Relational Algebra  Souhad M. Daraghma. Relational Query Languages Query languages: Allow manipulation and retrieval of data from a database. Relational.

Slides:



Advertisements
Similar presentations
COMP 5138 Relational Database Management Systems Semester 2, 2007 Lecture 5A Relational Algebra.
Advertisements

Relational Algebra Jianlin Feng School of Software SUN YAT-SEN UNIVERSITY courtesy of Joe Hellerstein for some slides.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke1 Relational Algebra Chapter 4, Part A.
D ATABASE S YSTEMS I R ELATIONAL A LGEBRA. 22 R ELATIONAL Q UERY L ANGUAGES Query languages (QL): Allow manipulation and retrieval of data from a database.
Relational Algebra Ch. 7.4 – 7.6 John Ortiz. Lecture 4Relational Algebra2 Relational Query Languages  Query languages: allow manipulation and retrieval.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke1 Relational Algebra Chapter 4, Part A Modified by Donghui Zhang.
INFS614, Fall 08 1 Relational Algebra Lecture 4. INFS614, Fall 08 2 Relational Query Languages v Query languages: Allow manipulation and retrieval of.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke1 Relational Algebra Chapter 4.
Relational Algebra Content based on Chapter 4 Database Management Systems, (Third Edition), by Raghu Ramakrishnan and Johannes Gehrke. McGraw Hill, 2003.
1 Relational Algebra & Calculus. 2 Relational Query Languages  Query languages: Allow manipulation and retrieval of data from a database.  Relational.
By relieving the brain of all unnecessary work, a good notation sets it free to concentrate on more advanced problems, and, in effect, increases the mental.
CMPT 354, Simon Fraser University, Fall 2008, Martin Ester 52 Database Systems I Relational Algebra.
By relieving the brain of all unnecessary work, a good notation sets it free to concentrate on more advanced problems, and, in effect, increases the mental.
By relieving the brain of all unnecessary work, a good notation sets it free to concentrate on more advanced problems, and, in effect, increases the mental.
FALL 2004CENG 351 File Structures and Data Managemnet1 Relational Algebra.
By relieving the brain of all unnecessary work, a good notation sets it free to concentrate on more advanced problems, and, in effect, increases the mental.
1 Relational Algebra. 2 Relational Query Languages Query languages: Allow manipulation and retrieval of data from a database. Relational model supports.
Lecture 4: Relational algebra
Database Management Systems, R. Ramakrishnan and J. Gehrke1 Relational Algebra Chapter 4, Part A.
Relational Algebra Chapter 4 - part I. 2 Relational Query Languages  Query languages: Allow manipulation and retrieval of data from a database.  Relational.
Introduction to Database Systems 1 Relational Algebra Relational Model: Topic 3.
Relational Algebra. Relational Query Languages n Query = “retrieval program” n Language examples: ù Theoretical : 1. Relational Algebra 2. Relational.
By relieving the brain of all unnecessary work, a good notation sets it free to concentrate on more advanced problems, and, in effect, increases the mental.
1 Relational Algebra and Calculus Yanlei Diao UMass Amherst Feb 1, 2007 Slides Courtesy of R. Ramakrishnan and J. Gehrke.
Rutgers University Relational Algebra 198:541 Rutgers University.
Relational Algebra Chapter 4 - part I. 2 Relational Query Languages  Query languages: Allow manipulation and retrieval of data from a database.  Relational.
CSCD343- Introduction to databases- A. Vaisman1 Relational Algebra.
Relational Algebra.
Relational Algebra, R. Ramakrishnan and J. Gehrke (with additions by Ch. Eick) 1 Relational Algebra.
1 Relational Algebra and Calculus Chapter 4. 2 Relational Query Languages  Query languages: Allow manipulation and retrieval of data from a database.
Lecture 05 Structured Query Language. 2 Father of Relational Model Edgar F. Codd ( ) PhD from U. of Michigan, Ann Arbor Received Turing Award.
1 Relational Algebra. 2 Relational Query Languages v Query languages: Allow manipulation and retrieval of data from a database. v Relational model supports.
CS 370 Database Systems Lecture 11 Relational Algebra.
Database Management Systems, R. Ramakrishnan and J. Gehrke1 Relational Algebra.
CS 4432query processing1 CS4432: Database Systems II Lecture #11 Professor Elke A. Rundensteiner.
1 Relational Algebra & Calculus Chapter 4, Part A (Relational Algebra)
1 Relational Algebra and Calculas Chapter 4, Part A.
1.1 CAS CS 460/660 Introduction to Database Systems Relational Algebra.
Database Management Systems 1 Raghu Ramakrishnan Relational Algebra Chpt 4 Xin Zhang.
Relational Algebra.
ICS 321 Fall 2011 The Relational Model of Data (i) Asst. Prof. Lipyeow Lim Information & Computer Science Department University of Hawaii at Manoa 8/29/20111Lipyeow.
1 Relational Algebra Chapter 4, Sections 4.1 – 4.2.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke1 Database Management Systems Chapter 4 Relational Algebra.
Database Management Systems 1 Raghu Ramakrishnan Relational Algebra Chpt 4 Xin Zhang.
CSCD34-Data Management Systems - A. Vaisman1 Relational Algebra.
IST 210 The Relational Language Todd S. Bacastow January 2004.
Database Management Systems, R. Ramakrishnan1 Relational Algebra Module 3, Lecture 1.
CMPT 258 Database Systems Relational Algebra (Chapter 4)
Relational Algebra p BIT DBMS II.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke1 Relational Algebra Chapter 4.
Database Management Systems 1 Raghu Ramakrishnan Relational Algebra Chpt 4 Jianping Fan.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke1 Relational Algebra Chapter 4, Part A.
1 CS122A: Introduction to Data Management Lecture #7 Relational Algebra I Instructor: Chen Li.
1 Relational Algebra. 2 Relational Query Languages  Query languages: Allow manipulation and retrieval of data from a database.  Relational model supports.
Relational Algebra. CENG 3512 Relational Query Languages Query languages: Allow manipulation and retrieval of data from a database. Relational model supports.
Relational Algebra & Calculus
Database Systems (資料庫系統)
Relational Algebra Chapter 4 1.
Relational Algebra Chapter 4, Part A
Relational Algebra 461 The slides for this text are organized into chapters. This lecture covers relational algebra, from Chapter 4. The relational calculus.
Relational Algebra.
Relational Algebra 1.
LECTURE 3: Relational Algebra
Relational Algebra Chapter 4 1.
Relational Algebra Chapter 4 - part I.
Relational Algebra Chapter 4, Sections 4.1 – 4.2
CENG 351 File Structures and Data Managemnet
Relational Algebra & Calculus
Relational Algebra Chapter 4 - part I.
Presentation transcript:

