2log (problem 3.3.6)

Average Accuracy: 54.2% → 100.0%

Time: 5.0s

Precision: binary64

Cost: 6592

Math

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

↓

\[\mathsf{log1p}\left(\frac{1}{N}\right) \]

FPCore

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

↓

(FPCore (N) :precision binary64 (log1p (/ 1.0 N)))

C

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

↓

double code(double N) {
	return log1p((1.0 / N));
}

Java

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

↓

public static double code(double N) {
	return Math.log1p((1.0 / N));
}

Python

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

↓

def code(N):
	return math.log1p((1.0 / N))

Julia

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

↓

function code(N)
	return log1p(Float64(1.0 / N))
end

Wolfram

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

↓

code[N_] := N[Log[1 + N[(1.0 / N), $MachinePrecision]], $MachinePrecision]

Derivation

1. Initial program 54.2%

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

2. Simplified 54.2%

   \[\leadsto \color{blue}{\mathsf{log1p}\left(N\right) - \log N} \]
   Proof

   [Start] 54.2	\[ \log \left(N + 1\right) - \log N \]
   +-commutative [=>] 54.2	\[ \log \color{blue}{\left(1 + N\right)} - \log N \]
   log1p-def [=>] 54.2	\[ \color{blue}{\mathsf{log1p}\left(N\right)} - \log N \]

3. Applied egg-rr 54.4%

   \[\leadsto \color{blue}{\mathsf{log1p}\left(\frac{N + 1}{N} - 1\right)} \]
   Proof

   [Start] 54.2	\[ \mathsf{log1p}\left(N\right) - \log N \]
   log1p-expm1-u [=>] 54.2	\[ \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\mathsf{log1p}\left(N\right) - \log N\right)\right)} \]
   expm1-udef [=>] 54.2	\[ \mathsf{log1p}\left(\color{blue}{e^{\mathsf{log1p}\left(N\right) - \log N} - 1}\right) \]
   exp-diff [=>] 54.2	\[ \mathsf{log1p}\left(\color{blue}{\frac{e^{\mathsf{log1p}\left(N\right)}}{e^{\log N}}} - 1\right) \]
   log1p-udef [=>] 54.2	\[ \mathsf{log1p}\left(\frac{e^{\color{blue}{\log \left(1 + N\right)}}}{e^{\log N}} - 1\right) \]
   add-exp-log [<=] 53.5	\[ \mathsf{log1p}\left(\frac{\color{blue}{1 + N}}{e^{\log N}} - 1\right) \]
   +-commutative [=>] 53.5	\[ \mathsf{log1p}\left(\frac{\color{blue}{N + 1}}{e^{\log N}} - 1\right) \]
   add-exp-log [<=] 54.4	\[ \mathsf{log1p}\left(\frac{N + 1}{\color{blue}{N}} - 1\right) \]

4. Taylor expanded in N around 0 100.0%

   \[\leadsto \mathsf{log1p}\left(\color{blue}{\frac{1}{N}}\right) \]

5. Final simplification 100.0%

   \[\leadsto \mathsf{log1p}\left(\frac{1}{N}\right) \]

Alternatives

Alternative 1
Accuracy	98.6%
Cost	6660

\[\begin{array}{l} \mathbf{if}\;N \leq 0.68:\\ \;\;\;\;-\log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1 + \frac{-0.5}{N}}{N}\\ \end{array} \]

Alternative 2
Accuracy	51.5%
Cost	192

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

Alternative 3
Accuracy	4.5%
Cost	64

\[N \]

Reproduce

herbie shell --seed 2023126 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))