NMSE Section 6.1 mentioned, B

Percentage Accurate: 78.7% → 99.1%
Time: 15.2s
Alternatives: 10
Speedup: 1.9×

Specification

?
\[\begin{array}{l} \\ \left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \end{array} \]
(FPCore (a b)
 :precision binary64
 (* (* (/ PI 2.0) (/ 1.0 (- (* b b) (* a a)))) (- (/ 1.0 a) (/ 1.0 b))))
double code(double a, double b) {
	return ((((double) M_PI) / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b));
}
public static double code(double a, double b) {
	return ((Math.PI / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b));
}
def code(a, b):
	return ((math.pi / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b))
function code(a, b)
	return Float64(Float64(Float64(pi / 2.0) * Float64(1.0 / Float64(Float64(b * b) - Float64(a * a)))) * Float64(Float64(1.0 / a) - Float64(1.0 / b)))
end
function tmp = code(a, b)
	tmp = ((pi / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b));
end
code[a_, b_] := N[(N[(N[(Pi / 2.0), $MachinePrecision] * N[(1.0 / N[(N[(b * b), $MachinePrecision] - N[(a * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(1.0 / a), $MachinePrecision] - N[(1.0 / b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right)
\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 10 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: 78.7% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \end{array} \]
(FPCore (a b)
 :precision binary64
 (* (* (/ PI 2.0) (/ 1.0 (- (* b b) (* a a)))) (- (/ 1.0 a) (/ 1.0 b))))
double code(double a, double b) {
	return ((((double) M_PI) / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b));
}
public static double code(double a, double b) {
	return ((Math.PI / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b));
}
def code(a, b):
	return ((math.pi / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b))
function code(a, b)
	return Float64(Float64(Float64(pi / 2.0) * Float64(1.0 / Float64(Float64(b * b) - Float64(a * a)))) * Float64(Float64(1.0 / a) - Float64(1.0 / b)))
end
function tmp = code(a, b)
	tmp = ((pi / 2.0) * (1.0 / ((b * b) - (a * a)))) * ((1.0 / a) - (1.0 / b));
end
code[a_, b_] := N[(N[(N[(Pi / 2.0), $MachinePrecision] * N[(1.0 / N[(N[(b * b), $MachinePrecision] - N[(a * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(1.0 / a), $MachinePrecision] - N[(1.0 / b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right)
\end{array}

Alternative 1: 99.1% accurate, 1.3× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;b \leq 8.5 \cdot 10^{+82}:\\ \;\;\;\;\frac{0.5}{a \cdot \left(b \cdot \left(a + b\right)\right)} \cdot \pi\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\pi \cdot 0.5}{a \cdot b}}{b}\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= b 8.5e+82)
   (* (/ 0.5 (* a (* b (+ a b)))) PI)
   (/ (/ (* PI 0.5) (* a b)) b)))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (b <= 8.5e+82) {
		tmp = (0.5 / (a * (b * (a + b)))) * ((double) M_PI);
	} else {
		tmp = ((((double) M_PI) * 0.5) / (a * b)) / b;
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (b <= 8.5e+82) {
		tmp = (0.5 / (a * (b * (a + b)))) * Math.PI;
	} else {
		tmp = ((Math.PI * 0.5) / (a * b)) / b;
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if b <= 8.5e+82:
		tmp = (0.5 / (a * (b * (a + b)))) * math.pi
	else:
		tmp = ((math.pi * 0.5) / (a * b)) / b
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (b <= 8.5e+82)
		tmp = Float64(Float64(0.5 / Float64(a * Float64(b * Float64(a + b)))) * pi);
	else
		tmp = Float64(Float64(Float64(pi * 0.5) / Float64(a * b)) / b);
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (b <= 8.5e+82)
		tmp = (0.5 / (a * (b * (a + b)))) * pi;
	else
		tmp = ((pi * 0.5) / (a * b)) / b;
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[b, 8.5e+82], N[(N[(0.5 / N[(a * N[(b * N[(a + b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * Pi), $MachinePrecision], N[(N[(N[(Pi * 0.5), $MachinePrecision] / N[(a * b), $MachinePrecision]), $MachinePrecision] / b), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;b \leq 8.5 \cdot 10^{+82}:\\
\;\;\;\;\frac{0.5}{a \cdot \left(b \cdot \left(a + b\right)\right)} \cdot \pi\\

