Let's say I have $k$ distributions, where $k$ is polynomially large, $D_1, D_2, \ldots, D_k$ such that each $D_i$ is computationally indistinguishable from the uniform distribution.

Is it true that the distribution $D_1 D_2 \ldots D_k$ is also computationally indistinguishable from $k$ copies of the uniform distribution?

This trivially holds if each $D_i$ is efficiently samplable. But let's say they are not.

Does the fact still remain true, by some clever way to bypass the samplability requirement?


1 Answer 1


This is a very interesting question. I looked around and found a paper called Computational Indistinguishability: A Sample Hierarchy by Goldreich and Sudan. This contains a proof that it doesn't hold.

  • $\begingroup$ Just a clarification. This paper talks about two distributions, and the setting when we are given $k$ samples from any one distribution out of the two. But, here, we are either given one sample each from $k$ different distributions (each computationally indistinguishable from the uniform), or we are given $k$ samples from the uniform distribution. Do you think the techniques that work for the first setting (that of the paper) also work for the second setting (that of my question)? $\endgroup$
    – BlackHat18
    Jul 22, 2021 at 13:58
  • $\begingroup$ There are also references in the paper to previous work that deals with the more basic question. My intuition says that this should translate to similar settings, but of course intuition always needs to be checked. $\endgroup$ Jul 22, 2021 at 18:39

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