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the fact that Alice is able to compute E(P(s)) does not guarantee she will indeed send E(P(s)) to Bob, rather than some completely unrelated value.
📖 事实是Alice可以计算出E(P(s))但不能保证Alice一定会发送E(P(s))给Bob, 而不是发送一下完全不相关的值.
For α∈F∗p [1], let us call a pair of elements (a,b) in G an α-pair if a,b≠0 and b=α⋅a.
The KC Test proceeds as follows.
- Bob chooses random α∈F∗p and a∈G. He computes b=α⋅a.
- He sends to Alice the "challenge" pair (a,b). Note that (a,b) is an α-pair.
- Alice must now respond with a different pair (a′,b′) that is also an α-pair.
- Bob accepts Alice’s response only if (a′,b′) is indeed an α-pair. (As he knows α he can check if b′=α⋅a′.)
📖 对于α∈F∗p, 我们称在G域中的元素(a,b)在满足a,b≠0 and b=α⋅a时为α-pair
系数测试知识流程如下所示:
- Bob从域中选择随机数α∈F∗p以及a∈G. 他计算了b=α⋅a.
- 他发送Alice挑战对(a,b). 注意到 (a,b)是一个α-pair.
- Alice必须使用不同的α-pair (a′,b′)来进行回应.
- Bob接受了Alice的回应如果(a′,b′)确实也是α-pair(我们知道α可以用于检查, 如果b′=α⋅a′)
📖 F∗p是Fp的无零域, 它与part I中提到的Z*p是一样的.