Examples.Basics.BasicTests:f2 from sbv-4.4

Percentage Accurate: 93.7% → 97.7%

Time: 2.2s

Alternatives: 4

Speedup: 0.8×

• 44.3% 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: 93.7% 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: 97.7% accurate, 0.1× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 1.45 \cdot 10^{+176}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (<= x 1.45e+176) (fma x x (* y (- y))) (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (x <= 1.45e+176) {
		tmp = fma(x, x, (y * -y));
	} else {
		tmp = x * x;
	}
	return tmp;
}

Julia

x = abs(x)
function code(x, y)
	tmp = 0.0
	if (x <= 1.45e+176)
		tmp = fma(x, x, Float64(y * Float64(-y)));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

Wolfram

NOTE: x should be positive before calling this function
code[x_, y_] := If[LessEqual[x, 1.45e+176], N[(x * x + N[(y * (-y)), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 1.4500000000000001e176

   1. Initial program 93.6%

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

      1. fma-neg 96.6%

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

      2. distribute-rgt-neg-in 96.6%

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

   3. Simplified 96.6%

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

   if 1.4500000000000001e176 < x

   1. Initial program 87.0%

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

   2. Taylor expanded in x around inf 100.0%

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

      1. unpow2 100.0%

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

   4. Simplified 100.0%

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

4. Final simplification 96.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 1.45 \cdot 10^{+176}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 2: 77.0% accurate, 0.4× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \cdot x \leq 6.2 \cdot 10^{-62} \lor \neg \left(x \cdot x \leq 1.35 \cdot 10^{+166}\right) \land x \cdot x \leq 4.2 \cdot 10^{+219}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (or (<= (* x x) 6.2e-62)
         (and (not (<= (* x x) 1.35e+166)) (<= (* x x) 4.2e+219)))
   (* y (- y))
   (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (((x * x) <= 6.2e-62) || (!((x * x) <= 1.35e+166) && ((x * x) <= 4.2e+219))) {
		tmp = y * -y;
	} else {
		tmp = x * x;
	}
	return tmp;
}

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (((x * x) <= 6.2d-62) .or. (.not. ((x * x) <= 1.35d+166)) .and. ((x * x) <= 4.2d+219)) then
        tmp = y * -y
    else
        tmp = x * x
    end if
    code = tmp
end function

Java

x = Math.abs(x);
public static double code(double x, double y) {
	double tmp;
	if (((x * x) <= 6.2e-62) || (!((x * x) <= 1.35e+166) && ((x * x) <= 4.2e+219))) {
		tmp = y * -y;
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y):
	tmp = 0
	if ((x * x) <= 6.2e-62) or (not ((x * x) <= 1.35e+166) and ((x * x) <= 4.2e+219)):
		tmp = y * -y
	else:
		tmp = x * x
	return tmp

Julia

x = abs(x)
function code(x, y)
	tmp = 0.0
	if ((Float64(x * x) <= 6.2e-62) || (!(Float64(x * x) <= 1.35e+166) && (Float64(x * x) <= 4.2e+219)))
		tmp = Float64(y * Float64(-y));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

MATLAB

x = abs(x)
function tmp_2 = code(x, y)
	tmp = 0.0;
	if (((x * x) <= 6.2e-62) || (~(((x * x) <= 1.35e+166)) && ((x * x) <= 4.2e+219)))
		tmp = y * -y;
	else
		tmp = x * x;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_, y_] := If[Or[LessEqual[N[(x * x), $MachinePrecision], 6.2e-62], And[N[Not[LessEqual[N[(x * x), $MachinePrecision], 1.35e+166]], $MachinePrecision], LessEqual[N[(x * x), $MachinePrecision], 4.2e+219]]], N[(y * (-y)), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 x x) < 6.1999999999999999e-62 or 1.35000000000000006e166 < (*.f64 x x) < 4.19999999999999976e219

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 83.2%

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

      1. unpow2 83.2%

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

      2. mul-1-neg 83.2%

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

      3. distribute-rgt-neg-in 83.2%

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

   4. Simplified 83.2%

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

   if 6.1999999999999999e-62 < (*.f64 x x) < 1.35000000000000006e166 or 4.19999999999999976e219 < (*.f64 x x)

   1. Initial program 85.2%

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

   2. Taylor expanded in x around inf 81.4%

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

      1. unpow2 81.4%

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

   4. Simplified 81.4%

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

4. Final simplification 82.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 6.2 \cdot 10^{-62} \lor \neg \left(x \cdot x \leq 1.35 \cdot 10^{+166}\right) \land x \cdot x \leq 4.2 \cdot 10^{+219}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 3: 96.5% accurate, 0.8× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 1.42 \cdot 10^{+150}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (<= x 1.42e+150) (- (* x x) (* y y)) (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (x <= 1.42e+150) {
		tmp = (x * x) - (y * y);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (x <= 1.42d+150) then
        tmp = (x * x) - (y * y)
    else
        tmp = x * x
    end if
    code = tmp
end function

Java

x = Math.abs(x);
public static double code(double x, double y) {
	double tmp;
	if (x <= 1.42e+150) {
		tmp = (x * x) - (y * y);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y):
	tmp = 0
	if x <= 1.42e+150:
		tmp = (x * x) - (y * y)
	else:
		tmp = x * x
	return tmp

Julia

x = abs(x)
function code(x, y)
	tmp = 0.0
	if (x <= 1.42e+150)
		tmp = Float64(Float64(x * x) - Float64(y * y));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

MATLAB

x = abs(x)
function tmp_2 = code(x, y)
	tmp = 0.0;
	if (x <= 1.42e+150)
		tmp = (x * x) - (y * y);
	else
		tmp = x * x;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_, y_] := If[LessEqual[x, 1.42e+150], N[(N[(x * x), $MachinePrecision] - N[(y * y), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 1.42e150

   1. Initial program 94.2%

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

   if 1.42e150 < x

   1. Initial program 84.8%

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

   2. Taylor expanded in x around inf 97.0%

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

      1. unpow2 97.0%

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

   4. Simplified 97.0%

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

4. Final simplification 94.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 1.42 \cdot 10^{+150}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 4: 53.2% accurate, 2.3× speedup

Math

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

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y) :precision binary64 (* x x))

C

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

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x * x
end function

Java

x = Math.abs(x);
public static double code(double x, double y) {
	return x * x;
}

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: x should be positive before calling this function
code[x_, y_] := N[(x * x), $MachinePrecision]

Derivation

1. Initial program 93.0%

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

2. Taylor expanded in x around inf 52.9%

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

   1. unpow2 52.9%

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

4. Simplified 52.9%

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

5. Final simplification 52.9%

   \[\leadsto x \cdot x \]

Reproduce

herbie shell --seed 2023228 
(FPCore (x y)
  :name "Examples.Basics.BasicTests:f2 from sbv-4.4"
  :precision binary64
  (- (* x x) (* y y)))