CS 255: Database System Principles slides: From Parse Trees to Logical Query Plans

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CS 255: Database System Principles slides: From Parse Trees to Logical Query Plans. By:- Arunesh Joshi Id:-006538558. Agenda. Conversion to Relational Algebra. Removing Sub queries From Conditions. Improving the Logical Query Plan. Grouping Associative/Commutative Operators. Parsing.

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CS 255: Database System Principlesslides: From Parse Trees to Logical Query Plans

By:- Arunesh Joshi

Id:-006538558

Agenda
• Conversion to Relational Algebra.
• Removing Sub queries From Conditions.
• Improving the Logical Query Plan.
• Grouping Associative/Commutative Operators.
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

• It replaces each reference to a view with a parse (sub)-tree that describes the view (i.e., a query)
• It does semantic checking:
• are relations and views mentioned in the schema?
• are attributes mentioned in the current scope?
• are attribute types correct?

Convert Parse Tree to Relational Algebra

• The complete algorithm depends on specific grammar, which determines forms of the parse trees
• Here is 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)))

SELECT movieTitle

FROM StarsIn, MovieStar

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

<Query>

<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 sub query
• Here is shown an example for IN operator and uncorrelated query (sub query 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

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

Example: Improved Logical Query Plan

title

starName=name

StarsIn name

birthdate LIKE ‘%1960’

MovieStar

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

Grouping Assoc/Comm Operators

• Sets up the logical QP for future optimization when physical QP is constructed: determine best order for doing a sequence of joins (or unions or intersections)

U D E F

U

D

E

F

U

A B C

A

B

C