Module Three Homework
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Mathematics
Date
Apr 3, 2024
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How M,MAT 260 – Cryptology Dejah Crews-Siler March 17, 2024 Section 5.4: Exercise Problem 5 Section 5.5: Computer Problem 1, 3 Section 6.6: Exercise Problems 2 ,6, 8 Section 6.7: Computer Problems 1 Section 5.4: Exercise Problem 5 Suppose we build an LFSR machine that works mod 3 instead of mod 2. It uses a recurrence of length 2 of the form
𝑥
𝑛+2
≡ ?
0
𝑥
𝑛
+ ?
1
𝑥
𝑛+1
(mod 3)
To generate the sequence 1,1,0,2,2,0,1,1. Set up and solve the matrix equation to find the coefficients ?
0
𝑎𝑛? ?
1
. ANSWER: 1 = [0,1] x [1] + [co, c] x [1] mod3 1 = [0,1] x [1] + [co, c] x [1] mod3
0 = [0,1] x [0] + [co, c] x [0] mod3
2 = [0,1] x [0] + [co, c] x [1] mod3
2 = [0,1] x [2] + [co, c] x [1] mod3
0 = [0,1] x [2] + [co, c] x [0] mod3
1 = [0,1] x [0] + [co, c] x [2] mod3
1 = [0,1] x [1] + [co, c] x [2] mod3
This implies that ?
0
= 1 and ?
1
= 1
Section 5.5: Computer Problem 1, 3 1.
The following sequence was generated by a linear feedback shift register. Determine the recurrence that generated it. 3.
The following sequence ciphertext was obtained by XORing an LSFR output with the plaintext. Suppose you know the plaintext starts Find the plaintext. 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0,
0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0,
0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1,
1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0,
1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1,
1, 1, 1, 1, 1
0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0,
1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0,
1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1
1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0
Section 6.6: Exercise Problems 2 ,6, 8 2.
The matrix mod 26 is not suitable for the matrix in a Hill cipher. Why? 6. Alice uses a Hill cipher with 3 x 3 matrix M that is invertible mod 26. Describe a chosen plaintext attack that will yield the entries of the matrix M. Explicitly say what plaintexts you will use.
- Plaintexts I chose are JKL MNO QRS which converts into plaintexts OPQ LMN IJK. These plaintext attack that were chosen give a specific plaintext which determines the use of the Hill cipher.
8.Suppose the matrix is used for an encryption matrix in a Hill cipher. Find two plaintexts that encrypt to the same ciphertext
. - Plaintext P1 = (
0
0
0
0
) ∙ (
1
2
3
4
)
= mod 26 (
0
0
)
- Plaintext P2 = (13 − 13) ∙ (
1
2
3
4
)
= mod 26 (
13
13
)
= mod 26 = (
0
0
)
- Both plaintext P1 and P2 encrypt to the same ciphertext AA
Section 6.7: Computer Problems 1 The following is the ciphertext of a Hill cipher Using the matrix zirkzwopjjoptfapuhfhadrq
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Decrypt. Proof:
(
25
8
25
22
17
10
14
15
9
9
19
5
14
15
0
15
) ∙ (
1
2
4
3
3
4
2
1
11
2
2
9
4
6
6
4
)
mod 26 = (
264
198
297
279
219
250
265
266
229
208
69
188
191
189
157
141
)
mod 26 = = (
10
7
11
10
8
9
10
10
8
8
2
7
7
7
6
5
)
= (
264
198
297
279
219
250
265
266
229
208
69
188
191
189
157
141
)
= (
?
?
?
?
?
?
?
?
?
?
𝐶
?
?
?
?
?
)