Relational Algebra Chapter 4 part I Relational Query
Relational Algebra Chapter 4 - part I
Relational Query Languages v Query languages: Allow manipulation and retrieval of data from a database. v Relational model supports simple powerful QLs: § § v 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. 2
Formal Relational Query Languages q Two mathematical Query Languages form the basis for “real” languages (e. g. SQL) and for implementation: q Relational Algebra: q. More operational, very useful for representing execution plans. q Relational Calculus: q. Lets users describe what they want, rather than how to compute it. (Non-operational, rather declarative. ) 3
Preliminaries A query is applied to relation instances. v The result of a query is also a relation instance. v § Schemas of input relations for a query are fixed ; query will run regardless of instance!) § The schema for result of given query is also fixed! Determined by definition of query language. (but 4
Preliminaries v Positional vs. named-field notation: § Positional field notation e. g. , S. 1 § Named field notation e. g. , v S. sid Pros/Cons: § Positional notation easier formal definitions, namedfield notation more readable. § Both used in SQL v. Assume that names of fields in query results are `inherited’ from names of fields in query input relations. 5
Example Instances Sailors Reserves R 1 Sailors S 2 6
Relational Algebra v Basic operations: § § § v Additional operations: § v 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. Intersection, join, division, renaming: essential, but (very!) useful. Since each operation returns a relation, be composed! (Algebra is “closed”. ) Not operations can 7
Selection Sailors S 2 8
Selection v v condition (R) Selects rows that satisfy selection condition. attribute op constant attribute op attribute Op is {<, >, <=, >=, =, =} v v No duplicates in result! Schema of result identical to schema of (only) input relation. 9
Selection v Result relation can be input for another relational algebra operation! (Operator composition. ) 10
Projection Sailors S 2 11
Projection v projectlist (R) v 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? ) v v 12
Union, Intersection, Set-Difference v All of these operations take two input relations, which must be union-compatible: § § v Same number of fields. `Corresponding’ fields have same type. What is the schema of result? 13
Example Instances : Union S 1 S 2 14
Difference Operation S 1 S 2 16
Intersection Operation S 1 S 2 18
Cross-Product (Cartesian Product) v S 1 R 1 : Each row of S 1 is paired with each row of R 1. Reserves R 1 Sailors S 1 20
Cross-Product (Cartesian Product) v S 1 R 1 : Result schema has one field per field of S 1 and R 1, with field names `inherited’ if possible. § Conflict: Both S 1 and R 1 have a field called sid. * Renaming operator: 21
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" 22
Condition Join / Theta Join v Condition Join: v Result schema same as that of cross-product. Fewer tuples than cross-product, more efficient. v 23
Equi. Join v 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. v An extra project: PROJECT ( THETA-JOIN) v 24
Natural Join v Natural Join: Equijoin on all common fields. 25
Division v Not supported as a primitive operator, useful for expressing queries like: but Find sailors who have reserved all boats. 26
Division v Let A have 2 fields x and y; B have only field y: § A/B = v A/B contains all x tuples (sailors) such that for every y tuple (boat) in B, there is an xy tuple in A. v 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. v A/B is the largest relation instance Q such that A. Q B 27
Division Example e. g. , A= B= A/B = all parts supplied by suppliers, relation parts suppliers who supply all parts listed in B 28
Examples of Division A/B B 1 B 2 B 3 A A/B 1 A/B 2 A/B 3 29
Expressing A/B Using Basic Operators v 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: 30
A few example queries 31
Find names of sailors who’ve reserved boat #103 Solution 1 Solution 2 Solution 3 32
Find names of sailors who’ve reserved a red boat Reserves R 1 Sailors S 1 Boats B 1 bid bname color 101 Interlake blue 103 Clipper red 33
Find names of sailors who’ve reserved a red boat v Information about boat color only available in Boats; so need an extra join: A more efficient solution: * A query optimizer can find this given the first solution! 34
Find sailors who’ve reserved a red or a green boat v 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? 35
Find sailors who’ve reserved a red and a green boat Previous approach won’t work! v Must identify sailors who reserved red boats, sailors who’ve reserved green boats, then find their intersection v 36
Find the names of sailors who’ve reserved all boats v v Uses division; schemas of the input relations to / must be carefully chosen: To find sailors who’ve reserved all ‘Interlake’ boats: . . . 37
Relational Algebra : Some More Operators Beyond Chapter 4
Generalized Projection v F 1, F 2, … (R) sname, (rating*2 as myrating) (Sailors) 39
Aggregation operators MIN, MAX, COUNT, SUM, AVG v AGGB (R) considers only non-null values of R. v R A B 1 2 3 4 1 null 1 3 MINB (R) SUMB (R) COUNTB (R) MINB (R) SUMB (R) 2 9 COUNTB (R) MAXB (R) 4 3 AVGB (R) COUNT* (R) 3 4 40
Aggregation Operators MIN, MAX, SUM, AVG must be on any 1 attribute. COUNT can be on any 1 attribute or COUNT* (R) v An aggregation operator returns a bag, not a single value ! But SQL allows treatment as a single value. v σB=MAXB (R) A B 3 4 41
Grouping Operator: GL, AL (R) v GL, AL (R) groups all attributes in GL, and performs the aggregation specified in AL. star. Name, MIN (year)→year, COUNT(title) →num (Stars. In) Stars. In title year star. Name SW 1 77 HF Matrix 99 KR 6 D&7 N 93 HF SW 2 79 HF Speed 94 KR star. Name year num HF 77 3 KR 94 2 42
Summary v The relational model has rigorously defined query languages that are simple and powerful. v Relational algebra is operational; useful as internal representation for query evaluation plans. v Several ways of expressing a given query; a query optimizer should choose most efficient version. 43
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