From: | Thomas Munro <thomas(dot)munro(at)gmail(dot)com> |
---|---|
To: | James Coleman <jtc331(at)gmail(dot)com> |
Cc: | Tomas Vondra <tomas(dot)vondra(at)2ndquadrant(dot)com>, PostgreSQL mailing lists <pgsql-bugs(at)lists(dot)postgresql(dot)org> |
Subject: | Re: BUG #16104: Invalid DSA Memory Alloc Request in Parallel Hash |
Date: | 2019-11-10 03:43:49 |
Message-ID: | CA+hUKGL0T7-Lsbp-dda5JWubAXEHMkL7j94-XW9ZhXDb4sn2+A@mail.gmail.com |
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Lists: | pgsql-bugs |
On Sun, Nov 10, 2019 at 3:25 PM James Coleman <jtc331(at)gmail(dot)com> wrote:
> So I should have run the earlier attached plan with VERBOSE, but
> here's the interesting thing: the parallel hash node's seq scan node
> outputs two columns: let's call them (from the redacted plan)
> items.system_id and items.account_id. The first (system_id) is both
> not null and unique; the second (account_id) definitely has massive
> skew. I'm not very up-to-speed on how the hash building works, but I
> would have (perhaps naïvely?) assumed that the first column being
> unique would make the hash keys very likely not to collide in any
> significantly skewed way. Am I missing something here?
Hrm. So the compound key is unique then. I was assuming up until now
that it had duplicates. The hashes of the individual keys are
combined (see ExecHashGetHashValue()), so assuming there is nothing
funky about the way citext gets hashed (and it's not jumping out at
me), your unique keys should give you uniform hash values and thus
partition size, and repartitioning should be an effective way of
reducing hash table size. So now it sounds like you have a simple
case of underestimation, but now I'm confused about how you got a
344MB hash table with work_mem = 150MB:
Buckets: 4194304 (originally 4194304) Batches: 32768 (originally
4096) Memory Usage: 344448kB
And I'm confused about what was different when it wanted the crazy
number of batches.
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