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

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Presentation on theme: "CPS216: Advanced Database Systems Notes 02:Query Processing (Overview) Shivnath Babu."— Presentation transcript:

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

2 Query Processing Declarative SQL Query  Query Plan Focus: Relational System (i.e., data is organized as tables, or relations) 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.

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

4 SQL Primer (contd.) Select From Where 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 We will focus on SPJ, or Select-Project-Join-Queries

5 RABC S CDE a11010x2 b12020y2 c21030z2 d23540x1 e34550y3 AnswerB D 2 x Select B,D From R,S Where R.A = “c”  S.E = 2  R.C=S.C

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

7 R X SR.AR.BR.CS.CS.DS.E a 1 10 10 x 2 a 1 10 20 y 2. c 2 10 10 x 2. Bingo! Got one... 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  R.A =“c”  S.E=2  R.C=S.C  X RS

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

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

11 Another idea:  B,D  R.A = “c”  S.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  ( R )  ( S ) C D E a 1 10 A B C C D E 10 x 2 b 1 20c 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 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 AC I1I1 I2I2 =“c” check=2? output: next tuple:

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

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

17 Example: Parse Tree SELECT FROM WHERE AND B R S R.A=“c” R.CS.C= D Select B,D From R,S Where R.A = “c”  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 parse Query rewriting Physical plan generation execute result SQL query parse tree logical query plan statistics physical query plan Initial logical plan “Best” logical plan Logical plan Rewrite rules

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

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

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

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

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

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

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

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

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 15.1Physical 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 Chapter 16 16.1Parsing 16.2Algebraic laws 16.3Parse tree  logical query plan 16.4Estimating result sizes 16.5-16.7 Cost based optimization Textbook outline (contd.)

31 Chapter 5 Relational Algebra Chapter 6 SQL Background Material

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


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