Data.Colour.Matrix:inverse from colour-2.3.3, B

Percentage Accurate: 91.1% → 97.0%

Time: 6.7s

Precision: binary64

Cost: 1737

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

• 27.7% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[\frac{x \cdot y - z \cdot t}{a} \]

↓

\[\begin{array}{l} [z, t] = \mathsf{sort}([z, t])\\ \\ \begin{array}{l} t_1 := x \cdot y - z \cdot t\\ \mathbf{if}\;t_1 \leq -\infty \lor \neg \left(t_1 \leq 10^{+269}\right):\\ \;\;\;\;y \cdot \frac{x}{a} - z \cdot \frac{t}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{t_1}{a}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y z t a) :precision binary64 (/ (- (* x y) (* z t)) a))

↓

NOTE: z and t should be sorted in increasing order before calling this function.
(FPCore (x y z t a)
 :precision binary64
 (let* ((t_1 (- (* x y) (* z t))))
   (if (or (<= t_1 (- INFINITY)) (not (<= t_1 1e+269)))
     (- (* y (/ x a)) (* z (/ t a)))
     (/ t_1 a))))

C

double code(double x, double y, double z, double t, double a) {
	return ((x * y) - (z * t)) / a;
}

↓

assert(z < t);
double code(double x, double y, double z, double t, double a) {
	double t_1 = (x * y) - (z * t);
	double tmp;
	if ((t_1 <= -((double) INFINITY)) || !(t_1 <= 1e+269)) {
		tmp = (y * (x / a)) - (z * (t / a));
	} else {
		tmp = t_1 / a;
	}
	return tmp;
}

Java

public static double code(double x, double y, double z, double t, double a) {
	return ((x * y) - (z * t)) / a;
}

↓

assert z < t;
public static double code(double x, double y, double z, double t, double a) {
	double t_1 = (x * y) - (z * t);
	double tmp;
	if ((t_1 <= -Double.POSITIVE_INFINITY) || !(t_1 <= 1e+269)) {
		tmp = (y * (x / a)) - (z * (t / a));
	} else {
		tmp = t_1 / a;
	}
	return tmp;
}

Python

def code(x, y, z, t, a):
	return ((x * y) - (z * t)) / a

↓

[z, t] = sort([z, t])
def code(x, y, z, t, a):
	t_1 = (x * y) - (z * t)
	tmp = 0
	if (t_1 <= -math.inf) or not (t_1 <= 1e+269):
		tmp = (y * (x / a)) - (z * (t / a))
	else:
		tmp = t_1 / a
	return tmp

Julia

function code(x, y, z, t, a)
	return Float64(Float64(Float64(x * y) - Float64(z * t)) / a)
end

↓

z, t = sort([z, t])
function code(x, y, z, t, a)
	t_1 = Float64(Float64(x * y) - Float64(z * t))
	tmp = 0.0
	if ((t_1 <= Float64(-Inf)) || !(t_1 <= 1e+269))
		tmp = Float64(Float64(y * Float64(x / a)) - Float64(z * Float64(t / a)));
	else
		tmp = Float64(t_1 / a);
	end
	return tmp
end

MATLAB

function tmp = code(x, y, z, t, a)
	tmp = ((x * y) - (z * t)) / a;
end

↓

z, t = num2cell(sort([z, t])){:}
function tmp_2 = code(x, y, z, t, a)
	t_1 = (x * y) - (z * t);
	tmp = 0.0;
	if ((t_1 <= -Inf) || ~((t_1 <= 1e+269)))
		tmp = (y * (x / a)) - (z * (t / a));
	else
		tmp = t_1 / a;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_, z_, t_, a_] := N[(N[(N[(x * y), $MachinePrecision] - N[(z * t), $MachinePrecision]), $MachinePrecision] / a), $MachinePrecision]

↓

NOTE: z and t should be sorted in increasing order before calling this function.
code[x_, y_, z_, t_, a_] := Block[{t$95$1 = N[(N[(x * y), $MachinePrecision] - N[(z * t), $MachinePrecision]), $MachinePrecision]}, If[Or[LessEqual[t$95$1, (-Infinity)], N[Not[LessEqual[t$95$1, 1e+269]], $MachinePrecision]], N[(N[(y * N[(x / a), $MachinePrecision]), $MachinePrecision] - N[(z * N[(t / a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$1 / a), $MachinePrecision]]]

Target

Original	91.1%
Target	91.6%
Herbie	97.0%

\[\begin{array}{l} \mathbf{if}\;z < -2.468684968699548 \cdot 10^{+170}:\\ \;\;\;\;\frac{y}{a} \cdot x - \frac{t}{a} \cdot z\\ \mathbf{elif}\;z < 6.309831121978371 \cdot 10^{-71}:\\ \;\;\;\;\frac{x \cdot y - z \cdot t}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{y}{a} \cdot x - \frac{t}{a} \cdot z\\ \end{array} \]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (*.f64 x y) (*.f64 z t)) < -inf.0 or 1e269 < (-.f64 (*.f64 x y) (*.f64 z t))

