On 26 Dec 2006 at 2:55p -0500, Tom Lane wrote:
> Kevin Hunter <hunteke@earlham.edu> writes:
>> A friend has asked me about creating a unique table for individual users
>> that sign up for his site. (In essence, each user who signs up would
>> essentially get a set of CREATE TABLE {users,friends,movies,eats}_<id> (
>> ... ); statements executed, the idea being to reduce the number of rows
>> that the DB needs to search or index/update in regards to a particular
>> user id.) The just seems ludicrous to me, because the database still
>> needs to find those tables from its internal structure, not to mention
>> that it just seems silly to me from a design perspective. Something
>> about unable to optimize any queries because not only is the WHERE
>> clause in flux, but so is the FROM clause.
>
>> Question: Could someone explain to me why this would be bad idea,
>> because I can't put into words why it is.
>
> I thought you did a fine job right there ;-). In essence this would be
> replacing one level of indexing with two, which is unlikely to be a win.
> If you have exactly M rows in each of N tables then theoretically your
> lookup costs would be about O(log(N) + log(M)), which is nominally the
> same as O(log(M*N)) which is the cost to index into one big table --- so
> at best you break even, and that's ignoring the fact that index search
> has a nonzero startup cost that'll be paid twice in the first case.
> But the real problem is that if the N tables contain different numbers
> of rows then you have an unevenly filled search tree, which is a net
> loss.

Hurm. If I remember my Algorithms/Data Structures course, that implies
that table lookup is implemented with a B-Tree . . . right? Since at
SQL preparation time the tables in the query are known, why couldn't you
use a hash lookup? In the above case, that would make it effectively
O(1 + log(M)) or O(log(M)). Granted, it's /still/ a bad idea because of
the next paragraph . . .

> Most DBMSes aren't really designed to scale to many thousands of tables
> anyway. In Postgres this would result in many thousands of files in
> the same database directory, which probably creates some filesystem
> lookup inefficiencies in addition to whatever might be inherent to
> Postgres.

So, still a bad idea, but I couldn't immediately think of why. Thank you.

> Partitioning is indeed something that is commonly done, but on a very
> coarse grain --- you might have a dozen or two active partitions, not
> thousands. The point of partitioning is either to spread a huge table
> across multiple filesystems (and how many filesystems have you got?)
> or else to make predictable removals of segments of the data cheap (for
> instance, dropping the oldest month's worth of data once a month, in a
> table where you only keep the last year or so's worth of data on-line).

Ah! I was missing where to hang/put partitioning in my head. Thank you
again!

> I can't see doing it on a per-user basis.

Perhaps not on a per-user basis, but he could certainly improve access
times by partitioning even coursely. I'll point him in that direction
Are there other, perhaps better ways to improve the access times? (Now
I'm curious just for my sake.) The best that I keep reading is just to
do as much in parallel as possible (i.e. multiple systems) and to use
Postgres ( ).

Thanks,

Kevin

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Kevin Hunter <hunteke@earlham.edu> writes:
> On 26 Dec 2006 at 2:55p -0500, Tom Lane wrote:
>> I thought you did a fine job right there ;-). In essence this would be
>> replacing one level of indexing with two, which is unlikely to be a win.
>> If you have exactly M rows in each of N tables then theoretically your
>> lookup costs would be about O(log(N) + log(M)), which is nominally the
>> same as O(log(M*N)) which is the cost to index into one big table --- so

> Hurm. If I remember my Algorithms/Data Structures course, that implies
> that table lookup is implemented with a B-Tree . . . right?

b-tree or something else with log-N behavior. I think it can be proved
that every search method is at best log-N once N gets large enough, but
of course that might be for N far beyond any practical values.

> Since at SQL preparation time the tables in the query are known, why
> couldn't you use a hash lookup?

The hash index implementation currently available in Postgres hasn't
ever been proven to beat our b-tree implementation, on any dimension.
This might be a matter of how much effort has been thrown at it, or
maybe there's some fundamental problem; but anyway you won't find a
lot of enthusiasm around here for moving the system-catalog indexes
to hashing...

>> I can't see doing it on a per-user basis.

> Perhaps not on a per-user basis, but he could certainly improve access
> times by partitioning even coursely. I'll point him in that direction
> Are there other, perhaps better ways to improve the access times?

Actually, I forgot to mention an interesting talk I heard last month,
wherein Casey Duncan explained how pandora.com scales across an
unreasonable number of users whose queries are mostly-but-not-always
localized. Essentially they partition their tables on the basis of
a hash of the user ID, where there are as many hash result values
as they have servers. The point still remains though that there
are far fewer partitions than users. I think what you want to take
away is that you choose the number of partitions based on physical
implementation numbers ("how many filesystems do we have today")
and not on logical numbers like "how many users do we have today".

regards, tom lane

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