Oracle Scratchpad

April 26, 2015

Not Exists

Filed under: Oracle,Troubleshooting — Jonathan Lewis @ 7:17 pm BST Apr 26,2015

Another question on a seemingly simple “not exists” query has appeared on OTN just a few days after my last post about the construct. There are two little differences between the actual form of the two queries that make it worth repeating the analysis.

The first query was of the form:

select from big_table
where  not exists (select exact_matching_row from small table);

while the new query is of the form:

select from big_table alias1
where not exists (select inexact_matching_row from big_table alias2)

In the absence of a complete explanation, we might guess that the intention of the first query is to: “check correctness of undeclared foreign key constraint” i.e. small_table is the parent table with unique values, and big_table is the child end with some data that may be invalid due to a recent bulk data load. (In my example for the previous posting I relaxed the uniqueness assumption in small_table to make the problem a little more expensive.)

Our guess for the second query ought to be different; we are using the same table twice, and we are checking for the non-existence of “imperfect matches”. This introduces two potential threats – first that the (possibly pseudo-)join between the two tables is between two large tables and therefore inherently likely to be expensive; second that we may be allowed to have very large numbers of “perfect matches” that will escalate the scale of the join quite dramatically. Here’s the actual query (second version) from the posting; this will make it easier to explain why the structure of the query introduces the second threat and requires us (as so often) to understand the data in order to optimise the query execution:

select  count(*)
from    ubl_stg.wk_sap_fat w1
where   not exists (
                select  1
                from    ubl_stg.wk_sap_fat w2
                where   w2.mes_mese_id =  w1.mes_mese_id
                and     w2.sistema     <> w1.sistema

Note the schema name ubl_stg – doesn’t that hint at “staging tables” for a data load.
Note the column name mes_mese_id – an “id” column, but clearly not one that’s supposed to be unique, so possibly a foreign key column that allows repetitions

The code is looking for, then counting, cases where for a given value of mes_mese_id there is only one corresponding value used for sistema. Since we don’t know the application we don’t know whether this count should be low or high relative to the number of rows in the table, nor do we know how many distinct values there might be of mes_mese_id – and these pieces of information are critical to identifying the best execution plan for the query.

The owner of this query told us that the table held 2 million rows which are “deleted and reloaded” every day, showed us that the (default) execution plan was a hash anti-join which took two hours to complete, told us that he (or she) didn’t think that an index would help because of the “not equal” predicate, and finally reported back that the problem was solved (no timing information supplied) by the addition of the /*+ no_unnest */ hint.

The question is – what has happened and why is there such a difference in performance ?

The first observation is that the 2 hours does seem an unreasonably long time for the query to run – and since it’s only a select statement it would be good to run it again and take a snapshot of the session statistics and session events (v$sesstat, v$session_event) for the session to see what the workload was and where the time was spent. Given the “deleted and reloaded” comment it’s possible that there may be some unexpected overhead due to some strange effects of read-consistency or delayed block cleanout, so we might also take a snapshot of tablespace or file I/O to check for lots of I/O on the undo tablespace (which might also show an I/O problem on the table’s tablespace or the user’s TEMP tablespace). The snapshots give us a very cheap, non-invasive option for getting some summary stats – but the tablespace/file stats are system-wide, of course, so may not tell us anything about our specific task, so we might even enable extended tracing (event 10046 level 8 / dbms_monitor with waits) to see in detail where we are losing time.

Since we don’t currently have the information we need to explain the two hours (it may have appeared by the time I post this note) it might be instructive to make some guesses about where the time could go in the hash anti-join, and why the /*+ no_unnest */ hint could make a difference.

Ignoring the possibility of strange undo/read-consistency/cleanout effects the first possiblity is simply that the hash join is large and turns into a multi-pass I/O thrash. The mes_mese_id column looks like it might be a number (id columns so often are) but the sistema column has the flavour of a reasonably large character column – so maybe our hash table has to be a couple of hundred megabytes – that could certainly be enough to spill to disc, though you’d have to have a really small PGA availability for it to turn into a multi-pass hash join.

