Rump's expression from Stadtherr's award speech

Percentage Accurate: 1.4% → 1.4%
Time: 1.3s
Alternatives: 6
Speedup: 1.0×

Specification

?
\[x = 77617 \land y = 33096\]
\[\begin{array}{l} \\ -0.8273960599468214 \end{array} \]
(FPCore (x y) :precision binary64 -0.8273960599468214)
double code(double x, double y) {
	return -0.8273960599468214;
}

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = -0.8273960599468214d0
end function

public static double code(double x, double y) {
	return -0.8273960599468214;
}

def code(x, y):
	return -0.8273960599468214

function code(x, y)
	return -0.8273960599468214
end

function tmp = code(x, y)
	tmp = -0.8273960599468214;
end

code[x_, y_] := -0.8273960599468214

\begin{array}{l}

\\
-0.8273960599468214
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 6 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 1.4% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \left(\left(333.75 \cdot {y}^{6} + \left(x \cdot x\right) \cdot \left(\left(\left(\left(\left(\left(11 \cdot x\right) \cdot x\right) \cdot y\right) \cdot y - {y}^{6}\right) - 121 \cdot {y}^{4}\right) - 2\right)\right) + 5.5 \cdot {y}^{8}\right) + \frac{x}{2 \cdot y} \end{array} \]
(FPCore (x y)
 :precision binary64
 (+
  (+
   (+
    (* 333.75 (pow y 6.0))
    (*
     (* x x)
     (-
      (- (- (* (* (* (* 11.0 x) x) y) y) (pow y 6.0)) (* 121.0 (pow y 4.0)))
      2.0)))
   (* 5.5 (pow y 8.0)))
  (/ x (* 2.0 y))))
double code(double x, double y) {
	return (((333.75 * pow(y, 6.0)) + ((x * x) * (((((((11.0 * x) * x) * y) * y) - pow(y, 6.0)) - (121.0 * pow(y, 4.0))) - 2.0))) + (5.5 * pow(y, 8.0))) + (x / (2.0 * y));
}

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = (((333.75d0 * (y ** 6.0d0)) + ((x * x) * (((((((11.0d0 * x) * x) * y) * y) - (y ** 6.0d0)) - (121.0d0 * (y ** 4.0d0))) - 2.0d0))) + (5.5d0 * (y ** 8.0d0))) + (x / (2.0d0 * y))
end function

public static double code(double x, double y) {
	return (((333.75 * Math.pow(y, 6.0)) + ((x * x) * (((((((11.0 * x) * x) * y) * y) - Math.pow(y, 6.0)) - (121.0 * Math.pow(y, 4.0))) - 2.0))) + (5.5 * Math.pow(y, 8.0))) + (x / (2.0 * y));
}

def code(x, y):
	return (((333.75 * math.pow(y, 6.0)) + ((x * x) * (((((((11.0 * x) * x) * y) * y) - math.pow(y, 6.0)) - (121.0 * math.pow(y, 4.0))) - 2.0))) + (5.5 * math.pow(y, 8.0))) + (x / (2.0 * y))

function code(x, y)
	return Float64(Float64(Float64(Float64(333.75 * (y ^ 6.0)) + Float64(Float64(x * x) * Float64(Float64(Float64(Float64(Float64(Float64(Float64(11.0 * x) * x) * y) * y) - (y ^ 6.0)) - Float64(121.0 * (y ^ 4.0))) - 2.0))) + Float64(5.5 * (y ^ 8.0))) + Float64(x / Float64(2.0 * y)))
end

function tmp = code(x, y)
	tmp = (((333.75 * (y ^ 6.0)) + ((x * x) * (((((((11.0 * x) * x) * y) * y) - (y ^ 6.0)) - (121.0 * (y ^ 4.0))) - 2.0))) + (5.5 * (y ^ 8.0))) + (x / (2.0 * y));
end

