1. 2019/8/30
Modified ElGamal Elliptic Curve Cryptosystem using Hexadecimal Representation
Author: Ziad E. Dawahdeh, Shahrul N. Yaakob and Ali Makki Sagheer
Publisher: Indian Journal of Science and Technology, 2015
Presenter: 柯懷貿
Date: 2019/03/27
Department of Computer Science and Information Engineering
National Cheng Kung University, Taiwan R.O.C. 1
CSIE CIAL Lab

2. ElGamal Assume x is Bob’s private key and y = g^x is Bob’s public key.
Before sending message M, Alice generates a random number k.
Alice calculates C1 = g^k, C2 = M*(y^k) and sends (C1, C2) to Bob.
Bob calculates C2 = M*(y^k) = M*((g^x)^k) = M*((g^k)^x) = M*(C1^x)
Since Bob knows C1 and x, C2 can be divided to get M.
在多維幾何空間中，我們可以用矩形表示規則。 這意味著規則的每個字段代表不同維度的覆蓋區域。
良好的數據包分類算法必須同時具有更快的速度和更少的存儲要求。
National Cheng Kung University CSIE Computer & Internet Architecture Lab

3. ElGamal on ECC (Maria Method)
Assume x is Bob’s private key and y = x*G is Bob’s public key.
Before sending message M, Alice generates a random number k and M’ = MG.
Alice calculates C1 = k*G, C2 = M+(k*y) and sends (C1, C2) to Bob.
Bob calculates C2 = M’+(k*y) = M’+(k*(x*G)) = M’+(x*(k*G)) = M’+(x*C1)
Since Bob knows C1 and x, C2 can be subtracted to get M’.
在多維幾何空間中，我們可以用矩形表示規則。 這意味著規則的每個字段代表不同維度的覆蓋區域。
良好的數據包分類算法必須同時具有更快的速度和更少的存儲要求。
National Cheng Kung University CSIE Computer & Internet Architecture Lab

4. Can it encrypt by private key & decrypt by public key?
Assume x is Bob’s private key and y = x*G is Bob’s public key.
Before sending message M, Bob generates a random number k and M’ = M*G.
Bob calculates C1 = k*G, C2 = M’+(k*x) and sends (C1, C2) to Alice.
Assume C2 = M’+(k*y) = M’+(k*(x*G)) = M’+(x*(k*G)) = M’+(x*C1)
Since Alice needs to know either k & y or x & C1, she cannot get M’.
在多維幾何空間中，我們可以用矩形表示規則。 這意味著規則的每個字段代表不同維度的覆蓋區域。
良好的數據包分類算法必須同時具有更快的速度和更少的存儲要求。
National Cheng Kung University CSIE Computer & Internet Architecture Lab

5. The Modified Encryption
The Modification of ElGamal Elliptic Curve Cryptosystem (MEGECC) depends on the speeding up of the computation on EGECC using hexadecimal ASCII values by reducing the number of doubling and addition operations.
First, we replace the random number k with sender’s own private key.
Assume xb is Bob’s private key and yb = xb*G is Bob’s public key. Alice will calculates C1 = xa*G , C2 = M+(xa*yb) while xa is Alice’s private key.
But C1 is Alice’s public key, ya, it’s no longer needed to send to Bob after exchanging public keys.
The precincts of all the AU-RSUs, where the OBUs and the AP-RSUs can communicate with each other safely, cover the street roads without any gap.
National Cheng Kung University CSIE Computer & Internet Architecture Lab

6. Hexadecimal Representation
To reduce the number of doubling and addition operations, convert each character in the message M into hexadecimal ASCII value of two digits (h1 h2) then separates the value into two values (h1, h2).
Moreover, convert each value of h1 and h2 to decimal values d1 and d2 respectively.
National Cheng Kung University CSIE Computer & Internet Architecture Lab

7. The Modified Decryption
Next, calculate P1 = d1*G, P2 = d2*G as encoded messages in ECC field.
Alice can send (C1, C2) to Bob where C1 = P1+xa*yb, C2 = P2+xa*yb.
Bob can get (P1, P2) by calculating C1-xa*yb = C1-xa*xb*G = C1-xb*(xa*G) = C1-xb*ya = P1 since Bob knows xb and ya. Then so as P2.
The solution of P1 = d1*G and P2 = d2*G is not difficult for Bob and will not take a long time because the largest value for d1 and d2 in decimal is 15.
National Cheng Kung University CSIE Computer & Internet Architecture Lab

8. Implementation Example
Assume following curve is used that {a, b, p ,G, n} = {1, 3, 31, (1, 6), 41}:
Assume Alice’s private key is 13 and public key is 13(1, 6) = (3, 23). Bob’s private key is 17 and public key is 17(1, 6) = (24, 5). xa*xb*G = 17(2, 23) = 13(24, 5) = (20, 5).
For “H”, (d1, d2) = (4, 8) and P1 = 4(1, 6) = (23, 17), P2 = 8(1, 6) = (18, 5).
Alice can send (C1, C2) to Bob where C1 = (23, 17)+(20, 5) = (4, 28), C2 = (18, 5)+(20, 5) = (24, 26).
National Cheng Kung University CSIE Computer & Internet Architecture Lab

9. Comparison
For “H”, (d1, d2) = (4, 8), calculate P1 = 4G = 2(2G) and P2 = 8G = 2(2(2G)), so the total operations are 2D+3D=5 operations.
Whereas, in Maria method “H” = 72 in ASCII, then calculate 72G = 2(2(2(2(2(2G))+G))) and the total operations are 6D+1A=7 operations.
National Cheng Kung University CSIE Computer & Internet Architecture Lab

10. Results Improvement percentage for "Hello" =
It is clear from following table that the proposed method is better than Maria method and the percentage of improvement increases when the plaintext size increases.
National Cheng Kung University CSIE Computer & Internet Architecture Lab