Examples.Basics.BasicTests:f2 from sbv-4.4

Percentage Accurate: 94.0% → 96.2%

Time: 2.4s

Alternatives: 4

Speedup: 0.6×

• 43.3% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ x \cdot x - y \cdot y \end{array} \]

FPCore

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

C

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

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = (x * x) - (y * y)
end function

Java

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

Python

def code(x, y):
	return (x * x) - (y * y)

Julia

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

MATLAB

function tmp = code(x, y)
	tmp = (x * x) - (y * y);
end

Wolfram

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

Initial Program: 94.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ x \cdot x - y \cdot y \end{array} \]

FPCore

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

C

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

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = (x * x) - (y * y)
end function

Java

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

Python

def code(x, y):
	return (x * x) - (y * y)

Julia

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

MATLAB

function tmp = code(x, y)
	tmp = (x * x) - (y * y);
end

Wolfram

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

Alternative 1: 96.2% accurate, 0.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \cdot x \leq 2.5 \cdot 10^{+267}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= (* x x) 2.5e+267) (- (* x x) (* y y)) (* x x)))

C

double code(double x, double y) {
	double tmp;
	if ((x * x) <= 2.5e+267) {
		tmp = (x * x) - (y * y);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if ((x * x) <= 2.5d+267) then
        tmp = (x * x) - (y * y)
    else
        tmp = x * x
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if ((x * x) <= 2.5e+267) {
		tmp = (x * x) - (y * y);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if (x * x) <= 2.5e+267:
		tmp = (x * x) - (y * y)
	else:
		tmp = x * x
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (Float64(x * x) <= 2.5e+267)
		tmp = Float64(Float64(x * x) - Float64(y * y));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((x * x) <= 2.5e+267)
		tmp = (x * x) - (y * y);
	else
		tmp = x * x;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[N[(x * x), $MachinePrecision], 2.5e+267], N[(N[(x * x), $MachinePrecision] - N[(y * y), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 x x) < 2.4999999999999999e267

   1. Initial program 100.0%

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

   if 2.4999999999999999e267 < (*.f64 x x)

   1. Initial program 76.9%

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

   2. Taylor expanded in x around inf 89.2%

      \[\leadsto \color{blue}{{x}^{2}} \]
   3. Step-by-step derivation

      1. unpow2 89.2%

         \[\leadsto \color{blue}{x \cdot x} \]

   4. Simplified 89.2%

      \[\leadsto \color{blue}{x \cdot x} \]
3. Recombined 2 regimes into one program.

4. Final simplification 97.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 2.5 \cdot 10^{+267}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 2: 97.0% accurate, 0.1× speedup

Math

\[\begin{array}{l} \\ \mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right) \end{array} \]

FPCore

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

C

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

Julia

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

Wolfram

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

Derivation

1. Initial program 94.1%

   \[x \cdot x - y \cdot y \]
2. Step-by-step derivation

   1. fma-neg 96.9%

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

   2. distribute-rgt-neg-in 96.9%

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

3. Simplified 96.9%

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

4. Final simplification 96.9%

   \[\leadsto \mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right) \]

Alternative 3: 77.9% accurate, 0.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \cdot x \leq 1.75 \cdot 10^{-28} \lor \neg \left(x \cdot x \leq 5.2 \cdot 10^{+84}\right) \land x \cdot x \leq 3.6 \cdot 10^{+130}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (or (<= (* x x) 1.75e-28)
         (and (not (<= (* x x) 5.2e+84)) (<= (* x x) 3.6e+130)))
   (* y (- y))
   (* x x)))

C

double code(double x, double y) {
	double tmp;
	if (((x * x) <= 1.75e-28) || (!((x * x) <= 5.2e+84) && ((x * x) <= 3.6e+130))) {
		tmp = y * -y;
	} else {
		tmp = x * x;
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (((x * x) <= 1.75d-28) .or. (.not. ((x * x) <= 5.2d+84)) .and. ((x * x) <= 3.6d+130)) then
        tmp = y * -y
    else
        tmp = x * x
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (((x * x) <= 1.75e-28) || (!((x * x) <= 5.2e+84) && ((x * x) <= 3.6e+130))) {
		tmp = y * -y;
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if ((x * x) <= 1.75e-28) or (not ((x * x) <= 5.2e+84) and ((x * x) <= 3.6e+130)):
		tmp = y * -y
	else:
		tmp = x * x
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if ((Float64(x * x) <= 1.75e-28) || (!(Float64(x * x) <= 5.2e+84) && (Float64(x * x) <= 3.6e+130)))
		tmp = Float64(y * Float64(-y));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (((x * x) <= 1.75e-28) || (~(((x * x) <= 5.2e+84)) && ((x * x) <= 3.6e+130)))
		tmp = y * -y;
	else
		tmp = x * x;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[Or[LessEqual[N[(x * x), $MachinePrecision], 1.75e-28], And[N[Not[LessEqual[N[(x * x), $MachinePrecision], 5.2e+84]], $MachinePrecision], LessEqual[N[(x * x), $MachinePrecision], 3.6e+130]]], N[(y * (-y)), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 x x) < 1.75e-28 or 5.2000000000000002e84 < (*.f64 x x) < 3.6000000000000001e130

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 85.7%

      \[\leadsto \color{blue}{-1 \cdot {y}^{2}} \]
   3. Step-by-step derivation

      1. unpow2 85.7%

         \[\leadsto -1 \cdot \color{blue}{\left(y \cdot y\right)} \]

      2. mul-1-neg 85.7%

         \[\leadsto \color{blue}{-y \cdot y} \]

      3. distribute-rgt-neg-in 85.7%

         \[\leadsto \color{blue}{y \cdot \left(-y\right)} \]

   4. Simplified 85.7%

      \[\leadsto \color{blue}{y \cdot \left(-y\right)} \]

   if 1.75e-28 < (*.f64 x x) < 5.2000000000000002e84 or 3.6000000000000001e130 < (*.f64 x x)

   1. Initial program 87.2%

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

   2. Taylor expanded in x around inf 80.4%

      \[\leadsto \color{blue}{{x}^{2}} \]
   3. Step-by-step derivation

      1. unpow2 80.4%

         \[\leadsto \color{blue}{x \cdot x} \]

   4. Simplified 80.4%

      \[\leadsto \color{blue}{x \cdot x} \]
3. Recombined 2 regimes into one program.

4. Final simplification 83.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 1.75 \cdot 10^{-28} \lor \neg \left(x \cdot x \leq 5.2 \cdot 10^{+84}\right) \land x \cdot x \leq 3.6 \cdot 10^{+130}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 4: 53.9% accurate, 2.3× speedup

Math

\[\begin{array}{l} \\ x \cdot x \end{array} \]

FPCore

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

C

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

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x * x
end function

Java

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

Python

def code(x, y):
	return x * x

Julia

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

MATLAB

function tmp = code(x, y)
	tmp = x * x;
end

Wolfram

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

Derivation

1. Initial program 94.1%

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

2. Taylor expanded in x around inf 53.3%

   \[\leadsto \color{blue}{{x}^{2}} \]
3. Step-by-step derivation

   1. unpow2 53.3%

      \[\leadsto \color{blue}{x \cdot x} \]

4. Simplified 53.3%

   \[\leadsto \color{blue}{x \cdot x} \]

5. Final simplification 53.3%

   \[\leadsto x \cdot x \]

Reproduce

herbie shell --seed 2023185 
(FPCore (x y)
  :name "Examples.Basics.BasicTests:f2 from sbv-4.4"
  :precision binary64
  (- (* x x) (* y y)))