The Problem

I am working on a problem that boils down to finding the closest representation of an arbitrary number ($x$) in the form:

$$x = A\times\frac{N}{D}$$

Where $A$ is a 32-bit integer, and $N$ and $D$ are 16-bit integers.

Typical inputs are real numbers between 1 and 1 billion.

My Attempts

I first attempted to use $N$ and $D$ to represent the fractional part of the number, but I can only correct for errors less than one part in 65 thousand, not very useful.

I then looked at naively testing all the possible combinations of $N$ and $D$ to find one with the lowest remainder, even when skipping like fractions (i.e., 2/4, 3/6, etc.), and skipping values of $N$ that would result in a fraction that is too large ($N \leq \frac{x}{D}$). I still need to search more than a billion combinations.

Some test cases I have found challenging; clearly, it's pretty easy to come up with these.

  • $1000\times \frac{255}{256}$
  • $ 123456 \times \frac{123}{456}$


This problem comes up as I am trying to configure a PLL to convert arbitrary input frequencies to a uniform output frequency with the lowest error possible.


Is there a much more efficient way to solve this problem? Does this kind of problem have a name that I can look up?

  • 1
    $\begingroup$ How is the real number given as input in your problem? An input to an algorithm should be a finite string. $\endgroup$
    – Vincenzo
    Jul 1, 2021 at 21:28
  • $\begingroup$ @Vincenzo Well, in my application, it would be a double-precision float. Here I use the term "Algorithm" in its colloquial sense. I also reviewed the tag description to see if this question was a good fit, "An algorithm is a sequence of well-defined steps that defines an abstract solution to a problem. Use this tag when your issue is related to design and analysis of algorithms". Based on that, I thought the tag was a good fit. $\endgroup$
    – Mark Omo
    Jul 1, 2021 at 21:35
  • $\begingroup$ I am looking for an algorithm that is better (ideally substantially) than brute force for finding the optimal set of integers $A$, $N$, $D$ that gives the lowest error. $\endgroup$
    – Mark Omo
    Jul 1, 2021 at 21:49

2 Answers 2


Generally speaking, this problem is called diophantine approximation.

However, your inputs are not just real numbers but double-precision floating point (aka double) numbers, which are in fact rational. Any double number has the form:

$$ (-1)^{sign} ( 1 . b_{51} b_{50} ... b_{0}) \times 2^{e - 1023} $$ where $b$ encodes the mantissa (52 bits) and $e'=e-1023$ encodes the exponent (11 bits). You mention that your typical inputs are at least 1, so your exponent $e'$ is always nonnegative. The mantissa is always an integer multiple of $2^{-52}$, and will be an integer multiple of $2^{-\ell}$ whenever the least significant nonzero bit of the mantissa is the $\ell$-th one.
The bottom line is that to achieve an exact representation you could do the following:

  • look at the binary representation of your floating point number $x$, isolating the mantissa bit sequence $b$ and the exponent $e'=e-1023$;
  • let $\ell$ be the index of the least significant nonzero bit of the mantissa $b$;
  • your number $x$ is an integer multiple of $1/D = 2^{e' - \ell}$, and no fraction with a smaller denominator than $D$ can represent it exactly;
  • divide $x$ by $1/D = 2^{e'-\ell}$ to find $N \times A$.

In the worst case, this may however not guarantee that $N$ and $A$ satisfy your size requirement, since $\ell- e'$ could be up to $52$, while you only have $32 + 16 = 48$ bits available for $N \times A$. A practical solution (but perhaps non-optimal) would be to round the least significant $52-48 = 4$ bits in the mantissa of $x$.

If you insist on finding the best approximation given a constraint on the largest denominator $D$, we need to turn to diophantine approximation techniques and continued fractions.

A fraction $A/B$ is a best rational approximation to real number $x$ if $|A - Bx| \le |C - Dx|$ for all integers $C$ and $D$ where $0 < D \le B$.

Then it is known that the best rational approximation of $x$ is a so-called convergent to $x$, which you can recover from the continued fraction expansion of $x$: $$ x = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \ldots}} $$ The $m$-th convergent $C_m$ is defined by the fraction $A_m/B_m$ where $A_k$ and $B_k$ are defined recursively by

$A_{-1}=1$, $A_0=a_0$, $A_{k} = a_{k} A_{k-1} + A_{k-2}$

$B_{-1}=0$, $B_0=1$, $B_{k} = a_{k} B_{k-1} + B_{k-2}$.

In other words, for each $k=0,1,2,\ldots$:

  • generate $a_k$ by setting $a_k = \lfloor x_k \rfloor$, $x_{k+1} = 1/(x_k - \lfloor x_k \rfloor)$, where $x_0 = x$;
  • compute $A_k$ and $B_k$ with the formulas above
  • the $k$-th convergent is $A_k/B_k$.

This will give you a sequence of closer and closer approximations of $x$, each of which is a best rational approximation. In your case, you want to stop when $B_k$ does not fit anymore into $16$ bits.

In your examples, we get the following:

First example.

$x = 1000 \times 255 / 256 \approx 996.09375$

$A_0 / B_0 = 996 / 1 \approx 996.0$

$A_1 / B_1 = 9961 / 10 \approx 996.1$

$A_2 / B_2 = 10957 / 11 \approx 996.0909090909091$

$A_3 / B_3 = 20918 / 21 \approx 996.0952380952381$

$A_4 / B_4 = 31875 / 32 \approx 996.09375$

Second example.

$x = 123456 \times 123 / 456 \approx 33300.63157894737$

$A_0 / B_0 = 33300 / 1 \approx 33300.0$

$A_1 / B_1 = 33301 / 1 \approx 33301.0$

$A_2 / B_2 = 66601 / 2 \approx 33300.5$

$A_3 / B_3 = 99902 / 3 \approx 33300.666666666664$

$A_4 / B_4 = 266405 / 8 \approx 33300.625$

$A_5 / B_5 = 632712 / 19 \approx 33300.63157894737$

  • $\begingroup$ Thanks! This seems to work well for my needs in practice. However, for around 0.08% of inputs, it does not find an optimal solution. For example, with the input $795082588 \times (1763/65225)$, the result is $988571985/46$, with an error of $1.5 \times 10^{-14}$. You can see my implementation here. $\endgroup$
    – Mark Omo
    Jul 3, 2021 at 0:39

The "best" rational approximations to a real number are given by it's continuous fraction, check that out.


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