Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Chapter 13: Query Optimization
©Silberschatz, Korth and Sudarshan1.2Database System Concepts - 6th Edition
Chapter 13: Query Optimization
 Introduction
 Transformation of Relational Expressions
 Catalog Information for Cost Estimation
 Statistical Information for Cost Estimation
 Cost-based optimization
 Dynamic Programming for Choosing Evaluation
Plans
 Materialized views
©Silberschatz, Korth and Sudarshan1.3Database System Concepts - 6th Edition
Introduction
 Alternative ways of evaluating a given query
 Equivalent expressions
 Different algorithms for each operation
©Silberschatz, Korth and Sudarshan1.4Database System Concepts - 6th Edition
Introduction (Cont.)
 An evaluation plan defines exactly what algorithm is used for each
operation, and how the execution of the operations is coordinated.
 Find out how to view query execution plans on your favorite database
©Silberschatz, Korth and Sudarshan1.5Database System Concepts - 6th Edition
Introduction (Cont.)
 Cost difference between evaluation plans for a query can be
enormous
 E.g. seconds vs. days in some cases
 Steps in cost-based query optimization
1. Generate logically equivalent expressions using equivalence
rules
2. Annotate resultant expressions to get alternative query plans
3. Choose the cheapest plan based on estimated cost
 Estimation of plan cost based on:
 Statistical information about relations. Examples:
 number of tuples, number of distinct values for an attribute
 Statistics estimation for intermediate results
 to compute cost of complex expressions
 Cost formulae for algorithms, computed using statistics
Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Generating Equivalent Expressions
©Silberschatz, Korth and Sudarshan1.7Database System Concepts - 6th Edition
Transformation of Relational Expressions
 Two relational algebra expressions are said to be equivalent if
the two expressions generate the same set of tuples on every
legal database instance
 Note: order of tuples is irrelevant
 we don’t care if they generate different results on databases
that violate integrity constraints
 In SQL, inputs and outputs are multisets of tuples
 Two expressions in the multiset version of the relational
algebra are said to be equivalent if the two expressions
generate the same multiset of tuples on every legal
database instance.
 An equivalence rule says that expressions of two forms are
equivalent
 Can replace expression of first form by second, or vice versa
©Silberschatz, Korth and Sudarshan1.8Database System Concepts - 6th Edition
Equivalence Rules
1. Conjunctive selection operations can be deconstructed into a
sequence of individual selections.
2. Selection operations are commutative.
3. Only the last in a sequence of projection operations is
needed, the others can be omitted.
4. Selections can be combined with Cartesian products and
theta joins.
a. (E1 X E2) = E1  E2
b. 1(E1 2 E2) = E1 1 2 E2
))(())(( 1221
EE   
))(()( 2121
EE   
)())))(((( 121
EE LLnLL  
©Silberschatz, Korth and Sudarshan1.9Database System Concepts - 6th Edition
Equivalence Rules (Cont.)
5. Theta-join operations (and natural joins) are commutative.
E1  E2 = E2  E1
6. (a) Natural join operations are associative:
(E1 E2) E3 = E1 (E2 E3)
(b) Theta joins are associative in the following manner:
(E1 1 E2) 2 3 E3 = E1 1 3 (E2 2 E3)
where 2 involves attributes from only E2 and E3.
©Silberschatz, Korth and Sudarshan1.10Database System Concepts - 6th Edition
Pictorial Depiction of Equivalence Rules
©Silberschatz, Korth and Sudarshan1.11Database System Concepts - 6th Edition
Equivalence Rules (Cont.)
7. The selection operation distributes over the theta join operation
under the following two conditions:
(a) When all the attributes in 0 involve only the attributes of one
of the expressions (E1) being joined.
0E1  E2) = (0(E1))  E2
(b) When  1 involves only the attributes of E1 and 2 involves
only the attributes of E2.
1 E1  E2) = (1(E1))  ( (E2))
©Silberschatz, Korth and Sudarshan1.12Database System Concepts - 6th Edition
Equivalence Rules (Cont.)
8. The projection operation distributes over the theta join operation
as follows:
(a) if  involves only attributes from L1  L2:
(b) Consider a join E1  E2.
 Let L1 and L2 be sets of attributes from E1 and E2,
respectively.
 Let L3 be attributes of E1 that are involved in join condition ,
but are not in L1  L2, and
 let L4 be attributes of E2 that are involved in join condition ,
but are not in L1  L2.
