1 / 40

Relational Algebra

Relational Algebra. Chapter 4 - part I. Relational Query Languages. Query languages : Allow manipulation and retrieval of data from a database. Relational model supports simple powerful QLs: Strong formal foundation based on logic. Allows for much optimization.

marged
Download Presentation

Relational Algebra

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Relational Algebra Chapter 4 - part I

  2. Relational Query Languages • Query languages: Allow manipulation and retrieval of data from a database. • Relational model supports simple powerful QLs: • Strong formal foundation based on logic. • Allows for much optimization. • Query Languages != Programming languages! • QLs not expected to be “Turing complete”. • QLs not intended to be used for complex calculations. • QLs support easy, efficient access to large data sets.

  3. Formal Relational Query Languages • Two mathematical Query Languages form the basis for “real” languages (e.g. SQL) and for implementation: • Relational Algebra: • More operational, very useful for representing execution plans. • Relational Calculus: • Lets users describe what they want, rather than how to compute it. (Non-operational, rather declarative.)

  4. Preliminaries • A query is applied to relation instances. • The result of a query is also a relation instance. • Schemasof input relations for a query are fixed ; (but query will run regardless of instance!) • The schema for result of given query is also fixed! Determined by definition of query language .

  5. Preliminaries • Positional vs. named-field notation: • Positional field notation e.g., S.1 • Named field notation e.g., S.sid • Pros/Cons: • Positional notation easier for formal definitions, named-field notation more readable. • Both used in SQL • Assume that names of fields in query results are `inherited’ from names of fields in query input relations.

  6. Example Instances Sailors Reserves R1 S1 Sailors S2

  7. Relational Algebra • Basic operations: • Selection ( ) Selects a subset of rows from relation. • Projection ( ) Deletes unwanted columns from relation. • Cartesian-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, but (very!) useful. • Since each operation returns a relation, operationscan be composed! (Algebra is “closed”.)

  8. Selection Sailors S2

  9. Selection • condition (R) • Selects rows that satisfy selection condition. attribute op constant attribute op attribute Op is {<,>,<=,>=, =, =} • No duplicates in result! • Schema of result identical to schema of (only) input relation.

  10. Selection • Result relation can be input for another relational algebra operation! (Operatorcomposition.)

  11. Projection Sailors S2

  12. Projection •  projectlist (R) • Deletes attributes that are not in projection list. • Schema of result = contains fields in projection list • Projection operator has to eliminate duplicates! (Why??) • Note: real systems typically don’t do duplicate elimination unless the user explicitly asks for it. (Why not?)

  13. Union, Intersection, Set-Difference • All of these operations take two input relations, which must be union-compatible: • Same number of fields. • `Corresponding’ fields have same type. • What is the schema of result?

  14. Example Instances : Union S1 S2

  15. Difference Operation S1 S2

  16. Intersection Operation S1 S2

  17. Cross-Product (Cartesian Product) • S1 R1 : Each row of S1 is paired with each row of R1. Reserves R1 Sailors S1

  18. Cross-Product (Cartesian Product) • S1 R1 : Result schema has one field per field of S1 and R1, with field names `inherited’ if possible. • Conflict: Both S1 and R1 have a field called sid. • Renaming operator:

  19. Why we need a Join Operator ? In many cases, Join = Cross-Product + Select + Project However : Cross-product is too large to materialize Apply Select and Project "On-the-fly"

  20. Condition Join / Theta Join • Condition Join: • Result schema same as that of cross-product. • Fewer tuples than cross-product, more efficient.

  21. EquiJoin • Equi-Join: A special case of condition join where the condition c contains only equalities. • Result schema similar to cross-product, but only one copy of fields for which equality is specified. • An extra project: PROJECT ( THETA-JOIN)

  22. Natural Join • Natural Join: Equijoin on all common fields.

  23. Division • Not supported as a primitive operator, but useful for expressing queries like: Find sailors who have reserved allboats.

  24. Division • Let A have 2 fields x and y; B have only field y: • A/B = • A/B contains all x tuples (sailors) such that for everyy tuple (boat) in B, there is an xy tuple in A. • If set of y values (boats) associated with an x value (sailor) in A contains all y values in B, then x value is in A/B. • A/B is the largest relation instance Q such that QB A.

  25. Division Example e.g., A = all parts supplied by suppliers, B= relation parts A/B = suppliers who supply all parts listed in B

  26. Examples of Division A/B B1 B2 B3 A/B1 A/B2 A/B3 A

  27. Expressing A/B Using Basic Operators • Idea: For A/B, compute all x values that are not `disqualified’ by some y value in B. • x value is disqualified if by attaching y value from B, • we obtain an xy tuple that is not in A. Disqualified x values: A/B:

  28. A few example queries

  29. Solution 1 Solution 2 Solution 3 Find names of sailors who’ve reserved boat #103

  30. Find names of sailors who’ve reserved a red boat Reserves R1 Sailors S1 Boats B1

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

  32. 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?) What happens if is replaced by in this query?

  33. Find sailors who’ve reserved a red and a green boat • Previous approach won’t work! • Must identify sailors who reserved red boats, sailors who’ve reserved green boats, then find their intersection

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

  35. Relational Algebra : Some More Operators Beyond Chapter 4

  36. Generalized Projection •  F1, F2, … (R)  sname, (rating*2 as myrating) (Sailors)

  37. Aggregation operators • MIN, MAX, COUNT, SUM, AVG • AGGB(R) considers only non-null values of R. SUMB (R) COUNTB (R) MINB (R) R AVGB (R) COUNT* (R) MAXB (R)

  38. Aggregation Operators • MIN, MAX, SUM, AVG must be on any 1 attribute. COUNT can be on any 1 attribute or COUNT* (R) • An aggregation operator returns a bag, not a single value ! But SQL allows treatment as a single value. σB=MAXB (R) (R)

  39. Grouping Operator: GL, AL (R) • GL, AL (R) groups all attributes in GL, and performs the aggregation specified in AL. starName, MIN (year)→year, COUNT(title) →num (StarsIn) StarsIn

  40. Summary • The relational model has rigorously defined query languages that are simple and powerful. • Relational algebra is operational; useful as internal representation for query evaluation plans. • Several ways of expressing a given query; a query optimizer should choose most efficient version.

More Related