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test(sql): cover cluster-key join cost factor
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‎tests/sqllogictests/suites/mode/standalone/explain/join_reorder/cluster_key_cost.test‎

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statement ok
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set ddl_column_type_nullable = 0;
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statement ok
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drop database if exists join_reorder_cluster_key_cost;
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statement ok
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create database join_reorder_cluster_key_cost;
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statement ok
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use join_reorder_cluster_key_cost;
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statement ok
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create table clustered_side(
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join_key int,
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filter_key int,
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payload int
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) cluster by(filter_key);
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statement ok
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insert into clustered_side
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select
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number % 100,
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number % 10,
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number
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from numbers(1000);
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statement ok
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create table plain_side(
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join_key int,
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filter_key int,
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payload int
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);
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statement ok
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insert into plain_side
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select
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number % 100,
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number % 10,
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number
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from numbers(1000);
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statement ok
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analyze table clustered_side;
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statement ok
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analyze table plain_side;
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# Isolate the cluster-key filter discount from TPCH data shape.
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# With the old formula disabled by factor 0, equal cardinality keeps the plain
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# table as build side. Factor 85 discounts the clustered-side filter and flips
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# the build side. Factor 100 is the neutral discount point and should not take
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# the discounted build-side path.
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statement ok
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set cost_factor_cluster_key = 0;
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query T
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explain join select *
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from clustered_side
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join plain_side on clustered_side.join_key = plain_side.join_key
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where clustered_side.filter_key = 1
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and plain_side.filter_key = 1;
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----
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HashJoin: INNER
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├── Build
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│ └── Scan: default.join_reorder_cluster_key_cost.plain_side (#1) (read rows: 1000)
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└── Probe
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└── Scan: default.join_reorder_cluster_key_cost.clustered_side (#0) (read rows: 1000)
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statement ok
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set cost_factor_cluster_key = 85;
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query T
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explain join select *
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from clustered_side
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join plain_side on clustered_side.join_key = plain_side.join_key
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where clustered_side.filter_key = 1
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and plain_side.filter_key = 1;
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----
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HashJoin: INNER
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├── Build
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│ └── Scan: default.join_reorder_cluster_key_cost.clustered_side (#0) (read rows: 1000)
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└── Probe
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└── Scan: default.join_reorder_cluster_key_cost.plain_side (#1) (read rows: 1000)
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statement ok
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set cost_factor_cluster_key = 100;
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query T
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explain join select *
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from clustered_side
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join plain_side on clustered_side.join_key = plain_side.join_key
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where clustered_side.filter_key = 1
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and plain_side.filter_key = 1;
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----
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HashJoin: INNER
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├── Build
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│ └── Scan: default.join_reorder_cluster_key_cost.plain_side (#1) (read rows: 1000)
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└── Probe
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└── Scan: default.join_reorder_cluster_key_cost.clustered_side (#0) (read rows: 1000)
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statement ok
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create or replace table a(k1 BIGINT, k2 BIGINT, v BIGINT) cluster by(k1, k2);
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statement ok
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insert into a
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select number, number, number
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from numbers(1000);
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statement ok
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create or replace table b(k1 BIGINT, k2 BIGINT, v BIGINT);
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statement ok
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insert into b
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select number, number, number
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from numbers(1000);
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statement ok
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create or replace table c(k1 BIGINT, k2 BIGINT, v BIGINT);
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statement ok
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insert into c
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select number, number, number
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from numbers(1000);
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statement ok
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analyze table a;
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statement ok
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analyze table b;
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statement ok
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analyze table c;
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statement ok
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set cost_factor_cluster_key = 85;
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# k1_k2_prefix from cluster_key_join_order.rs.
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query T
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explain join
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SELECT *
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FROM a
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JOIN b ON a.k1 = b.k1
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JOIN c ON a.k2 = c.k2;
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----
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HashJoin: INNER
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├── Build
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│ └── Scan: default.join_reorder_cluster_key_cost.b (#1) (read rows: 1000)
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└── Probe
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└── HashJoin: INNER
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├── Build
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│ └── Scan: default.join_reorder_cluster_key_cost.c (#2) (read rows: 1000)
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└── Probe
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└── Scan: default.join_reorder_cluster_key_cost.a (#0) (read rows: 1000)
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statement ok
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create or replace table a(k1 BIGINT, k2 BIGINT, v BIGINT) cluster by(k2, k1);
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statement ok
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insert into a
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select number, number, number
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from numbers(1000);
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statement ok
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analyze table a;
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# k2_k1_prefix from cluster_key_join_order.rs.
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query T
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explain join
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SELECT *
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FROM a
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JOIN b ON a.k1 = b.k1
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JOIN c ON a.k2 = c.k2;
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----
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HashJoin: INNER
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├── Build
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│ └── Scan: default.join_reorder_cluster_key_cost.c (#2) (read rows: 1000)
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└── Probe
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└── HashJoin: INNER
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├── Build
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│ └── Scan: default.join_reorder_cluster_key_cost.b (#1) (read rows: 1000)
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└── Probe
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└── Scan: default.join_reorder_cluster_key_cost.a (#0) (read rows: 1000)
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statement ok
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create or replace table a(k1 BIGINT, k2 BIGINT, v BIGINT) cluster by(k1, k2);
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statement ok
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insert into a
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select number, number, number
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from numbers(1000);
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statement ok
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analyze table a;
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# filter_preserves_cluster_keys from cluster_key_join_order.rs.
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query T
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explain join
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SELECT *
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FROM (SELECT * FROM a WHERE v >= 0) a
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JOIN b ON a.k1 = b.k1
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JOIN c ON a.k2 = c.k2;
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----
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HashJoin: INNER
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├── Build
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│ └── Scan: default.join_reorder_cluster_key_cost.b (#1) (read rows: 1000)
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└── Probe
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└── HashJoin: INNER
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├── Build
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│ └── Scan: default.join_reorder_cluster_key_cost.c (#2) (read rows: 1000)
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└── Probe
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└── Scan: default.join_reorder_cluster_key_cost.a (#0) (read rows: 1000)
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statement ok
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unset cost_factor_cluster_key;

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