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
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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