Equirectangular approximation to distance on a great circle

Percentage Accurate: 51.6% → 51.6%
Time: 3.1s
Alternatives: 1
Speedup: 1.0×

Specification

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\[\begin{array}{l} \\ \begin{array}{l} t_0 := \left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\\ R \cdot \sqrt{t\_0 \cdot t\_0 + \left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right)} \end{array} \end{array} \]
(FPCore (R lambda1 lambda2 phi1 phi2)
 :precision binary64
 (let* ((t_0 (* (- lambda1 lambda2) (cos (/ (+ phi1 phi2) 2.0)))))
   (* R (sqrt (+ (* t_0 t_0) (* (- phi1 phi2) (- phi1 phi2)))))))
double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	double t_0 = (lambda1 - lambda2) * cos(((phi1 + phi2) / 2.0));
	return R * sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))));
}

real(8) function code(r, lambda1, lambda2, phi1, phi2)
    real(8), intent (in) :: r
    real(8), intent (in) :: lambda1
    real(8), intent (in) :: lambda2
    real(8), intent (in) :: phi1
    real(8), intent (in) :: phi2
    real(8) :: t_0
    t_0 = (lambda1 - lambda2) * cos(((phi1 + phi2) / 2.0d0))
    code = r * sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))))
end function

public static double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	double t_0 = (lambda1 - lambda2) * Math.cos(((phi1 + phi2) / 2.0));
	return R * Math.sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))));
}

def code(R, lambda1, lambda2, phi1, phi2):
	t_0 = (lambda1 - lambda2) * math.cos(((phi1 + phi2) / 2.0))
	return R * math.sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))))

function code(R, lambda1, lambda2, phi1, phi2)
	t_0 = Float64(Float64(lambda1 - lambda2) * cos(Float64(Float64(phi1 + phi2) / 2.0)))
	return Float64(R * sqrt(Float64(Float64(t_0 * t_0) + Float64(Float64(phi1 - phi2) * Float64(phi1 - phi2)))))
end

function tmp = code(R, lambda1, lambda2, phi1, phi2)
	t_0 = (lambda1 - lambda2) * cos(((phi1 + phi2) / 2.0));
	tmp = R * sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))));
end

code[R_, lambda1_, lambda2_, phi1_, phi2_] := Block[{t$95$0 = N[(N[(lambda1 - lambda2), $MachinePrecision] * N[Cos[N[(N[(phi1 + phi2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, N[(R * N[Sqrt[N[(N[(t$95$0 * t$95$0), $MachinePrecision] + N[(N[(phi1 - phi2), $MachinePrecision] * N[(phi1 - phi2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]

\begin{array}{l}

\\
\begin{array}{l}
t_0 := \left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\\
R \cdot \sqrt{t\_0 \cdot t\_0 + \left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right)}
\end{array}
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 1 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 51.6% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\\ R \cdot \sqrt{t\_0 \cdot t\_0 + \left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right)} \end{array} \end{array} \]
(FPCore (R lambda1 lambda2 phi1 phi2)
 :precision binary64
 (let* ((t_0 (* (- lambda1 lambda2) (cos (/ (+ phi1 phi2) 2.0)))))
   (* R (sqrt (+ (* t_0 t_0) (* (- phi1 phi2) (- phi1 phi2)))))))
double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	double t_0 = (lambda1 - lambda2) * cos(((phi1 + phi2) / 2.0));
	return R * sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))));
}

real(8) function code(r, lambda1, lambda2, phi1, phi2)
    real(8), intent (in) :: r
    real(8), intent (in) :: lambda1
    real(8), intent (in) :: lambda2
    real(8), intent (in) :: phi1
    real(8), intent (in) :: phi2
    real(8) :: t_0
    t_0 = (lambda1 - lambda2) * cos(((phi1 + phi2) / 2.0d0))
    code = r * sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))))
end function

public static double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	double t_0 = (lambda1 - lambda2) * Math.cos(((phi1 + phi2) / 2.0));
	return R * Math.sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))));
}

def code(R, lambda1, lambda2, phi1, phi2):
	t_0 = (lambda1 - lambda2) * math.cos(((phi1 + phi2) / 2.0))
	return R * math.sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))))

function code(R, lambda1, lambda2, phi1, phi2)
	t_0 = Float64(Float64(lambda1 - lambda2) * cos(Float64(Float64(phi1 + phi2) / 2.0)))
	return Float64(R * sqrt(Float64(Float64(t_0 * t_0) + Float64(Float64(phi1 - phi2) * Float64(phi1 - phi2)))))
end

function tmp = code(R, lambda1, lambda2, phi1, phi2)
	t_0 = (lambda1 - lambda2) * cos(((phi1 + phi2) / 2.0));
	tmp = R * sqrt(((t_0 * t_0) + ((phi1 - phi2) * (phi1 - phi2))));
end

code[R_, lambda1_, lambda2_, phi1_, phi2_] := Block[{t$95$0 = N[(N[(lambda1 - lambda2), $MachinePrecision] * N[Cos[N[(N[(phi1 + phi2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, N[(R * N[Sqrt[N[(N[(t$95$0 * t$95$0), $MachinePrecision] + N[(N[(phi1 - phi2), $MachinePrecision] * N[(phi1 - phi2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]

\begin{array}{l}

\\
\begin{array}{l}
t_0 := \left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\\
R \cdot \sqrt{t\_0 \cdot t\_0 + \left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right)}
\end{array}
\end{array}

