3 ms·
No. You have a function F which maps inputs x to outputs y. You transform this function into new one F', that will map the encrypted inputs encrypt(x) to the e
by layoutIfNeeded 6y ago
No.
You have a function F which maps inputs x to outputs y. You transform this function into new one F', that will map the encrypted inputs encrypt(x) to the encrypted outputs encrypt(y), without knowing how to decrypt.
F(x) = y
F'(encrypt(x)) = encrypt(y)
- russfink 6y agoThis is correct. Additionally, the entity performing the computation does not get to learn the answer. This is an important distinction between a technique such as this, and software obfuscation.