Question

Consider a simple system with 8-bit block size. Assume the encryption (and decryption) to be a simple XOR of the key with the input. In other words, to encrypt x with k we simply perform $x \oplus k$ giving y. Similarly, to decrypt y, we perform $y \oplus k$ giving x. You are given the following 16-bit input ABCD in hexadecimal. You are provided IV as EF in hexadecimal. The key to be used (where appropriate) is BB in hexadecimal. Answer the following to compute the encrypted output using CBC. $y_1 = (x_1 \oplus IV) \oplus k = $ FF $y_2 = (x_2 \oplus y_1) \oplus k = $ 89 Output y = FF89

          Consider a simple system with 8-bit block size. Assume the encryption (and decryption) to be a simple XOR of the key with the input. In other words, to encrypt x with k we simply perform $x \oplus k$ giving y. Similarly, to decrypt y, we perform $y \oplus k$ giving x.
You are given the following 16-bit input ABCD in hexadecimal. You are provided IV as EF in hexadecimal. The key to be used (where appropriate) is BB in hexadecimal. Answer the following to compute the encrypted output using CBC.
$y_1 = (x_1 \oplus IV) \oplus k = $ FF
$y_2 = (x_2 \oplus y_1) \oplus k = $ 89
Output y = FF89

Consider a simple system with 8-bit block size. Assume the encryption (and decryption) to be a simple XOR of the key with the input. In other words, to encrypt x with k we simply perform x ⊕ k giving y. Similarly, to decrypt y, we perform y ⊕ k giving x.
You are given the following 16-bit input ABCD in hexadecimal. You are provided IV as EF in hexadecimal. The key to be used (where appropriate) is BB in hexadecimal. Answer the following to compute the encrypted output using CBC.
y1 = (x1 ⊕ IV) ⊕ k = FF
y2 = (x2 ⊕ y1) ⊕ k = 89
Output y = FF89

Added by Tamara L.

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