Another possibility is that the pattern in the data makes the hash join burn up a huge amount of CPU – that should be easy to see on a re-run.  If there are relatively few distinct sets of values for (mes_mese_id, sistema) and there are very few cases where a mes_mese_id is associated with more than one sistema, then a large fraction of the hash table probes would have to follow a very long chain of matches to the very end, and that would take a large amount of CPU.

Pursue that “long chain” hypothesis to a slight extreme – what if there’s one mes_mese_id that appears in 250,000 of the 2M rows, and the sistema value is the same for every one of those quarter million rows,  which would require Oracle to walk a chain of 250,000 elements 250,000 times, for a total of 62.5 billion pointers to follow and comparisions to make – how much CPU might that take.  Worse still,since having the same mes_mese_id (the hashing is only on the equality predicate) means the quarter of a million rows would have to go into the same hash bucket so we might end up doing a multipass operation because that one bucket was very much larger than anything Oracle had anticipated when it did its internal partitioning calculations for the hash table. (There are some related notes in this article and in chapter 12 of Cost Based Oracle – Fundamentals)

Why would the /*+ unnest */ hint address these performance problems ? My first guess would be that the OP may have created the index on (mes_mese_id, sistema), in which case a full scan of that index (or even a fast-full scan if the index were newly created, or even a tablescan if the data had been loaded in the right order) followed by the filter subquery being driven by an index range scan would result in a relatively efficient subquery being executed once per distinct value of (mes_mese_id, sistema) rather than once per row. This could be much more efficient than a really badly skewed data set doing the hash anti-join. In fact, even in the absence of the index, if the number of distinct combinations was really quite small, that many tablescans – if the table were cached, whether in Oracle or the filesystem or the SAN cache – might still be a lot faster than the hash anti-join. (We were told that the problem was solved – but not told how much time constituted a viable solution.)


Hand-waving and talk is fine – but a few modelled results might make it easier to comprehend, so here’s a way of generating a few versions of data sets and testing:

drop table t1 purge;

create table t1
with generator as (
        select  --+ materialize
                rownum id
        from dual
        connect by
                level <= 1e4
        rownum                                          id,
        --      Random generation nearly guaranteeing no random
        --      duplicates of (x,y) but quite a lot of mismatches
        --      Count 735,715 in 3.36 seconds.
        trunc(dbms_random.value(0,2e6))                 x,
        lpad(trunc(dbms_random.value(0,1e6)),64,'0')    y
        --      One specific pair repeated many times with no mismatch the rest
        --      as previously generated above. Times with different repeat counts:
        --       10,000            14 seconds
        --       20,000            52 seconds
        --       40,000           202 seconds
        --      CPU time quadruples as the count doubles (twice the number of
        --      probes walking twice the number of steps in the hash chain)
        --       80,000 =>        800 seconds
        --      160,000 =>      3,200 seconds
        --      240,000 =>      ca. 2 hours.
                when rownum <= 40000
                        then 2e6
                        else trunc(dbms_random.value(0,2e6))
        end                                             x,
                when rownum <= 40000
                                        trunc((rownum - 1)/1000) +
                                                case mod(rownum-1,1000) when 0 then 0 else 0 end,
        end                                             y
        --      2,000 distinct values repeated 1,000 times each
        --      Query result: 2,000,000 in 235 seconds.
        trunc((rownum - 1)/1000)                x,
                trunc((rownum - 1)/1000) +
                        case mod(rownum-1,1000) when 0 then 0 else 0 end,
        )                                       y
        generator       v1,
        generator       v2
        rownum <= 2e6

                ownname          => user,
                tabname          =>'T1',
                method_opt       => 'for all columns size 1'

from    t1 w1
where   not exists (
                select  1
                from    t1 w2
                where   w2.x = w1.x
                and     w2.y <> w1.y

As you can see I’ve changed the table and column names – this keeps them in line with the original SQL statement presented on OTN before we got the graphic display of a similar statement and plan. The SQL to create the data includes three variants and 5 sets of results (and 3 conjectures) running These results appeared when the optimizer took the hash anti-join execution plan, and also spilled to disc with a one-pass workarea operation that first extended to about 200MB of PGA then dropped back to about 8MB.

