Quotient of products

Percentage Accurate: 86.6% → 86.0%

Time: 3.6s

Alternatives: 5

Speedup: N/A×

• 20.6% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ \frac{a1 \cdot a2}{b1 \cdot b2} \end{array} \]

FPCore

(FPCore (a1 a2 b1 b2) :precision binary64 (/ (* a1 a2) (* b1 b2)))

C

double code(double a1, double a2, double b1, double b2) {
	return (a1 * a2) / (b1 * b2);
}

Fortran

real(8) function code(a1, a2, b1, b2)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: b1
    real(8), intent (in) :: b2
    code = (a1 * a2) / (b1 * b2)
end function

Java

public static double code(double a1, double a2, double b1, double b2) {
	return (a1 * a2) / (b1 * b2);
}

Python

def code(a1, a2, b1, b2):
	return (a1 * a2) / (b1 * b2)

Julia

function code(a1, a2, b1, b2)
	return Float64(Float64(a1 * a2) / Float64(b1 * b2))
end

MATLAB

function tmp = code(a1, a2, b1, b2)
	tmp = (a1 * a2) / (b1 * b2);
end

Wolfram

code[a1_, a2_, b1_, b2_] := N[(N[(a1 * a2), $MachinePrecision] / N[(b1 * b2), $MachinePrecision]), $MachinePrecision]

Initial Program: 86.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{a1 \cdot a2}{b1 \cdot b2} \end{array} \]

FPCore

(FPCore (a1 a2 b1 b2) :precision binary64 (/ (* a1 a2) (* b1 b2)))

C

double code(double a1, double a2, double b1, double b2) {
	return (a1 * a2) / (b1 * b2);
}

Fortran

real(8) function code(a1, a2, b1, b2)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: b1
    real(8), intent (in) :: b2
    code = (a1 * a2) / (b1 * b2)
end function

Java

public static double code(double a1, double a2, double b1, double b2) {
	return (a1 * a2) / (b1 * b2);
}

Python

def code(a1, a2, b1, b2):
	return (a1 * a2) / (b1 * b2)

Julia

function code(a1, a2, b1, b2)
	return Float64(Float64(a1 * a2) / Float64(b1 * b2))
end

MATLAB

function tmp = code(a1, a2, b1, b2)
	tmp = (a1 * a2) / (b1 * b2);
end

Wolfram

code[a1_, a2_, b1_, b2_] := N[(N[(a1 * a2), $MachinePrecision] / N[(b1 * b2), $MachinePrecision]), $MachinePrecision]

Alternative 1: 86.0% accurate, 0.6× speedup

Math

\[\begin{array}{l} [a1, a2] = \mathsf{sort}([a1, a2])\\ [b1, b2] = \mathsf{sort}([b1, b2])\\ \\ \begin{array}{l} \mathbf{if}\;a2 \leq -1.22 \cdot 10^{-138} \lor \neg \left(a2 \leq 3.8 \cdot 10^{-53}\right):\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b2}}{b1}\\ \end{array} \end{array} \]

FPCore

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 b1 b2)
 :precision binary64
 (if (or (<= a2 -1.22e-138) (not (<= a2 3.8e-53)))
   (* (/ a2 b1) (/ a1 b2))
   (* a1 (/ (/ a2 b2) b1))))

C

assert(a1 < a2);
assert(b1 < b2);
double code(double a1, double a2, double b1, double b2) {
	double tmp;
	if ((a2 <= -1.22e-138) || !(a2 <= 3.8e-53)) {
		tmp = (a2 / b1) * (a1 / b2);
	} else {
		tmp = a1 * ((a2 / b2) / b1);
	}
	return tmp;
}

Fortran

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, b1, b2)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: b1
    real(8), intent (in) :: b2
    real(8) :: tmp
    if ((a2 <= (-1.22d-138)) .or. (.not. (a2 <= 3.8d-53))) then
        tmp = (a2 / b1) * (a1 / b2)
    else
        tmp = a1 * ((a2 / b2) / b1)
    end if
    code = tmp
end function

Java

assert a1 < a2;
assert b1 < b2;
public static double code(double a1, double a2, double b1, double b2) {
	double tmp;
	if ((a2 <= -1.22e-138) || !(a2 <= 3.8e-53)) {
		tmp = (a2 / b1) * (a1 / b2);
	} else {
		tmp = a1 * ((a2 / b2) / b1);
	}
	return tmp;
}

