## Cryptography in Network Security-

Before you go through this article, make sure that you have gone through the previous article on Cryptography.

We have discussed-

• Cryptography is a method of storing and transmitting data in a particular form.
• Cryptography techniques are- ## Asymmetric Key Cryptography-

In this technique,

• Sender and receiver use different keys to encrypt and decrypt the message.
• It is called so because sender and receiver use different keys.
• It is also called as public key cryptography.

## Working-

The message exchange using public key cryptography involves the following steps- ## Step-01:

At sender side,

• Sender encrypts the message using receiver’s public key.
• The public key of receiver is publicly available and known to everyone.
• Encryption converts the message into a cipher text.
• This cipher text can be decrypted only using the receiver’s private key.

## Step-02:

• The cipher text is sent to the receiver over the communication channel.

## Step-03:

• Receiver decrypts the cipher text using his private key.
• The private key of the receiver is known only to the receiver.
• Using the public key, it is not possible for anyone to determine the receiver’s private key.
• After decryption, cipher text converts back into a readable format.

The advantages of public key cryptography are-

• It is more robust.
• It is less susceptible to third-party security breach attempts.

The disadvantages of public key cryptography are-

• It involves high computational requirements.
• It is slower than symmetric key cryptography.

## Number of Keys Required-

To use public key cryptography,

• Each individual requires two keys- one public key and one private key.
• For n individuals to communicate, number of keys required = 2 x n = 2n keys.

## Asymmetric Encryption Algorithms-

The famous asymmetric encryption algorithms are- 1. RSA Algorithm
2. Diffie-Hellman Key Exchange

## RSA Algorithm-

Let-

• Public key of the receiver = (e , n)
• Private key of the receiver = (d , n)

Then, RSA Algorithm works in the following steps-

## Step-01:

At sender side,

• Sender represents the message to be sent as an integer between 0 and n-1.
• Sender encrypts the message using the public key of receiver.
• It raises the plain text message ‘P’ to the eth power modulo n.
• This converts the message into cipher text ‘C’.

 C = Pe mod n

## Step-02:

• The cipher text ‘C’ is sent to the receiver over the communication channel.

## Step-03:

• Receiver decrypts the cipher text using his private key.
• It raises the cipher text ‘C’ to the dth power modulo n.
• This converts the cipher text back into the plain text ‘P’.

 P = Cd mod n

## NOTE-

### ‘e’ and ‘d’ must be multiplicative inverses modulo Ø(n)

P = Cd mod n

P = (Pe mod n)d mod n

P = Ped mod n

For this equation to be true, by Euler’s Theorem, we must have-

ed = 1 mod Ø(n)

OR

ed = kØ(n) + 1

Thus, e and d must be multiplicative inverses modulo Ø(n).

## Steps to Generate Public Key And Private Key-

An individual can generate his public key and private key using the following steps-

### Step-01:

Choose any two prime numbers p and q such that-

• They are different.
• They are very large.

### Step-02:

Calculate ‘n’ and toilent function Ø(n) where-

• n = p x q
• Ø(n) = (p-1) x (q-1)

### Step-03:

Choose any value of ‘e’ such that-

• 1 < e < Ø(n)
• gcd (e, Ø(n)) = 1

### Step-04:

Determine ‘d’ such that- • You already know the value of ‘e’ and Ø(n).
• Choose the least positive integer value of ‘k’ which gives the integer value of ‘d’ as a result.
• Use trial and error method.
• Start substituting different values of ‘k’ from 0.

## Problem-01:

In a RSA cryptosystem, a participant A uses two prime numbers p = 13 and q = 17 to generate her public and private keys. If the public key of A is 35, then the private key of A is _______.

## Solution-

Given-

• Prime numbers p = 13 and q = 17
• Public key = 35

### Step-01:

Calculate ‘n’ and toilent function Ø(n).

Value of n,

n = p x q

n = 13 x 17

∴ n = 221

Toilent function,

Ø(n) = (p-1) x (q-1)

Ø(n) = (13-1) x (17-1)

∴ Ø(n) = 192

### Step-02:

• We are already given the value of e = 35.
• Thus, public key = (e , n) = (35 , 221)

### Step-03:

Determine ‘d’ such that- Here,

• The least value of ‘k’ which gives the integer value of ‘d’ is k = 2.
• On substituting k = 2, we get d = 11.

Thus, private key of participant A = (d , n) = (11, 221).

## Problem-02:

In the RSA public key cryptosystem, the private and public keys are (e, n) and (d, n) respectively, where n = p x q and p and q are large primes. Besides, n is public and p and q are private. Let M be an integer such that 0 < M < n and f(n) = (p-1)(q-1).

Now consider the following equations-

I. M’ = Me mod n and M = (M’)d mod n

II. ed ≡ 1 mod n

III. ed = 1 mod f(n)

IV. M’ = Me mod f(n) and M = (M’)d mod f(n)

Which of the above equations correctly represent RSA cryptosystem?

1. I and II
2. I and III
3. II and IV
4. III and IV

## Solution-

Clearly, Option (B) is correct.

To gain better understanding about RSA Algorithm,

Watch this Video Lecture

Next Article- Diffie Hellman Key Exchange Algorithm

Get more notes and other study material of Computer Networks.

Watch video lectures by visiting our YouTube channel LearnVidFun.