//
// There is one bug in the value stream: r.Int63() may be so close
// to 1<<63 that the division rounds up to 1.0, and we've guaranteed
- // that the result is always less than 1.0. To fix that, we treat the
- // range as cyclic and map 1 back to 0. This is justified by observing
- // that while some of the values rounded down to 0, nothing was
- // rounding up to 0, so 0 was underrepresented in the results.
- // Mapping 1 back to zero restores some balance.
- // (The balance is not perfect because the implementation
- // returns denormalized numbers for very small r.Int63(),
- // and those steal from what would normally be 0 results.)
- // The remapping only happens 1/2⁵³ of the time, so most clients
+ // that the result is always less than 1.0.
+ //
+ // We tried to fix this by mapping 1.0 back to 0.0, but since float64
+ // values near 0 are much denser than near 1, mapping 1 to 0 caused
+ // a theoretically significant overshoot in the probability of returning 0.
+ // Instead of that, if we round up to 1, just try again.
+ // Getting 1 only happens 1/2⁵³ of the time, so most clients
// will not observe it anyway.
+again:
f := float64(r.Int63()) / (1 << 63)
if f == 1 {
- f = 0
+ goto again // resample; this branch is taken O(never)
}
return f
}
func (r *Rand) Float32() float32 {
// Same rationale as in Float64: we want to preserve the Go 1 value
// stream except we want to fix it not to return 1.0
- // There is a double rounding going on here, but the argument for
- // mapping 1 to 0 still applies: 0 was underrepresented before,
- // so mapping 1 to 0 doesn't cause too many 0s.
// This only happens 1/2²⁴ of the time (plus the 1/2⁵³ of the time in Float64).
+again:
f := float32(r.Float64())
if f == 1 {
- f = 0
+ goto again // resample; this branch is taken O(very rarely)
}
return f
}