For the remaining questions, consider a 4-bit block cipher, described in hexadecimal by the following table:
|
Plaintext
|
Ciphertext
|
Plaintext
|
Ciphertext
|
|
0
|
a
|
8
|
e
|
|
1
|
c
|
9
|
d
|
|
2
|
f
|
a
|
0
|
|
3
|
6
|
b
|
7
|
|
4
|
3
|
c
|
5
|
|
5
|
8
|
d
|
b
|
|
6
|
4
|
e
|
9
|
|
7
|
2
|
f
|
1
|
You can think of this as a simple substitution cipher for hexadecimal digits. There is no "key" other than the table itself.
For this question, you will perform encryption and decryption using the same cipher described above, but in CTR mode.
Encrypt the message face using the above block cipher in CTR mode, with a one-bit nonce of 1 and three bits of a counter that starts from 0.
Decrypt the message 46e2 using the above block cipher in CTR mode, with a one-bit nonce of 1 and three bits of a counter that starts from 0.
Was this easier or more difficult than encrypting/decrypting in CBC mode? Do you suspect any difference in security? Why or why not?