Re: If not readdir() then what?

[Neil Brown <neilb@suse.de>]

On Wednesday April 11, david.lang@digitalinsight.com wrote:
> On Thu, 12 Apr 2007, Neil Brown wrote:
> 
> > For the second.
> >  You say that you " would need at least 96 bits in order to make that
> >  guarantee; 64 bits of hash, plus a 32-bit count value in the hash
> >  collision chain".  I think 96 is a bit greedy.  Surely 48 bits of
> >  hash and 16 bits of collision-chain-position would plenty.  You would
> >  need 65537 entries before a collision was even possible, and
> >  billions before it was at all likely. (How big does a set of 48bit
> >  numbers have to get before the probability that "No subset of 65536
> >  numbers are all the same" drops below 0.95?)
> 
> Neil,
>    you can get a hash collision with two entries.

You need at least 65537 entries before there is any possibility of
collision between two
   "48-bit-hash ++ 16-bit-sequence-number"
objects where the 16-bit-sequence-number is chosen to be different from all
other 16 bit sequence numbers combined with the same 48 bit hash.

NeilBrown