Relational Algebra
- 1. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 1 Relational Algebra
- 2. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 2 Relational Query Languages Query languages: Allow manipulation and retrieval of data from a database. 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. Remark: There are new developments (e.g. SQL3) with the goal: SQL=PL
- 3. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 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, declarative.) Remark: Only relational algebra will be covered in COSC 3480
- 4. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 4 Why is Relational Algebra Important? As a theoretical foundation of the relational data model and query languages. It introduces a terminology that is important to talk about relational databases (e.g. join,…) As a language to specify plans that implement SQL queries (query optimization; implemetation of relational DBMS) Some people believe that knowing relational algebra makes it easy to write correct SQL queries.
- 5. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 5 Preliminaries A query is applied to relation instances, and the result of a query is also a relation instance. – Schemas of input relations for a query are fixed (but query will run regardless of instance!) – The schema for the result of a given query is also fixed! Determined by definition of query language constructs. Positional vs. named-field notation: – Positional notation easier for formal definitions, named-field notation more readable. – Both used in SQL
- 6. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 6 Example Instances sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0 sid bid day 22 101 10/10/96 58 103 11/12/96 R1 S1 S2 “Sailors” and “Reserves” relations for our examples. We’ll use positional or named field notation, assume that names of fields in query results are `inherited’ from names of fields in query input relations.
- 7. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 7 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, but (very!) useful. Since each operation returns a relation, operations can be composed! (Algebra is “closed”.)
- 8. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 8 Projection sname rating yuppy 9 lubber 8 guppy 5 rusty 10 sname rating S , ( )2 age 35.0 55.5 age S( )2 Deletes attributes that are not in projection list. Schema of result contains exactly the fields in the projection list, with the same names that they had in the (only) input relation. 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?)
- 9. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 9 Selection rating S 8 2( ) sid sname rating age 28 yuppy 9 35.0 58 rusty 10 35.0 sname rating yuppy 9 rusty 10 sname rating rating S , ( ( )) 8 2 Selects rows that satisfy selection condition. No duplicates in result! (Why?) Schema of result identical to schema of (only) input relation. Result relation can be the input for another relational algebra operation! (Operator composition.)
- 10. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 10 Union, Intersection, Set-Difference All of these operations take two input relations, which must be union-compatible: – Same number of fields. – `Corresponding’ fields have the same type. What is the schema of result? sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 44 guppy 5 35.0 28 yuppy 9 35.0 sid sname rating age 31 lubber 8 55.5 58 rusty 10 35.0 S S1 2 S S1 2 sid sname rating age 22 dustin 7 45.0 S S1 2
- 11. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 11 Cross-Product Each row of S1 is paired with each row of 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. ( ( , ), )C sid sid S R1 1 5 2 1 1 (sid) sname rating age (sid) bid day 22 dustin 7 45.0 22 101 10/10/96 22 dustin 7 45.0 58 103 11/12/96 31 lubber 8 55.5 22 101 10/10/96 31 lubber 8 55.5 58 103 11/12/96 58 rusty 10 35.0 22 101 10/10/96 58 rusty 10 35.0 58 103 11/12/96 Renaming operator:
- 12. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 12 Joins Condition Join: Result schema same as that of cross-product. Fewer tuples than cross-product, might be able to compute more efficiently Sometimes called a theta-join. R c S c R S ( ) (sid) sname rating age (sid) bid day 22 dustin 7 45.0 58 103 11/12/96 31 lubber 8 55.5 58 103 11/12/96 S R S sid R sid 1 1 1 1 . .
- 13. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 13 Joins 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. Natural Join: Equijoin on all common fields. sid sname rating age bid day 22 dustin 7 45.0 101 10/10/96 58 rusty 10 35.0 103 11/12/96 S R sid 1 1
- 14. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 14 Division Not supported as a primitive operator, but useful for expressing queries like: Find sailors who have reserved all boats. Let A have 2 fields, x and y; B have only field y: – A/B = – i.e., A/B contains all x tuples (sailors) such that for every y tuple (boat) in B, there is an xy tuple in A. – Or: If the set of y values (boats) associated with an x value (sailor) in A contains all y values in B, the x value is in A/B. In general, x and y can be any lists of fields; y is the list of fields in B, and x y is the list of fields of A. AyxxByyx ,|
- 15. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 15 Examples of Division A/B sno pno s1 p1 s1 p2 s1 p3 s1 p4 s2 p1 s2 p2 s3 p2 s4 p2 s4 p4 pno p2 pno p2 p4 pno p1 p2 p4 sno s1 s2 s3 s4 sno s1 s4 sno s1 A B1 B2 B3 A/B1 A/B2 A/B3
- 16. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 16 Expressing A/B Using Basic Operators Division is not essential op; just a useful shorthand. – (Also true of joins, but joins are so common that systems implement joins specially.) 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: x x A B A(( ( ) ) ) x A( ) all disqualified tuples
- 17. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 17 Find names of sailors who’ve reserved boat #103 Solution 1: sname bid serves Sailors(( Re ) ) 103 Solution 2: ( , Re )Temp serves bid 1 103 ( , )Temp Temp Sailors2 1 sname Temp( )2 Solution 3: sname bid serves Sailors( (Re )) 103
- 18. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 18 Find names of sailors who’ve reserved a red boat Information about boat color only available in Boats; so need an extra join: sname color red Boats serves Sailors(( ' ' ) Re ) A more efficient solution: sname sid bid color red Boats s Sailors( (( ' ' ) Re ) ) A query optimizer can find this given the first solution!
- 19. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 19 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: ( , ( ' ' ' ' ))Tempboats color red color green Boats sname Tempboats serves Sailors( Re ) Can also define Tempboats using union! (How?) What happens if is replaced by in this query?
- 20. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 20 Find sailors who’ve reserved a red and a green boat Previous approach won’t work! Must identify sailors who’ve reserved red boats, sailors who’ve reserved green boats, then find the intersection (note that sid is a key for Sailors): ( , (( ' ' ) Re ))Tempred sid color red Boats serves sname Tempred Tempgreen Sailors(( ) ) ( , (( ' ' ) Re ))Tempgreen sid color green Boats serves
- 21. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 21 Find the names of sailors who’ve reserved all boats Uses division; schemas of the input relations to / must be carefully chosen: ( , ( , Re ) / ( ))Tempsids sid bid serves bid Boats sname Tempsids Sailors( ) To find sailors who’ve reserved all ‘Interlake’ boats: / ( ' ' ) bid bname Interlake Boats .....
- 22. Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 22 Summary The relational model has rigorously defined query languages that are simple and powerful. Relational algebra is more operational; useful as internal representation for query evaluation plans. Several ways of expressing a given query; a query optimizer should choose the most efficient version.