Python

[a1, a2] = sort([a1, a2])
[b1, b2] = sort([b1, b2])
def code(a1, a2, b1, b2):
	tmp = 0
	if (a2 <= -1.22e-138) or not (a2 <= 3.8e-53):
		tmp = (a2 / b1) * (a1 / b2)
	else:
		tmp = a1 * ((a2 / b2) / b1)
	return tmp

Julia

a1, a2 = sort([a1, a2])
b1, b2 = sort([b1, b2])
function code(a1, a2, b1, b2)
	tmp = 0.0
	if ((a2 <= -1.22e-138) || !(a2 <= 3.8e-53))
		tmp = Float64(Float64(a2 / b1) * Float64(a1 / b2));
	else
		tmp = Float64(a1 * Float64(Float64(a2 / b2) / b1));
	end
	return tmp
end

MATLAB

a1, a2 = num2cell(sort([a1, a2])){:}
b1, b2 = num2cell(sort([b1, b2])){:}
function tmp_2 = code(a1, a2, b1, b2)
	tmp = 0.0;
	if ((a2 <= -1.22e-138) || ~((a2 <= 3.8e-53)))
		tmp = (a2 / b1) * (a1 / b2);
	else
		tmp = a1 * ((a2 / b2) / b1);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
code[a1_, a2_, b1_, b2_] := If[Or[LessEqual[a2, -1.22e-138], N[Not[LessEqual[a2, 3.8e-53]], $MachinePrecision]], N[(N[(a2 / b1), $MachinePrecision] * N[(a1 / b2), $MachinePrecision]), $MachinePrecision], N[(a1 * N[(N[(a2 / b2), $MachinePrecision] / b1), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if a2 < -1.22e-138 or 3.7999999999999998e-53 < a2

   1. Initial program 85.4%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
   2. Step-by-step derivation

      1. times-frac 91.4%

         \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

      2. associate-*l/ 88.0%

         \[\leadsto \color{blue}{\frac{a1 \cdot \frac{a2}{b2}}{b1}} \]

      3. associate-*r/ 90.3%

         \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

   3. Simplified 90.3%

      \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

   4. Taylor expanded in a1 around 0 85.4%

      \[\leadsto \color{blue}{\frac{a1 \cdot a2}{b1 \cdot b2}} \]
   5. Step-by-step derivation

      1. *-commutative 85.4%

         \[\leadsto \frac{\color{blue}{a2 \cdot a1}}{b1 \cdot b2} \]

      2. times-frac 90.3%

         \[\leadsto \color{blue}{\frac{a2}{b1} \cdot \frac{a1}{b2}} \]

   6. Simplified 90.3%

      \[\leadsto \color{blue}{\frac{a2}{b1} \cdot \frac{a1}{b2}} \]

   if -1.22e-138 < a2 < 3.7999999999999998e-53

   1. Initial program 83.5%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
   2. Step-by-step derivation

      1. times-frac 82.3%

         \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

      2. associate-*l/ 81.8%

         \[\leadsto \color{blue}{\frac{a1 \cdot \frac{a2}{b2}}{b1}} \]

      3. associate-*r/ 85.3%

         \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

   3. Simplified 85.3%

      \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 88.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a2 \leq -1.22 \cdot 10^{-138} \lor \neg \left(a2 \leq 3.8 \cdot 10^{-53}\right):\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b2}}{b1}\\ \end{array} \]

Alternative 2: 86.4% accurate, 0.6× speedup

Math

\[\begin{array}{l} [a1, a2] = \mathsf{sort}([a1, a2])\\ [b1, b2] = \mathsf{sort}([b1, b2])\\ \\ \begin{array}{l} \mathbf{if}\;b1 \leq -2.7 \cdot 10^{+161}:\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \mathbf{elif}\;b1 \leq -2.75 \cdot 10^{-132}:\\ \;\;\;\;a2 \cdot \frac{a1}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b2}}{b1}\\ \end{array} \end{array} \]

FPCore

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 b1 b2)
 :precision binary64
 (if (<= b1 -2.7e+161)
   (* (/ a2 b2) (/ a1 b1))
   (if (<= b1 -2.75e-132) (* a2 (/ a1 (* b1 b2))) (* a1 (/ (/ a2 b2) b1)))))

C

assert(a1 < a2);
assert(b1 < b2);
double code(double a1, double a2, double b1, double b2) {
	double tmp;
	if (b1 <= -2.7e+161) {
		tmp = (a2 / b2) * (a1 / b1);
	} else if (b1 <= -2.75e-132) {
		tmp = a2 * (a1 / (b1 * b2));
	} else {
		tmp = a1 * ((a2 / b2) / b1);
	}
	return tmp;
}

