expax (section 3.5)

Percentage Accurate: 64.5% → 100.0%

Time: 9.3s

Precision: binary64

Cost: 6592

• Unsound rule application detected in e-graph. Results may not be sound.

• 24.6% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[e^{a \cdot x} - 1 \]

↓

\[\mathsf{expm1}\left(a \cdot x\right) \]

FPCore

(FPCore (a x) :precision binary64 (- (exp (* a x)) 1.0))

↓

(FPCore (a x) :precision binary64 (expm1 (* a x)))

C

double code(double a, double x) {
	return exp((a * x)) - 1.0;
}

↓

double code(double a, double x) {
	return expm1((a * x));
}

Java

public static double code(double a, double x) {
	return Math.exp((a * x)) - 1.0;
}

↓

public static double code(double a, double x) {
	return Math.expm1((a * x));
}

Python

def code(a, x):
	return math.exp((a * x)) - 1.0

↓

def code(a, x):
	return math.expm1((a * x))

Julia

function code(a, x)
	return Float64(exp(Float64(a * x)) - 1.0)
end

↓

function code(a, x)
	return expm1(Float64(a * x))
end

Wolfram

code[a_, x_] := N[(N[Exp[N[(a * x), $MachinePrecision]], $MachinePrecision] - 1.0), $MachinePrecision]

↓

code[a_, x_] := N[(Exp[N[(a * x), $MachinePrecision]] - 1), $MachinePrecision]

Target

Original	64.5%
Target	99.8%
Herbie	100.0%

\[\begin{array}{l} \mathbf{if}\;\left|a \cdot x\right| < 0.1:\\ \;\;\;\;\left(a \cdot x\right) \cdot \left(1 + \left(\frac{a \cdot x}{2} + \frac{{\left(a \cdot x\right)}^{2}}{6}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \end{array} \]

Derivation

1. Initial program 66.4%

   \[e^{a \cdot x} - 1 \]

2. Simplified 99.9%

   \[\leadsto \color{blue}{\mathsf{expm1}\left(a \cdot x\right)} \]
   Step-by-step derivation

   [Start] 66.4	\[ e^{a \cdot x} - 1 \]
   expm1-def [=>] 99.9	\[ \color{blue}{\mathsf{expm1}\left(a \cdot x\right)} \]

3. Final simplification 99.9%

   \[\leadsto \mathsf{expm1}\left(a \cdot x\right) \]

Alternatives

Alternative 1
Accuracy	69.2%
Cost	1224

\[\begin{array}{l} \mathbf{if}\;a \cdot x \leq 5 \cdot 10^{-10}:\\ \;\;\;\;a \cdot x\\ \mathbf{elif}\;a \cdot x \leq 5 \cdot 10^{+73}:\\ \;\;\;\;a \cdot \left(x + a \cdot \left(x \cdot \left(x \cdot 0.5\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(0.5 \cdot \left(a \cdot a\right)\right) \cdot \left(x \cdot x\right)\\ \end{array} \]

Alternative 2
Accuracy	68.3%
Cost	836

\[\begin{array}{l} \mathbf{if}\;a \cdot x \leq 0.001:\\ \;\;\;\;a \cdot x\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(\left(a \cdot 0.5\right) \cdot \left(x \cdot x\right)\right)\\ \end{array} \]

Alternative 3
Accuracy	68.0%
Cost	836

\[\begin{array}{l} \mathbf{if}\;a \cdot x \leq 2 \cdot 10^{+41}:\\ \;\;\;\;a \cdot x\\ \mathbf{else}:\\ \;\;\;\;\left(0.5 \cdot \left(a \cdot a\right)\right) \cdot \left(x \cdot x\right)\\ \end{array} \]

Alternative 4
Accuracy	66.1%
Cost	836

\[\begin{array}{l} \mathbf{if}\;x \leq -1.22 \cdot 10^{-249}:\\ \;\;\;\;x \cdot \left(a + x \cdot \left(a \cdot \left(a \cdot 0.5\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(x \cdot \left(1 + a \cdot \left(x \cdot 0.5\right)\right)\right)\\ \end{array} \]

Alternative 5
Accuracy	66.2%
Cost	836

\[\begin{array}{l} \mathbf{if}\;x \leq -1.05 \cdot 10^{-247}:\\ \;\;\;\;x \cdot \left(a + x \cdot \left(a \cdot \left(a \cdot 0.5\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(x + \left(a \cdot x\right) \cdot \left(x \cdot 0.5\right)\right)\\ \end{array} \]

Alternative 6
Accuracy	58.1%
Cost	192

\[a \cdot x \]

Reproduce

herbie shell --seed 2023161 
(FPCore (a x)
  :name "expax (section 3.5)"
  :precision binary64

  :herbie-target
  (if (< (fabs (* a x)) 0.1) (* (* a x) (+ 1.0 (+ (/ (* a x) 2.0) (/ (pow (* a x) 2.0) 6.0)))) (- (exp (* a x)) 1.0))

  (- (exp (* a x)) 1.0))