Agfdhyk

An easy way to generate a random float64 in [0,1) is by generating a uniformly random int in [0,2⁵³) and dividing it by 2⁵³. This is essentially what rand.Float64() is doing.
However, not all possible float64 values between 0 and 1 can be generated this way: if the value is lower than 2⁻⁴ for example, the 4 last bits of the significand are always going to be 0. Or, put more simply, the naive method always returns multiples of 2⁻⁵³, and not all floating point numbers between 0 and 1 are multiples of 2⁻⁵³.

How do you generate a uniformly random float64 such as every possible value has a chance of being returned? (Here, uniformly random means over the real interval [0,1): conceptually, I want to pick a uniformly random real number between 0 and 1 and return the closest float.)

For context, I need this because I'm implementing this paper and the assumption "all possible values between 0 and 1 are represented" is essential for the result to hold.

Because the binary64 floating point numbers are not uniformly spaced, you cannot generate a uniform distribution which can return all possible values less that 1.

If you omit the requirement uniform you have to generate all representable multiples of the smallest positive denormal number 2^(-1074)and zero.

Porting this code (suggested in Severin's answer) is a possible option.

I think that it is equivalent to first generate the significand bits (by generating a random float in [1,2)), and then choose the exponent from a geometric distribution (it has a 0.5 chance of being -1, 0.25 of being -2, etc.).

// uniform returns a uniformly random float in [0,1).

func uniform() float64 {

  sig := rand.Uint64() % (1 << 52)

  return (1 + float64(i)/(1<<52)) / math.Pow(2, geometric())

}



// geometric returns a number picked from a geometric

// distribution of parameter 0.5.

func geometric() float64 {

  b := 1

  for rand.Uint64()%2 == 0 {

     b++

  }

  return b

}

We can probably make geometric() faster by using one of the LeadingZeros* functions from the bits package instead of doing one coin flip per bit.

Well, standard way, I believe, is to generate up to 1074bits integer and map it to the double. Beware, that your RNG should have internal state at least 1074bits long.

Reference implementation: http://xoshiro.di.unimi.it/random_real.c

Discussion about it: http://xoshiro.di.unimi.it/

Another good link: https://lemire.me/blog/2017/02/28/how-many-floating-point-numbers-are-in-the-interval-01/

You could use brute-force rejection sampling by generating 16 random bytes and using it only if it's a valid float64 in [0,1). This approach should give you a normal distribution of all values in that range with performance not much worse than other strategies based on simple benchmarking.

For example (Go Playground):

import "math/rand"



func randFloat64() float64 {

  for {

    f := math.Float64frombits(rand.Uint64())

    if f >= 0 && f < 1.0 {

      return f

    }

  }

}

If performance is critical then you could build an enormous lookup table containing only the valid numbers and choose a random location in the table. The table could be generated ahead of time in a similar fashion as above, by enumerating the bitfield and storing only valid numbers.

You can use rand Float64 function to return float in the [0, 1) range:

package main



import (

    "fmt"

    "math/rand"

)



func main() {

    fmt.Println(rand.Float64())

}

From the docs:

Float64 returns, as a float64, a pseudo-random number in [0.0,1.0) from the default Source.