CPS216: Advanced Database Systems Notes 02:Query Processing (Overview)

1. CPS216: Advanced Database SystemsNotes 02:Query Processing (Overview) Shivnath Babu

2. Query Processing Declarative SQL Query  Query Plan NOTE: You will not be tested on how well you know SQL. Understanding the SQL introduced in class will be sufficient (a primer follows). SQL is described in Chapter 6, GMUW. Focus: Relational System (i.e., data is organized as tables, or relations)

3. SQL Primer We will focus on SPJ, or Select-Project-Join Queries Select <attribute list> From <relation list> Where <condition list> Example Filter Query over R(A,B,C): Select B From R Where R.A = “c”  R.C > 10

4. SQL Primer (contd.) We will focus on SPJ, or Select-Project-Join-Queries Select <attribute list> From <relation list> Where <condition list> Example Join Query over R(A,B,C) and S(C,D,E): Select B, D From R, S Where R.A = “c”  S.E = 2  R.C = S.C

5. Answer B D 2 x R A B C S C D E a 1 10 10 x 2 b 1 20 20 y 2 c 2 10 30 z 2 d 2 35 40 x 1 e 3 45 50 y 3 Select B,D From R,S Where R.A = “c”  S.E = 2  R.C=S.C

6. How do we execute this query? Select B,D From R,S Where R.A = “c”  S.E = 2  R.C=S.C - Do Cartesian product - Select tuples - Do projection One idea

7. Bingo! Got one... R X S R.A R.B R.C S.C S.D S.E a 1 10 10 x 2 a 1 10 20 y 2 . . c 2 10 10 x 2 . . Select B,D From R,S Where R.A = “c” S.E = 2  R.C=S.C

8. Relational Algebra - can be used to describe plans Ex: Plan I B,D sR.A=“c” S.E=2  R.C=S.C X R S

9. Relational Algebra Primer (Chapter 5, GMUW) Select:sR.A=“c” R.C=10 Project: B,D Cartesian Product: R XS Natural Join: R S

10. Relational Algebra - can be used to describe plans Ex: Plan I B,D sR.A=“c” S.E=2  R.C=S.C X R S OR: B,D [sR.A=“c” S.E=2  R.C = S.C (RXS)]

11. Another idea: B,D sR.A = “c”sS.E = 2 R(A,B,C) S(C,D,E) Plan II natural join Select B,D From R,S Where R.A = “c”  S.E = 2  R.C=S.C

12. R S A B C s (R) s(S) C D E a 1 10 A B C C D E 10 x 2 b 1 20 c 2 10 10 x 2 20 y 2 c 2 10 20 y 2 30 z 2 d 2 35 30 z 2 40 x 1 e 3 45 50 y 3 Select B,D From R,S Where R.A = “c”  S.E = 2  R.C=S.C

13. Plan III Use R.A and S.C Indexes (1) Use R.A index to select R tuples with R.A = “c” (2) For each R.C value found, use S.C index to find matching tuples (3) Eliminate S tuples S.E  2 (4) Join matching R,S tuples, project B,D attributes, and place in result

14. =“c” <c,2,10> <10,x,2> check=2? output: <2,x> next tuple: <c,7,15> R S A B C C D E a 1 10 10 x 2 b 1 20 20 y 2 c 2 10 30 z 2 d 2 35 40 x 1 e 3 45 50 y 3 c 7 15 A C I1 I2

15. Overview of Query Processing SQL query parse parse tree Query rewriting Query Optimization statistics logical query plan Physical plan generation physical query plan execute Query Execution result

16. Example Query Select B,D From R,S Where R.A = “c”  R.C=S.C

17. Example: Parse Tree <Query> <SFW> SELECT <SelList> FROM <FromList> WHERE <Cond> <Attribute> <SelList> <RelName> <FromList> <Cond> AND <Cond> B <Attribute> R <RelName> <Attr> <Op> <Const> D S R.A = “c” Select B,D From R,S Where R.A = “c”  R.C=S.C <Attr> <Op> <Attr> R.C = S.C

18. Along with Parsing … • Semantic checks • Do the projected attributes exist in the relations in the From clause? • Ambiguous attributes? • Type checking, ex: R.A > 17.5 • Expand views

19. SQL query Initial logical plan parse Rewrite rules parse tree Query rewriting Logical plan statistics logical query plan Physical plan generation “Best” logical plan physical query plan execute result

20. Initial Logical Plan B,D Select B,D From R,S Where R.A = “c”  R.C=S.C R.A = “c” Λ R.C = S.C X R S Relational Algebra: B,D [sR.A=“c” R.C = S.C (RXS)]

21. Apply Rewrite Rule (1) B,D B,D R.C = S.C R.A = “c” Λ R.C = S.C R.A = “c” X X R S R S B,D [sR.C=S.C [R.A=“c”(R X S)]]

22. Apply Rewrite Rule (2) B,D B,D R.C = S.C R.C = S.C R.A = “c” X R.A = “c” S X R S R B,D [sR.C=S.C [R.A=“c”(R)] X S]

23. Apply Rewrite Rule (3) B,D B,D R.C = S.C Natural join R.A = “c” X S R.A = “c” S R R B,D [[R.A=“c”(R)] S]

24. SQL query Initial logical plan parse Rewrite rules parse tree Query rewriting Logical plan statistics logical query plan Physical plan generation “Best” logical plan physical query plan execute result

25. SQL query parse parse tree Query rewriting statistics Best logical query plan Physical plan generation Best physical query plan execute result

26. Physical Plan Generation B,D Project Natural join Hash join R.A = “c” S Index scan Table scan R R S Best logical plan

27. SQL query parse parse tree Enumerate possible physical plans Query rewriting statistics Best logical query plan Find the cost of each plan Physical plan generation Best physical query plan Pick plan with minimum cost execute result

28. Physical Plan Generation Logical Query Plan P1 P2 …. Pn C1 C2 …. Cn Pick minimum cost one Physical plans Costs

29. Textbook outline Chapter 15 15.1 Physical operators - Scan, Sort (Ch. 11.4), Indexes (Ch. 13) 15.2-15.6 Implementing operators + estimating their cost 15.8 Buffer Management 15.9 Parallel Processing

30. Textbook outline (contd.) Chapter 16 16.1 Parsing 16.2 Algebraic laws 16.3 Parse tree  logical query plan 16.4 Estimating result sizes 16.5-16.7 Cost based optimization

31. Background Material Chapter 5 Relational Algebra Chapter 6 SQL

32. SQL query parse parse tree Query rewriting statistics logical query plan Physical plan generation physical query plan execute result Query Processing - In class order 2; 16.1 3; 16.2,16.3 4; 16.4—16.7 1; 13, 15