Quotient of products

Average Accuracy: 82.3% → 96.0%

Time: 5.3s

Precision: binary64

Cost: 2513

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

Math

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

↓

\[\begin{array}{l} t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{if}\;t_0 \leq -1 \cdot 10^{+288}:\\ \;\;\;\;\frac{a1}{\frac{b1}{\frac{a2}{b2}}}\\ \mathbf{elif}\;t_0 \leq -2 \cdot 10^{-318} \lor \neg \left(t_0 \leq 5 \cdot 10^{-281}\right) \land t_0 \leq 5 \cdot 10^{+298}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \end{array} \]

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))))
   (if (<= t_0 -1e+288)
     (/ a1 (/ b1 (/ a2 b2)))
     (if (or (<= t_0 -2e-318) (and (not (<= t_0 5e-281)) (<= t_0 5e+298)))
       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 tmp;
	if (t_0 <= -1e+288) {
		tmp = a1 / (b1 / (a2 / b2));
	} else if ((t_0 <= -2e-318) || (!(t_0 <= 5e-281) && (t_0 <= 5e+298))) {
		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) :: tmp
    t_0 = (a1 * a2) / (b1 * b2)
    if (t_0 <= (-1d+288)) then
        tmp = a1 / (b1 / (a2 / b2))
    else if ((t_0 <= (-2d-318)) .or. (.not. (t_0 <= 5d-281)) .and. (t_0 <= 5d+298)) 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 tmp;
	if (t_0 <= -1e+288) {
		tmp = a1 / (b1 / (a2 / b2));
	} else if ((t_0 <= -2e-318) || (!(t_0 <= 5e-281) && (t_0 <= 5e+298))) {
		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)
	tmp = 0
	if t_0 <= -1e+288:
		tmp = a1 / (b1 / (a2 / b2))
	elif (t_0 <= -2e-318) or (not (t_0 <= 5e-281) and (t_0 <= 5e+298)):
		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))
	tmp = 0.0
	if (t_0 <= -1e+288)
		tmp = Float64(a1 / Float64(b1 / Float64(a2 / b2)));
	elseif ((t_0 <= -2e-318) || (!(t_0 <= 5e-281) && (t_0 <= 5e+298)))
		tmp = t_0;
	else
		tmp = Float64(Float64(a2 / b2) * Float64(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);
	tmp = 0.0;
	if (t_0 <= -1e+288)
		tmp = a1 / (b1 / (a2 / b2));
	elseif ((t_0 <= -2e-318) || (~((t_0 <= 5e-281)) && (t_0 <= 5e+298)))
		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]}, If[LessEqual[t$95$0, -1e+288], N[(a1 / N[(b1 / N[(a2 / b2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[Or[LessEqual[t$95$0, -2e-318], And[N[Not[LessEqual[t$95$0, 5e-281]], $MachinePrecision], LessEqual[t$95$0, 5e+298]]], t$95$0, N[(N[(a2 / b2), $MachinePrecision] * N[(a1 / b1), $MachinePrecision]), $MachinePrecision]]]]

Target

Original	82.3%
Target	82.7%
Herbie	96.0%

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

Derivation

1. Split input into 3 regimes

2. if (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < -1e288

   1. Initial program 14.1%

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

   2. Simplified 76.0%

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

      [Start] 14.1	\[ \frac{a1 \cdot a2}{b1 \cdot b2} \]
      times-frac [=>] 76.0	\[ \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

   3. Applied egg-rr 70.6%

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

      [Start] 76.0	\[ \frac{a1}{b1} \cdot \frac{a2}{b2} \]
      associate-*l/ [=>] 68.3	\[ \color{blue}{\frac{a1 \cdot \frac{a2}{b2}}{b1}} \]
      associate-/l* [=>] 70.6	\[ \color{blue}{\frac{a1}{\frac{b1}{\frac{a2}{b2}}}} \]

   if -1e288 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < -2.0000024e-318 or 4.9999999999999998e-281 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < 5.0000000000000003e298

