Sturmian word

A Sturmian word is a binary sequence, finite or infinite, that makes up the cutting sequence for a positive real number x, as shown in the picture.

The Sturmian word can be computed thus as an algorithm:

If $x>1$ , then it is the inverse of the Sturmian word for $1/x$ . So we have reduced to the case of $0<x\leq 1$ .
Iterate over $floor(1x),floor(2x),floor(3x),\dots$
If $kx$ is an integer, then the program terminates. Else, if $floor((k-1)x)=floor(kx)$ , then the program outputs 0, else, it outputs 10.

The problem:

Given a positive rational number ${\frac {m}{n}}$ , specified by two positive integers $m,n$ , output its entire Sturmian word.
Given a quadratic real number ${\frac {{\sqrt {a}}+m}{n}}$ , specified by three positive integers $a,m,n$ , where $a$ is not a perfect square, output the first $k$ letters of its Sturmian word when given a positive number $k$ .

(If the programming language can represent infinite data structures, then that works too.)

Stretch goal: calculate the Sturmian word for other kinds of definable real numbers, such as cubic roots.

The key difficulty is accurately calculating $floor(k{\sqrt {a}})$ for large $k$ . Floating point arithmetic would lose precision. One can either do this simply by directly searching for some integer $a'$ such that $a'^{2}\leq k^{2}a<(a'+1)^{2}$ , or by more trickly methods, such as the continued fraction approach.

First calculate the continued fraction representation of ${\sqrt {a}}$ , then obtaining the convergent sequence to ${\sqrt {a}}$ . Let ${\frac {m}{n}}$ be a convergent to ${\sqrt {a}}$ , such that $n\geq k$ , then since the convergent sequence is the best rational approximant for denominators up to that point, we know for sure that, if we write out ${\frac {0}{k}},{\frac {1}{k}},\dots$ , the sequence would stride right across the gap $(m/n,2x-m/n)$ . Thus, we can take the largest $l$ such that $l/k\leq m/n$ , and we would know for sure that $l=floor(k{\sqrt {a}})$ .