Examples.Basics.BasicTests:f2 from sbv-4.4

Percentage Accurate: 94.1% → 98.6%

Time: 2.2s

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: 94.1% 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: 98.6% accurate, 0.1× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 5 \cdot 10^{+253}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (<= x 5e+253) (fma x x (* y (- y))) (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (x <= 5e+253) {
		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+253)
		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+253], N[(x * x + N[(y * (-y)), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 4.9999999999999997e253

   1. Initial program 95.9%

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

      1. sqr-neg 95.9%

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

      2. cancel-sign-sub 95.9%

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

      3. fma-def 99.2%

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

   3. Simplified 99.2%

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

   if 4.9999999999999997e253 < x

   1. Initial program 72.7%

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

   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 99.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 5 \cdot 10^{+253}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 2: 75.6% accurate, 0.4× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \cdot x \leq 5.4 \cdot 10^{-138} \lor \neg \left(x \cdot x \leq 108000000\right) \land x \cdot x \leq 2.4 \cdot 10^{+117}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (or (<= (* x x) 5.4e-138)
         (and (not (<= (* x x) 108000000.0)) (<= (* x x) 2.4e+117)))
   (* y (- y))
   (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (((x * x) <= 5.4e-138) || (!((x * x) <= 108000000.0) && ((x * x) <= 2.4e+117))) {
		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) <= 5.4d-138) .or. (.not. ((x * x) <= 108000000.0d0)) .and. ((x * x) <= 2.4d+117)) 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) <= 5.4e-138) || (!((x * x) <= 108000000.0) && ((x * x) <= 2.4e+117))) {
		tmp = y * -y;
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y):
	tmp = 0
	if ((x * x) <= 5.4e-138) or (not ((x * x) <= 108000000.0) and ((x * x) <= 2.4e+117)):
		tmp = y * -y
	else:
		tmp = x * x
	return tmp

Julia

x = abs(x)
function code(x, y)
	tmp = 0.0
	if ((Float64(x * x) <= 5.4e-138) || (!(Float64(x * x) <= 108000000.0) && (Float64(x * x) <= 2.4e+117)))
		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) <= 5.4e-138) || (~(((x * x) <= 108000000.0)) && ((x * x) <= 2.4e+117)))
		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], 5.4e-138], And[N[Not[LessEqual[N[(x * x), $MachinePrecision], 108000000.0]], $MachinePrecision], LessEqual[N[(x * x), $MachinePrecision], 2.4e+117]]], N[(y * (-y)), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 x x) < 5.40000000000000057e-138 or 1.08e8 < (*.f64 x x) < 2.3999999999999999e117

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 87.8%

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

      1. unpow2 87.8%

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

      2. mul-1-neg 87.8%

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

      3. distribute-rgt-neg-in 87.8%

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

   4. Simplified 87.8%

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

   if 5.40000000000000057e-138 < (*.f64 x x) < 1.08e8 or 2.3999999999999999e117 < (*.f64 x x)

   1. Initial program 90.6%

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

   2. Taylor expanded in x around inf 78.2%

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

      1. unpow2 78.2%

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

   4. Simplified 78.2%

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

4. Final simplification 82.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 5.4 \cdot 10^{-138} \lor \neg \left(x \cdot x \leq 108000000\right) \land x \cdot x \leq 2.4 \cdot 10^{+117}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 3: 96.9% accurate, 0.6× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;y \cdot y \leq 5 \cdot 10^{+298}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (<= (* y y) 5e+298) (- (* x x) (* y y)) (* y (- y))))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if ((y * y) <= 5e+298) {
		tmp = (x * x) - (y * y);
	} else {
		tmp = y * -y;
	}
	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 ((y * y) <= 5d+298) then
        tmp = (x * x) - (y * y)
    else
        tmp = y * -y
    end if
    code = tmp
end function

Java

x = Math.abs(x);
public static double code(double x, double y) {
	double tmp;
	if ((y * y) <= 5e+298) {
		tmp = (x * x) - (y * y);
	} else {
		tmp = y * -y;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y):
	tmp = 0
	if (y * y) <= 5e+298:
		tmp = (x * x) - (y * y)
	else:
		tmp = y * -y
	return tmp

Julia

x = abs(x)
function code(x, y)
	tmp = 0.0
	if (Float64(y * y) <= 5e+298)
		tmp = Float64(Float64(x * x) - Float64(y * y));
	else
		tmp = Float64(y * Float64(-y));
	end
	return tmp
end

MATLAB

x = abs(x)
function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((y * y) <= 5e+298)
		tmp = (x * x) - (y * y);
	else
		tmp = y * -y;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_, y_] := If[LessEqual[N[(y * y), $MachinePrecision], 5e+298], N[(N[(x * x), $MachinePrecision] - N[(y * y), $MachinePrecision]), $MachinePrecision], N[(y * (-y)), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 y y) < 5.0000000000000003e298

   1. Initial program 100.0%

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

   if 5.0000000000000003e298 < (*.f64 y y)

   1. Initial program 77.2%

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

   2. Taylor expanded in x around 0 91.2%

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

      1. unpow2 91.2%

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

      2. mul-1-neg 91.2%

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

      3. distribute-rgt-neg-in 91.2%

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

   4. Simplified 91.2%

      \[\leadsto \color{blue}{y \cdot \left(-y\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 98.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \cdot y \leq 5 \cdot 10^{+298}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \end{array} \]

Alternative 4: 53.5% 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 94.9%

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

2. Taylor expanded in x around inf 54.7%

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

   1. unpow2 54.7%

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

4. Simplified 54.7%

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

5. Final simplification 54.7%

   \[\leadsto x \cdot x \]

Reproduce

herbie shell --seed 2023271 
(FPCore (x y)
  :name "Examples.Basics.BasicTests:f2 from sbv-4.4"
  :precision binary64
  (- (* x x) (* y y)))