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Oct 19, 2015 at 14:37 history edited e-sushi CC BY-SA 3.0
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Jul 28, 2015 at 18:59 comment added bmm6o @FlorianBourse the binary representability of the value being approximated is not relevant. Simpler tests like checking lsb or msb can be seen as a random estimation of 1/2. And of course the test must be able to evaluate how close the estimate is to pi, so it needs a sufficiently precise representation of the actual value.
Jul 28, 2015 at 18:21 comment added tylo I don't think that's relevant. But a 64 bit approximation is not nearly enough to estimate the quality of a RNG. If you think of applications which require a lot of random numbers (e.g. simulations), the required entropy might be higher.
Jul 28, 2015 at 8:26 answer added fgrieu timeline score: 7
Jul 27, 2015 at 22:49 comment added Paul Uszak Not sure that's relevant. It's unlikely that you'd ever be testing a file sufficiently large that the resultant pi approximation wouldn't fit into a 64 bit variable.
Jul 27, 2015 at 22:47 answer added Paul Uszak timeline score: 0
Jul 27, 2015 at 17:36 history tweeted twitter.com/#!/StackCrypto/status/625721493377392640
Mar 24, 2015 at 12:42 comment added Florian Bourse The point of computing an approximation to Pi compared to 0.5 is that Pi cannot be stored in a computer because it's binary notation doesn't terminate, whereas 0.5 is just written 0.1 in binary.
Mar 24, 2015 at 11:46 comment added yyyyyyy This is only very remotely related to cryptographic random number generators, but you're right: the circle is not inherently more useful for testing statistical randomness; it just happens to be one of the standard introductory examples for Monte Carlo simulations.
Mar 24, 2015 at 1:53 history asked Paul Uszak CC BY-SA 3.0