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Dominic van der Zypen
Dominic van der Zypen
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This question tries to get right what went wrong in an earlier question.

Let $\{0,1\}^\mathbb{Z}$ denote the set of all functions $x:\mathbb{Z}\to \{0,1\}$. Let $+$ denote addition modulo $2$ on $\{0,1\}$ (corresponding to ${\sf XOR}$ in computer science).

Define a function $f:\{0,1\}^\mathbb{Z} \to \{0,1\}^\mathbb{Z}$ by $x \mapsto f(x)$ where $f(x):\mathbb{Z}\to\{0,1\}$ is defined by $$f(x)(i) = x(i) \; + \;\max\{x(i-1), x(i+1)\}\ \text{ for all }i\in \mathbb{Z}.$$$$f(x)(i) = x(i-1) \; + \;\max\{x(i), x(i+1)\}\ \text{ for all }i\in \mathbb{Z}.$$ Inductively define $f^{(0)}(x) = x$, and $f^{(n+1)}(x) = f(f^{(n)}(x))$ for all integers $n\geq 0$.

Question. For what integers $n>1$ is there $x\in \{0,1\}^\mathbb{Z}$ such that $f^{(n)}(x) = x$ but $f^{(k)}(x) \neq x$ for all integers $k\geq 1$ with $k<n$?

This question tries to get right what went wrong in an earlier question.

Let $\{0,1\}^\mathbb{Z}$ denote the set of all functions $x:\mathbb{Z}\to \{0,1\}$. Let $+$ denote addition modulo $2$ on $\{0,1\}$ (corresponding to ${\sf XOR}$ in computer science).

Define a function $f:\{0,1\}^\mathbb{Z} \to \{0,1\}^\mathbb{Z}$ by $x \mapsto f(x)$ where $f(x):\mathbb{Z}\to\{0,1\}$ is defined by $$f(x)(i) = x(i) \; + \;\max\{x(i-1), x(i+1)\}\ \text{ for all }i\in \mathbb{Z}.$$ Inductively define $f^{(0)}(x) = x$, and $f^{(n+1)}(x) = f(f^{(n)}(x))$ for all integers $n\geq 0$.

Question. For what integers $n>1$ is there $x\in \{0,1\}^\mathbb{Z}$ such that $f^{(n)}(x) = x$ but $f^{(k)}(x) \neq x$ for all integers $k\geq 1$ with $k<n$?

This question tries to get right what went wrong in an earlier question.

Let $\{0,1\}^\mathbb{Z}$ denote the set of all functions $x:\mathbb{Z}\to \{0,1\}$. Let $+$ denote addition modulo $2$ on $\{0,1\}$ (corresponding to ${\sf XOR}$ in computer science).

Define a function $f:\{0,1\}^\mathbb{Z} \to \{0,1\}^\mathbb{Z}$ by $x \mapsto f(x)$ where $f(x):\mathbb{Z}\to\{0,1\}$ is defined by $$f(x)(i) = x(i-1) \; + \;\max\{x(i), x(i+1)\}\ \text{ for all }i\in \mathbb{Z}.$$ Inductively define $f^{(0)}(x) = x$, and $f^{(n+1)}(x) = f(f^{(n)}(x))$ for all integers $n\geq 0$.

Question. For what integers $n>1$ is there $x\in \{0,1\}^\mathbb{Z}$ such that $f^{(n)}(x) = x$ but $f^{(k)}(x) \neq x$ for all integers $k\geq 1$ with $k<n$?

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Dominic van der Zypen
Dominic van der Zypen
  • 48.9k
  • 8
  • 47
  • 162

"Rule 30" in the infinite setting

This question tries to get right what went wrong in an earlier question.

Let $\{0,1\}^\mathbb{Z}$ denote the set of all functions $x:\mathbb{Z}\to \{0,1\}$. Let $+$ denote addition modulo $2$ on $\{0,1\}$ (corresponding to ${\sf XOR}$ in computer science).

Define a function $f:\{0,1\}^\mathbb{Z} \to \{0,1\}^\mathbb{Z}$ by $x \mapsto f(x)$ where $f(x):\mathbb{Z}\to\{0,1\}$ is defined by $$f(x)(i) = x(i) \; + \;\max\{x(i-1), x(i+1)\}\ \text{ for all }i\in \mathbb{Z}.$$ Inductively define $f^{(0)}(x) = x$, and $f^{(n+1)}(x) = f(f^{(n)}(x))$ for all integers $n\geq 0$.

Question. For what integers $n>1$ is there $x\in \{0,1\}^\mathbb{Z}$ such that $f^{(n)}(x) = x$ but $f^{(k)}(x) \neq x$ for all integers $k\geq 1$ with $k<n$?