Assume you are using any correct plaintext padding method, such as those described in lecture, with a 128-bit (16-byte) block cipher. If you are sending a message that is 184 bytes long, how many padding bytes would you need to add? Justify your answer. 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.
Assume you are using any correct plaintext padding method, such as those described in lecture, with a 128-bit (16-byte) block cipher. If you are sending a message that is 184 bytes long, how many padding bytes would you need to add? Justify your answer. 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.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
- Assume you are using any correct plaintext padding method, such as those described in lecture, with a 128-bit (16-byte) block cipher. If you are sending a message that is 184 bytes long, how many padding bytes would you need to add? Justify your answer.
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.
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