I might have missed it earlier Greg but was the actual problem that this helps with? I was intrigued by the underlying problem.
Regards,
Greg
Dr Greg Low
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From: Greg Keogh via ozdotnet <ozdotnet@ozdotnet.com>
Sent: Tuesday, 5 July 2022 8:37 AM
To: ozDotNet <ozdotnet@ozdotnet.com>
Cc: Greg Keogh <gfkeogh@gmail.com>
Subject: Re: 53-bit double
uint u1 = [32 random bits];
uint u2 = [32 random bits];
uint a = u1 >> 5, b = u2 >> 6;
return (a * 67108864.0 + b) * (1.0 / 9007199254740992.0);
Can anyone explain this magic? Is this correct?
This bit trick is no longer important to decode. I was trying to figure out how to convert a UInt64 of random bits into a normalised double in the range [0,1) in such a way that all 2^53 possible floating values could be produced. Some
web sites suggested simply doing this:
uint u = [64 random bits]
double d = (u >> 11) * 1.11022302462515654e-16 // * 2-53
I was suspicious that subtle rounding problems might produce a defective output range, but a quick experiment in LINQPad that fed limit values into the calculation demonstrated that a complete set of significand bit patterns was generated
at the limits. This supplies good evidence that the above shift-and-multiply converts 53 of the 64 random bits into a complete possible double range.