1. Query Rewrite
Starburst Model (IBM)

2. DB2 Query Optimizer (Starburst)
Control Flow
Parsing and Semantic Checking
Data Flow
Query Graph
Model
Query Rewrite
Plan Optimization
Executable
Plan
Compile Time
Run Time
Query Evaluation System

3. Goal of Query Rewrite Make queries as declarative as possible:
Poorly expressed queries could force the optimizer into choosing suboptimal plans
Perform natural heuristics
For example, “predicate pushdown”

4. Components of Rewrite Engine
Rewrite rules (more later)
Rule engine
control strategies
sequential (rules are processed sequentially)
priority (higher priority rules are given a chance first)
statistical (next rule is chosen randomly based on a user defined probability distribution
budget
to avoid spending too much time on rewrites, the processing stops at a consistent state of QGM when the budget is exhausted
Search facility
browses through QGM providing the context for the rules to work on

5. Problem
How do we choose between competing incompatible transformations?
Optimal solution: apply cost analysis and pick the transformation leading to a cheaper plan
Practical solution (why?): generate multiple alternatives and send them to plan optimization phase (problems?)

6. Rewrite Rules: SELECT Merge
CREATE VIEW itpv AS
(SELECT DISTINCT itp.itemn, pur.vendn
FROM itp, pur
WHERE itp.ponum = pur.ponum AND
pur.odate > ’85’)
SELECT itm.itmn, itpv.vendn
FROM itm, itpv
WHERE itm.itemn = itpv.itemn AND
itm.itemn > ’01’ AND
itm.itemn < ’20’
SELECT DISTINCT itm.itmn, pur.vendn
FROM itm, itp, pur
WHERE itp.ponum = pur.ponum AND itm.itemn = itpv.itemn AND
pur.odate > ’85’ AND
itm.itemn > ’01’ AND
itm.itemn < ’20’
Speedup: 200 times

7. Rewrite Rules: Existential Subquery Merge
SELECT *
FROM itp
WHERE itm.itemn IN
( SELECT itl.itmn
FROM itl
WHERE itl.wkcen = ‘WK468’ AND
itl.locan = ‘L’)
SELECT DISTINCT itp.*
FROM itp, itl
WHERE itp.itmn = itl.itemn AND
itl.wkcen = ‘WK468’ AND
itl.locan = ‘L’
Speedup: 15 times

8. Rewrite Rules:Intersect to Exists
SELECT itemn
FROM wor
WHERE empno = ‘EMPN1279’
INTERSECT
SELECT itmn
FROM itl
WHERE entry_time = ‘9773’ AND
wkctr = ‘WK195’)
SELECT DISTINCT itemn
FROM wor, itl
WHERE empno = ‘EMPN1279’
entry_time = ‘9773’ AND
wkctr = ‘WK195’) AND
itl.itmn = wor.itemn
Speedup: 8 times

9. The Count Bug (cont.) parts(PNUM,QOH) supply(PNUM,QUAN,SHIPDATE)
Query: Find the part numbers of those parts whose quantities on hand equal the number of shipments of those parts before
select PNUM
from parts
where QOH = ( select count(SHIPDATE)
from supply
where supply.PNUM = parts.PNUM
and SHIPDATE < )

10. The Count Bug (cont.) select PNUM from parts
where QOH = ( select count(SHIPDATE)
from supply
where supply.PNUM = parts.PNUM
and SHIPDATE < )
temp (SUPPNUM,CT) =
(select PNUM, count(SHIPDATE)
from supply
where SHIPDATE < )
group by PNUM)
select PNUM
from parts, temp
where parts.QOH = temp.CT and
temp.PNUM = parts.PNUM

11. The Count Bug (cont.) Parts Supply PNUM QOH 3 6 10 1 8 PNUM QUAN
PNUM
QUAN
SHIPDATE 3 4
7-3-79 2 10 1
6-8-78 8 5
5-7-83
select PNUM
from parts
where QOH = ( select count(SHIPDATE)
from supply
where supply.PNUM = parts.PNUM
and SHIPDATE < )
Result
PNUM 10 8

12. The Count Bug (cont.) Parts Supply PNUM QOH 3 6 10 1 8 PNUM QUAN
PNUM
QUAN
SHIPDATE 3 4
7-3-79 2 10 1
6-8-78 8 5
5-7-83
Temp
temp (SUPPNUM,CT) =
(select PNUM, count(SHIPDATE)
from supply
where SHIPDATE < )
group by PNUM)
Suppnum CT 3 2 10 1

13. The Count Bug (cont.) Parts Temp PNUM QOH 3 6 10 1 8 SUPPNUM CT 3 2 10
SUPPNUM CT 3 2 10 1
Result
select PNUM
from parts, temp
where parts.QOH = temp.CT and
temp.PNUM = parts.PNUM
PNUM 10

14. The Count Bug – solution with outer joinsX A B Y B C E X Y A
null B C E
Null R S
R=+S

15. The Count Bug – solution with outer joins
temp (SUPPNUM,CT) =
(select parts.PNUM, count(SHIPDATE)
from parts, supply
where SHIPDATE < and
parts.PNUM =+ supply.PNUM
group by parts.PNUM)
parts.PNUM =+ supply.PNUM (for SHIPDATE < )
Parts.PNUM
Parts.QOH
Supply.PNUM
Supply.QUON
Supply.SHIPDATE 3 6 4
7-3-79 2 10 1
6-8-78 8
null