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I am reading a paper, NIST SP800-22 Revision 1a., about Randomness test in Computer system, which is from NIST, U.S. Department of Commerce.

Question

Does "Binary Logarithm" means "Binary Number System" in Computer Science?

Why I think

In that paper, 2.12.4 Test Description (Page 51), there is equation as below


$ (3) \ \ \ \mathrm{Compute} \ C_i^m = \frac{\sharp i}{n} $

$ \mathrm{For \ the \ example \ in \ this \ section},\\\ C^3_{000} = 0, \ C^3_{001} = 0.1, \ C^3_{010} = 0.3, \ C^3_{011} = 0.1, \ C^3_{100} = 0.1, \ C^3_{101} = 0.3, \ C^3_{110} = 0.1, \ C^3_{111} = 0 $



$ (4) \ \ \ \psi^{(m)}=\displaystyle \sum_{i=0}^{2^m-1} \pi_i log{\pi_i}, \ \mathrm{where} \ \pi_i=C^3_j, \ \mathrm{and} \ j=log_2i. $

$ \mathrm{For \ the \ example \ in \ this \ section},\\\ \psi^{(3)} = 0(log \ 0)+0.1(log \ 0.1)+0.3(log \ 0.3)+0.1(log \ 0.1)+0.1(log \ 0.1)+0.3(log \ 0.3)+0.1(log \ 0.1)+0(log \ 0)=-1.64341772. $


As you can see at Section (4), $j=log_2i$.

Why $\pi_3 = C^3_{log_23}=C_{011}^3$ ? Then, does "Binary Logarithm" means "Binary Number System" in Computing?

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No, 'binary logarithm' does not mean 'binary number system'; it just means 'logarithm with base 2'.

Purely based on your description, it looks as if the authors of the paper made a mistake and simply meant '$j$ is the binary representation of $i$' instead of '$j = \log_2 i$'.

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  • $\begingroup$ Thank you. Then This is just misprint.. :( $\endgroup$
    – TyeolRik
    May 31 at 12:53

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