In ECDH, when two person wants to share private key, they first select a point G on elliptic curve and after that, each of them pick a random integer a and b, respectively, and multiply with G. After the multiplication each of them shares aG and bG with each other and after that, they multiply again using their keys a(bG) and b(aG), respectively, and creates a shared key between themselves. However, if one person wants to communicate with a group of person (more than 2) using a shared key utilizing ECDH, how he can use this method? because each of the person may choose different integer while establishing the key.
How to employ ECDH in key exchange with a group of people?
hi
In ECDH, suppose three persons wants to share private key, where a, b, and c, secrete data, the first step or initia step l aG, bG , and cG, and then the second step
1- A sends to B , the value aG, B calculate baG, then B sends it(baG) to C then C calculate cbaG,
2- B sends to C bG to calculate the value cbG, then C send to A ( the value cbG) to calculate acbG.
3- C send to A cG to calculate the value acG, then A sends it to B ( the value acG), to calculate bacG.
done
Dear Salah Albermany
Thanks for your answer. But, that can be possible when a,b, and c can communicate with each other. I have provided a case in the image. Please check this image.
The definition on Wikipedia states it is a 2-party protocol. Adapting this to an [i > 2] party-scenario would make it something else.
Speaking as a non-expert on the subject, perhaps it's the elliptic nature of the curve that limits it to a 2-party solution? Have you explored this?
Maybe therein lies the answer to the 3DH problem. (Just a fun name off the top of my head)
Cheers
Jake
If you use only elliptic curve means only additive group operation to exchange keys Between more than 2 parties then you require more operation and more rounds of communication between each participant. The best way to solve this to employ paring based multi-party key exchange. This is one of the first papers on this topic https://www.semanticscholar.org/paper/A-One-Round-Protocol-for-Tripartite-Diffie%E2%80%93Hellman-Joux/845e96c20e5a5ff3b03f4caf72c3cb817a7fa542
Hi Dr. Anik Islam Abhi
Try with the following, they could help you.
http://ijns.jalaxy.com.tw/contents/ijns-v15-n4/ijns-2013-v15-n4-p256-264.pdf
https://crypto.stackexchange.com/questions/1025/can-one-generalize-the-diffie-hellman-key-exchange-to-three-or-more-parties
Sincerely
One possible protocol could be the following:
1. A, B, C each generate their private keys $x_{A}, x_{B}, x_{C}$.
2. A, B, C each calculate $y_{A}=g^{x_{A}}, y_{B}=g^{x_{B}}, y_{C}=g^{x_{C}}$.
3. A sends $y_{A}$ to $\mathrm{B}, \mathrm{B}$ sends $y_{B}$ to $\mathrm{C}, \mathrm{C}$ sends $y_{C}$ to $\mathrm{A}$.
4. A calculates $z_{C A}=y_{C} x_{A}, B$ calculates $z_{A B}=y_{A}^{x_{B}}, C$ calculates $z_{B C}=y_{B}^{x_{C}}$.
5. A sends $z_{C A}$ to $\mathrm{B}, \mathrm{B}$ sends $z_{A B}$ to $\mathrm{C}, \mathrm{C}$ sends $z_{B C}$ to $\mathrm{A}$.
6. A calculates $k_{B C A}=z_{B C} x_{A}, B$ calculates $k_{C A B}=z_{C A} x_{B}, C$ calculates $k_{A B C}=z_{A B}^{x_{C}}$
The above equality means that the three parties now know a common secret $k_{A B C}=k_{C A B}=k_{B C A}$
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