Kahan p13 Example 2

Percentage Accurate: 100.0% → 100.0%

Time: 14.1s

Precision: binary64

Cost: 3264

Math

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

↓

\[\begin{array}{l} t_1 := 2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\\ t_2 := t_1 \cdot t_1\\ \frac{1 + t_2}{2 + t_2} \end{array} \]

FPCore

(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 (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t))))) (t_2 (* t_1 t_1)))
   (/ (+ 1.0 t_2) (+ 2.0 t_2))))

C

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 = 2.0 - ((2.0 / t) / (1.0 + (1.0 / t)));
	double t_2 = t_1 * t_1;
	return (1.0 + t_2) / (2.0 + t_2);
}

Fortran

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

Java

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 = 2.0 - ((2.0 / t) / (1.0 + (1.0 / t)));
	double t_2 = t_1 * t_1;
	return (1.0 + t_2) / (2.0 + t_2);
}

Python

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 = 2.0 - ((2.0 / t) / (1.0 + (1.0 / t)))
	t_2 = t_1 * t_1
	return (1.0 + t_2) / (2.0 + t_2)

Julia

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

MATLAB

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

Wolfram

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[(2.0 - N[(N[(2.0 / t), $MachinePrecision] / N[(1.0 + N[(1.0 / t), $MachinePrecision]), $MachinePrecision]), $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]]]

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. Final simplification 100.0%

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

Alternatives

Alternative 1
Accuracy	100.0%
Cost	3264

\[\begin{array}{l} t_1 := 2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\\ t_2 := t_1 \cdot t_1\\ \frac{1 + t_2}{2 + t_2} \end{array} \]

Alternative 2
Accuracy	100.0%
Cost	3008

\[\begin{array}{l} t_1 := 2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\\ \frac{1 + t_1 \cdot t_1}{2 + t_1 \cdot \left(2 + \frac{-2}{1 + t}\right)} \end{array} \]

Alternative 3
Accuracy	100.0%
Cost	1728

\[\begin{array}{l} t_1 := \frac{\frac{4}{1 + t} + -8}{1 + t}\\ \frac{5 + t_1}{t_1 + 6} \end{array} \]

Alternative 4
Accuracy	99.4%
Cost	1609

\[\begin{array}{l} \mathbf{if}\;t \leq -0.6 \lor \neg \left(t \leq 0.56\right):\\ \;\;\;\;\frac{0.037037037037037035}{t \cdot t} + \left(0.8333333333333334 - \frac{0.2222222222222222}{t}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1 + \left(2 + \frac{-2}{1 + t}\right) \cdot \left(2 \cdot t\right)}{2 + 4 \cdot \left(t \cdot t\right)}\\ \end{array} \]

Alternative 5
Accuracy	97.8%
Cost	1344

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

Alternative 6
Accuracy	99.3%
Cost	1225

\[\begin{array}{l} t_1 := 4 \cdot \left(t \cdot t\right)\\ \mathbf{if}\;t \leq -0.65 \lor \neg \left(t \leq 0.42\right):\\ \;\;\;\;\frac{0.037037037037037035}{t \cdot t} + \left(0.8333333333333334 - \frac{0.2222222222222222}{t}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1 + t_1}{2 + t_1}\\ \end{array} \]

Alternative 7
Accuracy	99.3%
Cost	969

\[\begin{array}{l} \mathbf{if}\;t \leq -0.82 \lor \neg \left(t \leq 0.235\right):\\ \;\;\;\;\frac{0.037037037037037035}{t \cdot t} + \left(0.8333333333333334 - \frac{0.2222222222222222}{t}\right)\\ \mathbf{else}:\\ \;\;\;\;t \cdot t + 0.5\\ \end{array} \]

Alternative 8
Accuracy	99.2%
Cost	585

\[\begin{array}{l} \mathbf{if}\;t \leq -0.8 \lor \neg \left(t \leq 0.56\right):\\ \;\;\;\;0.8333333333333334 - \frac{0.2222222222222222}{t}\\ \mathbf{else}:\\ \;\;\;\;t \cdot t + 0.5\\ \end{array} \]

Alternative 9
Accuracy	98.6%
Cost	584

\[\begin{array}{l} \mathbf{if}\;t \leq -0.9:\\ \;\;\;\;0.8333333333333334\\ \mathbf{elif}\;t \leq 0.58:\\ \;\;\;\;t \cdot t + 0.5\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334\\ \end{array} \]

Alternative 10
Accuracy	98.5%
Cost	328

\[\begin{array}{l} \mathbf{if}\;t \leq -0.34:\\ \;\;\;\;0.8333333333333334\\ \mathbf{elif}\;t \leq 1:\\ \;\;\;\;0.5\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334\\ \end{array} \]

Alternative 11
Accuracy	58.9%
Cost	64

\[0.5 \]

Reproduce

herbie shell --seed 2023178 
(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))))))))