))(())(()( 2121 2121
EEEE LLLL   
)))(())((()( 2121 42312121
EEEE LLLLLLLL   
©Silberschatz, Korth and Sudarshan1.13Database System Concepts - 6th Edition
Equivalence Rules (Cont.)
9. The set operations union and intersection are commutative
E1  E2 = E2  E1
E1  E2 = E2  E1
 (set difference is not commutative).
10. Set union and intersection are associative.
(E1  E2)  E3 = E1  (E2  E3)
(E1  E2)  E3 = E1  (E2  E3)
11. The selection operation distributes over ,  and –.
 (E1 – E2) =  (E1) – (E2)
and similarly for  and  in place of –
Also:  (E1 – E2) = (E1) – E2
and similarly for  in place of –, but not for 
12. The projection operation distributes over union
L(E1  E2) = (L(E1))  (L(E2))
©Silberschatz, Korth and Sudarshan1.14Database System Concepts - 6th Edition
Transformation Example: Pushing Selections
 Query: Find the names of all instructors in the Music
department, along with the titles of the courses that they teach
 name, title(dept_name= “Music”
(instructor (teaches course_id, title (course))))
 Transformation using rule 7a.
 name, title((dept_name= “Music”(instructor))
(teaches course_id, title (course)))
 Performing the selection as early as possible reduces the size
of the relation to be joined.
©Silberschatz, Korth and Sudarshan1.15Database System Concepts - 6th Edition
Example with Multiple Transformations
 Query: Find the names of all instructors in the Music department
who have taught a course in 2009, along with the titles of the
courses that they taught
 name, title(dept_name= “Music”gear = 2009
(instructor (teaches course_id, title (course))))
 Transformation using join associatively (Rule 6a):
 name, title(dept_name= “Music”gear = 2009
((instructor teaches) course_id, title (course)))
 Second form provides an opportunity to apply the “perform
selections early” rule, resulting in the subexpression
dept_name = “Music” (instructor)  year = 2009 (teaches)
©Silberschatz, Korth and Sudarshan1.16Database System Concepts - 6th Edition
Multiple Transformations (Cont.)
©Silberschatz, Korth and Sudarshan1.17Database System Concepts - 6th Edition
Transformation Example: Pushing Projections
 Consider: name, title(dept_name= “Music” (instructor) teaches)
course_id, title (course))))
 When we compute
(dept_name = “Music” (instructor teaches)
we obtain a relation whose schema is:
(ID, name, dept_name, salary, course_id, sec_id, semester,
year)
 Push projections using equivalence rules 8a and 8b; eliminate
unneeded attributes from intermediate results to get:
name, title(name, course_id (
dept_name= “Music” (instructor) teaches))
course_id, title (course))))
 Performing the projection as early as possible reduces the size
of the relation to be joined.
©Silberschatz, Korth and Sudarshan1.18Database System Concepts - 6th Edition
Join Ordering Example
 For all relations r1, r2, and r3,
(r1 r2) r3 = r1 (r2 r3 )
(Join Associativity)
 If r2 r3 is quite large and r1 r2 is small, we choose
(r1 r2) r3
so that we compute and store a smaller temporary relation.
©Silberschatz, Korth and Sudarshan1.19Database System Concepts - 6th Edition
Join Ordering Example (Cont.)
 Consider the expression
name, title(dept_name= “Music” (instructor) teaches)
course_id, title (course))))
 Could compute teaches course_id, title (course) first, and
join result with
dept_name= “Music” (instructor)
but the result of the first join is likely to be a large relation.
 Only a small fraction of the university’s instructors are likely to
be from the Music department
 it is better to compute
dept_name= “Music” (instructor) teaches
first.
