linux-kernel - Re: If not readdir() then what?

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Date:	Thu, 12 Apr 2007 10:00:30 +1000
From:	Neil Brown <neilb@...e.de>
To:	David Lang <david.lang@...italinsight.com>
Cc:	Theodore Tso <tytso@....edu>,
	Jörn Engel <joern@...ybastard.org>,
	"H. Peter Anvin" <hpa@...or.com>,
	Christoph Hellwig <hch@...radead.org>,
	Ulrich Drepper <drepper@...il.com>,
	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>
Subject: Re: If not readdir() then what?

On Wednesday April 11, david.lang@...italinsight.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
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