if thats the case, how is a semi_nl useful? its doing a nested loop also.
There is alot of academic talk on semi-joins on the web, but I cant find an algorithm
anywhere. The basics of it appear to be generic.
>
> From: "Tanel Poder" <[EMAIL PROTECTED]>
> Date: 2003/11/03 Mon PM 12:49:25 EST
> To: Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
> Subject: Re: anyone have success with semi-joins?
>
> Hi!
>
> I don't know the exact algorithm either, but check the execution plans:
>
> SQL> set autot trace exp
> SQL> select name from obj$
> 2 where exists (select 1 from tab$ where obj$.obj# = tab$.obj#);
>
> Execution Plan
> ----------------------------------------------------------
> 0 SELECT STATEMENT Optimizer=CHOOSE
> 1 0 FILTER
> 2 1 TABLE ACCESS (FULL) OF 'OBJ$'
> 3 1 TABLE ACCESS (CLUSTER) OF 'TAB$'
> 4 3 INDEX (UNIQUE SCAN) OF 'I_OBJ#' (NON-UNIQUE)
>
>
>
> SQL> select /*+ HASH_SJ */ name from obj$
> 2 where exists (select 1 from tab$ where obj$.obj# = tab$.obj#);
>
> Execution Plan
> ----------------------------------------------------------
> 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=50 Card=8168 Bytes=3
> 51224)
>
> 1 0 HASH JOIN (SEMI) (Cost=50 Card=8168 Bytes=351224)
> 2 1 TABLE ACCESS (FULL) OF 'OBJ$' (Cost=25 Card=8168 Bytes=2
> 45040)
>
> 3 1 VIEW OF 'VW_SQ_1' (Cost=25 Card=8168 Bytes=106184)
> 4 3 TABLE ACCESS (FULL) OF 'TAB$' (Cost=25 Card=8168 Bytes
> =106184)
>
> You see in first execution plan we use FILTER for getting our results. FILTER has
> several meanings, but in this case it could be bounded nested loop (e.g. stopping
> for particular driving key when exists condition is satisfied). That means, for
> every row in obj$ we have to check separately whether there is any matching rows,
> this means lots of logical IOs.
>
> In second execution plan however, a hash join is done between obj$ and distinct obj#
> key values from tab$ (that's the reason why we're calling it a semi-join). Semi-join
> allow us to replace big amount of small nested loop IOs with few big reads and a
> hash join instead.
>
> Note that I don't know whether Oracle actually behaves this way, I'm just telling
> you how I think it is done, based only on brief analysis...
>
> Cheers,
> Tanel.
>
> ----- Original Message -----
> To: "Multiple recipients of list ORACLE-L" <[EMAIL PROTECTED]>
> Sent: Monday, November 03, 2003 5:04 PM
>
>
> > I picked up Mark Gurry's SQL pocket reference on Friday and it states that you can
> > radically improve performance of 'where exists' statements by using semi-joins
> > since it only returns the sub-query ones.
> >
> > Now logically you have to return the sub-query repeatedly, this would only come
> > into play if there are duplicates right?
> >
> > anyone know the algorithm for the semi-join? Not just the result? I cant find it
> > anywhere.
> >
> > anyone ever hint this or let oracle decide?
> >
> > --
> > Please see the official ORACLE-L FAQ: http://www.orafaq.net
> > --
> > Author: <[EMAIL PROTECTED]
> > INET: [EMAIL PROTECTED]
> >
> > Fat City Network Services -- 858-538-5051 http://www.fatcity.com
> > San Diego, California -- Mailing list and web hosting services
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> > also send the HELP command for other information (like subscribing).
> >
>
Hi!
I don't know the exact algorithm either, but check
the execution plans:
SQL> set autot trace exp
SQL> select name
from obj$
2 where exists (select 1 from tab$ where obj$.obj# =
tab$.obj#);
Execution
Plan
----------------------------------------------------------
0 SELECT STATEMENT
Optimizer=CHOOSE
1 0
FILTER
2 1 TABLE
ACCESS (FULL) OF 'OBJ$'
3
1 TABLE ACCESS (CLUSTER) OF 'TAB$'
4 3 INDEX (UNIQUE SCAN) OF
'I_OBJ#' (NON-UNIQUE)
SQL> select /*+ HASH_SJ */ name from
obj$
2 where exists (select 1 from tab$ where obj$.obj# =
tab$.obj#);
Execution
Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=50
Card=8168 Bytes=3
51224)
1
0 HASH JOIN (SEMI) (Cost=50 Card=8168 Bytes=351224)
2 1 TABLE ACCESS (FULL) OF 'OBJ$'
(Cost=25 Card=8168
Bytes=2
45040)
3
1 VIEW OF 'VW_SQ_1' (Cost=25 Card=8168
Bytes=106184)
4
3 TABLE ACCESS (FULL) OF 'TAB$' (Cost=25
Card=8168 Bytes
=106184)
You see in first execution plan we use FILTER for
getting our results. FILTER has several meanings, but in this case it could be
bounded nested loop (e.g. stopping for particular driving key when exists
condition is satisfied). That means, for every row in obj$ we have to check
separately whether there is any matching rows, this means lots of logical
IOs.
In second execution plan however, a hash join is
done between obj$ and distinct obj# key values from tab$
(that's the reason why we're calling it a semi-join). Semi-join allow us to replace big amount of small nested loop
IOs with few big reads and a hash join instead.
Note that I don't know whether Oracle actually
behaves this way, I'm just telling you how I think it is done, based
only on brief analysis...
Cheers,
Tanel.
----- Original Message -----
Sent: Monday, November 03, 2003 5:04
PM
Subject: anyone have success with
semi-joins?
> I picked up Mark Gurry's SQL pocket reference on Friday and it
states that you can radically improve performance of 'where exists' statements
by using semi-joins since it only returns the sub-query ones.
>
>
Now logically you have to return the sub-query repeatedly, this would only come
into play if there are duplicates right?
>
> anyone know the
algorithm for the semi-join? Not just the result? I cant find it anywhere.
>
> anyone ever hint this or let oracle decide?
>
>
--
> Please see the official ORACLE-L FAQ: http://www.orafaq.net
> --
> Author: <[EMAIL PROTECTED]
>
INET: [EMAIL PROTECTED]
>
>
Fat City Network Services -- 858-538-5051 http://www.fatcity.com
> San
Diego, California -- Mailing list and
web hosting services
>
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>