   1. Initial program 59.5%

      \[\frac{x \cdot y - z \cdot t}{a} \]

   2. Taylor expanded in x around 0 56.4%

      \[\leadsto \color{blue}{-1 \cdot \frac{t \cdot z}{a} + \frac{y \cdot x}{a}} \]

   3. Applied egg-rr 95.3%

      \[\leadsto \color{blue}{y \cdot \frac{x}{a} - \frac{t}{a} \cdot z} \]
      Step-by-step derivation

      [Start] 56.4%	\[ -1 \cdot \frac{t \cdot z}{a} + \frac{y \cdot x}{a} \]
      +-commutative [=>] 56.4%	\[ \color{blue}{\frac{y \cdot x}{a} + -1 \cdot \frac{t \cdot z}{a}} \]
      mul-1-neg [=>] 56.4%	\[ \frac{y \cdot x}{a} + \color{blue}{\left(-\frac{t \cdot z}{a}\right)} \]
      unsub-neg [=>] 56.4%	\[ \color{blue}{\frac{y \cdot x}{a} - \frac{t \cdot z}{a}} \]
      associate-*r/ [<=] 75.2%	\[ \color{blue}{y \cdot \frac{x}{a}} - \frac{t \cdot z}{a} \]
      associate-/l* [=>] 95.3%	\[ y \cdot \frac{x}{a} - \color{blue}{\frac{t}{\frac{a}{z}}} \]
      associate-/r/ [=>] 95.3%	\[ y \cdot \frac{x}{a} - \color{blue}{\frac{t}{a} \cdot z} \]

   if -inf.0 < (-.f64 (*.f64 x y) (*.f64 z t)) < 1e269

   1. Initial program 99.6%

      \[\frac{x \cdot y - z \cdot t}{a} \]
3. Recombined 2 regimes into one program.

4. Final simplification 98.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot y - z \cdot t \leq -\infty \lor \neg \left(x \cdot y - z \cdot t \leq 10^{+269}\right):\\ \;\;\;\;y \cdot \frac{x}{a} - z \cdot \frac{t}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y - z \cdot t}{a}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	97.0%
Cost	1737

\[\begin{array}{l} t_1 := x \cdot y - z \cdot t\\ \mathbf{if}\;t_1 \leq -\infty \lor \neg \left(t_1 \leq 10^{+269}\right):\\ \;\;\;\;y \cdot \frac{x}{a} - z \cdot \frac{t}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{t_1}{a}\\ \end{array} \]

Alternative 2
Accuracy	70.8%
Cost	1944

\[\begin{array}{l} t_1 := \frac{z}{a} \cdot \left(-t\right)\\ t_2 := \frac{x \cdot y}{a}\\ \mathbf{if}\;x \cdot y \leq -5 \cdot 10^{-62}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;x \cdot y \leq 2 \cdot 10^{-163}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;x \cdot y \leq 2 \cdot 10^{-120}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;x \cdot y \leq 2 \cdot 10^{-71}:\\ \;\;\;\;\frac{t}{a} \cdot \left(-z\right)\\ \mathbf{elif}\;x \cdot y \leq 10^{-24}:\\ \;\;\;\;x \cdot \frac{y}{a}\\ \mathbf{elif}\;x \cdot y \leq 5 \cdot 10^{+59}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\frac{y}{\frac{a}{x}}\\ \end{array} \]

Alternative 3
Accuracy	94.4%
Cost	1097

\[\begin{array}{l} \mathbf{if}\;z \cdot t \leq -\infty \lor \neg \left(z \cdot t \leq 10^{+190}\right):\\ \;\;\;\;\frac{t}{a} \cdot \left(-z\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y - z \cdot t}{a}\\ \end{array} \]

Alternative 4
Accuracy	68.5%
Cost	912

\[\begin{array}{l} t_1 := \frac{x \cdot y}{a}\\ t_2 := \frac{t}{a} \cdot \left(-z\right)\\ \mathbf{if}\;z \leq -2.6 \cdot 10^{+67}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;z \leq -2.7 \cdot 10^{-10}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;z \leq -1.5 \cdot 10^{-175}:\\ \;\;\;\;y \cdot \frac{x}{a}\\ \mathbf{elif}\;z \leq 7 \cdot 10^{-110}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;t_2\\ \end{array} \]

Alternative 5
Accuracy	51.3%
Cost	320

\[y \cdot \frac{x}{a} \]

Alternative 6
Accuracy	51.6%
Cost	320

\[x \cdot \frac{y}{a} \]

Alternative 7
Accuracy	51.5%
Cost	320

\[\frac{x}{\frac{a}{y}} \]

Reproduce

herbie shell --seed 2023167 
(FPCore (x y z t a)
  :name "Data.Colour.Matrix:inverse from colour-2.3.3, B"
  :precision binary64

  :herbie-target
  (if (< z -2.468684968699548e+170) (- (* (/ y a) x) (* (/ t a) z)) (if (< z 6.309831121978371e-71) (/ (- (* x y) (* z t)) a) (- (* (/ y a) x) (* (/ t a) z))))

  (/ (- (* x y) (* z t)) a))