Quotient of products

Average Error: 82.7% → 95.2%

Time: 7.4s

Precision: binary64

Cost: 2512.00

Math

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

↓

\[\begin{array}{l} t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\ t_1 := \frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{if}\;t_0 \leq -5 \cdot 10^{+289}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;t_0 \leq -1 \cdot 10^{-316}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;t_0 \leq 5 \cdot 10^{-301}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;t_0 \leq 5 \cdot 10^{+272}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a2}{\frac{b2}{a1}}}{b1}\\ \end{array} \]

FPCore

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

↓

(FPCore (a1 a2 b1 b2)
 :precision binary64
 (let* ((t_0 (/ (* a1 a2) (* b1 b2))) (t_1 (* (/ a1 b1) (/ a2 b2))))
   (if (<= t_0 -5e+289)
     t_1
     (if (<= t_0 -1e-316)
       t_0
       (if (<= t_0 5e-301)
         t_1
         (if (<= t_0 5e+272) t_0 (/ (/ a2 (/ b2 a1)) b1)))))))

C

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

↓

double code(double a1, double a2, double b1, double b2) {
	double t_0 = (a1 * a2) / (b1 * b2);
	double t_1 = (a1 / b1) * (a2 / b2);
	double tmp;
	if (t_0 <= -5e+289) {
		tmp = t_1;
	} else if (t_0 <= -1e-316) {
		tmp = t_0;
	} else if (t_0 <= 5e-301) {
		tmp = t_1;
	} else if (t_0 <= 5e+272) {
		tmp = t_0;
	} else {
		tmp = (a2 / (b2 / a1)) / b1;
	}
	return tmp;
}

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

↓

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) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = (a1 * a2) / (b1 * b2)
    t_1 = (a1 / b1) * (a2 / b2)
    if (t_0 <= (-5d+289)) then
        tmp = t_1
    else if (t_0 <= (-1d-316)) then
        tmp = t_0
    else if (t_0 <= 5d-301) then
        tmp = t_1
    else if (t_0 <= 5d+272) then
        tmp = t_0
    else
        tmp = (a2 / (b2 / a1)) / b1
    end if
    code = tmp
end function

Java

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

↓

public static double code(double a1, double a2, double b1, double b2) {
	double t_0 = (a1 * a2) / (b1 * b2);
	double t_1 = (a1 / b1) * (a2 / b2);
	double tmp;
	if (t_0 <= -5e+289) {
		tmp = t_1;
	} else if (t_0 <= -1e-316) {
		tmp = t_0;
	} else if (t_0 <= 5e-301) {
		tmp = t_1;
	} else if (t_0 <= 5e+272) {
		tmp = t_0;
	} else {
		tmp = (a2 / (b2 / a1)) / b1;
	}
	return tmp;
}

Python

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

↓

def code(a1, a2, b1, b2):
	t_0 = (a1 * a2) / (b1 * b2)
	t_1 = (a1 / b1) * (a2 / b2)
	tmp = 0
	if t_0 <= -5e+289:
		tmp = t_1
	elif t_0 <= -1e-316:
		tmp = t_0
	elif t_0 <= 5e-301:
		tmp = t_1
	elif t_0 <= 5e+272:
		tmp = t_0
	else:
		tmp = (a2 / (b2 / a1)) / b1
	return tmp

Julia

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

↓

function code(a1, a2, b1, b2)
	t_0 = Float64(Float64(a1 * a2) / Float64(b1 * b2))
	t_1 = Float64(Float64(a1 / b1) * Float64(a2 / b2))
	tmp = 0.0
	if (t_0 <= -5e+289)
		tmp = t_1;
	elseif (t_0 <= -1e-316)
		tmp = t_0;
	elseif (t_0 <= 5e-301)
		tmp = t_1;
	elseif (t_0 <= 5e+272)
		tmp = t_0;
	else
		tmp = Float64(Float64(a2 / Float64(b2 / a1)) / b1);
	end
	return tmp
end

MATLAB

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

↓

function tmp_2 = code(a1, a2, b1, b2)
	t_0 = (a1 * a2) / (b1 * b2);
	t_1 = (a1 / b1) * (a2 / b2);
	tmp = 0.0;
	if (t_0 <= -5e+289)
		tmp = t_1;
	elseif (t_0 <= -1e-316)
		tmp = t_0;
	elseif (t_0 <= 5e-301)
		tmp = t_1;
	elseif (t_0 <= 5e+272)
		tmp = t_0;
	else
		tmp = (a2 / (b2 / a1)) / b1;
	end
	tmp_2 = tmp;
end

