Re: Hash join order

On Jun 25, 2:13 pm, pratap.deshm..._at_gmail.com wrote:
> Hi,
>
> We are using Oracle 10.2.0.3.0. I have a simple query like this -
>
> select *
> from large_table, small_table
> where small_table.col = large_table.col (+)
>
> large_table has 140 million rows. Small table can have any number of
> rows.I am posting the execution plan for different scenarios where
> small table has different number of records.
>
> Case 1 - Small table has 180K records and large table has 140 million
> records. The execution plan shows that small table is accessed first
> followed by large_table and the large_table probes the hash table of
> small table
>
> ---------------------------------------------------------------------------­---------------------------------------------------
> | Id  | Operation               | Name                       | Rows  |
> Bytes |TempSpc| Cost (%CPU)| Time     | Pstart| Pstop |
> ---------------------------------------------------------------------------­---------------------------------------------------
> |   0 | SELECT STATEMENT        |                            |
> 180K|   122M|       |   272K  (3)| 01:03:42 |       |       |
> |*  1 |  HASH JOIN OUTER        |                            |
> 180K|   122M|   122M|   272K  (3)| 01:03:42 |       |       |
> |   2 |   PARTITION RANGE SINGLE|                            |
> 180K|   120M|       |  3079   (4)| 00:00:44 |     3 |     3 |
> |   3 |    PARTITION LIST ALL   |                            |
> 180K|   120M|       |  3079   (4)| 00:00:44 |     1 |    25 |
> |*  4 |     TABLE ACCESS FULL   | SMALL_TABLE                |
> 180K|   120M|       |  3079   (4)| 00:00:44 |    51 |    75 |
> |   5 |   INDEX FAST FULL SCAN  | LARGE_TABLE_INDEX          |
> 140M|  2007M|       |   105K  (3)| 00:24:40 |       |       |
> ---------------------------------------------------------------------------­---------------------------------------------------
>
> Case 2 - Large table has 140 million records and small table has 12
> million records. But this time the join order is reversed
>
> ---------------------------------------------------------------------------­---------------------------------------------------
> | Id  | Operation               | Name                       | Rows  |
> Bytes |TempSpc| Cost (%CPU)| Time     | Pstart| Pstop |
> ---------------------------------------------------------------------------­---------------------------------------------------
> |   0 | SELECT STATEMENT        |                            |
> 12M|  8234M|       |   815K  (2)| 03:10:13 |       |       |
> |*  1 |  HASH JOIN RIGHT OUTER  |                            |
> 12M|  8234M|  3614M|   815K  (2)| 03:10:13 |       |       |
> |   2 |   INDEX FAST FULL SCAN  | LARGE_TABLE_INDEX          |
> 140M|  2007M|       |   105K  (3)| 00:24:40 |       |       |
> |   3 |   PARTITION RANGE SINGLE|                            |
> 12M|  8062M|       |   200K  (4)| 00:46:50 |     1 |     1 |
> |   4 |    PARTITION LIST ALL   |                            |
> 12M|  8062M|       |   200K  (4)| 00:46:50 |     1 |    25 |
> |*  5 |     TABLE ACCESS FULL   | SMALL_TABLE                |
> 12M|  8062M|       |   200K  (4)| 00:46:50 |     1 |    25 |
> ---------------------------------------------------------------------------­---------------------------------------------------
>
> Question is - Why is Oracle reversing the join order in Case 2 when
> ideally it should be probing the large table into the hash of the
> small table?
>
> Regards,
> Pratap

Because at some point between 180K and 12M rows Oracle decides it is cheaper to read the whole small table into a temp segment than to probe 12M times with a right outer join. Do you disagree? Try a hint and see what happens.

jg

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