Relational Algebra Chapter 4 part I Relational Query

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Relational Algebra Chapter 4 - part I

Relational Algebra Chapter 4 - part I

Relational Query Languages v Query languages: Allow manipulation and retrieval of data from a

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 Two mathematical Query Languages form the basis for “real” languages

Formal Relational Query Languages Two mathematical Query Languages form the basis for “real” languages (e. g. SQL), and for implementation: q 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, rather declarative. ) q * Understanding Algebra & Calculus is key to * understanding SQL, query processing! 3

Preliminaries v A query is applied to relation instances, and the result of a

Preliminaries v 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 ; 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: v Positional field notation e. g. , S.

Preliminaries v Positional vs. named-field notation: v Positional field notation e. g. , S. 1 v v Named field notation e. g. , 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

Example Instances Sailors Reserves R 1 Sailors S 2 6

Relational Algebra v Basic operations: § § § v Additional operations: § v Selection

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 Sailors S 2 8

Selection v v condition (R) Selects rows that satisfy selection condition. attribute op constant

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.

Selection v Result relation can be input for another relational algebra operation! (Operator composition. ) 10

Projection Sailors S 2 11

Projection Sailors S 2 11

Projection v projectlist (R) v Deletes attributes that are not in projection list. Schema

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

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 Sailors S 1 Sailors S 2 14

Example Instances : Union Sailors S 1 Sailors S 2 14

Difference Operation Sailors S 1 Sailors S 2 16

Difference Operation Sailors S 1 Sailors S 2 16

Intersection Operation Sailors S 1 Sailors S 2 18

Intersection Operation Sailors S 1 Sailors S 2 18

Cross-Product (Cartesian Product) v v S 1 R 1 Each row of S 1

Cross-Product (Cartesian Product) v v S 1 R 1 Each row of S 1 is paired with each row of 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: 20

Joins (Why we need a Join? ) §In many cases, Join = Cross-Product +

Joins (Why we need a Join? ) §In many cases, Join = Cross-Product + Select + Project § However : Cross-product is too large to materialize Apply Select and Project "On-the-fly" 21

Condition Join / Theta Join v Condition Join: v Result schema same as that

Condition Join / Theta Join v Condition Join: v Result schema same as that of cross-product. Fewer tuples than cross-product, more efficient. v 22

Equi. Join v Equi-Join: A special case of condition join where the condition c

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 v 23

Natural Join v Natural Join: Equijoin on all common fields. 24

Natural Join v Natural Join: Equijoin on all common fields. 24

Division v v Not supported as a primitive operator, but useful for expressing queries

Division v v 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, then x value is in A/B. 25

Division § A/B is the largest relation instance Q such that Q B A.

Division § A/B is the largest relation instance Q such that Q B A. e. g. , A = all parts supplied by suppliers, B= relation parts A/B = suppliers who supply all parts listed in B 26

Examples of Division A/B B 1 B 2 B 3 A A/B 1 A/B

Examples of Division A/B B 1 B 2 B 3 A A/B 1 A/B 2 A/B 3 27

Expressing A/B Using Basic Operators v Division is not essential op; just a useful

Expressing A/B Using Basic Operators v Division is not essential op; just a useful shorthand. § v (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: all disqualified tuples 28

Find names of sailors who’ve reserved boat #103 Solution 1 Solution 2 Solution 3

Find names of sailors who’ve reserved boat #103 Solution 1 Solution 2 Solution 3 29

Find names of sailors who’ve reserved a red boat v Information about boat color

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! 30

Find sailors who’ve reserved a red or a green boat v Can identify all

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? 31

Find sailors who’ve reserved a red and a green boat v Previous approach won’t

Find sailors who’ve reserved a red and a green boat v 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): 32

Find the names of sailors who’ve reserved all boats v Uses division; schemas of

Find the names of sailors who’ve reserved all boats v Uses division; schemas of the input relations to / must be carefully chosen: v To find sailors who’ve reserved all ‘Interlake’ boats: . . . 33

Summary v The relational model has rigorously defined query languages that are simple and

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. 34

Example Instances Sailors S 1 Sailors S 2 35

Example Instances Sailors S 1 Sailors S 2 35