Question

Explain how Bob will decrypt the message. 
9) a) Solve the discrete logarithm problem equation using Baby-Step, Giant-Step algorithm. 

09 Jan 2026
Answer :
Word Count : 722
We are asked to solve (5^x \equiv 22 \pmod{47}) using the Baby-Step Giant-Step (BSGS) algorithm. Step 1: Determine (m = \lceil \sqrt{p-1} \rceil), where (p = 47). [ p-1 = 46, \quad \sqrt{46} \approx 6.78 \implies m = 7 ] Step 2: Compute the baby steps (5^j \mod 47) for (j = 0, 1, \dots, m-1). * (j=0: 5^0 \equiv 1 \pmod{47}) * (j=1: 5^1 \equiv 5 \pmod{47}) * (j=2: 5^2 \equiv 25 \pmod{47}) * (j=3: 5^3 \equiv 5^2 \cdot 5 \equiv 25 \cdot 5 = 125 \equiv 31 \pmod{47}) (since (125 - 2\cdot47 = 125-94=31)) * (j=4: 5^4 \equiv 31 \cdot 5 = 155 \equiv 155 - 3\cdot47 = 155-141=14 \pmod{47}) * (j=5: 5^5 \equiv 14 \cdot 5 = 70 \equiv 70-47=23 \pmod{47}) * (j=6: 5^6 \equiv 23 \cdot 5 = 115 \equiv 115 - 2\cdot47 = 115-94=21 \pmod{47}) Baby steps table: [ ______ _____ ____ _____ ____ ______ ________ ___ __________ _________ __________.
____ _______ ___ ________ ____ _______ _____ ______ ______ ______.
_____ ______ __________ ____ _____ _________ ____ __________ _______.
____ ____ ______ ___ __________ _________ ___ __________ _________ __________.
________ _____ ___ _______ ________ _________ ________ _________ ____ _________ ________.
__________ _______ ___ __________ ______ _______ ___.
_____ ____ ______ ________ _________ ____ ____ _________ ______ __________ _____ _____.
_______ ______ __________ __________ ___ ________ ________ ____.
______ ______ ______ ____ _________.
____ ___ ______ _______ _________ _____ _________ ____ __________.
_______ _________ ________ _________ ______ _______ ___ _________ ___ _______.
_________ ___ _________ ___ ________ ____ _________ ________ ____ __________.
______ ________ __________ ________ _____.
__________ ____ ____ ___ ______ __________ ___ ____ _____.
____ _______ ________ _________ __________ ________ ___ _______ _____.
_______ ____ ____ ___ _______.
____ ________ ________ ___ ______ ______ ________ ____.
_________ _____ ______ _______ _________ ___ __________ _______ _______.
____ __________ _______ _________ _____ ______.
____ __________ _______ _________ ______ ______ ____ _______ _____ _________.
__________ ___ ______ ______ ___ ______.
__________ ____ ____ ______ ___ __________ ______ _____ ___.
__________ ________ ______ _________ ________ __________ ____ _____.
______ _______ ___ _________ ______ _________ _______ ______ ________ ________ _____.
______ ______ ______ _____ ______ __________ ___ _______ _______ _______ ____ ____.
________ ______ _____ _______ __________ ___ _________ ___ _________ _______ _____.
_______ __________ _______ ________ _____ ______ ___ ____.
____ _______ ______ ___ _____ _________ _____.
__________ __________ ___ _________ _____ _________.
_______ ___ ____ _______ _____.
______ _________ __________ _________ ________ ______ _________ ____ _______.
_____ ______ __________ ________ _____ ___.
______ _________ ___ ________ _______ ______ _________ _________ __________ __________ _____ _______.
_____ ______ ____ ________ ____.
_____ _____ _____ _____ _____ ________ ____ __________ ___ ___.
______ __________ _________ ______ __________ ______ _______ ________ _________.
__________ __________ _________ _______ ________ ____ _____ _____ _____.
__________ __________ _______ ______ ______ _________.
___ _____ ___ ________ __________ _________ _________ _____ ______.
____ ___ _______ ______ ___.
________ ______ ________ ____ _________ _________.
______ ___ _____ _________ _____ ___ _________ ________ ______ _____ ________.
__________ _______ ___ _______ ____.
____ _________ _________ _________ _________ ________ _____ ________.
_________ ______ __________ _____ __________ _____ ________.
__________ ____ ________ ____ ______ ________ __________ ___ ________ _________ _____ _____.
__________ ________ _______ __________ ______ _______ ___.
_________ __________ _______ ______ _______ ________ _______ ________ ________ ____ _________.
______ ___ __________ ___ _________ _____ __________.
______ ______ ___ _______ _________ ____ _______ _____.
_____ ______ _____ ____ _________ _______ ______ ____ _______ __________ _______ ___.
____ _____ ___ ___ _________ _______ ______ _____ _______ _________ ______ _________.
___ _______ _________ ________ _____ _________ ______ ________ ___ ______ ______ ______.
______ ______ ___ ____ _______ ___ _______ ______ ___ _________ ______ __________.
______ _____ ______ _________ ______ _____ _________.
________ __________ _____ _____ ___ ____ _______ _____.
_________ _____ _______ _____ __________ _________ ____ _____.
_________ ______ __________ ___ ________ _________.
________ _______ ______ ______ ______ __________ ______ ______ _________.
________ ______ _____ _________ ____ ___ ______.
________ _______ ___ __________ ____ ___ ____ _________.
____ _________ ____ ____ __________ _______ _______ _________ ________ ________ ____.
____ ________ _________ ______ _________ _________ _________ __________ ____ _____ ________ ___.
__________ _______ ________ _________ ____ ______ ______ _______ _____ ______ ____ ___.
________ ___ ______ _____ _________ ____ ______.
________ ___ _________ _________ _________ ________.
_____ ____ ___ ___ _________ _____ ____.
______ _____ ____ __________.
Get Full Answer on WhatsApp
Related Questions

