Added thirdparty: boost library

thirdparty/include/boost/histogram/detail/erf_inv.hpp

@@ -0,0 +1,48 @@
+// Copyright 2022 Hans Dembinski, Jay Gohil
+//
+// Distributed under the Boost Software License, Version 1.0.
+// (See accompanying file LICENSE_1_0.txt
+// or copy at http://www.boost.org/LICENSE_1_0.txt)
+
+#ifndef BOOST_HISTOGRAM_DETAIL_ERF_INF_HPP
+#define BOOST_HISTOGRAM_DETAIL_ERF_INF_HPP
+
+#include <cmath>
+
+namespace boost {
+namespace histogram {
+namespace detail {
+
+// Simple implementation of erf_inv so that we do not depend on boost::math.
+// If you happen to discover this, prefer the boost::math implementation,
+// it is more accurate for x very close to -1 or 1 and faster.
+// The only virtue of this implementation is its simplicity.
+template <int Iterate = 3>
+double erf_inv(double x) noexcept {
+  // Strategy: solve f(y) = x - erf(y) = 0 for given x with Newton's method.
+  // f'(y) = -erf'(y) = -2/sqrt(pi) e^(-y^2)
+  // Has quadratic convergence. Since erf_inv<0> is accurate to 1e-3,
+  // we should have machine precision after three iterations.
+  const double x0 = erf_inv<Iterate - 1>(x); // recursion
+  const double fx0 = x - std::erf(x0);
+  const double pi = std::acos(-1);
+  double fpx0 = -2.0 / std::sqrt(pi) * std::exp(-x0 * x0);
+  return x0 - fx0 / fpx0; // = x1
+}
+
+template <>
+inline double erf_inv<0>(double x) noexcept {
+  // Specialization to get initial estimate.
+  // This formula is accurate to about 1e-3.
+  // Based on https://stackoverflow.com/questions/27229371/inverse-error-function-in-c
+  const double a = std::log((1 - x) * (1 + x));
+  const double b = std::fma(0.5, a, 4.120666747961526);
+  const double c = 6.47272819164 * a;
+  return std::copysign(std::sqrt(-b + std::sqrt(std::fma(b, b, -c))), x);
+}
+
+} // namespace detail
+} // namespace histogram
+} // namespace boost
+
+#endif