Hi,
Thank you very much for the explanation! I think the join order
benchmark I used [1] is somewhat representative, however, I probably
didn't use the most accurate cost estimation.
I find the cost from cheapest_total_path->total_cost is different
from the cost from queryDesc->planstate->total_cost. What I saw was
that GEQO tends to form paths with lower
cheapest_total_path->total_cost (aka the fitness of the children).
However, standard_join_search is more likely to produce a lower
queryDesc->planstate->total_cost, which is the cost we get using
explain.
I wonder why those two total costs are different? If the total_cost
from the planstate is more accurate, could we use that instead as the
fitness in geqo_eval?
[1] https://github.com/gregrahn/join-order-benchmark
Regards,
Donald Dong
> On May 7, 2019, at 4:44 PM, Tom Lane <tgl@sss.pgh.pa.us> wrote:
>
> Donald Dong <xdong@csumb.edu> writes:
>> I was expecting the plans generated by standard_join_search to have lower costs
>> than the plans from GEQO. But after the results I have from a join order
>> benchmark show that GEQO produces plans with lower costs most of the time!
>
>> I wonder what is causing this observation? From my understanding,
>> standard_join_search is doing a complete search. So I'm not sure how the GEQO
>> managed to do better than that.
>
> standard_join_search is *not* exhaustive; there's a heuristic that causes
> it not to consider clauseless joins unless it has to.
>
> For the most part, GEQO uses the same heuristic (cf desirable_join()),
> but given the right sort of query shape you can probably trick it into
> situations where it will be forced to use a clauseless join when the
> core code wouldn't. It'd still be surprising for that to come out with
> a lower cost estimate than a join order that obeys the heuristic,
> though. Clauseless joins are generally pretty awful.
>
> I'm a tad suspicious about the representativeness of your benchmark
> queries if you find this is happening "most of the time".
>
> regards, tom lane
