c) Alice wants to use the Digital Signature algorithm for signing messages. She chooses p = 83, q = 41, g = 2 and a = 3. Alice wants to sign the message M = 20. She chooses the secret value k = 8. Explain the procedure that Alice will use for computing the signature. What information will she send Bob?
We are given the following parameters for the Digital Signature Algorithm (DSA): * $p = 83$ (a prime number) * $q = 41$ (a prime divisor of $p - 1 = 82$) * $g = 2$ (generator of order $q$ mod $p$) * $a = 3$ (private key of Alice) * $M = 20$ (message to sign) * $k = 8$ (random secret per-signature integer) We will now follow the DSA signature generation ________ ________ ___ ______ __________ ____ _____ ____ _________.
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Find a recurrence that generates the sequence 110110110110110.
i) Prepare logarithm and antilogarithm tables as given in page 23 of block 1
b) i) Bob uses ElGamal cyrptosystem with parametersand the secret value x = 3. What values will he make public?
ii) Alice wants to send Bob the messageShe chooses k = 5. How will she compute the cipher text? What information does she send to Bob?
iii) Explain how Bob will decrypt the message
Find a recurrence that generates the sequence 110110110110110.
a) 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 726-
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.This matrix has determinant
5 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
a) Decrypt the message c = 23 that was encrypted using RSA algorithm with e = 43 and n = 77.
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 |