RSA (Rivest–Shamir–Adleman)

Detailed Explanations : How it works?

RSA works using a public key and a private key and its strength relies on the hardness of prime factorization (a set of prime numbers that you multiply together to get another number). The first step in the RSA algorithm involves generating the keys.

Generating key

Choosing two primes

This is done in generate_keypair(private_key, public_key) function. Before we call this function we choose two different prime numbers (p and q), one for our private key and the other for our public key. Lets imagine I pick 2 and 7, and then we pass it to our function, so:

p = 17
q = 19
105     public_key, private_key = generate_keypair(17, 19)

Calculating n

Now the first thing we do inside this function is to compute n and it's nothing more than p and q multiplied together. n is used as the modulus for the encrypting key and decryption key. So let't calculate our n:

61      # n = prime_one * prime_two
62      p_times_q = prime_one * prime_two #n
62      p_times_q = 17 * 19 # 323

Using Euler's totient function to calculate Phi (φ)

Next comes calculating Phi(n) (φ(n)) and it's used to calculate how many integers are less than or equal to n that do not share any common factor with n. What's interesting and important here, is that, calculating the phi function is hard, except in the case of prime numbers. Since prime numbers have no factors greater than one, the phi of any prime number, p is simply p - 1. Therefore:

# n = p * q
# φ(n) = φ(p) * φ(q)

In our code:

68      phi = (prime_one - 1) * (prime_two - 1)
68      phi = (17 -1) * (19 -1) # 288 is the number of integers that are coprime with 323

Choosing e

e is important for our encryption because e and n it will become our public_key. e must be picked randomly and have certain properties. Such as:

1 < e < φ(n) (in our case e must be between 1 and 288)
Must be coprime with n and φ(n) (in our case, must be coprime with 323 and 288) We do so here:

75      e = random.randrange(1, phi+1)
86      g = gcd(e, phi) # if this returns one, then e is coprime with phi

if g returns one then e is coprime with phi, so their greatest common divisor is 1, which means that 1 is the only positive integer (factor) that divides both of them, e and phi.

Calculating d

We use the Extended Euclid's Algorithm to generate d. This is a private exponent, it will be a part of our private key and this is used to undo the effect of e

98      d = multiplicative_inverse(e, phi)

Public key and Private key

With n, e and d, we can now get our public and private key pairs.

104     return ((e, p_times_q), (d, p_times_q))

Public key

The public key pair is (e, n). This is what the server sends us, its public and anyone can read it.

Private key

The private key pair is (d, n). This can not be shared with anyone. It will be used to decrypt messages encrypted with the public key.

Encrypting

For encrypting this is what we do: (m^e) % n This function will use the public key to encrypt our plain text. We'll encrypt the string "asd". Let's first get our e and n from our public key pair. Then, we loop through each character in our string and get the Unicode value using ord() function. With this we then exponentiate that value by e and finally get the remainder with mod n.

111     e, n = public_key
129     plain = [(ord(char) ** e) % n for char in ciphertext]

Our encrypted message should look like this:

147191111

Decrypting

In the decrypting process we use our private key, and the process in almost the same, except that, we use d our private exponent instead of e. So:

(encrypted_character^d) % n

Now we loop through each character performing our calculation. In the end we use chr() function to get our character back.

113     d, n = private_key
115     plain = [chr((char ** d) % n) for char in ciphertext]

lgrvs / RSA