\mathbf{else}:\\
\;\;\;\;\frac{\frac{\pi \cdot 0.5}{a \cdot b}}{b}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if b < 8.4999999999999995e82

    1. Initial program 81.5%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Applied egg-rr0

      \[\leadsto expr\]
    4. Taylor expanded in b around 0 0

      \[\leadsto expr\]
    5. Simplified0

      \[\leadsto expr\]
    6. Applied egg-rr0

      \[\leadsto expr\]

    if 8.4999999999999995e82 < b

    1. Initial program 49.7%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around inf 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Applied egg-rr0

      \[\leadsto expr\]
  3. Recombined 2 regimes into one program.
  4. Add Preprocessing

Alternative 2: 99.6% accurate, 1.2× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \frac{\frac{\frac{\frac{\pi}{a} - \frac{\pi}{b}}{2}}{b + a}}{b - a} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (/ (/ (/ (- (/ PI a) (/ PI b)) 2.0) (+ b a)) (- b a)))
assert(a < b);
double code(double a, double b) {
	return ((((((double) M_PI) / a) - (((double) M_PI) / b)) / 2.0) / (b + a)) / (b - a);
}
assert a < b;
public static double code(double a, double b) {
	return ((((Math.PI / a) - (Math.PI / b)) / 2.0) / (b + a)) / (b - a);
}
[a, b] = sort([a, b])
def code(a, b):
	return ((((math.pi / a) - (math.pi / b)) / 2.0) / (b + a)) / (b - a)
a, b = sort([a, b])
function code(a, b)
	return Float64(Float64(Float64(Float64(Float64(pi / a) - Float64(pi / b)) / 2.0) / Float64(b + a)) / Float64(b - a))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = ((((pi / a) - (pi / b)) / 2.0) / (b + a)) / (b - a);
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(N[(N[(N[(N[(Pi / a), $MachinePrecision] - N[(Pi / b), $MachinePrecision]), $MachinePrecision] / 2.0), $MachinePrecision] / N[(b + a), $MachinePrecision]), $MachinePrecision] / N[(b - a), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\frac{\frac{\frac{\frac{\pi}{a} - \frac{\pi}{b}}{2}}{b + a}}{b - a}
\end{array}
Derivation
  1. Initial program 75.1%

    \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
  2. Add Preprocessing
  3. Applied egg-rr0

    \[\leadsto expr\]
  4. Add Preprocessing

Alternative 3: 99.5% accurate, 1.2× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \frac{\frac{\pi}{a} - \frac{\pi}{b}}{b + a} \cdot \frac{0.5}{b - a} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (* (/ (- (/ PI a) (/ PI b)) (+ b a)) (/ 0.5 (- b a))))
assert(a < b);
double code(double a, double b) {
	return (((((double) M_PI) / a) - (((double) M_PI) / b)) / (b + a)) * (0.5 / (b - a));
}
assert a < b;
public static double code(double a, double b) {
	return (((Math.PI / a) - (Math.PI / b)) / (b + a)) * (0.5 / (b - a));
}
[a, b] = sort([a, b])
def code(a, b):
	return (((math.pi / a) - (math.pi / b)) / (b + a)) * (0.5 / (b - a))
a, b = sort([a, b])
function code(a, b)
	return Float64(Float64(Float64(Float64(pi / a) - Float64(pi / b)) / Float64(b + a)) * Float64(0.5 / Float64(b - a)))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = (((pi / a) - (pi / b)) / (b + a)) * (0.5 / (b - a));
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(N[(N[(N[(Pi / a), $MachinePrecision] - N[(Pi / b), $MachinePrecision]), $MachinePrecision] / N[(b + a), $MachinePrecision]), $MachinePrecision] * N[(0.5 / N[(b - a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\frac{\frac{\pi}{a} - \frac{\pi}{b}}{b + a} \cdot \frac{0.5}{b - a}
\end{array}
Derivation
  1. Initial program 75.1%