   1. Initial program 98.8%

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

   if -2.0000024e-318 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < 4.9999999999999998e-281 or 5.0000000000000003e298 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2))

   1. Initial program 65.9%

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

   2. Simplified 94.5%

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

      [Start] 65.9	\[ \frac{a1 \cdot a2}{b1 \cdot b2} \]
      times-frac [=>] 94.5	\[ \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

3. Recombined 3 regimes into one program.

4. Final simplification 96.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq -1 \cdot 10^{+288}:\\ \;\;\;\;\frac{a1}{\frac{b1}{\frac{a2}{b2}}}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq -2 \cdot 10^{-318} \lor \neg \left(\frac{a1 \cdot a2}{b1 \cdot b2} \leq 5 \cdot 10^{-281}\right) \land \frac{a1 \cdot a2}{b1 \cdot b2} \leq 5 \cdot 10^{+298}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	82.6%
Cost	977

\[\begin{array}{l} \mathbf{if}\;a2 \leq 8.5 \cdot 10^{+86}:\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \mathbf{elif}\;a2 \leq 6 \cdot 10^{+179}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b1}}{b2}\\ \mathbf{elif}\;a2 \leq 4.5 \cdot 10^{+213} \lor \neg \left(a2 \leq 1.02 \cdot 10^{+270}\right):\\ \;\;\;\;a2 \cdot \frac{\frac{a1}{b1}}{b2}\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{a2}{b1 \cdot b2}\\ \end{array} \]

Alternative 2
Accuracy	82.4%
Cost	977

\[\begin{array}{l} \mathbf{if}\;a2 \leq 4.3 \cdot 10^{+86}:\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \mathbf{elif}\;a2 \leq 1.3 \cdot 10^{+180}:\\ \;\;\;\;a1 \cdot \frac{\frac{a2}{b1}}{b2}\\ \mathbf{elif}\;a2 \leq 7.6 \cdot 10^{+214} \lor \neg \left(a2 \leq 7 \cdot 10^{+269}\right):\\ \;\;\;\;\frac{a2}{b2 \cdot \frac{b1}{a1}}\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{a2}{b1 \cdot b2}\\ \end{array} \]

Alternative 3
Accuracy	82.0%
Cost	844

\[\begin{array}{l} t_0 := \frac{a2}{b2} \cdot \frac{a1}{b1}\\ \mathbf{if}\;a1 \leq -3 \cdot 10^{-41}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;a1 \leq -5.8 \cdot 10^{-274}:\\ \;\;\;\;a1 \cdot \frac{a2}{b1 \cdot b2}\\ \mathbf{elif}\;a1 \leq 1.35 \cdot 10^{-251}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \frac{\frac{a1}{b1}}{b2}\\ \end{array} \]

Alternative 4
Accuracy	81.9%
Cost	713

\[\begin{array}{l} \mathbf{if}\;a1 \leq -8.4 \cdot 10^{-41} \lor \neg \left(a1 \leq -7.5 \cdot 10^{-275}\right):\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \mathbf{else}:\\ \;\;\;\;a1 \cdot \frac{a2}{b1 \cdot b2}\\ \end{array} \]

Alternative 5
Accuracy	81.6%
Cost	712

\[\begin{array}{l} \mathbf{if}\;a1 \leq -2.6 \cdot 10^{-41}:\\ \;\;\;\;\frac{a2}{b2} \cdot \frac{a1}{b1}\\ \mathbf{elif}\;a1 \leq 3.7 \cdot 10^{-191}:\\ \;\;\;\;a1 \cdot \frac{a2}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \end{array} \]

Alternative 6
Accuracy	82.7%
Cost	448

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

Reproduce

herbie shell --seed 2023146 
(FPCore (a1 a2 b1 b2)
  :name "Quotient of products"
  :precision binary64

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

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