(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
(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 passage describes a multi-step encryption and decryption process involving three parties: Aldebaran, Chandra, and Borealis.
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Aldebaran first encrypts a message 'm' using Borealis's public key (pkB). This encrypted message is then encrypted again using Chandra's public key (pkC). This double-encrypted message is referred to as 'cmsg'.
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Aldebaran also encrypts Borealis's public key using Chandra's public key. This encrypted key is referred to as 'cdest'.
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Aldebaran then broadcasts three pieces of information: Chandra's public key, the double-encrypted message, and the encrypted destination key.
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Chandra, upon seeing her public key in the broadcast, decrypts both 'cmsg' and 'cdest' using her private key (skC). This gives her the encrypted message (now only encrypted with Borealis's public key) and Borealis's public key. She then broadcasts these two pieces of information.
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Borealis, upon seeing their public key in the broadcast, decrypts the message using their private key (skB), finally obtaining the original message 'm'.
This process ensures that the message is securely transmitted from Aldebaran to Borealis, with Chandra acting as an intermediary. Only Borealis, with their private key, can ultimately decrypt and read the original message.
Similar Questions
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?
. 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).
(Sign-then-Double-Encrypt) Aldebaran computes σ = Sign(sk′ A, m), and cσ = Enc(pkC, Enc(pkB, σ)). Aldebaran sends cσ along with their usual broadcast, (pkC, cdest, cmsg). Chandra performs her usual steps, as well as decrypting to obtain c ′ σ = Dec(skC, cσ). She sends it along with her usual broadcast, (pkB, c′ msg) for Borealis. Lastly, Borealis, who will receives the message m, now also obtains σ = Dec(skB, c′ σ ). Borealis believes the message should have come from Aldebaran. He runs Verify(pk′ A, m, σ) and is satisfied only if the signature accepts
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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