Graphics.Rasterific.QuadraticFormula:discriminant from Rasterific-0.6.1

Percentage Accurate: 98.6% → 99.5%

Time: 3.2s

Alternatives: 5

Speedup: 0.1×

• 33.8% 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: 98.6% 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: 99.5% accurate, 0.1× speedup

Math

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

FPCore

NOTE: x, y, and z should be sorted in increasing order before calling this function.
(FPCore (x y z) :precision binary64 (fma (* z -4.0) y (* x x)))

C

assert(x < y && y < z);
double code(double x, double y, double z) {
	return fma((z * -4.0), y, (x * x));
}

Julia

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

Wolfram

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

Derivation

1. Initial program 97.7%

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

   1. sub-neg 97.7%

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

   2. +-commutative 97.7%

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

   3. distribute-lft-neg-in 97.7%

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

   4. distribute-rgt-neg-in 97.7%

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

   5. metadata-eval 97.7%

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

   6. associate-*r* 97.7%

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

   7. *-commutative 97.7%

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

   8. *-commutative 97.7%

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

   9. fma-def 98.8%

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

   10. pow2 98.8%

       \[\leadsto \mathsf{fma}\left(z \cdot -4, y, \color{blue}{{x}^{2}}\right) \]

3. Applied egg-rr 98.8%

   \[\leadsto \color{blue}{\mathsf{fma}\left(z \cdot -4, y, {x}^{2}\right)} \]
4. Step-by-step derivation

   1. unpow2 98.8%

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

5. Applied egg-rr 98.8%

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

6. Final simplification 98.8%

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

Alternative 2: 99.0% accurate, 0.1× speedup

Math

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

FPCore

NOTE: x, y, and z should be sorted in increasing order before calling this function.
(FPCore (x y z) :precision binary64 (fma x x (* (* z -4.0) y)))

C

assert(x < y && y < z);
double code(double x, double y, double z) {
	return fma(x, x, ((z * -4.0) * y));
}

Julia

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

Wolfram

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

Derivation

1. Initial program 97.7%

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

   1. fma-neg 98.4%

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

   2. associate-*l* 98.4%

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

   3. *-commutative 98.4%

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

   4. distribute-rgt-neg-in 98.4%

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

   5. distribute-rgt-neg-in 98.4%

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

   6. metadata-eval 98.4%

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

3. Simplified 98.4%

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

4. Final simplification 98.4%

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

Alternative 3: 98.5% accurate, 0.7× speedup

Math

\[\begin{array}{l} [x, y, z] = \mathsf{sort}([x, y, z])\\ \\ \begin{array}{l} \mathbf{if}\;x \cdot x \leq 5 \cdot 10^{+239}:\\ \;\;\;\;x \cdot x - z \cdot \left(y \cdot 4\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x, y, and z should be sorted in increasing order before calling this function.
(FPCore (x y z)
 :precision binary64
 (if (<= (* x x) 5e+239) (- (* x x) (* z (* y 4.0))) (* x x)))

C

assert(x < y && y < z);
double code(double x, double y, double z) {
	double tmp;
	if ((x * x) <= 5e+239) {
		tmp = (x * x) - (z * (y * 4.0));
	} else {
		tmp = x * x;
	}
	return tmp;
}

Fortran

NOTE: x, y, and z should be sorted in increasing order 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) <= 5d+239) then
        tmp = (x * x) - (z * (y * 4.0d0))
    else
        tmp = x * x
    end if
    code = tmp
end function

Java

assert x < y && y < z;
public static double code(double x, double y, double z) {
	double tmp;
	if ((x * x) <= 5e+239) {
		tmp = (x * x) - (z * (y * 4.0));
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

[x, y, z] = sort([x, y, z])
def code(x, y, z):
	tmp = 0
	if (x * x) <= 5e+239:
		tmp = (x * x) - (z * (y * 4.0))
	else:
		tmp = x * x
	return tmp

Julia

x, y, z = sort([x, y, z])
function code(x, y, z)
	tmp = 0.0
	if (Float64(x * x) <= 5e+239)
		tmp = Float64(Float64(x * x) - Float64(z * Float64(y * 4.0)));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

MATLAB

x, y, z = num2cell(sort([x, y, z])){:}
function tmp_2 = code(x, y, z)
	tmp = 0.0;
	if ((x * x) <= 5e+239)
		tmp = (x * x) - (z * (y * 4.0));
	else
		tmp = x * x;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x, y, and z should be sorted in increasing order before calling this function.
code[x_, y_, z_] := If[LessEqual[N[(x * x), $MachinePrecision], 5e+239], 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 (*.f64 x x) < 5.00000000000000007e239

