?

Average Accuracy: 82.1% → 91.9%
Time: 5.3s
Precision: binary64
Cost: 1488

?

\[\frac{a1 \cdot a2}{b1 \cdot b2} \]
\[\begin{array}{l} t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{if}\;b1 \cdot b2 \leq -\infty:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;b1 \cdot b2 \leq -1 \cdot 10^{-167}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;b1 \cdot b2 \leq 2 \cdot 10^{-209}:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \mathbf{elif}\;b1 \cdot b2 \leq 5 \cdot 10^{+283}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a2}{\frac{b1}{a1}}}{b2}\\ \end{array} \]
(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 (<= (* b1 b2) (- INFINITY))
     (* (/ a1 b1) (/ a2 b2))
     (if (<= (* b1 b2) -1e-167)
       t_0
       (if (<= (* b1 b2) 2e-209)
         (* (/ a2 b1) (/ a1 b2))
         (if (<= (* b1 b2) 5e+283) t_0 (/ (/ a2 (/ b1 a1)) b2)))))))
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 ((b1 * b2) <= -((double) INFINITY)) {
		tmp = (a1 / b1) * (a2 / b2);
	} else if ((b1 * b2) <= -1e-167) {
		tmp = t_0;
	} else if ((b1 * b2) <= 2e-209) {
		tmp = (a2 / b1) * (a1 / b2);
	} else if ((b1 * b2) <= 5e+283) {
		tmp = t_0;
	} else {
		tmp = (a2 / (b1 / a1)) / b2;
	}
	return tmp;
}
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 ((b1 * b2) <= -Double.POSITIVE_INFINITY) {
		tmp = (a1 / b1) * (a2 / b2);
	} else if ((b1 * b2) <= -1e-167) {
		tmp = t_0;
	} else if ((b1 * b2) <= 2e-209) {
		tmp = (a2 / b1) * (a1 / b2);
	} else if ((b1 * b2) <= 5e+283) {
		tmp = t_0;
	} else {
		tmp = (a2 / (b1 / a1)) / b2;
	}
	return tmp;
}
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 (b1 * b2) <= -math.inf:
		tmp = (a1 / b1) * (a2 / b2)
	elif (b1 * b2) <= -1e-167:
		tmp = t_0
	elif (b1 * b2) <= 2e-209:
		tmp = (a2 / b1) * (a1 / b2)
	elif (b1 * b2) <= 5e+283:
		tmp = t_0
	else:
		tmp = (a2 / (b1 / a1)) / b2
	return tmp
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 (Float64(b1 * b2) <= Float64(-Inf))
		tmp = Float64(Float64(a1 / b1) * Float64(a2 / b2));
	elseif (Float64(b1 * b2) <= -1e-167)
		tmp = t_0;
	elseif (Float64(b1 * b2) <= 2e-209)
		tmp = Float64(Float64(a2 / b1) * Float64(a1 / b2));
	elseif (Float64(b1 * b2) <= 5e+283)
		tmp = t_0;
	else
		tmp = Float64(Float64(a2 / Float64(b1 / a1)) / b2);
	end
	return tmp
end
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 ((b1 * b2) <= -Inf)
		tmp = (a1 / b1) * (a2 / b2);
	elseif ((b1 * b2) <= -1e-167)
		tmp = t_0;
	elseif ((b1 * b2) <= 2e-209)
		tmp = (a2 / b1) * (a1 / b2);
	elseif ((b1 * b2) <= 5e+283)
		tmp = t_0;
	else
		tmp = (a2 / (b1 / a1)) / b2;
	end
	tmp_2 = tmp;
end
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[N[(b1 * b2), $MachinePrecision], (-Infinity)], N[(N[(a1 / b1), $MachinePrecision] * N[(a2 / b2), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(b1 * b2), $MachinePrecision], -1e-167], t$95$0, If[LessEqual[N[(b1 * b2), $MachinePrecision], 2e-209], N[(N[(a2 / b1), $MachinePrecision] * N[(a1 / b2), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(b1 * b2), $MachinePrecision], 5e+283], t$95$0, N[(N[(a2 / N[(b1 / a1), $MachinePrecision]), $MachinePrecision] / b2), $MachinePrecision]]]]]]
\frac{a1 \cdot a2}{b1 \cdot b2}
\begin{array}{l}
t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\
\mathbf{if}\;b1 \cdot b2 \leq -\infty:\\
\;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\

\mathbf{elif}\;b1 \cdot b2 \leq -1 \cdot 10^{-167}:\\
\;\;\;\;t_0\\

\mathbf{elif}\;b1 \cdot b2 \leq 2 \cdot 10^{-209}:\\
\;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\

\mathbf{elif}\;b1 \cdot b2 \leq 5 \cdot 10^{+283}:\\
\;\;\;\;t_0\\

\mathbf{else}:\\
\;\;\;\;\frac{\frac{a2}{\frac{b1}{a1}}}{b2}\\


\end{array}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original82.1%
Target81.9%
Herbie91.9%
\[\frac{a1}{b1} \cdot \frac{a2}{b2} \]

Derivation?

