Toniolo and Linder, Equation (3b), real

?

Percentage Accurate: 93.7% → 99.6%
Time: 28.5s
Precision: binary64
Cost: 45184

?

\[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
↓
\[\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)\right) \cdot \sin th \]
(FPCore (kx ky th)
 :precision binary64
 (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))
↓
(FPCore (kx ky th)
 :precision binary64
 (* (log1p (expm1 (/ (sin ky) (hypot (sin ky) (sin kx))))) (sin th)))
double code(double kx, double ky, double th) {
	return (sin(ky) / sqrt((pow(sin(kx), 2.0) + pow(sin(ky), 2.0)))) * sin(th);
}
↓
double code(double kx, double ky, double th) {
	return log1p(expm1((sin(ky) / hypot(sin(ky), sin(kx))))) * sin(th);
}

public static double code(double kx, double ky, double th) {
	return (Math.sin(ky) / Math.sqrt((Math.pow(Math.sin(kx), 2.0) + Math.pow(Math.sin(ky), 2.0)))) * Math.sin(th);
}
↓
public static double code(double kx, double ky, double th) {
	return Math.log1p(Math.expm1((Math.sin(ky) / Math.hypot(Math.sin(ky), Math.sin(kx))))) * Math.sin(th);
}

def code(kx, ky, th):
	return (math.sin(ky) / math.sqrt((math.pow(math.sin(kx), 2.0) + math.pow(math.sin(ky), 2.0)))) * math.sin(th)
↓
def code(kx, ky, th):
	return math.log1p(math.expm1((math.sin(ky) / math.hypot(math.sin(ky), math.sin(kx))))) * math.sin(th)

function code(kx, ky, th)
	return Float64(Float64(sin(ky) / sqrt(Float64((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th))
end
↓
function code(kx, ky, th)
	return Float64(log1p(expm1(Float64(sin(ky) / hypot(sin(ky), sin(kx))))) * sin(th))
end

code[kx_, ky_, th_] := N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[(N[Power[N[Sin[kx], $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[Sin[ky], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]
↓
code[kx_, ky_, th_] := N[(N[Log[1 + N[(Exp[N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th
↓
\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)\right) \cdot \sin th

Local Percentage Accuracy?

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 20 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

The accuracy (vertical axis) and speed (horizontal axis) of each of Herbie's proposed alternatives. Up and to the right is better. Each dot represents an alternative program; the red square represents the initial program.

Bogosity?

Bogosity

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Initial program 95.1%

    \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

  2. Simplified99.7%

    \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]
    Step-by-step derivation

    [Start]95.1

    \[ \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

    +-commutative [=>]95.1

    \[ \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]

    unpow2 [=>]95.1

    \[ \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]

    unpow2 [=>]95.1

    \[ \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]

    hypot-def [=>]99.7

    \[ \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

  3. Applied egg-rr99.7%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)\right)} \cdot \sin th \]
    Step-by-step derivation

    [Start]99.7

    \[ \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th \]

    log1p-expm1-u [=>]99.7

    \[ \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)\right)} \cdot \sin th \]

  4. Final simplification99.7%

    \[\leadsto \mathsf{log1p}\left(\mathsf{expm1}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\right)\right) \cdot \sin th \]

