Graphics.Rasterific.QuadraticFormula:discriminant from Rasterific-0.6.1

Percentage Accurate: 98.4% → 99.6%

Time: 2.3s

Alternatives: 5

Speedup: 0.8×

• 33.7% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ x \cdot x - \left(y \cdot 4\right) \cdot z \end{array} \]

FPCore

(FPCore (x y z) :precision binary64 (- (* x x) (* (* y 4.0) z)))

C

double code(double x, double y, double z) {
	return (x * x) - ((y * 4.0) * z);
}

Fortran

real(8) function code(x, y, z)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8), intent (in) :: z
    code = (x * x) - ((y * 4.0d0) * z)
end function

Java

public static double code(double x, double y, double z) {
	return (x * x) - ((y * 4.0) * z);
}

Python

def code(x, y, z):
	return (x * x) - ((y * 4.0) * z)

Julia

function code(x, y, z)
	return Float64(Float64(x * x) - Float64(Float64(y * 4.0) * z))
end

MATLAB

function tmp = code(x, y, z)
	tmp = (x * x) - ((y * 4.0) * z);
end

Wolfram

code[x_, y_, z_] := N[(N[(x * x), $MachinePrecision] - N[(N[(y * 4.0), $MachinePrecision] * z), $MachinePrecision]), $MachinePrecision]

Initial Program: 98.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ x \cdot x - \left(y \cdot 4\right) \cdot z \end{array} \]

FPCore

(FPCore (x y z) :precision binary64 (- (* x x) (* (* y 4.0) z)))

C

double code(double x, double y, double z) {
	return (x * x) - ((y * 4.0) * z);
}

Fortran

real(8) function code(x, y, z)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8), intent (in) :: z
    code = (x * x) - ((y * 4.0d0) * z)
end function

Java

public static double code(double x, double y, double z) {
	return (x * x) - ((y * 4.0) * z);
}

Python

def code(x, y, z):
	return (x * x) - ((y * 4.0) * z)

Julia

function code(x, y, z)
	return Float64(Float64(x * x) - Float64(Float64(y * 4.0) * z))
end

MATLAB

function tmp = code(x, y, z)
	tmp = (x * x) - ((y * 4.0) * z);
end

Wolfram

code[x_, y_, z_] := N[(N[(x * x), $MachinePrecision] - N[(N[(y * 4.0), $MachinePrecision] * z), $MachinePrecision]), $MachinePrecision]

Alternative 1: 99.6% accurate, 0.1× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 10^{+196}:\\ \;\;\;\;\mathsf{fma}\left(x, x, z \cdot \left(y \cdot -4\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y z)
 :precision binary64
 (if (<= x 1e+196) (fma x x (* z (* y -4.0))) (* x x)))

C

x = abs(x);
double code(double x, double y, double z) {
	double tmp;
	if (x <= 1e+196) {
		tmp = fma(x, x, (z * (y * -4.0)));
	} else {
		tmp = x * x;
	}
	return tmp;
}

Julia

x = abs(x)
function code(x, y, z)
	tmp = 0.0
	if (x <= 1e+196)
		tmp = fma(x, x, Float64(z * Float64(y * -4.0)));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

Wolfram

NOTE: x should be positive before calling this function
code[x_, y_, z_] := If[LessEqual[x, 1e+196], N[(x * x + N[(z * N[(y * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 9.9999999999999995e195

   1. Initial program 99.1%

      \[x \cdot x - \left(y \cdot 4\right) \cdot z \]
   2. Step-by-step derivation

      1. fma-neg 99.6%

         \[\leadsto \color{blue}{\mathsf{fma}\left(x, x, -\left(y \cdot 4\right) \cdot z\right)} \]

      2. *-commutative 99.6%

         \[\leadsto \mathsf{fma}\left(x, x, -\color{blue}{z \cdot \left(y \cdot 4\right)}\right) \]

      3. distribute-rgt-neg-in 99.6%

         \[\leadsto \mathsf{fma}\left(x, x, \color{blue}{z \cdot \left(-y \cdot 4\right)}\right) \]

      4. distribute-rgt-neg-in 99.6%

         \[\leadsto \mathsf{fma}\left(x, x, z \cdot \color{blue}{\left(y \cdot \left(-4\right)\right)}\right) \]