Fortran

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, b1, b2)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: b1
    real(8), intent (in) :: b2
    real(8) :: tmp
    if (b1 <= (-2.7d+161)) then
        tmp = (a2 / b2) * (a1 / b1)
    else if (b1 <= (-2.75d-132)) then
        tmp = a2 * (a1 / (b1 * b2))
    else
        tmp = a1 * ((a2 / b2) / b1)
    end if
    code = tmp
end function

Java

assert a1 < a2;
assert b1 < b2;
public static double code(double a1, double a2, double b1, double b2) {
	double tmp;
	if (b1 <= -2.7e+161) {
		tmp = (a2 / b2) * (a1 / b1);
	} else if (b1 <= -2.75e-132) {
		tmp = a2 * (a1 / (b1 * b2));
	} else {
		tmp = a1 * ((a2 / b2) / b1);
	}
	return tmp;
}

Python

[a1, a2] = sort([a1, a2])
[b1, b2] = sort([b1, b2])
def code(a1, a2, b1, b2):
	tmp = 0
	if b1 <= -2.7e+161:
		tmp = (a2 / b2) * (a1 / b1)
	elif b1 <= -2.75e-132:
		tmp = a2 * (a1 / (b1 * b2))
	else:
		tmp = a1 * ((a2 / b2) / b1)
	return tmp

Julia

a1, a2 = sort([a1, a2])
b1, b2 = sort([b1, b2])
function code(a1, a2, b1, b2)
	tmp = 0.0
	if (b1 <= -2.7e+161)
		tmp = Float64(Float64(a2 / b2) * Float64(a1 / b1));
	elseif (b1 <= -2.75e-132)
		tmp = Float64(a2 * Float64(a1 / Float64(b1 * b2)));
	else
		tmp = Float64(a1 * Float64(Float64(a2 / b2) / b1));
	end
	return tmp
end

MATLAB

a1, a2 = num2cell(sort([a1, a2])){:}
b1, b2 = num2cell(sort([b1, b2])){:}
function tmp_2 = code(a1, a2, b1, b2)
	tmp = 0.0;
	if (b1 <= -2.7e+161)
		tmp = (a2 / b2) * (a1 / b1);
	elseif (b1 <= -2.75e-132)
		tmp = a2 * (a1 / (b1 * b2));
	else
		tmp = a1 * ((a2 / b2) / b1);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
code[a1_, a2_, b1_, b2_] := If[LessEqual[b1, -2.7e+161], N[(N[(a2 / b2), $MachinePrecision] * N[(a1 / b1), $MachinePrecision]), $MachinePrecision], If[LessEqual[b1, -2.75e-132], N[(a2 * N[(a1 / N[(b1 * b2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(a1 * N[(N[(a2 / b2), $MachinePrecision] / b1), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if b1 < -2.6999999999999998e161

   1. Initial program 69.9%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
   2. Step-by-step derivation

      1. times-frac 86.8%

         \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

   3. Simplified 86.8%

      \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

   if -2.6999999999999998e161 < b1 < -2.75e-132

   1. Initial program 90.0%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
   2. Step-by-step derivation

      1. associate-/l* 92.3%

         \[\leadsto \color{blue}{\frac{a1}{\frac{b1 \cdot b2}{a2}}} \]

      2. associate-/r/ 83.5%

         \[\leadsto \color{blue}{\frac{a1}{b1 \cdot b2} \cdot a2} \]

      3. *-commutative 83.5%

         \[\leadsto \frac{a1}{\color{blue}{b2 \cdot b1}} \cdot a2 \]

   3. Applied egg-rr 83.5%

      \[\leadsto \color{blue}{\frac{a1}{b2 \cdot b1} \cdot a2} \]

   if -2.75e-132 < b1

   1. Initial program 85.3%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
   2. Step-by-step derivation

      1. times-frac 88.8%

         \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

      2. associate-*l/ 87.7%

         \[\leadsto \color{blue}{\frac{a1 \cdot \frac{a2}{b2}}{b1}} \]

      3. associate-*r/ 91.9%

         \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

   3. Simplified 91.9%

      \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 89.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b1 \leq -2.7 \cdot 10^{+161}:\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \mathbf{elif}\;b1 \leq -2.75 \cdot 10^{-132}:\\ \;\;\;\;a2 \cdot \frac{a1}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b2}}{b1}\\ \end{array} \]

