Logo
ScienceStack
Table of Contents
Fetching ...
Sign In

A sequential solution to the density classification task using an intermediate alphabet

Pacôme Perrotin, Pedro Paulo Balbi, Eurico Ruivo

TL;DR

A sequential cellular automaton of radius 2 1 is presented as a solution to the density classification task that makes use of an intermediate alphabet, and converges to a clean fixed point with no remaining auxiliary or intermediate information.

Abstract

We present a sequential cellular automaton of radius 2 1 as a solution to the density classification task that makes use of an intermediate alphabet, and converges to a clean fixed point with no remaining auxiliary or intermediate information. We extend this solution to arbitrary finite alphabets and to configurations in higher dimensions.

A sequential solution to the density classification task using an intermediate alphabet

TL;DR

A sequential cellular automaton of radius 2 1 is presented as a solution to the density classification task that makes use of an intermediate alphabet, and converges to a clean fixed point with no remaining auxiliary or intermediate information.

Abstract

We present a sequential cellular automaton of radius 2 1 as a solution to the density classification task that makes use of an intermediate alphabet, and converges to a clean fixed point with no remaining auxiliary or intermediate information. We extend this solution to arbitrary finite alphabets and to configurations in higher dimensions.
Paper Structure (10 sections, 5 theorems, 3 figures, 1 table)

This paper contains 10 sections, 5 theorems, 3 figures, 1 table.

Table of Contents

  1. Introduction
  2. The problem and the general description of the solution
  3. The problem
  4. The overall idea of the solution
  5. Mechanics of the process
  6. Proofs
  7. Generalisations
  8. Larger alphabets
  9. Higher dimensions
  10. Closing words

Key Result

Lemma 1

Throughout any propagation phase, the count of $0$ and $1$ symbols in the configuration plus the symbols contained in the active memory are preserved.

Figures (3)

  • Figure 1: An execution of the proposed solution to a configuration of size $7$, in the form of a table. The first row contains the initial configuration, and the execution operates from top to bottom. The first propagation phase starts at line $2$, column $4$. This propagation phase is followed by a second one with internal counter $\bullet$, and together they remove two symbols $0$ and two symbols $1$ from the configuration, leaving only $0$ symbols. The third propagation phase ends with the active memory $\{0\}$, which leads to the convergence phase starting at line $5$.
  • Figure 2: An execution of the proposed solution to a configuration of size $13$, in the form of a table. The first row contains the initial configuration, and the execution operates from top to bottom. This execution contains a total of $7$ propagation phases, which is also the count of $1$ symbols in the starting configuration, plus one.
  • Figure 3: An execution of $F_{\{0, 1, 2\}, 2}$ on a $3$ by $3$ configuration, with the set $S$ of the input alphabet being $\{0, 1, 2\}$. This execution lasts for $7$ configurations, which are illustrated from left to right, top to bottom. Within each configuration, the sequential order is from left to right, and from top to bottom. The element at the bottom right of the figure indicates the coordinates of the cells; for instance, the neighbours of $(0, 0)$ are the cells $(2, 0)$, $(2, 2)$ and $(0, 0)$ itself. As all these cells have the value $0$ in the initial configuration, the cell $(0, 0)$ does not change value during the first update.

Theorems & Definitions (10)

  • Lemma 1
  • proof
  • Lemma 2
  • proof
  • Theorem 1
  • proof
  • Theorem 2
  • proof
  • Theorem 3
  • proof