RSA Algorithm
Introduction
Ron Rivest, Adi Shamir and Len Adleman have developed this algorithm (Rivest-Shamir-Adleman). It is a block cipher which converts plain text into cipher text and vice versa at receiver side.
RSA Algorithm Steps
Step-1: Select two prime numbers p and q where p ≠ q.
Step-2: Calculate n = p * q.
Step-3: Calculate Ф(n) = (p-1) * (q-1).
Step-4: Select e such that, e is relatively prime to Ф(n), i.e. (e, Ф(n)) = 1 and 1 < e < Ф(n)
Step-5: Calculate d = e -1 mod Ф(n) or ed = 1 mod Ф(n).
Step-6: Public key = {e, n}, private key = {d, n}.
Step-7: Find out cipher text using the formula,
C = Pe mod n where, P < n where C = Cipher text, P = Plain text, e = Encryption key and n=block size.
Step-8: P = Cd mod n. Plain text P can be obtain using the given formula. where, d = decryption key
RSA algorithm explanation with example step by step
Step – 1: Select two prime numbers p and q where p ≠ q.
Example, Two prime numbers p = 13, q = 11.
Step – 2: Calculate n = p * q.
Example, n = p * q = 13 * 11 = 143.
Step – 3: Calculate Ф(n) = (p-1) * (q-1).
Example, Ф(n) = (13 – 1) * (11 – 1) = 12 * 10 = 120.
Step – 4: Select e such that, e is relatively prime to Ф(n), i.e. (e, Ф(n)) = 1 and 1 < e < Ф(n).
Example, Select e = 13, gcd (13, 120) = 1.
Step – 5: Calculate d = e -1 mod Ф(n) or e * d = 1 mod Ф(n)
Example, Finding d: e * d mod Ф(n) = 1
13 * d mod 120 = 1
(How to find: d *e = 1 mod Ф(n)
d = ((Ф(n) * i) + 1) / e
d = (120 + 1) / 13 = 9.30 (∵ i = 1)
d = (240 + 1) / 13 = 18.53 (∵ i = 2)
d = (360 + 1) / 13 = 27.76 (∵ i = 3)
d = (480 + 1) / 13 = 37 (∵ i = 4))
Step – 6: Public key = {e, n}, private key = {d, n}.
Example, Public key = {13, 143} and private key = {37, 143}.
Step – 7: Find out cipher text using the formula, C = Pe mod n where, P < n.
Example, Plain text P = 13. (Where, P < n)
C = Pe mod n = 1313 mod 143 = 52.
Step – 8: P = Cd mod n. Plain text P can be obtain using the given formula.
Example, Cipher text C = 52
P = Cd mod n = 5237 mod 143 = 13.
Solution of exercise (Given in video)
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Nice info sir
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DeleteError in exercise 4 part 3:
ReplyDelete(p-1)(q-1) = (3-1)(11-1)= 2*10 = 20
ohhh...Thank you for suggestion...It was my mistake.... Now check this...It is changed..
DeleteThe of the given original message which is plain text is 59. the value after decrypting is 26 is differ from the plain text. can you correct it please
DeleteSolution is true... You can try...You also will get decrypted cipher text valus is 26. becuase,as per algorithm step no 7, the value of plain text must be P < n.
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