To summarise the results recorded in the SQL – if we use the term “bad data” to describe rows where more than one Y value appears for a given X value then:

  1. With a large number of distinct pairs and a lot of bad data: the anti-join is pretty fast at 3.36 seconds.
  2. With no bad data, a small number of distinct pairs (2,000) and lots of rows per pair (1,000): the anti-join takes 235 CPU seconds
  3. As for #1 above, but with one extreme “good” pair that appears a large number of times: CPU time is proportional to the square of the number of duplicates of this value

I didn’t actually test beyond 40,000 duplicates for the last case, but you can see the double/quadruple pattern very clearly and the CPU time would have hit 2 hours at around 240,000 identical copies of one (x,y) pair.

/*+ no_unnest */

So what happens if you try using the /*+ no_unnest */ hint ? The target here is that the more repetitive the data the smaller the number of times you may have to run the subquery; and if you can get the driving data ordered you can guarantee the smallest possible number of runs of the subquery. I haven’t worked through all the possibilities, but to give you the flavour of what can happen, when I added the /*+ no_unnest */ hint to the query (and ensured that the table would be cached rather than read using direct path reads into the PGA) the execution time for the test with 1,000 copies of 2,000 pairs took 181 seconds to do 2,001 tablescans (compared with 235 seconds to do 2 tablescans and a hash anti-join) with the following execution path:

| Id  | Operation           | Name | Rows  | Bytes | Cost (%CPU)| Time     |
|   0 | SELECT STATEMENT    |      |     1 |    69 |  8256K  (4)| 11:28:04 |
|   1 |  SORT AGGREGATE     |      |     1 |    69 |            |          |
|*  2 |   FILTER            |      |       |       |            |          |
|   3 |    TABLE ACCESS FULL| T1   |  2000K|   131M|  2877   (4)| 00:00:15 |
|*  4 |    TABLE ACCESS FULL| T1   |     2 |   138 |     4   (0)| 00:00:01 |

Predicate Information (identified by operation id):
   2 - filter( NOT EXISTS (SELECT /*+ NO_UNNEST NO_INDEX ("W2") */ 0
              FROM "T1" "W2" WHERE "W2"."X"=:B1 AND "W2"."Y"<>:B2))
   4 - filter("W2"."X"=:B1 AND "W2"."Y"<>:B2)

More significantly, when I created the optimum index the execution time dropped to 0.9 seconds – here’s the create index statement and subsequent plan – the extreme benefit appears because the data was effectively loaded in sorted order; if this had not been the case I would have forced an index full scan for the driving data set (with a “not null” predicate or constraint to make it possible for the optimizer to use the index to drive the query)

create index t1_i1 on t1(x,y) compress pctfree 0;

| Id  | Operation           | Name  | Rows  | Bytes | Cost (%CPU)| Time     |
|   0 | SELECT STATEMENT    |       |     1 |    69 |  6008K  (1)| 08:20:45 |
|   1 |  SORT AGGREGATE     |       |     1 |    69 |            |          |
|*  2 |   FILTER            |       |       |       |            |          |
|   3 |    TABLE ACCESS FULL| T1    |  2000K|   131M|  2877   (4)| 00:00:15 |
|*  4 |    INDEX RANGE SCAN | T1_I1 |     2 |   138 |     3   (0)| 00:00:01 |

Predicate Information (identified by operation id):
   2 - filter( NOT EXISTS (SELECT /*+ NO_UNNEST */ 0 FROM "T1" "W2"
              WHERE "W2"."X"=:B1 AND "W2"."Y"<>:B2))
   4 - access("W2"."X"=:B1)

Similarly, when I had the index in place and the 40,000 repetitions of a “good pair”, Oracle took a total of 9 seconds even though it had to run the subquery 1,960,000 times for the non-repetitive data (and once for the repetitive_pair). I have to say I was a little surprised at how rapidly it managed to get through that 2M subquery executions – but then I keep forgetting how ridiculously overpowered my new laptop is.

