英文:
Is there a simple method for square root of big.Rat?
问题
我需要找到一个大数的平方根。有没有一种方法可以在不丢失（已经存在的）精度的情况下进行计算？
例如，我可以将分子和分母转换为浮点数，获取平方根，然后再转换回来...
func ratSquareRoot(num *big.Rat) *big.Rat {
f, exact := num.Float64() //Yuck! Floats!
squareRoot := math.Sqrt(f)
var accuracy int64 = 10 ^ 15 //float64的有效数字位数
return big.NewRat(int64(squareRoot*float64(accuracy)), accuracy)
// ^ 这样做完全没有意义。而且可能没有很好地简化。
}
...但这样会消除使用有理数的所有精度。有没有更好的方法来做这个？
英文:
I need to find the square root of a big.Rat. Is there a way to do it without losing (already existing) accuracy?
For example, I could convert the numerator and denominator into floats, get the square root, and then convert it back...
func ratSquareRoot(num *big.Rat) *big.Rat {
f, exact := num.Float64() //Yuck! Floats!
squareRoot := math.Sqrt(f)
var accuracy int64 = 10 ^ 15 //Significant digits of precision for float64
return big.NewRat(int64(squareRoot*float64(accuracy)), accuracy)
// ^ This is now totally worthless. And also probably not simplified very well.
}
...but that would eliminate all of the accuracy of using a rational. Is there a better way of doing this?
答案1
得分: 2
big.Float类型具有.Sqrt(x)操作，并且可以明确定义所需的精度。我建议尝试使用它，并使用你问题中的相同操作将结果转换回Rat，只需操作big.Int值。
r := big.NewRat(1, 3)
var x big.Float
x.SetPrec(30) // 我没有正确理解'Prec'部分，比我更仔细地阅读文档并进行实验
x.SetRat(r)
var s big.Float
s.SetPrec(15)
s.Sqrt(&x)
r, _ = s.Rat(nil)
fmt.Println(x.String(), s.String())
fmt.Println(r.String(), float64(18919)/float64(32768))
英文:
The big.Float type has a .Sqrt(x) operation, and handles defining explicitly the precision you aim for. I'd try to use that and convert the result back to a Rat with the same operations in your question, only manipulating big.Int values.
r := big.NewRat(1, 3)
var x big.Float
x.SetPrec(30) // I didn't figure out the 'Prec' part correctly, read the docs more carefully than I did and experiement
x.SetRat(r)
var s big.Float
s.SetPrec(15)
s.Sqrt(&x)
r, _ = s.Rat(nil)
fmt.Println(x.String(), s.String())
fmt.Println(r.String(), float64(18919)/float64(32768))
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