CSCE 608 – 600 Database Systems

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CSCE 608 – 600 Database Systems. Chapter 15: Query Execution. 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

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### CSCE 608 – 600 Database Systems

Chapter 15: Query Execution

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
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)
Examples
• Suppose B(R) = 1000, T(R) = 20,000, there is an index on a 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 is not clustered and don\'t use index: 20,000 disk I/O\'s
• 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) = 20,000 (a is a key) and use index: 1 disk I/O
Using Indexes in Other Operations
• 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
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 in the attribute(s) Y

output the join of these tuples

Analysis of Index-Based Join
• To get all the blocks of R, either B(R) or T(R) disk I/O\'s are needed
• 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 is much smaller than S and V(S,Y) is large (i.e., not many tuples of S match)
• Suppose we want to join R(X,Y)and S(Y,Z).
• 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
Buffer Management
• The availability of blocks (buffers) of main memory is controlled by buffer manager.
• When a new buffer is needed, a replacement policy is used to decide which existing buffer should be returned to disk.
• If the number of buffers available for an operation cannot be predicted in advance, then the algorithm chosen must degrade gracefully as the number of buffers shrinks.
• 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.

### CSCE 608 - 600 Database Systems

Chapter 16: Query Compiler

### Query Compiler

Parsing

Logical Query Plan

SQL query

parse

parse tree

convert

logical query plan

execute

apply laws

statistics

Pi

“improved” l.q.p

pick best

estimate result sizes

{(P1,C1),(P2,C2)...}

l.q.p. +sizes

estimate costs

consider physical plans

{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
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
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.

SELECT <SelList> FROM <FromList> WHERE <Condition>

<Attribute> <RelName> <Attribute> LIKE <Pattern>

nameMovieStar birthDate‘%1960’

Example: Parse Tree

<Query>

<SFW>

SELECT <SelList> FROM <FromList> WHERE <Condition>

<Attribute> <RelName> <Tuple> IN <Query>

titleStarsIn <Attribute> ( <Query> )

starName <SFW>

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?
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
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
Conversion
• Suppose there are no subqueries.
• SELECT att-list FROM rel-list WHERE cond

is converted into

PROJatt-list(SELECTcond(PRODUCT(rel-list))), or

att-list(cond( X (rel-list)))

<Query>

SELECT movieTitle

FROM StarsIn, MovieStar

WHERE starName = name AND birthdate LIKE \'%1960\';

<SFW>

SELECT <SelList> FROM <FromList> WHERE <Condition>

<Attribute> <RelName> , <FromList> AND <Condition>

movieTitleStarsIn <RelName> <Attribute> LIKE <Pattern>

MovieStarbirthdate\'%1960\'

<Condition>

<Attribute> = <Attribute>

starName name

Equivalent Algebraic Expression Tree

movieTitle

starname = name AND birthdate LIKE \'%1960\'

X

StarsIn MovieStar

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

Example: Two-Argument Selection

title

StarsIn <condition>

<tuple> IN name

<attribute> birthdate LIKE ‘%1960’

starName MovieStar

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)
Rules for IN

C

R <Condition>

X

R 

t IN S

S

C is the condition that equates

attributes in t with corresponding

attributes in S

Example: Logical Query Plan

title

starName=name

StarsIn

name

birthdate LIKE ‘%1960’

MovieStar

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
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
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
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
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
Selection Splitting
• C1 AND C2 (R) = C1 ( C2 (R))
• C1 OR C2 (R) = (C1 (R)) Uset (C2 (R))

if R is a set

• C1 ( C2 (R)) = C2 ( C1 (R))
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 (R - S) = C (R) - S
• C (R - S) = C (R) - C (S)
• Push to at least one arg with all attributes mentioned in C:
• product, natural join, theta join, intersection
• e.g., C (R X S) = C (R) X S, if R has all the atts in C
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;

Remember the rule

C(R S) = C(R) S ?

The Straightforward Tree

starName,studioName

year=1996 StarsIn

Movie

starName,studioName

starName,studioName

starName,studioName

year=1996

year=1996 year=1996

year=1996 StarsIn

StarsIn

Movie

StarsIn

Movie

push selection

up tree

push selection

down tree

Movie

The Improved Logical Query Plan
Laws Involving 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 Ubag S) = L(R) UbagL(S)

But watch out for set union!

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

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
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  through bag union, bag difference or projection