Migdal et al, Equation (64)

Percentage Accurate: 99.6% → 99.6%
Time: 1.3s
Alternatives: 1
Speedup: 1.0×

Specification

?
\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{\cos th}{\sqrt{2}}\\ t\_1 \cdot \left(a1 \cdot a1\right) + t\_1 \cdot \left(a2 \cdot a2\right) \end{array} \end{array} \]
(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) {
	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
    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) {
	double t_1 = Math.cos(th) / Math.sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (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)
	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)
	t_1 = cos(th) / sqrt(2.0);
	tmp = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
end

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]]

\begin{array}{l}

\\
\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}
\end{array}

Sampling outcomes in binary64 precision:

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.

Accuracy vs Speed?

Herbie found 1 alternatives:

AlternativeAccuracySpeedup
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.

Initial Program: 99.6% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{\cos th}{\sqrt{2}}\\ t\_1 \cdot \left(a1 \cdot a1\right) + t\_1 \cdot \left(a2 \cdot a2\right) \end{array} \end{array} \]
(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) {
	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
    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) {
	double t_1 = Math.cos(th) / Math.sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (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)
	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)
	t_1 = cos(th) / sqrt(2.0);
	tmp = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
end

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]]

\begin{array}{l}

\\
\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}
\end{array}

Alternative 1: 99.6% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{\cos th}{\sqrt{2}}\\ \left(a2 \cdot a2\right) \cdot t\_1 + \left(a1 \cdot a1\right) \cdot t\_1 \end{array} \end{array} \]
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (/ (cos th) (sqrt 2.0))))
   (+ (* (* a2 a2) t_1) (* (* a1 a1) t_1))))
double code(double a1, double a2, double th) {
	double t_1 = cos(th) / sqrt(2.0);
	return ((a2 * a2) * t_1) + ((a1 * a1) * t_1);
}

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 = ((a2 * a2) * t_1) + ((a1 * a1) * t_1)
end function

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

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

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

function tmp = code(a1, a2, th)
	t_1 = cos(th) / sqrt(2.0);
	tmp = ((a2 * a2) * t_1) + ((a1 * a1) * t_1);
end

code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[Cos[th], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]}, N[(N[(N[(a2 * a2), $MachinePrecision] * t$95$1), $MachinePrecision] + N[(N[(a1 * a1), $MachinePrecision] * t$95$1), $MachinePrecision]), $MachinePrecision]]

\begin{array}{l}

\\
\begin{array}{l}
t_1 := \frac{\cos th}{\sqrt{2}}\\
\left(a2 \cdot a2\right) \cdot t\_1 + \left(a1 \cdot a1\right) \cdot t\_1
\end{array}
\end{array}
Derivation

  1. Initial program 99.5%

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

  3. Final simplification99.5%

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

Reproduce

?
herbie shell --seed 2024343 
(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))))