Question

a) Decrypt the ciphertext 101000111001 which was encrypted with the Toy block cipher once
using the key 101010010. Show all the steps. (5)
b) 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.

10 Mar 2024
Answer :
Word Count : 833
Let's break down both parts of your problem step by step: --- ### Part a) Decrypt the ciphertext `101000111001` using the Toy block cipher with the key `101010010`. #### Step 1: Understanding the Toy block cipher The Toy block cipher is a simple block cipher that often uses a fixed key and a series of substitutions or permutations. However, as the exact Toy block cipher algorithm isn’t standard and can vary, we will make assumptions about how the cipher works based on common small ciphers. Let’s assume: - The plaintext and ciphertext are both 12 bits in length. - The Toy cipher works with a substitution-permutation network (SPN) or similar technique. - The key size is 9 bits. In general, block ciphers use a process of substitution (replacing bits based on a key) and permutation (reordering the bits). ________ ______ ______ _________ _______ ___ ______ _____ ____.
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Related Questions

a) Considering the bytes 10001001 and 10101010 as elements of the field F_2[X]/\left \langle g(X) \right \rangle, where g(X) is the polynomial X^{8}+X^{4}+X^{3}+X+1, find their product and quotient.

b) Find a recurrence that generates the sequence 110110110110110.

Alice wants to use the ElGamal digital signature scheme with public parameters p = 47, α = 2, secret value\alpha =7 and \beta =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?

Text: "CBBGYAEBBFZCFEPXYAEBB", encrypted with affine cipher with key (7,2)

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 Z_{26} .

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 Z_{26} as follows:

                       Image ignouassignments-ignouacademy-com--p-ciphertext-36347

Next, we choose an inveritble 2×2 matrix with coefficients in Z_{26}, for example,A=\left [ \frac{3}{7}\, \frac{1}{4} \right ].

This matrix has determinant \bar{3},\bar{4}-\bar{7},\bar{1}=\bar{5}\, and\, \bar{5} and 5 is a unit in  Z_{26} with inverse \bar{21}. We write each pair of elements in Z_{26} as a column vector and multiply it by A:

A\left [ \frac{\bar{0}}{11} \right ]=\left [ \frac{\bar{11}}{18} \right ].A\left [ \frac{\bar{11}}{8} \right ]=\left [ \frac{\bar{15}}{5} \right ]....   

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 A^{1}=\bar{5}^{-1} \left [ \frac{\bar{4}}{-7}\:\, \: \frac{\bar{-1}}{3} \right ]=\bar{21}\left [ \frac{\bar{4}}{-7}\, \, \, \frac{\bar{-1}}{3} \right ]=\left [ \frac{\bar{6}}{9}\: \, \, \frac{\bar{5}}{11} \right ]  and we have 

\left [ \frac{\bar{6}}{9}\, \, \bar{\frac{5}{11}} \right ]\left [ \frac{\bar{11}}{18} \right ]=\left [ \frac{\bar{0}}{18} \right ].... 

Decrypt the text "TWDXHUJLUENN" which was encrypted using the Hill’s cipher with the matrix \left [ \frac{\bar{3}}{0} \, \, \, \frac{\bar1}{9}\right ]  as the encryption matrix

Let f(x)=x^{2}-x-1\in Z_5[x]. We represent the field F_2 by F_2[x]/(f(x)).LEt us write y=x=(f(x)). the table of values is given below:

i y^{i} Vector i y^{i} Vector
0 1 (0,0,0,1) 8 y^{2}+1 (0,1,0,1)
1 y (0,0,1,0) 9 y^{3}+y (1,0,1,0
2 y^{2} (0,0,1,0) 10 y^{3}+y+1 (0,1,1,1)
3 y^{3} (1,0,0,0) 11 y^{3}+y^{2}+y (1,1,1,0)
4 y+1 (1,0,0,0) 12 y^{3}+y^{2}+y+1 (1,1,1,1)
5 y^{2}+y (0,1,1,0) 13 y^{3}+y^{2}+1 (1,1,0,1)
6 y^{3}+y^{2} (1,1,0,0) 14 y^{3}+1 (1,0,0,1)
7 y^{3}+y+1 (1,0,1,1)      

 i) Prepare logarithm and antilogarithm tables as given in page 23 of block 1.

ii) Compute \frac{(y^{4}+y^{2})+(y^{3}+y+1)}{(1+y^{2}+y^{4})(1+y^{3})}\: and\: \frac{y^{2}(y^{2}+y+1)}{(y^{3}+y^{2})(1+y^{5})}  using the logarithm and antilogarithm tables.

Solve the discrete logarithm problem 5^{x}\equiv 22 (mod 47) using Baby-Step, Giant-Step algorithm.

Decrypt the message c = 23 that was encrypted using RSA algorithm with e = 43 and n = 77.

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