sqrt C (should all be same)

Percentage Accurate: 54.5% → 99.5%
Time: 2.6s
Alternatives: 2
Speedup: 1.0×

Specification

?
\[\begin{array}{l} \\ \sqrt{2 \cdot \left(x \cdot x\right)} \end{array} \]
(FPCore (x) :precision binary64 (sqrt (* 2.0 (* x x))))
double code(double x) {
	return sqrt((2.0 * (x * x)));
}

real(8) function code(x)
    real(8), intent (in) :: x
    code = sqrt((2.0d0 * (x * x)))
end function

public static double code(double x) {
	return Math.sqrt((2.0 * (x * x)));
}

def code(x):
	return math.sqrt((2.0 * (x * x)))

function code(x)
	return sqrt(Float64(2.0 * Float64(x * x)))
end

function tmp = code(x)
	tmp = sqrt((2.0 * (x * x)));
end

code[x_] := N[Sqrt[N[(2.0 * N[(x * x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]

\begin{array}{l}

\\
\sqrt{2 \cdot \left(x \cdot x\right)}
\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 2 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: 54.5% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \sqrt{2 \cdot \left(x \cdot x\right)} \end{array} \]
(FPCore (x) :precision binary64 (sqrt (* 2.0 (* x x))))
double code(double x) {
	return sqrt((2.0 * (x * x)));
}

real(8) function code(x)
    real(8), intent (in) :: x
    code = sqrt((2.0d0 * (x * x)))
end function

public static double code(double x) {
	return Math.sqrt((2.0 * (x * x)));
}

def code(x):
	return math.sqrt((2.0 * (x * x)))

function code(x)
	return sqrt(Float64(2.0 * Float64(x * x)))
end

function tmp = code(x)
	tmp = sqrt((2.0 * (x * x)));
end

code[x_] := N[Sqrt[N[(2.0 * N[(x * x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]

\begin{array}{l}

\\
\sqrt{2 \cdot \left(x \cdot x\right)}
\end{array}

Alternative 1: 99.5% accurate, 0.5× speedup?

\[\begin{array}{l} x = |x|\\ \\ \sqrt{2 \cdot x} \cdot \sqrt{x} \end{array} \]
NOTE: x should be positive before calling this function
(FPCore (x) :precision binary64 (* (sqrt (* 2.0 x)) (sqrt x)))
x = abs(x);
double code(double x) {
	return sqrt((2.0 * x)) * sqrt(x);
}

NOTE: x should be positive before calling this function
real(8) function code(x)
    real(8), intent (in) :: x
    code = sqrt((2.0d0 * x)) * sqrt(x)
end function

x = Math.abs(x);
public static double code(double x) {
	return Math.sqrt((2.0 * x)) * Math.sqrt(x);
}

x = abs(x)
def code(x):
	return math.sqrt((2.0 * x)) * math.sqrt(x)

x = abs(x)
function code(x)
	return Float64(sqrt(Float64(2.0 * x)) * sqrt(x))
end

x = abs(x)
function tmp = code(x)
	tmp = sqrt((2.0 * x)) * sqrt(x);
end

NOTE: x should be positive before calling this function
code[x_] := N[(N[Sqrt[N[(2.0 * x), $MachinePrecision]], $MachinePrecision] * N[Sqrt[x], $MachinePrecision]), $MachinePrecision]

\begin{array}{l}
x = |x|\\
\\
\sqrt{2 \cdot x} \cdot \sqrt{x}
\end{array}
Derivation

  1. Initial program 55.2%

    \[\sqrt{2 \cdot \left(x \cdot x\right)} \]
  2. Step-by-step derivation

    1. associate-*r*55.2%

      \[\leadsto \sqrt{\color{blue}{\left(2 \cdot x\right) \cdot x}} \]

    2. sqrt-prod50.5%

      \[\leadsto \color{blue}{\sqrt{2 \cdot x} \cdot \sqrt{x}} \]

  3. Applied egg-rr50.5%

    \[\leadsto \color{blue}{\sqrt{2 \cdot x} \cdot \sqrt{x}} \]

  4. Final simplification50.5%

    \[\leadsto \sqrt{2 \cdot x} \cdot \sqrt{x} \]

Alternative 2: 99.3% accurate, 1.0× speedup?

\[\begin{array}{l} x = |x|\\ \\ x \cdot \sqrt{2} \end{array} \]
NOTE: x should be positive before calling this function
(FPCore (x) :precision binary64 (* x (sqrt 2.0)))
x = abs(x);
double code(double x) {
	return x * sqrt(2.0);
}

NOTE: x should be positive before calling this function
real(8) function code(x)
    real(8), intent (in) :: x
    code = x * sqrt(2.0d0)
end function

x = Math.abs(x);
public static double code(double x) {
	return x * Math.sqrt(2.0);
}

x = abs(x)
def code(x):
	return x * math.sqrt(2.0)

x = abs(x)
function code(x)
	return Float64(x * sqrt(2.0))
end

x = abs(x)
function tmp = code(x)
	tmp = x * sqrt(2.0);
end

NOTE: x should be positive before calling this function
code[x_] := N[(x * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]

\begin{array}{l}
x = |x|\\
\\
x \cdot \sqrt{2}
\end{array}
Derivation

  1. Initial program 55.2%

    \[\sqrt{2 \cdot \left(x \cdot x\right)} \]

  2. Taylor expanded in x around 0 51.5%

    \[\leadsto \color{blue}{\sqrt{2} \cdot x} \]

  3. Final simplification51.5%

    \[\leadsto x \cdot \sqrt{2} \]

Reproduce

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herbie shell --seed 2023229 
(FPCore (x)
  :name "sqrt C (should all be same)"
  :precision binary64
  (sqrt (* 2.0 (* x x))))