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Superposition as memory: unlocking quantum automatic complexity

Imagine a lock with two states, locked and unlocked, which may be manipulated using two operations, called 0 and 1. Moreover, the only way to (with certainty) unlock using four operations is to do them in the sequence 0011, i.e., $0^n1^n$ where $n=2$. In this scenario one might think that the lock needs to be in certain further states after each operation, so that there is some memory of what has been done so far. Here we show that this memory can be entirely encoded in superpositions of the two basic states locked and unlocked, where, as dictated by quantum mechanics, the operations are given by unitary matrices. Moreover, we show using the Jordan–Schur lemma that a similar lock is not possible for $n=60$.

Details in the paper: Superposition as memory: unlocking quantum automatic complexity which is to appear in the Lecture Notes in Computer Science volume of the conference Unconventional Computation and Natural Computation (UCNC) 2017.


Few paths, fewer words, and the maximum probability of writing 001 in a two-state Hidden Markov Model being $8/27$

The aim of this note is to give the simplest possible non-trivial calculation of the parameters of a HMM that maximize the probability of emitting a certain string.

Let $\{0,1\}$ be our alphabet.
Let $p$ be the probability of emitting 1 in state $s_0$.
Let $q$ be the probability of emitting 1 in state $s_1$.
Let $\epsilon$ be the probability of transitioning from $s_0$ to $s_1$.
Let $\delta$ be the probability of transitioning from $s_1$ to $s_0$.
Let $S(t)$ be the state after transitioning $t$ times, a random variable.
The probability of emitting the string 001 when starting in state $s_0$ is then
f(p,q,\epsilon,\delta)=\Pr(001; S(1)=s_0=S(2))+\Pr(001; S(1)=s_0, S(2)=s_1)$$
$$+\Pr(001; S(1)=s_1, S(2)=s_0)+\Pr(001; S(1)=s_1=S(2))$$
$$=\overline p^2 p \overline\epsilon^2 + \overline p^2q\overline\epsilon\epsilon + \overline p\overline q p\epsilon\delta + \overline p\overline q q \epsilon\overline\delta.
Which choice of parameters $p, q, \epsilon, \delta$ will maximize this probability?
To answer this we first note that
$$\frac{\partial f}{\partial\delta}=0\iff p=1, q=1, \epsilon=0\text{ or }p=q.$$
Going through these possibilities we keep finding values of $f$ bounded above by $\overline p^2p\le 4/27$:

  1. $p=1$ immediately gives $f=0$.
  2. $q=1$ gives $f=\overline p^2 p \overline\epsilon^2 + \overline p^2\overline\epsilon\epsilon=\overline p^2p\overline\epsilon.$
  3. $\epsilon=0$ gives $f=\overline p^2 p.$
  4. $p=q$ gives $f=\overline p^2 p \overline\epsilon^2 + \overline p^2p\overline\epsilon\epsilon + \overline p^2 p\epsilon\delta + \overline p^2 p (\epsilon\overline\delta)=\overline p^2p(\overline\epsilon^2 + \overline\epsilon\epsilon + \epsilon\delta + \epsilon\overline\delta)=\overline p^2p.$

We next consider boundary values for $\delta$.

  1. $\delta=0$. We may assume $p=0$, since there is no use in considering a positive probability of emitting a 1 in state $s_0$ if there is no chance of ever returning to that state. Then upon calculation of $\partial f/\partial q$ we find that $\epsilon=2\overline q$, which gives $f=2q^2\overline q$. This is maximized at $q=2/3$ which corresponds to $\epsilon=2/3$ as well, and gives a value $f=8/27>1/4$.

    This $8/27$ is decomposable as a sum of two disjoint scenarios of probability $4/27$:

    1. One is that after writing the first 0 we stay in state $s_0$, write another 0, and then transition to state $s_1$ to write a 1.
    2. The other is that after writing the first 0 we move to state $s_1$, write the 2nd zero there, and stay there to write the 3rd letter, 1.
  2. $\delta=1$. Then $f=\overline p^2 p \overline\epsilon^2 + \overline p^2q\overline\epsilon\epsilon + \overline p\overline q p\epsilon
    =\overline p(\overline p p \overline\epsilon^2 + \overline pq\overline\epsilon\epsilon +\overline q p\epsilon)$.
    Then $0=\partial f/\partial q = \overline p\epsilon(\overline p\overline\epsilon – p)$ if $p=\frac{\overline\epsilon}{1+\overline\epsilon}$. This gives $f=\frac{\overline\epsilon}{(1+\overline\epsilon)^3}$ (turns out not to depend on $q$) which is maximized for $\epsilon=1/2$ with value $f=4/27$. So we consider boundary values for $q$:

    1. $q=0$. Then $f=\overline pp(\overline p\overline\epsilon^2+\epsilon)\le \frac14\cdot 1$.
    2. $q=1$. Then $f=\overline p^2p\overline\epsilon^2+\overline p^2\overline\epsilon\epsilon$ and $\partial f/\partial\epsilon=\overline p^2(1-2\epsilon-2p\overline\epsilon)=0$ if $\overline p=1/(2\overline\epsilon)$, which gives $f=\overline p/4\le 1/4$.

Now note how much easier this is if we only consider a single path. Then clearly $1/4$ is the maximum possible, via 3 different paths, because of the presence of terms of the form $a\overline a$.
Replacing such occurrences by $1/4$ we upper bound $f$ by
$$\frac14\left(\overline p \overline\epsilon^2 + \overline p^2q + \overline q\epsilon\delta + \overline p \epsilon\overline\delta\right).


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Few paths, fewer words: model selection with automatic structure functions

The paper “Kolmogorov structure functions for automatic complexity in computational statistics” appeared in the Lecture Notes in Computer Science proceedings of the conference COCOA 2014, Maui, Hawaii.
The paper then appeared in the journal Theoretical Computer Science 2015.

The ideas are implemented in the Structure function calculator.

A new paper Few paths, fewer words: model selection with automatic structure functions has been conditionally accepted for publication in Experimental Mathematics.

Some slides


Covering the computable sets

I participated in the workshop on Algorithmic Randomness in Singapore and the conference on Computability, Complexity and Randomness.

With host Frank Stephan and fellow participant Sebastiaan Terwijn we wrote a paper entitled Covering the recursive sets which appeared in Lecture Notes in Computer Science, Conference on Computability in Europe, 2015, and has now been published in Annals of Pure and Applied Logic.


A Conflict Between Some Semantic Conditions

Damir Dzhafarov, Stefan Kaufmann, Bjørn Kjos-Hanssen, Dave Ripley, et al., at the 2016 ASL Annual Meeting at UConn.


José Carmo and Andrew J.I. Jones have studied contrary-to-duties obligations in a series of papers.

They develop a logical framework for scenarios such as the following:

1. There ought to be no dog.
2. If there is a dog, there ought to be a fence.

One conjecture from Carmo and Jones 1997 was refuted in a rather technical way in my 1996 term paper at University of Oslo.
The conjecture stated that one could simply add the condition
(Z \in \pii(X)) \land
(Y \subseteq X) \land
(Y \cap Z \ne \emptyset ) \rightarrow (Z \in \pii(Y )) \tag{5e}
for the conditional obligation operator ob.
In a follow-up paper (2001) they argued that (5e) could be added by weakening some other conditions.
In a new paper, to appear in Studia Logica, and presented at the Association for Symbolic Logic Annual Meeting 2016 at UConn, I argue that (5d) and (5e) are in conflict with each other. The argument is a generalization and strengthening of the 1996 argument.