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CS4432: Database Systems II

CS4432: Database Systems II. Lecture #13. Professor Elke A. Rundensteiner. Query Processing. Query in SQL  Query Plan in Algebra. Example. Data: relation R (A, B, C) relation S (C, D, E) Query: SELECT B, D FROM R, S

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CS4432: Database Systems II

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  1. CS4432: Database Systems II Lecture #13 Professor Elke A. Rundensteiner query processing

  2. Query Processing Query in SQL  Query Plan in Algebra query processing

  3. Example Data: relation R (A, B, C) relation S (C, D, E) Query: SELECT B, D FROM R, S WHERE R.A = “c” and S.E = 2 and R.C=S.C query processing

  4. SELECT B, D FROM R, S WHERE R.A = “c” and S.E = 2 and R.C=S.C 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 query processing

  5. How do we execute query? - Form Cartesian product of all tables in FROM-clause - Select tuples that match WHERE-clause - Project columns that occur in SELECT-clause One idea query processing

  6. 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 . . query processing

  7. But ? • Performance would be unacceptable! • We need a better approach for: • reasoning about queries, • their execution orders and • their respective costs query processing

  8. Relational Algebra ? Query: SELECT B, D FROM R, S WHERE R.A = “c” and S.E = 2 and R.C=S.C query processing

  9. 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)] query processing

  10. R1 Example Instances “Sailors” and “Reserves” relations S1 S2 query processing

  11. Relational Algebra • Basic operations: • Selection ( ) Selects a subset of rows from relation. • Projection ( ) Deletes unwanted columns from relation. • Cross-product( ) Allows us to combine two relations. • Set-difference ( ) Tuples in reln. 1, but not in reln. 2. • Union( ) Tuples in reln. 1 and in reln. 2. • Additional operations: • Intersection, join, division, renaming: Not essential ! • Algebra is “closed”: Since each operation returns a relation, operationscan be composed ! query processing

  12. Projection query processing

  13. Selection query processing

  14. Union, Intersection, Set Difference • Operate on two union-compatible relations: • Same number of fields. • `Corresponding’ fields have same type. query processing

  15. Cross-Product • Each row of S1 is paired with each row of R1. • Conflict: Both S1 and R1 have a field called sid. • Renaming operator: query processing

  16. Joins • Condition Join : • Result schema same as that of cross-product. query processing

  17. Joins • Equi-Join: condition contains only equalities. • Result schema only one copy of fields for which equality is specified. • Natural Join: Equijoin on all common fields. query processing

  18. Division • Not primitive operator, but useful: Find sailors who have reserved allboats. • A has 2 fields x and y; B has only field y: • A/B = • i.e., A/B contains all x tuples (sailors) such that for everyy tuple (boat) in B, there is an xy tuple in A. query processing

  19. B1 B2 B3 A Examples of Division A/B A/B1 A/B2 A/B3 query processing

  20. Disqualified x values: A/B: all disqualified tuples Expressing A/B Using Basic Operators • Division is useful shorthand. • Idea: For A/B, compute all x values that are not `disqualified’ by some y value in B. query processing

  21. Solution 1: • Solution 2: • Solution 3: Find names of sailors who’ve reserved boat #103 query processing

  22. A more efficient solution: Find names of sailors who’ve reserved a red boat • Information about boat color only available in Boats; so need an extra join: A query optimizer can find this, given the first solution! query processing

  23. What happens if is replaced by in this query? Find sailors who’ve reserved a red or a green boat • Can identify all red or green boats, then find sailors who’ve reserved one of these boats: • Can also define Tempboats using union! (How?) query processing

  24. Find sailors who’ve reserved a red and a green boat • Must identify sailors who’ve reserved red boats, sailors who’ve reserved green boats, then find the intersection (sid is a key for Sailors): query processing

  25. Find names of sailors who’ve reserved all boats • Uses division; schemas of input relations to / must be carefully chosen: • To find sailors who’ve reserved all ‘Interlake’ boats: ..... query processing

  26. Relational Algebra representation used to describe plans... query processing

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