©Silberschatz, Korth and Sudarshan1.20Database System Concepts - 6th Edition
Enumeration of Equivalent Expressions
 Query optimizers use equivalence rules to systematically generate
expressions equivalent to the given expression
 Can generate all equivalent expressions as follows:
 Repeat
 apply all applicable equivalence rules on every subexpression of
every equivalent expression found so far
 add newly generated expressions to the set of equivalent
expressions
Until no new equivalent expressions are generated above
 The above approach is very expensive in space and time
 Two approaches
 Optimized plan generation based on transformation rules
 Special case approach for queries with only selections, projections
and joins
©Silberschatz, Korth and Sudarshan1.21Database System Concepts - 6th Edition
Implementing Transformation Based
Optimization
 Space requirements reduced by sharing common sub-expressions:
 when E1 is generated from E2 by an equivalence rule, usually only the top
level of the two are different, subtrees below are the same and can be
shared using pointers
 E.g. when applying join commutativity
 Same sub-expression may get generated multiple times
 Detect duplicate sub-expressions and share one copy
 Time requirements are reduced by not generating all expressions
 Dynamic programming
 We will study only the special case of dynamic programming for join
order optimization
E1 E2
©Silberschatz, Korth and Sudarshan1.22Database System Concepts - 6th Edition
Cost Estimation
 Cost of each operator computer as described in Chapter 12
 Need statistics of input relations
 E.g. number of tuples, sizes of tuples
 Inputs can be results of sub-expressions
 Need to estimate statistics of expression results
 To do so, we require additional statistics
 E.g. number of distinct values for an attribute
 More on cost estimation later
©Silberschatz, Korth and Sudarshan1.23Database System Concepts - 6th Edition
Choice of Evaluation Plans
 Must consider the interaction of evaluation techniques when choosing
evaluation plans
 choosing the cheapest algorithm for each operation independently
may not yield best overall algorithm. E.g.
 merge-join may be costlier than hash-join, but may provide a
sorted output which reduces the cost for an outer level
aggregation.
 nested-loop join may provide opportunity for pipelining
 Practical query optimizers incorporate elements of the following two
broad approaches:
1. Search all the plans and choose the best plan in a
cost-based fashion.
2. Uses heuristics to choose a plan.
©Silberschatz, Korth and Sudarshan1.24Database System Concepts - 6th Edition
Cost-Based Optimization
 Consider finding the best join-order for r1 r2 . . . rn.
 There are (2(n – 1))!/(n – 1)! different join orders for above expression.
With n = 7, the number is 665280, with n = 10, the number is greater
than 176 billion!
 No need to generate all the join orders. Using dynamic programming,
the least-cost join order for any subset of
{r1, r2, . . . rn} is computed only once and stored for future use.
©Silberschatz, Korth and Sudarshan1.25Database System Concepts - 6th Edition
Dynamic Programming in Optimization
 To find best join tree for a set of n relations:
 To find best plan for a set S of n relations, consider all possible
plans of the form: S1 (S – S1) where S1 is any non-empty
subset of S.
 Recursively compute costs for joining subsets of S to find the cost
of each plan. Choose the cheapest of the 2n – 2 alternatives.
 Base case for recursion: single relation access plan
 Apply all selections on Ri using best choice of indices on Ri
 When plan for any subset is computed, store it and reuse it when it
is required again, instead of recomputing it
 Dynamic programming
©Silberschatz, Korth and Sudarshan1.26Database System Concepts - 6th Edition
Join Order Optimization Algorithm
procedure findbestplan(S)
if (bestplan[S].cost  )
return bestplan[S]
// else bestplan[S] has not been computed earlier, compute it now
if (S contains only 1 relation)
set bestplan[S].plan and bestplan[S].cost based on the best way
of accessing S /* Using selections on S and indices on S */
else for each non-empty subset S1 of S such that S1  S
P1= findbestplan(S1)
P2= findbestplan(S - S1)
A = best algorithm for joining results of P1 and P2
cost = P1.cost + P2.cost + cost of A
if cost < bestplan[S].cost
bestplan[S].cost = cost
bestplan[S].plan = “execute P1.plan; execute P2.plan;
join results of P1 and P2 using A”
return bestplan[S]
* Some modifications to allow indexed nested loops joins on relations that have
selections (see book)
©Silberschatz, Korth and Sudarshan1.27Database System Concepts - 6th Edition
Left Deep Join Trees
 In left-deep join trees, the right-hand-side input for each join is
a relation, not the result of an intermediate join.
©Silberschatz, Korth and Sudarshan1.28Database System Concepts - 6th Edition
Cost of Optimization
 With dynamic programming time complexity of optimization with bushy
trees is O(3n).
 With n = 10, this number is 59000 instead of 176 billion!
 Space complexity is O(2n)
 To find best left-deep join tree for a set of n relations:
 Consider n alternatives with one relation as right-hand side input
and the other relations as left-hand side input.
 Modify optimization algorithm:
 Replace “for each non-empty subset S1 of S such that S1  S”
 By: for each relation r in S
let S1 = S – r .
