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Remove TODO.detail rtree now that item is fixed.

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Bruce Momjian 2005-06-24 04:42:58 +00:00
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From pgsql-bugs-owner+M10740=pgman=candle.pha.pa.us@postgresql.org Mon Jan 24 18:00:00 2005
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From: Andrew - Supernews <andrew+nonews@supernews.com>
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Subject: [BUGS] incorrect index behaviour with rtree on box values
Date: Mon, 24 Jan 2005 23:29:12 -0000
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Status: OR
Testcase:
create table boxtest (a box);
create index boxtest_idx on boxtest using rtree (a);
create function gen_data() returns void as '
begin for i in 1..200 loop
insert into boxtest
values (box(point((i*2-1)::float,0),point((i*2)::float,1)));
end loop;
return;
end;' language plpgsql;
select gen_data();
analyze boxtest;
set enable_seqscan = false;
set enable_bitmapscan = true;
set enable_indexscan = true;
select * from boxtest where a << '(3,0),(3,1)'::box;
set enable_seqscan = true;
set enable_bitmapscan = false;
set enable_indexscan = false;
select * from boxtest where a << '(3,0),(3,1)'::box;
Those two selects at the end should clearly return the same result, a
single row. In fact, what happens is that the second returns no rows at
all; I tested this on 7.4.6, but others have confirmed this on everything
from 7.3 to latest.
The problem is that the semantics of the &< and &> operators for the box
type are not what rtree needs for the "OverLeft" and "OverRight" slots of
the operator class. Specifically, what rtree needs is this:
if X << K or X &< K
then for all A where A is a union of values including X,
then A &< K
(the designation "&<" is of course arbitrary, what matters is what operator
is placed in the applicable slot of the opclass. Same goes for >> and &>.)
This is because rtree converts (see rtstrat.c) the original "Left" operator
to an "OverLeft" when comparing against internal nodes of the index, which
contain values which are the union of all values in their subtree. In the
testcase, the top node of the tree contains as its first entry a union
value of the form (184,1),(1,0), which the scan then rejects since
(184,1),(1,0) &< (3,0),(3,1) is false.
I can see three possible approaches to fixing this:
1) change the semantics of &< and &> to match rtree's expectations
2) replace &< and &> in the opclass with operators that behave as rtree
expects (this will have the side effect of rendering &< and &> un-indexable)
3) change rtree's behaviour in some way.
--
Andrew, Supernews
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From pgsql-bugs-owner+M10748=pgman=candle.pha.pa.us@postgresql.org Mon Jan 24 18:57:46 2005
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To: andrew@supernews.com
cc: pgsql-bugs@postgresql.org
Subject: Re: [BUGS] incorrect index behaviour with rtree on box values
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Comments: In-reply-to Andrew - Supernews <andrew+nonews@supernews.com>
message dated "Mon, 24 Jan 2005 23:29:12 +0000"
Date: Mon, 24 Jan 2005 19:09:41 -0500
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From: Tom Lane <tgl@sss.pgh.pa.us>
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Andrew - Supernews <andrew+nonews@supernews.com> writes:
> The problem is that the semantics of the &< and &> operators for the box
> type are not what rtree needs for the "OverLeft" and "OverRight" slots of
> the operator class.
This was observed nearly a year ago, see this thread:
http://archives.postgresql.org/pgsql-general/2004-03/msg01135.php
but apparently no one cares enough to fix it. Are you volunteering?
regards, tom lane
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From pgsql-bugs-owner+M10762=pgman=candle.pha.pa.us@postgresql.org Wed Jan 26 08:56:08 2005
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From: Andrew - Supernews <andrew+nonews@supernews.com>
X-Newsgroups: pgsql.bugs
Subject: Re: [BUGS] incorrect index behaviour with rtree on box values
Date: Wed, 26 Jan 2005 14:54:41 -0000
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On 2005-01-25, Tom Lane <tgl@sss.pgh.pa.us> wrote:
> Andrew - Supernews <andrew+nonews@supernews.com> writes:
>> The problem is that the semantics of the &< and &> operators for the box
>> type are not what rtree needs for the "OverLeft" and "OverRight" slots of
>> the operator class.
>
> This was observed nearly a year ago, see this thread:
> http://archives.postgresql.org/pgsql-general/2004-03/msg01135.php
>
> but apparently no one cares enough to fix it. Are you volunteering?
Possibly. I don't feel comfortable with changing anything specific to the
geometric operators, since (a) I don't actually use them (I discovered
this issue when adding rtree support to a type of my own) and (b) the
compatibility implications are obvious. But I think there is a solution
that involves only changes to the rtree strategy code.
Looking at the earlier discussion: it seems to have ended with the
conclusion that &< should mean "does not extend to the right of", which
matches the current implementation for box, but not for some other types.
So for box values, we seem (and someone please correct me if I'm wrong) to
have the following semantics:
a << b - a is strictly left of b, i.e. a.right < b.left
a &< b - a is no further right than b, i.e. a.right <= b.right
a &> b - a is no further left than b, i.e. a.left >= b.left
a >> b - a is strictly right of b, i.e. a.left > b.right
For rtree to work as apparently intended, it needs four more operators,
to use for inner nodes when the scan operator is one of the above four.
However, a small modification to the way that the internal scan key is
initialised should eliminate the requirement to explicitly specify these
operators, which strikes me as the solution which preserves maximum
compatibility. The four operators required are:
NOT (a &> b) (used when the scan operator is (a << b))
NOT (a >> b) (used when the scan operator is (a &< b))
NOT (a << b) (used when the scan operator is (a &> b))
NOT (a &< b) (used when the scan operator is (a >> b))
(This won't fix rtree on contrib/seg or contrib/cube, but those appear to be
broken already since they have different, and equally incorrect, definitions
of &> and &<. Fixing those would require slightly more complex operators,
such as NOT (a &> b OR a >> b) and so on. The more complex operators would
work for box too, so it might be worth using them anyway, but I don't yet
understand the scan key handling well enough to know if these can be
constructed rather than supplied in the opclass.)
Proof:
Let V be the scan key, i.e. the value we are searching for in the index.
Let U be a union over a set of values.
Let X be some value for which X OP V holds.
Consider an internal node entry with union U. We require that the following
holds: if U contains some value X where X OP V holds, then U OP' V must be
true. (But not the converse; U OP' V may be true even if no such X exists in
U. However, we wish it to be false as much as possible for efficiency.)
When OP is << :
X << V, therefore X.right < V.left, therefore X.left < V.left
therefore NOT (X &> V)
If U contains X, then U &> V is true iff U.left >= V.left
U.left <= min(E.left) for all elements E of U, and therefore for X if X in U
So if X in U, then U.left <= X.left < V.left, and therefore NOT (U &> V)
When OP is &< :
X &< V, therefore X.right <= V.right, therefore X.left <= V.right
therefore NOT (X >> V), and similar reasoning for U containing X as above.
--
Andrew, Supernews
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