Re: rand("state")

[Francesco Potorti`]

>> Or else you can see it as this: have a generator of period P, from which 
>> you have already taken N non-overlapping sequences of length L.  The 
>> probability that one more sequence of lenght L starting from a random 
>> point does not overlap with one of the already taken N sequences is 
>> equal to the probability that no taken sequence starts in an interval of 
>> length 2*L, that is, (1-N/P)^(2*L), under the assumption that N*L<<P, 
>> which is true in our case.
>>
>> If what I wrote is correct, then you have the probability of choosing a 
>> new good sequence, given that you have already chosen N.
>
>I don't have time right now to figure out if it is correct, but I will 
>add this:
>
>(1-N/P)^(2*L) = ((1-N/P)^(P/N))^((N/P)(2*L)) = exp(-2*L*N/P)
>
>That last equality is approximate, but it is very close whenever N/P is 
>small, and it is always small.

In fact, what you wrote is the probability of non collision in an Aloha
communications system.  Now, we should compute the probability of
generating N non-overlapping sequences, which is
prod_i=0^N-1 exp(-2*L*i/P) = exp(sum_i=0^N-1 -2*L*i/P)
                           = exp(-2*L/P*N*(N-1)/2)
                           = exp(-L/P*N*(N-1))
                           ~ exp(-L/P*N^2)
                           ~ 1-L/P*N^2
so, the probability of at least two sequences overlapping is, with
excellent approximation, L/P*N^2.  If we set P=10^6001, L=10^l, N=10^n,
the probability of overlapping is 10^(l+2*n-6001).  Nice result.

If I made no errors, of course.  Anyone willing to check this?

-- 
Francesco Potortì (ricercatore)        Voice: +39 050 315 3058 (op.2111)
ISTI - Area della ricerca CNR          Fax:   +39 050 313 8091
via G. Moruzzi 1, I-56124 Pisa         Email: address@hidden
Web: http://fly.isti.cnr.it/           Key:   fly.isti.cnr.it/public.key



-------------------------------------------------------------
Octave is freely available under the terms of the GNU GPL.

Octave's home on the web:  http://www.octave.org
How to fund new projects:  http://www.octave.org/funding.html
Subscription information:  http://www.octave.org/archive.html
-------------------------------------------------------------