2log (problem 3.3.6)

Average Error: 29.1 → 0.2

Time: 7.6s

Precision: binary64

Cost: 32836

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 10^{-8}:\\ \;\;\;\;\frac{1}{N} - \frac{0.5}{{N}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{N + 1}\right) \cdot 2 - \log N\\ \end{array} \]

FPCore

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

↓

(FPCore (N)
 :precision binary64
 (if (<= (- (log (+ N 1.0)) (log N)) 1e-8)
   (- (/ 1.0 N) (/ 0.5 (pow N 2.0)))
   (- (* (log (sqrt (+ N 1.0))) 2.0) (log N))))

C

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

↓

double code(double N) {
	double tmp;
	if ((log((N + 1.0)) - log(N)) <= 1e-8) {
		tmp = (1.0 / N) - (0.5 / pow(N, 2.0));
	} else {
		tmp = (log(sqrt((N + 1.0))) * 2.0) - log(N);
	}
	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) :: tmp
    if ((log((n + 1.0d0)) - log(n)) <= 1d-8) then
        tmp = (1.0d0 / n) - (0.5d0 / (n ** 2.0d0))
    else
        tmp = (log(sqrt((n + 1.0d0))) * 2.0d0) - log(n)
    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 tmp;
	if ((Math.log((N + 1.0)) - Math.log(N)) <= 1e-8) {
		tmp = (1.0 / N) - (0.5 / Math.pow(N, 2.0));
	} else {
		tmp = (Math.log(Math.sqrt((N + 1.0))) * 2.0) - Math.log(N);
	}
	return tmp;
}

Python

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

↓

def code(N):
	tmp = 0
	if (math.log((N + 1.0)) - math.log(N)) <= 1e-8:
		tmp = (1.0 / N) - (0.5 / math.pow(N, 2.0))
	else:
		tmp = (math.log(math.sqrt((N + 1.0))) * 2.0) - math.log(N)
	return tmp

Julia

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

↓

function code(N)
	tmp = 0.0
	if (Float64(log(Float64(N + 1.0)) - log(N)) <= 1e-8)
		tmp = Float64(Float64(1.0 / N) - Float64(0.5 / (N ^ 2.0)));
	else
		tmp = Float64(Float64(log(sqrt(Float64(N + 1.0))) * 2.0) - log(N));
	end
	return tmp
end

MATLAB

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

↓

function tmp_2 = code(N)
	tmp = 0.0;
	if ((log((N + 1.0)) - log(N)) <= 1e-8)
		tmp = (1.0 / N) - (0.5 / (N ^ 2.0));
	else
		tmp = (log(sqrt((N + 1.0))) * 2.0) - log(N);
	end
	tmp_2 = tmp;
end

Wolfram

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

↓

code[N_] := If[LessEqual[N[(N[Log[N[(N + 1.0), $MachinePrecision]], $MachinePrecision] - N[Log[N], $MachinePrecision]), $MachinePrecision], 1e-8], N[(N[(1.0 / N), $MachinePrecision] - N[(0.5 / N[Power[N, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Log[N[Sqrt[N[(N + 1.0), $MachinePrecision]], $MachinePrecision]], $MachinePrecision] * 2.0), $MachinePrecision] - N[Log[N], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N)) < 1e-8

   1. Initial program 60.0

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

   2. Taylor expanded in N around inf 0.0

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

   3. Taylor expanded in N around 0 0.0

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

   if 1e-8 < (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N))

   1. Initial program 0.3

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

   2. Applied egg-rr 0.3

      \[\leadsto \color{blue}{\log \left(\sqrt{N + 1}\right) \cdot 2} - \log N \]
3. Recombined 2 regimes into one program.

4. Final simplification 0.2

   \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 10^{-8}:\\ \;\;\;\;\frac{1}{N} - \frac{0.5}{{N}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{N + 1}\right) \cdot 2 - \log N\\ \end{array} \]

Alternatives

Alternative 1
Error	0.2
Cost	26308

\[\begin{array}{l} t_0 := \log \left(N + 1\right) - \log N\\ \mathbf{if}\;t_0 \leq 10^{-8}:\\ \;\;\;\;\frac{1}{N} - \frac{0.5}{{N}^{2}}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 2
Error	0.6
Cost	7044

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

Alternative 3
Error	0.9
Cost	6980

\[\begin{array}{l} \mathbf{if}\;N \leq 1:\\ \;\;\;\;\left(\left(N - \log N\right) + 2\right) + -2\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N}\\ \end{array} \]

Alternative 4
Error	0.9
Cost	6724

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

Alternative 5
Error	1.2
Cost	6660

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

Alternative 6
Error	31.2
Cost	192

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

Alternative 7
Error	61.1
Cost	64

\[N \]

Reproduce

herbie shell --seed 2023074 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))