Relational Algebra  Souhad M. Daraghma

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 logic. Allows for much optimization. Query Languages != programming languages! QLs not intended to be used for complex calculations. QLs support easy, efficient access to large data sets.

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. (Non-procedural, declarative.) We will concentrate on relational algebra in this course * Understanding Algebra & Calculus is key to understanding SQL, query processing!

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 over any legal instance) The schema for the result of a given query is also fixed. It is determined by the definitions of the query language constructs.

Basics of Querying A query is applied to relation instances, and the result of a query is also a relation instance.

Relational Algebra: 5 Basic Operations Selection (  ) Selects a subset of rows from relation (horizontal). Projection (  ) Retains only wanted columns from relation (vertical). Cross-product (  ) Allows us to combine two relations. Set-difference ( — ) Tuples in r1, but not in r2. Union (  ) Tuples in r1 or in r2. Since each operation returns a relation, operations can be composed! (Algebra is “closed”.)

R1 S1 S2 Boats Example Instances

Projection (  ) Examples: Retains only attributes that are in the “projection list”. Schema of result: exactly the fields in the projection list, with the same names that they had in the input relation. Projection operator has to eliminate duplicates (How do they arise? Why remove them?) Note: real systems typically don’t do duplicate elimination unless the user explicitly asks for it.

Projection (  ) S2 snamerating yuppy9 lubber8 guppy5 rusty10

Selection (  ) Selects rows that satisfy selection condition. Result is a relation. Schema of result is same as that of the input relation. Do we need to do duplicate elimination? sid sname rating age 28 yuppy lubber guppy rusty snamerating yuppy9 rusty10

Union, Intersection and Set-Difference Both of these operations take two input relations, which must be union-compatible: Two relations R and S are union-compatible if they have the same number of columns and corresponding columns have the same domains.