code[x_, y_] := N[(N[(N[(N[(333.75 * N[Power[y, 6.0], $MachinePrecision]), $MachinePrecision] + N[(N[(x * x), $MachinePrecision] * N[(N[(N[(N[(N[(N[(N[(11.0 * x), $MachinePrecision] * x), $MachinePrecision] * y), $MachinePrecision] * y), $MachinePrecision] - N[Power[y, 6.0], $MachinePrecision]), $MachinePrecision] - N[(121.0 * N[Power[y, 4.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(5.5 * N[Power[y, 8.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(x / N[(2.0 * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
\left(\left(333.75 \cdot {y}^{6} + \left(x \cdot x\right) \cdot \left(\left(\left(\left(\left(\left(11 \cdot x\right) \cdot x\right) \cdot y\right) \cdot y - {y}^{6}\right) - 121 \cdot {y}^{4}\right) - 2\right)\right) + 5.5 \cdot {y}^{8}\right) + \frac{x}{2 \cdot y}
\end{array}

Alternative 1: 1.4% accurate, 1.0× speedup?

\[\begin{array}{l} x = |x|\\ y = |y|\\ [x, y] = \mathsf{sort}([x, y])\\ \\ \left(\left(333.75 \cdot {y}^{6} + \left(x \cdot x\right) \cdot \left(\left(\left(\left(\left(\left(11 \cdot x\right) \cdot x\right) \cdot y\right) \cdot y - {y}^{6}\right) - 121 \cdot {y}^{4}\right) - 2\right)\right) + 5.5 \cdot {y}^{8}\right) + \frac{x}{2 \cdot y} \end{array} \]
NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (+
  (+
   (+
    (* 333.75 (pow y 6.0))
    (*
     (* x x)
     (-
      (- (- (* (* (* (* 11.0 x) x) y) y) (pow y 6.0)) (* 121.0 (pow y 4.0)))
      2.0)))
   (* 5.5 (pow y 8.0)))
  (/ x (* 2.0 y))))
x = abs(x);
y = abs(y);
assert(x < y);
double code(double x, double y) {
	return (((333.75 * pow(y, 6.0)) + ((x * x) * (((((((11.0 * x) * x) * y) * y) - pow(y, 6.0)) - (121.0 * pow(y, 4.0))) - 2.0))) + (5.5 * pow(y, 8.0))) + (x / (2.0 * y));
}

NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = (((333.75d0 * (y ** 6.0d0)) + ((x * x) * (((((((11.0d0 * x) * x) * y) * y) - (y ** 6.0d0)) - (121.0d0 * (y ** 4.0d0))) - 2.0d0))) + (5.5d0 * (y ** 8.0d0))) + (x / (2.0d0 * y))
end function

x = Math.abs(x);
y = Math.abs(y);
assert x < y;
public static double code(double x, double y) {
	return (((333.75 * Math.pow(y, 6.0)) + ((x * x) * (((((((11.0 * x) * x) * y) * y) - Math.pow(y, 6.0)) - (121.0 * Math.pow(y, 4.0))) - 2.0))) + (5.5 * Math.pow(y, 8.0))) + (x / (2.0 * y));
}

x = abs(x)
y = abs(y)
[x, y] = sort([x, y])
def code(x, y):
	return (((333.75 * math.pow(y, 6.0)) + ((x * x) * (((((((11.0 * x) * x) * y) * y) - math.pow(y, 6.0)) - (121.0 * math.pow(y, 4.0))) - 2.0))) + (5.5 * math.pow(y, 8.0))) + (x / (2.0 * y))

x = abs(x)
y = abs(y)
x, y = sort([x, y])
function code(x, y)
	return Float64(Float64(Float64(Float64(333.75 * (y ^ 6.0)) + Float64(Float64(x * x) * Float64(Float64(Float64(Float64(Float64(Float64(Float64(11.0 * x) * x) * y) * y) - (y ^ 6.0)) - Float64(121.0 * (y ^ 4.0))) - 2.0))) + Float64(5.5 * (y ^ 8.0))) + Float64(x / Float64(2.0 * y)))
end

x = abs(x)
y = abs(y)
x, y = num2cell(sort([x, y])){:}
function tmp = code(x, y)
	tmp = (((333.75 * (y ^ 6.0)) + ((x * x) * (((((((11.0 * x) * x) * y) * y) - (y ^ 6.0)) - (121.0 * (y ^ 4.0))) - 2.0))) + (5.5 * (y ^ 8.0))) + (x / (2.0 * y));
end

NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := N[(N[(N[(N[(333.75 * N[Power[y, 6.0], $MachinePrecision]), $MachinePrecision] + N[(N[(x * x), $MachinePrecision] * N[(N[(N[(N[(N[(N[(N[(11.0 * x), $MachinePrecision] * x), $MachinePrecision] * y), $MachinePrecision] * y), $MachinePrecision] - N[Power[y, 6.0], $MachinePrecision]), $MachinePrecision] - N[(121.0 * N[Power[y, 4.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(5.5 * N[Power[y, 8.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(x / N[(2.0 * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}
x = |x|\\
y = |y|\\
[x, y] = \mathsf{sort}([x, y])\\
\\
\left(\left(333.75 \cdot {y}^{6} + \left(x \cdot x\right) \cdot \left(\left(\left(\left(\left(\left(11 \cdot x\right) \cdot x\right) \cdot y\right) \cdot y - {y}^{6}\right) - 121 \cdot {y}^{4}\right) - 2\right)\right) + 5.5 \cdot {y}^{8}\right) + \frac{x}{2 \cdot y}
\end{array}
Derivation

  1. Initial program 9.2%

    \[\left(\left(333.75 \cdot {y}^{6} + \left(x \cdot x\right) \cdot \left(\left(\left(\left(\left(\left(11 \cdot x\right) \cdot x\right) \cdot y\right) \cdot y - {y}^{6}\right) - 121 \cdot {y}^{4}\right) - 2\right)\right) + 5.5 \cdot {y}^{8}\right) + \frac{x}{2 \cdot y} \]

Alternative 2: 1.4% accurate, 0.8× speedup?

\[\begin{array}{l} x = |x|\\ y = |y|\\ [x, y] = \mathsf{sort}([x, y])\\ \\ \mathsf{fma}\left(333.75, {y}^{6}, \left(x \cdot x\right) \cdot \left(\left(y \cdot \left(y \cdot \left(x \cdot \left(x \cdot 11\right)\right)\right) - {y}^{6}\right) - \left(121 \cdot {y}^{4} + 2\right)\right)\right) + \left(5.5 \cdot {y}^{8} + \frac{x}{y \cdot 2}\right) \end{array} \]
NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (+
  (fma
   333.75
   (pow y 6.0)
   (*
    (* x x)
    (-
     (- (* y (* y (* x (* x 11.0)))) (pow y 6.0))
     (+ (* 121.0 (pow y 4.0)) 2.0))))
  (+ (* 5.5 (pow y 8.0)) (/ x (* y 2.0)))))
x = abs(x);
y = abs(y);
assert(x < y);
double code(double x, double y) {
	return fma(333.75, pow(y, 6.0), ((x * x) * (((y * (y * (x * (x * 11.0)))) - pow(y, 6.0)) - ((121.0 * pow(y, 4.0)) + 2.0)))) + ((5.5 * pow(y, 8.0)) + (x / (y * 2.0)));
}

x = abs(x)
y = abs(y)
x, y = sort([x, y])
function code(x, y)
	return Float64(fma(333.75, (y ^ 6.0), Float64(Float64(x * x) * Float64(Float64(Float64(y * Float64(y * Float64(x * Float64(x * 11.0)))) - (y ^ 6.0)) - Float64(Float64(121.0 * (y ^ 4.0)) + 2.0)))) + Float64(Float64(5.5 * (y ^ 8.0)) + Float64(x / Float64(y * 2.0))))
end

NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := N[(N[(333.75 * N[Power[y, 6.0], $MachinePrecision] + N[(N[(x * x), $MachinePrecision] * N[(N[(N[(y * N[(y * N[(x * N[(x * 11.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - N[Power[y, 6.0], $MachinePrecision]), $MachinePrecision] - N[(N[(121.0 * N[Power[y, 4.0], $MachinePrecision]), $MachinePrecision] + 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(N[(5.5 * N[Power[y, 8.0], $MachinePrecision]), $MachinePrecision] + N[(x / N[(y * 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}
x = |x|\\
y = |y|\\
[x, y] = \mathsf{sort}([x, y])\\
\\
\mathsf{fma}\left(333.75, {y}^{6}, \left(x \cdot x\right) \cdot \left(\left(y \cdot \left(y \cdot \left(x \cdot \left(x \cdot 11\right)\right)\right) - {y}^{6}\right) - \left(121 \cdot {y}^{4} + 2\right)\right)\right) + \left(5.5 \cdot {y}^{8} + \frac{x}{y \cdot 2}\right)
\end{array}
Derivation

  1. Initial program 9.2%

    \[\mathsf{fma}\left(333.75, {y}^{6}, \left(x \cdot x\right) \cdot \left(\left(y \cdot \left(y \cdot \left(x \cdot \left(x \cdot 11\right)\right)\right) - {y}^{6}\right) - \left(121 \cdot {y}^{4} + 2\right)\right)\right) + \left(5.5 \cdot {y}^{8} + \frac{x}{y \cdot 2}\right) \]

Alternative 3: 1.4% accurate, 0.5× speedup?

\[\begin{array}{l} x = |x|\\ y = |y|\\ [x, y] = \mathsf{sort}([x, y])\\ \\ \mathsf{fma}\left(333.75, {y}^{6}, x \cdot \left(x \cdot \left(\mathsf{fma}\left(y, \left(y \cdot 11\right) \cdot \left(x \cdot x\right), -{y}^{6}\right) - \mathsf{fma}\left(121, {y}^{4}, 2\right)\right)\right)\right) + \mathsf{fma}\left(5.5, {y}^{8}, \frac{x}{y \cdot 2}\right) \end{array} \]
NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (+
  (fma
   333.75
   (pow y 6.0)
   (*
    x
    (*
     x
     (-
      (fma y (* (* y 11.0) (* x x)) (- (pow y 6.0)))
      (fma 121.0 (pow y 4.0) 2.0)))))
  (fma 5.5 (pow y 8.0) (/ x (* y 2.0)))))
x = abs(x);
y = abs(y);
assert(x < y);
double code(double x, double y) {
	return fma(333.75, pow(y, 6.0), (x * (x * (fma(y, ((y * 11.0) * (x * x)), -pow(y, 6.0)) - fma(121.0, pow(y, 4.0), 2.0))))) + fma(5.5, pow(y, 8.0), (x / (y * 2.0)));
}

x = abs(x)
y = abs(y)
x, y = sort([x, y])
function code(x, y)
	return Float64(fma(333.75, (y ^ 6.0), Float64(x * Float64(x * Float64(fma(y, Float64(Float64(y * 11.0) * Float64(x * x)), Float64(-(y ^ 6.0))) - fma(121.0, (y ^ 4.0), 2.0))))) + fma(5.5, (y ^ 8.0), Float64(x / Float64(y * 2.0))))
end

NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := N[(N[(333.75 * N[Power[y, 6.0], $MachinePrecision] + N[(x * N[(x * N[(N[(y * N[(N[(y * 11.0), $MachinePrecision] * N[(x * x), $MachinePrecision]), $MachinePrecision] + (-N[Power[y, 6.0], $MachinePrecision])), $MachinePrecision] - N[(121.0 * N[Power[y, 4.0], $MachinePrecision] + 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(5.5 * N[Power[y, 8.0], $MachinePrecision] + N[(x / N[(y * 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}
x = |x|\\
y = |y|\\
[x, y] = \mathsf{sort}([x, y])\\
\\
\mathsf{fma}\left(333.75, {y}^{6}, x \cdot \left(x \cdot \left(\mathsf{fma}\left(y, \left(y \cdot 11\right) \cdot \left(x \cdot x\right), -{y}^{6}\right) - \mathsf{fma}\left(121, {y}^{4}, 2\right)\right)\right)\right) + \mathsf{fma}\left(5.5, {y}^{8}, \frac{x}{y \cdot 2}\right)
\end{array}
Derivation

