Hybrid argument without efficient samplability

BlackHat18

11/21/22, 10:46 AM

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 + 0

indistinguishability

Score:3

Crypto

Yehuda Lindell

11/22/22, 1:10 AM

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.

0 + 2

BlackHat18

11/22/22, 1:58 PM

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)?

Yehuda Lindell

11/22/22, 6:39 PM

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.

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: Hybrid argument without efficient samplability

TH: อาร์กิวเมนต์ไฮบริดโดยไม่มีการสุ่มตัวอย่างที่มีประสิทธิภาพ

RO: Argument hibrid fără eșantionare eficientă

RU: Гибридный аргумент без эффективной выборки

VI: Đối số lai mà không có khả năng lấy mẫu hiệu quả

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