[aklug] Re: random bits vs random Hex

On 5/28/13 6:42 PM, Dan Wolf wrote:
> Any math types out there?
>
> Suppose you have a source of "random" bits.(0's and 1's)....... assume all
> the bits generated meet whichever tests for "randomness" you use,
>
> Now if you peel off the bits in groups of four you now have hex digits.
>
> ie; Binary = 1100 1000 0000 1110 0101 0001 1100 1010
>
> Hex = C 8 0 E 5 1 C
> A
>
> So the question is ;
>
> Are the binary digits A) More random than the Hex
> digits
>
> B) Less
> random than the Hex digits
>
> C) Equally
> random to the Hex digits
>
> D)
> Randomness tests would be different for Binary bits from Hex digits
>
> E) Other

> I apologize in advance if there is an obvious answer here but I have torn
> thru my copies of Knuth and can't find anything like a communitive or
> associative law for random numbers

Short answer: They contain the same amount of information, so they are
both "equally random", for whatever definition of "random" you are
using. (Which does not imply that either string is random at all.)

Ghetto sketch of proof: "Random" in the sense you're using it usually
means something close to "Digit X_n+1 is i.i.d. WRT Digit X_n this
sequence"; in other words, whether bit 2 is 0 or 1 has bugger-all to do
with whether bit 1 is 0 or 1. The independence of any future bits, or
bit strings, is implied in that definition. Since hex digits are
nybbles ("4-bit chunks"), saying that "The value of hex digit Y_1 is
independent of hex digit Y_2" is equivalent to saying "The value of
4-bit chunk X_(n..n+4) is independent of the value of 4-bit chunk
X_(m..m+4)", which is trivially true if the independence of arbitrary
pairs of single bits is a given.

Also, ultimately, keep in mind that your randomness tests are *only*
operating on binary digits; hex representation is only for
human-readable convenience. (The processor doan know from know hex, dude.)

-- 
"proof by intimidation"
-SD
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