Compound Interest

Percentage Accurate: 33.9% → 99.6%

Time: 38.3s

Precision: binary64

Cost: 20740

• could not determine a ground truth

  1. i = 1.5004564603928934e-240
  2. n = -1.593288660425601e+149

Math

\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]

↓

\[\begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))

↓

(FPCore (i n)
 :precision binary64
 (if (<= (/ (+ (pow (+ 1.0 (/ i n)) n) -1.0) (/ i n)) INFINITY)
   (/ (* (expm1 (* n (log1p (/ i n)))) 100.0) (/ i n))
   (* 100.0 (/ i (/ i n)))))

C

double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

↓

double code(double i, double n) {
	double tmp;
	if (((pow((1.0 + (i / n)), n) + -1.0) / (i / n)) <= ((double) INFINITY)) {
		tmp = (expm1((n * log1p((i / n)))) * 100.0) / (i / n);
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	return 100.0 * ((Math.pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

↓

public static double code(double i, double n) {
	double tmp;
	if (((Math.pow((1.0 + (i / n)), n) + -1.0) / (i / n)) <= Double.POSITIVE_INFINITY) {
		tmp = (Math.expm1((n * Math.log1p((i / n)))) * 100.0) / (i / n);
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Python

def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) - 1.0) / (i / n))

↓

def code(i, n):
	tmp = 0
	if ((math.pow((1.0 + (i / n)), n) + -1.0) / (i / n)) <= math.inf:
		tmp = (math.expm1((n * math.log1p((i / n)))) * 100.0) / (i / n)
	else:
		tmp = 100.0 * (i / (i / n))
	return tmp

Julia

function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) - 1.0) / Float64(i / n)))
end

↓

function code(i, n)
	tmp = 0.0
	if (Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0) / Float64(i / n)) <= Inf)
		tmp = Float64(Float64(expm1(Float64(n * log1p(Float64(i / n)))) * 100.0) / Float64(i / n));
	else
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	end
	return tmp
end

Wolfram

code[i_, n_] := N[(100.0 * N[(N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] - 1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

↓

code[i_, n_] := If[LessEqual[N[(N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], Infinity], N[(N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] * 100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Target

Original	33.9%
Target	23.8%
Herbie	99.6%

\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}} \]

Derivation

1. Split input into 2 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < +inf.0

   1. Initial program 34.7%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]

   2. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}} \]
      Step-by-step derivation

      [Start] 34.7%	\[ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
      associate-*r/ [=>] 34.7%	\[ \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]
      *-commutative [=>] 34.7%	\[ \frac{\color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right) \cdot 100}}{\frac{i}{n}} \]
      pow-to-exp [=>] 34.6%	\[ \frac{\left(\color{blue}{e^{\log \left(1 + \frac{i}{n}\right) \cdot n}} - 1\right) \cdot 100}{\frac{i}{n}} \]
      expm1-def [=>] 56.2%	\[ \frac{\color{blue}{\mathsf{expm1}\left(\log \left(1 + \frac{i}{n}\right) \cdot n\right)} \cdot 100}{\frac{i}{n}} \]
      add-log-exp [=>] 34.6%	\[ \frac{\mathsf{expm1}\left(\color{blue}{\log \left(e^{\log \left(1 + \frac{i}{n}\right) \cdot n}\right)}\right) \cdot 100}{\frac{i}{n}} \]
      pow-to-exp [<=] 34.6%	\[ \frac{\mathsf{expm1}\left(\log \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n}\right)}\right) \cdot 100}{\frac{i}{n}} \]
      log-pow [=>] 56.2%	\[ \frac{\mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right) \cdot 100}{\frac{i}{n}} \]
      log1p-udef [<=] 99.5%	\[ \frac{\mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right) \cdot 100}{\frac{i}{n}} \]

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

   1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]

   2. Taylor expanded in i around 0 100.0%

      \[\leadsto 100 \cdot \frac{\color{blue}{i}}{\frac{i}{n}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 99.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	99.6%
Cost	20740

\[\begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]

Alternative 2
Accuracy	99.2%
Cost	20740

\[\begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;n \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{100}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]

Alternative 3
Accuracy	99.2%
Cost	20740

\[\begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{n}{\frac{i}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)} \cdot 0.01}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]

Alternative 4
Accuracy	99.2%
Cost	20740

\[\begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{n}{\frac{i}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]

Alternative 5
Accuracy	84.6%
Cost	13636

\[\begin{array}{l} \mathbf{if}\;i \leq 1.95 \cdot 10^{-24}:\\ \;\;\;\;n \cdot \frac{1}{\mathsf{fma}\left(0.01, i \cdot \left(\frac{0.5}{n} + -0.5\right), 0.01\right)}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n \cdot n}{\frac{i}{\log i - \log n}}\\ \end{array} \]

Alternative 6
Accuracy	81.4%
Cost	7364

\[\begin{array}{l} \mathbf{if}\;n \leq 1.5:\\ \;\;\;\;n \cdot \frac{1}{\mathsf{fma}\left(0.01, i \cdot \left(\frac{0.5}{n} + -0.5\right), 0.01\right)}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \end{array} \]

Alternative 7
Accuracy	81.4%
Cost	6980

\[\begin{array}{l} \mathbf{if}\;n \leq 1:\\ \;\;\;\;\frac{n}{0.01 + 0.01 \cdot \left(i \cdot \left(\frac{0.5}{n} + -0.5\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \end{array} \]

Alternative 8
Accuracy	80.3%
Cost	832

\[\frac{n}{0.01 \cdot \left(1 + i \cdot \left(\frac{0.5}{n} + -0.5\right)\right)} \]

Alternative 9
Accuracy	80.3%
Cost	832

\[\frac{n}{0.01 + 0.01 \cdot \left(i \cdot \left(\frac{0.5}{n} + -0.5\right)\right)} \]

Alternative 10
Accuracy	71.9%
Cost	713

\[\begin{array}{l} \mathbf{if}\;i \leq -1.1 \cdot 10^{-54} \lor \neg \left(i \leq 4.8 \cdot 10^{-83}\right):\\ \;\;\;\;\frac{\left(n \cdot n\right) \cdot 200}{i}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]

Alternative 11
Accuracy	80.2%
Cost	576

\[\frac{n}{0.01 + \frac{i}{n} \cdot 0.005} \]

Alternative 12
Accuracy	61.5%
Cost	448

\[100 \cdot \frac{i}{\frac{i}{n}} \]

Alternative 13
Accuracy	43.9%
Cost	192

\[n \cdot 100 \]

Reproduce

herbie shell --seed 2023277 
(FPCore (i n)
  :name "Compound Interest"
  :precision binary64

  :herbie-target
  (* 100.0 (/ (- (exp (* n (if (== (+ 1.0 (/ i n)) 1.0) (/ i n) (/ (* (/ i n) (log (+ 1.0 (/ i n)))) (- (+ (/ i n) 1.0) 1.0))))) 1.0) (/ i n)))

  (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))