    \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
  2. Add Preprocessing
  3. Applied egg-rr0

    \[\leadsto expr\]
  4. Add Preprocessing

Alternative 4: 89.5% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;b \leq 2.1 \cdot 10^{-22}:\\ \;\;\;\;\frac{\frac{0.5}{a \cdot b}}{\frac{a}{\pi}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\pi \cdot 0.5}{b}}{a \cdot b}\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= b 2.1e-22) (/ (/ 0.5 (* a b)) (/ a PI)) (/ (/ (* PI 0.5) b) (* a b))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (b <= 2.1e-22) {
		tmp = (0.5 / (a * b)) / (a / ((double) M_PI));
	} else {
		tmp = ((((double) M_PI) * 0.5) / b) / (a * b);
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (b <= 2.1e-22) {
		tmp = (0.5 / (a * b)) / (a / Math.PI);
	} else {
		tmp = ((Math.PI * 0.5) / b) / (a * b);
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if b <= 2.1e-22:
		tmp = (0.5 / (a * b)) / (a / math.pi)
	else:
		tmp = ((math.pi * 0.5) / b) / (a * b)
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (b <= 2.1e-22)
		tmp = Float64(Float64(0.5 / Float64(a * b)) / Float64(a / pi));
	else
		tmp = Float64(Float64(Float64(pi * 0.5) / b) / Float64(a * b));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (b <= 2.1e-22)
		tmp = (0.5 / (a * b)) / (a / pi);
	else
		tmp = ((pi * 0.5) / b) / (a * b);
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[b, 2.1e-22], N[(N[(0.5 / N[(a * b), $MachinePrecision]), $MachinePrecision] / N[(a / Pi), $MachinePrecision]), $MachinePrecision], N[(N[(N[(Pi * 0.5), $MachinePrecision] / b), $MachinePrecision] / N[(a * b), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;b \leq 2.1 \cdot 10^{-22}:\\
\;\;\;\;\frac{\frac{0.5}{a \cdot b}}{\frac{a}{\pi}}\\

\mathbf{else}:\\
\;\;\;\;\frac{\frac{\pi \cdot 0.5}{b}}{a \cdot b}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if b < 2.10000000000000008e-22

    1. Initial program 78.5%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Taylor expanded in a around inf 0

      \[\leadsto expr\]
    6. Simplified0

      \[\leadsto expr\]
    7. Applied egg-rr0

      \[\leadsto expr\]

    if 2.10000000000000008e-22 < b

    1. Initial program 67.6%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around inf 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Applied egg-rr0

      \[\leadsto expr\]
    6. Applied egg-rr0

      \[\leadsto expr\]
  3. Recombined 2 regimes into one program.
  4. Add Preprocessing

Alternative 5: 89.4% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;b \leq 2 \cdot 10^{-22}:\\ \;\;\;\;\frac{\frac{0.5}{a \cdot b}}{\frac{a}{\pi}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{0.5}{\frac{b}{\pi}}}{a \cdot b}\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= b 2e-22) (/ (/ 0.5 (* a b)) (/ a PI)) (/ (/ 0.5 (/ b PI)) (* a b))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (b <= 2e-22) {
		tmp = (0.5 / (a * b)) / (a / ((double) M_PI));
	} else {
		tmp = (0.5 / (b / ((double) M_PI))) / (a * b);
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (b <= 2e-22) {
		tmp = (0.5 / (a * b)) / (a / Math.PI);
	} else {
		tmp = (0.5 / (b / Math.PI)) / (a * b);
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if b <= 2e-22:
		tmp = (0.5 / (a * b)) / (a / math.pi)
	else:
		tmp = (0.5 / (b / math.pi)) / (a * b)
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (b <= 2e-22)
		tmp = Float64(Float64(0.5 / Float64(a * b)) / Float64(a / pi));
	else
		tmp = Float64(Float64(0.5 / Float64(b / pi)) / Float64(a * b));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (b <= 2e-22)
		tmp = (0.5 / (a * b)) / (a / pi);
	else
		tmp = (0.5 / (b / pi)) / (a * b);
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[b, 2e-22], N[(N[(0.5 / N[(a * b), $MachinePrecision]), $MachinePrecision] / N[(a / Pi), $MachinePrecision]), $MachinePrecision], N[(N[(0.5 / N[(b / Pi), $MachinePrecision]), $MachinePrecision] / N[(a * b), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;b \leq 2 \cdot 10^{-22}:\\
\;\;\;\;\frac{\frac{0.5}{a \cdot b}}{\frac{a}{\pi}}\\

