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 in quadratic territory.
Andrei
