Examples.Basics.BasicTests:f2 from sbv-4.4

Percentage Accurate: 94.0% → 98.4%

Time: 2.6s

Alternatives: 4

Speedup: 0.8×

• 43.7% 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: 98.4% accurate, 0.1× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 5 \cdot 10^{+198}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (<= x 5e+198) (fma x x (* y (- y))) (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (x <= 5e+198) {
		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 <= 5e+198)
		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, 5e+198], N[(x * x + N[(y * (-y)), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 5.00000000000000049e198

   1. Initial program 94.5%

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

      1. sqr-neg 94.5%

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

      2. cancel-sign-sub 94.5%

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

      3. fma-def 97.9%

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

   3. Simplified 97.9%

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

   if 5.00000000000000049e198 < x

   1. Initial program 70.0%

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

   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 98.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 5 \cdot 10^{+198}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 2: 96.6% accurate, 0.8× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 1.05 \cdot 10^{+150}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (<= x 1.05e+150) (- (* x x) (* y y)) (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (x <= 1.05e+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 <= 1.05d+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 <= 1.05e+150) {
		tmp = (x * x) - (y * y);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y):
	tmp = 0
	if x <= 1.05e+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 <= 1.05e+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 <= 1.05e+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, 1.05e+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 < 1.04999999999999999e150

   1. Initial program 97.3%

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

   if 1.04999999999999999e150 < x

   1. Initial program 62.9%

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

   2. Taylor expanded in x around inf 82.9%

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

      1. unpow2 82.9%

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

   4. Simplified 82.9%

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

4. Final simplification 95.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 1.05 \cdot 10^{+150}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 3: 78.4% accurate, 0.9× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \cdot x \leq 1.3 \cdot 10^{+39}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (<= (* x x) 1.3e+39) (* y (- y)) (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if ((x * x) <= 1.3e+39) {
		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) <= 1.3d+39) 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) <= 1.3e+39) {
		tmp = y * -y;
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y):
	tmp = 0
	if (x * x) <= 1.3e+39:
		tmp = y * -y
	else:
		tmp = x * x
	return tmp

Julia

x = abs(x)
function code(x, y)
	tmp = 0.0
	if (Float64(x * x) <= 1.3e+39)
		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) <= 1.3e+39)
		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[LessEqual[N[(x * x), $MachinePrecision], 1.3e+39], N[(y * (-y)), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 x x) < 1.3e39

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 82.1%

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

      1. unpow2 82.1%

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

      2. mul-1-neg 82.1%

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

      3. distribute-rgt-neg-in 82.1%

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

   4. Simplified 82.1%

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

   if 1.3e39 < (*.f64 x x)

   1. Initial program 85.2%

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

   2. Taylor expanded in x around inf 78.6%

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

      1. unpow2 78.6%

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

   4. Simplified 78.6%

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

4. Final simplification 80.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 1.3 \cdot 10^{+39}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 4: 53.4% 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 92.6%

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

2. Taylor expanded in x around inf 54.6%

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

   1. unpow2 54.6%

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

4. Simplified 54.6%

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

5. Final simplification 54.6%

   \[\leadsto x \cdot x \]

Reproduce

herbie shell --seed 2023274 
(FPCore (x y)
  :name "Examples.Basics.BasicTests:f2 from sbv-4.4"
  :precision binary64
  (- (* x x) (* y y)))