Re: Planner issue on sorting joining of two tables with limit

>
> Well, no, because that plan wouldn't produce the specified ordering;
> or at least it would be a lucky coincidence if it did. It's only
> sorting on t1.value.
>
I just don't find why it is coincidence. I think that such plan will always
produce result ordered by two columns, because such nested index scan always
produce this result.
Let's consider some simple example in order to illustrate how this plan
works.

t1
id | value
---+------
1 | 0.1
2 | 0.3
3 | 0.2

t2
id | id1 | value
---+-----+------
1 | 2 | 0.2
2 | 1 | 0.9
3 | 2 | 0.6
4 | 1 | 0.7
5 | 1 | 0.4
6 | 3 | 0.2

1) The outer index scan will find the row of t1 with least value using
test1_value_idx. It will be row (1, 0.1)
2) The inner index scan will find all the rows in t2 where id1 = 1 using
test2_id1_value_idx. But index test2_id1_value_idx have second order by
value column and the index scan result will be ordered by value. That's why
inner index scan will find rows (5, 1, 0.4), (4, 1, 0.7) and (2, 1, 0.9).
And following query output will be produced:
value1 | value2
--------+-------
0.1 | 0.4
0.1 | 0.7
0.1 | 0.9
3) The outer index scan will find the row of t1 with the second value using
test1_value_idx. It will be row (3, 0.2)
4) The inner index scan will find all the rows in t2 where id1 = 3 using
test2_id1_value_idx. This row is (6, 3, 0.2). The following query output
will be produced:
value1 | value2
--------+-------
0.2 | 0.2
5) The outer index scan will find the row of t1 with the third value using
test1_value_idx. It will be row (2, 0.3)
6) The inner index scan will find all the rows in t2 where id1 = 2 using
test2_id1_value_idx. These rows are (1, 2, 0.2) and (3, 2, 0.6). The
following query output will be produced:
value1 | value2
--------+-------
0.3 | 0.2
0.3 | 0.6

And the whole query result is:
value1 | value2
--------+-------
0.1 | 0.4
0.1 | 0.7
0.1 | 0.9
0.2 | 0.2
0.3 | 0.2
0.3 | 0.6

And this result is really ordered by t1.value, t2.value.
I can't find error in my reasoning :)

The query without limit produce similar plan.

EXPLAIN SELECT t1.value AS value1, t2.value AS value2 FROM test1 t1 JOIN
test2 t2 ON t2.id1 = t1.id ORDER BY t1.value

Nested Loop (cost=0.00..62109.86 rows=995025 width=16)
-> Index Scan using test1_value_idx on test1 t1 (cost=0.00..19.19
rows=200 width=12)
-> Index Scan using test2_id1_value_idx on test2 t2 (cost=0.00..248.27
rows=4975 width=12)
Index Cond: (t2.id1 = t1.id)

And I checked that the result is ordered by t1.value and t2.value.

2010/4/26 Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>

> =?KOI8-R?B?68/Sz9TLz9cg4czFy9PBzsTS?= <aekorotkov(at)gmail(dot)com> writes:
> > So PostgreSQL planner can produce the plan I need but it doesn't produce
> > this plan when I specify particular second ordering column.
>
> Well, no, because that plan wouldn't produce the specified ordering;
> or at least it would be a lucky coincidence if it did. It's only
> sorting on t1.value.
>
> > So is there any
> > way to make planner produce desired plan when particular second ordering
> > column is specified?
>
> Not when the ordering columns come from two different tables. (If they
> were in the same table then scanning a two-column index could produce
> the demanded sort order.) I don't see any way to satisfy this query
> without an explicit sort step, which means it has to look at the whole
> join output.
>
> If you're willing to make assumptions like "the required 10 rows will be
> within the first 100 t1.value rows" then you could nest an ORDER BY
> t1.value LIMIT 100 query inside something that did an ORDER BY with both
> columns. But this fails if you have too many duplicate t1.value values,
> and your test case suggests that you might have a lot of them. In any
> case it would stop being fast if you make the inner LIMIT very large.
>
> regards, tom lane
>