Relational Algebra Calculus Chapter 4 Part A Relational
Relational Algebra & Calculus Chapter 4, Part A (Relational Algebra) 1
Relational Query Languages Query languages: Allow manipulation and retrieval of data from a database. v Relational model supports simple, powerful QLs: v § § 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 v Two mathematical Query Languages form the basis for “real” languages (e. g. SQL), and for implementation: § Relational Algebra: More operational (procedural), useful for representing execution plans. § Relational Calculus: Allows users to describe what they want, rather than how to compute it: Nonoperational, declarative. 3
Preliminaries v A query is applied to relation instances, and the result of a query is also a relation instance. § § v 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 formal definitions, named-field notation more readable. Both used in SQL 4
R 1 Example Instances v v “Sailors” and “Reserves” S 1 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. S 2 5
Relational Algebra v Basic operations: § § § v Additional operations: § v 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”. 6
Projection v v v 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 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 (by DISTINCT). Why not? 7
Selection v v v Selects rows that satisfy selection condition. No duplicates in result! Why? Schema of result identical to schema of input relation. What is Operator composition? Selection is distributive over binary operators Selection is commutative 8
Union, Intersection, Set. Difference v v 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? 9
Cross-Product (Cartesian Product) Each row of S 1 is paired with each row of R 1. v 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. v § Renaming operator: 10
Joins: used to combine relations v Condition Join: Result schema same as that of cross-product. v Fewer tuples than cross-product, might be able to compute more efficiently v Sometimes called a theta-join. v 11
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 Natural Join: Equijoin on all common fields. v 12
Properties of join Selecting power: can join be used for selection? v Is join commutative? = ? v Is join associative? v Join and projection perform complementary functions v Lossless and lossy decomposition v 13
Exercises Given relational schema: Sailors (sid, sname, rating, age) Reservation (sid, bid, date) Boats (bid, bname, color) 1) 2) 3) 4) 5) Find names of sailors who’ve reserved boat #103 Find names of sailors who’ve reserved a red boat Find sailors who’ve reserved a red or a green boat Find sailors who’ve reserved a red and a green boat Find the names of sailors who’ve reserved all boats 14
Division Not supported as a primitive operator, but useful for expressing queries like: Find sailors who have reserved all boats. v Let A have 2 fields, x and y; B have only field y: § A/B = v § § v 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. 15
Examples of Division A/B B 1 B 2 B 3 A A/B 1 A/B 2 A/B 3 16
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 17
Exercises Given relational schema: Sailors (sid, sname, rating, age) Reservation (sid, bid, date) Boats (bid, bname, color) 1) 2) 3) 4) 5) Find names of sailors who’ve reserved boat #103 Find names of sailors who’ve reserved a red boat Find sailors who’ve reserved a red or a green boat Find sailors who’ve reserved a red and a green boat Find the names of sailors who’ve reserved all boats 18
Exercise Given relational schema: Sailors (sid, sname, rating, age) Reservation (sid, bid, date) Boats (bid, bname, color) 1) 2) 3) Find all sialors with a rating above 7. Find the names and ages of sailors with a rating above 7 Find the sailor name, boat id, and reservation date for each reservation 19
More Examples Given instances : S 3 of Sailors R 2 of Reservation B 1 of Boats 20
More Examples (Q 1) Find the names of sailors who have reserved boat 103. Or … 21
More Examples (Q 1) Find the names of sailors who have reserved boat 103. 22
More Examples (Q 2) Find the names of sailors who have reserved a red boat. 23
More Examples (Q 3) Find the colors of boats reserved by Lubber. 24
More Examples (Q 4) Find the names of sailors who have reserved at least one boat. 25
More Examples (Q 5) Find the names of sailors who have reserved a read or a green boat. 26
More Examples (Q 5) Find the names of sailors who have reserved a read or a green boat. 27
More Examples (Q 6) Find the names of sailors who have reserved a read and a green boat. Or… 28
More Examples (Q 6) Find the names of sailors who have reserved a read and a green boat. Or… 29
More Examples (Q 7) Find the names of sailors who have reserved at least two boats. 30
More Examples (Q 8) Find the sids of sailors with age over 20 who have not reserved a red boats. 31
More Examples (Q 9) Find the names of sailors who have not reserved all boats. *Using division operation. 32
More Examples (Q 10) Find the names of sailors who have not reserved all boats called interlake. *Using division operation. 33
Summary of Relational Algebra The relational model has rigorously defined query languages that are simple and powerful. v Relational algebra is more operational; useful as internal representation for query evaluation plans. v Several ways of expressing a given query; a query optimizer should choose the most efficient version. v 34
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