Equirectangular approximation to distance on a great circle

?

Percentage Accurate: 60.5% → 99.8%
Time: 23.2s
Precision: binary64
Cost: 46016

?

\[R \cdot \sqrt{\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\right) \cdot \left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\right) + \left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right)} \]
\[R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \mathsf{fma}\left(\cos \phi_2, \cos \phi_1, \sin \phi_1 \cdot \left(-\sin \phi_2\right)\right)}, \phi_1 - \phi_2\right) \]
(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)
    (sqrt
     (+
      0.5
      (* 0.5 (fma (cos phi2) (cos phi1) (* (sin phi1) (- (sin phi2))))))))
   (- phi1 phi2))))
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) * sqrt((0.5 + (0.5 * fma(cos(phi2), cos(phi1), (sin(phi1) * -sin(phi2))))))), (phi1 - phi2));
}
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) * sqrt(Float64(0.5 + Float64(0.5 * fma(cos(phi2), cos(phi1), Float64(sin(phi1) * Float64(-sin(phi2)))))))), Float64(phi1 - phi2)))
end
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[Sqrt[N[(0.5 + N[(0.5 * N[(N[Cos[phi2], $MachinePrecision] * N[Cos[phi1], $MachinePrecision] + N[(N[Sin[phi1], $MachinePrecision] * (-N[Sin[phi2], $MachinePrecision])), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] ^ 2 + N[(phi1 - phi2), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]
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 \sqrt{0.5 + 0.5 \cdot \mathsf{fma}\left(\cos \phi_2, \cos \phi_1, \sin \phi_1 \cdot \left(-\sin \phi_2\right)\right)}, \phi_1 - \phi_2\right)

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.

Herbie found 12 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

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.

Bogosity?

Bogosity

Derivation?

  1. Initial program 58.5%

    \[R \cdot \sqrt{\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\right) \cdot \left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right)\right) + \left(\phi_1 - \phi_2\right) \cdot \left(\phi_1 - \phi_2\right)} \]
  2. Simplified96.0%

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

    [Start]58.5%

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

    hypot-def [=>]96.0%

    \[ R \cdot \color{blue}{\mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right), \phi_1 - \phi_2\right)} \]
  3. Applied egg-rr96.0%

    \[\leadsto R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \color{blue}{\sqrt{0.5 + 0.5 \cdot \cos \left(\phi_1 + \phi_2\right)}}, \phi_1 - \phi_2\right) \]
    Step-by-step derivation

    [Start]96.0%

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

    add-sqr-sqrt [=>]64.3%

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

    sqrt-unprod [=>]96.0%

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

    sqr-cos-a [=>]96.0%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{\color{blue}{0.5 + 0.5 \cdot \cos \left(2 \cdot \frac{\phi_1 + \phi_2}{2}\right)}}, \phi_1 - \phi_2\right) \]

    cos-2 [=>]96.0%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \color{blue}{\left(\cos \left(\frac{\phi_1 + \phi_2}{2}\right) \cdot \cos \left(\frac{\phi_1 + \phi_2}{2}\right) - \sin \left(\frac{\phi_1 + \phi_2}{2}\right) \cdot \sin \left(\frac{\phi_1 + \phi_2}{2}\right)\right)}}, \phi_1 - \phi_2\right) \]

    cos-sum [<=]96.0%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \color{blue}{\cos \left(\frac{\phi_1 + \phi_2}{2} + \frac{\phi_1 + \phi_2}{2}\right)}}, \phi_1 - \phi_2\right) \]

    add-log-exp [=>]29.3%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \cos \left(\color{blue}{\log \left(e^{\frac{\phi_1 + \phi_2}{2}}\right)} + \frac{\phi_1 + \phi_2}{2}\right)}, \phi_1 - \phi_2\right) \]

    add-log-exp [=>]29.3%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \cos \left(\log \left(e^{\frac{\phi_1 + \phi_2}{2}}\right) + \color{blue}{\log \left(e^{\frac{\phi_1 + \phi_2}{2}}\right)}\right)}, \phi_1 - \phi_2\right) \]

    sum-log [=>]29.3%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \cos \color{blue}{\log \left(e^{\frac{\phi_1 + \phi_2}{2}} \cdot e^{\frac{\phi_1 + \phi_2}{2}}\right)}}, \phi_1 - \phi_2\right) \]

    exp-sqrt [=>]29.3%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \cos \log \left(\color{blue}{\sqrt{e^{\phi_1 + \phi_2}}} \cdot e^{\frac{\phi_1 + \phi_2}{2}}\right)}, \phi_1 - \phi_2\right) \]

    exp-sqrt [=>]29.3%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \cos \log \left(\sqrt{e^{\phi_1 + \phi_2}} \cdot \color{blue}{\sqrt{e^{\phi_1 + \phi_2}}}\right)}, \phi_1 - \phi_2\right) \]

    add-sqr-sqrt [<=]29.3%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \cos \log \color{blue}{\left(e^{\phi_1 + \phi_2}\right)}}, \phi_1 - \phi_2\right) \]

    add-log-exp [<=]96.0%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \cos \color{blue}{\left(\phi_1 + \phi_2\right)}}, \phi_1 - \phi_2\right) \]
  4. Simplified96.0%

    \[\leadsto R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \color{blue}{\sqrt{0.5 + 0.5 \cdot \cos \left(\phi_2 + \phi_1\right)}}, \phi_1 - \phi_2\right) \]
    Step-by-step derivation

