by Janos Kramar 1114 days ago | Jessica Taylor and Patrick LaVictoire like this | link | parent It may still be possible to get a unique (up to scaling) invariant measure (with infinite sum) over the UTMs by invoking something like the Krein-Rutman theorem and applying it to the Markov chain’s transition operator. I haven’t yet verified that all the conditions hold. This measure would then be an encoding-invariant way to compare UTMs’ “intrinsic complexity” in the sense of “number of bits needed to simulate”.

 by Janos Kramar 1101 days ago | Patrick LaVictoire likes this | link Consider the function $$a(M_1,M_2)=2^{-d(M_1,M_2)-d(M_2,M_1)}$$ where $$d(M_1,M_2)=\min\left(|x|\middle|x\in\{0,1\}^*:\forall y\in\{0,1\}^*: M_1(xy)=M_2(y)\mbox{ unless neither of these halts}\right)$$. The reversible Markov chain with transition probabilities $$p(M_1,M_2)=\frac{a(M_1,M_2)}{\sum_{M'_2}a(M_1,M'_2)}$$ has a bounded positive invariant measure $$\mu(M)=\sum_{M'}a(M,M')$$. Of course, as the post showed, the total measure is infinite. Also, because the chain is reversible and transient, the invariant measure is far from unique - indeed, for any machine $$M_0$$, the measure $$\mu(M)=p^{(0)}(M,M_0)+2\sum_{n=1}^\infty p^{(n)}(M,M_0)$$ will be a bounded positive invariant measure. It seems tempting (to me) to try to get a probability measure by modding out the output-permutations (that the post uses to show this isn’t possible for the full set of UTMs). To this end, consider the set of UTMs that have no output. (These will be unaffected by the output-permutations.) We can try to use the induced sub-digraph on these to build a probability measure $$\mu$$. The measure of each UTM should be a function of the rooted edge-labeled digraph $$G_M$$ rooted at that UTM. The most natural topology on rooted edge-labeled infinite digraphs is the one generated by the sets $$\{G:G'\mbox{ is isomorphic to an induced rooted edge-labeled subgraph of G}\}$$ where $$G'$$ ranges over finite rooted edge-labeled digraphs - we could hope that $$\mu$$ is continuous according to this topology. Unfortunately, this can’t work: if $$\mu(M)>0$$ then $$\mu^{-1}((\frac{1}{2}\mu(M),\infty))$$ must be open, and so it must contain some finite intersection of the generating sets; however, every such intersection that’s nonempty (as this one is) contains infinitely many UTMs, so the total measure must be infinite as well. reply
 by Janos Kramar 1100 days ago | Patrick LaVictoire likes this | link In order to understand what the measure $$\mu$$ that was constructed from $$d$$ will reward, here’s the sort of machine that comes close to $$\sup_M\mu(M)=3$$: Let $$M_0$$ be an arbitrary UTM. Now consider the function $$r(n)=n-2^{\lfloor \lg n \rfloor}$$ (or, really, any function $$r:\mathbb{N}^+\rightarrow\mathbb{N}^0$$ with $$r(n)2,x_{|x|-1}=x_{r(|x|-1)},x_{|x|-2}=x_{r(|x|-2)}\}$$. (The indices here are zero-based.) Choose $$x_0\in L$$ such that $$x_0$$ has no proper prefix in $$L$$. Then, construct the UTM $$M$$ that does: repeat: s := "" while s not in L: # if there is no next character, halt s := s + readchar() if s == x0: break M0() This $$M$$ will have $$\mu(M)>3-2^{-|x_0|}+d(M_0,M)2^{-|x_0|-d(M_0,M)}$$. $$M$$ here is optimized for building up internal states (that are then UTMs that are efficiently encoded), while also being very easy to reset from these internal states; in other words being easy to “encode” from the UTMs it efficiently encodes, using at most 2 bits (an average of $$\frac{1+\sqrt{5}}{2}$$). This is somewhat interesting, but clearly doesn’t capture the kind of computational expressivity we’re primarily interested in. reply
 by Janos Kramar 1097 days ago | Patrick LaVictoire likes this | link These results are still a bit unsatisfying. The first half constructs an invariant measure which is then shown to be unsatisfactory because UTMs can rank arbitrarily high while only being good at encoding variations of themselves. This is mostly the case because the chain is transient; if it was positive recurrent then the measure would be finite, and UTMs ranking high would have to be good at encoding (and being encoded by) the average UTM rather than just a select family of UTMs. The second half looks at whether we can get better results (ie a probability measure) by restricting our attention to output-free “UTMs” (though I misspoke; these are not actually UTMs but rather universal semidecidable languages (we can call them USDLs)). It concludes that we can’t if the measure will be continuous on the given digraph - however, this is an awkward notion of continuity: a low-complexity USDL whose behavior is tweaked very slightly but in a complex way may be very close in the given topology, but should have measure much lower than the starting USDL. So I consider this question unanswered. reply

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