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Relational Algebra Relational Model: Topic 3 Introduction to Database Systems 1 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. Introduction to Database Systems 2 Example Instances R1 sid 22 58 “Sailors” and “Reserves” sid S1 relations for our examples. 22 31 58 S2 sid 28 31 44 58 Introduction to Database Systems 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 3 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. Operations can be composed! (Algebra is “closed”.) Introduction to Database Systems 4 Projection Deletes attributes that are not in projection list. Schema of result? Duplicates! sname rating yuppy lubber guppy rusty 9 8 5 10 sname,rating(S2) age 35.0 55.5 age(S2) Introduction to Database Systems 5 Selection sid sname rating age 28 yuppy 9 35.0 58 rusty 10 35.0 rating 8(S2) Selects rows that satisfy selection condition. Duplicates? Schema of result? sname rating yuppy 9 rusty 10 sname,rating( rating 8(S2)) Introduction to Database Systems 6 Union, Intersection, Set-Difference sid sname rating age Inputs: union-compatible: – Same number/type of fields. Schema of result? Duplicates? sid sname rating age 22 dustin 7 45.0 S1 S2 Introduction to Database Systems 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 7 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. (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: Introduction to Database Systems (C(1 sid1, 5 sid2), S1 R1) 8 Joins Condition Join: R c S c ( R S) (sid) sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 S1 (sid) bid 58 103 58 103 S1.sid R1.sid day 11/12/96 11/12/96 R1 Result schema ? Fewer tuples than cross-product, might be able to compute more efficiently Introduction to Database Systems 9 Joins Equi-Join: A special case of condition join where the condition c contains only equalities. sid sname rating age bid day 22 dustin 7 45.0 101 10/10/96 58 rusty 10 35.0 103 11/12/96 S1 R1 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. Introduction to Database Systems 10 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 = x | x, y A y 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. Introduction to Database Systems 11 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 Introduction to Database Systems pno p2 B1 pno p2 p4 B2 pno p1 p2 p4 B3 sno s1 s2 s3 s4 sno s1 s4 sno s1 A/B1 A/B2 A/B3 12 Expressing A/B Using Basic Operators Division is not essential operator; just a useful shorthand. 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 ( A) Introduction to Database Systems x (( x ( A) B) A) all disqualified x values 13 Find names of sailors who’ve reserved boat #103 Solution 1: Solution 2: sname (( bid 103 (Temp1, Re serves) Sailors) bid 103 Re serves) ( Temp2, Temp1 Sailors) sname (Temp2) Solution 3: sname ( Introduction to Database Systems bid 103 (Re serves Sailors)) 14 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 ( sid (( bid color ' red ' Boats) Re s) Sailors) A query optimizer can find this Introduction to Database Systems 15 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? Introduction to Database Systems 16 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) Introduction to Database Systems 17 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 ( Introduction to Database Systems bname ' Interlake' Boats) 18 Expressive Power Codd’s Theorem: Every Relational Algebra query can be expressed as a safe query in DRC / TRC; the converse is also true. Relational Completeness: A “relationally complete” query language (e.g., SQL) can express every query that is expressible in relational algebra/calculus. Introduction to Database Systems 19 A Limitation of the Algebra For any particular instance of Edges, there is an R.A. expression to compute transitive closure. (What is it?) There’s no R.A. for transitive closure of an arbitrary instance of Edges. (Why?) Edges From a a c y To b c y z Introduction to Database Systems a b z c y 20