Quotient of products

Percentage Accurate: 85.6% → 97.2%

Time: 4.0s

Precision: binary64

Cost: 2512

• Unsound rule application detected in e-graph. Results may not be sound.

• 20.3% of points produce a very large (infinite) output. You may want to add a precondition.

Math

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

↓

\[\begin{array}{l} [a1, a2] = \mathsf{sort}([a1, a2]) \\ [b1, b2] = \mathsf{sort}([b1, b2])\\ \end{array} \begin{array}{l} t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{if}\;t_0 \leq -\infty:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;t_0 \leq -1 \cdot 10^{-290}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;t_0 \leq 0:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \mathbf{elif}\;t_0 \leq 5 \cdot 10^{+306}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a2}{b1}}{\frac{b2}{a1}}\\ \end{array} \]

FPCore

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

↓

;; Ensure these are sorted, for example in Racket, do
(match-define (list a1 a2) (sort (a1 a2) <))
(match-define (list b1 b2) (sort (b1 b2) <))

(FPCore (a1 a2 b1 b2)
 :precision binary64
 (let* ((t_0 (/ (* a1 a2) (* b1 b2))))
   (if (<= t_0 (- INFINITY))
     (* (/ a1 b1) (/ a2 b2))
     (if (<= t_0 -1e-290)
       t_0
       (if (<= t_0 0.0)
         (* (/ a2 b1) (/ a1 b2))
         (if (<= t_0 5e+306) t_0 (/ (/ a2 b1) (/ b2 a1))))))))

C

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

↓

// Ensure these are sorted
assert(a1 < a2);
assert(b1 < b2);

double code(double a1, double a2, double b1, double b2) {
	double t_0 = (a1 * a2) / (b1 * b2);
	double tmp;
	if (t_0 <= -((double) INFINITY)) {
		tmp = (a1 / b1) * (a2 / b2);
	} else if (t_0 <= -1e-290) {
		tmp = t_0;
	} else if (t_0 <= 0.0) {
		tmp = (a2 / b1) * (a1 / b2);
	} else if (t_0 <= 5e+306) {
		tmp = t_0;
	} else {
		tmp = (a2 / b1) / (b2 / a1);
	}
	return tmp;
}

Java

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

↓

// Ensure these are sorted
assert a1 < a2;
assert 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 <= -Double.POSITIVE_INFINITY) {
		tmp = (a1 / b1) * (a2 / b2);
	} else if (t_0 <= -1e-290) {
		tmp = t_0;
	} else if (t_0 <= 0.0) {
		tmp = (a2 / b1) * (a1 / b2);
	} else if (t_0 <= 5e+306) {
		tmp = t_0;
	} else {
		tmp = (a2 / b1) / (b2 / a1);
	}
	return tmp;
}

Python

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

↓

[a1, a2] = sort([a1, a2])
[b1, b2] = sort([b1, b2])

def code(a1, a2, b1, b2):
	t_0 = (a1 * a2) / (b1 * b2)
	tmp = 0
	if t_0 <= -math.inf:
		tmp = (a1 / b1) * (a2 / b2)
	elif t_0 <= -1e-290:
		tmp = t_0
	elif t_0 <= 0.0:
		tmp = (a2 / b1) * (a1 / b2)
	elif t_0 <= 5e+306:
		tmp = t_0
	else:
		tmp = (a2 / b1) / (b2 / a1)
	return tmp

Julia

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

↓

a1, a2 = sort([a1, a2])
b1, b2 = sort([b1, b2])

function code(a1, a2, b1, b2)
	t_0 = Float64(Float64(a1 * a2) / Float64(b1 * b2))
	tmp = 0.0
	if (t_0 <= Float64(-Inf))
		tmp = Float64(Float64(a1 / b1) * Float64(a2 / b2));
	elseif (t_0 <= -1e-290)
		tmp = t_0;
	elseif (t_0 <= 0.0)
		tmp = Float64(Float64(a2 / b1) * Float64(a1 / b2));
	elseif (t_0 <= 5e+306)
		tmp = t_0;
	else
		tmp = Float64(Float64(a2 / b1) / Float64(b2 / a1));
	end
	return tmp
end

MATLAB

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

↓

a1, a2 = num2cell(sort([a1, a2])){:}
b1, b2 = num2cell(sort([b1, b2])){:}

function tmp_2 = code(a1, a2, b1, b2)
	t_0 = (a1 * a2) / (b1 * b2);
	tmp = 0.0;
	if (t_0 <= -Inf)
		tmp = (a1 / b1) * (a2 / b2);
	elseif (t_0 <= -1e-290)
		tmp = t_0;
	elseif (t_0 <= 0.0)
		tmp = (a2 / b1) * (a1 / b2);
	elseif (t_0 <= 5e+306)
		tmp = t_0;
	else
		tmp = (a2 / b1) / (b2 / a1);
	end
	tmp_2 = tmp;
end

Wolfram

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

↓

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_] := Block[{t$95$0 = N[(N[(a1 * a2), $MachinePrecision] / N[(b1 * b2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$0, (-Infinity)], N[(N[(a1 / b1), $MachinePrecision] * N[(a2 / b2), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$0, -1e-290], t$95$0, If[LessEqual[t$95$0, 0.0], N[(N[(a2 / b1), $MachinePrecision] * N[(a1 / b2), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$0, 5e+306], t$95$0, N[(N[(a2 / b1), $MachinePrecision] / N[(b2 / a1), $MachinePrecision]), $MachinePrecision]]]]]]