Alternative 1: 51.6% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \cos \left(\frac{\phi_2 + \phi_1}{2}\right) \cdot \left(\lambda_1 - \lambda_2\right)\\ \sqrt{\left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right) + t\_0 \cdot t\_0} \cdot R \end{array} \end{array} \]
(FPCore (R lambda1 lambda2 phi1 phi2)
 :precision binary64
 (let* ((t_0 (* (cos (/ (+ phi2 phi1) 2.0)) (- lambda1 lambda2))))
   (* (sqrt (+ (* (- phi1 phi2) (- phi1 phi2)) (* t_0 t_0))) R)))
double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	double t_0 = cos(((phi2 + phi1) / 2.0)) * (lambda1 - lambda2);
	return sqrt((((phi1 - phi2) * (phi1 - phi2)) + (t_0 * t_0))) * R;
}

real(8) function code(r, lambda1, lambda2, phi1, phi2)
    real(8), intent (in) :: r
    real(8), intent (in) :: lambda1
    real(8), intent (in) :: lambda2
    real(8), intent (in) :: phi1
    real(8), intent (in) :: phi2
    real(8) :: t_0
    t_0 = cos(((phi2 + phi1) / 2.0d0)) * (lambda1 - lambda2)
    code = sqrt((((phi1 - phi2) * (phi1 - phi2)) + (t_0 * t_0))) * r
end function

public static double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	double t_0 = Math.cos(((phi2 + phi1) / 2.0)) * (lambda1 - lambda2);
	return Math.sqrt((((phi1 - phi2) * (phi1 - phi2)) + (t_0 * t_0))) * R;
}

def code(R, lambda1, lambda2, phi1, phi2):
	t_0 = math.cos(((phi2 + phi1) / 2.0)) * (lambda1 - lambda2)
	return math.sqrt((((phi1 - phi2) * (phi1 - phi2)) + (t_0 * t_0))) * R

function code(R, lambda1, lambda2, phi1, phi2)
	t_0 = Float64(cos(Float64(Float64(phi2 + phi1) / 2.0)) * Float64(lambda1 - lambda2))
	return Float64(sqrt(Float64(Float64(Float64(phi1 - phi2) * Float64(phi1 - phi2)) + Float64(t_0 * t_0))) * R)
end

function tmp = code(R, lambda1, lambda2, phi1, phi2)
	t_0 = cos(((phi2 + phi1) / 2.0)) * (lambda1 - lambda2);
	tmp = sqrt((((phi1 - phi2) * (phi1 - phi2)) + (t_0 * t_0))) * R;
end

code[R_, lambda1_, lambda2_, phi1_, phi2_] := Block[{t$95$0 = N[(N[Cos[N[(N[(phi2 + phi1), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision] * N[(lambda1 - lambda2), $MachinePrecision]), $MachinePrecision]}, N[(N[Sqrt[N[(N[(N[(phi1 - phi2), $MachinePrecision] * N[(phi1 - phi2), $MachinePrecision]), $MachinePrecision] + N[(t$95$0 * t$95$0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * R), $MachinePrecision]]

\begin{array}{l}

\\
\begin{array}{l}
t_0 := \cos \left(\frac{\phi_2 + \phi_1}{2}\right) \cdot \left(\lambda_1 - \lambda_2\right)\\
\sqrt{\left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right) + t\_0 \cdot t\_0} \cdot R
\end{array}
\end{array}
Derivation

  1. Initial program 58.7%

    \[R \cdot \sqrt{\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\right) \cdot \left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\right) + \left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right)} \]
  2. Add Preprocessing

  3. Final simplification58.7%

    \[\leadsto \sqrt{\left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right) + \left(\cos \left(\frac{\phi_2 + \phi_1}{2}\right) \cdot \left(\lambda_1 - \lambda_2\right)\right) \cdot \left(\cos \left(\frac{\phi_2 + \phi_1}{2}\right) \cdot \left(\lambda_1 - \lambda_2\right)\right)} \cdot R \]
  4. Add Preprocessing

Reproduce

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herbie shell --seed 2024343 
(FPCore (R lambda1 lambda2 phi1 phi2)
  :name "Equirectangular approximation to distance on a great circle"
  :precision binary64
  (* R (sqrt (+ (* (* (- lambda1 lambda2) (cos (/ (+ phi1 phi2) 2.0))) (* (- lambda1 lambda2) (cos (/ (+ phi1 phi2) 2.0)))) (* (- phi1 phi2) (- phi1 phi2))))))