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El_Gamal Cryptography

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IntroductionIntroduction

El_Gamal is a public-key cryptosystem El_Gamal is a public-key cryptosystem techniquetechnique

was designed by Dr. Taher Elgamal .was designed by Dr. Taher Elgamal .

El_GamalEl_Gamal depends on the one way depends on the one way function, means that the encryption and function, means that the encryption and decryption are done in separate functions.decryption are done in separate functions.

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IntroductionIntroduction

The encryption process requires two The encryption process requires two modular exponentiations (extra modular exponentiations (extra time).time).

A disadvantage of El_Gamal encryption is A disadvantage of El_Gamal encryption is that there is message expansion by a that there is message expansion by a factor of 2. That is, the ciphertext is twice factor of 2. That is, the ciphertext is twice as long as the corresponding plaintext.as long as the corresponding plaintext.

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Key generationKey generation

Receiver A must do the following:Receiver A must do the following:

1- Generate a large random prime number (p)1- Generate a large random prime number (p)

2- Choose a generator number (a) 2- Choose a generator number (a) {show in slide 7 }{show in slide 7 }

3- Choose an integer (x) less than (p-2) ,as 3- Choose an integer (x) less than (p-2) ,as secretsecret

number.number.

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Key generationKey generation

4- 4- Compute (d) whereCompute (d) where

d= ad= axx mod p mod p

5-5- Determine the public key (p, a, d) Determine the public key (p, a, d)

and the private key (x)and the private key (x)

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Key generationKey generation

Example :Example :

let p = 11 and a = 2 and x = 5let p = 11 and a = 2 and x = 5

calculate d = 2calculate d = 255 mod 11 = 10 mod 11 = 10

public key = (11 ,2 ,10)public key = (11 ,2 ,10)

private key = (5)private key = (5)

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Generator numberGenerator number

How to test (a) generator or not :How to test (a) generator or not :

1- (a) must be between 1 and p-11- (a) must be between 1 and p-1

2- Find Ø = p-12- Find Ø = p-1

3- Find the all factors of Ø {f1,f2,….,fn} – 3- Find the all factors of Ø {f1,f2,….,fn} – { 1 }{ 1 }

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Generator numberGenerator number

4- Find {q1,q2,…..,qn} where 4- Find {q1,q2,…..,qn} where

qi = fiqi = fi

for the redundant factors for the redundant factors

qi = fiqi = fifreqfreq

5- (a) generator number if and only if 5- (a) generator number if and only if

wi= awi= a Ø/qi Ø/qi mode p <> 1 , for all qi mode p <> 1 , for all qi

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Generator numberGenerator number

Example 1Example 1 : : let p= 11 , a=2 ,test a is generator number let p= 11 , a=2 ,test a is generator number

or not ? or not ? sol:sol: Ø= p-1 = 10 , factors of 10 = {2 , 5}Ø= p-1 = 10 , factors of 10 = {2 , 5} q1 = 2 ,q2 = 5q1 = 2 ,q2 = 5 w1 = 2w1 = 210/210/2 mod 11 = 10 <> 1mod 11 = 10 <> 1 w2 = 2w2 = 210/510/5 mod 11 = 4 <> 1 mod 11 = 4 <> 1 i.e a generator number .i.e a generator number .

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Generator numberGenerator number

Example 2Example 2 : : let p= 11 , a=3 ,test a is generator number let p= 11 , a=3 ,test a is generator number

or not ? or not ? sol:sol: Ø= p-1 = 10 , factors of 10 = {2 , 5}Ø= p-1 = 10 , factors of 10 = {2 , 5} q1 = 2 ,q2 = 5q1 = 2 ,q2 = 5 w1 = 3w1 = 310/210/2 mod 11 = 1== 1mod 11 = 1== 1 w2 = 3w2 = 310/510/5 mod 11 = 9 <> 1 mod 11 = 9 <> 1 i.e a not generator number .i.e a not generator number .

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Generator numberGenerator number

Example 3 :Example 3 : let p= 41 , a=2 ,test a is generator number let p= 41 , a=2 ,test a is generator number

or not ? or not ? sol:sol: Ø= p-1 = 40 , factors of 40 = {2 , 2 , 2, 5}Ø= p-1 = 40 , factors of 40 = {2 , 2 , 2, 5} q1 = 2q1 = 211 = 2 ,q2 = 2 = 2 ,q2 = 222 = 4,q3 = 2 = 4,q3 = 233 = 8 = 8 q4 = 5q4 = 5 w1 = 2w1 = 240/240/2 mod 41 = 0.98 <> 1 mod 41 = 0.98 <> 1 w2 = 2w2 = 240/440/4 mod 41 = 40 <> 1 mod 41 = 40 <> 1

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Generator numberGenerator number

w2 = 2w2 = 240/840/8 mod 41 = 32 <> 1 mod 41 = 32 <> 1

w2 = 2w2 = 240/540/5 mod 41 = 10 <> 1 mod 41 = 10 <> 1

i.e a generator numberi.e a generator number

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Encryption Encryption

Sender B must do the following :Sender B must do the following :

1- Obtain the public key (p , a , d ) from1- Obtain the public key (p , a , d ) from

the receiver A.the receiver A.

