Maksimov and Kolovsky, Equation (3)

?

Percentage Accurate: 73.2% → 88.2%
Time: 22.5s
Precision: binary64
Cost: 20352

?

\[\left(\left(-2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)\right) \cdot \sqrt{1 + {\left(\frac{U}{\left(2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)}\right)}^{2}} \]
↓
\[\begin{array}{l} t_0 := \cos \left(\frac{K}{2}\right)\\ \left(J \cdot \left(-2 \cdot t_0\right)\right) \cdot \mathsf{hypot}\left(1, \frac{U}{J \cdot \left(2 \cdot t_0\right)}\right) \end{array} \]
(FPCore (J K U)
 :precision binary64
 (*
  (* (* -2.0 J) (cos (/ K 2.0)))
  (sqrt (+ 1.0 (pow (/ U (* (* 2.0 J) (cos (/ K 2.0)))) 2.0)))))
↓
(FPCore (J K U)
 :precision binary64
 (let* ((t_0 (cos (/ K 2.0))))
   (* (* J (* -2.0 t_0)) (hypot 1.0 (/ U (* J (* 2.0 t_0)))))))
double code(double J, double K, double U) {
	return ((-2.0 * J) * cos((K / 2.0))) * sqrt((1.0 + pow((U / ((2.0 * J) * cos((K / 2.0)))), 2.0)));
}
↓
double code(double J, double K, double U) {
	double t_0 = cos((K / 2.0));
	return (J * (-2.0 * t_0)) * hypot(1.0, (U / (J * (2.0 * t_0))));
}

public static double code(double J, double K, double U) {
	return ((-2.0 * J) * Math.cos((K / 2.0))) * Math.sqrt((1.0 + Math.pow((U / ((2.0 * J) * Math.cos((K / 2.0)))), 2.0)));
}
↓
public static double code(double J, double K, double U) {
	double t_0 = Math.cos((K / 2.0));
	return (J * (-2.0 * t_0)) * Math.hypot(1.0, (U / (J * (2.0 * t_0))));
}

def code(J, K, U):
	return ((-2.0 * J) * math.cos((K / 2.0))) * math.sqrt((1.0 + math.pow((U / ((2.0 * J) * math.cos((K / 2.0)))), 2.0)))
↓
def code(J, K, U):
	t_0 = math.cos((K / 2.0))
	return (J * (-2.0 * t_0)) * math.hypot(1.0, (U / (J * (2.0 * t_0))))

function code(J, K, U)
	return Float64(Float64(Float64(-2.0 * J) * cos(Float64(K / 2.0))) * sqrt(Float64(1.0 + (Float64(U / Float64(Float64(2.0 * J) * cos(Float64(K / 2.0)))) ^ 2.0))))
end
↓
function code(J, K, U)
	t_0 = cos(Float64(K / 2.0))
	return Float64(Float64(J * Float64(-2.0 * t_0)) * hypot(1.0, Float64(U / Float64(J * Float64(2.0 * t_0)))))
end

function tmp = code(J, K, U)
	tmp = ((-2.0 * J) * cos((K / 2.0))) * sqrt((1.0 + ((U / ((2.0 * J) * cos((K / 2.0)))) ^ 2.0)));
end
↓
function tmp = code(J, K, U)
	t_0 = cos((K / 2.0));
	tmp = (J * (-2.0 * t_0)) * hypot(1.0, (U / (J * (2.0 * t_0))));
end

code[J_, K_, U_] := N[(N[(N[(-2.0 * J), $MachinePrecision] * N[Cos[N[(K / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(1.0 + N[Power[N[(U / N[(N[(2.0 * J), $MachinePrecision] * N[Cos[N[(K / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
↓
code[J_, K_, U_] := Block[{t$95$0 = N[Cos[N[(K / 2.0), $MachinePrecision]], $MachinePrecision]}, N[(N[(J * N[(-2.0 * t$95$0), $MachinePrecision]), $MachinePrecision] * N[Sqrt[1.0 ^ 2 + N[(U / N[(J * N[(2.0 * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]]

\left(\left(-2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)\right) \cdot \sqrt{1 + {\left(\frac{U}{\left(2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)}\right)}^{2}}
↓
\begin{array}{l}
t_0 := \cos \left(\frac{K}{2}\right)\\
\left(J \cdot \left(-2 \cdot t_0\right)\right) \cdot \mathsf{hypot}\left(1, \frac{U}{J \cdot \left(2 \cdot t_0\right)}\right)
\end{array}