Alternatives

Alternative 1
Accuracy61.5%
Cost39049
\[\begin{array}{l} \mathbf{if}\;\sin th \leq -0.04 \lor \neg \left(\sin th \leq 10^{-5}\right):\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \end{array} \]
Alternative 2
Accuracy73.3%
Cost39048
\[\begin{array}{l} t_1 := \mathsf{hypot}\left(\sin ky, \sin kx\right)\\ \mathbf{if}\;\sin ky \leq -0.0004:\\ \;\;\;\;\frac{\sin ky}{t_1} \cdot th\\ \mathbf{elif}\;\sin ky \leq 2 \cdot 10^{-9}:\\ \;\;\;\;\frac{ky \cdot \sin th}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 3
Accuracy44.9%
Cost32584
\[\begin{array}{l} \mathbf{if}\;\sin ky \leq -2 \cdot 10^{-66}:\\ \;\;\;\;\sqrt{{\sin th}^{2}}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-115}:\\ \;\;\;\;\frac{\sin ky \cdot \sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 4
Accuracy44.9%
Cost32584
\[\begin{array}{l} t_1 := \sin ky \cdot \sin th\\ \mathbf{if}\;\sin ky \leq -2 \cdot 10^{-66}:\\ \;\;\;\;\sqrt{{\sin th}^{2}}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-115}:\\ \;\;\;\;\frac{t_1}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\frac{t_1}{\sin ky}\\ \end{array} \]
Alternative 5
Accuracy99.6%
Cost32384
\[\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
Alternative 6
Accuracy99.7%
Cost32384
\[\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th \]
Alternative 7
Accuracy44.9%
Cost26184
\[\begin{array}{l} \mathbf{if}\;\sin ky \leq -2 \cdot 10^{-66}:\\ \;\;\;\;\sqrt{{\sin th}^{2}}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-115}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 8
Accuracy33.7%
Cost19784
\[\begin{array}{l} \mathbf{if}\;\sin ky \leq -1 \cdot 10^{-218}:\\ \;\;\;\;\frac{ky \cdot th}{\sin kx}\\ \mathbf{elif}\;\sin ky \leq 10^{-132}:\\ \;\;\;\;\frac{ky \cdot \sin th}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 9
Accuracy34.6%
Cost19784
\[\begin{array}{l} \mathbf{if}\;\sin ky \leq -1 \cdot 10^{-218}:\\ \;\;\;\;\sin ky \cdot \frac{th}{\sin kx}\\ \mathbf{elif}\;\sin ky \leq 10^{-132}:\\ \;\;\;\;\frac{ky \cdot \sin th}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 10
Accuracy40.7%
Cost19652
\[\begin{array}{l} \mathbf{if}\;\sin ky \leq 5 \cdot 10^{-115}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 11
Accuracy40.7%
Cost19652
\[\begin{array}{l} \mathbf{if}\;\sin ky \leq 5 \cdot 10^{-115}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 12
Accuracy39.9%
Cost19652
\[\begin{array}{l} \mathbf{if}\;\sin ky \leq 5 \cdot 10^{-115}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 13
Accuracy34.6%
Cost7116
\[\begin{array}{l} \mathbf{if}\;ky \leq -1.32:\\ \;\;\;\;\sin th\\ \mathbf{elif}\;ky \leq -9.2 \cdot 10^{-219}:\\ \;\;\;\;th \cdot \frac{ky}{\sin kx}\\ \mathbf{elif}\;ky \leq 6.5 \cdot 10^{-133}:\\ \;\;\;\;\frac{ky}{\frac{kx}{\sin th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 14
Accuracy34.3%
Cost7116
\[\begin{array}{l} \mathbf{if}\;ky \leq -1.32:\\ \;\;\;\;\sin th\\ \mathbf{elif}\;ky \leq -1.5 \cdot 10^{-218}:\\ \;\;\;\;\frac{ky \cdot th}{\sin kx}\\ \mathbf{elif}\;ky \leq 7.5 \cdot 10^{-125}:\\ \;\;\;\;\frac{ky}{\frac{kx}{\sin th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 15
Accuracy32.5%
Cost6984
\[\begin{array}{l} \mathbf{if}\;ky \leq -1.32:\\ \;\;\;\;\sin th\\ \mathbf{elif}\;ky \leq 2.05 \cdot 10^{-219}:\\ \;\;\;\;th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 16
Accuracy30.3%
Cost6728
\[\begin{array}{l} \mathbf{if}\;ky \leq -1.32:\\ \;\;\;\;\sin th\\ \mathbf{elif}\;ky \leq 7.8 \cdot 10^{-209}:\\ \;\;\;\;\frac{ky \cdot th}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
Alternative 17
Accuracy21.6%
Cost584
\[\begin{array}{l} \mathbf{if}\;ky \leq -4 \cdot 10^{-11}:\\ \;\;\;\;th\\ \mathbf{elif}\;ky \leq 9.5 \cdot 10^{-209}:\\ \;\;\;\;th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;th\\ \end{array} \]
Alternative 18
Accuracy20.8%
Cost584
\[\begin{array}{l} \mathbf{if}\;ky \leq -3.4 \cdot 10^{-13}:\\ \;\;\;\;th\\ \mathbf{elif}\;ky \leq 9.5 \cdot 10^{-209}:\\ \;\;\;\;\frac{ky \cdot th}{kx}\\ \mathbf{else}:\\ \;\;\;\;th\\ \end{array} \]
Alternative 19
Accuracy14.0%
Cost64
\[th \]

Reproduce?

herbie shell --seed 2023161 
(FPCore (kx ky th)
  :name "Toniolo and Linder, Equation (3b), real"
  :precision binary64
  (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))