 01010000 01001000 01000101 01010010 01010100 01000101 01011000 01010100 01111111 01000010 01001100 01001111 01000011 01001011

Any help is appreciated thank you.

2. [40 pts] For the given plaintext, what would be the state (intermediate cipher) after the first round of AES? Show the value of Co. A ciphertext block right after the first key addition layer (01000011 01001001 01010000 01001000 01000101 01010010 01010100 01000101

Expert Answer

 

In BASIC, where BYTE is each decimal byte value:

HI=INT(BYTE/16)
LOW=BYTE-HIGH*16
HINYBBLE$=BINARY$(HI*4+1)
LOWNYBBLE$=BINARY$(LOW*4+1)

Have BINARY$ a string of length 64 containing the text “0000 0001 0010 0011” etc (without spaces).

In Assembler, it’s a bit easier. Just a loop with LSR BYTE, then output a 0 or 1 to the text file depending on the Carry status.

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