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Published 20 hours ago • verified publisher icon FredericJacobs

Is it Function Secure PRF ?

Open PhanDuc opened this issue 7 years ago • 0 comments

Hi Jacobs,

If i have a function F (k , F(k, 0)) = 0 This is definite of secure PRF Let F: K x X -> Y be a PRF Funs[X,Y]: the set of all functions from X to Y S_{F} = { F(k,・) s.t. k ∈ K} ⊆ Funs[X,Y]

So my function is secure PRF or not? I really appreciate any help you can provide.

PhanDuc avatar May 22 '17 06:05 PhanDuc