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

- Jason House <jason.james.house gmail.com> Feb 14 2009
- Jason House <jason.james.house gmail.com> Feb 14 2009

Andrei Alexandrescu 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 in quadratic territory. Andrei

Retrying when 90% full gives you a geometric series for the number of tries: 1+0.1+0.1^2+0.1^3+... Ignoring the math trick to get 1/(1-p), it's easy to see it's 1.111111... You're firmly in linear territory.

Feb 14 2009

Jason House Wrote:Andrei Alexandrescu 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 in quadratic territory. Andrei

Retrying when 90% full gives you a geometric series for the number of tries: 1+0.1+0.1^2+0.1^3+... Ignoring the math trick to get 1/(1-p), it's easy to see it's 1.111111... You're firmly in linear territory.

Ugh, I should not post when tired. p=0.9, not 0.1! 1/(1-0.9)=10. It's still linear, but won't be as nice as my prior post implied. Sorry.

Feb 14 2009