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42 Pages

### RelationalAlgebra

Course: CSC 319, Fall 2009
School: DePaul
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Word Count: 1612

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Algebra Relational Manipulating Databases To access information in a database we use a query Ex: How many customers have the first name = `John'? Good query writing follows a formal model called relational algebra Relational Algebra Relational algebra: a collection of mathematical operations that manipulate tables Familiarity with relational algebra helps understanding the logic behind...

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Algebra Relational Manipulating Databases To access information in a database we use a query Ex: How many customers have the first name = `John'? Good query writing follows a formal model called relational algebra Relational Algebra Relational algebra: a collection of mathematical operations that manipulate tables Familiarity with relational algebra helps understanding the logic behind complex queries and ease the way for writing them Tables and Queries Recall: a table is a set of rows/records having the same number and types of attributes When you send a query to the database, it Finds the appropriate rows of information in the stored tables Performs the requested operations on the data Represents the results in a new temporary table Delivers the table of results to the user Example Ex: How many customers have the first name = `John'? The database creates a table containing all customers whose first name is `John' and returns the table to the user Basic types of queries There are 4 basic types of queries A projection operation produces a result table with A selection operation produces a result table with Only some of the columns of its input table. A join or product operation produces a result table by All of the columns of the input table Only those rows of its input table that satisfy some criteria. Combining the columns of two input tables. A set operation produces a result table by Combining rows from one or the other of its input tables Projection operation: A projection query selects some of the columns of the input table project T onto (attribute1, attribute2, ...) Relational algebra form: attribute1, attribute2,... (T) Example firstName,lastName(Customer) lastName Block Hamilton Harrison Breaux firstName Jane Cherry Katherine Carroll street 345 Randolph Circle 3230 Dade St. 103 Landis Hall 76 Main St. 9501 Lafayette St. 123 Main St. 14325 N. Bankside St. accountId 101 102 103 104 106 111 201 444 city Apopka Dade City Bratt Apopka Houma Apopka Godfrey state FL FL FL FL LA FL IL zipcode 30458 30555 30457 30458 44099 30458 43580 32306 balance \$0.00 \$3.00 \$31.00 \$35.00 \$0.00 \$0.00 \$0.00 \$10.55 Morehouse Anita Doe Greaves Doe Jane Joseph Jane Cawthon Dorm, room 642 Tallahassee FL Example ... Notice that the result firstName table has fewer rows lastName Morehouse Block Breaux Hamilton Harrison Doe Greaves Duplicate rows have Jane been removed because the Carroll attributes do not Cherry contain a key Jane Joseph Anita Catherine Storing the temporary results T' attribute1, We can store the result of a query in a table T' as follows: attribute2,... (T) This will create a table T' with attributes: attribute1, attribute2, ... containing the result of the query Selection queries: A selection query selects rows that match a selection criteria from a table Relational algebra form <condition>(T) Each row is checked to see if it satisfies the condition and selected accordingly Example lastName=`Doe'(Customer) firstName Jane Jane lastName Doe Doe street 123 Main St. city Apopka state FL zipcode 34331 32306 balance 0.00 10.55 accountId 111 444 Cawthon Dorm, Tallahassee FL room 642 Complex selection criteria The selection criterion can be any boolean expression containing operators like: and, or, =, , <, >, , , etc ... Example T ssn='376-77-0099' ssn 145090967 245114554 376770099 376770099 145090967 376770099 376770099 and date < '01-mar-2002' (TimeCard) storeId 3 3 5 5 3 5 5 paid yes yes yes yes yes yes yes date startTime endTime 12:00 12:00 22:00 22:00 12:00 14:00 19:00 01/ 14/ 2002 8:15 01/ 14/ 2002 8:15 02/ 23/ 2002 14:00 03/ 21/ 2002 14:00 01/ 16/ 2002 8:15 01/ 03/ 2002 10:00 01/ 03/ 2002 15:00 Product queries: X The product query takes two tables and produce a table which is the cross product of the two, i.e., combines every row of one table with every row of other table R(A1, A2 , ... , An) S(B1, B2 , ... , Bm)= Q(A1, A2 , ... , An, B1, B2 , ... , Bm) Relational algebra form: R S Example Employee TimeCard Employee.ssn 145090967 245114554 376770099 479980098 579988778 145090967 245114554 376770099 479980098 579988778 145090967 245114554 lastName Uno Toulouse Threat Fortune firstName Jane Jie Ayisha Julian TimeCard.ssn 145090967 145090967 145090967 145090967 145090967 145090967 145090967 145090967 145090967 145090967 245114554 245114554 date startTime endTime 1/ 14/ 2002 8:15:00 AM 1/ 14/ 2002 8:15:00 AM 1/ 14/ 2002 8:15:00 AM 1/ 14/ 2002 8:15:00 AM 1/ 14/ 2002 8:15:00 AM 1/ 16/ 2002 8:15:00 AM 1/ 16/ 2002 8:15:00 AM 1/ 16/ 2002 8:15:00 AM 1/ 16/ 2002 8:15:00 AM 1/ 16/ 2002 8:15:00 AM 1/ 14/ 2002 8:15:00 AM 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 12:00:00 PM 3 storeId paid Yes Yes Yes Yes Fivozinsky Bruce Uno Toulouse Threat Fortune Jane Jie Ayisha Julian Yes Yes Yes Yes Yes Fivozinsky Bruce Uno Toulouse Jane Jie Yes Yes 1/ 14/ 2002 8:15:00 AM Yes Product queries ... If two attributes in two tables T and R have the same name, we prefix them with the relation name: T.