2log (problem 3.3.6)

Percentage Accurate: 53.8% → 99.7%

Time: 4.6s

Alternatives: 7

Speedup: 2.0×

Specification

Math

\[\begin{array}{l} \\ \log \left(N + 1\right) - \log N \end{array} \]

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 53.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \log \left(N + 1\right) - \log N \end{array} \]

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 99.7% accurate, 0.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 5 \cdot 10^{-9}:\\ \;\;\;\;\frac{1}{N} - \frac{0.5}{{N}^{2}}\\ \mathbf{else}:\\ \;\;\;\;-\log \left(\frac{N}{N + 1}\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (N)
 :precision binary64
 (if (<= (- (log (+ N 1.0)) (log N)) 5e-9)
   (- (/ 1.0 N) (/ 0.5 (pow N 2.0)))
   (- (log (/ N (+ N 1.0))))))

C

double code(double N) {
	double tmp;
	if ((log((N + 1.0)) - log(N)) <= 5e-9) {
		tmp = (1.0 / N) - (0.5 / pow(N, 2.0));
	} else {
		tmp = -log((N / (N + 1.0)));
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((log((n + 1.0d0)) - log(n)) <= 5d-9) then
        tmp = (1.0d0 / n) - (0.5d0 / (n ** 2.0d0))
    else
        tmp = -log((n / (n + 1.0d0)))
    end if
    code = tmp
end function

Java

public static double code(double N) {
	double tmp;
	if ((Math.log((N + 1.0)) - Math.log(N)) <= 5e-9) {
		tmp = (1.0 / N) - (0.5 / Math.pow(N, 2.0));
	} else {
		tmp = -Math.log((N / (N + 1.0)));
	}
	return tmp;
}

Python

def code(N):
	tmp = 0
	if (math.log((N + 1.0)) - math.log(N)) <= 5e-9:
		tmp = (1.0 / N) - (0.5 / math.pow(N, 2.0))
	else:
		tmp = -math.log((N / (N + 1.0)))
	return tmp

Julia

function code(N)
	tmp = 0.0
	if (Float64(log(Float64(N + 1.0)) - log(N)) <= 5e-9)
		tmp = Float64(Float64(1.0 / N) - Float64(0.5 / (N ^ 2.0)));
	else
		tmp = Float64(-log(Float64(N / Float64(N + 1.0))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(N)
	tmp = 0.0;
	if ((log((N + 1.0)) - log(N)) <= 5e-9)
		tmp = (1.0 / N) - (0.5 / (N ^ 2.0));
	else
		tmp = -log((N / (N + 1.0)));
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := If[LessEqual[N[(N[Log[N[(N + 1.0), $MachinePrecision]], $MachinePrecision] - N[Log[N], $MachinePrecision]), $MachinePrecision], 5e-9], N[(N[(1.0 / N), $MachinePrecision] - N[(0.5 / N[Power[N, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], (-N[Log[N[(N / N[(N + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N)) < 5.0000000000000001e-9

   1. Initial program 6.5%

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

   2. Taylor expanded in N around inf 100.0%

      \[\leadsto \color{blue}{\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}} \]
   3. Step-by-step derivation

      1. associate-*r/ 100.0%

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

      2. metadata-eval 100.0%

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

   4. Simplified 100.0%

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

   if 5.0000000000000001e-9 < (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N))

   1. Initial program 99.8%

      \[\log \left(N + 1\right) - \log N \]
   2. Step-by-step derivation

      1. diff-log 99.9%

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

      2. clear-num 99.9%

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

      3. log-rec 99.9%

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

   3. Applied egg-rr 99.9%

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

4. Final simplification 99.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 5 \cdot 10^{-9}:\\ \;\;\;\;\frac{1}{N} - \frac{0.5}{{N}^{2}}\\ \mathbf{else}:\\ \;\;\;\;-\log \left(\frac{N}{N + 1}\right)\\ \end{array} \]

Alternative 2: 99.5% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;N \leq 250000000:\\ \;\;\;\;-\log \left(\frac{N}{N + 1}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N}\\ \end{array} \end{array} \]

FPCore

(FPCore (N)
 :precision binary64
 (if (<= N 250000000.0) (- (log (/ N (+ N 1.0)))) (/ 1.0 N)))

C

double code(double N) {
	double tmp;
	if (N <= 250000000.0) {
		tmp = -log((N / (N + 1.0)));
	} else {
		tmp = 1.0 / N;
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 250000000.0d0) then
        tmp = -log((n / (n + 1.0d0)))
    else
        tmp = 1.0d0 / n
    end if
    code = tmp
end function

