On 2017/10/26 20:34, Robert Haas wrote:
> On Thu, Oct 26, 2017 at 1:17 PM, Amit Langote
> <langote_amit...@lab.ntt.co.jp> wrote:
>> It can perhaps taught to not make that conclusion by taking into account
>> the default partition's partition constraint, which includes constraint
>> inherited from the parent, viz. 1 <= col1 < 50001.  To do that, it might
>> be possible to summon up predtest.c's powers to conclude from the default
>> partition's partition constraint that it cannot contain any keys < 1, but
>> then we'll have to frame up a clause expression describing the latter.
>> Generating such a clause expression can be a bit daunting for a
>> multi-column key.   So, I haven't yet tried really hard to implement this.
>>  Any thoughts on that?
> 
> 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.

> Whatever we're going to do we should be able to do without
> that - keeping it in the form of btree-strategy + value.  It doesn't
> seem that hard.  Suppose we're asked to select partitions from tprt
> subject to (<, 10000).  Well, we determine that some of the tprt_1
> partitions may be relevant, so we tell tprt_1 to select partitions
> subject to (>=, 1, <, 10000).  We know to do that because we know that
> 10000 < 50000 and we know to include >= 1 because we haven't got any
> lower bound currently at all.  What's the problem?

Hmm, that's interesting.  With the approach that the patch currently
takes, (>= 1) wouldn't be passed down when selecting the partitions of
tprt_1.  The source of values (+ btree strategy) to use to select
partitions is the same original set of clauses for all partitioned tables
in the tree, as the patch currently implements it.  Nothing would get
added to that set (>= 1, as in this example), nor subtracted (such as
clauses that are trivially true).

I will think about this approach in general and to solve this problem in
particular.

> In some sense it's tempting to say that this case just doesn't matter
> very much; after all, subpartitioning on the same column used to
> partition at the top level is arguably lame.  But if we can get it
> right in a relatively straightforward manner then let's do it.

Yeah, I tend to agree.

Thanks for the input.

Regards,
Amit



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