Linear.Matrix:fromQuaternion from linear-1.19.1.3, A

Percentage Accurate: 95.6% → 100.0%

Time: 3.0s

Precision: binary64

Cost: 448

• 27.8% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[2 \cdot \left(x \cdot x - x \cdot y\right) \]

↓

\[\left(x - y\right) \cdot \left(x \cdot 2\right) \]

FPCore

(FPCore (x y) :precision binary64 (* 2.0 (- (* x x) (* x y))))

↓

(FPCore (x y) :precision binary64 (* (- x y) (* x 2.0)))

C

double code(double x, double y) {
	return 2.0 * ((x * x) - (x * y));
}

↓

double code(double x, double y) {
	return (x - y) * (x * 2.0);
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = 2.0d0 * ((x * x) - (x * y))
end function

↓

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = (x - y) * (x * 2.0d0)
end function

Java

public static double code(double x, double y) {
	return 2.0 * ((x * x) - (x * y));
}

↓

public static double code(double x, double y) {
	return (x - y) * (x * 2.0);
}

Python

def code(x, y):
	return 2.0 * ((x * x) - (x * y))

↓

def code(x, y):
	return (x - y) * (x * 2.0)

Julia

function code(x, y)
	return Float64(2.0 * Float64(Float64(x * x) - Float64(x * y)))
end

↓

function code(x, y)
	return Float64(Float64(x - y) * Float64(x * 2.0))
end

MATLAB

function tmp = code(x, y)
	tmp = 2.0 * ((x * x) - (x * y));
end

↓

function tmp = code(x, y)
	tmp = (x - y) * (x * 2.0);
end

Wolfram

code[x_, y_] := N[(2.0 * N[(N[(x * x), $MachinePrecision] - N[(x * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

↓

code[x_, y_] := N[(N[(x - y), $MachinePrecision] * N[(x * 2.0), $MachinePrecision]), $MachinePrecision]

Target

Original	95.6%
Target	100.0%
Herbie	100.0%

\[\left(x \cdot 2\right) \cdot \left(x - y\right) \]

Derivation

1. Initial program 96.9%

   \[2 \cdot \left(x \cdot x - x \cdot y\right) \]

2. Simplified 100.0%

   \[\leadsto \color{blue}{\left(x - y\right) \cdot \left(2 \cdot x\right)} \]
   Step-by-step derivation

   [Start] 96.9%	\[ 2 \cdot \left(x \cdot x - x \cdot y\right) \]
   distribute-lft-out-- [=>] 100.0%	\[ 2 \cdot \color{blue}{\left(x \cdot \left(x - y\right)\right)} \]
   associate-*r* [=>] 100.0%	\[ \color{blue}{\left(2 \cdot x\right) \cdot \left(x - y\right)} \]
   *-commutative [=>] 100.0%	\[ \color{blue}{\left(x - y\right) \cdot \left(2 \cdot x\right)} \]

3. Final simplification 100.0%

   \[\leadsto \left(x - y\right) \cdot \left(x \cdot 2\right) \]

Alternatives

Alternative 1
Accuracy	100.0%
Cost	448

\[\left(x - y\right) \cdot \left(x \cdot 2\right) \]

Alternative 2
Accuracy	84.7%
Cost	585

\[\begin{array}{l} \mathbf{if}\;y \leq -2.3 \cdot 10^{+31} \lor \neg \left(y \leq 2.1 \cdot 10^{-47}\right):\\ \;\;\;\;y \cdot \left(x \cdot -2\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(x + x\right)\\ \end{array} \]

Alternative 3
Accuracy	59.7%
Cost	320

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

Reproduce

herbie shell --seed 2023272 
(FPCore (x y)
  :name "Linear.Matrix:fromQuaternion from linear-1.19.1.3, A"
  :precision binary64

  :herbie-target
  (* (* x 2.0) (- x y))

  (* 2.0 (- (* x x) (* x y))))