Migdal et al, Equation (64)

?

Percentage Accurate: 99.5% → 99.5%
Time: 16.5s
Precision: binary64
Cost: 26560

?

\[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
↓
\[\begin{array}{l} t_1 := \frac{\cos th}{\sqrt{2}}\\ t_1 \cdot \left(a1 \cdot a1\right) + t_1 \cdot \left(a2 \cdot a2\right) \end{array} \]
(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
 (let* ((t_1 (/ (cos th) (sqrt 2.0))))
   (+ (* t_1 (* a1 a1)) (* t_1 (* a2 a2)))))
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) {
	double t_1 = cos(th) / sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}

real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = ((cos(th) / sqrt(2.0d0)) * (a1 * a1)) + ((cos(th) / sqrt(2.0d0)) * (a2 * a2))
end function
↓
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    t_1 = cos(th) / sqrt(2.0d0)
    code = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))
end function

public static double code(double a1, double a2, double th) {
	return ((Math.cos(th) / Math.sqrt(2.0)) * (a1 * a1)) + ((Math.cos(th) / Math.sqrt(2.0)) * (a2 * a2));
}
↓
public static double code(double a1, double a2, double th) {
	double t_1 = Math.cos(th) / Math.sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}

def code(a1, a2, th):
	return ((math.cos(th) / math.sqrt(2.0)) * (a1 * a1)) + ((math.cos(th) / math.sqrt(2.0)) * (a2 * a2))
↓
def code(a1, a2, th):
	t_1 = math.cos(th) / math.sqrt(2.0)
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))

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)
	t_1 = Float64(cos(th) / sqrt(2.0))
	return Float64(Float64(t_1 * Float64(a1 * a1)) + Float64(t_1 * Float64(a2 * a2)))
end

function tmp = code(a1, a2, th)
	tmp = ((cos(th) / sqrt(2.0)) * (a1 * a1)) + ((cos(th) / sqrt(2.0)) * (a2 * a2));
end
↓
function tmp = code(a1, a2, th)
	t_1 = cos(th) / sqrt(2.0);
	tmp = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
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_] := Block[{t$95$1 = N[(N[Cos[th], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]}, N[(N[(t$95$1 * N[(a1 * a1), $MachinePrecision]), $MachinePrecision] + N[(t$95$1 * N[(a2 * a2), $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)
↓
\begin{array}{l}
t_1 := \frac{\cos th}{\sqrt{2}}\\
t_1 \cdot \left(a1 \cdot a1\right) + t_1 \cdot \left(a2 \cdot a2\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 13 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 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. Final simplification99.6%

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

Alternatives

Alternative 1
Accuracy99.5%
Cost26560
\[\begin{array}{l} t_1 := \frac{\cos th}{\sqrt{2}}\\ t_1 \cdot \left(a1 \cdot a1\right) + t_1 \cdot \left(a2 \cdot a2\right) \end{array} \]
Alternative 2
Accuracy69.6%
Cost13512
\[\begin{array}{l} \mathbf{if}\;a1 \leq -2.6 \cdot 10^{+32}:\\ \;\;\;\;\cos th \cdot \left(a1 \cdot \left(a1 \cdot \sqrt{0.5}\right)\right)\\ \mathbf{elif}\;a1 \leq -1.25 \cdot 10^{-114}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \frac{a2}{\frac{\sqrt{2}}{\cos th}}\\ \end{array} \]
Alternative 3
Accuracy69.6%
Cost13512
\[\begin{array}{l} \mathbf{if}\;a1 \leq -2.65 \cdot 10^{+32}:\\ \;\;\;\;\cos th \cdot \frac{a1}{\frac{\sqrt{2}}{a1}}\\ \mathbf{elif}\;a1 \leq -1.05 \cdot 10^{-114}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \frac{a2}{\frac{\sqrt{2}}{\cos th}}\\ \end{array} \]
Alternative 4
Accuracy69.6%
Cost13512
\[\begin{array}{l} \mathbf{if}\;a1 \leq -1.06 \cdot 10^{+33}:\\ \;\;\;\;\cos th \cdot \frac{a1}{\frac{\sqrt{2}}{a1}}\\ \mathbf{elif}\;a1 \leq -9.2 \cdot 10^{-116}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \mathbf{else}:\\ \;\;\;\;\frac{\cos th \cdot \left(a2 \cdot a2\right)}{\sqrt{2}}\\ \end{array} \]
Alternative 5
Accuracy99.5%
Cost13504
\[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
Alternative 6
Accuracy68.9%
Cost13380
\[\begin{array}{l} \mathbf{if}\;a2 \leq 1.15 \cdot 10^{-81}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \frac{1}{\sqrt{2}}\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \frac{a2}{\frac{\sqrt{2}}{\cos th}}\\ \end{array} \]
Alternative 7
Accuracy66.7%
Cost7364
\[\begin{array}{l} \mathbf{if}\;a2 \leq 2.05 \cdot 10^{+146}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(a2 \cdot a2\right) \cdot \left(1 + -0.5 \cdot \left(th \cdot th\right)\right)}{\sqrt{2}}\\ \end{array} \]
Alternative 8
Accuracy47.2%
Cost6980
\[\begin{array}{l} \mathbf{if}\;a1 \leq -7 \cdot 10^{-53}:\\ \;\;\;\;\frac{1}{\frac{\sqrt{2}}{a1 \cdot a1}}\\ \mathbf{else}:\\ \;\;\;\;\left(a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \end{array} \]
Alternative 9
Accuracy66.6%
Cost6976
\[\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \sqrt{0.5} \]
Alternative 10
Accuracy47.2%
Cost6852
\[\begin{array}{l} \mathbf{if}\;a1 \leq -6.5 \cdot 10^{-57}:\\ \;\;\;\;a1 \cdot \left(a1 \cdot \sqrt{0.5}\right)\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \left(a2 \cdot \sqrt{0.5}\right)\\ \end{array} \]
Alternative 11
Accuracy47.2%
Cost6852
\[\begin{array}{l} \mathbf{if}\;a1 \leq -5.6 \cdot 10^{-47}:\\ \;\;\;\;a1 \cdot \left(a1 \cdot \sqrt{0.5}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \end{array} \]
Alternative 12
Accuracy47.2%
Cost6852
\[\begin{array}{l} \mathbf{if}\;a1 \leq -1.65 \cdot 10^{-51}:\\ \;\;\;\;\frac{a1}{\frac{\sqrt{2}}{a1}}\\ \mathbf{else}:\\ \;\;\;\;\left(a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \end{array} \]
Alternative 13
Accuracy39.8%
Cost6720
\[a1 \cdot \left(a1 \cdot \sqrt{0.5}\right) \]

Reproduce?

herbie shell --seed 2023255 
(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))))