Download presentation

Presentation is loading. Please wait.

Published byVirgil Sullivan Modified over 4 years ago

2
Introduction Requirements for RSA Ingredients for RSA RSA Algorithm RSA Example Problems on RSA

3
Developed by Rivest, Shamir, and Adleman in 1977. Public-key encryption technique : - Public-key/two-key/asymmetric cryptography involves the use of two keys: - a public key, which may be known by anybody, and can be used to encrypt messages, and verify signatures. - a private key, known only to the recipient, used to decrypt messages, and sign (create) signatures. - is asymmetric because those who encrypt messages or verify signatures cannot decrypt messages or create signatures.

4
Block cipher scheme - A block cipher is one in which a block of plaintext is treated as a whole and used to produce a ciphertext block of equal length. - Each block having a binary value less than some number n (0 to n-1). - Typical size of block is 1024 bits or 309 decimal digits. - The block size is i bits, where 2^i < n <= 2^(i+1)

5
Encryption and Decryption are of the following form: Plaintext M Ciphertext C C=M e mod n M=C d mod n = (M e ) d mod n= M ed mod n - Both sender and receiver must know the value of n. - The sender knows the value of e, and only the receiver knows the value of d. - Public key = {e, n} - Private key = {d, n}

6
We need to find a relationship of the form M ed mod n - This relationship holds if e and d are multiplicative inverse modulo ø(n), where ø(n) is the Euler totient function. - Euler totient function ø(n), defined as number of the integers less than n and relatively prime to n. - The relationship between e and d can be expressed as e.d mod ø(n) =1 - d(and therefore e) is relatively prime to ø(n).{by the rules of moduler arithmetic} gcd(ø(n),d)=1

7
For this algorithm to be satisfactory for public key encryption, the following requirements must be met: - It is possible to find values of e, d, n such that M ed mod n = M for all M<n. - It is relatively easy to calculate M e mod n and C d mod n for all values of M<n. - It is infeasible to determine d given e and n.

8
p, q two prime numbers (private, chosen) n=pq (public, calculated) e, with gcd(ø(n),e)=1; 1<e<ø(n) (public, chosen) e.d=1 mod ø(n) (private, calculated)

9
Key Generation Select two large primes at random - p, q Calculate n=p.q Calculate ø(n)=(p-1)(q-1) Selecting at random the encryption key e where 1<e<ø(n), gcd(e,ø(n))=1 Solve following equation to find decryption key d e.d=1 mod ø(n) and 0≤d≤n Publish their Public key: PU={e,n} keep secret Private key: PR={d,n}

10
Encryption Plaintext M, 0≤M<n Ciphertext C=M e mod n Decryption Ciphertext C Plaintext M=C d mod n

11
1. Select primes: p=17 & q=11 2. Compute n = pq =17×11=187 3. Compute ø(n)=(p–1)(q-1)=16×10=160 4. Select e : gcd(e,160)=1; choose e=7 5. Determine d: de=1 mod 160 and d < 160 Value is d=23 since 23×7=161= 1×160+1 6. Publish public key PU={7,187} 7. Keep secret private key PR={23,187}

12
Given message M = 88 Encryption: C = 88 7 mod 187 = 11 Decryption: M = 11 23 mod 187 = 88

13
1. Perform encryption and decryption using RSA algorithm, for the following: ① p = 3; q = 11, e = 7; M = 5(Ans. d=3, C=14) ② p = 5; q = 11, e = 3; M = 9(Ans. d=27, C=14) 2. In a public-key system using RSA, you intercept the ciphertext C = 10 sent to a user whose public key is e = 5, n = 35. What is the plaintext M?(Ans. M=5)

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google