   1. Initial program 100.0%

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

   if 5.00000000000000007e239 < (*.f64 x x)

   1. Initial program 92.5%

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

   2. Taylor expanded in x around inf 97.5%

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

      1. unpow2 96.3%

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

   4. Applied egg-rr 97.5%

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

4. Final simplification 99.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 5 \cdot 10^{+239}:\\ \;\;\;\;x \cdot x - z \cdot \left(y \cdot 4\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 4: 84.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} [x, y, z] = \mathsf{sort}([x, y, z])\\ \\ \begin{array}{l} \mathbf{if}\;x \cdot x \leq 2.3 \cdot 10^{+46}:\\ \;\;\;\;-4 \cdot \left(z \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \end{array} \]

FPCore

NOTE: x, y, and z should be sorted in increasing order before calling this function.
(FPCore (x y z)
 :precision binary64
 (if (<= (* x x) 2.3e+46) (* -4.0 (* z y)) (* x x)))

C

assert(x < y && y < z);
double code(double x, double y, double z) {
	double tmp;
	if ((x * x) <= 2.3e+46) {
		tmp = -4.0 * (z * y);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Fortran

NOTE: x, y, and z should be sorted in increasing order 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) <= 2.3d+46) then
        tmp = (-4.0d0) * (z * y)
    else
        tmp = x * x
    end if
    code = tmp
end function

Java

assert x < y && y < z;
public static double code(double x, double y, double z) {
	double tmp;
	if ((x * x) <= 2.3e+46) {
		tmp = -4.0 * (z * y);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Python

[x, y, z] = sort([x, y, z])
def code(x, y, z):
	tmp = 0
	if (x * x) <= 2.3e+46:
		tmp = -4.0 * (z * y)
	else:
		tmp = x * x
	return tmp

Julia

x, y, z = sort([x, y, z])
function code(x, y, z)
	tmp = 0.0
	if (Float64(x * x) <= 2.3e+46)
		tmp = Float64(-4.0 * Float64(z * y));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

MATLAB

x, y, z = num2cell(sort([x, y, z])){:}
function tmp_2 = code(x, y, z)
	tmp = 0.0;
	if ((x * x) <= 2.3e+46)
		tmp = -4.0 * (z * y);
	else
		tmp = x * x;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x, y, and z should be sorted in increasing order before calling this function.
code[x_, y_, z_] := If[LessEqual[N[(x * x), $MachinePrecision], 2.3e+46], N[(-4.0 * N[(z * y), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 x x) < 2.3000000000000001e46

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 87.8%

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

   if 2.3000000000000001e46 < (*.f64 x x)

   1. Initial program 95.0%

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

   2. Taylor expanded in x around inf 91.1%

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

      1. unpow2 97.5%

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

   4. Applied egg-rr 91.1%

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

4. Final simplification 89.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot x \leq 2.3 \cdot 10^{+46}:\\ \;\;\;\;-4 \cdot \left(z \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]

Alternative 5: 52.6% accurate, 3.0× speedup

Math

\[\begin{array}{l} [x, y, z] = \mathsf{sort}([x, y, z])\\ \\ x \cdot x \end{array} \]

FPCore

NOTE: x, y, and z should be sorted in increasing order before calling this function.
(FPCore (x y z) :precision binary64 (* x x))

C

assert(x < y && y < z);
double code(double x, double y, double z) {
	return x * x;
}

Fortran

NOTE: x, y, and z should be sorted in increasing order 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

assert x < y && y < z;
public static double code(double x, double y, double z) {
	return x * x;
}

Python

[x, y, z] = sort([x, y, z])
def code(x, y, z):
	return x * x

Julia

x, y, z = sort([x, y, z])
function code(x, y, z)
	return Float64(x * x)
end

MATLAB

x, y, z = num2cell(sort([x, y, z])){:}
function tmp = code(x, y, z)
	tmp = x * x;
end

Wolfram

NOTE: x, y, and z should be sorted in increasing order before calling this function.
code[x_, y_, z_] := N[(x * x), $MachinePrecision]

Derivation

1. Initial program 97.7%

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

2. Taylor expanded in x around inf 54.6%

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

   1. unpow2 98.8%

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

4. Applied egg-rr 54.6%

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

5. Final simplification 54.6%

   \[\leadsto x \cdot x \]

Reproduce

herbie shell --seed 2023332 
(FPCore (x y z)
  :name "Graphics.Rasterific.QuadraticFormula:discriminant from Rasterific-0.6.1"
  :precision binary64
  (- (* x x) (* (* y 4.0) z)))