Score:2

is $F_{k_{1}}(m)||F_{k_{2}}(F_{k_{1}}(m))$ always a PRF? when F is a PRF

vn flag

is $F_{k_{1}}(m)||F_{k_{2}}(F_{k_{1}}(m))$ always a PRF? when F is a PRF

As an intuition it seems to me that the answer is "NO" as the two halves of the output depends on each other

Ievgeni avatar
cn flag
Is it homework?
Doron Bruder avatar
vn flag
NOPE, philosophy thinking at midnight. @levgeni Im not even sure if it's provable.
pe flag
What happens when two $F_{k_1}$ outputs collide, and what is the probability of that happening?
Geoffroy Couteau avatar
cn flag
Not only this seems to be a secure PRF to me, I believe that even if you replace $F_{k_2}$ by a *weak* PRF (I.e. a PRF only guaranteed to look random on random inputs), the whole thing is still a PRF. Try writing the security proof, it's just two hybrids! (for the case with two PRFs at least - with a weak PRF it looks much harder and much less clear)
Mark avatar
ng flag
@GeoffroyCouteau It is unclear that it works when $F$ is a weak PRF for me. You could imagine hard-coding a particular point $F_k(a) = b$ into a PRF. This will still be a weak PRF, as the probability that $a$ is uniformly chosen will be negligible. Applying this construction to the weak PRF will not yield a PRF though.
Geoffroy Couteau avatar
cn flag
I was talking about using two different PRFs, a strong PRF F for the $F_{k_1}$ part, and a weak PRF F' for the $F_{k_2}$ part.
Geoffroy Couteau avatar
cn flag
@Mark But if both parts are replaced by a weak PRF, then clearly (and provably) the full construction is not in general a strong PRF.
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