    [Start]96.0%

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

    +-commutative [=>]96.0%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \cos \color{blue}{\left(\phi_2 + \phi_1\right)}}, \phi_1 - \phi_2\right) \]
  5. Applied egg-rr99.9%

    \[\leadsto R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \color{blue}{\mathsf{fma}\left(\cos \phi_2, \cos \phi_1, -\sin \phi_2 \cdot \sin \phi_1\right)}}, \phi_1 - \phi_2\right) \]
    Step-by-step derivation

    [Start]96.0%

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

    cos-sum [=>]99.9%

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

    fma-neg [=>]99.9%

    \[ R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \color{blue}{\mathsf{fma}\left(\cos \phi_2, \cos \phi_1, -\sin \phi_2 \cdot \sin \phi_1\right)}}, \phi_1 - \phi_2\right) \]
  6. Final simplification99.9%

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

Alternatives

Alternative 1
Accuracy99.8%
Cost46016
\[R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \mathsf{fma}\left(\cos \phi_2, \cos \phi_1, \sin \phi_1 \cdot \left(-\sin \phi_2\right)\right)}, \phi_1 - \phi_2\right) \]
Alternative 2
Accuracy88.2%
Cost39748
\[\begin{array}{l} \mathbf{if}\;\lambda_1 \leq -1.65 \cdot 10^{-137}:\\ \;\;\;\;R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_2 + \phi_1}{2}\right), \phi_1 - \phi_2\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \mathsf{hypot}\left(\lambda_2 \cdot \left(-\sqrt{0.5 + 0.5 \cdot \left(\cos \phi_2 \cdot \cos \phi_1 - \sin \phi_2 \cdot \sin \phi_1\right)}\right), \phi_1 - \phi_2\right)\\ \end{array} \]
Alternative 3
Accuracy99.8%
Cost39680
\[R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \sqrt{0.5 + 0.5 \cdot \left(\cos \phi_2 \cdot \cos \phi_1 - \sin \phi_2 \cdot \sin \phi_1\right)}, \phi_1 - \phi_2\right) \]
Alternative 4
Accuracy93.4%
Cost13700
\[\begin{array}{l} \mathbf{if}\;\phi_2 \leq 9.5:\\ \;\;\;\;R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(0.5 \cdot \phi_1\right), \phi_1 - \phi_2\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(0.5 \cdot \phi_2\right), \phi_1 - \phi_2\right)\\ \end{array} \]
Alternative 5
Accuracy96.2%
Cost13696
\[R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(\frac{\phi_2 + \phi_1}{2}\right), \phi_1 - \phi_2\right) \]
Alternative 6
Accuracy91.0%
Cost13568
\[R \cdot \mathsf{hypot}\left(\left(\lambda_1 - \lambda_2\right) \cdot \cos \left(0.5 \cdot \phi_1\right), \phi_1 - \phi_2\right) \]
Alternative 7
Accuracy36.2%
Cost7369
\[\begin{array}{l} \mathbf{if}\;\phi_2 \leq -2.3 \cdot 10^{-38} \lor \neg \left(\phi_2 \leq 200000000\right):\\ \;\;\;\;R \cdot \left(\phi_2 - \phi_1\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(0.5 \cdot \left(\phi_2 + \phi_1\right)\right) \cdot \left(R \cdot \left(\lambda_2 - \lambda_1\right)\right)\\ \end{array} \]
Alternative 8
Accuracy36.1%
Cost7241
\[\begin{array}{l} \mathbf{if}\;\phi_2 \leq -3.1 \cdot 10^{-38} \lor \neg \left(\phi_2 \leq 2900000\right):\\ \;\;\;\;R \cdot \left(\phi_2 - \phi_1\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(0.5 \cdot \phi_1\right) \cdot \left(R \cdot \left(\lambda_2 - \lambda_1\right)\right)\\ \end{array} \]
Alternative 9
Accuracy33.4%
Cost585
\[\begin{array}{l} \mathbf{if}\;\phi_2 \leq -4 \cdot 10^{-39} \lor \neg \left(\phi_2 \leq 250000000\right):\\ \;\;\;\;R \cdot \left(\phi_2 - \phi_1\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \left(\lambda_2 - \lambda_1\right)\\ \end{array} \]
Alternative 10
Accuracy32.8%
Cost584
\[\begin{array}{l} \mathbf{if}\;\phi_2 \leq -3.1 \cdot 10^{-38}:\\ \;\;\;\;R \cdot \left(-\phi_1\right)\\ \mathbf{elif}\;\phi_2 \leq 2.8 \cdot 10^{+15}:\\ \;\;\;\;R \cdot \left(\lambda_2 - \lambda_1\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \phi_2\\ \end{array} \]
Alternative 11
Accuracy28.5%
Cost388
\[\begin{array}{l} \mathbf{if}\;\phi_1 \leq -1 \cdot 10^{+71}:\\ \;\;\;\;R \cdot \left(-\phi_1\right)\\ \mathbf{else}:\\ \;\;\;\;R \cdot \phi_2\\ \end{array} \]
Alternative 12
Accuracy17.5%
Cost192
\[R \cdot \phi_2 \]

Reproduce?

herbie shell --seed 2023272 
(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))))))