CSCE 310 / 608 Database Systems

Chapter 15: Query Execution 1

Index-Based Algorithms

 The existence of an index is especially helpful for selection, and helps others  Clustered relation : tuples are packed into the minimum number of blocks  Clustering index : all tuples with the same value for the index's search key are packed into the minimum number of blocks

2

Index-Based Selection

 Without an index, selection takes

B(R),

or even

T(R),

disk I/O's.

 To select all tuples with attribute

a

equal to value

v

, when there is an index on

a

:  search the index for value

v

and get pointers to exactly the blocks containing the desired tuples  If index is clustering, then number of disk I/O's is about

B(R)/V(R,a)

3

Examples

 Suppose

B(R)

index on

a

= 1000,

T(R)

= 20,000, there is an and we want to select all tuples with

a

= 0.

     If

R

is clustered and don't use index: 1000 disk I/O's If

R

I/O's is not clustered and don't use index: 20,000 disk If

V(R,a)

= 100, index is clustering, and use index: 1000/100 = 10 disk I/O's (on average) If

V(R,a)

= 10, R is not clustered, index is non clustering, and use index: 20,000/10 = 2000 disk I/O's (on average) If

V(R,a)

I/O = 20,000 (

a

is a key) and use index: 1 disk

4

Using Indexes in Other Operations

1.

2.

If the index is a B-tree, can efficiently select tuples with indexed attribute in a range  If selection is on a complex condition such as

"a = v

AND …", first do the index-based algorithm to get tuples satisfying

"a = v"

.

Such splitting is part of the job of the query optimizer

5

Index-Based Join Algorithm

 Consider natural join of

R(X,Y)

and

S(Y,Z).

 Suppose

S

has an index on

Y

.

for each block of

R

for each tuple

t

in the current block use index on

S

to find tuples of

S

that match

t

the attribute(s)

Y

in output the join of these tuples

6

Analysis of Index-Based Join

 To get all the blocks of

R

, either

B(R)

disk I/O's are needed or

T(R)

 For each tuple of

R

, there are on average

T(S)/V(S,Y)

matching tuples of

S



T(R)*T(S)/V(S,Y)

disk I/O's if index is not clustering 

T(R)*B(S)/V(S,Y)

disk I/O's if index is clustering  This method is efficient if

R S

and

V(S,Y)

is much smaller than is large (i.e., not many tuples of

S

match)

7

Join Using a Sorted Index

 Suppose we want to join

R(X,Y)

S(Y,Z).

and  Suppose we have a sorted index (e.g., B-tree) on Y for R and S:  do sort-join but  no need to sort the indexed relations first

8

Buffer Management

    The availability of blocks (buffers) of main memory is controlled by buffer manager .

When a new buffer is needed, a replacement policy used to decide which existing buffer should be returned to disk.

is If the number of buffers available for an operation cannot be predicted in advance, then the algorithm chosen must degrade gracefully buffers shrinks.

as the number of If the number of buffers available is not large enough for a two-pass algorithm, then there are generalizations to algorithms that use three or more passes.

9

CSCE 608 - 600 Database Systems

Chapter 16: Query Compiler 10

Query Compiler

Parsing Logical Query Plan

11

SQL query parse parse tree convert logical query plan apply laws “ improved ” l.q.p

estimate result sizes l.q.p. +sizes consider physical plans statistics answer execute Pi pick best {(P1,C1),(P2,C2)...} estimate costs {P1,P2,…..}

Outline

    Convert SQL query to a parse tree  Semantic checking: attributes, relation names, types Convert to a logical query plan (relational algebra expression)  deal with subqueries Improve the logical query plan  use algebraic transformations  group together certain operators  evaluate logical plan based on estimated size of relations Convert to a physical query plan  search the space of physical plans   choose order of operations complete the physical query plan

13

Parsing

 Goal is to convert a text string containing a query into a parse tree data structure:  leaves form the text string (broken into lexical elements)  internal nodes are syntactic categories  Uses standard algorithmic techniques from compilers  given a grammar for the language (e.g., SQL), process the string and build the tree

14

Example: SQL query

SELECT title FROM StarsIn WHERE starName IN ( SELECT name FROM MovieStar WHERE birthdate LIKE ‘ %1960 ’ ); (Find the movies with stars born in 1960) Assume we have a simplified grammar for SQL.

15

Example: Parse Tree

SELECT FROM WHERE IN title StarsIn starName ( ) SELECT FROM WHERE LIKE name MovieSta r birthDate ‘ %1960

16

’

The Preprocessor

 replaces each reference to a view with a parse (sub)-tree that describes the view (i.e., a query)  does semantic checking:  are relations and views mentioned in the schema?

 are attributes mentioned in the current scope?

 are attribute types correct?

17

Outline

    Convert SQL query to a parse tree  Semantic checking: attributes, relation names, types Convert to a logical query plan (relational algebra expression)  deal with subqueries Improve the logical query plan    use algebraic transformations group together certain operators evaluate logical plan based on estimated size of relations Convert to a physical query plan    search the space of physical plans choose order of operations complete the physical query plan

18

Convert Parse Tree to Relational Algebra  Complete algorithm depends on specific grammar, which determines forms of the parse trees  Here give a flavor of the approach

19

Conversion

 Suppose there are no subqueries.

