User:Pandar Mayur/sandbox

====Caesar shift cryptography and RSA algorithm combination ====
(1). Introduction.

• The world of this time is digital.
  .

(2). Problem.

• we need to protect our message and data.
  

(3). Solution

• Our algorithm is a little bits difference. We encode message by Caesar shift algorithm and pass the shift key to the receiver by RSA algorithm.
  

3.1 Algorithm steps.

• In this first steps we encode message by shift Caesar cryptography. Caesar cipher cryptography is the one of the oldest cryptography.
  
• Caesar cipher Method named after Julius Caesar.
  
• Who used it with shift of three to protect messages of military significance.
  

Caption 1

Cryptography

• For example we have a message "Mayur"{12,0,24,20,17} is shifted by 3. So, it's become "Pdbxu"{15,3,1,23,20}
• above message "Mayur" is called plaintext. and "Pdbxu" is called ciphertext.
• Now we will public this encoded message.
• Next we will move to RSA algorithm.

3.2 RSA algorithm

• In the third step receiver need to decode this encoded message.for this sender send a shift key value by the RSA algorithm.
• The RSA cryptosystem, invented by Ron Rivest, Adi Shamir, and Len Adleman[18], was first publicized in the August 1977 issue of Scientific American. The cryptosystem is most commonly used for providing privacy and ensuring authenticity of digital data. (Boneh et al., 1999)
• In such a cryptosystem, the encryption key is public and it is different from the decryption key which is kept secret (private).(Rivest, n.d.)

3.3 RSA algorithm steps.

1. Choose large prime number p and q.
   
2. Calculate n = p × q.
   
3. Choose e ≠ 1 such that gcd(e, (p - 1)(q - 1)) = 1
   
4. Compute d = e^(-1)×mod((p - 1)(q - 1))
   
5. Public e & n (it called public key)
   
6. d is secret(private) key.
   
7. Compute y = x^e×mod(n)
   
8. When compute the y and d send it to the receiver.
   
9. Receiver compute z=y^d×mod(n)
   

• Now we need to prove that this z is shift key.So, we need to prove that z = x

Proof

 y^d = x^(ed)×mod(n)

 gcd(e, (p - 1)(q - 1)) = 1

 ed = 1 + mod((p -1)(q - 1))

 every K such that

 ed = 1 + K ×(p -1) × (q - 1)

 y^d = x^(ed) = x(1 + K × (p - 1) × (q - 1))mod(n)

     = x × x(k × (p - 1) ×(q -1))mod(n)

 gcd(x, p) = 1

 x^(p -1) =1 × mod(p)(format little therom)
 
 gcd(x, q) = 1

 x^(q -1) = 1 × mod(q)

 y^d =x × (x^(p - 1))^(K × (q - 1)) = x × mod(p)

 y^d = x ^ (x^(q - 1))^(K × (p -1)) = x × mod(q)

 y^d =x × mod(p × q)

 y^d = x × mod(n)

 So, y^d = z then z = x×mod(n)

 So,z = x 

• After reach shift key value to the receiver. Receiver decode the message that is public.For decode he/she use the complement of encoded shift key. For example above We encode Mayur{12,0,24,20,17} to Pdbxu{15,3,1,23,20}. So, this is hear we decode from Pdbxu{15,3,1,23,20} to Mayur.{12,0,24,20,17}
  

4. Example.

• Now we give A message called "I am Mayur"{8,0,12,12,0,24,20,17} and it send to the Receiver.
  
• we shifted it by 3. so it's Become "Ldppdbxu"{11,3,15,15,3,1,23,20}.
  
• Now we move the RSA algorithm to send the shift key value(x).
  
• We choose p = 5 and q = 7. So,
  

 n = p × q = 5 ×7 = 35

 e is gcd(e, (p - 1) × (q - 1)) = 1

 gcd(e, 4 × 6) = 1

 So, we choose e = 5

• Now public e & n.
  
• Find
  

 d = e^(-1)×mod((P - 1) × (q - 1))

 d = 5^(-1) × mod(4 × 6)

 d = 5(-11) × mod(24)

 d = 5

• calculate
  

 y = x^e × mod(n)

 y = 35 × mod(35)

 y = 243 × mod(35)

 y = 33

• I Send y and d to receiver.
  
• I calculate the value of which is our shifting key.
  

 z = y^d × mod(n)

   = (33)^5 × mod(35)

   = 39135393 × mod(35)

 z = 3

• So, our decode message got by encoded message shift by -3(complement of). our encoded message is "Ldppdbxu"{11,3,15,15,3,1,23,20}

 it shifted by -3 so, it's become "I am Mayur"{8,0,12,12,0,24,20,17}.