<attribute> Ex: Employee.ssn, TimeCard.ssn Remark. Many of the resulting rows in the previous example don't make sense Join queries: In the previous table we are only interested in the rows that match: rows with Employee.ssn = TimeCard.ssn We are interested in the query: Employee.ssn=TimeCrad.ssn(Employee x TimeCard) Join operations ... A join query is a cross product with a restriction on the result rows Typical join condition is equality of attributes The join condition determines which rows match Only matching rows are in the result It table is called equijoin R <condition>S Relational algebra form: Example Some rows from the table Employee Employee.ssn=TimeCard.ssn TimeCard Employee. ssn 145090967 145090967 245114554 376770099 last Name Uno Uno Toulouse Threat first Name Jane Jane Jie Ayisha TimeCard. ssn date 145090967 145090967 245114554 376770099 start Time storeId 3 3 3 5 paid no no no no end Time 12:00 12:00 12:00 22:00 01/ 14/ 2002 8:15 01/ 16/ 2002 8:15 01/ 14/ 2002 8:15 02/ 23/ 2002 14:00 Natural join: * Frequently, when doing an equijoin, the attributes have the same name A natural join is an equijoin with an equality condition on the common attributes: Employee ssn TimeCard Employee * TimeCard In natural join the common attributes appear once Queries with multiple joins Consider the VideoRental schema, and suppose we want to retrieve for every currentlyrented video, the renter's account number, video number, rental date, due date, title of the movie, and cost Solution accountId, videoId, dateRented, dateDue, title, cost ((Rental videoId Video) movieId Movie) Combining operations Suppose we want to find the following info. For customer with account = 113, find all the videos that he is renting: For each video, find the video number, the title of the movie, and the due date Solution videoId, title, dateDue (( accountId=113(Rental) videoId Video) movieId Movie) Or: T1= accountId=113(Rental) T2=T1 videoId Video T3 = T2 movieId Movie T4 = videoId, title, dateDue ( T3) More examples List all comedy movies that were rented on December 21, 2001. For every movie list the customer's name, movie title, and date returned Solution T1= daterented=`December 21 2001'(PreviousRental) T2=T1 videoId Video T3= genre=`comedy' (Movie) T4= T2 movieId T3 T5= T4 accountId Customer T6 = firstName, lastName, title, dateReturned ( T5) Set operations Set operations include: Union, intersection, and difference Relational algebra form: , , - Set operations can be applied to any tables with the same shape (compatible) The same order and type of attributes Attribute names do not have to agree Set operations If R and S are two compatible tables: R S is the table that contains the set of rows that are either in R or in S R S is the table that contains the set of rows that are both in R and S R S is the table that contains the set of rows that are in R but not in S Example of Retrieve all the videos that are currently or were previously rented EverRented= Rental PreviousRental Example of Retrieve the video id of all the videos that are currently rented and have been rented at least once before Veterans= videoId, ( Rental) videoId, ( PreviousRental) Example of Retrieve the video id of all the videos that are currently rented and have never been rented before FirstTime= videoId, ( Rental) videoId, ( PreviousRental) Aggregate functions Not all queries can be expressed using the basic operations described previously. What if we want to compute the average salary of all employees? Aggregate functions What if we want to count the number of employees in each department? For such queries, we use aggregate functions. Relational algebra form <grouping attributes> <function list> (T) Aggregate functions ... The function list includes: average, sum, count, maximum, minimum The result of the query will be a table containing the results The attributes consist of the grouping attributes + function parameters Examples Ex1: compute the average salary of all the employees Average(salary)(Employee) The resulting table contains one attribute: Average_Salary and one value Ex2: compute the number of employees in each department DNO Count(ssn) (Employee) The resulting table contains two attributes: DNO and Count_ssn. There is a row for every dept. containing the DNO value and the number of employees Renaming attributes It is sometimes convenient to rename the attributes in the resulting relation: R(DEPTNUM, NUM_EMPL) DNO Count(ssn) (Employee) Recursive operations Compute all the employees supervised by ``Pinochio'' Compute all the emplyees supervised by ``Pinochio'' at level two Compute all the employees supervisod by ``Pinochio'' at any level!!! Answers A1: Pinochio_ssn < ssn ( fname=`pinochio'(Employee)) Result1 ssn (Pinochio ssn=superssn Employee) A2: Result2 ssn (Result1 ssn=superssn Employee) Result Result1 Result2 A3: is not supported by standard relational algebra Outer Join Left Outer Join Ex: list the employee names and also the name of the department they manage in case it exists Right Outer Join Full Outer Join Examples from (EmpDeptProj schema) List everybody who makes more than \$30000. List names of everybody working for the research department. List employees with a dependent. List employees that have a daughter. List employees without dependents. List employees working on a project in Houston. List all supervisors. List names of all managers. List names of managers with at least one dependent Examples from (EmpDeptProj schema ...) For every project located in `Chicago', list the project number, the controlling department number, the department manager's last name, address, and birthdate. Make a list of project numbers for projects involving an employee whose first name is `Pinochio' either as a worker on the project, or as a manager of the department that controls the project. Find the names of all employees who are directly supervised by `Isaac Newton' For each department, retrieve the department name and average salary of its employees. Retrieve the average salary of all female employees For each project, list the project name and the total number of hours spent on the project.
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