ElGamal cryptography Algorithm Encryption Decryption Solved Example in CNS by Vidya Mahesh Huddar

Welcome back. In this video, I will discuss the concept of Elgamal cryptography algorithm. So first we'll understand what is Elgal cryptography algorithm. Elgamal is one of the important aymmetric key cryptography technique based on the discrete algorithm problem. So it was proposed by Taher Algalam in 1985. In this algorithm, the sender encrypt the message using public key and the receiver decrypt the message by using private key. Let us understand the working of Elgamal algorithm. The whole process contains the three steps uh those are key generation, encryption and decryption. In the first step that is in the key generation this step is performed by the Bob who is the receiver. So here Bob select P that is a very large prime number and he select E1 that is a primitive root and he select the D which is a secret key and by using P E1 and D he will calculate E2 by using this formula that is E2 is equal to E1 to D mod of P. After computation the public key become E1, E2P and private key become D. The public key is shared openly with the Alice but the private key is kept secret only. So next step is a encryption. This step is performed by the Alice who wants to send a secure message. Alice encrypt the plain text P. In this first choose the random number. So in this Alice number that is R. By choosing the random number even if the same message is encrypted multiple times different cipher text are generated using Bob's public key Alice compute two cipher text values. A first cipher text component that is C1 is equal to E1 to R mod of P and second cipher text component that is C2 is equal to E2 to R into P mod of P. Uh here P is a prime number and E1 is a primitive root and E2 is already calculated by the Bob. So after computation Alice sends the cipher text that is C1 and C2 with the secure communication channel. So last step is a decryption. This step is performed by the Bob after receiving the cipher text using the private key that is D. Bob recover the original plain text. To recover the version of plain text we need to use this formula that is P is equal to C2 into C1 to D rest to minus1 mod of P. To understand this process we will take one simple example. In that first step is key generation. This is done by the Bob. So here Bob chooses prime number which is equal to 23. E1 is equal to 5 D is equal to 6. So after that he will compute E2 by using this formula as I already discussed this one. So E1D mod of P here we know all the value just we need to substitute those values here. Once you substitute those values here E2 is equal to 5 to 6 23. So 5 to 6 is nothing but 15,625 more 23. Once you solve this one we'll get E2 is equal to 8. Once you calculate the E2, the Bob got the public key as 5823 and private key as D which is equal to 6. So Bob publish the public key and keeps the private key secret only. Next step is encryption. This is done by the Alice. Uh suppose Alice want to send a plain text that is P is equal to 10. To perform the encryption, Alice has to select the random number. In this case, we have selected R is equal to 3. So here she has to calculate cipher text components. Those are C1 and C2. C1 can be calculated by using this formula as I already discussed. So here we know the value of E1, R and P. Just we need to substitute those values here. So once you substitute those values C1 will become 5 to 3 23 which is equal to 125 more 23. Once you solve this one we'll get C1 is equal to 10. So similarly we

need to calculate the C2 value. This can be calculated by using this formula. C2 is equal to P that is a plain text multiplied with E2 to R mod P. This is a prime number. So here also we need to substitute the required values and we need to simplify this one. Once you do this one we will get C2 is equal to 14. So finally we will get the cipher text as C1 and C2 which is equal to 10A 14. So Alice will send this message. The final step is decryption. This is done by the Bob. So Bob receives the cipher text as a 10A 14. So using the private key that is d is = 6 uh the Bob has to recover the plain text. text we need to use the following formula that is p = c2 into c1dus1 mod of p. So first we will calculate c1d c1 is 10 to 6 mod 23. 10 to 6 is nothing but 10 lakh mod 23. So once you do this one you can use calculator for this one. We'll get a answer as a 6. So next one is C1 to 2D rest -1 mod of P. So C1D mod of P is a 6 to -1 mod of 23. 6 -1 mod of 23 which is equivalent to 6 into n 23 which is equal to 1. So here we need to check for which n value we'll get uh this one is equal to 1. So if you goes on changing the n value. So once you put n is equal to 4. So 6 into 4 that is 24. 24 mod 23 which is equal to 1. So this condition is satisfied. It means that n is equal to 4. It means that c1 d2d minus1 which is equal to 4. So we know the c2 value. We have already calculate this one. So we need to substitute those values here. So 14 into this 4 23. So once you solve this one, we'll get a p is equal to 56 23 which is equal to 10. So this 10 is nothing but a plain text. So it means that the encryption and decryption is done successfully. So this is how we can encrypt and decrypt the message by using Elgamal algorithm. I hope the concept is clear. If you like the video, do like and share with your friends. Press the subscribe button for more videos. Press the bell icon for regular updates. Thank you for watching.