  1. Initial program 3.1%

    \[\mathsf{fma}\left(333.75, {y}^{6}, x \cdot \left(x \cdot \left(\mathsf{fma}\left(y, \left(y \cdot 11\right) \cdot \left(x \cdot x\right), -{y}^{6}\right) - \mathsf{fma}\left(121, {y}^{4}, 2\right)\right)\right)\right) + \mathsf{fma}\left(5.5, {y}^{8}, \frac{x}{y \cdot 2}\right) \]

Alternative 4: 1.4% accurate, 0.4× speedup?

\[\begin{array}{l} x = |x|\\ y = |y|\\ [x, y] = \mathsf{sort}([x, y])\\ \\ \mathsf{fma}\left(333.75, {y}^{6}, \mathsf{fma}\left(x \cdot x, \mathsf{fma}\left(y, y \cdot \left(x \cdot \left(x \cdot 11\right)\right), \mathsf{fma}\left(-1, {y}^{6}, \mathsf{fma}\left({y}^{4}, -121, -2\right)\right)\right), \mathsf{fma}\left(5.5, {y}^{8}, \frac{x}{y \cdot 2}\right)\right)\right) \end{array} \]
NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (fma
  333.75
  (pow y 6.0)
  (fma
   (* x x)
   (fma
    y
    (* y (* x (* x 11.0)))
    (fma -1.0 (pow y 6.0) (fma (pow y 4.0) -121.0 -2.0)))
   (fma 5.5 (pow y 8.0) (/ x (* y 2.0))))))
x = abs(x);
y = abs(y);
assert(x < y);
double code(double x, double y) {
	return fma(333.75, pow(y, 6.0), fma((x * x), fma(y, (y * (x * (x * 11.0))), fma(-1.0, pow(y, 6.0), fma(pow(y, 4.0), -121.0, -2.0))), fma(5.5, pow(y, 8.0), (x / (y * 2.0)))));
}

x = abs(x)
y = abs(y)
x, y = sort([x, y])
function code(x, y)
	return fma(333.75, (y ^ 6.0), fma(Float64(x * x), fma(y, Float64(y * Float64(x * Float64(x * 11.0))), fma(-1.0, (y ^ 6.0), fma((y ^ 4.0), -121.0, -2.0))), fma(5.5, (y ^ 8.0), Float64(x / Float64(y * 2.0)))))
end

NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := N[(333.75 * N[Power[y, 6.0], $MachinePrecision] + N[(N[(x * x), $MachinePrecision] * N[(y * N[(y * N[(x * N[(x * 11.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(-1.0 * N[Power[y, 6.0], $MachinePrecision] + N[(N[Power[y, 4.0], $MachinePrecision] * -121.0 + -2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(5.5 * N[Power[y, 8.0], $MachinePrecision] + N[(x / N[(y * 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}
x = |x|\\
y = |y|\\
[x, y] = \mathsf{sort}([x, y])\\
\\
\mathsf{fma}\left(333.75, {y}^{6}, \mathsf{fma}\left(x \cdot x, \mathsf{fma}\left(y, y \cdot \left(x \cdot \left(x \cdot 11\right)\right), \mathsf{fma}\left(-1, {y}^{6}, \mathsf{fma}\left({y}^{4}, -121, -2\right)\right)\right), \mathsf{fma}\left(5.5, {y}^{8}, \frac{x}{y \cdot 2}\right)\right)\right)
\end{array}
Derivation

