RSA Encryption Algorithm - Javatpoint
RSA encryption algorithm is a type of public-key encryption algorithm. To better understand RSA, lets first understand what is public-key encryption algorithm.
Public Key encryption algorithm is also called the Asymmetric algorithm. Asymmetric algorithms are those algorithms in which sender and receiver use different keys for encryption and decryption. Each sender is assigned a pair of keys:
The Public key is used for encryption, and the Private Key is used for decryption. Decryption cannot be done using a public key. The two keys are linked, but the private key cannot be derived from the public key. The public key is well known, but the private key is secret and it is known only to the user who owns the key. It means that everybody can send a message to the user using user's public key. But only the user can decrypt the message using his private key.
RSA is the most common public-key algorithm, named after its inventors Rivest, Shamir, and Adelman (RSA).
RSA algorithm uses the following procedure to generate public and private keys:
This example shows how we can encrypt plaintext 9 using the RSA public-key encryption algorithm. This example uses prime numbers 7 and 11 to generate the public and private keys.
Explanation:
Step 1: Select two large prime numbers, p, and q.
p = 7
q = 11
Step 2: Multiply these numbers to find n = p x q, where n is called the modulus for encryption and decryption.
First, we calculate
n = p x q
n = 7 x 11
n = 77
Step 3: Choose a number e less that n, such that n is relatively prime to (p - 1) x (q -1). It means that e and (p - 1) x (q - 1) have no common factor except 1. Choose "e" such that 1 Second, we calculate φ (n) = (p - 1) x (q-1) φ (n) = (7 - 1) x (11 - 1) φ (n) = 6 x 10 φ (n) = 60 Let us now choose relative prime e of 60 as 7. Thus the public key is Step 4: A plaintext message m is encrypted using public key To find ciphertext from the plain text following formula is used to get ciphertext C. C = me mod n C = 97 mod 77 C = 37 Step 5: The private key is De mod {(p - 1) x (q - 1)} = 1 7d mod 60 = 1, which gives d = 43 The private key is Step 6: A ciphertext message c is decrypted using private key m = cd mod n m = 3743 mod 77 m = 9 In this example, Plain text = 9 and the ciphertext = 37 In an RSA cryptosystem, a particular A uses two prime numbers, 13 and 17, to generate the public and private keys. If the public of A is 35. Then the private key of A is ……………?. Explanation: Step 1: in the first step, select two large prime numbers, p and q. p = 13 q = 17 Step 2: Multiply these numbers to find n = p x q, where n is called the modulus for encryption and decryption. First, we calculate n = p x q n = 13 x 17 n = 221 Step 3: Choose a number e less that n, such that n is relatively prime to (p - 1) x (q -1). It means that e and (p - 1) x (q - 1) have no common factor except 1. Choose "e" such that 1 Second, we calculate φ (n) = (p - 1) x (q-1) φ (n) = (13 - 1) x (17 - 1) φ (n) = 12 x 16 φ (n) = 192 g.c.d (35, 192) = 1 Step 3: To determine the private key, we use the following formula to calculate the d such that: Calculate d = de mod φ (n) = 1 d = d x 35 mod 192 = 1 d = (1 + k.φ (n))/e [let k =0, 1, 2, 3………………] Put k = 0 d = (1 + 0 x 192)/35 d = 1/35 Put k = 1 d = (1 + 1 x 192)/35 d = 193/35 Put k = 2 d = (1 + 2 x 192)/35 d = 385/35 d = 11 The private key is Hence, private key i.e. d = 11 A RSA cryptosystem uses two prime numbers 3 and 13 to generate the public key= 3 and the private key = 7. What is the value of cipher text for a plain text? Explanation: Step 1: In the first step, select two large prime numbers, p and q. p = 3 q = 13 Step 2: Multiply these numbers to find n = p x q, where n is called the modulus for encryption and decryption. First, we calculate n = p x q n = 3 x 13 n = 39 Step 3: If n = p x q, then the public key is To find ciphertext from the plain text following formula is used to get ciphertext C. C = me mod n C = 53 mod 39 C = 125 mod 39 C = 8 Hence, the ciphertext generated from plain text, C = 8. A RSA cryptosystem uses two prime numbers, 3 and 11, to generate private key = 7. What is the value of ciphertext for a plain text 5 using the RSA public-key encryption algorithm? Explanation: Step 1: in the first step, select two large prime numbers, p and q. p = 3 q = 11 Step 2: Multiply these numbers to find n = p x q, where n is called the modulus for encryption and decryption. First, we calculate n = p x q n = 3 x 11 n = 33 Step 3: Choose a number e less that n, such that n is relatively prime to (p - 1) x (q -1). It means that e and (p - 1) x (q - 1) have no common factor except 1. Choose "e" such that 1< e < φ (n), e is prime to φ (n), gcd (e, d (n)) =1. Second, we calculate φ (n) = (p - 1) x (q-1) φ (n) = (3 - 1) x (11 - 1) φ (n) = 2 x 10 φ (n) = 20 Step 4: To determine the public key, we use the following formula to calculate the d such that: Calculate e x d = 1 mod φ (n) e x 7 = 1 mod 20 e x 7 = 1 mod 20 e = (1 + k. φ (n))/ d [let k =0, 1, 2, 3………………] Put k = 0 e = (1 + 0 x 20) / 7 e = 1/7 Put k = 1 e = (1 + 1 x 20) / 7 e = 21/7 e = 3 The public key is Hence, public key i.e. e = 3 PostgreSQL Solr MongoDB Gimp Verilog Teradata PhoneGap Gmail Vue.js PLC Illustrator Aptitude Reasoning Verbal A. Interview Company AI AWS Selenium Cloud Hadoop ReactJS D. Science Angular 7 Blockchain Git ML DevOps DBMS DS DAA OS C. Network Compiler D. COA D. Math. E. Hacking C. Graphics Software E. Web Tech. Cyber Sec. Automata C C++ Java .Net Python Programs Control S. Data Mining Hindi100 Lyricsia Website Development Android Development Website Designing Digital Marketing Summer Training Industrial Training College Campus Training Address: G-13, 2nd Floor, Sec-3 Noida, UP, 201301, India Contact No: 0120-4256464, 9990449935 © Copyright 2011-2018 www.javatpoint.com. All rights reserved. Developed by JavaTpoint.
