## digitalmars.D - Re: random cover of a range

```Andrei Alexandrescu Wrote:

Bill Baxter wrote:
On Sat, Feb 14, 2009 at 1:03 PM, Andrei Alexandrescu
<SeeWebsiteForEmail erdani.org <mailto:SeeWebsiteForEmail erdani.org>>
wrote:

bearophile wrote:

Andrei Alexandrescu:

Say at some point there are k available (not taken) slots out of
"n". There is a k/n chance that a random selection finds an
unoccupied slot. The average number of random trials needed
to find
an unoccupied slot is proportional to 1/(k/n) = n/k. So the
total
number of random trials to span the entire array is quadratic.
Multiplying that by 0.9 leaves it quadratic.

It's like in hashing: if you want to fill 99% of the available space
in a hash, then you take ages to find empty slots. But if you
fill it
only at 75-90%, then on average you need only one or two tries to
find an empty slot. So your time is linear, with a small
multiplicative constant. When the slots start to get mostly
full, you
change algorithm, copying the empty slots elsewhere.

Well I don't buy it. If you make a point, you need to be more
precise than such hand-waving. It's not like in hashing. It's like
in the algorithm we discuss. If you make a clear point that your
performance is better than O(n*n) by stopping at 90% then make it. I
didn't go through much formalism, but my napkin says you're firmly

Well he has a point that the number of trials required to find an empty
depends not on the absolute number of empty items, but only the ratio of
empties to fulls.   Even your own claim about average number of trials
was n/k -- not sure how you got that though.

If you toss a N-side dice hoping for a specific face to show up (and
stopping afterwards), how many times do you have to toss it on average?
I recall (without being sure) that you need to toss it a number of times
proportional to N. Could anyone confirm or deny?

avg_trials(N) = sum(t = 1-inf) { 1/N * ((N-1)/N)^(t-1) }

where t is the number of trials.

Experimentally, it converges to N
```
Feb 14 2009