Compound Interest

Average Accuracy: 26.1% → 99.2%

Time: 18.0s

Precision: binary64

Cost: 21768

Math

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

↓

\[\begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n} + -1\\ t_1 := \frac{t_0}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq 0:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;t_1 \leq 4000000000:\\ \;\;\;\;\frac{\left(n \cdot 100\right) \cdot t_0}{i}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \end{array} \]

FPCore

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

↓

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ (pow (+ 1.0 (/ i n)) n) -1.0)) (t_1 (/ t_0 (/ i n))))
   (if (<= t_1 0.0)
     (* 100.0 (/ (expm1 (* n (log1p (/ i n)))) (/ i n)))
     (if (<= t_1 4000000000.0)
       (/ (* (* n 100.0) t_0) i)
       (* 100.0 (/ n (+ 1.0 (* i -0.5))))))))

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 t_0 = pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (expm1((n * log1p((i / n)))) / (i / n));
	} else if (t_1 <= 4000000000.0) {
		tmp = ((n * 100.0) * t_0) / i;
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	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 t_0 = Math.pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (Math.expm1((n * Math.log1p((i / n)))) / (i / n));
	} else if (t_1 <= 4000000000.0) {
		tmp = ((n * 100.0) * t_0) / i;
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	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):
	t_0 = math.pow((1.0 + (i / n)), n) + -1.0
	t_1 = t_0 / (i / n)
	tmp = 0
	if t_1 <= 0.0:
		tmp = 100.0 * (math.expm1((n * math.log1p((i / n)))) / (i / n))
	elif t_1 <= 4000000000.0:
		tmp = ((n * 100.0) * t_0) / i
	else:
		tmp = 100.0 * (n / (1.0 + (i * -0.5)))
	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)
	t_0 = Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0)
	t_1 = Float64(t_0 / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(expm1(Float64(n * log1p(Float64(i / n)))) / Float64(i / n)));
	elseif (t_1 <= 4000000000.0)
		tmp = Float64(Float64(Float64(n * 100.0) * t_0) / i);
	else
		tmp = Float64(100.0 * Float64(n / Float64(1.0 + Float64(i * -0.5))));
	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_] := Block[{t$95$0 = N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, 4000000000.0], N[(N[(N[(n * 100.0), $MachinePrecision] * t$95$0), $MachinePrecision] / i), $MachinePrecision], N[(100.0 * N[(n / N[(1.0 + N[(i * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Target

Original	26.1%
Target	26.7%
Herbie	99.2%

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

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 0.0

   1. Initial program 28.6%

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

   2. Applied egg-rr 99.6%

      \[\leadsto 100 \cdot \frac{\color{blue}{1 \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]

   3. Simplified 99.6%

      \[\leadsto 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]
      Proof

      [Start] 99.6	\[ 100 \cdot \frac{1 \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}} \]
      *-lft-identity [=>] 99.6	\[ 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]

   if 0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 4e9

   1. Initial program 95.8%

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

   2. Simplified 95.9%

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

      [Start] 95.8	\[ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
      associate-/r/ [=>] 95.9	\[ 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]
      *-commutative [=>] 95.9	\[ 100 \cdot \color{blue}{\left(n \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]
      *-rgt-identity [<=] 95.9	\[ 100 \cdot \left(\color{blue}{\left(n \cdot 1\right)} \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \]
      associate-*l* [=>] 95.9	\[ 100 \cdot \color{blue}{\left(n \cdot \left(1 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)\right)} \]
      *-lft-identity [=>] 95.9	\[ 100 \cdot \left(n \cdot \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}}\right) \]
      sub-neg [=>] 95.9	\[ 100 \cdot \left(n \cdot \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)}}{i}\right) \]
      metadata-eval [=>] 95.9	\[ 100 \cdot \left(n \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-1}}{i}\right) \]

   3. Applied egg-rr 96.0%

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

   if 4e9 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

   1. Initial program 1.2%

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

   2. Taylor expanded in n around inf 2.7%

      \[\leadsto \color{blue}{100 \cdot \frac{n \cdot \left(e^{i} - 1\right)}{i}} \]

