Logical inductors in multistable situations.

Donald Hobson

I was reading about logical induction at

https://intelligence.org/files/LogicalInduction.pdf

and understand how it resolves paradoxical self reference, but I'm not sure what the inductor will do in situations where multiple stable solutions exist.

Let $f : [0, 1] \to [0, 1]$

If $f$ is continuous then it must have a fixed point. Even if it has finitely many discontinuities, it must have an "almost fixed" point. An $x$ such that $\forall ϵ > 0 : i n f_{y \in (x - ϵ, x)} f (y) \leq x \leq {sup}_{y \in (x, x + ϵ)} f (y)$

However some $f$ have multiple such points.

f (x) = {\begin{matrix} 0 & x < \frac{1}{2} 1 & x \geq \frac{1}{2} \end{matrix}

Has "almost fixed" points at $0$ , $\frac{1}{2}$ and $1$ .

A similar continuous $f$ is

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 0 & x \leq \frac{1}{3} 3 x - 1 & \frac{1}{3} \leq x \leq \frac{2}{3} 1 & x \geq \frac{2}{3} \end{matrix}

With

f (x) = x

Having every point fixed.

Consider $ϕ_{n} =" f (E_{n} (ϕ_{n})) "$

These functions make $ϕ_{n}$ the logical inductor version of "this statement is true". Multiple values can be consistently applied to this logically uncertain variable. None of the possible values allow a money pump, so the technique of showing that some behaviour would make the market exploitable that is used repeatedly in the paper don't work here.

Is the value of $E_{n} (ϕ_{n})$ uniquely defined or does it depend on the implementation details of the logical inductor? Does it tend to a limit as $n \to \infty$ ? Is there a sense in which

f (x) = {\begin{matrix} 0.83 & 0.82 \leq x \leq 0.84 0.1 & e l s e \end{matrix}

causes $E_{n} (ϕ_{n})$ has a stronger attractor to $0.1$ than it does to $0.83$ ?

Can $E_{n} (ϕ_{n})$ be 0.6 where

f (x) = {\begin{matrix} 0.6 & x = 0.6 0.1 & e l s e \end{matrix}

because the smallest variation would force it to be $0.1$ ?

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jessicata

Jan 04, 2019

Different logical inductors will give different probabilities for each $ϕ_{n}$ . The logical induction criterion does not require any answer in particular.

Any particular deterministic algorithm for finding a logical inductor (such as the one in the paper) will yield a logical inductor that gives particular probabilities for these statements, which are close to fixed points in the limit. The algorithm in the paper is parameterized over some measure on Turing machines, and will give different answers depending on this measure. You could analyze which measures would lead to which fixed points, but this doesn't seem very interesting.

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[-]Gurkenglas4y 1

I see no almost fixed point for the function that is 1 until 0.5 and 0 after.

[-]Donald Hobson4y 1

0.5 is the almost fixed point. Its the point where $f (x) - x$ goes from being positive to negative. If you take a sequence of continuous functions $f_{n} (x)$ that converge pointwise to $f (x)$ then there will exist a sequence $y_{n}$ such that $f_{n} (y_{n}) = y_{n}$ and ${lim}_{n \to \infty} y_{n} = 0.5$ .

[-]Gurkenglas4y 1

That definition makes more sense than the one in the question. :)

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Logical inductors in multistable situations.

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Jan 04, 2019

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Logical inductors in multistable situations.

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Jan 04, 2019

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