Bill Ancalagon the black - 1 year ago 69

Swift Question

Refer to

Fastest Inverse Square Root on iPhone

I need do a "Fastest Inverse Square Root" on iPhone iOS Swift, which is supposed to be faster than

`1/sqrt(float)`

How do I do it?

In embedded C programming, it is:

`// Fast inverse square-root`

// See: http://en.wikipedia.org/wiki/Fast_inverse_square_root

func invSqrt(x: Float) -> Float {

var halfx : Float = 0.5 * x

var y : Float = x

long i = *(long*)&y

i = 0x5f3759df - (i>>1)

y = *(float*)&i

y = y * (1.5 - (halfx * y * y))

return y

}

Answer Source

The only tricky part is how to do the forced conversions between floating
point numbers and integer types, and the easiest way is to use
`memcpy()`

:

```
// Fast inverse square-root
// See: http://en.wikipedia.org/wiki/Fast_inverse_square_root
func invSqrt(x: Float) -> Float {
let halfx = 0.5 * x
var y = x
var i : Int32 = 0
memcpy(&i, &y, 4)
i = 0x5f3759df - (i >> 1)
memcpy(&y, &i, 4)
y = y * (1.5 - (halfx * y * y))
return y
}
```

I made some performance tests on an iPhone 6s with 1.000.000 random
floating point numbers in the range 0 ... 1000, and it turned out
that `invSqrt(x)`

is about 40% faster than `1.0/sqrt(x)`

.

The maximal relative error was below 0.176%, confirming the bound in the Wikipedia article.

I also made a test with `vvrsqrtf`

from the
Accelerate framework, but this was actually *slower* than
calling `1.0/sqrt(x)`

, at least when called with single floating
point numbers.

As of **Swift 3,** `memcpy()`

can be replaced by the `bitPattern:`

method of `Float`

and the corresponding constructor from `UInt32`

:

```
func invSqrt(x: Float) -> Float {
let halfx = 0.5 * x
var i = x.bitPattern
i = 0x5f3759df - (i >> 1)
var y = Float(bitPattern: i)
y = y * (1.5 - (halfx * y * y))
return y
}
```