'''Originally Posted By: cox.michaelj6''' Below is the solution for the 3rd problem on practice midterm 2. <br><br>By: Michael Cox, Rohit Krishnan<br><br>3) In a 3 out of 4 Shamir's sharing scheme, give 4 parties 4 unique points on a parabola. Any 3 of the parties may work together to generate a system of 3 equations which may then be used to solve the equation of the parabola (and therefore find the Secret, S, which is identified by the y-intercept). An example follows:<br><br>Alice has (3,1), Bob has (2,3), Charlie has (1,1), and Daisy has (4,-5). If Alice, Bob, and Daisy work together, they generate the following system of equations from y = ax^2 + bx + c and their unique points:<br>Alice: 1=a(3^2)+b(3)+c<br>Bob: 3=a(2^2)+b(2)+c<br>Daisy: -5=a(4^2)+b(4)+c<br><br>The solution to this system of equations is a=-2, b=8, c=-5, which identifies the parabola y=-2x^2 + 8x + -5. At the y-intercept (S), x=0, and y = -5.<br><br>Any combination of 3 (or 4) of the 4 parties may work together to solve the resulting system of equations to determine the secret, S, but 2 parties working together without the other 2 is not enough to identify the equation.