Alternative 3: 86.3% accurate, 0.8× speedup

Math

\[\begin{array}{l} [a1, a2] = \mathsf{sort}([a1, a2])\\ [b1, b2] = \mathsf{sort}([b1, b2])\\ \\ \begin{array}{l} \mathbf{if}\;b1 \leq -2.5 \cdot 10^{-176}:\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b2}}{b1}\\ \end{array} \end{array} \]

FPCore

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 b1 b2)
 :precision binary64
 (if (<= b1 -2.5e-176) (* (/ a2 b2) (/ a1 b1)) (* a1 (/ (/ a2 b2) b1))))

C

assert(a1 < a2);
assert(b1 < b2);
double code(double a1, double a2, double b1, double b2) {
	double tmp;
	if (b1 <= -2.5e-176) {
		tmp = (a2 / b2) * (a1 / b1);
	} else {
		tmp = a1 * ((a2 / b2) / b1);
	}
	return tmp;
}

Fortran

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, b1, b2)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: b1
    real(8), intent (in) :: b2
    real(8) :: tmp
    if (b1 <= (-2.5d-176)) then
        tmp = (a2 / b2) * (a1 / b1)
    else
        tmp = a1 * ((a2 / b2) / b1)
    end if
    code = tmp
end function

Java

assert a1 < a2;
assert b1 < b2;
public static double code(double a1, double a2, double b1, double b2) {
	double tmp;
	if (b1 <= -2.5e-176) {
		tmp = (a2 / b2) * (a1 / b1);
	} else {
		tmp = a1 * ((a2 / b2) / b1);
	}
	return tmp;
}

Python

[a1, a2] = sort([a1, a2])
[b1, b2] = sort([b1, b2])
def code(a1, a2, b1, b2):
	tmp = 0
	if b1 <= -2.5e-176:
		tmp = (a2 / b2) * (a1 / b1)
	else:
		tmp = a1 * ((a2 / b2) / b1)
	return tmp

Julia

a1, a2 = sort([a1, a2])
b1, b2 = sort([b1, b2])
function code(a1, a2, b1, b2)
	tmp = 0.0
	if (b1 <= -2.5e-176)
		tmp = Float64(Float64(a2 / b2) * Float64(a1 / b1));
	else
		tmp = Float64(a1 * Float64(Float64(a2 / b2) / b1));
	end
	return tmp
end

MATLAB

a1, a2 = num2cell(sort([a1, a2])){:}
b1, b2 = num2cell(sort([b1, b2])){:}
function tmp_2 = code(a1, a2, b1, b2)
	tmp = 0.0;
	if (b1 <= -2.5e-176)
		tmp = (a2 / b2) * (a1 / b1);
	else
		tmp = a1 * ((a2 / b2) / b1);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
code[a1_, a2_, b1_, b2_] := If[LessEqual[b1, -2.5e-176], N[(N[(a2 / b2), $MachinePrecision] * N[(a1 / b1), $MachinePrecision]), $MachinePrecision], N[(a1 * N[(N[(a2 / b2), $MachinePrecision] / b1), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if b1 < -2.5e-176

   1. Initial program 84.2%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
   2. Step-by-step derivation

      1. times-frac 85.8%

         \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

   3. Simplified 85.8%

      \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

   if -2.5e-176 < b1

   1. Initial program 85.1%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
   2. Step-by-step derivation

      1. times-frac 89.8%

         \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

      2. associate-*l/ 87.5%

         \[\leadsto \color{blue}{\frac{a1 \cdot \frac{a2}{b2}}{b1}} \]

      3. associate-*r/ 91.9%

         \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

   3. Simplified 91.9%

      \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 89.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b1 \leq -2.5 \cdot 10^{-176}:\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b2}}{b1}\\ \end{array} \]

Alternative 4: 86.5% accurate, 0.8× speedup

Math

\[\begin{array}{l} [a1, a2] = \mathsf{sort}([a1, a2])\\ [b1, b2] = \mathsf{sort}([b1, b2])\\ \\ \begin{array}{l} \mathbf{if}\;a2 \leq 2.1 \cdot 10^{-55}:\\ \;\;\;\;\frac{a2}{\frac{b1}{a1} \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \end{array} \end{array} \]

FPCore

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 b1 b2)
 :precision binary64
 (if (<= a2 2.1e-55) (/ a2 (* (/ b1 a1) b2)) (* (/ a2 b1) (/ a1 b2))))

