Kahan p13 Example 3

Average Accuracy: 100.0% → 100.0%

Time: 14.4s

Precision: binary64

Cost: 1344

Math

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

↓

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

FPCore

(FPCore (t)
 :precision binary64
 (-
  1.0
  (/
   1.0
   (+
    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 (+ 1.0 t)))))
   (+ 1.0 (/ -1.0 (+ 2.0 (* t_1 t_1))))))

C

double code(double t) {
	return 1.0 - (1.0 / (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 / (1.0 + t));
	return 1.0 + (-1.0 / (2.0 + (t_1 * t_1)));
}

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[t_] := N[(1.0 - N[(1.0 / 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]), $MachinePrecision]

↓

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

Derivation

1. Initial program 100.0%

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

2. Applied egg-rr 100.0%

   \[\leadsto 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \color{blue}{\left(e^{\mathsf{log1p}\left(\frac{2}{t \cdot \left(1 + \frac{1}{t}\right)}\right)} - 1\right)}\right)} \]
   Proof

   [Start] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)} \]
   expm1-log1p-u [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)\right)}\right)} \]
   expm1-udef [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \color{blue}{\left(e^{\mathsf{log1p}\left(\frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)} - 1\right)}\right)} \]
   associate-/l/ [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \left(e^{\mathsf{log1p}\left(\color{blue}{\frac{2}{\left(1 + \frac{1}{t}\right) \cdot t}}\right)} - 1\right)\right)} \]
   *-commutative [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \left(e^{\mathsf{log1p}\left(\frac{2}{\color{blue}{t \cdot \left(1 + \frac{1}{t}\right)}}\right)} - 1\right)\right)} \]

3. Simplified 100.0%

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

   [Start] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \left(e^{\mathsf{log1p}\left(\frac{2}{t \cdot \left(1 + \frac{1}{t}\right)}\right)} - 1\right)\right)} \]
   expm1-def [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{2}{t \cdot \left(1 + \frac{1}{t}\right)}\right)\right)}\right)} \]
   expm1-log1p [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \color{blue}{\frac{2}{t \cdot \left(1 + \frac{1}{t}\right)}}\right)} \]
   distribute-rgt-in [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{2}{\color{blue}{1 \cdot t + \frac{1}{t} \cdot t}}\right)} \]
   *-lft-identity [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{2}{\color{blue}{t} + \frac{1}{t} \cdot t}\right)} \]
   lft-mult-inverse [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{2}{t + \color{blue}{1}}\right)} \]

4. Applied egg-rr 100.0%

   \[\leadsto 1 - \frac{1}{2 + \left(2 - \color{blue}{\left(e^{\mathsf{log1p}\left(\frac{2}{t \cdot \left(1 + \frac{1}{t}\right)}\right)} - 1\right)}\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   Proof

   [Start] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   expm1-log1p-u [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)\right)}\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   expm1-udef [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \color{blue}{\left(e^{\mathsf{log1p}\left(\frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)} - 1\right)}\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   associate-/l/ [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \left(e^{\mathsf{log1p}\left(\color{blue}{\frac{2}{\left(1 + \frac{1}{t}\right) \cdot t}}\right)} - 1\right)\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   *-commutative [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \left(e^{\mathsf{log1p}\left(\frac{2}{\color{blue}{t \cdot \left(1 + \frac{1}{t}\right)}}\right)} - 1\right)\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]

5. Simplified 100.0%

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

   [Start] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \left(e^{\mathsf{log1p}\left(\frac{2}{t \cdot \left(1 + \frac{1}{t}\right)}\right)} - 1\right)\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   expm1-def [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{2}{t \cdot \left(1 + \frac{1}{t}\right)}\right)\right)}\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   expm1-log1p [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \color{blue}{\frac{2}{t \cdot \left(1 + \frac{1}{t}\right)}}\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   distribute-rgt-in [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{2}{\color{blue}{1 \cdot t + \frac{1}{t} \cdot t}}\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   *-lft-identity [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{2}{\color{blue}{t} + \frac{1}{t} \cdot t}\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]
   lft-mult-inverse [=>] 100.0	\[ 1 - \frac{1}{2 + \left(2 - \frac{2}{t + \color{blue}{1}}\right) \cdot \left(2 - \frac{2}{t + 1}\right)} \]

