sqrt(2*pi)

Percentage Accurate: 100.0% → 100.0%
Time: 4.7s
Alternatives: 1
Speedup: 1.0×

Specification

?
\[0 \leq pi \land pi \leq 1\]
\[\begin{array}{l} \\ \sqrt{2 \cdot pi} \end{array} \]
(FPCore (pi) :precision binary64 (sqrt (* 2.0 pi)))
double code(double pi) {
	return sqrt((2.0 * pi));
}

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

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

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

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

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

code[pi_] := N[Sqrt[N[(2.0 * pi), $MachinePrecision]], $MachinePrecision]

\begin{array}{l}

\\
\sqrt{2 \cdot pi}
\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 1 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: 100.0% accurate, 1.0× speedup?

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

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

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

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

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

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

code[pi_] := N[Sqrt[N[(2.0 * pi), $MachinePrecision]], $MachinePrecision]

\begin{array}{l}

\\
\sqrt{2 \cdot pi}
\end{array}

Alternative 1: 100.0% accurate, 1.0× speedup?

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

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

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

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

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

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

code[pi_] := N[Sqrt[N[(2.0 * pi), $MachinePrecision]], $MachinePrecision]

\begin{array}{l}

\\
\sqrt{2 \cdot pi}
\end{array}
Derivation

  1. Initial program 100.0%

    \[\sqrt{2 \cdot pi} \]
  2. Add Preprocessing
  3. Add Preprocessing

Reproduce

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herbie shell --seed 1 
(FPCore (pi)
  :name "sqrt(2*pi)"
  :precision binary64
  :pre (and (<= 0.0 pi) (<= pi 1.0))
  (sqrt (* 2.0 pi)))