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CSCE 608 – 600 Database Systems Download Presentation ## CSCE 608 – 600 Database Systems

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

2. 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

3. 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)

4. 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

5. 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

6. 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

7. 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)

8. Join Using a Sorted Index • 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

9. 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.

10. CSCE 608 - 600 Database Systems Chapter 16: Query Compiler

11. Query Compiler Parsing Logical Query Plan

12. SQL query parse parse tree convert answer 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,…..}

13. 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

14. 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

15. 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.

16. 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>

17. 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?

18. 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

19. 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

20. 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)))

21. <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

22. Equivalent Algebraic Expression Tree movieTitle  starname = name AND birthdate LIKE '%1960' X StarsIn MovieStar

23. 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

24. Example: Two-Argument Selection title  StarsIn <condition> <tuple> IN name <attribute> birthdate LIKE ‘%1960’ starName MovieStar

25. 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)

26. 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

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

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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))

34. 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

35. 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;

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

37. 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

38. 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 Ubag S) = L(R) UbagL(S) But watch out for set union!

39. 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

40. 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

41. 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