Equirectangular approximation to distance on a great circle

Average Accuracy: 39.6% → 94.5%

Time: 18.9s

Precision: binary64

Cost: 13696

Math

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

↓

\[R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right), \phi_1 - \phi_2\right) \]

FPCore

(FPCore (R lambda1 lambda2 phi1 phi2)
 :precision binary64
 (*
  R
  (sqrt
   (+
    (*
     (* (- lambda1 lambda2) (cos (/ (+ phi1 phi2) 2.0)))
     (* (- lambda1 lambda2) (cos (/ (+ phi1 phi2) 2.0))))
    (* (- phi1 phi2) (- phi1 phi2))))))

↓

(FPCore (R lambda1 lambda2 phi1 phi2)
 :precision binary64
 (*
  R
  (hypot (* (- lambda1 lambda2) (cos (/ (+ phi1 phi2) 2.0))) (- phi1 phi2))))

C

double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	return R * sqrt(((((lambda1 - lambda2) * cos(((phi1 + phi2) / 2.0))) * ((lambda1 - lambda2) * cos(((phi1 + phi2) / 2.0)))) + ((phi1 - phi2) * (phi1 - phi2))));
}

↓

double code(double R, double lambda1, double lambda2, double phi1, double phi2) {
	return R * hypot(((lambda1 - lambda2) * cos(((phi1 + phi2) / 2.0))), (phi1 - phi2));
}

Java

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

↓

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

Python

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

↓

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

Julia

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

↓

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

MATLAB

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

↓

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

Wolfram

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

↓

code[R_, lambda1_, lambda2_, phi1_, phi2_] := N[(R * N[Sqrt[N[(N[(lambda1 - lambda2), $MachinePrecision] * N[Cos[N[(N[(phi1 + phi2), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] ^ 2 + N[(phi1 - phi2), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 39.6%

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

2. Simplified 94.5%

   \[\leadsto \color{blue}{R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right), \phi_1 - \phi_2\right)} \]
   Proof

   [Start] 39.6	\[ R \cdot \sqrt{\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\right) \cdot \left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\right) + \left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right)} \]
   hypot-def [=>] 94.5	\[ R \cdot \color{blue}{\mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right), \phi_1 - \phi_2\right)} \]

3. Final simplification 94.5%

   \[\leadsto R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right), \phi_1 - \phi_2\right) \]

Alternatives

Alternative 1
Accuracy	90.1%
Cost	13700

\[\begin{array}{l} \mathbf{if}\;\phi_1 \leq -2800:\\ \;\;\;\;R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right), \phi_1 - \phi_2\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\phi_2 \cdot 0.5\right), \phi_1 - \phi_2\right)\\ \end{array} \]

Alternative 2
Accuracy	79.2%
Cost	13636

\[\begin{array}{l} \mathbf{if}\;\lambda_2 \leq 1.9 \cdot 10^{+181}:\\ \;\;\;\;R \cdot \mathsf{hypot}\left(\lambda_1 - \lambda_2, \phi_1 - \phi_2\right)\\ \mathbf{else}:\\ \;\;\;\;\lambda_2 \cdot \left(R \cdot \sqrt{0.5 + 0.5 \cdot \cos \left(\phi_1 + \phi_2\right)}\right)\\ \end{array} \]

Alternative 3
Accuracy	86.7%
Cost	13568

\[R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\phi_1 \cdot 0.5\right), \phi_1 - \phi_2\right) \]

Alternative 4
Accuracy	62.9%
Cost	6916

\[\begin{array}{l} \mathbf{if}\;\phi_1 \leq -2.25 \cdot 10^{+39}:\\ \;\;\;\;R \cdot \left(\phi_2 - \phi_1\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \mathsf{hypot}\left(\phi_2, \lambda_1 - \lambda_2\right)\\ \end{array} \]

Alternative 5
Accuracy	78.5%
Cost	6912

\[R \cdot \mathsf{hypot}\left(\lambda_1 - \lambda_2, \phi_1 - \phi_2\right) \]

Alternative 6
Accuracy	24.0%
Cost	388

\[\begin{array}{l} \mathbf{if}\;\phi_1 \leq -3.5 \cdot 10^{+41}:\\ \;\;\;\;R \cdot \left(-\phi_1\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \phi_2\\ \end{array} \]

Alternative 7
Accuracy	27.0%
Cost	320

\[R \cdot \left(\phi_2 - \phi_1\right) \]

Alternative 8
Accuracy	14.8%
Cost	192

\[R \cdot \phi_2 \]

Reproduce

herbie shell --seed 2023138 
(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))))))