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Using finite automata to compute the base-$b$ representation of the golden ratio and other quadratic irrationals

Aaron Barnoff, Curtis Bright, Jeffrey Shallit

Abstract

We show that the $n$'th digit of the base-$b$ representation of the golden ratio is a finite-state function of the Zeckendorf representation of $b^n$, and hence can be computed by a finite automaton. Similar results can be proven for any quadratic irrational. We use a satisfiability (SAT) solver to prove, in some cases, that the automata we construct are minimal.

Using finite automata to compute the base-$b$ representation of the golden ratio and other quadratic irrationals

Abstract

We show that the 'th digit of the base- representation of the golden ratio is a finite-state function of the Zeckendorf representation of , and hence can be computed by a finite automaton. Similar results can be proven for any quadratic irrational. We use a satisfiability (SAT) solver to prove, in some cases, that the automata we construct are minimal.
Paper Structure (17 sections, 1 theorem, 6 equations, 9 figures, 5 tables)

This paper contains 17 sections, 1 theorem, 6 equations, 9 figures, 5 tables.

Key Result

Theorem 1

For all integers $b \geq 2$, there exists a DFAO $A_b$ that, on input the Zeckendorf representation of $b^n$, computes the $n\mspace{-2mu}$'th digit to the right of the point in the base-$b$ representation of $\varphi$.

Figures (9)

  • Figure 1: Synchronized automaton for $\lfloor n \varphi \rfloor$. The inputs are the Zeckendorf representation of $n$ and $x$, in parallel.
  • Figure 2: Automaton for the $n$'th bit to the right of the binary point of $\varphi$. States are labeled in the form $a/b$, where $a$ is the state number and $b$ is the output. The input is the Zeckendorf representation of $2^n$, and the output is $b$ when the last state reached is labeled $a/b$.
  • Figure 3: Automaton for the $n$'th bit to the right of the point of $\varphi$ in base $3$.
  • Figure 4: Automaton for the $n$'th bit to the right of the binary point of $\sqrt{2}$. Input is $2^n$ in Pell representation.
  • Figure 5: Synchronized automaton for $\lfloor n \alpha \rfloor$ for $\alpha=\sqrt{3}+1$.
  • ...and 4 more figures

Theorems & Definitions (2)

  • Theorem 1
  • proof