Java

public static double code(double N) {
	double tmp;
	if (N <= 250000000.0) {
		tmp = -Math.log((N / (N + 1.0)));
	} else {
		tmp = 1.0 / N;
	}
	return tmp;
}

Python

def code(N):
	tmp = 0
	if N <= 250000000.0:
		tmp = -math.log((N / (N + 1.0)))
	else:
		tmp = 1.0 / N
	return tmp

Julia

function code(N)
	tmp = 0.0
	if (N <= 250000000.0)
		tmp = Float64(-log(Float64(N / Float64(N + 1.0))));
	else
		tmp = Float64(1.0 / N);
	end
	return tmp
end

MATLAB

function tmp_2 = code(N)
	tmp = 0.0;
	if (N <= 250000000.0)
		tmp = -log((N / (N + 1.0)));
	else
		tmp = 1.0 / N;
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := If[LessEqual[N, 250000000.0], (-N[Log[N[(N / N[(N + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), N[(1.0 / N), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if N < 2.5e8

   1. Initial program 99.5%

      \[\log \left(N + 1\right) - \log N \]
   2. Step-by-step derivation

      1. diff-log 99.6%

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

      2. clear-num 99.6%

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

      3. log-rec 99.6%

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

   3. Applied egg-rr 99.6%

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

   if 2.5e8 < N

   1. Initial program 6.1%

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

   2. Taylor expanded in N around inf 99.7%

      \[\leadsto \color{blue}{\frac{1}{N}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 99.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;N \leq 250000000:\\ \;\;\;\;-\log \left(\frac{N}{N + 1}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N}\\ \end{array} \]

Alternative 3: 99.5% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;N \leq 95000000:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N}\\ \end{array} \end{array} \]

FPCore

(FPCore (N)
 :precision binary64
 (if (<= N 95000000.0) (log (/ (+ N 1.0) N)) (/ 1.0 N)))

C

double code(double N) {
	double tmp;
	if (N <= 95000000.0) {
		tmp = log(((N + 1.0) / N));
	} else {
		tmp = 1.0 / N;
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 95000000.0d0) then
        tmp = log(((n + 1.0d0) / n))
    else
        tmp = 1.0d0 / n
    end if
    code = tmp
end function

Java

public static double code(double N) {
	double tmp;
	if (N <= 95000000.0) {
		tmp = Math.log(((N + 1.0) / N));
	} else {
		tmp = 1.0 / N;
	}
	return tmp;
}

Python

def code(N):
	tmp = 0
	if N <= 95000000.0:
		tmp = math.log(((N + 1.0) / N))
	else:
		tmp = 1.0 / N
	return tmp

Julia

function code(N)
	tmp = 0.0
	if (N <= 95000000.0)
		tmp = log(Float64(Float64(N + 1.0) / N));
	else
		tmp = Float64(1.0 / N);
	end
	return tmp
end

MATLAB

function tmp_2 = code(N)
	tmp = 0.0;
	if (N <= 95000000.0)
		tmp = log(((N + 1.0) / N));
	else
		tmp = 1.0 / N;
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := If[LessEqual[N, 95000000.0], N[Log[N[(N[(N + 1.0), $MachinePrecision] / N), $MachinePrecision]], $MachinePrecision], N[(1.0 / N), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if N < 9.5e7

   1. Initial program 99.8%

      \[\log \left(N + 1\right) - \log N \]
   2. Step-by-step derivation

      1. diff-log 99.9%

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

   3. Applied egg-rr 99.9%

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

   if 9.5e7 < N

   1. Initial program 6.5%

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

   2. Taylor expanded in N around inf 99.4%

      \[\leadsto \color{blue}{\frac{1}{N}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 99.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;N \leq 95000000:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N}\\ \end{array} \]

Alternative 4: 98.5% accurate, 2.0× speedup

Math

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

FPCore

(FPCore (N) :precision binary64 (if (<= N 1.0) (- N (log N)) (/ 1.0 N)))

C

double code(double N) {
	double tmp;
	if (N <= 1.0) {
		tmp = N - log(N);
	} else {
		tmp = 1.0 / N;
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 1.0d0) then
        tmp = n - log(n)
    else
        tmp = 1.0d0 / n
    end if
    code = tmp
end function

Java

public static double code(double N) {
	double tmp;
	if (N <= 1.0) {
		tmp = N - Math.log(N);
	} else {
		tmp = 1.0 / N;
	}
	return tmp;
}

Python

def code(N):
	tmp = 0
	if N <= 1.0:
		tmp = N - math.log(N)
	else:
		tmp = 1.0 / N
	return tmp

Julia

function code(N)
	tmp = 0.0
	if (N <= 1.0)
		tmp = Float64(N - log(N));
	else
		tmp = Float64(1.0 / N);
	end
	return tmp
end

