Decidability of plane edge coloring with three colors

TL;DR

This paper proves that the problem of determining whether a set of Wang tiles with three colors can tile the plane is decidable, extending known results for two and five or more colors.

Contribution

It establishes the decidability of plane edge coloring with three colors and classifies minimal cycle generators and maximal non-cycle generators.

Findings

Decidability of the three-color edge coloring problem is proven.

The set of minimal cycle generators contains 787,605 members.

The set of maximal non-cycle generators is characterized and used to determine tiling possibilities.

Abstract

This investigation studies the decidability problem of plane edge coloring with three symbols. In the edge coloring (or Wang tiles) of a plane, unit squares with colored edges that have one of $p$ colors are arranged side by side such that the touching edges of the adjacent tiles have the same colors. Given a basic set $B$ of Wang tiles, the decision problem is to find an algorithm to determine whether or not $\Sigma(B)\neq\emptyset$, where $\Sigma(B)$ is the set of all global patterns on $\mathbb{Z}^{2}$ that can be constructed from the Wang tiles in $B$. When $p\geq 5$, the problem is known to be undecidable. When $p=2$, the problem is decidable. This study proves that when $p=3$, the problem is also decidable. $\mathcal{P}(B)$ is the set of all periodic patterns on $\mathbb{Z}^{2}$ that can be generated by the tiles in $B$. If $\mathcal{P}(B)\neq\emptyset$, then $B$ has a subset $B'$ of minimal cycle generators such that $\mathcal{P}(B')\neq\emptyset$ and $\mathcal{P}(B")=\emptyset$ for $B"\subsetneqq B'$. This study demonstrates that the set $\mathcal{C}(3)$ of all minimal cycle generators contains $787,605$ members that can be classified into $2,906$ equivalence classes. $\mathcal{N}(3)$ is the set of all maximal non-cycle generators: if $B\in \mathcal{N}(3)$, then $\mathcal{P}(B)=\emptyset$ and $\mathcal{P}(\tilde{B})\neq\emptyset$ for $\tilde{B}\supsetneqq B$. The problem is shown to be decidable by proving that $B\in \mathcal{N}(3)$ implies $\Sigma(B)=\emptyset$. Consequently, $\Sigma(B)\neq\emptyset$ if and only if $\mathcal{P}(B)\neq\emptyset$.