Kahan p13 Example 1

Percentage Accurate: 99.9% → 99.9%
Time: 2.2s
Alternatives: 1
Speedup: 1.0×

Specification

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\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{2 \cdot t}{1 + t}\\ t_2 := t\_1 \cdot t\_1\\ \frac{1 + t\_2}{2 + t\_2} \end{array} \end{array} \]
(FPCore (t)
 :precision binary64
 (let* ((t_1 (/ (* 2.0 t) (+ 1.0 t))) (t_2 (* t_1 t_1)))
   (/ (+ 1.0 t_2) (+ 2.0 t_2))))
double code(double t) {
	double t_1 = (2.0 * t) / (1.0 + t);
	double t_2 = t_1 * t_1;
	return (1.0 + t_2) / (2.0 + t_2);
}

real(8) function code(t)
    real(8), intent (in) :: t
    real(8) :: t_1
    real(8) :: t_2
    t_1 = (2.0d0 * t) / (1.0d0 + t)
    t_2 = t_1 * t_1
    code = (1.0d0 + t_2) / (2.0d0 + t_2)
end function

public static double code(double t) {
	double t_1 = (2.0 * t) / (1.0 + t);
	double t_2 = t_1 * t_1;
	return (1.0 + t_2) / (2.0 + t_2);
}

def code(t):
	t_1 = (2.0 * t) / (1.0 + t)
	t_2 = t_1 * t_1
	return (1.0 + t_2) / (2.0 + t_2)

function code(t)
	t_1 = Float64(Float64(2.0 * t) / Float64(1.0 + t))
	t_2 = Float64(t_1 * t_1)
	return Float64(Float64(1.0 + t_2) / Float64(2.0 + t_2))
end

function tmp = code(t)
	t_1 = (2.0 * t) / (1.0 + t);
	t_2 = t_1 * t_1;
	tmp = (1.0 + t_2) / (2.0 + t_2);
end

code[t_] := Block[{t$95$1 = N[(N[(2.0 * t), $MachinePrecision] / N[(1.0 + t), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(t$95$1 * t$95$1), $MachinePrecision]}, N[(N[(1.0 + t$95$2), $MachinePrecision] / N[(2.0 + t$95$2), $MachinePrecision]), $MachinePrecision]]]

\begin{array}{l}

\\
\begin{array}{l}
t_1 := \frac{2 \cdot t}{1 + t}\\
t_2 := t\_1 \cdot t\_1\\
\frac{1 + t\_2}{2 + t\_2}
\end{array}
\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: 99.9% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{2 \cdot t}{1 + t}\\ t_2 := t\_1 \cdot t\_1\\ \frac{1 + t\_2}{2 + t\_2} \end{array} \end{array} \]
(FPCore (t)
 :precision binary64
 (let* ((t_1 (/ (* 2.0 t) (+ 1.0 t))) (t_2 (* t_1 t_1)))
   (/ (+ 1.0 t_2) (+ 2.0 t_2))))
double code(double t) {
	double t_1 = (2.0 * t) / (1.0 + t);
	double t_2 = t_1 * t_1;
	return (1.0 + t_2) / (2.0 + t_2);
}

real(8) function code(t)
    real(8), intent (in) :: t
    real(8) :: t_1
    real(8) :: t_2
    t_1 = (2.0d0 * t) / (1.0d0 + t)
    t_2 = t_1 * t_1
    code = (1.0d0 + t_2) / (2.0d0 + t_2)
end function

public static double code(double t) {
	double t_1 = (2.0 * t) / (1.0 + t);
	double t_2 = t_1 * t_1;
	return (1.0 + t_2) / (2.0 + t_2);
}

def code(t):
	t_1 = (2.0 * t) / (1.0 + t)
	t_2 = t_1 * t_1
	return (1.0 + t_2) / (2.0 + t_2)

