As in our 1 solution above this implies that the

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Unformatted text preview: that the secret is equal to 3 is 10 . Another way to prove this is by assuming that Bob’s share is equal to 10, and then computing what the secret is (again using Property 1 and Lagrange interpolation). You’ll find that the secret is 9 in this case, which means that when the secret is 3, Bob’s share is not equal to 10. As in our 1 solution above, this implies that the probability that the secret is equal to 3 is 10 . Both of these solutions implicitly make use of the following observation: since Alice already has 2 shares, Bob’s share uniquely determines the value of the secret (P(0)). This means that there is a bijection between all possible values of Bob’s share and all possible values of the secret. This is due to property 2: once 2 values of a degree 2 polynomial are known, the third point uniquely determines the polynomial. CS 70, Fall 2013, Midterm #2 5 Name: SID: 4. (15 points) Euler. One way to think of Eulerian tour is as follows: an Eulerian tour is a way of drawing the graph by tracing each edge exactly once and return to the starting point, without lifting the pencil. Now suppose we wish to draw a graph that does not have an Eulerian tour, but is connected and has exactly m vertices of odd degree (m must be even). It is still possible to draw the graph without lifting our pencil and return to our starting point, but...
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This document was uploaded on 01/28/2014.

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