      5. metadata-eval 99.6%

         \[\leadsto \mathsf{fma}\left(x, x, z \cdot \left(y \cdot \color{blue}{-4}\right)\right) \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\mathsf{fma}\left(x, x, z \cdot \left(y \cdot -4\right)\right)} \]

   if 9.9999999999999995e195 < x

   1. Initial program 86.7%

      \[x \cdot x - \left(y \cdot 4\right) \cdot z \]

   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 99.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 10^{+196}:\\ \;\;\;\;\mathsf{fma}\left(x, x, z \cdot \left(y \cdot -4\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 2: 99.5% accurate, 0.1× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \mathsf{fma}\left(y, z \cdot -4, x \cdot x\right) \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y z) :precision binary64 (fma y (* z -4.0) (* x x)))

C

x = abs(x);
double code(double x, double y, double z) {
	return fma(y, (z * -4.0), (x * x));
}

Julia

x = abs(x)
function code(x, y, z)
	return fma(y, Float64(z * -4.0), Float64(x * x))
end

Wolfram

NOTE: x should be positive before calling this function
code[x_, y_, z_] := N[(y * N[(z * -4.0), $MachinePrecision] + N[(x * x), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 97.6%

   \[x \cdot x - \left(y \cdot 4\right) \cdot z \]
2. Step-by-step derivation

   1. sub-neg 97.6%

      \[\leadsto \color{blue}{x \cdot x + \left(-\left(y \cdot 4\right) \cdot z\right)} \]

   2. +-commutative 97.6%

      \[\leadsto \color{blue}{\left(-\left(y \cdot 4\right) \cdot z\right) + x \cdot x} \]

   3. associate-*l* 97.6%

      \[\leadsto \left(-\color{blue}{y \cdot \left(4 \cdot z\right)}\right) + x \cdot x \]

   4. distribute-rgt-neg-in 97.6%

      \[\leadsto \color{blue}{y \cdot \left(-4 \cdot z\right)} + x \cdot x \]

   5. fma-def 99.2%

      \[\leadsto \color{blue}{\mathsf{fma}\left(y, -4 \cdot z, x \cdot x\right)} \]

   6. *-commutative 99.2%

      \[\leadsto \mathsf{fma}\left(y, -\color{blue}{z \cdot 4}, x \cdot x\right) \]

   7. distribute-rgt-neg-in 99.2%

      \[\leadsto \mathsf{fma}\left(y, \color{blue}{z \cdot \left(-4\right)}, x \cdot x\right) \]

   8. metadata-eval 99.2%

      \[\leadsto \mathsf{fma}\left(y, z \cdot \color{blue}{-4}, x \cdot x\right) \]

3. Simplified 99.2%

   \[\leadsto \color{blue}{\mathsf{fma}\left(y, z \cdot -4, x \cdot x\right)} \]

4. Final simplification 99.2%

   \[\leadsto \mathsf{fma}\left(y, z \cdot -4, x \cdot x\right) \]

Alternative 3: 99.3% accurate, 0.8× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 6.8 \cdot 10^{+145}:\\ \;\;\;\;x \cdot x - z \cdot \left(y \cdot 4\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y z)
 :precision binary64
 (if (<= x 6.8e+145) (- (* x x) (* z (* y 4.0))) (* x x)))

C

x = abs(x);
double code(double x, double y, double z) {
	double tmp;
	if (x <= 6.8e+145) {
		tmp = (x * x) - (z * (y * 4.0));
	} else {
		tmp = x * x;
	}
	return tmp;
}

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x, y, z)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8), intent (in) :: z
    real(8) :: tmp
    if (x <= 6.8d+145) then
        tmp = (x * x) - (z * (y * 4.0d0))
    else
        tmp = x * x
    end if
    code = tmp
end function

Java

x = Math.abs(x);
public static double code(double x, double y, double z) {
	double tmp;
	if (x <= 6.8e+145) {
		tmp = (x * x) - (z * (y * 4.0));
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y, z):
	tmp = 0
	if x <= 6.8e+145:
		tmp = (x * x) - (z * (y * 4.0))
	else:
		tmp = x * x
	return tmp