 If only left-deep trees are considered, time complexity of finding best join
order is O(n 2n)
 Space complexity remains at O(2n)
 Cost-based optimization is expensive, but worthwhile for queries on
large datasets (typical queries have small n, generally < 10)
©Silberschatz, Korth and Sudarshan1.29Database System Concepts - 6th Edition
Cost Based Optimization with Equivalence Rules
 Physical equivalence rules allow logical query plan to be converted
to physical query plan specifying what algorithms are used for each
operation.
 Efficient optimizer based on equivalent rules depends on
 A space efficient representation of expressions which avoids
making multiple copies of subexpressions
 Efficient techniques for detecting duplicate derivations of
expressions
 A form of dynamic programming based on memorization, which
stores the best plan for a subexpression the first time it is
optimized, and reuses in on repeated optimization calls on same
subexpression
 Cost-based pruning techniques that avoid generating all plans
 Pioneered by the Volcano project and implemented in the SQL Server
optimizer
©Silberschatz, Korth and Sudarshan1.30Database System Concepts - 6th Edition
Heuristic Optimization
 Cost-based optimization is expensive, even with dynamic programming.
 Systems may use heuristics to reduce the number of choices that must
be made in a cost-based fashion.
 Heuristic optimization transforms the query-tree by using a set of rules
that typically (but not in all cases) improve execution performance:
 Perform selection early (reduces the number of tuples)
 Perform projection early (reduces the number of attributes)
 Perform most restrictive selection and join operations (i.e. with
smallest result size) before other similar operations.
 Some systems use only heuristics, others combine heuristics with
partial cost-based optimization.
©Silberschatz, Korth and Sudarshan1.31Database System Concepts - 6th Edition
Structure of Query Optimizers
 Many optimizers considers only left-deep join orders.
 Plus heuristics to push selections and projections down the query
tree
 Reduces optimization complexity and generates plans amenable to
pipelined evaluation.
 Heuristic optimization used in some versions of Oracle:
 Repeatedly pick “best” relation to join next
 Starting from each of n starting points. Pick best among these
 Intricacies of SQL complicate query optimization
 E.g. nested subqueries
©Silberschatz, Korth and Sudarshan1.32Database System Concepts - 6th Edition
Structure of Query Optimizers (Cont.)
 Some query optimizers integrate heuristic selection and the generation of
alternative access plans.
 Frequently used approach
 heuristic rewriting of nested block structure and aggregation
 followed by cost-based join-order optimization for each block
 Some optimizers (e.g. SQL Server) apply transformations to entire query
and do not depend on block structure
 Optimization cost budget to stop optimization early (if cost of plan is
less than cost of optimization)
 Plan caching to reuse previously computed plan if query is resubmitted
 Even with different constants in query
 Even with the use of heuristics, cost-based query optimization imposes a
substantial overhead.
 But is worth it for expensive queries
 Optimizers often use simple heuristics for very cheap queries, and
perform exhaustive enumeration for more expensive queries
Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Statistics for Cost Estimation
©Silberschatz, Korth and Sudarshan1.34Database System Concepts - 6th Edition
Statistical Information for Cost Estimation
 nr: number of tuples in a relation r.
 br: number of blocks containing tuples of r.
 lr: size of a tuple of r.
 fr: blocking factor of r – i.e., the number of tuples of r that fit into one block.
 V(A, r): number of distinct values that appear in r for attribute A; same as
the size of A(r).
 If tuples of r are stored together physically in a file, then:











rf
rn
rb
©Silberschatz, Korth and Sudarshan1.35Database System Concepts - 6th Edition
Selection Size Estimation
 A=v(r)
 nr / V(A,r) : number of records that will satisfy the selection
 Equality condition on a key attribute: size estimate = 1
 AV(r) (case of A  V(r) is symmetric)
 Let c denote the estimated number of tuples satisfying the condition.
 If min(A,r) and max(A,r) are available in catalog
 c = 0 if v < min(A,r)
 c =
 In absence of statistical information c is assumed to be nr / 2.
),min(),max(
),min(
.
rArA
rAv
nr


©Silberschatz, Korth and Sudarshan1.36Database System Concepts - 6th Edition
Size Estimation of Complex Selections
 The selectivity of a condition i is the probability that a tuple in the
relation r satisfies i .
 If si is the number of satisfying tuples in r, the selectivity of i is
given by si /nr.
