From: | Claudio Freire <klaussfreire(at)gmail(dot)com> |
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To: | Peter Geoghegan <pg(at)heroku(dot)com> |
Cc: | Robert Haas <robertmhaas(at)gmail(dot)com>, Heikki Linnakangas <hlinnakangas(at)vmware(dot)com>, Stephen Frost <sfrost(at)snowman(dot)net>, Greg Stark <stark(at)mit(dot)edu>, Noah Misch <noah(at)leadboat(dot)com>, Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>, Thom Brown <thom(at)linux(dot)com>, Pg Hackers <pgsql-hackers(at)postgresql(dot)org> |
Subject: | Re: B-Tree support function number 3 (strxfrm() optimization) |
Date: | 2014-07-14 18:03:44 |
Message-ID: | CAGTBQpY4Qbunj+kYc5hRin3jWP4uovmTbgcKW-VY0LKxH9Ggxg@mail.gmail.com |
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On Mon, Jul 14, 2014 at 2:53 PM, Peter Geoghegan <pg(at)heroku(dot)com> wrote:
> My concern is that it won't be worth it to do the extra work,
> particularly given that I already have 8 bytes to work with. Supposing
> I only had 4 bytes to work with (as researchers writing [2] may have
> only had in 1994), that would leave me with a relatively small number
> of distinct normalized keys in many representative cases. For example,
> I'd have a mere 40,665 distinct normalized keys in the case of my
> "cities" database, rather than 243,782 (out of a set of 317,102 rows)
> for 8 bytes of storage. But if I double that to 16 bytes (which might
> be taken as a proxy for what a good compression scheme could get me),
> I only get a modest improvement - 273,795 distinct keys. To be fair,
> that's in no small part because there are only 275,330 distinct city
> names overall (and so most dups get away with a cheap memcmp() on
> their tie-breaker), but this is a reasonably organic, representative
> dataset.
Are those numbers measured on MAC's strxfrm?
That was the one with suboptimal entropy on the first 8 bytes.
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