function code(t)
	t_1 = Float64(Float64(2.0 * t) / Float64(1.0 + t))
	t_2 = Float64(t_1 * t_1)
	return Float64(Float64(1.0 + t_2) / Float64(2.0 + t_2))
end

function tmp = code(t)
	t_1 = (2.0 * t) / (1.0 + t);
	t_2 = t_1 * t_1;
	tmp = (1.0 + t_2) / (2.0 + t_2);
end

code[t_] := Block[{t$95$1 = N[(N[(2.0 * t), $MachinePrecision] / N[(1.0 + t), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(t$95$1 * t$95$1), $MachinePrecision]}, N[(N[(1.0 + t$95$2), $MachinePrecision] / N[(2.0 + t$95$2), $MachinePrecision]), $MachinePrecision]]]

\begin{array}{l}

\\
\begin{array}{l}
t_1 := \frac{2 \cdot t}{1 + t}\\
t_2 := t\_1 \cdot t\_1\\
\frac{1 + t\_2}{2 + t\_2}
\end{array}
\end{array}

Alternative 1: 99.9% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{2 \cdot t}{1 + t}\\ t_2 := t\_1 \cdot t\_1\\ \frac{1 + t\_2}{2 + t\_2} \end{array} \end{array} \]
(FPCore (t)
 :precision binary64
 (let* ((t_1 (/ (* 2.0 t) (+ 1.0 t))) (t_2 (* t_1 t_1)))
   (/ (+ 1.0 t_2) (+ 2.0 t_2))))
double code(double t) {
	double t_1 = (2.0 * t) / (1.0 + t);
	double t_2 = t_1 * t_1;
	return (1.0 + t_2) / (2.0 + t_2);
}

real(8) function code(t)
    real(8), intent (in) :: t
    real(8) :: t_1
    real(8) :: t_2
    t_1 = (2.0d0 * t) / (1.0d0 + t)
    t_2 = t_1 * t_1
    code = (1.0d0 + t_2) / (2.0d0 + t_2)
end function

public static double code(double t) {
	double t_1 = (2.0 * t) / (1.0 + t);
	double t_2 = t_1 * t_1;
	return (1.0 + t_2) / (2.0 + t_2);
}

def code(t):
	t_1 = (2.0 * t) / (1.0 + t)
	t_2 = t_1 * t_1
	return (1.0 + t_2) / (2.0 + t_2)

function code(t)
	t_1 = Float64(Float64(2.0 * t) / Float64(1.0 + t))
	t_2 = Float64(t_1 * t_1)
	return Float64(Float64(1.0 + t_2) / Float64(2.0 + t_2))
end

function tmp = code(t)
	t_1 = (2.0 * t) / (1.0 + t);
	t_2 = t_1 * t_1;
	tmp = (1.0 + t_2) / (2.0 + t_2);
end

code[t_] := Block[{t$95$1 = N[(N[(2.0 * t), $MachinePrecision] / N[(1.0 + t), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(t$95$1 * t$95$1), $MachinePrecision]}, N[(N[(1.0 + t$95$2), $MachinePrecision] / N[(2.0 + t$95$2), $MachinePrecision]), $MachinePrecision]]]

\begin{array}{l}

\\
\begin{array}{l}
t_1 := \frac{2 \cdot t}{1 + t}\\
t_2 := t\_1 \cdot t\_1\\
\frac{1 + t\_2}{2 + t\_2}
\end{array}
\end{array}
Derivation

  1. Initial program 100.0%

    \[\frac{1 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}}{2 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}} \]
  2. Add Preprocessing

Reproduce

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herbie shell --seed 2024185 -o setup:simplify
(FPCore (t)
  :name "Kahan p13 Example 1"
  :precision binary64
  (/ (+ 1.0 (* (/ (* 2.0 t) (+ 1.0 t)) (/ (* 2.0 t) (+ 1.0 t)))) (+ 2.0 (* (/ (* 2.0 t) (+ 1.0 t)) (/ (* 2.0 t) (+ 1.0 t))))))