 

Find the keys for the second round.
6) a) Considering the bytes 10001001 and 10101010 as elements of the field equation, where g(X) is the polynomial X8 + X4 + X3 + X + 1, find their product and quotient. 

 Decrypt the message equation that was encrypted using RSA algorithm with equation and equation. 

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". 
   b) 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.

 

 Alice wants to send Bob the message equation. She chooses equation. How will she compute the cipher text? What information does she send to 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:

 

equation\underline{\overline{\begin{aligned}AL&\quad(0,11)\\LI&\quad(11,8)\\SW&\quad(18,22)\\EL&\quad(4,11)\\LX&\quad(11,23)\end{aligned}}}">

 

Next, we choose an invertible equation matrix with coefficients in equation, for example, equation.
This matrix has determinant equation and equation is a unit in equation with inverse equation. We write each pair of elements in equation as a column vector and multiply it by A:
equation
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 equation and we have
equation
Decrypt the text "TWDXHUJLUENN" which was encrypted using the Hill’s cipher with the matrix equation as the encryption matrix.

) Let equation. We represent the field equation by equation. Let us write equation. The table of values is given below:
 

 


i
 

  


γi
 

 Vector  


i
 

  


γi
 

 Vector
0  


1
 

  


(0,0,0,1)
 

 8  


γ2+1
 

  


(0,1,0,1)
 
1  


γ
 

  


(0,0,1,0)
 

 9  


γ3+γ
 

  


(1,0,1,0)
 
2  


γ2
 

  


(0,1,0,0)
 

 10  


γ2+γ+1
 

  


(0,1,1,1)
 
3  


γ3
 

  


(1,0,0,0)
 

 11  


γ3+γ2+γ
 

  


(1,1,1,0)
 
4  


γ+1
 

  


(0,0,1,1)
 

 12  


γ3+γ2+γ+1
 

  


(1,1,1,1)
 
5  


γ2+γ
 

  


(0,1,1,0)
 

 13  


γ3+γ2+1
 

  


(1,1,0,1)
 
6  


γ3+γ2
 

  


(1,1,0,0)
 

 14  


γ3+1
 

  


(1,0,0,1)
 
7  


γ3+γ+1
 

  


(1,0,1,1)
 

      

 

 i) Prepare logarithm and antilogarithm tables as given in page 23 of block 1. 
    ii) Compute equation and equation using the logarithm and antilogarithm tables

IGNOU Solved Assignments Assignments Ignou Projects Project Previous Year Solved Exam Papers Notes Notes Sample Question Papers Sample Question Papers
Login Cart
IGNOU NEWS
Assignment Submission Last Date Extended Till 30 June 2026 Click Here★★★IGNOU June 2026 TEE Date Sheet Released Click Here★★★
Top
📞
Call Support Instant phone assistance Explain how Bob will decrypt the message. 9) a) Solve the
→
🟢
WhatsApp Chat Fast live messaging
→
Email Us Business enquiries & support
→