\mathbf{else}:\\
\;\;\;\;\frac{\frac{0.5}{\frac{b}{\pi}}}{a \cdot b}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if b < 2.0000000000000001e-22

    1. Initial program 78.5%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Taylor expanded in a around inf 0

      \[\leadsto expr\]
    6. Simplified0

      \[\leadsto expr\]
    7. Applied egg-rr0

      \[\leadsto expr\]

    if 2.0000000000000001e-22 < b

    1. Initial program 67.6%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around inf 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Applied egg-rr0

      \[\leadsto expr\]
    6. Taylor expanded in b around inf 0

      \[\leadsto expr\]
    7. Simplified0

      \[\leadsto expr\]
    8. Applied egg-rr0

      \[\leadsto expr\]
  3. Recombined 2 regimes into one program.
  4. Add Preprocessing

Alternative 6: 89.4% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;b \leq 2.5 \cdot 10^{-22}:\\ \;\;\;\;\frac{\frac{0.5}{a \cdot b}}{\frac{a}{\pi}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{0.5}{a}}{b} \cdot \frac{\pi}{b}\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= b 2.5e-22) (/ (/ 0.5 (* a b)) (/ a PI)) (* (/ (/ 0.5 a) b) (/ PI b))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (b <= 2.5e-22) {
		tmp = (0.5 / (a * b)) / (a / ((double) M_PI));
	} else {
		tmp = ((0.5 / a) / b) * (((double) M_PI) / b);
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (b <= 2.5e-22) {
		tmp = (0.5 / (a * b)) / (a / Math.PI);
	} else {
		tmp = ((0.5 / a) / b) * (Math.PI / b);
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if b <= 2.5e-22:
		tmp = (0.5 / (a * b)) / (a / math.pi)
	else:
		tmp = ((0.5 / a) / b) * (math.pi / b)
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (b <= 2.5e-22)
		tmp = Float64(Float64(0.5 / Float64(a * b)) / Float64(a / pi));
	else
		tmp = Float64(Float64(Float64(0.5 / a) / b) * Float64(pi / b));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (b <= 2.5e-22)
		tmp = (0.5 / (a * b)) / (a / pi);
	else
		tmp = ((0.5 / a) / b) * (pi / b);
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[b, 2.5e-22], N[(N[(0.5 / N[(a * b), $MachinePrecision]), $MachinePrecision] / N[(a / Pi), $MachinePrecision]), $MachinePrecision], N[(N[(N[(0.5 / a), $MachinePrecision] / b), $MachinePrecision] * N[(Pi / b), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;b \leq 2.5 \cdot 10^{-22}:\\
\;\;\;\;\frac{\frac{0.5}{a \cdot b}}{\frac{a}{\pi}}\\

\mathbf{else}:\\
\;\;\;\;\frac{\frac{0.5}{a}}{b} \cdot \frac{\pi}{b}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if b < 2.49999999999999977e-22

    1. Initial program 78.5%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Taylor expanded in a around inf 0

      \[\leadsto expr\]
    6. Simplified0

      \[\leadsto expr\]
    7. Applied egg-rr0

      \[\leadsto expr\]

    if 2.49999999999999977e-22 < b

    1. Initial program 67.6%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around inf 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Applied egg-rr0

      \[\leadsto expr\]
    6. Taylor expanded in b around inf 0

      \[\leadsto expr\]
    7. Simplified0

      \[\leadsto expr\]
  3. Recombined 2 regimes into one program.
  4. Add Preprocessing

