random cover of a range

[Jason House]

Andrei Alexandrescu Wrote:

> Jason House wrote:
> > Andrei Alexandrescu Wrote: 
> 
> > No.  Your wording sounds like you're doing
> > stuff that's way off, but the resulting math is correct.  My
> > calculation would be based on the average length of a sequence of 1's
> > (k/(n-k)).  That means the work is 1+k/(n-k) = n/(n-k).
> 
> Well my wording means this: in an array of length n with k "holes"
> randomly distributed, the probability one slot is a a no-hole is
> (n-k)/n. What we want is to find the first no-hole starting from a
> random position in the array. How many steps do we do on average? That
> is the same as the average number of steps of rolling a fair dice with
> (n-k) faces until we obtain a particular face. And the average number of
> steps is IIRC 1/p = n/(n-k).
> 
> > Given that O(n*log(n)) is the theoretical best you can do, having a
> > result that is < O(n*log(n)) is highly suspect.  The sum 1/1 + 1/2 +
> > 1/3 + 1/4 + 1/5 + ... is in fact O(log(n)).
> 
> Ever so pedantic. :o) I meant "<=" because I wasn't aware of the best
> theoretical bound. Do you have a pointer? Thanks.

I was thinking of the theoretical bound on sorting being n*log(n) but that does not apply in this case.

The bound on sum(1/x) is pretty simple. The discrete sampling of 1/n can be made into either an over or under approximation of the integral depending on how you shift the starting points. That means sum(1/x) = ln(n)+C, where is 0 < C < 1


> I have the feeling there is something clever to do after half of the
> array was covered. After that, the probability of a random element being
> uncovered falls below 0.5.
> 
> I also have the feeling that interesting things can be done if the
> length of the range has certain values, such as the period of certain
> generators with certain parameters. I don't have the time to look into
> that now. Anyone versed in e.g. linear congruential generators with a
> given period?
> 
> One way or another, I'll add RandomCover to std.random. Thanks Leonardo,
> Denis, Steve, and Jason.
> 
> 
> Andrei