RSA Algorithm Practical : - 1 Outcomes Working of RSA algorithm Example 4/26/2023 PRESENTATION TITLE 2 Introduction RSA algorithm is an asymmetric cryptography algorithm. RSA stands for Rivest – Shamir – Adleman. The idea of RSA is based on the fact that it is difficult to factorize a large integer. The public key consists of two numbers where one number is a multiplication of two large prime numbers. And private key is also derived from the same two prime numbers. So if somebody can factorize the large number, the private key is compromised. Therefore encryption strength totally lies on the key size and if we double or triple the key size, the strength of encryption increases exponentially. RSA keys can be typically 1024 or 2048 bits long. 4/26/2023 PRESENTATION TITLE 3 RSA Alogrithm 4/26/2023 PRESENTATION TITLE 4 Working of RSA Algorithm Key Generation Select p,q (p and q both prime) Calculate n = p*q Calculate φ(n) = (p - 1) * (q - 1) 4/26/2023 Select integer e gcd(φ(n),e)=1; 1 < e < φ(n) Calculate d d = e mod φ (n) = 1 Public Key KU = {e,n} Private Key KR = {d,n} PRESENTATION TITLE 6 Encryption 4/26/2023 Plaintext M<n Ciphertext C = me mod n PRESENTATION TITLE 7 Decryption 4/26/2023 Ciphertext C Plaintext M = cd mod n PRESENTATION TITLE 8 Example of RSA Algorithm Example Select p = 3 and q = 11 (both are prime numbers) Calculate n = p*q = 3 * 11 = 33. Calculate φ(n) = (p-1).(q-1) = 2 x 10 = 20. Select e such that e is relatively prime to φ(n) = 20 and less than φ(n) we choose e = 7. Determine d such that d = e mod φ (n) = 1 ( d= e (mod 20) and d < 20. The resulting keys are public key PU = {7,33} and private key PR = {3,33}. The example shows the use of these keys for a plaintext input of M = 9. For encryption, we need to calculate C. RSA Algorithm 10 Example C = me mod n = 5 ^7 mod 33 = 3380 mod 33 = 14 M = cd mod n = 14^3 mod 33 = 2744 mod 33 =5 RSA Algorithm 11 Thank you Aayushi Patel Enrollment : 206140316022 Semester 6th Department : IT