EECE 433 Chapter 4 Part 1, Spring 2012

Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke Relational Algebra Chapter 4, Part A Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 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 intended to be used for complex calculations.  QLs support easy, efficient access to large data sets. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 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.) Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke Preliminaries  A relation is a set of tuples. Therefore we assume no repetitions in tuples.  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  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 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 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. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 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 or in reln. 2.  Additional operations:  Intersection, join, division, renaming.  Since each operation returns a relation, operations can be composed!     Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke Projection (1/2) age 35.0 55.5 age S ( ) 2  Selects desired attributes (Deletes attributes that are not) in projection list.  Example: sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0 What does it produce? Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke Projection (2/2) sname rating yuppy 9 lubber 8 guppy 5 rusty 10 sname rating S , ( ) 2  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??) sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0 What does it produce? Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke Selection (1/2)  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 sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0 What does it produce? What does it produce? Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 Selection (2/2)  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.) Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 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 S 1 2  S S 1 2  sid sname rating age 22 dustin 7 45.0 S 1 S 2  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 S1 S2 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 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 R 1 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: sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 sid bid day 22 101 10/10/96 58 103 11/12/96 S1 R1 S1 X R1 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 Joins  Condition Join:  Result schema same as that of cross-product.  Fewer tuples than cross-product, might be able to compute more efficiently 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  . .  sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 sid bid day 22 101 10/10/96 58 103 11/12/96 S1 R1 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 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 sid R 1 1  sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 sid bid day 22 101 10/10/96 58 103 11/12/96 S1 R1 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 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.   x x y A y B | ,      Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 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 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 EXAMPLES Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 Assume following relations/schemas sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 sid bid day 22 101 10/10/96 58 103 11/12/96 Sailors Boats bid bname bcolor 78 XYZ red 98 ABC green Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 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 Sailors 2 1   sname Temp ( ) 2  Solution 3:  sname  bid serves Sailors ( (Re )) 103  Which one is more efficient? Why? Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 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! Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 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?   Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 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   Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 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 ) ( Sailors sname Tempsids     To find sailors who’ve reserved all ‘Interlake’ boats: / ( ' ' )  bid  bname Interlake Boats ..... Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 Find the names of sailors who have reserved at least two boats. )) Re ( , , , (Re serves sid sname bid Sailors servations     ) Re 2),Re 6,2 5,2 4 ,1 3,1 2,1 1( (Re servations servationsX bid sname sid bid sname sid airs servationp        ) )2 (Re 1 ( )2 1 1 ( servationpairs bid bid sid sid sname       Hint: create res x res, then pick. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 Find the sids of sailors with age over 20 who have not reserved a red boat. ) Re ) ' ' (( ) 20 ( serves color red Boats sid Sailors age sid          Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 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.