. For each approach, state Secure or Insecure, and explain why that approach does or does not achieve the two desired notions of confidentiality described above.(e) Aldebaran computes cmsg = Enc(pkC, Enc(pkB, m)), cdest = Enc(pkC, pkB) and then broadcasts (pkC, cmsg, cdest). Chandra observes the broadcast containing her public key. She then computes c ′ msg = Dec(skC, cmsg), pkdest = Dec(skC, cdest), and broadcasts (pkdest, c′ msg). Lastly, Borealis observes a broadcast containing their public key, and obtains the message as m = Dec(skB, c′ msg).
Question
. For each approach, state Secure or Insecure, and explain why that approach does or does not achieve the two desired notions of confidentiality described above.(e) Aldebaran computes cmsg = Enc(pkC, Enc(pkB, m)), cdest = Enc(pkC, pkB) and then broadcasts (pkC, cmsg, cdest). Chandra observes the broadcast containing her public key. She then computes c ′ msg = Dec(skC, cmsg), pkdest = Dec(skC, cdest), and broadcasts (pkdest, c′ msg). Lastly, Borealis observes a broadcast containing their public key, and obtains the message as m = Dec(skB, c′ msg).
Solution
This approach is Secure.
The reason is that it achieves both desired notions of confidentiality. The first notion of confidentiality is that an adversary should not be able to learn anything about the message. In this approach, the message is encrypted twice, first with Borealis's public key and then with Chandra's public key. An adversary who does not have access to either of these private keys cannot decrypt the message and therefore cannot learn anything about it.
The second notion of confidentiality is that an adversary should not be able to learn who the intended recipient of the message is. In this approach, the recipient's public key is also encrypted with Chandra's public key. An adversary who does not have access to Chandra's private key cannot decrypt the recipient's public key and therefore cannot determine who the intended recipient is.
Therefore, this approach is secure as it achieves both desired notions of confidentiality.
Similar Questions
For each approach, state Secure or Insecure, and explain why that approach does or does not achieve the two desired notions of confidentiality described above.(a) Aldebaran computes c = Enc(pkB , m) and broadcasts (pkB , c). Borealis observes the broadcast containing their public key and obtains the message as m = Dec(skB , c). (b) Aldebaran computes cmsg = Enc(pkC , m), cdest = Enc(pkC , pkB ) and broadcasts (pkC , cmsg, cdest). Chandra observes the broadcast containing her public key. She then decrypts the des-tination address as pkdest = Dec(skC , cdest) and broadcasts (pkdest, cmsg). Borealis then obtains the message as m = Dec(skB , cmsg).
Aldebaran computes cmsg = Enc(pkC , m), cdest = Enc(pkC , pkB ) and broadcasts (pkC , cmsg, cdest). Chandra observes the broadcast containing her public key. She then computes m = Dec(skC , cmsg) and pkdest = Dec(skC , cdest). Lastly, she re-encrypts c′ = Enc(pkdest, m) and broadcasts (pkdest, c′). Borealis identifies their public key in the broadcast and obtains the message m = Dec(skB , c′). state Secure or Insecure, and explain why that approach does or does not achieve the two desired notions of confidentiality described above.
Aldebaran computes cmsg = Enc(pkC, m), cdest = Enc(pkC, pkB) and broadcasts(pkC, cmsg, cdest). Chandra observes the broadcast containing her public key. She then decrypts the destination address as pkdest = Dec(skC, cdest) and broadcasts (pkdest, cmsg). Borealis then obtains the message as m = Dec(skB, cmsg).Is it secure?
) The consortium decide to implement the final approach described in question 1, using Elgamal public key encryption with the following parameters: (p, g) = (103, 5). Aldebaran’s public key is pkA = 51, Borealis’ public key is pkB = 55 and Chandra’s public key is pkC = 38. Some time later, Chandra receives a different broadcast (38, cmsg, cdest) where cdest = (55, 10) and cmsg = (c1, c2) = ((101, 28),(90, 94)). i. (2 marks) Confirm whether or not Chandra’s public key corresponds to her secret key skC = 22. ii. (5 marks) Who is the final intended recipient of the message? (Hint: compute the Elgamal decryption Dec(skC, cdest) and compare with the known public keys.) iii. (6 marks) Hence, what does Chandra broadcast? (Hint: compute the Elgamal decryptions Dec(skC, c1) and Dec(skC, c2))
Some time later, Chandra receives a different broadcast (38, cmsg, cdest) where cdest = (55, 10) and cmsg = (c1, c2) = ((101, 28),(90, 94)). i. (2 marks) Confirm whether or not Chandra’s public key corresponds to her secret key skC = 22. ii. (5 marks) Who is the final intended recipient of the message? (Hint: compute the Elgamal decryption Dec(skC, cdest) and compare with the known public keys.) iii. (6 marks) Hence, what does Chandra broadcast? (Hint: compute the Elgamal decryptions Dec(skC, c1) and Dec(skC, c2))
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