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Re: Random array (fwd)
Forwarded message:
> Date: Thu, 29 Oct 1998 17:33:29 +0100
> From: Anonymous <[email protected]>
> Subject: Re: Random array
> A couple responders have said there is no SWAP_TIMES
> which would work. But I don't understand why the
> following wouldn't work:
>
> First of all, make sure that x and y can't be equal, since
> there's no point in swapping an element with itself. This
> should add a negligible amount of time.
>
> x=getrand();
> do
> y=getrand();
> while (x == y);
>
> Now, simply calculate how many SWAP_TIMES it would take for it
> to be equally as likely that an element would not be touched as
> it would to land in its original spot.
>
> (255/256)^(t*2) = 1/256
>
> or
>
> t = ln(1/256)/(2*ln(255/256))
>
> or
>
> 708.3955
>
>
> Another anonymous wrote:
>
> In the following snippet of pseudo-code, what should the value of
> SWAP_TIMES be to make the array A[] random, assuming
> that getrand() returned a truly random integer between
> 0 and 255
>
> A[256];
>
> for(i=0;i<SWAP_TIMES;i++){
> x=getrand();
> y=getrand();
> swap(A[x],A[y]);
> }
>
> Thanks
>
Actualy the optimal value is even easier to calculate.
Given an initial array, A(m), and the desire to randomly swap the contents
until there is sufficient entropy to pass the various statistical tests
leaves us with:
Given m initial element and we are swapping them two at a time then we
need to execute at most m/2 swaps to randomize the array.
So, the optimal SWAP_TIMES ends up being, in this case, 128.
This of course doesn't touch on the new rule above about testing for x=y.
In which case you could simply do it over again. Now given that there are
m elements and the odds of selecting any given one is 1/n and the odds of
selecting it twice at one time is 1/n^2.
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