An alternating colouring function on strings

Abstract

An alternating colouring function is defined on strings over the alphabet $\{0, 1\}$. It divides the strings in colourable and non-colourable ones. The points in the subshift of finite type defined by forbidding all non-colourable strings of a certain length alternate between states of one colour and states of the other colour. In other words, the points in the 2nd power shifts all have the same colour. The number $K_n$ of non-colourable strings of length $n \ge 2$ is shown to be $2 \cdot (J_{n-2} + 1)$ where $J$ is the sequence of Jacobsthal numbers. The number of sources and sinks in the de Bruijn graph of dimension $n \ge 3$ with non-colourable edges removed is shown each to be $K_n - 4$.

Figures (13)

• Figure \zarabicfigure: The $1$-dimensional de Bruijn graph. The vertex $\bm 0$ and the edge $\bm{10}$ are coloured red indicating that $\phi \p \bm 0 = \phi \p{\bm{10}} = -1$ while the vertex $\bm 1$ and the edge $\bm{01}$ are coloured blue indicating $\phi \p \bm 1 = \phi \p{\bm{01}} = 1$. The edges $\bm{00}, \bm{11}$ and the background are coloured green because $\xi \p \bm 0 = \xi \p \bm 1 = \phi \p{\bm{00}} = \phi \p{\bm{11}} = 0$.
• Figure \zarabicfigure: The $2$-dimensional de Bruijn graph. The vertices and edges are coloured red if $\phi = -1$, green if $\phi = 0$ and blue if $\phi = 1$. The background of the vertex $\bm{10}$ is painted red because $\xi \p{\bm{10}} = -1$. In the centre the background is green because $\xi \p{\bm{00}} = \xi \p{\bm{11}} = 0$ and for the vertex $\bm{01}$ it is blue because $\xi \p{\bm{01}} = 1$.
• Figure \zarabicfigure: The $3$-dimensional de Bruijn graph. The vertices and edges are coloured red if $\phi = -1$, green if $\phi = 0$ and blue if $\phi = 1$. The background of the vertices is painted red where $\xi = -1$, green where $\xi = 0$ and blue where $\xi = 1$. Note that since the vertices have odd length, none of them is green.
• Figure \zarabicfigure: The $4$-dimensional de Bruijn graph. The vertices and edges are coloured red if $\phi = -1$, green if $\phi = 0$ and blue if $\phi = 1$. The background of the vertices is painted red where $\xi = -1$, green where $\xi = 0$ and blue where $\xi = 1$. Note that since the edges have odd length, none of them is green while the green vertices are exactly those that also have green background.
• Figure \zarabicfigure: The $1$-dimensional de Bruijn graph. The edges $\bm{01}$ and $\bm{10}$ are coloured blue and red respectively indicating that $\psi \p{\bm{01}} = 1 \land \psi \p{\bm{10}} = -1$. The other two edges and the vertices are coloured green because $\psi$ is $0$ there.

Theorems & Definitions (146)

• Remark
• Example
• Definition
• Remark
• Lemma \zarabicsection.\zarabicthrm
• proof
• Corollary \zarabicsection.\zarabicthrm
• proof
• Corollary \zarabicsection.\zarabicthrm
• proof