math.cube on real

Percentage Accurate: 99.9% → 100.0%

Time: 1.8s

Precision: binary64

Cost: 6528

• 33.5% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[\left(x \cdot x\right) \cdot x \]

↓

\[{x}^{3} \]

FPCore

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

↓

(FPCore (x) :precision binary64 (pow x 3.0))

C

double code(double x) {
	return (x * x) * x;
}

↓

double code(double x) {
	return pow(x, 3.0);
}

Fortran

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

↓

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

Java

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

↓

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

Python

def code(x):
	return (x * x) * x

↓

def code(x):
	return math.pow(x, 3.0)

Julia

function code(x)
	return Float64(Float64(x * x) * x)
end

↓

function code(x)
	return x ^ 3.0
end

MATLAB

function tmp = code(x)
	tmp = (x * x) * x;
end

↓

function tmp = code(x)
	tmp = x ^ 3.0;
end

Wolfram

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

↓

code[x_] := N[Power[x, 3.0], $MachinePrecision]

Target

Original	99.9%
Target	100.0%
Herbie	100.0%

\[{x}^{3} \]

Derivation

1. Initial program 99.8%

   \[\left(x \cdot x\right) \cdot x \]

2. Simplified 100.0%

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

   [Start] 99.8	\[ \left(x \cdot x\right) \cdot x \]
   unpow3 [<=] 100.0	\[ \color{blue}{{x}^{3}} \]

3. Final simplification 100.0%

   \[\leadsto {x}^{3} \]

Alternatives

Alternative 1
Accuracy	99.9%
Cost	320

\[x \cdot \left(x \cdot x\right) \]

Reproduce

herbie shell --seed 2023160 
(FPCore (x)
  :name "math.cube on real"
  :precision binary64

  :herbie-target
  (pow x 3.0)

  (* (* x x) x))