Boost C++ Libraries

...one of the most highly regarded and expertly designed C++ library projects in the world. Herb Sutter and Andrei Alexandrescu, C++ Coding Standards

This is the documentation for a snapshot of the master branch, built from commit 493d7c2b6c.
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Simple Continued Fractions

#include <boost/math/tools/simple_continued_fraction.hpp>
namespace boost::math::tools {

template<typename Real, typename Z = int64_t>
class simple_continued_fraction {
public:
    simple_continued_fraction(Real x);

    Real khinchin_geometric_mean() const;

    Real khinchin_harmonic_mean() const;

    template<typename T, typename Z_>
    friend std::ostream& operator<<(std::ostream& out, simple_continued_fraction<T, Z>& scf);
};
}

The simple_continued_fraction class provided by Boost affords the ability to convert a floating point number into a simple continued fraction. In addition, we can answer a few questions about the number in question using this representation.

Here's a minimal working example:

using boost::math::constants::pi;
using boost::math::tools::simple_continued_fraction;
auto cfrac = simple_continued_fraction(pi<long double>());
std::cout << "π ≈ " << cfrac << "\n";
// Prints:
// π ≈ [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2]

The class computes partial denominators while simultaneously computing convergents with the modified Lentz's algorithm. Once a convergent is within a few ulps of the input value, the computation stops.

Note that every floating point number is a rational number, and this exact rational can be exactly converted to a finite continued fraction. This is perfectly sensible behavior, but we do not do it here. This is because when examining known values like π, it creates a large number of incorrect partial denominators, even if every bit of the binary representation is correct.

It may be the case the a few incorrect partial convergents is harmless, but we compute continued fractions because we would like to do something with them. One sensible thing to do it to ask whether the number is in some sense "random"; a question that can be partially answered by computing the Khinchin geometric mean

and Khinchin harmonic mean

If these approach Khinchin's constant K0 and K-1 as the number of partial denominators goes to infinity, then our number is "uninteresting" with respect to the characterization. These violations are washed out if too many incorrect partial denominators are included in the expansion.

Note: The convergence of these means to the Khinchin limit is exceedingly slow; we've used 30,000 decimal digits of π and only found two digits of agreement with K0. However, clear violations of are obvious, such as the continued fraction expansion of √2, whose Khinchin geometric mean is precisely 2.


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