Using Three Breeds Of Rsa

Published Date: 02 Nov 2017

In public key cryptography, one of the most famous and 36 years aged system called as RSA. Since it is well known to the people, the algorithm has got various attacks such as Wiener, Boneh, etc., To overcome some security issues of RSA especially hacking the code in terms of deriving ‘d’ private key, there are various algorithms are evolved with the variance of RSA, such as Twin RSA, Rebalance RSA, etc., which has been designed and implemented. In this paper, we take the opportunity to analyze and demonstrate with at least three breeds of RSA and pasteurize their corresponding pros and cons with respect to time and other essential factors as testing parameters. In addition to that, we come up with some suggestion and challenging outcomes to create another optimized form of RSA.

Keywords: cryptography, RSA, Breeds of RSA.

Introduction:

In this paper, we aimed to come up with some new ideas to frame a new RSA breed (i.e.) optimized form of RSA algorithm. To do this We need to know about RSA ,what are the pros and cons of it, what are the security threats and issues, and what are the measures can be included to overcome the threats and issues .To understand the concepts clearly, first we start with some existing papers of RSA attack algorithm. They describe about the number of ways to crack the RSA and discuss how it can be cracked. So, with this analysis we come to know the weak areas of RSA. After understanding the weak parts of RSA, it is possible for us to frame a new and a strong optimized RSA [4].

RSA Algorithm:

1. In first step randomly take two different prime numbers p and q.

2. Calculate the |n|=p*q

3. Compute the value of Ѳ (n) = (p-1) (q-1)

4. Select some random value e, and 1< e < n and gcd (e, Ѳ (n)) =1

5. know calculate e * d = 1 mod Ѳ (n)

6. Find entity for a public and private key.

The pair of (e, n) is for public key and (d, n) for private key

RSA Attack:

Obtain entity A public key (e, n)

Calculate q i and r i using following steps

f = (e/n)

q0 =|f|, ro =f-q0 , m=0

For i =1, 2, 3…..

{ m = m+1;

qi = |1/ri-1|

ri = (1/ri-1)-qi

if (qi = qi-1 ) break;

}

f = (q0, q1………. qn)

Calculate k/ (d*g) as follows:

for i=2,3….m (Obtain m from step 2)

{if ( i is even) (follow algorithm 1 for the value q0 ,q1………..qi-1,qi+1)

if ( i is odd) (follow algorithm 1 for the value q0 ,q1………..qi)

}

Compute d= (d*g)/g.

Algorithm:

To reconstruct from its expansion staring from qo and let ni and di = i=0, 1……m is a sequence of numerators and denominator, so f can be computed as follows:

This is the simple way to attack RSA algorithm. Through this algorithm we can easily understand the holes in RSA .Public key cryptography method used to encrypt messages with a secret key which can be read only by receiver. There are two related keys--a key pair. A public key is available to anyone who might want to send you a message. A second, private key is kept secret, so that only receiver know it. It is the first public key cryptography named after Rivest, Shamir and Aldemen [1].

RSA algorithm has three steps: key generation, encryption and decryption [1]. There pair of related keys - A public key that is known to everyone for used of encrypting a text/value/message. Another is private key, which is known to only receiver and used to decrypt the messages. In this paper we did survey on modified RSA over existing RSA algorithm.

Cryptography

The art of protecting information by encrypting it into unreadable format which is called as cipher text [5]. Only those who is having a secret key can decrypt the message into plain text. Cryptography system is classified into Symmetric key cryptography that both sender and receiver uses a single key, and Public key cryptography also called as Asymmetric key Cryptography that uses two keys , a public key known to everyone and a private key known by only receiver[5].

Encryption and Decryption

Encryption is the process of Translating data into secret code. Information is encrypted to prevent it from strangers and only receiver can read the message. Decryption is the process of translating secret code to original data i.e. decoding with the help of secret key.

Co - Prime Numbers

A set of numbers which do not have any other common factor other than 1 are co- prime numbers. For example 8 and 9 have no other common factor other than 1 so they are co- prime numbers. No prime number divides both the numbers of a co- prime set [1]. One of the interesting property of co-prime is multiplicative inverse. According [1] to which if x and y are two co-prime numbers and y has a multiplicative inverse modulo x, then there exists an integer z such that y * z = 1 * mod (x).

RSA Algorithm

RSA algorithm [1] is named after Rivest, Shamir and Adleman, who are its creators. This algorithm is a public key cryptography algorithm which uses two different keys, public key for encryption and private key for decryption. This algorithm provides a security for transmission of messages. RSA works better if long keys are used in this case it becomes difficult to figure out the factors of n [1]. The following example gives the concept of RSA algorithm and its operations.

RSA algorithm consists of three steps:

(a) First step is key generation:

The RSA public key cryptosystem involves two keys, public key for encryption and for decryption of messages, private key. Following steps are used for generating key:

1) Choose two distinct prime numbers p and q

For example p= 211 and q = 233

2) Compute n = p q

n = 211 * 233

n = 49136

3) Compute phi (n) = (p-1) (q-1) where phi is Euler's totient function.