6. Final simplification 100.0%

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

Alternatives

Alternative 1
Accuracy	100.0%
Cost	1216

\[1 + \frac{-1}{\frac{1}{-1 - t} \cdot \left(8 + \frac{4}{-1 - t}\right) + 6} \]

Alternative 2
Accuracy	99.3%
Cost	1092

\[\begin{array}{l} \mathbf{if}\;t \leq -0.52:\\ \;\;\;\;1 + \frac{-1}{6 + \frac{-8 + \frac{4}{t}}{1 + t}}\\ \mathbf{elif}\;t \leq 0.44:\\ \;\;\;\;1 + \frac{-1}{2 + 4 \cdot \left(t \cdot t\right)}\\ \mathbf{else}:\\ \;\;\;\;1 + \left(-0.16666666666666666 + \frac{-0.2222222222222222 - \frac{-0.037037037037037035}{t}}{t}\right)\\ \end{array} \]

Alternative 3
Accuracy	100.0%
Cost	1088

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

Alternative 4
Accuracy	99.3%
Cost	969

\[\begin{array}{l} \mathbf{if}\;t \leq -0.8 \lor \neg \left(t \leq 0.235\right):\\ \;\;\;\;1 + \left(-0.16666666666666666 + \frac{-0.2222222222222222 - \frac{-0.037037037037037035}{t}}{t}\right)\\ \mathbf{else}:\\ \;\;\;\;t \cdot t + 0.5\\ \end{array} \]

Alternative 5
Accuracy	99.3%
Cost	969

\[\begin{array}{l} \mathbf{if}\;t \leq -0.65 \lor \neg \left(t \leq 0.44\right):\\ \;\;\;\;1 + \left(-0.16666666666666666 + \frac{-0.2222222222222222 - \frac{-0.037037037037037035}{t}}{t}\right)\\ \mathbf{else}:\\ \;\;\;\;1 + \frac{-1}{2 + 4 \cdot \left(t \cdot t\right)}\\ \end{array} \]

Alternative 6
Accuracy	99.2%
Cost	712

\[\begin{array}{l} \mathbf{if}\;t \leq -0.78:\\ \;\;\;\;\left(1 + \frac{-0.2222222222222222}{t}\right) + -0.16666666666666666\\ \mathbf{elif}\;t \leq 0.38:\\ \;\;\;\;t \cdot t + 0.5\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{1.2 + \frac{0.32}{t}}\\ \end{array} \]

Alternative 7
Accuracy	99.1%
Cost	585

\[\begin{array}{l} \mathbf{if}\;t \leq -0.78 \lor \neg \left(t \leq 0.58\right):\\ \;\;\;\;0.8333333333333334 + \frac{-0.2222222222222222}{t}\\ \mathbf{else}:\\ \;\;\;\;t \cdot t + 0.5\\ \end{array} \]

Alternative 8
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 9
Accuracy	99.1%
Cost	584

\[\begin{array}{l} \mathbf{if}\;t \leq -0.78:\\ \;\;\;\;\left(1 + \frac{-0.2222222222222222}{t}\right) + -0.16666666666666666\\ \mathbf{elif}\;t \leq 0.58:\\ \;\;\;\;t \cdot t + 0.5\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334 + \frac{-0.2222222222222222}{t}\\ \end{array} \]

Alternative 10
Accuracy	98.4%
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	60.2%
Cost	64

\[0.5 \]

Reproduce

herbie shell --seed 2023138 
(FPCore (t)
  :name "Kahan p13 Example 3"
  :precision binary64
  (- 1.0 (/ 1.0 (+ 2.0 (* (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t)))) (- 2.0 (/ (/ 2.0 t) (+ 1.0 (/ 1.0 t)))))))))