2- Choose an integer k such that :2- Choose an integer k such that :

1 < k < p-2 1 < k < p-2

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EncryptionEncryption

3- Represent the plaintext as an integer m 3- Represent the plaintext as an integer m where where 0 < m < p-1 0 < m < p-1

4- compute (y) as follows :4- compute (y) as follows :

y = ay = akk mod p mod p

5- compute (z) as follows :5- compute (z) as follows :

z = (dz = (dkk * m ) mod p * m ) mod p

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EncryptionEncryption

6- Find the ciphertext (C) as follows :6- Find the ciphertext (C) as follows :

C= ( y , z )C= ( y , z )

7- The sender B send C to The receiver A .7- The sender B send C to The receiver A .

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DecryptionDecryption

Receiver A must do the following :Receiver A must do the following :

1- Obtain the ciphertext (C) from B .1- Obtain the ciphertext (C) from B .

2- compute (r) as follows :2- compute (r) as follows : r = yr = yp-1-xp-1-x mod p mod p

3- Recover the plaintext as follows:3- Recover the plaintext as follows: m = ( r * z ) mod p m = ( r * z ) mod p

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ExampleExample

Let p = 11 and a generator number = 2Let p = 11 and a generator number = 2

and select integer number x = 5and select integer number x = 5

calculate d = 2calculate d = 255 mod 11 = 10 mod 11 = 10

Then Then

public key = ( 11 , 2 , 10)public key = ( 11 , 2 , 10)

private key = (5)private key = (5)

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ExampleExample

Plaintext = AgePlaintext = Age

Represent the plaintext as integer value as Represent the plaintext as integer value as follows:follows:

The new plaintext = ( 1 7 5 )The new plaintext = ( 1 7 5 )

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ExampleExample

Encryption (sender):Encryption (sender):

y = ay = akk mod p , z = (d mod p , z = (dkk * m ) mod p * m ) mod p

Choose an random integer value k = 6Choose an random integer value k = 6

yyAA = 2 = 266 mod 11 = 9 mod 11 = 9

zzAA = (10 = (1066*1) mod 11 = 1*1) mod 11 = 1

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ExampleExample

Choose an random integer value k = 4Choose an random integer value k = 4 yygg = 2 = 244 mod 11 = 5 mod 11 = 5 zzgg = (10 = (1044*7) mod 11 = 7*7) mod 11 = 7

Choose an random integer value k = 7Choose an random integer value k = 7 yyee = 2 = 277 mod 11 = 7 mod 11 = 7 zzee = (10 = (1077*5) mod 11 = 6*5) mod 11 = 6

Ciphertext = (9,1) (5,7) (7,6) Ciphertext = (9,1) (5,7) (7,6)

The sender B send the ciphertext to the receiver A.The sender B send the ciphertext to the receiver A.

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ExampleExample

The receiver decrypt the ciphertext as follows :The receiver decrypt the ciphertext as follows :

Compute (r) and (m) whereCompute (r) and (m) where

r = yr = yp-1-xp-1-x mod p , m = ( r * z ) mod p mod p , m = ( r * z ) mod p

r1= 9r1= 911-1-511-1-5 mod 11 = 1 mod 11 = 1

m1= (1*1) mod 11= 1m1= (1*1) mod 11= 1

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ExampleExample

r1= 5r1= 511-1-511-1-5 mod 11 = 1 mod 11 = 1

m2 = ( 1 * 7 ) mod 11 = 7m2 = ( 1 * 7 ) mod 11 = 7

r1= 7r1= 711-1-511-1-5 mod 11 = 10 mod 11 = 10

m3 = ( 10 * 6 ) mod 11 = 5m3 = ( 10 * 6 ) mod 11 = 5

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ExampleExample

The receiver find the plaintext ( 1 7 5 )The receiver find the plaintext ( 1 7 5 )

Convert the plaintext to letters = AgeConvert the plaintext to letters = Age

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End End

For more information about For more information about EL_Gamal cryptosystem please visit :EL_Gamal cryptosystem please visit :

http://en.wikipedia.org/wiki/Taher_ElGamalhttp://en.wikipedia.org/wiki/Taher_ElGamal