Phi (n) = (211 - 1) * (233 - 1)

Phi (n) = 210* 232

Phi (n) = 48720

4) Choose an integer e such that 1<e<phi (n) and gcd (phi (n)) = 1. i.e. e and phi(n) are co-prime.

e is the public key exponent.

1 < e < 48720

Let e = 83 as its co- prime to 48720

5) Determine d as d = e ^ -1(mod phi (n)).

d is the private key exponent.

e * d = 1 * mod ( phi (n))

So, 83 * d = 1*mod (48720)

d = 587

Thus Public key: (n, e) = (49163, 83)

Private key: (n, d) = (49163, 587)

(b) Encryption

Encryption means transforming the messages into coded form using key.

Let’s take an example:

The message m = 12

C = (m ^ e mod (n))

C = (12 ^ 83 mod (49163))

C = 46930

(c) Decryption

Decryption is transforming the coded message into readable form.

Formula to decrypt message is

M = (c ^ d mod (n))

M = (46930 ^ 83 mod (49163))

M = 12

Thus we saw the message which was sent is same we get after the decryption using RSA algorithm.

Limitations of RSA algorithm

Some of the limitations are stated as follows:

Its factors can be found out easily by hit and trial as n value is transmitted in public key. Thus security of RSA algorithm gets reduced.

The intruder can easily come to know about the message by encrypting a plaintext using public key

Modified RSA

The Modified RSA is similar to RSA algorithm. Some changes are made in the Modified RSA to provide security.

Let’s understand the algorithm with an example

1) Choose two prime numbers p and q randomly

p = 5 and q = 7

2) Compute n = p q

n = 5 * 7

3) Compute phi (n) = (p - 1) (q - 1)

Phi (n) = 4 * 6

Phi (n) = 24

4) Compute kp where lg n < kp < n i.e. kp is public key exponent and co-prime of n.

lg 35 < kp < 35

5.129, kp < 35

kp = 6

5) Compute d

If p > q

n-p < d < n and d is co-prime to n

Formula

kp * ks mod ( d) = 1 where ks is private key exponent

ks is found by

ks * kp = 1 * mod ( d )

In the example q > p

Therefore 35 - 7 < d < 35

28 < d < 35

kp * ks mod ( d) =1

6 * ks mod (d) =1

Let d = 29 and ks = 5

6) Send the following public key: (d)

private key : (d, ks)

public key = 29

private key = (29 , 5)

Examples:-

1) NUMERICAL VALUE :

2) TEXT VALUE :

Encryption

In above screenshot, "this is a sample" stored in file sample.txt and after performing RSA encryption data is converted into cypher text given in the in centric rectangle. RSA take 0.000second to encrypt the above sample.txt file.

Decryption

Thus we saw the message which was sent is same we get after the decryption using RSA Modified algorithm. The Modified RSA is more secure than RSA algorithm. In the modified algorithm it is difficult to know the public key exponent d as n value is not transmitted in public key. So difficult to read encrypted message and it also overcome the limitation of RSA algorithm.

Sample.txt document contain 57 different byte value. The entropy [probability distribution of this source] of the whole document is 5.78 (maximum possible entropy 8.00) wich is much less then maximum value.

A modified RSA cryptosystem based on 'n' prime number

A modified RSA cryptosystem based on 'n' prime numbers is similar to RSA algorithm. The changes in this algorithm is using 'n' prime numbers. Using 'n' prime numbers doesn't break so easily. This technique provides more efficiency and reliability over the networks [3].

Let’s understand the algorithm with an example

Here we have taken four prime numbers

p=2, q=3, r=5, s=11

1) Calculate n=p*q*r*s

n= 2*3*5*11

n= 330

2) Calculate f (n) = (p-1) (q-1) (r-1) (s-1)

f (858) = (2-1)(3-1)(5-1)(11-1)

f (858) = 1*2*4*10

= 80

3) Select e such that 1< e<240

f (n) must not be divisible by e

Let e = 3

4) Select d, multiplicative of e (mod f (n))

d=27 the public key is (n, e) i.e. (330, 3)

the private key is (n, d) i.e. ( 330, 27)

Given message m = 15

Encryption:

C = [m e (mod n)]

C = [153 (mod 330)]

C = 75

Decryption:

M = [cd (mod n)]

M = [75 ^ 27 (mod 330)]

M = 15

Thus we saw the message which was sent is same we get after the decryption using modified RSA cryptosystem based on 'n' prime numbers. This technique provides a maximum security for data. In this technique 'n' prime number is used so that it cannot be easily decompose. This technique provides a security over the network.

CONCLUSION

Modified RSA approach is more secure as compare to traditional simple RSA. Exponent d (public key) can be found only if knowing the values p and q.Entropy value means probability distribution value of source is less than maximum possible entropy.at the destination end or for attacker it difficult to find the value of d, hence the cypher text cannot be read easily

Public Key ( e )

Private key ( d )

Encryption

Decryption

RSA

( n , e )

( n , d )

C = (m ^ e mod (n))

M = (c ^ d mod (n))

Modified RSA

(d)

(d, ks)

C = m ^ kp mod ( d )

M = [c ^ ks mod ( d )] ^ ½

RSA with ‘n’ Prime numbers

( n , e )

( n , d )

C = (m ^ e mod (n))

M = (c ^ d mod (n))