2log (problem 3.3.6)

?

Percentage Accurate: 54.1% → 99.9%
Time: 12.3s
Precision: binary64
Cost: 27076

?

\[\log \left(N + 1\right) - \log N \]
↓
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\left(\frac{0.3333333333333333}{{N}^{3}} + \frac{1 + \frac{-0.5}{N}}{N}\right) - \frac{0.25}{{N}^{4}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array} \]
(FPCore (N) :precision binary64 (- (log (+ N 1.0)) (log N)))
↓
(FPCore (N)
 :precision binary64
 (if (<= (- (log (+ N 1.0)) (log N)) 2e-5)
   (-
    (+ (/ 0.3333333333333333 (pow N 3.0)) (/ (+ 1.0 (/ -0.5 N)) N))
    (/ 0.25 (pow N 4.0)))
   (log (/ (+ N 1.0) N))))
double code(double N) {
	return log((N + 1.0)) - log(N);
}
↓
double code(double N) {
	double tmp;
	if ((log((N + 1.0)) - log(N)) <= 2e-5) {
		tmp = ((0.3333333333333333 / pow(N, 3.0)) + ((1.0 + (-0.5 / N)) / N)) - (0.25 / pow(N, 4.0));
	} else {
		tmp = log(((N + 1.0) / N));
	}
	return tmp;
}

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)) <= 2d-5) then
        tmp = ((0.3333333333333333d0 / (n ** 3.0d0)) + ((1.0d0 + ((-0.5d0) / n)) / n)) - (0.25d0 / (n ** 4.0d0))
    else
        tmp = log(((n + 1.0d0) / n))
    end if
    code = tmp
end function

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)) <= 2e-5) {
		tmp = ((0.3333333333333333 / Math.pow(N, 3.0)) + ((1.0 + (-0.5 / N)) / N)) - (0.25 / Math.pow(N, 4.0));
	} else {
		tmp = Math.log(((N + 1.0) / N));
	}
	return tmp;
}

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)) <= 2e-5:
		tmp = ((0.3333333333333333 / math.pow(N, 3.0)) + ((1.0 + (-0.5 / N)) / N)) - (0.25 / math.pow(N, 4.0))
	else:
		tmp = math.log(((N + 1.0) / N))
	return tmp

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)) <= 2e-5)
		tmp = Float64(Float64(Float64(0.3333333333333333 / (N ^ 3.0)) + Float64(Float64(1.0 + Float64(-0.5 / N)) / N)) - Float64(0.25 / (N ^ 4.0)));
	else
		tmp = log(Float64(Float64(N + 1.0) / N));
	end
	return tmp
end

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)) <= 2e-5)
		tmp = ((0.3333333333333333 / (N ^ 3.0)) + ((1.0 + (-0.5 / N)) / N)) - (0.25 / (N ^ 4.0));
	else
		tmp = log(((N + 1.0) / N));
	end
	tmp_2 = tmp;
end

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], 2e-5], N[(N[(N[(0.3333333333333333 / N[Power[N, 3.0], $MachinePrecision]), $MachinePrecision] + N[(N[(1.0 + N[(-0.5 / N), $MachinePrecision]), $MachinePrecision] / N), $MachinePrecision]), $MachinePrecision] - N[(0.25 / N[Power[N, 4.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Log[N[(N[(N + 1.0), $MachinePrecision] / N), $MachinePrecision]], $MachinePrecision]]

\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;\log \left(N + 1\right) - \log N \leq 2 \cdot 10^{-5}:\\
\;\;\;\;\left(\frac{0.3333333333333333}{{N}^{3}} + \frac{1 + \frac{-0.5}{N}}{N}\right) - \frac{0.25}{{N}^{4}}\\

\mathbf{else}:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\


\end{array}

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Herbie found 6 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Bogosity?

Bogosity

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Split input into 2 regimes

  2. if (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N)) < 2.00000000000000016e-5

    1. Initial program 7.3%

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

    2. Simplified7.3%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(N\right) - \log N} \]
      Step-by-step derivation

      [Start]7.3%

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

      +-commutative [=>]7.3%

      \[ \log \color{blue}{\left(1 + N\right)} - \log N \]

      log1p-def [=>]7.3%

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

    3. Taylor expanded in N around inf 100.0%

      \[\leadsto \color{blue}{\left(\frac{1}{N} + 0.3333333333333333 \cdot \frac{1}{{N}^{3}}\right) - \left(0.25 \cdot \frac{1}{{N}^{4}} + 0.5 \cdot \frac{1}{{N}^{2}}\right)} \]

    4. Simplified100.0%

      \[\leadsto \color{blue}{\left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \left(\frac{0.5}{N \cdot N} + \frac{0.25}{{N}^{4}}\right)} \]
      Step-by-step derivation

      [Start]100.0%

      \[ \left(\frac{1}{N} + 0.3333333333333333 \cdot \frac{1}{{N}^{3}}\right) - \left(0.25 \cdot \frac{1}{{N}^{4}} + 0.5 \cdot \frac{1}{{N}^{2}}\right) \]

      associate-*r/ [=>]100.0%

      \[ \left(\frac{1}{N} + \color{blue}{\frac{0.3333333333333333 \cdot 1}{{N}^{3}}}\right) - \left(0.25 \cdot \frac{1}{{N}^{4}} + 0.5 \cdot \frac{1}{{N}^{2}}\right) \]

      metadata-eval [=>]100.0%

      \[ \left(\frac{1}{N} + \frac{\color{blue}{0.3333333333333333}}{{N}^{3}}\right) - \left(0.25 \cdot \frac{1}{{N}^{4}} + 0.5 \cdot \frac{1}{{N}^{2}}\right) \]