 Conjunction: 1 2. . .  n (r). Assuming indepdence, estimate of
tuples in the result is:
 Disjunction:1 2 . . .  n (r). Estimated number of tuples:
 Negation: (r). Estimated number of tuples:
nr – size((r))
n
r
n
r
n
sss
n


...21






 )1(...)1()1(1 21
r
n
rr
r
n
s
n
s
n
s
n
©Silberschatz, Korth and Sudarshan1.37Database System Concepts - 6th Edition
Join Operation: Running Example
Running example:
student takes
Catalog information for join examples:
 nstudent = 5,000.
 fstudent = 50, which implies that
bstudent =5000/50 = 100.
 ntakes = 10000.
 ftakes = 25, which implies that
btakes = 10000/25 = 400.
 V(ID, takes) = 2500, which implies that on average, each student
who has taken a course has taken 4 courses.
 Attribute ID in takes is a foreign key referencing student.
 V(ID, student) = 5000 (primary key!)
©Silberschatz, Korth and Sudarshan1.38Database System Concepts - 6th Edition
Estimation of the Size of Joins
 The Cartesian product r  s contains nr .ns tuples; each tuple occupies
sr + ss bytes.
 If R  S = , then r s is the same as r  s.
 If R  S is a key for R, then a tuple of s will join with at most one tuple
from r
 therefore, the number of tuples in r s is no greater than the
number of tuples in s.
 If R  S in S is a foreign key in S referencing R, then the number of
tuples in r s is exactly the same as the number of tuples in s.
 The case for R  S being a foreign key referencing S is
symmetric.
 In the example query student takes, ID in takes is a foreign key
referencing student
 hence, the result has exactly ntakes tuples, which is 10000
©Silberschatz, Korth and Sudarshan1.39Database System Concepts - 6th Edition
Estimation of the Size of Joins (Cont.)
 If R  S = {A} is not a key for R or S.
If we assume that every tuple t in R produces tuples in R S, the
number of tuples in R S is estimated to be:
If the reverse is true, the estimate obtained will be:
The lower of these two estimates is probably the more accurate one.
),( sAV
nn sr 
),( rAV
nn sr 
©Silberschatz, Korth and Sudarshan1.40Database System Concepts - 6th Edition
Estimation of the Size of Joins (Cont.)
 Compute the size estimates for takes  student without using
information about foreign keys:
 V(ID, takes) = 2,500, and
V(ID, student) = 5,000
 The two estimates are
 5,000 * 10,000/2,500 = 20,000 and
 5,000 * 10,000/5,000 = 10,000
 We choose the lower estimate, which in this case, is the same as
our earlier computation using foreign keys.
©Silberschatz, Korth and Sudarshan1.41Database System Concepts - 6th Edition
Size Estimation for Other Operations
 Projection: estimated size of A(r) = V(A,r)
 Aggregation : estimated size of AGF(r) = V(A,r)
 Set operations
 For unions/intersections of selections on the same relation:
rewrite and use size estimate for selections
 E.g. 1 (r)  2 (r) can be rewritten as 1  2 (r)
 For operations on different relations:
 estimated size of r  s = size of r + size of s.
 estimated size of r  s = minimum size of r and size of s.
 estimated size of r – s = r.
 All the three estimates may be quite inaccurate, but provide
upper bounds on the sizes.
©Silberschatz, Korth and Sudarshan1.42Database System Concepts - 6th Edition
Size Estimation (Cont.)
 Outer join:
 Estimated size of r s = size of r s + size of r
 Case of right outer join is symmetric
 Estimated size of r s = size of r s + size of r + size of s
©Silberschatz, Korth and Sudarshan1.43Database System Concepts - 6th Edition
Estimation of Number of Distinct Values
Selections:  (r)
 If  forces A to take a specified value: V(A, (r)) = 1.
 e.g., A = 3
 If  forces A to take on one of a specified set of values:
V(A, (r)) = number of specified values.
 (e.g., (A = 1  A = 3  A = 4 )),
 If the selection condition  is of the form A op r
estimated V(A, (r)) = V(A.r) * s
 where s is the selectivity of the selection.
 In all the other cases: use approximate estimate of
min(V(A,r), n (r) )
 More accurate estimate can be got using probability theory, but
this one works fine generally
Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Additional Optimization Techniques
 Nested Subqueries
 Materialized Views
©Silberschatz, Korth and Sudarshan1.45Database System Concepts - 6th Edition
Optimizing Nested Subqueries**
 Nested query example:
select name
from instructor
where exists (select *
from teaches
where instructor.ID = teaches.ID and teaches.year = 2007)
 SQL conceptually treats nested subqueries in the where clause as
functions that take parameters and return a single value or set of values
 Parameters are variables from outer level query that are used in the
nested subquery; such variables are called correlation variables
 Conceptually, nested subquery is executed once for each tuple in the
cross-product generated by the outer level from clause
 Such evaluation is called correlated evaluation
 Note: other conditions in where clause may be used to compute a join
(instead of a cross-product) before executing the nested subquery
©Silberschatz, Korth and Sudarshan1.46Database System Concepts - 6th Edition
Optimizing Nested Subqueries (Cont.)