Target

Original	85.6%
Target	86.5%
Herbie	97.2%

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

Derivation

1. Split input into 4 regimes

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

   1. Initial program 78.3%

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

   2. Simplified 99.8%

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

      [Start] 78.3	\[ \frac{a1 \cdot a2}{b1 \cdot b2} \]
      times-frac [=>] 99.8	\[ \color{blue}{\frac{a1}{b1} \cdot \frac{a2}{b2}} \]

   if -inf.0 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < -1.0000000000000001e-290 or -0.0 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < 4.99999999999999993e306

   1. Initial program 98.0%

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

   if -1.0000000000000001e-290 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2)) < -0.0

   1. Initial program 84.9%

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

   2. Simplified 90.8%

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

      [Start] 84.9	\[ \frac{a1 \cdot a2}{b1 \cdot b2} \]
      associate-/l* [=>] 93.8	\[ \color{blue}{\frac{a1}{\frac{b1 \cdot b2}{a2}}} \]
      *-commutative [=>] 93.8	\[ \frac{a1}{\frac{\color{blue}{b2 \cdot b1}}{a2}} \]
      associate-/l* [=>] 90.8	\[ \frac{a1}{\color{blue}{\frac{b2}{\frac{a2}{b1}}}} \]

   3. Taylor expanded in a1 around 0 84.9%

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

   4. Simplified 95.8%

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

      [Start] 84.9	\[ \frac{a1 \cdot a2}{b2 \cdot b1} \]
      times-frac [=>] 95.8	\[ \color{blue}{\frac{a1}{b2} \cdot \frac{a2}{b1}} \]
      *-commutative [=>] 95.8	\[ \color{blue}{\frac{a2}{b1} \cdot \frac{a1}{b2}} \]

   if 4.99999999999999993e306 < (/.f64 (*.f64 a1 a2) (*.f64 b1 b2))

   1. Initial program 59.7%

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

   2. Simplified 95.0%

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

      [Start] 59.7	\[ \frac{a1 \cdot a2}{b1 \cdot b2} \]
      associate-/l* [=>] 70.2	\[ \color{blue}{\frac{a1}{\frac{b1 \cdot b2}{a2}}} \]
      *-commutative [=>] 70.2	\[ \frac{a1}{\frac{\color{blue}{b2 \cdot b1}}{a2}} \]
      associate-/l* [=>] 95.0	\[ \frac{a1}{\color{blue}{\frac{b2}{\frac{a2}{b1}}}} \]

   3. Taylor expanded in a1 around 0 59.7%

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

   4. Simplified 97.3%

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

      [Start] 59.7	\[ \frac{a1 \cdot a2}{b2 \cdot b1} \]
      times-frac [=>] 97.3	\[ \color{blue}{\frac{a1}{b2} \cdot \frac{a2}{b1}} \]
      *-commutative [=>] 97.3	\[ \color{blue}{\frac{a2}{b1} \cdot \frac{a1}{b2}} \]

   5. Applied egg-rr 97.4%

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

      [Start] 97.3	\[ \frac{a2}{b1} \cdot \frac{a1}{b2} \]
      clear-num [=>] 97.3	\[ \frac{a2}{b1} \cdot \color{blue}{\frac{1}{\frac{b2}{a1}}} \]
      un-div-inv [=>] 97.4	\[ \color{blue}{\frac{\frac{a2}{b1}}{\frac{b2}{a1}}} \]

3. Recombined 4 regimes into one program.

4. Final simplification 97.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq -\infty:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq -1 \cdot 10^{-290}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq 0:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \mathbf{elif}\;\frac{a1 \cdot a2}{b1 \cdot b2} \leq 5 \cdot 10^{+306}:\\ \;\;\;\;\frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a2}{b1}}{\frac{b2}{a1}}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	97.2%
Cost	2513

\[\begin{array}{l} t_0 := \frac{a1 \cdot a2}{b1 \cdot b2}\\ \mathbf{if}\;t_0 \leq -\infty:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \mathbf{elif}\;t_0 \leq -1 \cdot 10^{-290} \lor \neg \left(t_0 \leq 0\right) \land t_0 \leq 5 \cdot 10^{+306}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \end{array} \]

Alternative 2
Accuracy	91.7%
Cost	1229

\[\begin{array}{l} \mathbf{if}\;b1 \cdot b2 \leq -2 \cdot 10^{-266} \lor \neg \left(b1 \cdot b2 \leq 2 \cdot 10^{-306}\right) \land b1 \cdot b2 \leq 5 \cdot 10^{+228}:\\ \;\;\;\;a2 \cdot \frac{a1}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a1}{b1} \cdot \frac{a2}{b2}\\ \end{array} \]

Alternative 3
Accuracy	91.9%
Cost	1229

\[\begin{array}{l} \mathbf{if}\;b1 \cdot b2 \leq -2 \cdot 10^{-176} \lor \neg \left(b1 \cdot b2 \leq 8 \cdot 10^{-204}\right) \land b1 \cdot b2 \leq 1.2 \cdot 10^{+254}:\\ \;\;\;\;a2 \cdot \frac{a1}{b1 \cdot b2}\\ \mathbf{else}:\\ \;\;\;\;\frac{a2}{b1} \cdot \frac{a1}{b2}\\ \end{array} \]

Alternative 4
Accuracy	86.5%
Cost	448

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

Reproduce

herbie shell --seed 2023160 
(FPCore (a1 a2 b1 b2)
  :name "Quotient of products"
  :precision binary64

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

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