   3. Simplified 78.3%

      \[\leadsto \color{blue}{\frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}} \cdot 100} \]
      Proof

      [Start] 2.7	\[ 100 \cdot \frac{n \cdot \left(e^{i} - 1\right)}{i} \]
      *-commutative [=>] 2.7	\[ \color{blue}{\frac{n \cdot \left(e^{i} - 1\right)}{i} \cdot 100} \]
      associate-/l* [=>] 2.7	\[ \color{blue}{\frac{n}{\frac{i}{e^{i} - 1}}} \cdot 100 \]
      expm1-def [=>] 78.3	\[ \frac{n}{\frac{i}{\color{blue}{\mathsf{expm1}\left(i\right)}}} \cdot 100 \]

   4. Taylor expanded in i around 0 98.2%

      \[\leadsto \frac{n}{\color{blue}{1 + -0.5 \cdot i}} \cdot 100 \]

   5. Simplified 98.2%

      \[\leadsto \frac{n}{\color{blue}{1 + i \cdot -0.5}} \cdot 100 \]
      Proof

      [Start] 98.2	\[ \frac{n}{1 + -0.5 \cdot i} \cdot 100 \]
      *-commutative [=>] 98.2	\[ \frac{n}{1 + \color{blue}{i \cdot -0.5}} \cdot 100 \]

3. Recombined 3 regimes into one program.

4. Final simplification 99.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 4000000000:\\ \;\;\;\;\frac{\left(n \cdot 100\right) \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} + -1\right)}{i}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	98.2%
Cost	21768

\[\begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n} + -1\\ t_1 := \frac{t_0}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{i}\right)\\ \mathbf{elif}\;t_1 \leq 4000000000:\\ \;\;\;\;\frac{\left(n \cdot 100\right) \cdot t_0}{i}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \end{array} \]

Alternative 2
Accuracy	82.0%
Cost	7244

\[\begin{array}{l} t_0 := 100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{if}\;n \leq -4.2 \cdot 10^{-225}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 2 \cdot 10^{-238}:\\ \;\;\;\;0\\ \mathbf{elif}\;n \leq 5 \cdot 10^{+19}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot \left(-0.5 + i \cdot 0.08333333333333333\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 3
Accuracy	82.1%
Cost	7244

\[\begin{array}{l} t_0 := 100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{if}\;n \leq -4.2 \cdot 10^{-227}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 1.9 \cdot 10^{-240}:\\ \;\;\;\;0\\ \mathbf{elif}\;n \leq 5 \cdot 10^{+19}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot \left(-0.5 + i \cdot 0.08333333333333333\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 4
Accuracy	71.5%
Cost	1229

\[\begin{array}{l} \mathbf{if}\;n \leq -4.8 \cdot 10^{-30}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \mathbf{elif}\;n \leq -4.4 \cdot 10^{-227} \lor \neg \left(n \leq 1.4 \cdot 10^{-236}\right):\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot \left(-0.5 + i \cdot 0.08333333333333333\right)}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 5
Accuracy	69.1%
Cost	841

\[\begin{array}{l} \mathbf{if}\;n \leq -6.8 \cdot 10^{-227} \lor \neg \left(n \leq 4 \cdot 10^{-218}\right):\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 6
Accuracy	69.1%
Cost	713

\[\begin{array}{l} \mathbf{if}\;n \leq -2.05 \cdot 10^{-219} \lor \neg \left(n \leq 4 \cdot 10^{-218}\right):\\ \;\;\;\;\frac{n}{0.01 + i \cdot -0.005}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 7
Accuracy	62.2%
Cost	456

\[\begin{array}{l} \mathbf{if}\;n \leq -3.5 \cdot 10^{-85}:\\ \;\;\;\;n \cdot 100\\ \mathbf{elif}\;n \leq 2.6 \cdot 10^{-204}:\\ \;\;\;\;0\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]

Alternative 8
Accuracy	21.0%
Cost	64

\[0 \]

Reproduce

herbie shell --seed 2023141 
(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))))