net 311d information securityinformation security tutorial 3 : asymmetric ciphers (rsa) networks and...

NET 311D

INFORMATION SECURITY

TUTORIAL 3 : Asymmetric Ciphers (RSA)

Networks and

Communication

Department

1

A Symmetric-Key Cryptography

(Public-Key Cryptography)

Asymmetric-key (public key cryptography) uses two

keys: one private and one public.

There are two algorithms:

RSA

ELGAMAL

RSA

The most common public key algorithm is RSA,

named for its inventors Rivest, Shamir, and Adelman.

It uses two numbers:

e public key

d private

Networks and Communication Systems Department - Prepared by : TA. Anfal AlHazzaa

RSA – Step 1:selecting keys

1. Choose two very large prime numbers p and q

2. Find n n=p x q

3. Calculates another number ф = (p -1) X (q -1)

4. Choose random encryption number e ,

1< e< ф

gcd(e, ф)=1 (coprime)

(You can use Euclidian algorithm to help you find correct e)

5. Calculate d so that d X e mod ф = 1 (using table method)

Note : Announce e and n to the public; keep ф and d secret.


RSA – Step 2: Encryption (Sender)

Where P : plain text

n= pxq

C : cipher text

e: public key

Restriction

P < n , if not, the plaintext needs to be divided into blocks to make P less than n.


RSA – Step 3: Decryption (Receiver)

Where P : plain text

n= pxq

C : cipher text

d: private key


RSA: Example 1

Perform encryption and decryption using the RSA

algorithm, for the following:

p = 3; q = 11, e = 7; M = 5



1. p=3 ; q =11

2. n=p x q = 3 x 11 =33 n=33

3. ф = (p -1) X (q -1) = (3-1) x (11-1)= 2 x10 = 20 ф = 20

4. e (which is mentioned in question ) e=7

1< e< ф






Euclidian algorithm

7 20

14

6

2

7

6

1

1

6

6

0

1

1. Divide the ф by e

2. Divide the divisor of the previous division

by the reminder of the previous division

3. Repeat step 2 until the reminder become 0

STOP


Rewrite the previous operations as the following:

Dividend= (divisor x quotient )+ Reminder

20=(7 x 2) + 6

7= (6 x 1) + 1

6=(6 x 1) +0

This is the gcd(20,7)

Euclidian algorithm cont.


Table method

1. As default in all situations , put the values of a 1 and b0 in 1st row and

the values of a 0 and b 1in 2nd row

2. 1st row : c= ф and g= empty

3. 2nd row : c= e and g= (c1 /c2) (take only the integer part)

(20/7)=2.8 2

Row # a b c g

1 1 0 20 -

2 0 1 7 2


Table method cont.

3. 3rd row:

a3= a1-a2*g2 1-0*2=1

b3= b1-b2*g2 0-1*2=-2

c3= c1-c2*g2 20-7*2=6 ( it’s the reminder of 20/7)

g3= (c2 /c3) (take only the integer part) (7/6)=1.16 1

Row # a b c g

1 1 0 20 -

2 0 1 7 2

3 1 -2 6 1

In 𝑟𝑜𝑤𝑛

𝑐𝑜𝑙𝑢𝑚𝑛𝑛 = 𝑐𝑜𝑙𝑢𝑚𝑛𝑛−2 − 𝑐𝑜𝑙𝑢𝑚𝑛𝑛−1 x 𝑔𝑛−1

except 𝑔𝑛 = 𝑐𝑛−1 / 𝑐𝑛 (take only the integer part)


Table method cont.

4. 4th row:

a4= a2-a3*g3 0-1*1=-1

b4= b2-b3*g3 1-(-2)*1=3


g4= (c3 /c4) (take only the integer part) (6/1)=6

Row # a b c g

1 1 0 20 -

2 0 1 7 2

3 1 -2 6 1

4 -1 3 1 6

In 𝑟𝑜𝑤𝑛




Table method cont.

Stop when the value of c become 1

Then the latest value of b is the value of d d=3

Row # a b c g

1 1 0 20 -

2 0 1 7 2

3 1 -2 6 1

4 -1 3 1 6

Important notes : 1. If the value of b is negative d= ф + b

2. If the value of b> ф d= b mod ф

To confirm your answer: a*ф + b*e = gcd (ф,e)

In our example : -20 + 21 = 1

1=1 it’s correct



1. p=3 ; q =11

2. n=33

3. ф = 20

4. e=7

5. d=3




P=M= 5 , e=7 ,n=33

C = 57 mod (33)

C = [(54 mod 33) * (52 mod 33) * (51 mod 33)] mod 33

C = (31 * 25 * 5) mod 33

C = 3875 mod 33

C=14



C=14 , d=3 ,n=33

P=143 mod 33

P = [(142 mod 33)*(141 mod 33)] mod 33

P= 434 mod 33

P=5


RSA: Example 2

Encrypt the following message “YES” By Using RSA

algorithm.

Assume that

p=5, q =11, e= 3

Each character will be encoded according to this

algorithm (A - Z) (00 – 25)

Find d , then encrypt each character individually



1. p=5 ; q =11

2. n=p x q = 5 x 11 =55 55

3. ф = (p -1) X (q -1) = (5-1) x (11-1)= 4 x10 = 40 ф = 40

4. e (which is mentioned in question ) e=3

1< e< ф






Euclidian algorithm

3 40

39

1

13

3

3

0

3

1. Divide the ф by e

2. Divide the divisor of the previous division

by the reminder of the previous division

3. Repeat step 2 until the reminder become 0 STOP


Rewrite the previous operations as the following:

Dividend= (divisor x quotient )+ Reminder

40=(13x 3) + 1

3= (3 x 1) + 0

This is the gcd(40,3)

Euclidian algorithm cont.


Table method

1. As default in all situations , put the values of a 1 and b0 in 1st row and

the values of a 0 and b 1in 2nd row

2. 1st row : c= ф and g= empty

3. 2nd row : c= e and g= (c1 /c2) (take only the integer part)

(40/3)=13.3 13

Row # a b c g

1 1 0 40 -

2 0 1 3 13


Table method cont.

3. 3rd row:

a3= a1-a2*g2 1-0*13=1

b3= b1-b2*g2 0-1*13=-13


g3= (c2 /c3) (take only the integer part) (3/1)=3

Row # a b c g

1 1 0 40 -

2 0 1 3 13

3 1 -13 1 3

In 𝑟𝑜𝑤𝑛




Table method cont.

Stop when the value of c become 1

Then the latest value of b is the value of d b= -13 (It’s negative )

Based on note 1 d= ф+b 40+(-13)= 27 d= 27

Row # a b c g

1 1 0 40 -

2 0 1 3 13

3 1 -13 1 3

Important notes : 1. If the value of b is negative d= ф + b

2. If the value of b> ф d= b mod ф

To confirm your answer: a*ф + b*e = gcd (ф,e)

In our example : 40+ (-39) = 1

1=1 it’s correct



1. p=5 ; q =11

2. n=55

3. ф = 40

4. e=3

5. d=27




P=M=YES(must be encoded before) , e=3 ,n=55

Y=24

C = 243 mod (55)

C= 19

E=4

C = 43 mod (55)

C= 9

S=18

C = 183 mod (55)

C= 2


RSA – Step 2: Encryption (Sender) cont.

24-Feb-15

27



C=19

P = 1927 mod (55)

P= 24

C=9

P = 927 mod (55)

P= 4

C=2

P = 227 mod (55)

P= 18


net 311d information securityinformation security tutorial 3 : asymmetric ciphers (rsa) networks and...

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