Compute CBC$-$MAC $($pages $325-526$ in textbook$)$ for a message of $24$ bits, $$A$1$A$2$A$3($in Hexa$).$

Assume a block size of $8$ bits with an IV $=$ D$3($hexa$)$ and key $=$ E$4($hexa$).$

Assume the encryption $($and decryption$)$ to be as follows:

If plaintext is LT$||$RT

Key is LK$||$RK$,$ where LC$,$ RC$,$ LT$,$ and RT are each $4$ bits, then

Ciphertext $=$ LC$||$RC

LC $=$ LK XOR RT

RC $=$ RK XOR LT

Plaintext and ciphertext are each $8$ bits. Similarly, to decrypt ciphertext, we perform exactly the reverse operation

LT $=$ RC XOR RK

RT $=$ LC XOR LK$.$

Hint: Divide the message into blocks of $8$ bits each; XOR each block with the previous cipher output; then encrypt this with the key.

For the first block, XOR it with IV$.$ Details in pages $325-326$ Ch$-12$ of the textbook.