Examples.Basics.BasicTests:f2 from sbv-4.4

Percentage Accurate: 94.1% → 98.5%

Time: 4.1s

Alternatives: 4

Speedup: 0.8×

• 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: 98.5% accurate, 0.1× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 3.5 \cdot 10^{+217}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (<= x 3.5e+217) (fma x x (* y (- y))) (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (x <= 3.5e+217) {
		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 <= 3.5e+217)
		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, 3.5e+217], N[(x * x + N[(y * (-y)), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 3.4999999999999998e217

   1. Initial program 96.6%

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

      1. fma-neg 99.2%

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

      2. distribute-rgt-neg-in 99.2%

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

   3. Simplified 99.2%

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

   if 3.4999999999999998e217 < x

   1. Initial program 58.8%

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

   2. Taylor expanded in x around inf 94.1%

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

      1. unpow2 94.1%

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

   4. Simplified 94.1%

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

4. Final simplification 98.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 3.5 \cdot 10^{+217}:\\ \;\;\;\;\mathsf{fma}\left(x, x, y \cdot \left(-y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 2: 76.7% accurate, 0.4× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \cdot x \leq 1.05 \cdot 10^{-127} \lor \neg \left(x \cdot x \leq 1.5 \cdot 10^{-5}\right) \land x \cdot x \leq 1.85 \cdot 10^{+82}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (or (<= (* x x) 1.05e-127)
         (and (not (<= (* x x) 1.5e-5)) (<= (* x x) 1.85e+82)))
   (* y (- y))
   (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (((x * x) <= 1.05e-127) || (!((x * x) <= 1.5e-5) && ((x * x) <= 1.85e+82))) {
		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) <= 1.05d-127) .or. (.not. ((x * x) <= 1.5d-5)) .and. ((x * x) <= 1.85d+82)) 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) <= 1.05e-127) || (!((x * x) <= 1.5e-5) && ((x * x) <= 1.85e+82))) {
		tmp = y * -y;
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y):
	tmp = 0
	if ((x * x) <= 1.05e-127) or (not ((x * x) <= 1.5e-5) and ((x * x) <= 1.85e+82)):
		tmp = y * -y
	else:
		tmp = x * x
	return tmp

Julia

x = abs(x)
function code(x, y)
	tmp = 0.0
	if ((Float64(x * x) <= 1.05e-127) || (!(Float64(x * x) <= 1.5e-5) && (Float64(x * x) <= 1.85e+82)))
		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) <= 1.05e-127) || (~(((x * x) <= 1.5e-5)) && ((x * x) <= 1.85e+82)))
		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], 1.05e-127], And[N[Not[LessEqual[N[(x * x), $MachinePrecision], 1.5e-5]], $MachinePrecision], LessEqual[N[(x * x), $MachinePrecision], 1.85e+82]]], N[(y * (-y)), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 x x) < 1.05000000000000005e-127 or 1.50000000000000004e-5 < (*.f64 x x) < 1.8500000000000001e82

   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 1.05000000000000005e-127 < (*.f64 x x) < 1.50000000000000004e-5 or 1.8500000000000001e82 < (*.f64 x x)

   1. Initial program 89.0%

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

   2. Taylor expanded in x around inf 75.4%

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

      1. unpow2 75.4%

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

   4. Simplified 75.4%

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

4. Final simplification 80.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 1.05 \cdot 10^{-127} \lor \neg \left(x \cdot x \leq 1.5 \cdot 10^{-5}\right) \land x \cdot x \leq 1.85 \cdot 10^{+82}:\\ \;\;\;\;y \cdot \left(-y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 3: 97.1% accurate, 0.8× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 1.32 \cdot 10^{+154}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y)
 :precision binary64
 (if (<= x 1.32e+154) (- (* x x) (* y y)) (* x x)))

C

x = abs(x);
double code(double x, double y) {
	double tmp;
	if (x <= 1.32e+154) {
		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 <= 1.32d+154) 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 <= 1.32e+154) {
		tmp = (x * x) - (y * y);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y):
	tmp = 0
	if x <= 1.32e+154:
		tmp = (x * x) - (y * y)
	else:
		tmp = x * x
	return tmp

Julia

x = abs(x)
function code(x, y)
	tmp = 0.0
	if (x <= 1.32e+154)
		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 <= 1.32e+154)
		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, 1.32e+154], N[(N[(x * x), $MachinePrecision] - N[(y * y), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 1.31999999999999998e154

   1. Initial program 97.4%

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

   if 1.31999999999999998e154 < x

   1. Initial program 64.0%

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

   2. Taylor expanded in x around inf 88.0%

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

      1. unpow2 88.0%

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

   4. Simplified 88.0%

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

4. Final simplification 96.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 1.32 \cdot 10^{+154}:\\ \;\;\;\;x \cdot x - y \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 4: 53.6% 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 94.1%

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

2. Taylor expanded in x around inf 51.5%

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

   1. unpow2 51.5%

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

4. Simplified 51.5%

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

5. Final simplification 51.5%

   \[\leadsto x \cdot x \]

Reproduce

herbie shell --seed 2023257 
(FPCore (x y)
  :name "Examples.Basics.BasicTests:f2 from sbv-4.4"
  :precision binary64
  (- (* x x) (* y y)))