Wolfram

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

↓

code[a1_, a2_, b1_, b2_] := Block[{t$95$0 = N[(N[(a1 * a2), $MachinePrecision] / N[(b1 * b2), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(a1 / b1), $MachinePrecision] * N[(a2 / b2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$0, -5e+289], t$95$1, If[LessEqual[t$95$0, -1e-316], t$95$0, If[LessEqual[t$95$0, 5e-301], t$95$1, If[LessEqual[t$95$0, 5e+272], t$95$0, N[(N[(a2 / N[(b2 / a1), $MachinePrecision]), $MachinePrecision] / b1), $MachinePrecision]]]]]]]

Target

Original	82.7%
Target	82.2%
Herbie	95.2%

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

Derivation

1. Split input into 3 regimes

2. if (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < -5.00000000000000031e289 or -9.999999837e-317 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < 5.00000000000000013e-301

   1. Initial program 73.1

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

   2. Simplified 94.4

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

      [Start] 73.1	\[ \frac{a1 \cdot a2}{b1 \cdot b2} \]
      times-frac [=>] 94.4	\[ \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

   if -5.00000000000000031e289 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < -9.999999837e-317 or 5.00000000000000013e-301 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < 4.99999999999999973e272

   1. Initial program 98.8

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

   if 4.99999999999999973e272 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2))

   1. Initial program 15.9

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

   2. Simplified 32.4

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

      [Start] 15.9	\[ \frac{a1 \cdot a2}{b1 \cdot b2} \]
      associate-*l/ [<=] 32.4	\[ \color{blue}{\frac{a1}{b1 \cdot b2} \cdot a2} \]
      *-commutative [=>] 32.4	\[ \color{blue}{a2 \cdot \frac{a1}{b1 \cdot b2}} \]

   3. Applied egg-rr 74.4

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

4. Final simplification 95.2

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq -5 \cdot 10^{+289}:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq -1 \cdot 10^{-316}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq 5 \cdot 10^{-301}:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq 5 \cdot 10^{+272}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a2}{\frac{b2}{a1}}}{b1}\\ \end{array} \]

Alternatives

Alternative 1
Error	95.2%
Cost	2512.00

\[\begin{array}{l} t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\ t_1 := \frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{if}\;t_0 \leq -5 \cdot 10^{+289}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;t_0 \leq -1 \cdot 10^{-316}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;t_0 \leq 5 \cdot 10^{-301}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;t_0 \leq 10^{+262}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b2}}{b1}\\ \end{array} \]

Alternative 2
Error	91.3%
Cost	1490.00

\[\begin{array}{l} \mathbf{if}\;b1 \cdot b2 \leq -\infty \lor \neg \left(b1 \cdot b2 \leq -2 \cdot 10^{-142}\right) \land \left(b1 \cdot b2 \leq 2 \cdot 10^{-185} \lor \neg \left(b1 \cdot b2 \leq 5 \cdot 10^{+260}\right)\right):\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \frac{a1}{b1 \cdot b2}\\ \end{array} \]

Alternative 3
Error	90.5%
Cost	1488.00

\[\begin{array}{l} t_0 := a2 \cdot \frac{a1}{b1 \cdot b2}\\ t_1 := \frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{if}\;b1 \cdot b2 \leq -\infty:\\ \;\;\;\;t_1\\ \mathbf{elif}\;b1 \cdot b2 \leq -2 \cdot 10^{-142}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;b1 \cdot b2 \leq 2 \cdot 10^{-185}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;b1 \cdot b2 \leq 5 \cdot 10^{+158}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b1}}{b2}\\ \end{array} \]

Alternative 4
Error	90.4%
Cost	1488.00

\[\begin{array}{l} t_0 := a2 \cdot \frac{a1}{b1 \cdot b2}\\ t_1 := \frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{if}\;b1 \cdot b2 \leq -\infty:\\ \;\;\;\;t_1\\ \mathbf{elif}\;b1 \cdot b2 \leq -2 \cdot 10^{-142}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;b1 \cdot b2 \leq 2 \cdot 10^{-185}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;b1 \cdot b2 \leq 10^{+173}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b2}}{b1}\\ \end{array} \]

Alternative 5
Error	82.5%
Cost	448.00

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

Reproduce

herbie shell --seed 2023093 
(FPCore (a1 a2 b1 b2)
  :name "Quotient of products"
  :precision binary64

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

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