Examples.Basics.BasicTests:f2 from sbv-4.4

Percentage Accurate: 93.8% → 96.8%

Time: 3.1s

Precision: binary64

Cost: 7044

• 43.8% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[x \cdot x - y \cdot y \]

↓

\[\begin{array}{l} \mathbf{if}\;x \cdot x \leq 5 \cdot 10^{+304}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (- (* x x) (* y y)))

↓

(FPCore (x y)
 :precision binary64
 (if (<= (* x x) 5e+304) (fma x x (* y (- y))) (* x x)))

C

double code(double x, double y) {
	return (x * x) - (y * y);
}

↓

double code(double x, double y) {
	double tmp;
	if ((x * x) <= 5e+304) {
		tmp = fma(x, x, (y * -y));
	} else {
		tmp = x * x;
	}
	return tmp;
}

Julia

function code(x, y)
	return Float64(Float64(x * x) - Float64(y * y))
end

↓

function code(x, y)
	tmp = 0.0
	if (Float64(x * x) <= 5e+304)
		tmp = fma(x, x, Float64(y * Float64(-y)));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

Wolfram

code[x_, y_] := N[(N[(x * x), $MachinePrecision] - N[(y * y), $MachinePrecision]), $MachinePrecision]

↓

code[x_, y_] := If[LessEqual[N[(x * x), $MachinePrecision], 5e+304], N[(x * x + N[(y * (-y)), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 x x) < 4.9999999999999997e304

   1. Initial program 100.0%

      \[x \cdot x - y \cdot y \]

   2. Simplified 100.0%

      \[\leadsto \color{blue}{\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)} \]
      Step-by-step derivation

      [Start] 100.0%	\[ x \cdot x - y \cdot y \]
      fma-neg [=>] 100.0%	\[ \color{blue}{\mathsf{fma}\left(x, x, -y \cdot y\right)} \]
      distribute-rgt-neg-in [=>] 100.0%	\[ \mathsf{fma}\left(x, x, \color{blue}{y \cdot \left(-y\right)}\right) \]

   if 4.9999999999999997e304 < (*.f64 x x)

   1. Initial program 85.5%

      \[x \cdot x - y \cdot y \]

   2. Taylor expanded in x around inf 96.8%

      \[\leadsto \color{blue}{{x}^{2}} \]

   3. Simplified 96.8%

      \[\leadsto \color{blue}{x \cdot x} \]
      Step-by-step derivation

      [Start] 96.8%	\[ {x}^{2} \]
      unpow2 [=>] 96.8%	\[ \color{blue}{x \cdot x} \]

3. Recombined 2 regimes into one program.

4. Final simplification 99.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 5 \cdot 10^{+304}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	96.8%
Cost	7044

\[\begin{array}{l} \mathbf{if}\;x \cdot x \leq 5 \cdot 10^{+304}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 2
Accuracy	77.3%
Cost	1037

\[\begin{array}{l} \mathbf{if}\;x \cdot x \leq 6.8 \cdot 10^{-64} \lor \neg \left(x \cdot x \leq 0.15\right) \land x \cdot x \leq 8.6 \cdot 10^{+130}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 3
Accuracy	96.8%
Cost	708

\[\begin{array}{l} \mathbf{if}\;x \cdot x \leq 10^{+306}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 4
Accuracy	53.8%
Cost	192

\[x \cdot x \]

Reproduce

herbie shell --seed 2023243 
(FPCore (x y)
  :name "Examples.Basics.BasicTests:f2 from sbv-4.4"
  :precision binary64
  (- (* x x) (* y y)))