MATLAB

function tmp_2 = code(N)
	tmp = 0.0;
	if (N <= 1.0)
		tmp = N - log(N);
	else
		tmp = 1.0 / N;
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := If[LessEqual[N, 1.0], N[(N - N[Log[N], $MachinePrecision]), $MachinePrecision], N[(1.0 / N), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if N < 1

   1. Initial program 99.9%

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

   2. Taylor expanded in N around 0 98.5%

      \[\leadsto \color{blue}{N + -1 \cdot \log N} \]
   3. Step-by-step derivation

      1. neg-mul-1 98.5%

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

      2. unsub-neg 98.5%

         \[\leadsto \color{blue}{N - \log N} \]

   4. Simplified 98.5%

      \[\leadsto \color{blue}{N - \log N} \]

   if 1 < N

   1. Initial program 7.1%

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

   2. Taylor expanded in N around inf 99.0%

      \[\leadsto \color{blue}{\frac{1}{N}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 98.7%

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

Alternative 5: 98.1% accurate, 2.0× speedup

Math

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

FPCore

(FPCore (N) :precision binary64 (if (<= N 0.55) (- (log N)) (/ 1.0 N)))

C

double code(double N) {
	double tmp;
	if (N <= 0.55) {
		tmp = -log(N);
	} else {
		tmp = 1.0 / N;
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 0.55d0) then
        tmp = -log(n)
    else
        tmp = 1.0d0 / n
    end if
    code = tmp
end function

Java

public static double code(double N) {
	double tmp;
	if (N <= 0.55) {
		tmp = -Math.log(N);
	} else {
		tmp = 1.0 / N;
	}
	return tmp;
}

Python

def code(N):
	tmp = 0
	if N <= 0.55:
		tmp = -math.log(N)
	else:
		tmp = 1.0 / N
	return tmp

Julia

function code(N)
	tmp = 0.0
	if (N <= 0.55)
		tmp = Float64(-log(N));
	else
		tmp = Float64(1.0 / N);
	end
	return tmp
end

MATLAB

function tmp_2 = code(N)
	tmp = 0.0;
	if (N <= 0.55)
		tmp = -log(N);
	else
		tmp = 1.0 / N;
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := If[LessEqual[N, 0.55], (-N[Log[N], $MachinePrecision]), N[(1.0 / N), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if N < 0.55000000000000004

   1. Initial program 99.9%

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

   2. Taylor expanded in N around 0 97.5%

      \[\leadsto \color{blue}{-1 \cdot \log N} \]
   3. Step-by-step derivation

      1. neg-mul-1 97.5%

         \[\leadsto \color{blue}{-\log N} \]

   4. Simplified 97.5%

      \[\leadsto \color{blue}{-\log N} \]

   if 0.55000000000000004 < N

   1. Initial program 7.1%

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

   2. Taylor expanded in N around inf 99.0%

      \[\leadsto \color{blue}{\frac{1}{N}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 98.3%

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

Alternative 6: 51.9% accurate, 68.3× speedup

Math

\[\begin{array}{l} \\ \frac{1}{N} \end{array} \]

FPCore

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

C

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

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    code = 1.0d0 / n
end function

Java

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

Python

def code(N):
	return 1.0 / N

Julia

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

MATLAB

function tmp = code(N)
	tmp = 1.0 / N;
end

Wolfram

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

Derivation

1. Initial program 52.1%

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

2. Taylor expanded in N around inf 53.7%

   \[\leadsto \color{blue}{\frac{1}{N}} \]

3. Final simplification 53.7%

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

Alternative 7: 4.6% accurate, 205.0× speedup

Math

\[\begin{array}{l} \\ N \end{array} \]

FPCore

(FPCore (N) :precision binary64 N)

C

double code(double N) {
	return N;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    code = n
end function

Java

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

Python

def code(N):
	return N

Julia

function code(N)
	return N
end

MATLAB

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

Wolfram

code[N_] := N

Derivation

1. Initial program 52.1%

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

2. Taylor expanded in N around 0 49.6%

   \[\leadsto \color{blue}{N + -1 \cdot \log N} \]
3. Step-by-step derivation

   1. neg-mul-1 49.6%

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

   2. unsub-neg 49.6%

      \[\leadsto \color{blue}{N - \log N} \]

4. Simplified 49.6%

   \[\leadsto \color{blue}{N - \log N} \]

5. Taylor expanded in N around inf 4.5%

   \[\leadsto \color{blue}{N} \]

6. Final simplification 4.5%

   \[\leadsto N \]

Reproduce

herbie shell --seed 2023318 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))