Decrypt the message c = 23 that was encrypted using RSA algorithm with e = 43 and n = 77.
See Answer →Apply runs test to the following sequence:
1001101000010000101111011
0111010010110110010011010
0110011100001100100111000
1100001101010111101001110
0010001111000001101010010
1000110100000110100101101
1110001001
See Answer →Apply poker test to the following sequence with level of significane α = 0.05. (4) 1001101000010000101111011 01110100101101100100110.
See Answer →Apply the frequency test, serial test and autocorrelation test to the following sequence at level of significance α = 0.05:
011001110000110010011100.
See Answer →Find a recurrence that generates the sequence 110110110110110.
See Answer →Considering the bytes 10001001 and 10101010 as elements of the field F2[X]/hg(X)i, where g(X) is the polynomial X 8 +X 4 +X 3 +X +1, find their product and quotient.
See Answer →A 64 bit key for the DES is given below
11000111 10000101
11110111 11000001
11111011 10101011
10011101 10010001
i) Check whether the key is error free using the parity bits.
ii) Find the keys for the second round
See Answer →Decrypt the ciphertext 101000111001 which was encrypted with the Toy block cipher once using the key 101010010. Show all the steps.
See Answer →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.
Use Miller-Rabin test to check whether 75521 is a strong pseuodprime to the base 2.
See Answer →Find the inverse of 13 (mod 51) using extended euclidean algorithm.
See Answer →Another version of the columnar transposition cipher is the cipher using a key word. In this cipher, we encrypt as follows: Given a key word, we remove all the duplicate characters in the key word. For example, if the key word is ‘SECRET’, we remove the second ‘E’ and use ‘SECRT’ as the key word. To encrypt, we form a table as follows: In the first row, we write down the key word. In the following rows, we write the plaintext. Suppose we want to encrypt the text ‘ATTACKATDAWN’. We make a table as follows:
| S | E | C | R | T |
| A | T | T | A | C |
| K | A | T | D | A |
| W | N | X | X | X |
Then we read off the columns in alphabetical order. We first read the column under ‘C’, followed by the columns under ‘E’, ‘R’, ‘S’ and ‘T’. We get the cipher text TTX TAN ADX AKW CAX. To decrypt, we reverse the process. Note that, since we know the length of the keyword, we can find the length of the columns by dividing the length of the message by the length of the keyword.
Given the ciphertext ‘HNDWUEOESSRORUTXLARFASUXTINOOGFNEGASTORX’ and the key word ‘LANCE’, find the plaintext.
See Answer →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"
See Answer →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)
|
i) Prepare logarithm and antilogarithm tables as given in page 23 of block 1.
ii) Compute (γ 4+γ 2 )+ γ 3+γ+1/(1+γ 2+γ 4)(1+γ 3) and γ 2 (γ 2+γ+1) / (γ 3+γ 2)(1+γ 5) using the logarithm an
antilogarithm tables.
See Answer →Let f (x) = x3 − x − 1 ∈ Z5[x]. Find the product of x2 + 2x + 1 + ( f (x)) and x2 + 3x − 1 + ( f (x)) using the algorithm in page 23, block 1. You should show all the steps as in example 11, page 22, block 1.
See Answer →प्रकट और अन्तर्हित भूमिकाओं के बीच अंतर कीजिए।
See Answer →सामाजिक नेटवर्क क्या है?
See Answer →सामाजिक संस्था की अवधारणा की व्याख्या कीजिए।
See Answer →