      +-commutative [=>]100.0%

      \[ \left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \color{blue}{\left(0.5 \cdot \frac{1}{{N}^{2}} + 0.25 \cdot \frac{1}{{N}^{4}}\right)} \]

      associate-*r/ [=>]100.0%

      \[ \left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \left(\color{blue}{\frac{0.5 \cdot 1}{{N}^{2}}} + 0.25 \cdot \frac{1}{{N}^{4}}\right) \]

      metadata-eval [=>]100.0%

      \[ \left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \left(\frac{\color{blue}{0.5}}{{N}^{2}} + 0.25 \cdot \frac{1}{{N}^{4}}\right) \]

      unpow2 [=>]100.0%

      \[ \left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \left(\frac{0.5}{\color{blue}{N \cdot N}} + 0.25 \cdot \frac{1}{{N}^{4}}\right) \]

      associate-*r/ [=>]100.0%

      \[ \left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \left(\frac{0.5}{N \cdot N} + \color{blue}{\frac{0.25 \cdot 1}{{N}^{4}}}\right) \]

      metadata-eval [=>]100.0%

      \[ \left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \left(\frac{0.5}{N \cdot N} + \frac{\color{blue}{0.25}}{{N}^{4}}\right) \]

    5. Taylor expanded in N around 0 100.0%

      \[\leadsto \color{blue}{\left(\frac{1}{N} + 0.3333333333333333 \cdot \frac{1}{{N}^{3}}\right) - \left(0.25 \cdot \frac{1}{{N}^{4}} + 0.5 \cdot \frac{1}{{N}^{2}}\right)} \]

    6. Simplified100.0%

      \[\leadsto \color{blue}{\left(\frac{0.3333333333333333}{{N}^{3}} + \frac{1 + \frac{-0.5}{N}}{N}\right) - \frac{0.25}{{N}^{4}}} \]
      Step-by-step derivation

      [Start]100.0%

      \[ \left(\frac{1}{N} + 0.3333333333333333 \cdot \frac{1}{{N}^{3}}\right) - \left(0.25 \cdot \frac{1}{{N}^{4}} + 0.5 \cdot \frac{1}{{N}^{2}}\right) \]

      +-commutative [=>]100.0%

      \[ \left(\frac{1}{N} + 0.3333333333333333 \cdot \frac{1}{{N}^{3}}\right) - \color{blue}{\left(0.5 \cdot \frac{1}{{N}^{2}} + 0.25 \cdot \frac{1}{{N}^{4}}\right)} \]

      associate--r+ [=>]100.0%

      \[ \color{blue}{\left(\left(\frac{1}{N} + 0.3333333333333333 \cdot \frac{1}{{N}^{3}}\right) - 0.5 \cdot \frac{1}{{N}^{2}}\right) - 0.25 \cdot \frac{1}{{N}^{4}}} \]

    if 2.00000000000000016e-5 < (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N))

    1. Initial program 99.9%

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

    2. Simplified99.9%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(N\right) - \log N} \]
      Step-by-step derivation

      [Start]99.9%

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

      +-commutative [=>]99.9%

      \[ \log \color{blue}{\left(1 + N\right)} - \log N \]

      log1p-def [=>]99.9%

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

    3. Applied egg-rr100.0%

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]
      Step-by-step derivation

      [Start]99.9%

      \[ \mathsf{log1p}\left(N\right) - \log N \]

      log1p-udef [=>]99.9%

      \[ \color{blue}{\log \left(1 + N\right)} - \log N \]

      diff-log [=>]100.0%

      \[ \color{blue}{\log \left(\frac{1 + N}{N}\right)} \]

      +-commutative [=>]100.0%

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

  4. Final simplification100.0%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\left(\frac{0.3333333333333333}{{N}^{3}} + \frac{1 + \frac{-0.5}{N}}{N}\right) - \frac{0.25}{{N}^{4}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array} \]

Alternatives

Alternative 1
Accuracy99.9%
Cost27076
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\left(\frac{0.3333333333333333}{{N}^{3}} + \frac{1 + \frac{-0.5}{N}}{N}\right) - \frac{0.25}{{N}^{4}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array} \]
Alternative 2
Accuracy99.9%
Cost20356
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\frac{0.3333333333333333}{{N}^{3}} + \frac{1 + \frac{-0.5}{N}}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array} \]
Alternative 3
Accuracy99.8%
Cost6852
\[\begin{array}{l} \mathbf{if}\;N \leq 155000:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \frac{0.5}{N}}{N}\\ \end{array} \]
Alternative 4
Accuracy98.9%
Cost6724
\[\begin{array}{l} \mathbf{if}\;N \leq 0.9:\\ \;\;\;\;N - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \frac{0.5}{N}}{N}\\ \end{array} \]
Alternative 5
Accuracy98.5%
Cost6660
\[\begin{array}{l} \mathbf{if}\;N \leq 0.68:\\ \;\;\;\;-\log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \frac{0.5}{N}}{N}\\ \end{array} \]
Alternative 6
Accuracy51.6%
Cost192
\[\frac{1}{N} \]

Reproduce?

herbie shell --seed 2023277 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))