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Example 11.3 We illustrate the construction in Figure 11.4. Suppose K is the key. The value K is given to each of the three input wires of the final or gate. Next, we consider the and gate corresponding to the clause
Thus, every participant receives two elements of Lets prove that the scheme is perfect. First, we verify that each basis subset can compute K. The authorized subset {P1, P2, P4} can compute The subset {P1, P3, P4} can compute Finally, the subset {P2, P3} can compute
Thus any authorized subset can compute K, so we turn our attention to the unauthorized subsets. Note that we do not need to look at all the unauthorized subsets. For, if B1 and B2 are both unauthorized subsets, In each case, it is easy to see that K cannot be computed, either because some necessary piece of random information is missing, or because all the shares possessed by the subset are random. For example, the subset {P1, P2} possesses only the random values a1, b1, a2, c1. As another example, the subset {P3, P4} possesses the shares b2, K - c1, K - a1 - a2, K - b1 - b2. Since the values of c1, a1, a2, and b1 are unknown random values, K cannot be computed. In each possible case, an unauthorized subset has no information about the value of K. We can obtain a different scheme realizing the same access structure by using a different circuit. We illustrate by returning again to the access structure of Example 11.2. Example 11.4 Suppose we convert the formula (11.1) to the so-called conjunctive normal form: (The reader can verify that this formula is equivalent to the formula (11.1).) If we implement the scheme using the circuit corresponding to formula (11.2), then we obtain the following:
We leave the details for the reader to check. We now prove that the monotone circuit construction always produces a perfect secret sharing scheme.
Copyright © CRC Press LLC
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