Bill Baxter wrote:
I found a pdf[1] that has a summary of the probabilities associated with finding empty slots in the context of hashing. If I'm reading it right, if you have a load factor (fraction full) of f, then it's expected it will take you 1/(1-f) trials to get an empty if you just continue picking randomly one after the other.--bb [1] http://www.cse.unsw.edu.au/~cs2011/lect/2711_HashProbe-4.pdf
Thank you, that's exactly the explanation I was looking for. Andrei
