Find the inverse of 13 (mod 51) using extended euclidean algorithm.
To find the inverse of 13 modulo 51 using the Extended Euclidean Algorithm, we need to solve for \( x \) in the equation:
\[
13x \equiv 1 \mod 51
\]
This is equivalent to finding \( x \) such that \( 13x - 51k = 1 \), where \( k \) is an integer. We can apply the Extended Euclidean Algorithm to solve this.
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Apply the frequency test, serial test and autocorrelation test to the following sequence at level of significance α = 0.05:
011001110000110010011100.
Let f (x) = x4 + x + 1 ∈ F2[x]. We represent the field F24 by F2[x]/( f (x)). Let us write γ = x + ( f (x)). The table of values is given below:
| i | γ i | Vector | i | γ i | Vector |
| 0 1 2 3 4 5 6 7 | 1 γ γ 2 γ 3 γ +1 γ 2 +γ γ 3 +γ 2 γ 3 +γ +1 | (0,0,0,1) (0,0,1,0) (0,1,0,0) (1,0,0,0) (0,0,1,1) (0,1,1,0) (1,1,0,0) (1,0,1,1) | 8 9 10 11 12 13 14
| γ 2 +1 γ 3 +γ γ 2 +γ +1 γ 3 +γ 2 +γ γ 3 +γ 2 +γ +1 γ 3 +γ 2 +1 γ 3 +1
| (0,1,0,1) (1,0,1,0) (0,1,1,1) (1,1,1,0) (1,1,1,1) (1,1,0,1) (1,0,0,1)
|
a) Decrypt each of the following cipher texts:
i) Text: "CBBGYAEBBFZCFEPXYAEBB", encrypted with affine cipher with key (7,2).
ii) Text:"KSTYZKESLNZUV", encrypted with Vigenère cipher with key "RESULT".
Find a recurrence that generates the sequence 110110110110110.
b) Alice wants to use the ElGamal digital signature scheme with public parameters secret value a = 7 and β = 34. She wants to sign the message M = 20 and send it to Bob. She chooses k = 5 as the secret value. Explain the procedure that Alice will use for computing the signature of the message. What information will she send Bob
Use Miller-Rabin test to check whether 75521 is a strong pseuodprime to the base 2.
In this exercise, we introduce you to Hill cipher. In this cipher, we convert our message to numbers, just as in affine cipher. However, instead of encrypting character by character, we encrypt pairs of characters by multiplying them with an invertible matrix with co-efficients in Z26.
Here is an example: Suppose we want to ENCRYPT "ALLISWELL". Since we require the plaintext to have even number of characters, we pad the message with the character ‘X’. We break up the message into pairs of characters AL, LI, SW, EL and LX. We convert each pair of characters into a pair elements in Z26 as follows:
Next, we choose an inveritble 2×2 matrix with coefficients in Z26, for example, A = 
This matrix has determinant 
is a unit in Z26 with inverse 
We write each pair of elements in Z26 as a column vector and multiply it by A:
We then convert each pair of numbers to a pair of characters and write them down. In this example, we get the cipher text "LSPFYGXUEN" corresponding to the plain text "ALLWELL".
To decrypt, we convert pairs of characters to pairs of numbers and multiply by
Decrypt the text "TWDXHUJLUENN" which was encrypted using the Hill’s cipher with the matrix 
as the encryption matrix.