Linear.Matrix:fromQuaternion from linear-1.19.1.3, B

Percentage Accurate: 95.3% → 100.0%

Time: 3.1s

Precision: binary64

Cost: 448

• 28.6% of points produce a very large (infinite) output. You may want to add a precondition.

Math

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

↓

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

FPCore

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

↓

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

C

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

↓

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

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 = 2.0d0 * (x * (x + y))
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 2.0 * (x * (x + y));
}

Python

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

↓

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

Julia

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

↓

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

MATLAB

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

↓

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

Wolfram

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

↓

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

Target

Original	95.3%
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%

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

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

3. Final simplification 100.0%

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

Alternatives

Alternative 1
Accuracy	100.0%
Cost	448

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

Alternative 2
Accuracy	83.0%
Cost	850

\[\begin{array}{l} \mathbf{if}\;y \leq -7.2 \cdot 10^{-64} \lor \neg \left(y \leq 2.1 \cdot 10^{+22}\right) \land \left(y \leq 1.85 \cdot 10^{+39} \lor \neg \left(y \leq 2.9 \cdot 10^{+56}\right)\right):\\ \;\;\;\;y \cdot \left(x + x\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(x + x\right)\\ \end{array} \]

Alternative 3
Accuracy	58.6%
Cost	320

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

Reproduce

herbie shell --seed 2023263 
(FPCore (x y)
  :name "Linear.Matrix:fromQuaternion from linear-1.19.1.3, B"
  :precision binary64

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

  (* 2.0 (+ (* x x) (* x y))))