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From: Jie Zhang <jzhang@greenplum.com>
To: Tom Lane <tgl@sss.pgh.pa.us>
Cc: Mark Kirkwood <markir@paradise.net.nz>
Cc: Josh Berkus <josh@agliodbs.com>
Cc: Gavin Sherry <swm@linuxworld.com.au>
Cc: pgsql-hackers@postgresql.org, "Luke Lonergan" <LLonergan@greenplum.com>
Subject: Re: On-disk bitmap index patch
Date: Thu, 27 Jul 2006 09:13:21 -0700
Message-ID: <C0EE3431.9116%jzhang@greenplum.com> (raw)
In-Reply-To: <748.1153983035@sss.pgh.pa.us>



On 7/26/06 11:50 PM, "Tom Lane" <tgl@sss.pgh.pa.us> wrote:

> "Jie Zhang" <jzhang@greenplum.com> writes:
>> On 7/26/06 10:14 PM, "Tom Lane" <tgl@sss.pgh.pa.us> wrote:
>>> ... A nonuniform distribution would probably mean that some
>>> of the bitmaps compress better-than-expected and others worse.  I have
>>> no idea how to model that and guess what the overall result is ...
> 
>> The paper "Optimizing Bitmap Indices With Efficient Compression" by Kesheng
>> Wu et al gave an approximate answer for this question. Assume that there are
>> c distinct values. Let the i-th value has a probability of p_i, the number
>> of rows r, and the word size w. then the total size of the compressed bitmap
>> index is about (N/(w-1))(c- \sum(1-p_i)^(2w-2) - \sum(p_i)^(2w-2)), where in
>> both \sum's, i is from 1 to c.
> 
> Hm, but that's still begging the question no?  It's still assuming that
> any one value is uniformly distributed.  ISTM the cases that would break
> my simplistic calculation involve clustering of particular values, such
> that some areas of the bitmap are all-zero while other areas have lots
> of ones.

Yes, you are right -- each value is still uniformly distributed. But this
will be the worst case in terms of the size of a bitmap vector. As for how
to model the size of a bitmap vector for an non-uniformly distributed value,
that's a good question. I don't really know. But we do know the best case
and the worse case.





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