本次澳洲代写是密码学的一个assignment
Instructions
This is an individual assignment. The aim of the assignment is that the student applies
concepts and methods studied in weeks 2-11 to solve problems on secret sharing and
applications of elliptic curves to cryptography.
The assignment has a value of 151 points and is worth 40% of the unit marks. It
consists of four problems that are to be solved.
GUILLERMO PINEDA-VILLAVICENCIO
Problems
(1) This question is about secret sharing.
(a) You set up a (3; 29) Shamir threshold scheme, working modulo the prime 211.
Three of the shares are (1; 4); (2; 5), and (3; 6). Another share is (4; x), but
the part denoted by x is unreadable. Find the correct value of x, the relevant
polynomial, and the message. Justify all your steps.
(b) In a (4; 31) Shamir threshold scheme, working mod the prime 223. The shares
(1; 8); (2; 16); (3; 32), and (4; 64) were given to Alice, Bob, Jerry, and Charles.
Calculate the corresponding interpolation polynomial p(x) modulo 223; that
is, write p(x) = a0+a1x+a2x2+a3x3 with a0; a1; a2; a3 2 Z223. Also, identify
the secret.
(c) Verify the solutions of Parts (a) and (b) in sagemath.
Part (a)
The student receives 26 marks if all the steps of the computation are correct
and he/she gives an answer. This includes 5 marks for setting up a matrix
equation, 10 marks for giving the adjugate matrix of a relevant matrix mod
211, 2 marks for the computation of a relevant determinant, 3 marks for the
computation of a relevant inverse mod 211, and 6 marks for giving the correct
answers. For different level of correctness the students receives between 25
and 0 marks.
Part (b)
The student receives 20 marks for each correctly justified step in his/her
answer. This includes 18 marks for the computation of the polynomial mod
223. The student needs to justify each step to get the full marks.
For different level of correctness the students receives between 19 and 0
marks.
Part (c)
In each case, the student receives 3 marks if a correct sagemath code is
provided. For different levels of correctness, the student receives between 2
and 0 marks.