Alternative 7: 89.4% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;b \leq 2.9 \cdot 10^{-22}:\\ \;\;\;\;\frac{\pi}{a \cdot b} \cdot \frac{0.5}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{0.5}{a}}{b} \cdot \frac{\pi}{b}\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= b 2.9e-22) (* (/ PI (* a b)) (/ 0.5 a)) (* (/ (/ 0.5 a) b) (/ PI b))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (b <= 2.9e-22) {
		tmp = (((double) M_PI) / (a * b)) * (0.5 / a);
	} else {
		tmp = ((0.5 / a) / b) * (((double) M_PI) / b);
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (b <= 2.9e-22) {
		tmp = (Math.PI / (a * b)) * (0.5 / a);
	} else {
		tmp = ((0.5 / a) / b) * (Math.PI / b);
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if b <= 2.9e-22:
		tmp = (math.pi / (a * b)) * (0.5 / a)
	else:
		tmp = ((0.5 / a) / b) * (math.pi / b)
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (b <= 2.9e-22)
		tmp = Float64(Float64(pi / Float64(a * b)) * Float64(0.5 / a));
	else
		tmp = Float64(Float64(Float64(0.5 / a) / b) * Float64(pi / b));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (b <= 2.9e-22)
		tmp = (pi / (a * b)) * (0.5 / a);
	else
		tmp = ((0.5 / a) / b) * (pi / b);
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[b, 2.9e-22], N[(N[(Pi / N[(a * b), $MachinePrecision]), $MachinePrecision] * N[(0.5 / a), $MachinePrecision]), $MachinePrecision], N[(N[(N[(0.5 / a), $MachinePrecision] / b), $MachinePrecision] * N[(Pi / b), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;b \leq 2.9 \cdot 10^{-22}:\\
\;\;\;\;\frac{\pi}{a \cdot b} \cdot \frac{0.5}{a}\\

\mathbf{else}:\\
\;\;\;\;\frac{\frac{0.5}{a}}{b} \cdot \frac{\pi}{b}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if b < 2.9000000000000002e-22

    1. Initial program 78.5%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Taylor expanded in a around inf 0

      \[\leadsto expr\]
    6. Simplified0

      \[\leadsto expr\]
    7. Applied egg-rr0

      \[\leadsto expr\]

    if 2.9000000000000002e-22 < b

    1. Initial program 67.6%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around inf 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Applied egg-rr0

      \[\leadsto expr\]
    6. Taylor expanded in b around inf 0

      \[\leadsto expr\]
    7. Simplified0

      \[\leadsto expr\]
  3. Recombined 2 regimes into one program.
  4. Add Preprocessing

Alternative 8: 89.2% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;b \leq 2.3 \cdot 10^{-22}:\\ \;\;\;\;\frac{\pi}{a \cdot b} \cdot \frac{0.5}{a}\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \frac{0.5}{b \cdot \left(a \cdot b\right)}\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= b 2.3e-22) (* (/ PI (* a b)) (/ 0.5 a)) (* PI (/ 0.5 (* b (* a b))))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (b <= 2.3e-22) {
		tmp = (((double) M_PI) / (a * b)) * (0.5 / a);
	} else {
		tmp = ((double) M_PI) * (0.5 / (b * (a * b)));
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (b <= 2.3e-22) {
		tmp = (Math.PI / (a * b)) * (0.5 / a);
	} else {
		tmp = Math.PI * (0.5 / (b * (a * b)));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if b <= 2.3e-22:
		tmp = (math.pi / (a * b)) * (0.5 / a)
	else:
		tmp = math.pi * (0.5 / (b * (a * b)))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (b <= 2.3e-22)
		tmp = Float64(Float64(pi / Float64(a * b)) * Float64(0.5 / a));
	else
		tmp = Float64(pi * Float64(0.5 / Float64(b * Float64(a * b))));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (b <= 2.3e-22)
		tmp = (pi / (a * b)) * (0.5 / a);
	else
		tmp = pi * (0.5 / (b * (a * b)));
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[b, 2.3e-22], N[(N[(Pi / N[(a * b), $MachinePrecision]), $MachinePrecision] * N[(0.5 / a), $MachinePrecision]), $MachinePrecision], N[(Pi * N[(0.5 / N[(b * N[(a * b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;b \leq 2.3 \cdot 10^{-22}:\\
\;\;\;\;\frac{\pi}{a \cdot b} \cdot \frac{0.5}{a}\\