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 10 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

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Initial program 72.4%

    \[\left(\left(-2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)\right) \cdot \sqrt{1 + {\left(\frac{U}{\left(2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)}\right)}^{2}} \]

  2. Simplified89.5%

    \[\leadsto \color{blue}{\left(J \cdot \left(-2 \cdot \cos \left(\frac{K}{2}\right)\right)\right) \cdot \mathsf{hypot}\left(1, \frac{U}{J \cdot \left(2 \cdot \cos \left(\frac{K}{2}\right)\right)}\right)} \]
    Step-by-step derivation

    [Start]72.4%

    \[ \left(\left(-2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)\right) \cdot \sqrt{1 + {\left(\frac{U}{\left(2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)}\right)}^{2}} \]

    *-commutative [=>]72.4%

    \[ \left(\color{blue}{\left(J \cdot -2\right)} \cdot \cos \left(\frac{K}{2}\right)\right) \cdot \sqrt{1 + {\left(\frac{U}{\left(2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)}\right)}^{2}} \]

    associate-*l* [=>]72.4%

    \[ \color{blue}{\left(J \cdot \left(-2 \cdot \cos \left(\frac{K}{2}\right)\right)\right)} \cdot \sqrt{1 + {\left(\frac{U}{\left(2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)}\right)}^{2}} \]

    unpow2 [=>]72.5%

    \[ \left(J \cdot \left(-2 \cdot \cos \left(\frac{K}{2}\right)\right)\right) \cdot \sqrt{1 + \color{blue}{\frac{U}{\left(2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)} \cdot \frac{U}{\left(2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)}}} \]

    hypot-1-def [=>]89.5%

    \[ \left(J \cdot \left(-2 \cdot \cos \left(\frac{K}{2}\right)\right)\right) \cdot \color{blue}{\mathsf{hypot}\left(1, \frac{U}{\left(2 \cdot J\right) \cdot \cos \left(\frac{K}{2}\right)}\right)} \]

    *-commutative [=>]89.5%

    \[ \left(J \cdot \left(-2 \cdot \cos \left(\frac{K}{2}\right)\right)\right) \cdot \mathsf{hypot}\left(1, \frac{U}{\color{blue}{\left(J \cdot 2\right)} \cdot \cos \left(\frac{K}{2}\right)}\right) \]

    associate-*l* [=>]89.5%

    \[ \left(J \cdot \left(-2 \cdot \cos \left(\frac{K}{2}\right)\right)\right) \cdot \mathsf{hypot}\left(1, \frac{U}{\color{blue}{J \cdot \left(2 \cdot \cos \left(\frac{K}{2}\right)\right)}}\right) \]

  3. Final simplification89.5%

    \[\leadsto \left(J \cdot \left(-2 \cdot \cos \left(\frac{K}{2}\right)\right)\right) \cdot \mathsf{hypot}\left(1, \frac{U}{J \cdot \left(2 \cdot \cos \left(\frac{K}{2}\right)\right)}\right) \]

