math/rand: restore Go 1.2 value stream for Float32, Float64

CL 22730043 fixed a bug in these functions: they could
return 1.0 despite documentation saying otherwise.
But the fix changed the values returned in the non-buggy case too,
which might invalidate programs depending on a particular
stream when using rand.Seed(0) or when passing their own
Source to rand.New.

The example test says:
        // These tests serve as an example but also make sure we don't change
        // the output of the random number generator when given a fixed seed.
so I think there is some justification for thinking we have
promised not to change the values. In any case, there's no point in
changing the values gratuitously: we can easily fix this bug without
changing the values, and so we should.

That CL just changed the test values too, which defeats the
stated purpose, but it was just a comment.
Add an explicit regression test, which might be
a clearer signal next time that we don't want to change
the values.

Fixes #6721. (again)
Fixes #8013.

LGTM=r
R=iant, r
CC=golang-codereviews
https://golang.org/cl/95460049

src/pkg/math/rand/example_test.go

@@ -83,8 +83,8 @@ func Example_rand() {
 	// Perm generates a random permutation of the numbers [0, n).
 	show("Perm", r.Perm(5), r.Perm(5), r.Perm(5))
 	// Output:
-	// Float32     0.73793465          0.38461488          0.9940225
-	// Float64     0.6919607852308565  0.29140004584133117 0.2262092163027547
+	// Float32     0.2635776           0.6358173           0.6718283
+	// Float64     0.628605430454327   0.4504798828572669  0.9562755949377957
 	// ExpFloat64  0.3362240648200941  1.4256072328483647  0.24354758816173044
 	// NormFloat64 0.17233959114940064 1.577014951434847   0.04259129641113857
 	// Int31       1501292890          1486668269          182840835

src/pkg/math/rand/rand.go

@@ -101,10 +101,46 @@ func (r *Rand) Intn(n int) int {
 }
 
 // Float64 returns, as a float64, a pseudo-random number in [0.0,1.0).
-func (r *Rand) Float64() float64 { return float64(r.Int63n(1<<53)) / (1 << 53) }
+func (r *Rand) Float64() float64 {
+	// A clearer, simpler implementation would be:
+	//	return float64(r.Int63n(1<<53)) / (1<<53)
+	// However, Go 1 shipped with
+	//	return float64(r.Int63()) / (1 << 63)
+	// and we want to preserve that value stream.
+	//
+	// 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
+	// will not observe it anyway.
+	f := float64(r.Int63()) / (1 << 63)
+	if f == 1 {
+		f = 0
+	}
+	return f
+}
 
 // Float32 returns, as a float32, a pseudo-random number in [0.0,1.0).
-func (r *Rand) Float32() float32 { return float32(r.Int31n(1<<24)) / (1 << 24) }
+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).
+	f := float32(r.Float64())
+	if f == 1 {
+		f = 0
+	}
+	return f
+}
 
 // Perm returns, as a slice of n ints, a pseudo-random permutation of the integers [0,n).
 func (r *Rand) Perm(n int) []int {