Comments on: Whirlygig Verification and rngtest analysis http://warmcat.com/_wp/2009/05/21/whirlygig-verification-and-rngtest-analysis/ Embedded and desktop Linux Thu, 18 Feb 2010 12:20:42 +0000 http://wordpress.org/?v=2.8.6 hourly 1 By: Whirlygig, Hardware RNG per Linux | bruno trani dot info http://warmcat.com/_wp/2009/05/21/whirlygig-verification-and-rngtest-analysis/comment-page-1/#comment-3638 Whirlygig, Hardware RNG per Linux | bruno trani dot info Thu, 18 Feb 2010 12:04:32 +0000 http://warmcat.com/_wp/?p=54#comment-3638 [...] il software per realizzare il tutto da soli. Per confermare la validità dell’approccio sono verificati che i dati generati fosse effettivamente casuali ed il risultato è più che [...] [...] il software per realizzare il tutto da soli. Per confermare la validità dell’approccio sono verificati che i dati generati fosse effettivamente casuali ed il risultato è più che [...]

]]> By: sysadmn http://warmcat.com/_wp/2009/05/21/whirlygig-verification-and-rngtest-analysis/comment-page-1/#comment-3633 sysadmn Wed, 10 Feb 2010 16:52:11 +0000 http://warmcat.com/_wp/?p=54#comment-3633 Re: Monobit A detailed discussion of the tests is at http://www.random.org/analysis/Analysis2005.pdf. It is presented in the context of testing random.org's randomness, but the results apply to WhirlyGig as well. I'm not a statistician, but here's what I think is happening. The test maps 0 -> -1, and 1->1. Sum all the bits in the block, and then normalize (take the absolute value, so the error "too many 0's" is the same as "too many 1's", then divide by the square root of the number of bits in the block). This normalized value is put in the "complementary error function" (erfc) to get a P-value. erfc(x) is the area under the normal curve above x (far right). (Actually, you have to divide the nomalized value by sqrt(2)...) I think it is interpreted like this: If you conclude the sum you got did not come from a random block, you'll be wrong p % of the time. So what's 275? If you work through the math, for your block size of 20000 bits, 275 gives a P-value of 0.05. So if you see a sum of 275 or more, and you reject the block as not random, you'll be wrong 5% of the time. Re: Monobit
A detailed discussion of the tests is at http://www.random.org/analysis/Analysis2005.pdf. It is presented in the context of testing random.org’s randomness, but the results apply to WhirlyGig as well.

I’m not a statistician, but here’s what I think is happening.
The test maps 0 -> -1, and 1->1. Sum all the bits in the block, and then normalize (take the absolute value, so the error “too many 0’s” is the same as “too many 1’s”, then divide by the square root of the number of bits in the block).

This normalized value is put in the “complementary error function” (erfc) to get a P-value. erfc(x) is the area under the normal curve above x (far right). (Actually, you have to divide the nomalized value by sqrt(2)…) I think it is interpreted like this: If you conclude the sum you got did not come from a random block, you’ll be wrong p % of the time.

So what’s 275? If you work through the math, for your block size of 20000 bits, 275 gives a P-value of 0.05. So if you see a sum of 275 or more, and you reject the block as not random, you’ll be wrong 5% of the time.

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