Oops, that's not quite right. I guess you're looking for the expected distance
between random continuous points in [0,1]^n not {0,1}^n. Sqrt(n/2) should be
an upper bound in any case. If you figure out the expected squared distance of
any single coordinate you can use the analysis in my last email.
On Oct 19, 2011, at 12:57 PM, Federico Castanedo wrote:
> what about this:
>
> http://www.wisdom.weizmann.ac.il/~oded/p_aver-metric.html
>
> HTW
>
> 2011/10/19 Sean Owen <[email protected]>
>
>> (And when I do the simulation correctly, I get a better answer: sqrt(n/6) )
>>
>> On Wed, Oct 19, 2011 at 5:21 PM, Sean Owen <[email protected]> wrote:
>>> Hmm. Not knowing the analytics answer I just wrote a simulation.
>>> sqrt(n / 3) looks like a shockingly good fit for the average distance
>>> between two randomly chosen points in the n-dimensional hypercube.
>>>
>>> Accident? error? known result? Seems clear that something like sqrt(n)
>>> would be a better factor than n. But, indeed, there are yet more
>>> possibilities with exponential functions.
>>
