Do we know if $\mathsf{ZPP}\subseteq\mathsf{UP}$ known or is there oracles against the hypothesis?

  • $\begingroup$ It is correct. We know that $CoRP \subset UP$ because unique-Sat is Conp-hard and we know the fact that RP is in NP so $ZPP = RP \cap coRP$ is in UP because it is closed under complement. $\endgroup$ Jan 27, 2021 at 10:09
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    $\begingroup$ @MohsenGhorbani $\mathsf{ZPP}\subseteq\mathsf{coRP}\subseteq\mathsf{coNP}\subseteq\mathsf{US}$ but where does $\mathsf{UP}$ come in? $\endgroup$
    – Turbo
    Jan 27, 2021 at 10:15
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    $\begingroup$ I thought UP = US‌. $\endgroup$ Jan 27, 2021 at 10:19

1 Answer 1


There is an oracle $\mathcal O$ such that $\textsf{ZPP}^{\mathcal O}\not\subseteq \textsf{UP}^{\mathcal O}$. The inclusion is not known, and it would therefore require non-relativizing techniques to resolve the question (and probably even non-algebrizing techniques). On the other hand, the inclusion $\textsf{ZPP}\subseteq \textsf{UP}$ is conjectured to be true, because the Derandomization Hypothesis implies $\textsf{P}=\textsf{ZPP}=\textsf{BPP}$, and since $\textsf{P}\subseteq \textsf{UP}$, we get $\textsf{ZPP}\subseteq \textsf{UP}$.

Oracle sketch. The oracle $\mathcal O$ is constructed in phases. For the $n$-th phase, we fix the strings in $\mathcal O\cap \{0,1\}^n$. For a yes-input, we take a random subset $S\subseteq \{0,1\}^{n-1}$, and set $\mathcal O\cap\{0,1\}^n:=0S$. For a no-input, we set $\mathcal O\cap\{0,1\}^n:=1S$.

The language $\mathcal L$ which now has $\mathcal L\in \textsf{ZPP}^{\mathcal O}\setminus \textsf{UP}^{\mathcal O}$ asks the question: on input $1^n$, is it true that (1) $|\mathcal L\cap 0\{0,1\}^{n-1}|>\frac{1}{4}2^{n}$, and (2) not $|\mathcal L\cap 1\{0,1\}^{n-1}|>\frac{1}{4}2^n$? Because of the way we constructed the oracle, for any $n$ we always have either both (1) and (2) are true, or neither (1) nor (2), so it suffices for an $\textsf{RP}$ algorithm to check whether (1) holds, and for a $\textsf{co-RP}$ algorithm to check whether (2) holds. Hence $\mathcal L\in \textsf{RP}^{\mathcal O}\cap \textsf{co-RP}^{\mathcal O}=\textsf{ZPP}^{\mathcal O}$.

The details of diagonalizing against all $\textsf{UP}$-Turing Machines are omitted, but are standard.

  • $\begingroup$ @LieuweVinkhujzen Can you provide a reference for 'standard'? $\endgroup$
    – Turbo
    Jan 31, 2021 at 11:04
  • $\begingroup$ @1.. A good reference is Computational Complexity: A Modern Approach, Chapter 3, by Arora & Barak. Second, if you have the time, the discovery of NP-Intermediate problems by Richard Ladner is instructive: On the structure of polynomial-time reducibility. For a good modern background, you can read the first few pages of Algebrization: A new barrier in Complexity Theory by Scott Aaronson and Avi Widgerson. $\endgroup$ Jan 31, 2021 at 11:33
  • $\begingroup$ I meant 'The details of diagonalizing against all'. What does it imply in the context in oracles? $\endgroup$
    – Turbo
    Jan 31, 2021 at 12:10

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