Let y = f(x) and f'(y) = x
Imagine Bob runs a f' cracking service. Imagine Alice has y and wants x. Alice may
or may not know f' however she wishes to take advantage of Bob's f' cracking service
to obtain x. But she doesn't want Bob to know x. Yet she wants Bob to compute it
for her.
Imagine there is a blinding function b, and an unblinding function
b'. Alice sends Bob b(y). Bob produces z=f'(b(y)). Alice extracts x =
b'(b).
Has this been done for RSA etc?
Is it possible to find blinding functions of this nature for any
function in number theory?
Cheers,
Julian.