  1. Initial program 9.2%

    \[\mathsf{fma}\left(333.75, {y}^{6}, \mathsf{fma}\left(x \cdot x, \mathsf{fma}\left(y, y \cdot \left(x \cdot \left(x \cdot 11\right)\right), \mathsf{fma}\left(-1, {y}^{6}, \mathsf{fma}\left({y}^{4}, -121, -2\right)\right)\right), \mathsf{fma}\left(5.5, {y}^{8}, \frac{x}{y \cdot 2}\right)\right)\right) \]

Alternative 5: 1.4% accurate, 0.5× speedup?

\[\begin{array}{l} x = |x|\\ y = |y|\\ [x, y] = \mathsf{sort}([x, y])\\ \\ \mathsf{fma}\left(333.75, {y}^{6}, \mathsf{fma}\left(x \cdot x, \mathsf{fma}\left(y \cdot 11, y \cdot \left(x \cdot x\right), \mathsf{fma}\left({y}^{4}, -121, -2\right)\right) - {y}^{6}, \mathsf{fma}\left(5.5, {y}^{8}, \frac{\frac{x}{2}}{y}\right)\right)\right) \end{array} \]
NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (fma
  333.75
  (pow y 6.0)
  (fma
   (* x x)
   (- (fma (* y 11.0) (* y (* x x)) (fma (pow y 4.0) -121.0 -2.0)) (pow y 6.0))
   (fma 5.5 (pow y 8.0) (/ (/ x 2.0) y)))))
x = abs(x);
y = abs(y);
assert(x < y);
double code(double x, double y) {
	return fma(333.75, pow(y, 6.0), fma((x * x), (fma((y * 11.0), (y * (x * x)), fma(pow(y, 4.0), -121.0, -2.0)) - pow(y, 6.0)), fma(5.5, pow(y, 8.0), ((x / 2.0) / y))));
}

x = abs(x)
y = abs(y)
x, y = sort([x, y])
function code(x, y)
	return fma(333.75, (y ^ 6.0), fma(Float64(x * x), Float64(fma(Float64(y * 11.0), Float64(y * Float64(x * x)), fma((y ^ 4.0), -121.0, -2.0)) - (y ^ 6.0)), fma(5.5, (y ^ 8.0), Float64(Float64(x / 2.0) / y))))
end

NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := N[(333.75 * N[Power[y, 6.0], $MachinePrecision] + N[(N[(x * x), $MachinePrecision] * N[(N[(N[(y * 11.0), $MachinePrecision] * N[(y * N[(x * x), $MachinePrecision]), $MachinePrecision] + N[(N[Power[y, 4.0], $MachinePrecision] * -121.0 + -2.0), $MachinePrecision]), $MachinePrecision] - N[Power[y, 6.0], $MachinePrecision]), $MachinePrecision] + N[(5.5 * N[Power[y, 8.0], $MachinePrecision] + N[(N[(x / 2.0), $MachinePrecision] / y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}
x = |x|\\
y = |y|\\
[x, y] = \mathsf{sort}([x, y])\\
\\
\mathsf{fma}\left(333.75, {y}^{6}, \mathsf{fma}\left(x \cdot x, \mathsf{fma}\left(y \cdot 11, y \cdot \left(x \cdot x\right), \mathsf{fma}\left({y}^{4}, -121, -2\right)\right) - {y}^{6}, \mathsf{fma}\left(5.5, {y}^{8}, \frac{\frac{x}{2}}{y}\right)\right)\right)
\end{array}
Derivation

  1. Initial program 1.5%

    \[\mathsf{fma}\left(333.75, {y}^{6}, \mathsf{fma}\left(x \cdot x, \mathsf{fma}\left(y \cdot 11, y \cdot \left(x \cdot x\right), \mathsf{fma}\left({y}^{4}, -121, -2\right)\right) - {y}^{6}, \mathsf{fma}\left(5.5, {y}^{8}, \frac{\frac{x}{2}}{y}\right)\right)\right) \]

