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Relational Algebra Chapter 4 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 1 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 algebra/logic. Allows for much optimization. Query Languages != programming languages! QLs not expected to be “Turing complete”. QLs not intended to be used for calculations. QLs support easy, efficient access to large data sets. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 2 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. (Nonoperational, declarative.) We’ll skip this for now. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 3 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 The schema for the result of a given query is also fixed! Determined by definition of query language constructs. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 4 Preliminaries Positional vs. named-attribute notation: Positional notation • • Named-attribute notation • • Ex: Sailor(1,2,3,4) easier for formal definitions Ex: Sailor(sid, sname, rating,age) more readable Advantages/disadvantages of one over the other? Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 5 Example Instances R1 sid 22 58 “Sailors” and “Reserves” sid S1 relations for our examples. 22 We’ll use positional or named field notation, 31 assume that names of fields 58 in query results are `inherited’ from names of S2 sid fields in query input 28 relations. 31 44 58 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke bid day 101 10/10/96 103 11/12/96 sname rating age dustin 7 45.0 lubber 8 55.5 rusty 10 35.0 sname rating age yuppy 9 35.0 lubber 8 55.5 guppy 5 35.0 rusty 10 35.0 6 Algebra In math, algebraic operations like +, -, x, /. Operate on numbers: input are numbers, output are numbers. Can also do Boolean algebra on sets, using union, intersect, difference. Focus on algebraic identities, e.g. x (y+z) = xy + xz. (Relational algebra lies between propositional and 1st-order logic.) Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 7 Relational Algebra Every operator takes one or two relation instances A relational algebra expression is recursively defined to be a relation A combination of relations is a relation Result is also a relation Can apply operator to • Relation from database • Relation as a result of another operator Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 8 Relational Algebra Operations Basic operations: Additional 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. Intersection, join, division, renaming: Not essential, but (very!) useful. Since each operation returns a relation, operations can be composed! (Algebra is “closed”.) Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 9 Projection 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?) sname rating yuppy lubber guppy rusty 9 8 5 10 sname,rating(S2) Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke age 35.0 55.5 age(S2) 10 Selection Selects rows that satisfy selection condition. No duplicates in result! (Why?) Schema of result identical to schema of (only) input relation. Selection conditions: simple conditions comparing attribute values (variables) and / or constants or complex conditions that combine simple conditions using logical connectives AND and OR. sid sname rating age 28 yuppy 9 35.0 58 rusty 10 35.0 rating 8(S2) sname rating yuppy 9 rusty 10 sname,rating( rating 8(S2)) Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 11 Union, Intersection, Set-Difference sid sname rating age 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 22 dustin rating age 7 45.0 22 31 58 44 28 dustin lubber rusty guppy yuppy 7 8 10 5 9 45.0 55.5 35.0 35.0 35.0 S1 S2 sid sname rating age 31 lubber 8 55.5 58 rusty 10 35.0 S1 S2 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke S1 S2 12 Exercise on Union Num shape ber holes 1 round 2 2 3 square 4 rectangle 8 Num shape ber 4 round 5 6 Blue blocks (BB) bottom top Stacked(S) 4 2 4 6 6 2 holes 2 square 4 rectangle 8 Yellow blocks(YB) 1. Which tables are unioncompatible? 2. What is the result of the possible unions? Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 13 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. (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: (C(1 sid1, 5 sid 2), S1 R1) Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 14 Exercise on Cross-Product Num shape ber holes 1 round 2 2 3 square 4 rectangle 8 Num shape ber 4 round 5 6 holes 2 square 4 rectangle 8 Blue blocks (BB) bottom top Stacked(S) 4 2 4 6 6 2 1. Write down 2 tuples in BB x S. 2. What is the cardinality of BB x S? Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 15 Joins R c S c ( R S) Condition Join: (sid) 22 31 sname dustin lubber rating age 7 45.0 8 55.5 S1 (sid) 58 58 S1.sid R1.sid bid 103 103 day 11/12/96 11/12/96 R1 Result schema same as that of cross-product. Fewer tuples than cross-product, might be able to compute more efficiently. How? Sometimes called a theta-join. Π-σ-x = SQL in a nutshell. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 16 Exercise on Join Num shape ber holes 1 round 2 2 3 square 4 rectangle 8 Num shape ber 4 round 5 6 Blue blocks (BB) BB holes 2 square 4 rectangle 8 Yellow blocks(YB) BB.holes YB.holes YB Write down 2 tuples in this join. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 17 Joins Equi-Join: A special case of condition join where the condition c contains only equalities. sid 22 58 S1>< sname dustin rusty rating age 7 45.0 10 35.0 bid 101 103 day 10/10/96 11/12/96 R1 R.sid S.sid Result schema similar to cross-product, but only one copy of fields for which equality is specified. Natural Join: Equijoin on all common fields. Without specified, condition A>< B means the natural join of A and B. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 18 Example for Natural Join Num shape ber 1 round 2 3 holes 2 square 4 rectangle 8 Blue blocks (BB) shape holes round 2 square 4 rectangle 8 Yellow blocks(YB) What is the natural join of BB and YB? Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 19 Find names of sailors who’ve reserved boat #103 Solution 1: Solution 2: sname(( bid 103 (Temp1, Reserves) Sailors) bid 103 Re serves) ( Temp2, Temp1 Sailors) sname (Temp2) Solution 3: sname ( bid 103 (Re serves Sailors)) Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 20 Find names of sailors who’ve reserved a red boat Information about boat color only available in Boats; so need an extra join: sname (( Boats) Re serves Sailors) color ' red ' A more efficient solution: sname ( (( Boats) Re s) Sailors) sid bid color ' red ' A query optimizer can find this, given the first solution! Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 21 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 Re serves Sailors) 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 22 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): (Tempred, sid (Tempgreen, (( sid color ' red ' (( Boats) Re serves)) color ' green' Boats) Re serves)) sname((Tempred Tempgreen) Sailors) Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 23 Division Not supported as a primitive operator, but useful for expressing queries like: Find sailors who have reserved all boats. Typical set-up: A has 2 fields (x,y) that are foreign key pointers, B has 1 matching field (y). Then A/B returns the set of x’s that match all y values in B. Example: A = Friend(x,y). B = set of 354 students. Then A/B returns the set of all x’s that are friends with all 354 students. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 24 Examples of Division A/B sno s1 s1 s1 s1 s2 s2 s3 s4 s4 pno p1 p2 p3 p4 p1 p2 p2 p2 p4 A pno p2 p4 pno p2 B1 B2 pno p1 p2 p4 B3 sno s1 s2 s3 s4 sno s1 s4 sno s1 A/B1 A/B2 A/B3 Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 25 Find the names of sailors who’ve reserved all boats Uses division; schemas of the input relations to / must be carefully chosen: (Tempsids, ( sid, bid Re serves) / ( bid Boats)) sname (Tempsids Sailors) To find sailors who’ve reserved all ‘Interlake’ boats: ..... / bid ( bname ' Interlake' Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke Boats) 26 Division in General Let A have 2 fields, x and y; B have only field y: A/B = {x: for all y in B. the tuple xy is in A}. 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. Then A/B returns the set of all x-tuples such that for every y-tuple in B, the tuple (x,y) is in A. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 27 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. Book has lots of query examples. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke 28