Alternatives

Alternative 1
Accuracy88.2%
Cost20352
\[\begin{array}{l} t_0 := \cos \left(\frac{K}{2}\right)\\ \left(J \cdot \left(-2 \cdot t_0\right)\right) \cdot \mathsf{hypot}\left(1, \frac{U}{J \cdot \left(2 \cdot t_0\right)}\right) \end{array} \]
Alternative 2
Accuracy70.0%
Cost27341
\[\begin{array}{l} t_0 := \cos \left(\frac{K}{2}\right)\\ t_1 := \cos \left(K \cdot 0.5\right)\\ \mathbf{if}\;t_0 \leq 0.02:\\ \;\;\;\;J \cdot \left(-2 \cdot t_1\right)\\ \mathbf{elif}\;t_0 \leq 0.58 \lor \neg \left(t_0 \leq 0.975\right):\\ \;\;\;\;J \cdot \left(-2 \cdot \mathsf{hypot}\left(1, 0.5 \cdot \frac{U}{J}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;-0.25 \cdot \frac{U}{\frac{J}{U}} + -2 \cdot \left(J \cdot t_1\right)\\ \end{array} \]
Alternative 3
Accuracy88.2%
Cost20352
\[\begin{array}{l} t_0 := \cos \left(\frac{K}{2}\right)\\ J \cdot \left(t_0 \cdot \left(-2 \cdot \mathsf{hypot}\left(1, \frac{U}{J \cdot \left(2 \cdot t_0\right)}\right)\right)\right) \end{array} \]
Alternative 4
Accuracy73.3%
Cost13960
\[\begin{array}{l} t_0 := \mathsf{hypot}\left(1, 0.5 \cdot \frac{U}{J}\right)\\ \mathbf{if}\;U \leq -2.3 \cdot 10^{+262}:\\ \;\;\;\;U\\ \mathbf{elif}\;U \leq 2.8 \cdot 10^{+118}:\\ \;\;\;\;\left(\cos \left(\frac{K}{2}\right) \cdot \left(J \cdot -2\right)\right) \cdot t_0\\ \mathbf{else}:\\ \;\;\;\;J \cdot \left(-2 \cdot t_0\right)\\ \end{array} \]
Alternative 5
Accuracy69.6%
Cost7305
\[\begin{array}{l} \mathbf{if}\;K \leq -2.2 \cdot 10^{-6} \lor \neg \left(K \leq 420\right):\\ \;\;\;\;J \cdot \left(-2 \cdot \cos \left(K \cdot 0.5\right)\right)\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(J \cdot \mathsf{hypot}\left(1, U \cdot \frac{0.5}{J}\right)\right)\\ \end{array} \]
Alternative 6
Accuracy69.7%
Cost7305
\[\begin{array}{l} \mathbf{if}\;K \leq -2.2 \cdot 10^{-6} \lor \neg \left(K \leq 430\right):\\ \;\;\;\;J \cdot \left(-2 \cdot \cos \left(K \cdot 0.5\right)\right)\\ \mathbf{else}:\\ \;\;\;\;J \cdot \left(-2 \cdot \mathsf{hypot}\left(1, 0.5 \cdot \frac{U}{J}\right)\right)\\ \end{array} \]
Alternative 7
Accuracy58.6%
Cost7244
\[\begin{array}{l} \mathbf{if}\;U \leq -3.8 \cdot 10^{+222}:\\ \;\;\;\;U\\ \mathbf{elif}\;U \leq -6 \cdot 10^{+84}:\\ \;\;\;\;-U\\ \mathbf{elif}\;U \leq 3.9 \cdot 10^{+86}:\\ \;\;\;\;J \cdot \left(-2 \cdot \cos \left(K \cdot 0.5\right)\right)\\ \mathbf{elif}\;U \leq 1.12 \cdot 10^{+204}:\\ \;\;\;\;-U\\ \mathbf{elif}\;U \leq 1.45 \cdot 10^{+232}:\\ \;\;\;\;U\\ \mathbf{else}:\\ \;\;\;\;-U\\ \end{array} \]
Alternative 8
Accuracy37.7%
Cost788
\[\begin{array}{l} \mathbf{if}\;U \leq -1 \cdot 10^{+222}:\\ \;\;\;\;U\\ \mathbf{elif}\;U \leq -1.76 \cdot 10^{-152}:\\ \;\;\;\;-U\\ \mathbf{elif}\;U \leq 1.28 \cdot 10^{+48}:\\ \;\;\;\;J \cdot -2\\ \mathbf{elif}\;U \leq 9.8 \cdot 10^{+201}:\\ \;\;\;\;-U\\ \mathbf{elif}\;U \leq 1.05 \cdot 10^{+232}:\\ \;\;\;\;U\\ \mathbf{else}:\\ \;\;\;\;-U\\ \end{array} \]
Alternative 9
Accuracy25.8%
Cost260
\[\begin{array}{l} \mathbf{if}\;J \leq 7.2 \cdot 10^{-178}:\\ \;\;\;\;U\\ \mathbf{else}:\\ \;\;\;\;-U\\ \end{array} \]
Alternative 10
Accuracy26.1%
Cost64
\[U \]

Reproduce?

herbie shell --seed 2023255 
(FPCore (J K U)
  :name "Maksimov and Kolovsky, Equation (3)"
  :precision binary64
  (* (* (* -2.0 J) (cos (/ K 2.0))) (sqrt (+ 1.0 (pow (/ U (* (* 2.0 J) (cos (/ K 2.0)))) 2.0)))))