  1. Split input into 4 regimes
  2. if (*.f64 b1 b2) < -inf.0

    1. Initial program 64.2%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
    2. Simplified95.9%

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

      [Start]64.2

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

      times-frac [=>]95.9

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

    if -inf.0 < (*.f64 b1 b2) < -1e-167 or 2.0000000000000001e-209 < (*.f64 b1 b2) < 5.0000000000000004e283

    1. Initial program 92.7%

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

    if -1e-167 < (*.f64 b1 b2) < 2.0000000000000001e-209

    1. Initial program 50.9%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
    2. Applied egg-rr84.4%

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

      [Start]50.9

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

      *-commutative [=>]50.9

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

      times-frac [=>]84.4

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

    if 5.0000000000000004e283 < (*.f64 b1 b2)

    1. Initial program 69.0%

      \[\frac{a1 \cdot a2}{b1 \cdot b2} \]
    2. Simplified95.4%

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

      [Start]69.0

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

      associate-/r* [=>]91.3

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

      *-commutative [=>]91.3

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

      associate-/l* [=>]95.4

      \[ \frac{\color{blue}{\frac{a2}{\frac{b1}{a1}}}}{b2} \]
  3. Recombined 4 regimes into one program.
  4. Final simplification91.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;b1 \cdot b2 \leq -\infty:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;b1 \cdot b2 \leq -1 \cdot 10^{-167}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{elif}\;b1 \cdot b2 \leq 2 \cdot 10^{-209}:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \mathbf{elif}\;b1 \cdot b2 \leq 5 \cdot 10^{+283}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a2}{\frac{b1}{a1}}}{b2}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy91.4%
Cost1490
\[\begin{array}{l} \mathbf{if}\;b1 \cdot b2 \leq -1 \cdot 10^{+283} \lor \neg \left(b1 \cdot b2 \leq -2 \cdot 10^{-188}\right) \land \left(b1 \cdot b2 \leq 4 \cdot 10^{-310} \lor \neg \left(b1 \cdot b2 \leq 2 \cdot 10^{+231}\right)\right):\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \frac{a1}{b1 \cdot b2}\\ \end{array} \]
Alternative 2
Accuracy91.0%
Cost1490
\[\begin{array}{l} \mathbf{if}\;b1 \cdot b2 \leq -2 \cdot 10^{+130} \lor \neg \left(b1 \cdot b2 \leq -1 \cdot 10^{-167} \lor \neg \left(b1 \cdot b2 \leq 5 \cdot 10^{-268}\right) \land b1 \cdot b2 \leq 4 \cdot 10^{+145}\right):\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \frac{a1}{b1 \cdot b2}\\ \end{array} \]
Alternative 3
Accuracy91.9%
Cost1489
\[\begin{array}{l} \mathbf{if}\;b1 \cdot b2 \leq -\infty:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;b1 \cdot b2 \leq -1 \cdot 10^{-167} \lor \neg \left(b1 \cdot b2 \leq 2 \cdot 10^{-209}\right) \land b1 \cdot b2 \leq 10^{+288}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \end{array} \]
Alternative 4
Accuracy92.0%
Cost1488
\[\begin{array}{l} t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{if}\;b1 \cdot b2 \leq -\infty:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;b1 \cdot b2 \leq -1 \cdot 10^{-167}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;b1 \cdot b2 \leq 2 \cdot 10^{-209}:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \mathbf{elif}\;b1 \cdot b2 \leq 2 \cdot 10^{+231}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{a1 \cdot \frac{a2}{b2}}{b1}\\ \end{array} \]
Alternative 5
Accuracy81.5%
Cost448
\[a2 \cdot \frac{a1}{b1 \cdot b2} \]

Error

Reproduce?

herbie shell --seed 2023137 
(FPCore (a1 a2 b1 b2)
  :name "Quotient of products"
  :precision binary64

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

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