 Correlated evaluation may be quite inefficient since
 a large number of calls may be made to the nested query
 there may be unnecessary random I/O as a result
 SQL optimizers attempt to transform nested subqueries to joins where
possible, enabling use of efficient join techniques
 E.g.: earlier nested query can be rewritten as
select name
from instructor, teaches
where instructor.ID = teaches.ID and teaches.year = 2007
 Note: the two queries generate different numbers of duplicates (why?)
 teaches can have duplicate IDs
 Can be modified to handle duplicates correctly as we will see
 In general, it is not possible/straightforward to move the entire nested
subquery from clause into the outer level query from clause
 A temporary relation is created instead, and used in body of outer
level query
©Silberschatz, Korth and Sudarshan1.47Database System Concepts - 6th Edition
Optimizing Nested Subqueries (Cont.)
In general, SQL queries of the form below can be rewritten as shown
 Rewrite: select …
from L1
where P1 and exists (select *
from L2
where P2)
 To: create table t1 as
select distinct V
from L2
where P2
1
select …
from L1, t1
where P1 and P2
2
 P2
1 contains predicates in P2 that do not involve any correlation
variables
 P2
2 reintroduces predicates involving correlation variables, with
relations renamed appropriately
 V contains all attributes used in predicates with correlation
variables
©Silberschatz, Korth and Sudarshan1.48Database System Concepts - 6th Edition
Optimizing Nested Subqueries (Cont.)
 In our example, the original nested query would be transformed to
create table t1 as
select distinct ID
from teaches
where year = 2007
select name
from instructor, t1
where t1.ID = instructor.ID
 The process of replacing a nested query by a query with a join (possibly
with a temporary relation) is called decorrelation.
 Decorrelation is more complicated when
 the nested subquery uses aggregation, or
 when the result of the nested subquery is used to test for equality, or
 when the condition linking the nested subquery to the other
query is not exists,
 and so on.
©Silberschatz, Korth and Sudarshan1.49Database System Concepts - 6th Edition
Materialized Views**
 A materialized view is a view whose contents are computed and
stored.
 Consider the view
create view department_total_salary(dept_name, total_salary) as
select dept_name, sum(salary)
from instructor
group by dept_name
 Materializing the above view would be very useful if the total salary by
department is required frequently
 Saves the effort of finding multiple tuples and adding up their
amounts
Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Additional Optimization Techniques
©Silberschatz, Korth and Sudarshan1.51Database System Concepts - 6th Edition
Top-K Queries
 Top-K queries
select *
from r, s
where r.B = s.B
order by r.A ascending
limit 10
 Alternative 1: Indexed nested loops join with r as outer
 Alternative 2: estimate highest r.A value in result and add
selection (and r.A <= H) to where clause
 If < 10 results, retry with larger H
©Silberschatz, Korth and Sudarshan1.52Database System Concepts - 6th Edition
Optimization of Updates
 Halloween problem
update R set A = 5 * A
where A > 10
 If index on A is used to find tuples satisfying A > 10, and tuples
updated immediately, same tuple may be found (and updated)
multiple times
 Solution 1: Always defer updates
 collect the updates (old and new values of tuples) and update
relation and indices in second pass
 Drawback: extra overhead even if e.g. update is only on R.B,
not on attributes in selection condition
 Solution 2: Defer only if required
 Perform immediate update if update does not affect attributes
in where clause, and deferred updates otherwise.
©Silberschatz, Korth and Sudarshan1.53Database System Concepts - 6th Edition
Join Minimization
 Join minimization
select r.A, r.B
from r, s
where r.B = s.B
 Check if join with s is redundant, drop it
 E.g. join condition is on foreign key from r to s, no selection on s
 Other sufficient conditions possible
select r.A, s1.B
from r, s as s1, s as s2
where r.B=s1.B and r.B = s2.B and s1.A < 20 and s2.A < 10
 join with s2 is redundant and can be dropped (along with
selection on s2)
 Lots of research in this area since 70s/80s!
Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
End of Chapter