In clause_is_computable_at(), test required_relids for clone clauses.
Use the clause's required_relids not clause_relids for testing whether it is computable at the current join level, if it is a clone clause generated by deconstruct_distribute_oj_quals(). Arguably, this is more correct and we should do it for all clauses; that would at least remove the handwavy claim that we are doing it to save cycles compared to inspecting Vars individually. However, attempting to do that exposes that we are not being careful to compute an accurate value for required_relids in all cases. I'm unsure whether it's a good idea to attempt to do that for v16, or leave it as future clean-up. In the meantime, this quick hack demonstrably fixes some cases, so let's squeeze it in for beta1. Patch by me, but great thanks to Richard Guo for investigation and testing. The new test cases are all modeled on his examples. Discussion: https://postgr.es/m/CAMbWs4-_vwkBij4XOQ5ukxUvLgwTm0kS5_DO9CicUeKbEfKjUw@mail.gmail.com
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@ -544,13 +544,24 @@ clause_is_computable_at(PlannerInfo *root,
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RestrictInfo *rinfo,
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Relids eval_relids)
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{
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Relids clause_relids = rinfo->clause_relids;
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Relids clause_relids;
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ListCell *lc;
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/* Nothing to do if no outer joins have been performed yet. */
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if (!bms_overlap(eval_relids, root->outer_join_rels))
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return true;
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/*
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* For an ordinary qual clause, we consider the actual clause_relids as
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* explained above. However, it's possible for multiple members of a
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* group of clone quals to have the same clause_relids, so for clones use
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* the required_relids instead to ensure we select just one of them.
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*/
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if (rinfo->has_clone || rinfo->is_clone)
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clause_relids = rinfo->required_relids;
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else
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clause_relids = rinfo->clause_relids;
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foreach(lc, root->join_info_list)
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{
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SpecialJoinInfo *sjinfo = (SpecialJoinInfo *) lfirst(lc);
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@ -2500,6 +2500,74 @@ select * from int4_tbl t1
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-> Seq Scan on int4_tbl t4
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(12 rows)
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explain (costs off)
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select * from int4_tbl t1
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left join (int4_tbl t2 left join int4_tbl t3 on t2.f1 > 0) on t2.f1 > 1
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left join int4_tbl t4 on t2.f1 > 2 and t3.f1 > 3
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where t1.f1 = coalesce(t2.f1, 1);
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QUERY PLAN
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----------------------------------------------------
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Nested Loop Left Join
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Join Filter: ((t2.f1 > 2) AND (t3.f1 > 3))
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-> Nested Loop Left Join
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Join Filter: (t2.f1 > 0)
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-> Nested Loop Left Join
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Filter: (t1.f1 = COALESCE(t2.f1, 1))
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-> Seq Scan on int4_tbl t1
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-> Materialize
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-> Seq Scan on int4_tbl t2
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Filter: (f1 > 1)
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-> Seq Scan on int4_tbl t3
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-> Materialize
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-> Seq Scan on int4_tbl t4
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(13 rows)
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explain (costs off)
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select * from int4_tbl t1
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left join ((select t2.f1 from int4_tbl t2
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left join int4_tbl t3 on t2.f1 > 0
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where t3.f1 is null) s
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left join tenk1 t4 on s.f1 > 1)
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on s.f1 = t1.f1;
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QUERY PLAN
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-------------------------------------------------
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Nested Loop Left Join
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Join Filter: (t2.f1 > 1)
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-> Hash Right Join
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Hash Cond: (t2.f1 = t1.f1)
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-> Nested Loop Left Join
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Join Filter: (t2.f1 > 0)
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Filter: (t3.f1 IS NULL)
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-> Seq Scan on int4_tbl t2
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-> Materialize
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-> Seq Scan on int4_tbl t3
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-> Hash
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-> Seq Scan on int4_tbl t1
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-> Seq Scan on tenk1 t4
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(13 rows)
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explain (costs off)
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select * from onek t1
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left join onek t2 on t1.unique1 = t2.unique1
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left join onek t3 on t2.unique1 = t3.unique1
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left join onek t4 on t3.unique1 = t4.unique1 and t2.unique2 = t4.unique2;
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QUERY PLAN
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------------------------------------------------------------------------
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Hash Left Join
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Hash Cond: ((t3.unique1 = t4.unique1) AND (t2.unique2 = t4.unique2))
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-> Hash Left Join
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Hash Cond: (t2.unique1 = t3.unique1)
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-> Hash Left Join
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Hash Cond: (t1.unique1 = t2.unique1)
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-> Seq Scan on onek t1
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-> Hash
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-> Seq Scan on onek t2
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-> Hash
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-> Seq Scan on onek t3
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-> Hash
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-> Seq Scan on onek t4
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(13 rows)
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--
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-- check a case where we formerly got confused by conflicting sort orders
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-- in redundant merge join path keys
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@ -488,6 +488,26 @@ select * from int4_tbl t1
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left join int4_tbl t3 on t2.f1 = t3.f1
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left join int4_tbl t4 on t3.f1 != t4.f1;
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explain (costs off)
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select * from int4_tbl t1
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left join (int4_tbl t2 left join int4_tbl t3 on t2.f1 > 0) on t2.f1 > 1
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left join int4_tbl t4 on t2.f1 > 2 and t3.f1 > 3
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where t1.f1 = coalesce(t2.f1, 1);
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explain (costs off)
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select * from int4_tbl t1
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left join ((select t2.f1 from int4_tbl t2
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left join int4_tbl t3 on t2.f1 > 0
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where t3.f1 is null) s
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left join tenk1 t4 on s.f1 > 1)
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on s.f1 = t1.f1;
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explain (costs off)
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select * from onek t1
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left join onek t2 on t1.unique1 = t2.unique1
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left join onek t3 on t2.unique1 = t3.unique1
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left join onek t4 on t3.unique1 = t4.unique1 and t2.unique2 = t4.unique2;
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--
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-- check a case where we formerly got confused by conflicting sort orders
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-- in redundant merge join path keys
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