Quotient of products

Average Accuracy: 81.7% → 94.2%

Time: 4.5s

Precision: binary64

Cost: 2512

Math

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

↓

\[\begin{array}{l} t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\ t_1 := \frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{if}\;t_0 \leq -\infty:\\ \;\;\;\;t_1\\ \mathbf{elif}\;t_0 \leq -1 \cdot 10^{-247}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;t_0 \leq 10^{-285}:\\ \;\;\;\;\frac{\frac{a2}{\frac{b1}{a1}}}{b2}\\ \mathbf{elif}\;t_0 \leq 2 \cdot 10^{+256}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \]

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 (- INFINITY))
     t_1
     (if (<= t_0 -1e-247)
       t_0
       (if (<= t_0 1e-285)
         (/ (/ a2 (/ b1 a1)) b2)
         (if (<= t_0 2e+256) t_0 t_1))))))

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 <= -((double) INFINITY)) {
		tmp = t_1;
	} else if (t_0 <= -1e-247) {
		tmp = t_0;
	} else if (t_0 <= 1e-285) {
		tmp = (a2 / (b1 / a1)) / b2;
	} else if (t_0 <= 2e+256) {
		tmp = t_0;
	} else {
		tmp = t_1;
	}
	return tmp;
}

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 <= -Double.POSITIVE_INFINITY) {
		tmp = t_1;
	} else if (t_0 <= -1e-247) {
		tmp = t_0;
	} else if (t_0 <= 1e-285) {
		tmp = (a2 / (b1 / a1)) / b2;
	} else if (t_0 <= 2e+256) {
		tmp = t_0;
	} else {
		tmp = t_1;
	}
	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 <= -math.inf:
		tmp = t_1
	elif t_0 <= -1e-247:
		tmp = t_0
	elif t_0 <= 1e-285:
		tmp = (a2 / (b1 / a1)) / b2
	elif t_0 <= 2e+256:
		tmp = t_0
	else:
		tmp = t_1
	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 <= Float64(-Inf))
		tmp = t_1;
	elseif (t_0 <= -1e-247)
		tmp = t_0;
	elseif (t_0 <= 1e-285)
		tmp = Float64(Float64(a2 / Float64(b1 / a1)) / b2);
	elseif (t_0 <= 2e+256)
		tmp = t_0;
	else
		tmp = t_1;
	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 <= -Inf)
		tmp = t_1;
	elseif (t_0 <= -1e-247)
		tmp = t_0;
	elseif (t_0 <= 1e-285)
		tmp = (a2 / (b1 / a1)) / b2;
	elseif (t_0 <= 2e+256)
		tmp = t_0;
	else
		tmp = t_1;
	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, (-Infinity)], t$95$1, If[LessEqual[t$95$0, -1e-247], t$95$0, If[LessEqual[t$95$0, 1e-285], N[(N[(a2 / N[(b1 / a1), $MachinePrecision]), $MachinePrecision] / b2), $MachinePrecision], If[LessEqual[t$95$0, 2e+256], t$95$0, t$95$1]]]]]]

Target

Original	81.7%
Target	82.3%
Herbie	94.2%

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

Derivation

1. Split input into 3 regimes

2. if (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < -inf.0 or 2.0000000000000001e256 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2))

   1. Initial program 14.8%

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

   2. Simplified 83.3%

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

      [Start] 14.8	\[ \frac{a1 \cdot a2}{b1 \cdot b2} \]
      times-frac [=>] 83.3	\[ \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

   if -inf.0 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < -1e-247 or 1.00000000000000007e-285 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < 2.0000000000000001e256

   1. Initial program 98.7%

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

   if -1e-247 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < 1.00000000000000007e-285

   1. Initial program 80.8%

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

   2. Simplified 91.5%

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

      [Start] 80.8	\[ \frac{a1 \cdot a2}{b1 \cdot b2} \]
      associate-/r* [=>] 88.4	\[ \color{blue}{\frac{\frac{a1 \cdot a2}{b1}}{b2}} \]
      *-commutative [=>] 88.4	\[ \frac{\frac{\color{blue}{a2 \cdot a1}}{b1}}{b2} \]
      associate-/l* [=>] 91.5	\[ \frac{\color{blue}{\frac{a2}{\frac{b1}{a1}}}}{b2} \]

3. Recombined 3 regimes into one program.

4. Final simplification 94.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq -\infty:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq -1 \cdot 10^{-247}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq 10^{-285}:\\ \;\;\;\;\frac{\frac{a2}{\frac{b1}{a1}}}{b2}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq 2 \cdot 10^{+256}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	94.9%
Cost	2514

\[\begin{array}{l} t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{if}\;t_0 \leq -\infty \lor \neg \left(t_0 \leq -1 \cdot 10^{-253}\right) \land \left(t_0 \leq 10^{-285} \lor \neg \left(t_0 \leq 2 \cdot 10^{+256}\right)\right):\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 2
Accuracy	91.5%
Cost	1490

\[\begin{array}{l} \mathbf{if}\;b1 \cdot b2 \leq -5 \cdot 10^{+177} \lor \neg \left(b1 \cdot b2 \leq -1 \cdot 10^{-245} \lor \neg \left(b1 \cdot b2 \leq 10^{-248}\right) \land b1 \cdot b2 \leq 10^{+186}\right):\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{else}:\\ \;\;\;\;a2 \cdot \frac{a1}{b1 \cdot b2}\\ \end{array} \]

Alternative 3
Accuracy	91.5%
Cost	1489

\[\begin{array}{l} t_0 := \frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{if}\;b1 \cdot b2 \leq -5 \cdot 10^{+177}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;b1 \cdot b2 \leq -1 \cdot 10^{-245}:\\ \;\;\;\;a2 \cdot \frac{a1}{b1 \cdot b2}\\ \mathbf{elif}\;b1 \cdot b2 \leq 0 \lor \neg \left(b1 \cdot b2 \leq 5 \cdot 10^{+228}\right):\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{a1}{\frac{b1 \cdot b2}{a2}}\\ \end{array} \]

Alternative 4
Accuracy	82.1%
Cost	448

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

Reproduce

herbie shell --seed 2023135 
(FPCore (a1 a2 b1 b2)
  :name "Quotient of products"
  :precision binary64

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

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