Alternative 6: 1.4% accurate, 0.5× speedup?

\[\begin{array}{l} x = |x|\\ y = |y|\\ [x, y] = \mathsf{sort}([x, y])\\ \\ \mathsf{fma}\left(333.75, {y}^{6}, \mathsf{fma}\left(x \cdot x, y \cdot \left(y \cdot \left(x \cdot \left(x \cdot 11\right)\right)\right) - \mathsf{fma}\left({y}^{4}, \mathsf{fma}\left(y, y, 121\right), 2\right), \mathsf{fma}\left(5.5, {y}^{8}, \frac{x}{y \cdot 2}\right)\right)\right) \end{array} \]
NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (fma
  333.75
  (pow y 6.0)
  (fma
   (* x x)
   (- (* y (* y (* x (* x 11.0)))) (fma (pow y 4.0) (fma y y 121.0) 2.0))
   (fma 5.5 (pow y 8.0) (/ x (* y 2.0))))))
x = abs(x);
y = abs(y);
assert(x < y);
double code(double x, double y) {
	return fma(333.75, pow(y, 6.0), fma((x * x), ((y * (y * (x * (x * 11.0)))) - fma(pow(y, 4.0), fma(y, y, 121.0), 2.0)), fma(5.5, pow(y, 8.0), (x / (y * 2.0)))));
}

x = abs(x)
y = abs(y)
x, y = sort([x, y])
function code(x, y)
	return fma(333.75, (y ^ 6.0), fma(Float64(x * x), Float64(Float64(y * Float64(y * Float64(x * Float64(x * 11.0)))) - fma((y ^ 4.0), fma(y, y, 121.0), 2.0)), fma(5.5, (y ^ 8.0), Float64(x / Float64(y * 2.0)))))
end

NOTE: x should be positive before calling this function
NOTE: y should be positive before calling this function
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := N[(333.75 * N[Power[y, 6.0], $MachinePrecision] + N[(N[(x * x), $MachinePrecision] * N[(N[(y * N[(y * N[(x * N[(x * 11.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - N[(N[Power[y, 4.0], $MachinePrecision] * N[(y * y + 121.0), $MachinePrecision] + 2.0), $MachinePrecision]), $MachinePrecision] + N[(5.5 * N[Power[y, 8.0], $MachinePrecision] + N[(x / N[(y * 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}
x = |x|\\
y = |y|\\
[x, y] = \mathsf{sort}([x, y])\\
\\
\mathsf{fma}\left(333.75, {y}^{6}, \mathsf{fma}\left(x \cdot x, y \cdot \left(y \cdot \left(x \cdot \left(x \cdot 11\right)\right)\right) - \mathsf{fma}\left({y}^{4}, \mathsf{fma}\left(y, y, 121\right), 2\right), \mathsf{fma}\left(5.5, {y}^{8}, \frac{x}{y \cdot 2}\right)\right)\right)
\end{array}
Derivation

  1. Initial program 1.5%

    \[\mathsf{fma}\left(333.75, {y}^{6}, \mathsf{fma}\left(x \cdot x, y \cdot \left(y \cdot \left(x \cdot \left(x \cdot 11\right)\right)\right) - \mathsf{fma}\left({y}^{4}, \mathsf{fma}\left(y, y, 121\right), 2\right), \mathsf{fma}\left(5.5, {y}^{8}, \frac{x}{y \cdot 2}\right)\right)\right) \]

Reproduce

?
herbie shell --seed 2023275 
(FPCore (x y)
  :name "Rump's expression from Stadtherr's award speech"
  :precision binary64
  :pre (and (== x 77617.0) (== y 33096.0))
  (+ (+ (+ (* 333.75 (pow y 6.0)) (* (* x x) (- (- (- (* (* (* (* 11.0 x) x) y) y) (pow y 6.0)) (* 121.0 (pow y 4.0))) 2.0))) (* 5.5 (pow y 8.0))) (/ x (* 2.0 y))))