Kahan p13 Example 1

Average Error: 0.0 → 0.0

Time: 11.9s

Precision: binary64

Cost: 20992

Math

\[\frac{1 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}}{2 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}} \]

↓

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

FPCore

(FPCore (t)
 :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))))))

↓

(FPCore (t)
 :precision binary64
 (let* ((t_1 (/ (* 2.0 t) (+ 1.0 t))))
   (/
    (+ 1.0 (log (exp (pow (* 2.0 (/ t (+ 1.0 t))) 2.0))))
    (+ 2.0 (* t_1 t_1)))))

C

double code(double t) {
	return (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))));
}

↓

double code(double t) {
	double t_1 = (2.0 * t) / (1.0 + t);
	return (1.0 + log(exp(pow((2.0 * (t / (1.0 + t))), 2.0)))) / (2.0 + (t_1 * t_1));
}

Fortran

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

↓

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

Java

public static double code(double t) {
	return (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))));
}

↓

public static double code(double t) {
	double t_1 = (2.0 * t) / (1.0 + t);
	return (1.0 + Math.log(Math.exp(Math.pow((2.0 * (t / (1.0 + t))), 2.0)))) / (2.0 + (t_1 * t_1));
}

Python

def code(t):
	return (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))))

↓

def code(t):
	t_1 = (2.0 * t) / (1.0 + t)
	return (1.0 + math.log(math.exp(math.pow((2.0 * (t / (1.0 + t))), 2.0)))) / (2.0 + (t_1 * t_1))

Julia

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

↓

function code(t)
	t_1 = Float64(Float64(2.0 * t) / Float64(1.0 + t))
	return Float64(Float64(1.0 + log(exp((Float64(2.0 * Float64(t / Float64(1.0 + t))) ^ 2.0)))) / Float64(2.0 + Float64(t_1 * t_1)))
end

MATLAB

function tmp = code(t)
	tmp = (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))));
end

↓

function tmp = code(t)
	t_1 = (2.0 * t) / (1.0 + t);
	tmp = (1.0 + log(exp(((2.0 * (t / (1.0 + t))) ^ 2.0)))) / (2.0 + (t_1 * t_1));
end

Wolfram

code[t_] := N[(N[(1.0 + N[(N[(N[(2.0 * t), $MachinePrecision] / N[(1.0 + t), $MachinePrecision]), $MachinePrecision] * N[(N[(2.0 * t), $MachinePrecision] / N[(1.0 + t), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(2.0 + N[(N[(N[(2.0 * t), $MachinePrecision] / N[(1.0 + t), $MachinePrecision]), $MachinePrecision] * N[(N[(2.0 * t), $MachinePrecision] / N[(1.0 + t), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

↓

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

Derivation

1. Initial program 0.0

   \[\frac{1 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}}{2 + \frac{2 \cdot t}{1 + t} \cdot \frac{2 \cdot t}{1 + t}} \]

2. Applied egg-rr 0.0

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

3. Final simplification 0.0

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

Alternatives

Alternative 1
Error	0.9
Cost	2248

\[\begin{array}{l} t_1 := \frac{\frac{4 \cdot \left(t \cdot t\right)}{1 + t}}{1 + t}\\ \mathbf{if}\;t \leq -4.295573961651241 \cdot 10^{+169}:\\ \;\;\;\;0.8333333333333334\\ \mathbf{elif}\;t \leq 228096.3490592989:\\ \;\;\;\;\frac{1 + t_1}{2 + t_1}\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334\\ \end{array} \]

Alternative 2
Error	0.0
Cost	2240

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

Alternative 3
Error	0.7
Cost	1864

\[\begin{array}{l} \mathbf{if}\;t \leq -1606770371.0761223:\\ \;\;\;\;\left(1 + \frac{-0.2222222222222222}{t}\right) + -0.16666666666666666\\ \mathbf{elif}\;t \leq 0.021398597363671983:\\ \;\;\;\;\frac{1 + \left(2 \cdot t\right) \cdot \left(t \cdot \left(2 + t \cdot \left(-2 + 2 \cdot t\right)\right)\right)}{2 + 4 \cdot \left(t \cdot t\right)}\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334 + \left(\frac{-0.2222222222222222}{t} + \frac{0.037037037037037035}{t \cdot t}\right)\\ \end{array} \]

Alternative 4
Error	0.7
Cost	1480

\[\begin{array}{l} \mathbf{if}\;t \leq -1606770371.0761223:\\ \;\;\;\;\left(1 + \frac{-0.2222222222222222}{t}\right) + -0.16666666666666666\\ \mathbf{elif}\;t \leq 0.021398597363671983:\\ \;\;\;\;\frac{1 + \frac{t \cdot \left(t \cdot 4\right)}{1 + t}}{2 + 4 \cdot \left(t \cdot t\right)}\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334 + \left(\frac{-0.2222222222222222}{t} + \frac{0.037037037037037035}{t \cdot t}\right)\\ \end{array} \]

Alternative 5
Error	0.7
Cost	1224

\[\begin{array}{l} t_1 := 4 \cdot \left(t \cdot t\right)\\ \mathbf{if}\;t \leq -1606770371.0761223:\\ \;\;\;\;\left(1 + \frac{-0.2222222222222222}{t}\right) + -0.16666666666666666\\ \mathbf{elif}\;t \leq 0.021398597363671983:\\ \;\;\;\;\frac{1 + t_1}{2 + t_1}\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334 + \left(\frac{-0.2222222222222222}{t} + \frac{0.037037037037037035}{t \cdot t}\right)\\ \end{array} \]

Alternative 6
Error	0.8
Cost	968

\[\begin{array}{l} \mathbf{if}\;t \leq -1606770371.0761223:\\ \;\;\;\;\left(1 + \frac{-0.2222222222222222}{t}\right) + -0.16666666666666666\\ \mathbf{elif}\;t \leq 0.021398597363671983:\\ \;\;\;\;0.5\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334 + \left(\frac{-0.2222222222222222}{t} + \frac{0.037037037037037035}{t \cdot t}\right)\\ \end{array} \]

Alternative 7
Error	0.9
Cost	584

\[\begin{array}{l} t_1 := 0.8333333333333334 + \frac{-0.2222222222222222}{t}\\ \mathbf{if}\;t \leq -1606770371.0761223:\\ \;\;\;\;t_1\\ \mathbf{elif}\;t \leq 0.021398597363671983:\\ \;\;\;\;0.5\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \]

Alternative 8
Error	0.9
Cost	584

\[\begin{array}{l} \mathbf{if}\;t \leq -1606770371.0761223:\\ \;\;\;\;\left(1 + \frac{-0.2222222222222222}{t}\right) + -0.16666666666666666\\ \mathbf{elif}\;t \leq 0.021398597363671983:\\ \;\;\;\;0.5\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334 + \frac{-0.2222222222222222}{t}\\ \end{array} \]

Alternative 9
Error	1.1
Cost	328

\[\begin{array}{l} \mathbf{if}\;t \leq -1606770371.0761223:\\ \;\;\;\;0.8333333333333334\\ \mathbf{elif}\;t \leq 0.021398597363671983:\\ \;\;\;\;0.5\\ \mathbf{else}:\\ \;\;\;\;0.8333333333333334\\ \end{array} \]

Alternative 10
Error	26.4
Cost	64

\[0.8333333333333334 \]

Reproduce

herbie shell --seed 2022297 
(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))))))