Обсуждение: Underestimated number of output rows with an aggregate function

Hi all,

Working on the emaj extension (for the curious ones, 
https://emaj.readthedocs.io/en/latest/ and 
https://github.com/dalibo/emaj), I recently faced a performance problem 
when querying and aggregating data changes. A query with 3 CTE has a O^2 
behavior (https://explain.dalibo.com/plan/1ded242d4ebf3gch#plan). I have 
found a workaround by setting enable_nestloop to FALSE. But this has 
drawbacks. So I want to better understand the issue.

During my analysis, I realized that the output rows estimate of the 
second CTE is really bad, leading to a bad plan for the next CTE.

I reproduced the issue in a very small test case with a simplified 
query. Attached is a shell script and its output.

A simple table is created, filled and analyzed.

The simplified statement is:
  WITH keys AS (
    SELECT c1, min(seq) AS seq FROM perf GROUP BY c1
    )
  SELECT tbl.*
    FROM perf tbl JOIN keys ON (keys.c1 = tbl.c1 AND keys.seq = tbl.seq);

Its plan is:
  Hash Join (cost=958.00..1569.00 rows=1 width=262) (actual 
time=18.516..30.702 rows=10000 loops=1)
    Output: tbl.c1, tbl.seq, tbl.c2
    Inner Unique: true
    Hash Cond: ((tbl.c1 = perf.c1) AND (tbl.seq = (min(perf.seq))))
    Buffers: shared hit=856
    ->  Seq Scan on public.perf tbl  (cost=0.00..548.00 rows=12000 
width=262) (actual time=0.007..2.323 rows=12000 loops=1)
          Output: tbl.c1, tbl.seq, tbl.c2
          Buffers: shared hit=428
    ->  Hash  (cost=808.00..808.00 rows=10000 width=8) (actual 
time=18.480..18.484 rows=10000 loops=1)
          Output: perf.c1, (min(perf.seq))
          Buckets: 16384  Batches: 1  Memory Usage: 519kB
          Buffers: shared hit=428
          ->  HashAggregate  (cost=608.00..708.00 rows=10000 width=8) 
(actual time=10.688..14.321 rows=10000 loops=1)
                Output: perf.c1, min(perf.seq)
                Group Key: perf.c1
                Batches: 1  Memory Usage: 1425kB
                Buffers: shared hit=428
                ->  Seq Scan on public.perf (cost=0.00..548.00 
rows=12000 width=8) (actual time=0.002..2.330 rows=12000 loops=1)
                      Output: perf.c1, perf.seq, perf.c2
                      Buffers: shared hit=428

It globally looks good to me, with 2 sequential scans and a hash join.
But the number of returned rows estimate is always 1, while it actually 
depends on the data content (here 10000).

For the hash join node, the plan shows a "Inner Unique: true" property. 
I wonder if this is normal. It look likes the optimizer doesn't take 
into account the presence of the GROUP BY clause in its estimate.

I reproduce the case with all supported postgres versions.

Thanks by advance for any explanation.
Philippe.

Philippe BEAUDOIN <phb.emaj@free.fr> writes:
> During my analysis, I realized that the output rows estimate of the 
> second CTE is really bad, leading to a bad plan for the next CTE.
> I reproduced the issue in a very small test case with a simplified 
> query. Attached is a shell script and its output.

Yeah.  If you try it you'll see that the estimates for the
"keys.c1 = tbl.c1" and "keys.seq = tbl.seq" clauses are spot-on
individually.  The problem is that the planner assumes that they
are independent clauses, so it multiplies those selectivities together.
In reality, because seq is already unique, the condition on c1 adds
no additional selectivity.

If seq is guaranteed unique in your real application, you could just
drop the condition on c1.  Otherwise I'm not sure about a good
answer.  In principle creating extended stats on c1 and seq should
help, but I think we don't yet apply those for join clauses.

A partial answer could be to defeat application of the table's
statistics by writing

          JOIN keys ON (keys.c1 = tbl.c1+0 AND keys.seq = tbl.seq+0)

For me this gives an output estimate of 3000 rows, which is still not
great but should at least prevent choice of an insane plan at the
next join level.  However, it pessimizes the plan for this query
itself a little bit (about doubling the runtime).

> For the hash join node, the plan shows a "Inner Unique: true" property. 
> I wonder if this is normal.

Sure.  The output of the WITH is visibly unique on c1.

            regards, tom lane

Philippe BEAUDOIN <phb.emaj@free.fr> writes:

During my analysis, I realized that the output rows estimate of the 
second CTE is really bad, leading to a bad plan for the next CTE.
I reproduced the issue in a very small test case with a simplified 
query. Attached is a shell script and its output.

Yeah.  If you try it you'll see that the estimates for the
"keys.c1 = tbl.c1" and "keys.seq = tbl.seq" clauses are spot-on
individually.  The problem is that the planner assumes that they
are independent clauses, so it multiplies those selectivities together.
In reality, because seq is already unique, the condition on c1 adds
no additional selectivity.

If seq is guaranteed unique in your real application, you could just
drop the condition on c1.  Otherwise I'm not sure about a good
answer.  In principle creating extended stats on c1 and seq should
help, but I think we don't yet apply those for join clauses.

A partial answer could be to defeat application of the table's
statistics by writing
          JOIN keys ON (keys.c1 = tbl.c1+0 AND keys.seq = tbl.seq+0)

For me this gives an output estimate of 3000 rows, which is still not
great but should at least prevent choice of an insane plan at the
next join level.  However, it pessimizes the plan for this query
itself a little bit (about doubling the runtime).

Thanks for the trick (and the quick answer). In the test case, it effectively brings a pretty good plan.

Unfortunately, as these statements are generated and depend on the base table structure, the issue remains for some of them (but not all). So, for the moment at least, I keep the previous workaround (disabling nested loops).

For the hash join node, the plan shows a "Inner Unique: true" property. 
I wonder if this is normal.

Sure.  The output of the WITH is visibly unique on c1.

			regards, tom lane