Kahan p13 Example 2

Average Error: 100.0% → 100.0%

Time: 23.5s

Precision: binary64

Cost: 2240.00

Math

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

↓

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

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 (+ 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 / (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) / (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 / (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 / (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(-2.0 / 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 / (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[(-2.0 / N[(1.0 + t), $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. Applied egg-rr 100.0

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

3. Simplified 100.0

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

   [Start] 100.0	\[ \frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 + \frac{-2}{t + t \cdot \frac{1}{t}}\right)} \]
   rgt-mult-inverse [=>] 100.0	\[ \frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)}{2 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 + \frac{-2}{t + \color{blue}{1}}\right)} \]

4. Applied egg-rr 100.0

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

5. Simplified 100.0

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

   [Start] 100.0	\[ \frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)}{2 + \left(2 + \frac{-2}{t + t \cdot \frac{1}{t}}\right) \cdot \left(2 + \frac{-2}{t + 1}\right)} \]
   rgt-mult-inverse [=>] 100.0	\[ \frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right)}{2 + \left(2 + \frac{-2}{t + \color{blue}{1}}\right) \cdot \left(2 + \frac{-2}{t + 1}\right)} \]

6. Applied egg-rr 100.0

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

7. Simplified 100.0

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

   [Start] 100.0	\[ \frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 + \frac{-2}{t + t \cdot \frac{1}{t}}\right)}{2 + \left(2 + \frac{-2}{t + 1}\right) \cdot \left(2 + \frac{-2}{t + 1}\right)} \]
   rgt-mult-inverse [=>] 100.0	\[ \frac{1 + \left(2 - \frac{\frac{2}{t}}{1 + \frac{1}{t}}\right) \cdot \left(2 + \frac{-2}{t + \color{blue}{1}}\right)}{2 + \left(2 + \frac{-2}{t + 1}\right) \cdot \left(2 + \frac{-2}{t + 1}\right)} \]

8. Applied egg-rr 100.0

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

9. Simplified 100.0

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

   [Start] 100.0	\[ \frac{1 + \left(2 + \frac{-2}{t + t \cdot \frac{1}{t}}\right) \cdot \left(2 + \frac{-2}{t + 1}\right)}{2 + \left(2 + \frac{-2}{t + 1}\right) \cdot \left(2 + \frac{-2}{t + 1}\right)} \]
   rgt-mult-inverse [=>] 100.0	\[ \frac{1 + \left(2 + \frac{-2}{t + \color{blue}{1}}\right) \cdot \left(2 + \frac{-2}{t + 1}\right)}{2 + \left(2 + \frac{-2}{t + 1}\right) \cdot \left(2 + \frac{-2}{t + 1}\right)} \]

10. Final simplification 100.0

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

Alternatives

Alternative 1
Error	100.0%
Cost	1728.00

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

Alternative 2
Error	99.4%
Cost	1604.00

\[\begin{array}{l} \mathbf{if}\;t \leq -1:\\ \;\;\;\;\frac{\left(5 + \frac{12}{t \cdot t}\right) + \frac{-8}{t}}{6 + \frac{-8 + \frac{4}{t}}{1 + t}}\\ \mathbf{elif}\;t \leq 1.8:\\ \;\;\;\;\frac{-1 + t \cdot \left(-4 + \frac{4}{1 + t}\right)}{-2 + t \cdot \left(t \cdot -4\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{5 + \frac{-8}{t}}{6 + \frac{-8}{t}}\\ \end{array} \]

Alternative 3
Error	99.3%
Cost	1480.00

\[\begin{array}{l} \mathbf{if}\;t \leq -0.56:\\ \;\;\;\;0.8333333333333334 + \frac{-0.2222222222222222}{t}\\ \mathbf{elif}\;t \leq 1.8:\\ \;\;\;\;\frac{-1 + t \cdot \left(-4 + \frac{4}{1 + t}\right)}{-2 + t \cdot \left(t \cdot -4\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{5 + \frac{-8}{t}}{6 + \frac{-8}{t}}\\ \end{array} \]

Alternative 4
Error	99.2%
Cost	1224.00

\[\begin{array}{l} t_1 := 4 \cdot \left(t \cdot t\right)\\ \mathbf{if}\;t \leq -0.6:\\ \;\;\;\;0.8333333333333334 + \frac{-0.2222222222222222}{t}\\ \mathbf{elif}\;t \leq 2.1:\\ \;\;\;\;\frac{1 + t_1}{2 + t_1}\\ \mathbf{else}:\\ \;\;\;\;\frac{5 + \frac{-8}{t}}{6 + \frac{-8}{t}}\\ \end{array} \]

Alternative 5
Error	99.2%
Cost	968.00

\[\begin{array}{l} \mathbf{if}\;t \leq -0.8:\\ \;\;\;\;0.8333333333333334 + \frac{-0.2222222222222222}{t}\\ \mathbf{elif}\;t \leq 1.7:\\ \;\;\;\;t \cdot t + 0.5\\ \mathbf{else}:\\ \;\;\;\;\frac{5 + \frac{-8}{t}}{6 + \frac{-8}{t}}\\ \end{array} \]

Alternative 6
Error	99.3%
Cost	585.00

\[\begin{array}{l} \mathbf{if}\;t \leq -0.8 \lor \neg \left(t \leq 0.55\right):\\ \;\;\;\;0.8333333333333334 + \frac{-0.2222222222222222}{t}\\ \mathbf{else}:\\ \;\;\;\;t \cdot t + 0.5\\ \end{array} \]

Alternative 7
Error	98.7%
Cost	584.00

\[\begin{array}{l} \mathbf{if}\;t \leq -0.92:\\ \;\;\;\;0.8333333333333334\\ \mathbf{elif}\;t \leq 0.58:\\ \;\;\;\;t \cdot t + 0.5\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334\\ \end{array} \]

Alternative 8
Error	98.6%
Cost	328.00

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

Alternative 9
Error	59.3%
Cost	64.00

\[0.5 \]

Reproduce

herbie shell --seed 2023093 
(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))))))))