Kahan p13 Example 2

?

Percentage Accurate: 100.0% → 100.0%
Time: 9.0s
Precision: binary64
Cost: 1728

?

\[\frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)} \]
↓
\[\begin{array}{l} t_1 := \frac{\frac{4}{1 + t} + -8}{1 + t}\\ \frac{5 + t_1}{t_1 + 6} \end{array} \]
(FPCore (t)
 :precision binary64
 (/
  (+
   1.0
   (*
    (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t))))
    (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t))))))
  (+
   2.0
   (*
    (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t))))
    (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t))))))))
↓
(FPCore (t)
 :precision binary64
 (let* ((t_1 (/ (+ (/ 4.0 (+ 1.0 t)) -8.0) (+ 1.0 t))))
   (/ (+ 5.0 t_1) (+ t_1 6.0))))
double code(double t) {
	return (1.0 + ((2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))) * (2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))))) / (2.0 + ((2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))) * (2.0 - ((2.0 / t) / (1.0 + (1.0 / t))))));
}
↓
double code(double t) {
	double t_1 = ((4.0 / (1.0 + t)) + -8.0) / (1.0 + t);
	return (5.0 + t_1) / (t_1 + 6.0);
}

real(8) function code(t)
    real(8), intent (in) :: t
    code = (1.0d0 + ((2.0d0 - ((2.0d0 / t) / (1.0d0 + (1.0d0 / t)))) * (2.0d0 - ((2.0d0 / t) / (1.0d0 + (1.0d0 / t)))))) / (2.0d0 + ((2.0d0 - ((2.0d0 / t) / (1.0d0 + (1.0d0 / t)))) * (2.0d0 - ((2.0d0 / t) / (1.0d0 + (1.0d0 / t))))))
end function
↓
real(8) function code(t)
    real(8), intent (in) :: t
    real(8) :: t_1
    t_1 = ((4.0d0 / (1.0d0 + t)) + (-8.0d0)) / (1.0d0 + t)
    code = (5.0d0 + t_1) / (t_1 + 6.0d0)
end function

public static double code(double t) {
	return (1.0 + ((2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))) * (2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))))) / (2.0 + ((2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))) * (2.0 - ((2.0 / t) / (1.0 + (1.0 / t))))));
}
↓
public static double code(double t) {
	double t_1 = ((4.0 / (1.0 + t)) + -8.0) / (1.0 + t);
	return (5.0 + t_1) / (t_1 + 6.0);
}

def code(t):
	return (1.0 + ((2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))) * (2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))))) / (2.0 + ((2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))) * (2.0 - ((2.0 / t) / (1.0 + (1.0 / t))))))
↓
def code(t):
	t_1 = ((4.0 / (1.0 + t)) + -8.0) / (1.0 + t)
	return (5.0 + t_1) / (t_1 + 6.0)

function code(t)
	return Float64(Float64(1.0 + Float64(Float64(2.0 - Float64(Float64(2.0 / t) / Float64(1.0 + Float64(1.0 / t)))) * Float64(2.0 - Float64(Float64(2.0 / t) / Float64(1.0 + Float64(1.0 / t)))))) / Float64(2.0 + Float64(Float64(2.0 - Float64(Float64(2.0 / t) / Float64(1.0 + Float64(1.0 / t)))) * Float64(2.0 - Float64(Float64(2.0 / t) / Float64(1.0 + Float64(1.0 / t)))))))
end
↓
function code(t)
	t_1 = Float64(Float64(Float64(4.0 / Float64(1.0 + t)) + -8.0) / Float64(1.0 + t))
	return Float64(Float64(5.0 + t_1) / Float64(t_1 + 6.0))
end

function tmp = code(t)
	tmp = (1.0 + ((2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))) * (2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))))) / (2.0 + ((2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))) * (2.0 - ((2.0 / t) / (1.0 + (1.0 / t))))));
end
↓
function tmp = code(t)
	t_1 = ((4.0 / (1.0 + t)) + -8.0) / (1.0 + t);
	tmp = (5.0 + t_1) / (t_1 + 6.0);
end

