Week 3 Tutorial Sheet
Symmetric Key Cryptography
IMPORTANT NOTES: Study lecture materials at least 1 hour and prepare Q1–3 prior to the tutorial session. Those questions will be discussed in the tutorial.
1. What is the difference between an unconditionally secure cipher and a computationally secure cipher?
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An encryption scheme is unconditionally secure if the ciphertext generated by the scheme does not contain enough information to determine uniquely the corresponding plaintext, no matter how much ciphertext is available.
An encryption scheme is said to be computationally secure if:
(a) the cost of breaking the cipher exceeds the value of the encrypted information, and (b) the time required to break the cipher exceeds the useful lifetime of the information.
2. What is a monoalphabetic cipher and what is a key in this cipher?
A monoalphabetic substitution cipher maps a plaintext alphabet to a ciphertext alphabet, so that each letter of the plaintext alphabet maps to a single unique letter of the ciphertext alphabet. The key in this cipher is a table that maps each of the letters of the 26 plaintext alphabet a,…,z to their corresponding ciphertext letters.
3. Vigenère cipher is a cipher similar to the one-time pad, uses the values of the letters of a secret key to shift the letters of a plaintext. However, unlike the one-time pad, it repeatedly uses the same short key to encrypt arbitrarily long plaintexts. This is done by repeating the secret key word to create a key which is as long as the message and then adding the values of the letters of the key to the letters of the plaintext. The result will be calculated mod 26 to make sure the ciphertext will also be comprised of letters. Using the Vigenère cipher, encrypt the word explanation using the key leg.
Note: You can use the foloowing values assigned for each letter: ABCDEFGHIJKLM
0 1 2 3 4 5 6 7 8 9 10 11 12 NOPQRSTUVWXYZ
13 14 15 16 17 18 19 20 21 22 23 24 25
𝐶=(𝑃+𝐾) mod26
K: legleglegle 11 4 6 11 4 6 11 4 6
P: explanation 4 23 15 11 0 13 0 19 8 C:PBVWETLXOZR15 1 21224191123142517
4. Block cipher: The SubBytes in AES round operation is performed as follows. Write the input byte by byte in hexadecimal. For each byte represented in hexadecimal, use the first hexadecimal digit to select the row and uses the second hexadecimal digit to select the column, then replace the input by the corresponding byte from the SBox specified by Table 1. For example, if the input byte is 53 in hexadecimal, then the SubBytes result is ED in hexadecimal. Given the input EA 36 56 78 in hexadecimal, write the result of SubBytes in hexadecimal.
11 4 14 13
FIT2093 Week 3 Tutorial Sheet
SubBytes(EA 36 56 78) = 87 05 B1 BC
5. CBC Block Cipher Mode: For the CBC block cipher mode shown in Figure 6.4:
(a) Identify which decrypted plaintext blocks 𝑃𝑥 will be corrupted if there is an error in the
received ciphertext block 𝐶4.
(b) Assuming that the ciphertext contains 𝑁 blocks, and there was a bit-flip error in the sender
side plaintext block 𝑃3, identify through how many ciphertext blocks this error is propagated.
(a) The question assumes that there was an error in block 𝐶4 of the transmitted ciphertext. From Fig. 6.4, ciphertext block 𝐶𝑖 is used as input to the XOR function when obtaining
plaintext blocks 𝑃𝑖 and 𝑃𝑖+1. Therefore, a transmission error in block 𝐶4 will corrupt blocks 𝑃4 and 𝑃5 of the decrypted plaintext, but will not propagate to any of the other blocks.
FIT2093 Week 3 Tutorial Sheet
(b) The question assumes that the ciphertext contains 𝑁 blocks, and that there was a bit error in the source version of 𝑃3.
From Fig. 6.4, ciphertext block 𝐶𝑖 is generated by XORing plaintext block 𝑃𝑖 with ciphertext block 𝐶𝑖−1. Therefore, a bit error in block 𝑃3 will affect ciphertext block 𝐶3, which in turn will affect ciphertext block 𝐶4 and so forth, and therefore the error will propagate through all remaining ciphertext blocks. Thus, 𝑁 − 2 ciphertext block will be corrupted.
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