 SELECT

att-list

FROM

rel-list

WHERE

cond

is converted into PROJ

att-list

(SELECT

cond

(PRODUCT(

rel-list

))), or 

att-list

( 

cond

( X (

rel-list

)))

20

SELECT movieTitle FROM StarsIn, MovieStar WHERE starName = name AND birthdate LIKE '%1960'; SELECT FROM WHERE , AND movieTitle StarsIn LIKE MovieStar starName name birthdate = '%1960'

Equivalent Algebraic Expression Tree

 movieTitle  starname = name AND birthdate LIKE '%1960' X StarsIn MovieStar

22

Handling Subqueries

 Recall the (equivalent) query: SELECT title FROM StarsIn WHERE starName IN ( SELECT name FROM MovieStar WHERE birthdate LIKE ‘ %1960 ’ );  Use an intermediate format called two argument selection

23

Example: Two-Argument Selection  title  StarsIn IN  name  birthdate LIKE ‘ %1960 ’ starName MovieStar

24

Converting Two-Argument Selection  To continue the conversion, we need rules for replacing two-argument selection with a relational algebra expression  Different rules depending on the nature of the subquery  Here show example for IN operator and uncorrelated query (subquery computes a relation independent of the tuple being tested)

25

Rules for IN

 R t IN S  C X R  S C is the condition that equates attributes in t with corresponding attributes in S

26

Example: Logical Query Plan  title  starName=name  StarsIn   name  birthdate LIKE ‘ %1960 ’ MovieStar

27

What if Subquery is Correlated?

 Example is when subquery refers to the current tuple of the outer scope that is being tested  More complicated to deal with, since subquery cannot be translated in isolation  Need to incorporate external attributes in the translation  Some details are in textbook

28

Outline

    Convert SQL query to a parse tree  Semantic checking: attributes, relation names, types Convert to a logical query plan (relational algebra expression)  deal with subqueries Improve the logical query plan  use algebraic transformations   group together certain operators evaluate logical plan based on estimated size of relations Convert to a physical query plan  search the space of physical plans   choose order of operations complete the physical query plan

29

Improving the Logical Query Plan

 There are numerous algebraic laws concerning relational algebra operations  By applying them to a logical query plan judiciously, we can get an equivalent query plan that can be executed more efficiently  Next we'll survey some of these laws

30

Associative and Commutative Operations  product  natural join  set and bag union  set and bag intersection 

associative:

(A op B) op C = A op (B op C) 

commutative:

A op B = B op A

31

Laws Involving Selection

 Selections usually reduce the size of the relation  Usually good to do selections early, i.e., "push them down the tree"  Also can be helpful to break up a complex selection into parts

32

Selection Splitting

  C1 AND C2 (R) =  C1 (  C2 (R))   C1 OR C2 (R) = ( if R is a set  C1 (R)) U set (  C2 (R))   C1 (  C2 (R)) =  C2 (  C1 (R))

33

Selection and Binary Operators

   Must push selection to both arguments:   C (R U S) =  C (R) U  C (S) Must push to first arg, optional for 2nd:     C C (R - S) = (R - S) =   C C (R) - S (R)  C (S) Push to at least one arg with all attributes mentioned in C:   product, natural join, theta join, intersection e.g.,  in C C (R X S) =  C (R) X S, if R has all the atts

34

Pushing Selection

Up

the Tree

 Suppose we have relations  StarsIn(title,year,starName)  Movie(title,year,len,inColor,studioName)  and a view  CREATE VIEW MoviesOf1996 AS SELECT * FROM Movie WHERE year = 1996;  and the query  SELECT starName, studioName FROM MoviesOf1996 NATURAL JOIN StarsIn;

35

The Straightforward Tree

 starName,studioName  year=1996 StarsIn Movie Remember the rule  C (R S) =  C (R) S ?

36

The Improved Logical Query Plan

 starName,studioName  starName,studioName  starName,studioName  year=1996 Movie StarsIn push selection up tree  year=1996 Movie  year=1996  year=1996 Movie StarsIn push selection down tree StarsIn

37

Laws Involving Projections

    Consider adding in additional projections Adding a projection lower in the tree can improve performance, since often tuple size is reduced  Usually not as helpful as pushing selections down If a projection is inserted in the tree, then none of the eliminated attributes can appear above this point in the tree  Ex:  L (R X S) =  L (  M (R) X  N (S)), where M (resp. N) is all attributes of R (resp. S) that are used in L Another example:   L (R U bag S) =  L (R) U bag  L (S) But watch out for set union!

38

Push Projection Below Selection?

 Rule:  L (  C (R)) =  L (  C (  M (R))) where M is all attributes used by L or C  But is it a good idea?

SELECT starName FROM StarsIn WHERE movieYear = 1996;  starName   starName movieYear=1996  movieYear=1996  starName,movieYear StarsIn StarsIn

39

Joins and Products

 Recall from the definitions of relational algebra:   R C S =  C (R X S) (theta join) R S =  L (  C (R X S)) (natural join) where C equates same-name attributes in R and S, and L includes all attributes of R and S dropping duplicates  To improve a logical query plan, replace a product followed by a selection with a join  Join algorithms are usually faster than doing product followed by selection

40

Duplicate Elimination

 Moving  down the tree is potentially beneficial as it can reduce the size of intermediate relations  Can be eliminated if argument has no duplicates    a relation with a primary key  a relation resulting from a grouping operator Legal to push  through product, join, selection, and bag intersection  Ex:  (R X S) =  (R) X  (S) Cannot push  or projection through bag union, bag difference

41