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The Farey sequence $\mathcal{F}_n$ is the list of all fractions in increasing order (in lowest terms) from $0$ to $1$, having denominator at most $n$. My question is, given some $a/b\in\mathcal{F}_n$ for known $n$, what is the most efficient method to find the fractions in $\mathcal{F}_n$ that are adjacent to $a/b$?

Here is one method. Suppose $\frac{a_0}{b_0}<\frac{a}{b}<\frac{a_1}{b_1}$ are consecutive terms of $\mathcal{F}_n$ for a known $n$. We want to find $a_0/b_0$ and $a_1/b_1$. By definition of $\mathcal{F}_n$, we want to minimize the differences $\frac{a}{b}-\frac{a_0}{b_0}$ and $\frac{a_1}{b_1}-\frac{a}{b}$. There is a well known property of Farey sequences that states if $a/b < a'/b'$ are adjacent terms in $\mathcal{F}_n$, then $\frac{a'}{b'}-\frac{a}{b}=\frac{1}{bb'}$. Then, the differences can be written as $$\frac{a}{b}-\frac{a_0}{b_0}=\frac{ab_0-a_0b}{bb_0}=\frac{1}{bb_0}, \frac{a_1}{b_1}-\frac{a}{b}=\frac{a_1b-ab_1}{b_1b}=\frac{1}{b_1b}.$$ Thus, $ab_0-a_0b=1$ and $a_1b-ab_1=1$. Since $\gcd(a,b)=1$ we can find $(a_0,b_0)$ and $(a_1, b_1)$ due to Bézout's Identity and the extended Euclidean algorithm.

Alternately, we could have drawn a tree diagram of sorts and used bounding arguments to arrive at the solutions.

Are there any other (maybe more efficient?) methods of finding $(a_0, b_0)$ and $(a_1,b_1)$? How can we find the computational complexities of such methods? Is there a method known to be most efficient? Any ideas are appreciated.

tc1729
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  • The first method that came to my mind is the extended Euclidean algorithm, as you said ... and the Euclidean algorithm is lightning fast in practice. – Greg Martin Nov 24 '13 at 08:42

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Here is a slow way of computing the neighbors of $a/b$.

For any $m$ we can ask look at the fractions $0 \leq \frac{1}{m} \leq \frac{2}{m} \leq \dots \leq \frac{m-1}{m} \leq 1$ and ask where $\frac{a}{b}$ fits in here.

We wish to minimize $\frac{a}{b} - \frac{k}{m} = \frac{am-bk}{bm} $ This is just an application of the Euclidean algorithm.

For some $0 \leq r < b$, $am = k b + r $. For a computer k = am/b where we use integer division.

Using the Euclidean algorithm we can speed of the process of long division. We will get some expansion

$ \frac{a}{b} = \cfrac{1}{c_1 + \cfrac{1}{c_2 + \dots } } = [c_1, c_2, \dots ]$

The digits are $b_0=1, b_1 = c_1, b_2 = c_2c_1 + 1, b_3 = c_3b_2 + b_1,\dots $

The algorithmic efficiency of the Euclidean algorithm is discussed on Wikipedia. The number $T(a,b)$ of steps on the Euclidean algorithm for $a/b$ is very chaotic.

For two random $m$-bit numbers $0 < a,b < 2^m$, the number of steps is proportional to $m$.

cactus314
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