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

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Message-ID: <461D6DE9.9070601@zytor.com>
Date:	Wed, 11 Apr 2007 16:23:21 -0700
From:	"H. Peter Anvin" <hpa@...or.com>
To:	David Lang <david.lang@...italinsight.com>
CC:	Neil Brown <neilb@...e.de>, Theodore Tso <tytso@....edu>,
	Jörn Engel <joern@...ybastard.org>,
	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?

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?)
> 
> Neil,
>   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.)

	-hpa
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