2log (problem 3.3.6)

Average Error: 29.5 → 0.4

Time: 9.1s

Precision: binary64

Cost: 26884

Math

\[\log \left(N + 1\right) - \log N \]

↓

\[\begin{array}{l} t_0 := \log \left(N + 1\right) - \log N\\ \mathbf{if}\;t_0 \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\left(\frac{0.5}{{N}^{2}} + \frac{1}{N}\right) + \left(-\frac{1}{{N}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

FPCore

(FPCore (N) :precision binary64 (- (log (+ N 1.0)) (log N)))

↓

(FPCore (N)
 :precision binary64
 (let* ((t_0 (- (log (+ N 1.0)) (log N))))
   (if (<= t_0 5e-12)
     (+ (+ (/ 0.5 (pow N 2.0)) (/ 1.0 N)) (- (/ 1.0 (pow N 2.0))))
     t_0)))

C

double code(double N) {
	return log((N + 1.0)) - log(N);
}

↓

double code(double N) {
	double t_0 = log((N + 1.0)) - log(N);
	double tmp;
	if (t_0 <= 5e-12) {
		tmp = ((0.5 / pow(N, 2.0)) + (1.0 / N)) + -(1.0 / pow(N, 2.0));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    code = log((n + 1.0d0)) - log(n)
end function

↓

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: tmp
    t_0 = log((n + 1.0d0)) - log(n)
    if (t_0 <= 5d-12) then
        tmp = ((0.5d0 / (n ** 2.0d0)) + (1.0d0 / n)) + -(1.0d0 / (n ** 2.0d0))
    else
        tmp = t_0
    end if
    code = tmp
end function

Java

public static double code(double N) {
	return Math.log((N + 1.0)) - Math.log(N);
}

↓

public static double code(double N) {
	double t_0 = Math.log((N + 1.0)) - Math.log(N);
	double tmp;
	if (t_0 <= 5e-12) {
		tmp = ((0.5 / Math.pow(N, 2.0)) + (1.0 / N)) + -(1.0 / Math.pow(N, 2.0));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(N):
	return math.log((N + 1.0)) - math.log(N)

↓

def code(N):
	t_0 = math.log((N + 1.0)) - math.log(N)
	tmp = 0
	if t_0 <= 5e-12:
		tmp = ((0.5 / math.pow(N, 2.0)) + (1.0 / N)) + -(1.0 / math.pow(N, 2.0))
	else:
		tmp = t_0
	return tmp

Julia

function code(N)
	return Float64(log(Float64(N + 1.0)) - log(N))
end

↓

function code(N)
	t_0 = Float64(log(Float64(N + 1.0)) - log(N))
	tmp = 0.0
	if (t_0 <= 5e-12)
		tmp = Float64(Float64(Float64(0.5 / (N ^ 2.0)) + Float64(1.0 / N)) + Float64(-Float64(1.0 / (N ^ 2.0))));
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp = code(N)
	tmp = log((N + 1.0)) - log(N);
end

↓

function tmp_2 = code(N)
	t_0 = log((N + 1.0)) - log(N);
	tmp = 0.0;
	if (t_0 <= 5e-12)
		tmp = ((0.5 / (N ^ 2.0)) + (1.0 / N)) + -(1.0 / (N ^ 2.0));
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := N[(N[Log[N[(N + 1.0), $MachinePrecision]], $MachinePrecision] - N[Log[N], $MachinePrecision]), $MachinePrecision]

↓

code[N_] := Block[{t$95$0 = N[(N[Log[N[(N + 1.0), $MachinePrecision]], $MachinePrecision] - N[Log[N], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$0, 5e-12], N[(N[(N[(0.5 / N[Power[N, 2.0], $MachinePrecision]), $MachinePrecision] + N[(1.0 / N), $MachinePrecision]), $MachinePrecision] + (-N[(1.0 / N[Power[N, 2.0], $MachinePrecision]), $MachinePrecision])), $MachinePrecision], t$95$0]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N)) < 4.9999999999999997e-12

   1. Initial program 60.3

      \[\log \left(N + 1\right) - \log N \]

   2. Taylor expanded in N around inf 0.0

      \[\leadsto \color{blue}{\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}} \]

   3. Applied egg-rr 0.0

      \[\leadsto \color{blue}{\left(\frac{1}{{N}^{2}} \cdot 0.5 + \frac{1}{N}\right) + \left(-\frac{1}{{N}^{2}}\right)} \]

   4. Taylor expanded in N around 0 0.0

      \[\leadsto \left(\color{blue}{\frac{0.5}{{N}^{2}}} + \frac{1}{N}\right) + \left(-\frac{1}{{N}^{2}}\right) \]

   if 4.9999999999999997e-12 < (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N))

   1. Initial program 0.7

      \[\log \left(N + 1\right) - \log N \]
3. Recombined 2 regimes into one program.

4. Final simplification 0.4

   \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\left(\frac{0.5}{{N}^{2}} + \frac{1}{N}\right) + \left(-\frac{1}{{N}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(N + 1\right) - \log N\\ \end{array} \]

Alternatives

Alternative 1
Error	0.4
Cost	26308

\[\begin{array}{l} t_0 := \log \left(N + 1\right) - \log N\\ \mathbf{if}\;t_0 \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{1}{N} - \frac{0.5}{{N}^{2}}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 2
Error	0.6
Cost	7044

\[\begin{array}{l} \mathbf{if}\;N \leq 0.9:\\ \;\;\;\;N - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} - \frac{0.5}{{N}^{2}}\\ \end{array} \]

Alternative 3
Error	0.9
Cost	6724

\[\begin{array}{l} \mathbf{if}\;N \leq 1:\\ \;\;\;\;N - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N}\\ \end{array} \]

Alternative 4
Error	1.2
Cost	6660

\[\begin{array}{l} \mathbf{if}\;N \leq 0.55:\\ \;\;\;\;-\log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N}\\ \end{array} \]

Alternative 5
Error	30.8
Cost	192

\[\frac{1}{N} \]

Reproduce

herbie shell --seed 2023092 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))