Julia

x = abs(x)
function code(x, y, z)
	tmp = 0.0
	if (x <= 6.8e+145)
		tmp = Float64(Float64(x * x) - Float64(z * Float64(y * 4.0)));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

MATLAB

x = abs(x)
function tmp_2 = code(x, y, z)
	tmp = 0.0;
	if (x <= 6.8e+145)
		tmp = (x * x) - (z * (y * 4.0));
	else
		tmp = x * x;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_, y_, z_] := If[LessEqual[x, 6.8e+145], N[(N[(x * x), $MachinePrecision] - N[(z * N[(y * 4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 6.7999999999999998e145

   1. Initial program 99.5%

      \[x \cdot x - \left(y \cdot 4\right) \cdot z \]

   if 6.7999999999999998e145 < x

   1. Initial program 86.8%

      \[x \cdot x - \left(y \cdot 4\right) \cdot z \]

   2. Taylor expanded in x around inf 97.4%

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

      1. unpow2 97.4%

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

   4. Simplified 97.4%

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

4. Final simplification 99.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 6.8 \cdot 10^{+145}:\\ \;\;\;\;x \cdot x - z \cdot \left(y \cdot 4\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 4: 83.9% accurate, 1.0× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \cdot x \leq 1.8 \cdot 10^{-107}:\\ \;\;\;\;-4 \cdot \left(y \cdot z\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x y z)
 :precision binary64
 (if (<= (* x x) 1.8e-107) (* -4.0 (* y z)) (* x x)))

C

x = abs(x);
double code(double x, double y, double z) {
	double tmp;
	if ((x * x) <= 1.8e-107) {
		tmp = -4.0 * (y * z);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x, y, z)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8), intent (in) :: z
    real(8) :: tmp
    if ((x * x) <= 1.8d-107) then
        tmp = (-4.0d0) * (y * z)
    else
        tmp = x * x
    end if
    code = tmp
end function

Java

x = Math.abs(x);
public static double code(double x, double y, double z) {
	double tmp;
	if ((x * x) <= 1.8e-107) {
		tmp = -4.0 * (y * z);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

x = abs(x)
def code(x, y, z):
	tmp = 0
	if (x * x) <= 1.8e-107:
		tmp = -4.0 * (y * z)
	else:
		tmp = x * x
	return tmp

Julia

x = abs(x)
function code(x, y, z)
	tmp = 0.0
	if (Float64(x * x) <= 1.8e-107)
		tmp = Float64(-4.0 * Float64(y * z));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

MATLAB

x = abs(x)
function tmp_2 = code(x, y, z)
	tmp = 0.0;
	if ((x * x) <= 1.8e-107)
		tmp = -4.0 * (y * z);
	else
		tmp = x * x;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_, y_, z_] := If[LessEqual[N[(x * x), $MachinePrecision], 1.8e-107], N[(-4.0 * N[(y * z), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 x x) < 1.79999999999999988e-107

   1. Initial program 100.0%

      \[x \cdot x - \left(y \cdot 4\right) \cdot z \]

   2. Taylor expanded in x around 0 91.3%

      \[\leadsto \color{blue}{-4 \cdot \left(y \cdot z\right)} \]

   if 1.79999999999999988e-107 < (*.f64 x x)

   1. Initial program 96.0%

      \[x \cdot x - \left(y \cdot 4\right) \cdot z \]

   2. Taylor expanded in x around inf 82.4%

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

      1. unpow2 82.4%

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

   4. Simplified 82.4%

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

4. Final simplification 86.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 1.8 \cdot 10^{-107}:\\ \;\;\;\;-4 \cdot \left(y \cdot z\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 5: 52.7% accurate, 3.0× 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 z) :precision binary64 (* x x))

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 97.6%

   \[x \cdot x - \left(y \cdot 4\right) \cdot z \]

2. Taylor expanded in x around inf 55.6%

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

   1. unpow2 55.6%

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

4. Simplified 55.6%

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

5. Final simplification 55.6%

   \[\leadsto x \cdot x \]

Reproduce

herbie shell --seed 2023221 
(FPCore (x y z)
  :name "Graphics.Rasterific.QuadraticFormula:discriminant from Rasterific-0.6.1"
  :precision binary64
  (- (* x x) (* (* y 4.0) z)))