Consider any language $L$. Define $s(L) \in {\lbrace 0, 1 \rbrace}^\omega$ (an infinite sequence of bits) by the recursive formula


Here $\chi_L$ is the characteristic function of $L$ i.e. $\chi_L(w)=1$ for $w \in L$, $\chi_L(w)=0$ for $w \notin L$

A language $U$ is called a "universal (closed) predictor" when

$$\forall L \in \mathsf{P} \, \forall n>>0:s(L)_n=\chi_U(s(L)_{<n})$$

It is easy to see $U \notin \mathsf{P}$ by considering $L = U^c$. However, $U$ can be recursive. To give an example, consider the language decided by the following algorithm $A$. Given input $w$, $A$ runs all possible programs in shortlex order, allowing each to execute for time $t(|w|)$ where $t$ is a function of superpolynomial growth. Once it reaches a program $R$ that outputs $w$ plus one or more bits and doesn't halt, $A$ outputs the first bit $R$ outputted after $w$. It easy easy to see that (under mild conditions on $t$) $A$ always halts and the language it decides is a universal predictor. $A's$ time complexity is approximately $2^nt(n)$

Given $a \in {\lbrace 0, 1 \rbrace}^\omega$, define $s(L, a)$ by

$$s(L, a)_{2n} = \chi_L(s(L, a)_{<2n})$$ $$s(L, a)_{2n + 1} = a_n$$

A language $V$ is called a "universal open predictor" when

  1. $\forall w \in V : |w|$ is even
  2. $\forall L \in \mathsf{P}, a \in {\lbrace 0, 1 \rbrace}^\omega \, \forall n>>0:s(L, a)_{2n}=\chi_V(s(L, a)_{<2n})$

[I am using 0-based indices so $|s_{<2n}| = 2n$]

Again it is easy to see $V \notin \mathsf{P}$ but $V$ can be in $\mathsf{E}$

Is there $V$ a universal open predictor s.t. $\mathsf{P}^V=\mathsf{NP}^V$?

I'm especially interested in having either a specific example of such $V$ or a proof such $V$ doesn't exist under reasonable assumptions such as $\mathsf{P} \ne \mathsf{NP}$

The question might seem strange, so I'll briefly outline my motivation for it. I'm interested in AIXI-like models of aritifical intelligence. Here $L$ plays the role of the environment, which I assume to be efficiently computable, and $a$ plays the role of the actions of the agent itself. Given a positive answer for my question, it is possible to construct an agent efficiently computable relative to $V$ which optimizes a given efficiently computable utility function $u$ by choosing its future actions s.t. $u$ is maximized assuming the environment behaves according to the prediction of $V$

  • $\begingroup$ In terms of relativization point of view, V=PSPACE works. $\endgroup$
    – Tayfun Pay
    Feb 16, 2013 at 22:37

1 Answer 1


I am not familiar with the notion of universal predictor, and I did not follow everything you wrote; in particular, I did not follow your sketch of the proof of existence of a universal predictor in E. But assuming that there exists a universal open predictor that belongs to E, the answer to your question is positive. And I am afraid that you will probably be disappointed by the reason.

Edit in revision 2: I changed the construction in response to the added restriction in revision 2 of the question, but the general idea is the same: mix a hard language into a universal open predictor in such a way that the definition of universal open predictor does not notice the difference.

Let T = {0|x|11x: x∈{0,1}*}. Note that every word in T has an even length. An important property of T is that no word in T is a proper prefix of another word in T. This implies that if the symmetric difference between two languages V and W is contained in T, then for every infinite sequence s∈{0,1}ω, there is at most one n such that χV(s<2n)≠χW(s<2n), and in particular V is a universal open predictor if and only if W is a universal open predictor.

It is easy to see that there exists an EXPSPACE-complete language which is a subset of T. Let L be such a language. Let V be a universal open predictor that belongs to E, and hence also to EXPSPACE. Define a language W = L ∪ (VT). Because V is a universal open predictor and the symmetric difference between V and W is contained in T, W is also a universal open predictor. It is easy to see that W is EXPSPACE-complete, and therefore PW = NPW = EXPSPACE. This concludes that W satisfies the desired property.

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    $\begingroup$ Ha! Good point, but you caught me on a technicality. I corrected the definition to state that a universal open predictor has only even-length words. "Universal predictor" is my own terminology. The concept is inspired by arxiv.org/abs/cs/0606070 but Legg avoids time complexity considerations $\endgroup$
    – Vanessa
    Feb 16, 2013 at 22:00
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    $\begingroup$ @Squark: Hmm, I guess that the idea of mixing in a hard language works also with the revised definition and therefore I guess that this is not just a technicality, but I have to think more. $\endgroup$ Feb 16, 2013 at 22:45
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    $\begingroup$ @Squark: I thought more. :) $\endgroup$ Feb 16, 2013 at 23:08
  • $\begingroup$ OK, this is not a technicality. Thx! $\endgroup$
    – Vanessa
    Feb 17, 2013 at 20:31

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