Finding the closest integer fraction to a given random real between 0..1, given ranges of numerator and denominator

Question

Given two ranges of positive integers x: [1 ... n] and y: [1 ... m] and random real R from 0 to 1, I need to find the pair of elements (i,j) from x and y such that x_i / y_j is closest to R.

What is the most efficient way to find this pair?

Answer 1

Using Farey sequence

This is a simple and mathematically beautiful algorithm to solve this: run a binary search, where on each iteration the next number is given by the mediant formula (below). By the properties of the Farey sequence that number is the one with the smallest denominator within that interval. Consequently this sequence will always converge and never 'miss' a valid solution.

In pseudocode:

input: m, n, R

a_num = 0, a_denom = 1
b_num = 1, b_denom = 1

repeat:
    -- interestingly c_num/c_denom is already in reduced form
    c_num = a_num + b_num
    c_denom = a_denom + b_denom

    -- if the numbers are too big, return the closest of a and b
    if c_num > n or c_denom > m then
        if R - a_num/a_denom < b_num/b_denom - R then
            return a_num, a_denom
        else
            return b_num, b_denom

    -- adjust the interval:
    if c_num/c_denom < R then
        a_num = c_num, a_denom = c_denom
    else
        b_num = c_num, b_denom = c_denom

goto repeat

Even though it's fast on average (my educated guess that it's O(log max(m,n))), it can still be slow if R is close to a fraction with a small denominator. For example finding an approximation to 1/1000000 with m = n = 1000000 will take a million iterations.

Answer 2

The standard approach to approximating reals with rationals is computing the continued fraction series (see [1]). Put a limit on the nominator and denominator while computing parts of the series, and the last value before you break the limits is a fraction very close to your real number.

This will find a very good approximation very fast, but I'm not sure this will always find a closest approximation. It is known that

any convergent [partial value of the continued fraction expansion] is nearer to the continued fraction than any other fraction whose denominator is less than that of the convergent

but there may be approximations with larger denominator (still below your limit) that are better approximations, but are not convergents.

[1] http://en.wikipedia.org/wiki/Continued_fraction

Answer 3

Given that R is a real number such that 0 <= R <= 1, integers x: [1 ... n] and integers y: [1 ... m]. It is assumed that n <= m, since if n > m then x[n]/y[m] will be greater than 1, which cannot be the closest approximation to R.

Therefore, the best approximation of R with the denominator d will be either floor(R*d) / d or ceil(R*d) / d.

The problem can be solved in O(m) time and O(1) space (in Python):

from __future__ import division
from random import random
from math import floor

def fractionize(R, n, d):
    error = abs(n/d - R)
    return (n, d, error)  # (numerator, denominator, absolute difference to R)

def better(a, b):
    return a if a[2] < b[2] else b

def approximate(R, n, m):
    best = (0, 1, R)
    for d in xrange(1, m+1):
        n1 = min(n, int(floor(R * d)))
        n2 = min(n, n1 + 1) # ceil(R*d)
        best = better(best, fractionize(R, n1, d))
        best = better(best, fractionize(R, n2, d))
    return best

if __name__ == '__main__': 
    def main():
        R = random()
        n = 30
        m = 100
        print R, approximate(R, n, m)
    main()