\mathbf{else}:\\
\;\;\;\;\pi \cdot \frac{0.5}{b \cdot \left(a \cdot b\right)}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if b < 2.2999999999999998e-22

    1. Initial program 78.5%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Taylor expanded in a around inf 0

      \[\leadsto expr\]
    6. Simplified0

      \[\leadsto expr\]
    7. Applied egg-rr0

      \[\leadsto expr\]

    if 2.2999999999999998e-22 < b

    1. Initial program 67.6%

      \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around inf 0

      \[\leadsto expr\]
    4. Simplified0

      \[\leadsto expr\]
    5. Applied egg-rr0

      \[\leadsto expr\]
  3. Recombined 2 regimes into one program.
  4. Add Preprocessing

Alternative 9: 99.0% accurate, 1.9× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \frac{0.5}{\frac{a + b}{\frac{\pi}{a \cdot b}}} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (/ 0.5 (/ (+ a b) (/ PI (* a b)))))
assert(a < b);
double code(double a, double b) {
	return 0.5 / ((a + b) / (((double) M_PI) / (a * b)));
}
assert a < b;
public static double code(double a, double b) {
	return 0.5 / ((a + b) / (Math.PI / (a * b)));
}
[a, b] = sort([a, b])
def code(a, b):
	return 0.5 / ((a + b) / (math.pi / (a * b)))
a, b = sort([a, b])
function code(a, b)
	return Float64(0.5 / Float64(Float64(a + b) / Float64(pi / Float64(a * b))))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = 0.5 / ((a + b) / (pi / (a * b)));
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(0.5 / N[(N[(a + b), $MachinePrecision] / N[(Pi / N[(a * b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\frac{0.5}{\frac{a + b}{\frac{\pi}{a \cdot b}}}
\end{array}
Derivation
  1. Initial program 75.1%

    \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
  2. Add Preprocessing
  3. Applied egg-rr0

    \[\leadsto expr\]
  4. Taylor expanded in b around 0 0

    \[\leadsto expr\]
  5. Simplified0

    \[\leadsto expr\]
  6. Applied egg-rr0

    \[\leadsto expr\]
  7. Add Preprocessing

Alternative 10: 62.2% accurate, 2.3× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \pi \cdot \frac{0.5}{b \cdot \left(a \cdot b\right)} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (* PI (/ 0.5 (* b (* a b)))))
assert(a < b);
double code(double a, double b) {
	return ((double) M_PI) * (0.5 / (b * (a * b)));
}
assert a < b;
public static double code(double a, double b) {
	return Math.PI * (0.5 / (b * (a * b)));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.pi * (0.5 / (b * (a * b)))
a, b = sort([a, b])
function code(a, b)
	return Float64(pi * Float64(0.5 / Float64(b * Float64(a * b))))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = pi * (0.5 / (b * (a * b)));
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(Pi * N[(0.5 / N[(b * N[(a * b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\pi \cdot \frac{0.5}{b \cdot \left(a \cdot b\right)}
\end{array}
Derivation
  1. Initial program 75.1%

    \[\left(\frac{\pi}{2} \cdot \frac{1}{b \cdot b - a \cdot a}\right) \cdot \left(\frac{1}{a} - \frac{1}{b}\right) \]
  2. Add Preprocessing
  3. Taylor expanded in b around inf 0

    \[\leadsto expr\]
  4. Simplified0

    \[\leadsto expr\]
  5. Applied egg-rr0

    \[\leadsto expr\]
  6. Add Preprocessing

Reproduce

?
herbie shell --seed 2024110 
(FPCore (a b)
  :name "NMSE Section 6.1 mentioned, B"
  :precision binary64
  (* (* (/ PI 2.0) (/ 1.0 (- (* b b) (* a a)))) (- (/ 1.0 a) (/ 1.0 b))))