Example: not union-compatible (different number of columns) Anne Bob Chris aaa bbb ccc Tom Sam Steve

Example: not union-compatible (different domains for the second column) Anne Bob Chris aaa bbb ccc Tom Sam Steve

Example: union-compatible Anne Bob Chris Tom Sam Steve

Union S1S2

Intersection S1 S2

Set Difference S1 S2 S2 – S1

Cross-Product S1 X R1 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. S1R1

Cross-Product S1 X R1 R1 S1

Cross-Product S1 X R1 R1 S1

Cross-Product S1 R1 S1 X R1

Cross-Product S1 R1 S1 X R1

Cross-Product S1 R1 S1 X R1

Cross-Product S1 R1 S1 X R1

Cross-Product S1 R1 S1 X R1

Cross-Product S1 R1 S1 X R1

Both S1 and R1 have a field called sid. Which may cause a conflict when referring to columns S1 X R1

Cross-Product S1  R1: Each row of S1 paired with each row of R1. Q: How many rows in the result? Result schema has one field per field of S1 and R1, with field names `inherited’ if possible. May have a naming conflict: Both S1 and R1 have a field with the same name. In this case, can use the renaming operator:

Renaming Operator sid1sid2 Takes a relation schema and gives a new name to the schema and the columns S1 X R1 C

Compound Operator: Intersection In addition to the 5 basic operators, there are several additional “Compound Operators” These add no computational power to the language, but are useful shorthands. Can be expressed solely with the basic ops. Intersection takes two input relations, which must be union-compatible. Q: How to express it using basic operators? R  S = R  (R  S)

Compound Operator: Join Joins are compound operators involving cross product, selection, and (sometimes) projection. Most common type of join is a “natural join” (often just called “join”). R S conceptually is: Compute R  S Select rows where attributes that appear in both relations have equal values Project all unique attributes and one copy of each of the common ones.

Joins Condition Join : Result schema same as that of cross-product. Fewer tuples than cross-product, might be able to compute more efficiently Sometimes called a theta-join. Equi-Join : A special case of condition join where the condition c contains only equalities.

Joins S1R1

S1 R1 S1 X R1

S1 R1

S1 R1

S1 R1 Example: Natural join

S1 R1 Example: Natural join

S1 R1 Example: Natural join

S1 R1 Example: Natural join

Outer Join An extension of the join operation that avoids loss of information. Computes the join and then adds tuples form one relation that does not match tuples in the other relation to the result of the join. Uses null values: null signifies that the value is unknown or does not exist All comparisons involving null are (roughly speaking) false by definition.

Outer Join – Example Relation loan customer_name loan_number Jones Smith Hayes Jones Smith Hayes L-170 L-230 L-155 L-170 L-230 L loan_number amount L-170 L-230 L-260 L-170 L-230 L-260 branch_name Downtown Redwood Perryridge Downtown Redwood Perryridge Relation borrower

Outer Join – Example Inner Join loan Borrower loan_number amount L-170 L-230 L-170 L customer_name Jones Smith Jones Smith branch_name Downtown Redwood Downtown Redwood Jones Smith null Jones Smith null loan_number amount L-170 L-230 L-260 L-170 L-230 L customer_name branch_name Downtown Redwood Perryridge Downtown Redwood Perryridge n Left Outer Join loan Borrower

Outer Join – Example loan_number amount L-170 L-230 L-155 L-170 L-230 L null null customer_name Jones Smith Hayes Jones Smith Hayes branch_name Downtown Redwood null Downtown Redwood null loan_number amount L-170 L-230 L-260 L-155 L-170 L-230 L-260 L null null customer_name Jones Smith null Hayes Jones Smith null Hayes branch_name Downtown Redwood Perryridge null Downtown Redwood Perryridge null n Full Outer Join loan borrower n Right Outer Join loan borrower

Compound Operator: Division Useful for expressing “for all” queries like: Find sids of sailors who have reserved all boats. For A/B, attributes of B must be subset of attrs of A. May need to “project” to make this happen. E.g., let A have 2 fields, x and y; B have only field y: A/B contains all tuples (x) such that for every y tuple in B, there is an xy tuple in A.

Examples of Division A/B A B1 B2 B3 A/B1 A/B2A/B3 Q for intuition: What is (R/S)  S?

Expressing A/B Using Basic Operators Division is not essential op; just a useful shorthand. (Also true of joins, but joins are so common that systems implement joins specially.) Idea: For A(x,y)/B(y), 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: Disqualified x values