C

assert(a1 < a2);
assert(b1 < b2);
double code(double a1, double a2, double b1, double b2) {
	double tmp;
	if (a2 <= 2.1e-55) {
		tmp = a2 / ((b1 / a1) * b2);
	} else {
		tmp = (a2 / b1) * (a1 / b2);
	}
	return tmp;
}

Fortran

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, b1, b2)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: b1
    real(8), intent (in) :: b2
    real(8) :: tmp
    if (a2 <= 2.1d-55) then
        tmp = a2 / ((b1 / a1) * b2)
    else
        tmp = (a2 / b1) * (a1 / b2)
    end if
    code = tmp
end function

Java

assert a1 < a2;
assert b1 < b2;
public static double code(double a1, double a2, double b1, double b2) {
	double tmp;
	if (a2 <= 2.1e-55) {
		tmp = a2 / ((b1 / a1) * b2);
	} else {
		tmp = (a2 / b1) * (a1 / b2);
	}
	return tmp;
}

Python

[a1, a2] = sort([a1, a2])
[b1, b2] = sort([b1, b2])
def code(a1, a2, b1, b2):
	tmp = 0
	if a2 <= 2.1e-55:
		tmp = a2 / ((b1 / a1) * b2)
	else:
		tmp = (a2 / b1) * (a1 / b2)
	return tmp

Julia

a1, a2 = sort([a1, a2])
b1, b2 = sort([b1, b2])
function code(a1, a2, b1, b2)
	tmp = 0.0
	if (a2 <= 2.1e-55)
		tmp = Float64(a2 / Float64(Float64(b1 / a1) * b2));
	else
		tmp = Float64(Float64(a2 / b1) * Float64(a1 / b2));
	end
	return tmp
end

MATLAB

a1, a2 = num2cell(sort([a1, a2])){:}
b1, b2 = num2cell(sort([b1, b2])){:}
function tmp_2 = code(a1, a2, b1, b2)
	tmp = 0.0;
	if (a2 <= 2.1e-55)
		tmp = a2 / ((b1 / a1) * b2);
	else
		tmp = (a2 / b1) * (a1 / b2);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
code[a1_, a2_, b1_, b2_] := If[LessEqual[a2, 2.1e-55], N[(a2 / N[(N[(b1 / a1), $MachinePrecision] * b2), $MachinePrecision]), $MachinePrecision], N[(N[(a2 / b1), $MachinePrecision] * N[(a1 / b2), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if a2 < 2.1000000000000002e-55

   1. Initial program 85.5%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
   2. Step-by-step derivation

      1. times-frac 86.8%

         \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

      2. associate-*l/ 84.0%

         \[\leadsto \color{blue}{\frac{a1 \cdot \frac{a2}{b2}}{b1}} \]

      3. associate-*r/ 87.8%

         \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

   3. Simplified 87.8%

      \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]
   4. Step-by-step derivation

      1. associate-*r/ 84.0%

         \[\leadsto \color{blue}{\frac{a1 \cdot \frac{a2}{b2}}{b1}} \]

      2. associate-*l/ 86.8%

         \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

      3. clear-num 86.7%

         \[\leadsto \color{blue}{\frac{1}{\frac{b1}{a1}}} \cdot \frac{a2}{b2} \]

      4. frac-times 87.0%

         \[\leadsto \color{blue}{\frac{1 \cdot a2}{\frac{b1}{a1} \cdot b2}} \]

      5. *-un-lft-identity 87.0%

         \[\leadsto \frac{\color{blue}{a2}}{\frac{b1}{a1} \cdot b2} \]

   5. Applied egg-rr 87.0%

      \[\leadsto \color{blue}{\frac{a2}{\frac{b1}{a1} \cdot b2}} \]

   if 2.1000000000000002e-55 < a2

   1. Initial program 82.6%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
   2. Step-by-step derivation

      1. times-frac 91.9%

         \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

      2. associate-*l/ 90.8%

         \[\leadsto \color{blue}{\frac{a1 \cdot \frac{a2}{b2}}{b1}} \]

      3. associate-*r/ 90.6%

         \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

   3. Simplified 90.6%

      \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

   4. Taylor expanded in a1 around 0 82.6%

      \[\leadsto \color{blue}{\frac{a1 \cdot a2}{b1 \cdot b2}} \]
   5. Step-by-step derivation

