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applied cryptography - protocols, algorithms, and source code in c

This scheme presented in 14681469 allows alice and bob

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Unformatted text preview: U·V0. The shadows are the products U·Vi, where i is a number from 1 to n. Any m shadows can be used to solve the m * m system of linear equations, where the unknowns are the coefficients of U. UV0 can be computed from U. Any m -1 shadows cannot solve the system of linear equations and therefore cannot recover the secret. Advanced Threshold Schemes The previous examples illustrate only the simplest threshold schemes: Divide a secret into n shadows such that any m can be used to recover the secret. These algorithms can be used to create far more complicated schemes. The following examples will use Shamir’s algorithm, although any of the others will work. To create a scheme in which one person is more important than another, give that person more shadows. If it takes five shadows to recreate a secret and one person has three shadows while everyone else has only one, then that person and two other people can recreate the secret. Without that person, it takes five to recreate the secret. Two or more people could get multiple shadows. Each person could have a different number of shadows. No matter how the shadows are distributed, any m of them can be used to reconstruct the secret. Someone with m -1 shadows, be it one person or a roomful of people, cannot do it. In other types of schemes, imagine a scenario with two hostile delegations. You can share the secret so that two people from the 7 in Delegation A and 3 people from the 12 in Delegation B are required to reconstruct the secret. Make a polynomial of degree 3 that is the product of a linear expression and a quadratic expression. Give everyone from Delegation A a shadow that is the result of an evaluation of the linear equation; give everyone from Delegation B a shadow that is the evaluation of the quadratic equation. Any two shadows from Delegation A can be used to reconstruct the linear equation, but no matter how many other shadows the group has, they cannot get any information about the secret. The same is true for Delegation B: They can get three shadows together to reconstruct the quadratic equation, but they cannot get any more information necessary to reconstruct the secret. Only when the two delegations share their equations can they be multiplied...
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