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?
Question
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?
Solution
The security of this protocol depends on several factors:
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The strength of the encryption algorithm: If a strong encryption algorithm is used, it would be computationally infeasible for an attacker to decrypt the messages without the corresponding private keys.
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The security of the private keys: If the private keys are kept secure and are not leaked, then the protocol is secure. However, if an attacker can obtain the private keys, then they can decrypt the messages.
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The security of the broadcast channel: If the broadcast channel is secure and cannot be eavesdropped on, then the protocol is secure. However, if an attacker can intercept the broadcasts, then they can obtain the encrypted messages.
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The authenticity of the public keys: If an attacker can trick Aldebaran into using a public key that they control, then they can decrypt the messages. This is known as a man-in-the-middle attack.
In conclusion, the protocol can be secure, but it depends on the security of the encryption algorithm, the private keys, the broadcast channel, and the authenticity of the public keys.
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 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.
. 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.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 identifies their public key in the broadcast and obtains the message m. 1 (d) 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′ ).
(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
) 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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