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Explain RSA algorithm with an example.
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1.Most widely accepted and implemented general purpose approach to public key encryption developed by Rivest-Shamir and Adleman (RSA) at MIT university.

2.RSA scheme is block cipher in which the plaintext and ciphertext are integers between 0 and n-1 for same n.

3.Typical size of n is 1024 bits. i.e n<2.

4.Description of Algorithm:

  • The scheme developed by Rivest, Shamir and Adleman makes use of an expression with exponentials.
  • Plaintext is encrypted in block having a binary value than same number n.
  • Block Size log2(n)

    If block size=1 bits then, 21n2i+1

  • Encryption and decryption are of following form for same plaintext M and ciphertext C.

    C=Memodn

    M=Cdmodn

    M=(Me)dmodn

    M=Medmodn

  • 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.
  • Thus this is a public key encryption algorithm with a public key of PU= {c, n} and private key of PR= {d, n}.

5.RSA algorithm as shown below:

a) Key Genration :

  • Select p,q…….. p and q both are the prime numbers, p≠q.
  • Calculate n=p×q
  • Calculate q(n) = (p-1) (q-1)
  • Select integer….g(d ( (n), e)) =1 & 1< e < (n)
  • Calculate d; d= e-1 mod (n)
  • Public Key, PU= {e, n}
  • Private Key, PR ={d,n}

    b) Encryption :

  • Plaintext : m<n< p="">

  • Ciphertext: C

    c) Decryption:

  • Ciphertext: C

  • Plaintext : M= Cd mod n
  • Note 1 : (n) -> Euler’s totient function
  • Note 2: Relationship between C and d is expressed as:

    ed (mod (n))=1

    ed = 1 mod (n)

    d = e1 mod (n)

6.Example:

  • Key Generation :

    1. Select 2 prime numbers -> p=17 and q=11
    2. Calculate n = p×q =17 ×11=187
    3. Calculate = 16 × 10= 160 Select ‘e’ such that e is relatively prime to (n)=160 and e <
    4. Determine d such that :

      de =1 mod (n)

      d × 7 = 1 mod 160

161

d=e1  mod  (n)[161/7=  

div.(d)23  and remainder (mod) =1d=23

  1. Then the resulting keys are public key :

    PU = {7, 187 }

    PR = {23, 187 }

    Let M=88 for encryption

    C=887mod(187)88mod187=88882mod187=7744mod187=77884mod187=59969536mod187=132

    887mod187 =(884mod187)×(882mod187)×(88mod187)mod187=(132×77×88)mod187=894432mod187=11

  • For Decryption :

    M=Cdmod187=1123mod187111mod187=11112mod187=121114mod187=14641/187=55118mod187=214358881mod187=331123mod187 =(118mod187×118mod187×114mod187×112mod187×111mod187)mod187=(33×33×55×81×11)mod187=79720245mod187=88

Figure 5.4 Solution of Above example

Figure 5.4 Solution of Above example

Public key is (n,e) Private key is (n,d)


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