(15 marks) One of the approaches below is ”secure”. 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.(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). (c) 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) and broadcasts (pkdest, m). Borealis identi-fies their public key in the broadcast and obtains the message m.
Question
(15 marks) One of the approaches below is ”secure”. 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.(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). (c) 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) and broadcasts (pkdest, m). Borealis identi-fies their public key in the broadcast and obtains the message m.
Solution
(b) Secure: This approach is secure because it achieves both notions of confidentiality. The first notion of confidentiality is that an adversary cannot learn anything about the message. This is achieved because the message is encrypted with Chandra's public key, and only Chandra has the corresponding private key to decrypt it. The second notion of confidentiality is that an adversary cannot learn who the intended recipient of the message is. This is achieved because the destination address is also encrypted with Chandra's public key. Only Chandra can decrypt the destination address to reveal Borealis's public key. Therefore, an adversary cannot determine who the intended recipient is.
(c) Insecure: This approach is insecure because it does not achieve the second notion of confidentiality. While the message is initially encrypted with Chandra's public key, Chandra then decrypts the message and broadcasts it along with the destination address. This means that any adversary can see the decrypted message when Chandra broadcasts it. Therefore, an adversary can learn the contents of the message, violating the second notion of confidentiality. The first notion of confidentiality is maintained because the adversary still cannot determine who the intended recipient is.
Similar Questions
(15 marks) One of the approaches below is ”secure”. 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).
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?
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.
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))
) 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))
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