Timeline for Why calculate pi to estimate randomness?
Current License: CC BY-SA 3.0
10 events
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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 |