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

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Message-ID: <20070411233326.GD4893@lazybastard.org>
Date:	Thu, 12 Apr 2007 01:33:26 +0200
From:	Jörn Engel <joern@...ybastard.org>
To:	"H. Peter Anvin" <hpa@...or.com>
Cc:	David Lang <david.lang@...italinsight.com>,
	Neil Brown <neilb@...e.de>, Theodore Tso <tytso@....edu>,
	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 Wed, 11 April 2007 16:23:21 -0700, H. Peter Anvin wrote:
> David Lang 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?)
> >
> >  you can get a hash collision with two entries.
> 
> Yes, but the probability is 2^-n for an n-bit hash, assuming it's 
> uniformly distributed.
> 
> The probability approaches 1/2 as the number of entries hashes 
> approaches 2^(n/2) (birthday number.)

I believe you are both barking up the wrong tree.  Neil proposed a 16bit
collision chain.  With that, it takes 65537 entries before a collision
chain overflow is possible.

Calling a collision chain overflow "collision" is inviting confusion, of
course. :)

Jörn

-- 
The competent programmer is fully aware of the strictly limited size of
his own skull; therefore he approaches the programming task in full
humility, and among other things he avoids clever tricks like the plague.
-- Edsger W. Dijkstra
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