(Encrypt-then-Sign-then-Encrypt) Recall that the original crypto-system, Aldebaran computes the double encryption cmsg = Enc(pkC, c′ msg) where c ′ msg = Enc(pkB, m). In this approach, Aldebaran computes σ = Sign(sk′ A, c′ msg), and then encrypts cσ = Enc(pkC, σ). Aldebaran sends cσ along with their usual broadcast. Chandra performs her usual steps, as well as decrypting to obtain σ = Dec(skC, cσ). She broadcasts σ along with her usual broadcast (pkB, c′ msg) for Borealis. Lastly, Borealis, believes the message should have come from Aldebaran. He runs Verify(pk′ A, c′ msg, σ), and decrypts c ′ msg to obtain m only if the signature accepts.
Question
(Encrypt-then-Sign-then-Encrypt) Recall that the original crypto-system, Aldebaran computes the double encryption cmsg = Enc(pkC, c′ msg) where c ′ msg = Enc(pkB, m). In this approach, Aldebaran computes σ = Sign(sk′ A, c′ msg), and then encrypts cσ = Enc(pkC, σ). Aldebaran sends cσ along with their usual broadcast. Chandra performs her usual steps, as well as decrypting to obtain σ = Dec(skC, cσ). She broadcasts σ along with her usual broadcast (pkB, c′ msg) for Borealis. Lastly, Borealis, believes the message should have come from Aldebaran. He runs Verify(pk′ A, c′ msg, σ), and decrypts c ′ msg to obtain m only if the signature accepts.
Solution
The process described is a cryptographic protocol known as "Encrypt-then-Sign-then-Encrypt". Here's a step-by-step breakdown:
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Aldebaran, the sender, first encrypts the message 'm' using Borealis's public key (pkB). This results in an encrypted message c'msg = Enc(pkB, m).
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Aldebaran then signs this encrypted message using their own private key (sk'A). This results in a signature σ = Sign(sk'A, c'msg).
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Aldebaran then encrypts this signature using Chandra's public key (pkC). This results in an encrypted signature cσ = Enc(pkC, σ).
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Aldebaran sends this encrypted signature along with their usual broadcast.
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Chandra, the intermediary, performs her usual steps and also decrypts the encrypted signature using her private key (skC) to obtain the original signature σ = Dec(skC, cσ).
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Chandra then broadcasts this signature along with her usual broadcast (pkB, c'msg) to Borealis.
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Borealis, the receiver, believes the message should have come from Aldebaran. He verifies the signature using Aldebaran's public key (pk'A), the encrypted message, and the signature: Verify(pk'A, c'msg, σ).
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If the signature is verified, Borealis decrypts the encrypted message using his private key to obtain the original message 'm'.
This protocol ensures that the message is securely transmitted from Aldebaran to Borealis via Chandra, with the assurance that the message indeed came from Aldebaran (as verified by the signature).
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.
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.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′ ).
) 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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