Rosa's Benchmark

Percentage Accurate: 99.8% → 99.8%

Time: 6.5s

Precision: binary64

Cost: 6912

• 33.6% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[0.954929658551372 \cdot x - 0.12900613773279798 \cdot \left(\left(x \cdot x\right) \cdot x\right) \]

↓

\[0.954929658551372 \cdot x - {x}^{3} \cdot 0.12900613773279798 \]

FPCore

(FPCore (x)
 :precision binary64
 (- (* 0.954929658551372 x) (* 0.12900613773279798 (* (* x x) x))))

↓

(FPCore (x)
 :precision binary64
 (- (* 0.954929658551372 x) (* (pow x 3.0) 0.12900613773279798)))

C

double code(double x) {
	return (0.954929658551372 * x) - (0.12900613773279798 * ((x * x) * x));
}

↓

double code(double x) {
	return (0.954929658551372 * x) - (pow(x, 3.0) * 0.12900613773279798);
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = (0.954929658551372d0 * x) - (0.12900613773279798d0 * ((x * x) * x))
end function

↓

real(8) function code(x)
    real(8), intent (in) :: x
    code = (0.954929658551372d0 * x) - ((x ** 3.0d0) * 0.12900613773279798d0)
end function

Java

public static double code(double x) {
	return (0.954929658551372 * x) - (0.12900613773279798 * ((x * x) * x));
}

↓

public static double code(double x) {
	return (0.954929658551372 * x) - (Math.pow(x, 3.0) * 0.12900613773279798);
}

Python

def code(x):
	return (0.954929658551372 * x) - (0.12900613773279798 * ((x * x) * x))

↓

def code(x):
	return (0.954929658551372 * x) - (math.pow(x, 3.0) * 0.12900613773279798)

Julia

function code(x)
	return Float64(Float64(0.954929658551372 * x) - Float64(0.12900613773279798 * Float64(Float64(x * x) * x)))
end

↓

function code(x)
	return Float64(Float64(0.954929658551372 * x) - Float64((x ^ 3.0) * 0.12900613773279798))
end

MATLAB

function tmp = code(x)
	tmp = (0.954929658551372 * x) - (0.12900613773279798 * ((x * x) * x));
end

↓

function tmp = code(x)
	tmp = (0.954929658551372 * x) - ((x ^ 3.0) * 0.12900613773279798);
end

Wolfram

code[x_] := N[(N[(0.954929658551372 * x), $MachinePrecision] - N[(0.12900613773279798 * N[(N[(x * x), $MachinePrecision] * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

↓

code[x_] := N[(N[(0.954929658551372 * x), $MachinePrecision] - N[(N[Power[x, 3.0], $MachinePrecision] * 0.12900613773279798), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.9%

   \[0.954929658551372 \cdot x - 0.12900613773279798 \cdot \left(\left(x \cdot x\right) \cdot x\right) \]

2. Taylor expanded in x around 0 99.9%

   \[\leadsto 0.954929658551372 \cdot x - \color{blue}{0.12900613773279798 \cdot {x}^{3}} \]

3. Simplified 99.9%

   \[\leadsto 0.954929658551372 \cdot x - \color{blue}{{x}^{3} \cdot 0.12900613773279798} \]
   Step-by-step derivation

   [Start] 99.9%	\[ 0.954929658551372 \cdot x - 0.12900613773279798 \cdot {x}^{3} \]
   *-commutative [=>] 99.9%	\[ 0.954929658551372 \cdot x - \color{blue}{{x}^{3} \cdot 0.12900613773279798} \]

4. Final simplification 99.9%

   \[\leadsto 0.954929658551372 \cdot x - {x}^{3} \cdot 0.12900613773279798 \]

Alternatives

Alternative 1
Accuracy	99.8%
Cost	6912

\[0.954929658551372 \cdot x - {x}^{3} \cdot 0.12900613773279798 \]

Alternative 2
Accuracy	98.9%
Cost	713

\[\begin{array}{l} \mathbf{if}\;x \leq -2.7 \lor \neg \left(x \leq 2.75\right):\\ \;\;\;\;x \cdot \left(\left(x \cdot x\right) \cdot -0.12900613773279798\right)\\ \mathbf{else}:\\ \;\;\;\;0.954929658551372 \cdot x\\ \end{array} \]

Alternative 3
Accuracy	99.8%
Cost	704

\[0.954929658551372 \cdot x - 0.12900613773279798 \cdot \left(x \cdot \left(x \cdot x\right)\right) \]

Alternative 4
Accuracy	99.8%
Cost	576

\[x \cdot \left(0.954929658551372 + \left(x \cdot x\right) \cdot -0.12900613773279798\right) \]

Alternative 5
Accuracy	5.1%
Cost	192

\[x \cdot -0.954929658551372 \]

Alternative 6
Accuracy	50.2%
Cost	192

\[0.954929658551372 \cdot x \]

Reproduce

herbie shell --seed 2023245 
(FPCore (x)
  :name "Rosa's Benchmark"
  :precision binary64
  (- (* 0.954929658551372 x) (* 0.12900613773279798 (* (* x x) x))))