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std::sqrt(std::valarray) - cppreference.com

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template< class T > valarray<T> sqrt( const valarray<T>& va );

For each element in va computes the square root of the value of the element.

Parameters

va - value array to apply the operation to

Return value

Value array containing square roots of the values in va.

Notes

Unqualified function (sqrt) is used to perform the computation. If such function is not available, std::sqrt is used due to argument-dependent lookup.

The function can be implemented with the return type different from std::valarray. In this case, the replacement type has the following properties:

  • All const member functions of std::valarray are provided.
  • std::valarray, std::slice_array, std::gslice_array, std::mask_array and std::indirect_array can be constructed from the replacement type.
  • For every function taking a const std::valarray<T>& except begin() and end()(since C++11), identical functions taking the replacement types shall be added;
  • For every function taking two const std::valarray<T>& arguments, identical functions taking every combination of const std::valarray<T>& and replacement types shall be added.
  • The return type does not add more than two levels of template nesting over the most deeply-nested argument type.

Possible implementation

template<class T>
valarray<T> sqrt(const valarray<T>& va)
{
    valarray<T> other = va;
    for (T& i : other)
        i = sqrt(i);

    return other; // proxy object may be returned
}

Example

Finds all three roots (two of which can be complex conjugates) of several Cubic equations at once.

#include <cassert>
#include <complex>
#include <cstddef>
#include <iostream>
#include <numbers>
#include <valarray>

using CD = std::complex<double>;
using VA = std::valarray<CD>;

// return all n complex roots out of a given complex number x
VA root(CD x, unsigned n)
{
    const double mag = std::pow(std::abs(x), 1.0 / n);
    const double step = 2.0 * std::numbers::pi / n;
    double phase = std::arg(x) / n;
    VA v(n);
    for (std::size_t i{}; i != n; ++i, phase += step)
        v[i] = std::polar(mag, phase);
    return v;
}

// return n complex roots of each element in v; in the output valarray first
// goes the sequence of all n roots of v[0], then all n roots of v[1], etc.
VA root(VA v, unsigned n)
{
    VA o(v.size() * n);
    VA t(n);
    for (std::size_t i = 0; i != v.size(); ++i)
    {
        t = root(v[i], n);
        for (unsigned j = 0; j != n; ++j)
            o[n * i + j] = t[j];
    }
    return o;
}

// floating-point numbers comparator that tolerates given rounding error
inline bool is_equ(CD x, CD y, double tolerance = 0.000'000'001)
{
    return std::abs(std::abs(x) - std::abs(y)) < tolerance;
}

int main()
{
    // input coefficients for polynomial x³ + p·x + q
    const VA p{1, 2, 3, 4, 5, 6, 7, 8};
    const VA q{1, 2, 3, 4, 5, 6, 7, 8};

    // the solver
    const VA d = std::sqrt(std::pow(q / 2, 2) + std::pow(p / 3, 3));
    const VA u = root(-q / 2 + d, 3);
    const VA n = root(-q / 2 - d, 3);

    // allocate memory for roots: 3 * number of input cubic polynomials
    VA x[3];
    for (std::size_t t = 0; t != 3; ++t)
        x[t].resize(p.size());

    auto is_proper_root = [](CD a, CD b, CD p) { return is_equ(a * b + p / 3.0, 0.0); };

    // sieve out 6 out of 9 generated roots, leaving only 3 proper roots (per polynomial)
    for (std::size_t i = 0; i != p.size(); ++i)
        for (std::size_t j = 0, r = 0; j != 3; ++j)
            for (std::size_t k = 0; k != 3; ++k)
                if (is_proper_root(u[3 * i + j], n[3 * i + k], p[i]))
                    x[r++][i] = u[3 * i + j] + n[3 * i + k];

    std::cout << "Depressed cubic equation:   Root 1: \t\t Root 2: \t\t Root 3:\n";
    for (std::size_t i = 0; i != p.size(); ++i)
    {
        std::cout << "x³ + " << p[i] << "·x + " << q[i] << " = 0  "
                  << std::fixed << x[0][i] << "  " << x[1][i] << "  " << x[2][i]
                  << std::defaultfloat << '\n';

        assert(is_equ(std::pow(x[0][i], 3) + x[0][i] * p[i] + q[i], 0.0));
        assert(is_equ(std::pow(x[1][i], 3) + x[1][i] * p[i] + q[i], 0.0));
        assert(is_equ(std::pow(x[2][i], 3) + x[2][i] * p[i] + q[i], 0.0));
    }
}

Output: 

Depressed cubic equation:   Root 1:              Root 2:                 Root 3:
x³ + (1,0)·x + (1,0) = 0  (-0.682328,0.000000)  (0.341164,1.161541)  (0.341164,-1.161541)
x³ + (2,0)·x + (2,0) = 0  (-0.770917,0.000000)  (0.385458,1.563885)  (0.385458,-1.563885)
x³ + (3,0)·x + (3,0) = 0  (-0.817732,0.000000)  (0.408866,1.871233)  (0.408866,-1.871233)
x³ + (4,0)·x + (4,0) = 0  (-0.847708,0.000000)  (0.423854,2.130483)  (0.423854,-2.130483)
x³ + (5,0)·x + (5,0) = 0  (-0.868830,0.000000)  (0.434415,2.359269)  (0.434415,-2.359269)
x³ + (6,0)·x + (6,0) = 0  (-0.884622,0.000000)  (0.442311,2.566499)  (0.442311,-2.566499)
x³ + (7,0)·x + (7,0) = 0  (-0.896922,0.000000)  (0.448461,2.757418)  (0.448461,-2.757418)
x³ + (8,0)·x + (8,0) = 0  (-0.906795,0.000000)  (0.453398,2.935423)  (0.453398,-2.935423)

See also