|From:||Amit Langote <Langote_Amit_f8(at)lab(dot)ntt(dot)co(dot)jp>|
|To:||Robert Haas <robertmhaas(at)gmail(dot)com>|
|Cc:||Beena Emerson <memissemerson(at)gmail(dot)com>, Rajkumar Raghuwanshi <rajkumar(dot)raghuwanshi(at)enterprisedb(dot)com>, David Rowley <david(dot)rowley(at)2ndquadrant(dot)com>, Dilip Kumar <dilipbalaut(at)gmail(dot)com>, Pg Hackers <pgsql-hackers(at)postgresql(dot)org>|
|Subject:||Re: path toward faster partition pruning|
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On 2017/10/27 13:57, Robert Haas wrote:
> On Fri, Oct 27, 2017 at 3:17 AM, Amit Langote
> <Langote_Amit_f8(at)lab(dot)ntt(dot)co(dot)jp> wrote:
>>> I don't think we really want to get into theorem-proving here, because
>>> it's slow.
>> Just to be clear, I'm saying we could use theorem-proving (if at all) just
>> for the default partition.
> I don't really see why it should be needed there either. We've got
> all the bounds in order, so we should know where there are any gaps
> that are covered by the default partition in the range we care about.
Sorry, I forgot to add: "...just for the default partition, for cases like
the one in Beena's example."
In normal cases, default partition selection doesn't require any
theorem-proving. It proceeds in a straightforward manner more or less
like what you said it should.
After thinking more on it, I think there is a rather straightforward trick
to implement the idea you mentioned to get this working for the case
presented in Beena's example, which works as follows:
For any non-root partitioned tables, we add the list of its partition
constraint clauses to the query-provided list of clauses and use the whole
list to drive the partition-pruning algorithm. So, when partition-pruning
runs for tprt_1, along with (< 10000) which the original query provides,
we also have (>= 1) which comes from the partition constraint of tprt_1
(which is >= 1 and < 50000). Note that there exists a trick in the new
code for the (< 50000) coming from the constraint to be overridden by the
more restrictive (< 10000) coming from the original query.
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