Migdal et al, Equation (64)

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Percentage Accurate: 99.5% → 99.6%
Time: 15.0s
Precision: binary64
Cost: 19776

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\[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
↓
\[\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}} \]
(FPCore (a1 a2 th)
 :precision binary64
 (+
  (* (/ (cos th) (sqrt 2.0)) (* a1 a1))
  (* (/ (cos th) (sqrt 2.0)) (* a2 a2))))
↓
(FPCore (a1 a2 th)
 :precision binary64
 (* (cos th) (/ (fma a1 a1 (* a2 a2)) (sqrt 2.0))))
double code(double a1, double a2, double th) {
	return ((cos(th) / sqrt(2.0)) * (a1 * a1)) + ((cos(th) / sqrt(2.0)) * (a2 * a2));
}
↓
double code(double a1, double a2, double th) {
	return cos(th) * (fma(a1, a1, (a2 * a2)) / sqrt(2.0));
}

function code(a1, a2, th)
	return Float64(Float64(Float64(cos(th) / sqrt(2.0)) * Float64(a1 * a1)) + Float64(Float64(cos(th) / sqrt(2.0)) * Float64(a2 * a2)))
end
↓
function code(a1, a2, th)
	return Float64(cos(th) * Float64(fma(a1, a1, Float64(a2 * a2)) / sqrt(2.0)))
end

code[a1_, a2_, th_] := N[(N[(N[(N[Cos[th], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision] * N[(a1 * a1), $MachinePrecision]), $MachinePrecision] + N[(N[(N[Cos[th], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision] * N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
↓
code[a1_, a2_, th_] := N[(N[Cos[th], $MachinePrecision] * N[(N[(a1 * a1 + N[(a2 * a2), $MachinePrecision]), $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right)
↓
\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}

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 16 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

Derivation?

  1. Initial program 99.6%

    \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]

  2. Simplified99.6%

    \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
    Step-by-step derivation

    [Start]99.6

    \[ \frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]

    distribute-lft-out [=>]99.6

    \[ \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

    associate-*l/ [=>]99.6

    \[ \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]

    associate-*r/ [<=]99.6

    \[ \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]

    fma-def [=>]99.6

    \[ \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]

  3. Final simplification99.6%

    \[\leadsto \cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}} \]

Alternatives

Alternative 1
Accuracy69.2%
Cost13512
\[\begin{array}{l} \mathbf{if}\;a1 \leq -1.9 \cdot 10^{-15}:\\ \;\;\;\;a1 \cdot \frac{\cos th \cdot a1}{\sqrt{2}}\\ \mathbf{elif}\;a1 \leq -4.1 \cdot 10^{-125}:\\ \;\;\;\;\frac{a2 \cdot a2 + a1 \cdot a1}{\sqrt{2}}\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \left(\cos th \cdot \left(a2 \cdot \sqrt{0.5}\right)\right)\\ \end{array} \]
Alternative 2
Accuracy69.1%
Cost13512
\[\begin{array}{l} \mathbf{if}\;a1 \leq -8.8 \cdot 10^{-17}:\\ \;\;\;\;\cos th \cdot \frac{a1 \cdot a1}{\sqrt{2}}\\ \mathbf{elif}\;a1 \leq -2.55 \cdot 10^{-124}:\\ \;\;\;\;\frac{a2 \cdot a2 + a1 \cdot a1}{\sqrt{2}}\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \left(\cos th \cdot \left(a2 \cdot \sqrt{0.5}\right)\right)\\ \end{array} \]
Alternative 3
Accuracy99.6%
Cost13504
\[\sqrt{0.5} \cdot \left(\cos th \cdot \left(a2 \cdot a2 + a1 \cdot a1\right)\right) \]
Alternative 4
Accuracy99.6%
Cost13504
\[\left(\cos th \cdot \sqrt{0.5}\right) \cdot \left(a2 \cdot a2 + a1 \cdot a1\right) \]
Alternative 5
Accuracy99.5%
Cost13504
\[\frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2 + a1 \cdot a1\right) \]
Alternative 6
Accuracy99.6%
Cost13504
\[\frac{a2 \cdot a2 + a1 \cdot a1}{\frac{\sqrt{2}}{\cos th}} \]
Alternative 7
Accuracy65.4%
Cost13380
\[\begin{array}{l} \mathbf{if}\;a2 \leq 1.45 \cdot 10^{-122}:\\ \;\;\;\;a1 \cdot \frac{\cos th \cdot a1}{\sqrt{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2 \cdot a2 + a1 \cdot a1}{\sqrt{2}}\\ \end{array} \]
Alternative 8
Accuracy47.6%
Cost6980
\[\begin{array}{l} \mathbf{if}\;a2 \leq 1900000000:\\ \;\;\;\;\frac{a1 \cdot a1}{\sqrt{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a2}{\sqrt{2}}}{\frac{1}{a2}}\\ \end{array} \]
Alternative 9
Accuracy66.5%
Cost6976
\[\frac{a2 \cdot a2 + a1 \cdot a1}{\sqrt{2}} \]
Alternative 10
Accuracy47.6%
Cost6852
\[\begin{array}{l} \mathbf{if}\;a2 \leq 1900000000:\\ \;\;\;\;\left(a1 \cdot a1\right) \cdot \sqrt{0.5}\\ \mathbf{else}:\\ \;\;\;\;\left(a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \end{array} \]
Alternative 11
Accuracy47.6%
Cost6852
\[\begin{array}{l} \mathbf{if}\;a2 \leq 1250000000:\\ \;\;\;\;a1 \cdot \frac{a1}{\sqrt{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \end{array} \]
Alternative 12
Accuracy47.6%
Cost6852
\[\begin{array}{l} \mathbf{if}\;a2 \leq 2200000000:\\ \;\;\;\;a1 \cdot \frac{a1}{\sqrt{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{\frac{\sqrt{2}}{a2}}\\ \end{array} \]
Alternative 13
Accuracy47.6%
Cost6852
\[\begin{array}{l} \mathbf{if}\;a2 \leq 4400000000:\\ \;\;\;\;\frac{a1 \cdot a1}{\sqrt{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{\frac{\sqrt{2}}{a2}}\\ \end{array} \]
Alternative 14
Accuracy47.6%
Cost6852
\[\begin{array}{l} \mathbf{if}\;a2 \leq 5100000000:\\ \;\;\;\;\frac{a1 \cdot a1}{\sqrt{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2 \cdot a2}{\sqrt{2}}\\ \end{array} \]
Alternative 15
Accuracy39.7%
Cost6720
\[\left(a1 \cdot a1\right) \cdot \sqrt{0.5} \]

Reproduce?

herbie shell --seed 2023161 
(FPCore (a1 a2 th)
  :name "Migdal et al, Equation (64)"
  :precision binary64
  (+ (* (/ (cos th) (sqrt 2.0)) (* a1 a1)) (* (/ (cos th) (sqrt 2.0)) (* a2 a2))))