Suppose that you receive a digital certificate that contains M and (h(M)]CA, where M = (Alice, Alice's public key) and "CA" is Certificate Authority. Assuming that you trust the CA, after verifying the signature on the certificate, what do you then know about the identity of the sender of the certificate? Remember that: [Xleob indicates encryption via Bob's private key to X (signature) h(X) indicates the cryptographic hash function applied to X O The sender is Alice O The sender is Trudy I cannot deduce the identity of the sender from the certificate The sender is the Certificate Authority

I just want to make sure my work is correct. Also, try to provide an explaination if possible.

Can anyone please explain to me why we have such equations below in part b) and c)? They are the solutions to the questions, but I can't really understand why and how to get that. Many thanks. ==== QUESTION ==== In the Shamir secret sharing scheme, we distribute a secret among q different users as follows. If our secret is a message (m1, . . . , mk) from V (k, q) then, we encode it as a codeword of the Reed-Solomon RSk(q) and give one coordinate to each user. In this problem, we will use q = 7, k = 4 and the parity check matrix H4 below for RS4(7). H4 = [1 1 1 1 1 1 1, 0 1 2 3 4 5 6, 0 1 4 2 2 4 1] a) Write a generator matrix for RS4(7) b) A new secret is selected and user #1 receives share value 0, user #2 receives share value 6 and user #3 receives share value 1 and are collaborating to discover the new secret. Explain why they can’t recover the secret with only this information. c) Now, suppose users #1, #2 and #3 (as in the previous item) discover, in addition to the values of…

Consider the following encryption scheme for a block cipher (using the notationfrequently used in class).y1 = ek(x1), y2 = ek(x1 ⊕ x2), y3 = ek(x1 ⊕ x2 ⊕ x3), etc.a. Describe the corresponding decryption scheme.b. Can encryption and/or decryption be run in parallel for this scheme?c. Taking into account that the operation of XOR is cheap, determinewhether or not this scheme improves on ECB mode or not