Toniolo and Linder, Equation (3b), real

Percentage Accurate: 94.2% → 99.7%

Time: 22.8s

Precision: binary64

Cost: 32384

Math

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

↓

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

FPCore

(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
 (* (/ (sin ky) (hypot (sin ky) (sin kx))) (sin th)))

C

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 (sin(ky) / hypot(sin(ky), sin(kx))) * sin(th);
}

Java

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.sin(ky) / Math.hypot(Math.sin(ky), Math.sin(kx))) * Math.sin(th);
}

Python

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.sin(ky) / math.hypot(math.sin(ky), math.sin(kx))) * math.sin(th)

Julia

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(Float64(sin(ky) / hypot(sin(ky), sin(kx))) * sin(th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / sqrt(((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th);
end

↓

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * sin(th);
end

Wolfram

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[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 92.8%

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

2. Simplified 99.7%

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

   [Start] 92.8%	\[ \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   +-commutative [=>] 92.8%	\[ \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \cdot \sin th \]
   unpow2 [=>] 92.8%	\[ \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \cdot \sin th \]
   unpow2 [=>] 92.8%	\[ \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \cdot \sin th \]
   hypot-def [=>] 99.7%	\[ \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \cdot \sin th \]

3. Final simplification 99.7%

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

Alternatives

Alternative 1
Accuracy	99.7%
Cost	32384

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

Alternative 2
Accuracy	47.4%
Cost	52048

\[\begin{array}{l} t_1 := \left|\frac{\sin ky}{\frac{\sin kx}{\sin th}}\right|\\ \mathbf{if}\;\sin ky \leq -1 \cdot 10^{-6}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\sin ky \leq -2 \cdot 10^{-243}:\\ \;\;\;\;\frac{ky \cdot th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\\ \mathbf{elif}\;\sin ky \leq -4 \cdot 10^{-305}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-14}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 3
Accuracy	42.0%
Cost	46096

\[\begin{array}{l} \mathbf{if}\;\sin kx \leq -0.05:\\ \;\;\;\;\frac{ky \cdot th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\\ \mathbf{elif}\;\sin kx \leq 4 \cdot 10^{-189}:\\ \;\;\;\;\frac{\sin ky \cdot \sin th}{\sin ky}\\ \mathbf{elif}\;\sin kx \leq 10^{-139}:\\ \;\;\;\;\frac{\sin th}{\frac{\sin kx}{ky} + \left(0.25 \cdot \left(ky \cdot kx\right) + 0.5 \cdot \frac{ky}{kx}\right)}\\ \mathbf{elif}\;\sin kx \leq 2 \cdot 10^{-73}:\\ \;\;\;\;\frac{\sin th}{1 + \frac{0.5 \cdot \left(kx \cdot kx\right)}{{\sin ky}^{2}}}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \]

Alternative 4
Accuracy	44.4%
Cost	46096

\[\begin{array}{l} \mathbf{if}\;\sin kx \leq -1 \cdot 10^{-93}:\\ \;\;\;\;\left|\frac{\sin ky}{\frac{\sin kx}{\sin th}}\right|\\ \mathbf{elif}\;\sin kx \leq 4 \cdot 10^{-189}:\\ \;\;\;\;\sin th\\ \mathbf{elif}\;\sin kx \leq 10^{-139}:\\ \;\;\;\;\frac{\sin th}{\frac{\sin kx}{ky} + \left(0.25 \cdot \left(ky \cdot kx\right) + 0.5 \cdot \frac{ky}{kx}\right)}\\ \mathbf{elif}\;\sin kx \leq 2 \cdot 10^{-73}:\\ \;\;\;\;\frac{\sin th}{1 + \frac{0.5 \cdot \left(kx \cdot kx\right)}{{\sin ky}^{2}}}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \]

Alternative 5
Accuracy	42.0%
Cost	45648

\[\begin{array}{l} t_1 := \frac{\sin ky \cdot \sin th}{\sin ky}\\ \mathbf{if}\;\sin kx \leq -0.05:\\ \;\;\;\;\frac{ky \cdot th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\\ \mathbf{elif}\;\sin kx \leq 4 \cdot 10^{-189}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\sin kx \leq 10^{-139}:\\ \;\;\;\;\frac{\sin th}{\frac{\sin kx}{ky} + \left(0.25 \cdot \left(ky \cdot kx\right) + 0.5 \cdot \frac{ky}{kx}\right)}\\ \mathbf{elif}\;\sin kx \leq 2 \cdot 10^{-73}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \]

Alternative 6
Accuracy	99.6%
Cost	32384

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

Alternative 7
Accuracy	75.0%
Cost	26249

\[\begin{array}{l} t_1 := \mathsf{hypot}\left(\sin ky, \sin kx\right)\\ \mathbf{if}\;th \leq -0.00076 \lor \neg \left(th \leq 3.3\right):\\ \;\;\;\;\frac{ky \cdot \sin th}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky}{t_1} \cdot th\\ \end{array} \]

Alternative 8
Accuracy	61.4%
Cost	26248

\[\begin{array}{l} \mathbf{if}\;th \leq -1.16:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{elif}\;th \leq 0.21:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \]

Alternative 9
Accuracy	40.7%
Cost	26052

\[\begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-36}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 10
Accuracy	39.6%
Cost	19652

\[\begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-36}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 11
Accuracy	33.4%
Cost	13252

\[\begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-183}:\\ \;\;\;\;\frac{\sin th}{\frac{kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 12
Accuracy	33.1%
Cost	6984

\[\begin{array}{l} \mathbf{if}\;ky \leq -390000:\\ \;\;\;\;\sin th\\ \mathbf{elif}\;ky \leq 2.1 \cdot 10^{-181}:\\ \;\;\;\;\sin th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 13
Accuracy	30.8%
Cost	6728

\[\begin{array}{l} \mathbf{if}\;ky \leq -390000:\\ \;\;\;\;\sin th\\ \mathbf{elif}\;ky \leq 9.2 \cdot 10^{-185}:\\ \;\;\;\;th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 14
Accuracy	21.3%
Cost	584

\[\begin{array}{l} \mathbf{if}\;ky \leq -2.2 \cdot 10^{+40}:\\ \;\;\;\;th\\ \mathbf{elif}\;ky \leq 1.9 \cdot 10^{-136}:\\ \;\;\;\;th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;th\\ \end{array} \]

Alternative 15
Accuracy	13.4%
Cost	64

\[th \]

Reproduce

herbie shell --seed 2023272 
(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)))