number generator

Duncan Smith buzzard at urubu.freeserve.co.uk
Wed Mar 14 02:25:42 CET 2007


Hendrik van Rooyen wrote:
>  "Nick Craig-Wood" <nick at craig-wood.com> wrote:
> 
> 
>>Paul Rubin <http> wrote:
>>
>>> The fencepost method still seems to be simplest:
>>>
>>>     t = sorted(random.sample(xrange(1,50), 4))
>>>     print [(j-i) for i,j in zip([0]+t, t+[50])]
>>
>>Mmm, nice.
>>
>>Here is another effort which is easier to reason about the
>>distribution produced but not as efficient.
>>
>>def real(N, M):
>>    while 1:
>>        t = [ random.random() for i in range(N) ]
>>        factor = M / sum(t)
>>        t = [ int(round(x * factor)) for x in t]
>>        if sum(t) == M:
>>            break
>>        print "again"
>>    assert len(t) == N
>>    assert sum(t) == M
>>    return t
>>
>>It goes round the while loop on average 0.5 times.
>>
>>If 0 isn't required then just test for it and go around the loop again
>>if found.  That of course skews the distribution in difficult to
>>calculate ways!
>>
> 
> 
> I have been wondering about the following as this thread unrolled:
> 
> Is it possible to devise a test that can distinguish between sets
> of:
> 
> - five random numbers that add to 50, and
> - four random numbers and a fudge number that add to 50?
> 
> My stats are way too small and rusty to attempt to answer 
> the question, but it seems intuitively a very difficult thing.
> 
> - Hendrik
> 

Yes, if the generating processes yield numbers from different
probability mass functions.  You could simply look at the likelihood
ratio.  Otherwise, the likelihood ratio will be 1.

Duncan



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