With these figures in mind you can appreciate that if the OP had lots of pairs with tens of thousands of repetitions, then even without creating the index on the table, the query time might drop from 2 hours for the hash anti-join to “a few minutes” for the filter subquery with a time that was good enough to look like “problem solved”.


If you have a few “long hash chains” in the build table – i.e. rows from the “first” table in the join that hash to the same value – then the amount of work Oracle has to do to check a single row from the probe (“second”) table that matches on the hash value can become significant. If a large number of rows from the probe table hit a long hash chain then the CPU time for the whole join can climb dramatically and you may want to force Oracle away from the hash join.

If the the long chains are the result of a skewed distribution where a small number of values appear very frequently in a table with a large number of distinct values that each appears infrequently then the optimizer may not notice the threat and may choose the hash plan when there is a much less resource-intensive alternative available.


There was another interesting observations I made while doing the experiments relating to whether the chains are due to hash collisions or require exact matches in the data – but I’ve spent an hour on desgining and running tests, and nearly 4 hours writing up the results so far. I need to do a few more tests to work out whether I’m seeing a very clever optimisation or a lucky coincidence in a certain scenario – so I’m going to save that for another day.




  1. Jonathan,
    thank you for the explanation and the creation of the model. I think this could be again a situation in which it would be interesting to have an additional information in the rowsource statistics showing the size of the intermediate set created by the HASH Join before the filtering takes place – something like the AE (Actually Evaluated) rows from

    An additional question: you write: “If the the long chains are the result of a skewed distribution where a small number of values appear very frequently in a table with a large number of distinct values that each appears infrequently then the optimizer may not notice the threat and may choose the hash plan when there is a much less resource-intensive alternative available.” Does this mean that you actually saw situations in which the optimizer choosed the no_unnest/filter strategy without hinting? Taking a look at the costing it seems that the optimizer regards this approach to be quite expensive – at least much more expensive than the Hash Join.



    Comment by Martin Preiss — April 28, 2015 @ 9:09 am BST Apr 28,2015 | Reply

    • Martin,

      Thanks for the link – it’s funny how the same problem can come up simultaneously: I’m surprised I didn’t notice Randolf’s article when it came out. I’ve upvoted the suggestion, and added a comment to it about the 3 numbers you need for the hash join (and the two numbers you need for even an index range scan).

      Regarding not hinting – as usualy it comes down to the accumulation of errors in cardinality estimates. The filter subquery is more expensive than the hash join if it has to happen enough times; but if the optimizer thinks it’s not going to happen often it’s more likely to appear: modify my code to remove the no_unnest hint in the subquery and add a cardinality(w1 100) to the main query block and Oracle will choose the filter subquery when the index is in place.

      Comment by Jonathan Lewis — April 28, 2015 @ 9:50 am BST Apr 28,2015 | Reply

      • thank you for the comments and additions (especially the third number for Hash Joins: though this makes the choice of a fitting name even more difficult…) – here and on OTN. Regarding the costing: in with your default values I got an adaptive plan potentially switching between HJ and NL (though the number of iterations was high above the inflection point so I also got the Hash Join all the time).

        Comment by Martin Preiss — April 28, 2015 @ 3:44 pm BST Apr 28,2015 | Reply

  2. […] whole thing about “not exists” subqueries can run and run. In the previous episode I walked through some ideas of how the following query might perform depending on the data, the […]

    Pingback by Not Exists | Oracle Scratchpad — April 29, 2015 @ 8:21 pm BST Apr 29,2015 | Reply

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