code[t_] := N[(N[(1.0 + N[(N[(2.0 - N[(N[(2.0 / t), $MachinePrecision] / N[(1.0 + N[(1.0 / t), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 - N[(N[(2.0 / t), $MachinePrecision] / N[(1.0 + N[(1.0 / t), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(2.0 + N[(N[(2.0 - N[(N[(2.0 / t), $MachinePrecision] / N[(1.0 + N[(1.0 / t), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 - N[(N[(2.0 / t), $MachinePrecision] / N[(1.0 + N[(1.0 / t), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
↓
code[t_] := Block[{t$95$1 = N[(N[(N[(4.0 / N[(1.0 + t), $MachinePrecision]), $MachinePrecision] + -8.0), $MachinePrecision] / N[(1.0 + t), $MachinePrecision]), $MachinePrecision]}, N[(N[(5.0 + t$95$1), $MachinePrecision] / N[(t$95$1 + 6.0), $MachinePrecision]), $MachinePrecision]]

\frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)}
↓
\begin{array}{l}
t_1 := \frac{\frac{4}{1 + t} + -8}{1 + t}\\
\frac{5 + t_1}{t_1 + 6}
\end{array}

Local Percentage Accuracy?

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.

Herbie found 6 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

The accuracy (vertical axis) and speed (horizontal axis) of each of Herbie's proposed alternatives. Up and to the right is better. Each dot represents an alternative program; the red square represents the initial program.

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Initial program 100.0%

    \[\frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)} \]

  2. Simplified100.0%

    \[\leadsto \color{blue}{\frac{5 + \frac{\frac{4}{1 + t} + -8}{1 + t}}{6 + \frac{\frac{4}{1 + t} + -8}{1 + t}}} \]
    Step-by-step derivation

    [Start]100.0

    \[ \frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)} \]

  3. Final simplification100.0%

    \[\leadsto \frac{5 + \frac{\frac{4}{1 + t} + -8}{1 + t}}{\frac{\frac{4}{1 + t} + -8}{1 + t} + 6} \]

Alternatives

Alternative 1
Accuracy99.3%
Cost969
\[\begin{array}{l} \mathbf{if}\;t \leq -0.82 \lor \neg \left(t \leq 0.235\right):\\ \;\;\;\;\frac{0.037037037037037035}{t \cdot t} + \left(0.8333333333333334 - \frac{0.2222222222222222}{t}\right)\\ \mathbf{else}:\\ \;\;\;\;t \cdot t + 0.5\\ \end{array} \]
Alternative 2
Accuracy99.2%
Cost585
\[\begin{array}{l} \mathbf{if}\;t \leq -0.78 \lor \neg \left(t \leq 0.58\right):\\ \;\;\;\;0.8333333333333334 - \frac{0.2222222222222222}{t}\\ \mathbf{else}:\\ \;\;\;\;t \cdot t + 0.5\\ \end{array} \]
Alternative 3
Accuracy98.5%
Cost584
\[\begin{array}{l} \mathbf{if}\;t \leq -0.92:\\ \;\;\;\;0.8333333333333334\\ \mathbf{elif}\;t \leq 0.58:\\ \;\;\;\;t \cdot t + 0.5\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334\\ \end{array} \]
Alternative 4
Accuracy98.4%
Cost328
\[\begin{array}{l} \mathbf{if}\;t \leq -0.33:\\ \;\;\;\;0.8333333333333334\\ \mathbf{elif}\;t \leq 1:\\ \;\;\;\;0.5\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334\\ \end{array} \]
Alternative 5
Accuracy58.8%
Cost64
\[0.5 \]

Reproduce?

herbie shell --seed 2023160 
(FPCore (t)
  :name "Kahan p13 Example 2"
  :precision binary64
  (/ (+ 1.0 (* (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t)))) (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t)))))) (+ 2.0 (* (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t)))) (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t))))))))