      1. *-commutative 82.6%

         \[\leadsto \frac{\color{blue}{a2 \cdot a1}}{b1 \cdot b2} \]

      2. times-frac 95.2%

         \[\leadsto \color{blue}{\frac{a2}{b1} \cdot \frac{a1}{b2}} \]

   6. Simplified 95.2%

      \[\leadsto \color{blue}{\frac{a2}{b1} \cdot \frac{a1}{b2}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 89.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a2 \leq 2.1 \cdot 10^{-55}:\\ \;\;\;\;\frac{a2}{\frac{b1}{a1} \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \end{array} \]

Alternative 5: 86.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} [a1, a2] = \mathsf{sort}([a1, a2])\\ [b1, b2] = \mathsf{sort}([b1, b2])\\ \\ a1 \cdot \frac{\frac{a2}{b2}}{b1} \end{array} \]

FPCore

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 b1 b2) :precision binary64 (* a1 (/ (/ a2 b2) b1)))

C

assert(a1 < a2);
assert(b1 < b2);
double code(double a1, double a2, double b1, double b2) {
	return a1 * ((a2 / b2) / b1);
}

Fortran

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, b1, b2)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: b1
    real(8), intent (in) :: b2
    code = a1 * ((a2 / b2) / b1)
end function

Java

assert a1 < a2;
assert b1 < b2;
public static double code(double a1, double a2, double b1, double b2) {
	return a1 * ((a2 / b2) / b1);
}

Python

[a1, a2] = sort([a1, a2])
[b1, b2] = sort([b1, b2])
def code(a1, a2, b1, b2):
	return a1 * ((a2 / b2) / b1)

Julia

a1, a2 = sort([a1, a2])
b1, b2 = sort([b1, b2])
function code(a1, a2, b1, b2)
	return Float64(a1 * Float64(Float64(a2 / b2) / b1))
end

MATLAB

a1, a2 = num2cell(sort([a1, a2])){:}
b1, b2 = num2cell(sort([b1, b2])){:}
function tmp = code(a1, a2, b1, b2)
	tmp = a1 * ((a2 / b2) / b1);
end

Wolfram

NOTE: a1 and a2 should be sorted in increasing order before calling this function.
NOTE: b1 and b2 should be sorted in increasing order before calling this function.
code[a1_, a2_, b1_, b2_] := N[(a1 * N[(N[(a2 / b2), $MachinePrecision] / b1), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 84.7%

   \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
2. Step-by-step derivation

   1. times-frac 88.1%

      \[\leadsto \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

   2. associate-*l/ 85.8%

      \[\leadsto \color{blue}{\frac{a1 \cdot \frac{a2}{b2}}{b1}} \]

   3. associate-*r/ 88.5%

      \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

3. Simplified 88.5%

   \[\leadsto \color{blue}{a1 \cdot \frac{\frac{a2}{b2}}{b1}} \]

4. Final simplification 88.5%

   \[\leadsto a1 \cdot \frac{\frac{a2}{b2}}{b1} \]

Developer target: 86.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{a1}{b1} \cdot \frac{a2}{b2} \end{array} \]

FPCore

(FPCore (a1 a2 b1 b2) :precision binary64 (* (/ a1 b1) (/ a2 b2)))

C

double code(double a1, double a2, double b1, double b2) {
	return (a1 / b1) * (a2 / b2);
}

Fortran

real(8) function code(a1, a2, b1, b2)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: b1
    real(8), intent (in) :: b2
    code = (a1 / b1) * (a2 / b2)
end function

Java

public static double code(double a1, double a2, double b1, double b2) {
	return (a1 / b1) * (a2 / b2);
}

Python

def code(a1, a2, b1, b2):
	return (a1 / b1) * (a2 / b2)

Julia

function code(a1, a2, b1, b2)
	return Float64(Float64(a1 / b1) * Float64(a2 / b2))
end

MATLAB

function tmp = code(a1, a2, b1, b2)
	tmp = (a1 / b1) * (a2 / b2);
end

Wolfram

code[a1_, a2_, b1_, b2_] := N[(N[(a1 / b1), $MachinePrecision] * N[(a2 / b2), $MachinePrecision]), $MachinePrecision]

Reproduce

herbie shell --seed 2023313 
(FPCore (a1 a2 b1 b2)
  :name "Quotient of products"
  :precision binary64

  :herbie-target
  (* (/ a1 b1) (/ a2 b2))

  (/ (* a1 a2) (* b1 b2)))