1 lecture no 16 asma ahmad relational algebra i march 10, 2011 database systems
TRANSCRIPT
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Relational AlgebraRelational Query LanguagesFormal Relational Query LanguagesOperations in Relational AlgebraSelectionProjectionCross ProductRenamingUnion, Intersection, Set DifferenceJoin and its typesDivisionGrouping & Aggregate FunctionsExamplesSummary
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Relational Query Languages Query languages: allow manipulation and
retrieval of data from a database. Relational QLs are simple & powerful.
Strong formal foundation based on logic. Allows for much optimization.
Query languages != programming languages! Not intended for complex calculations. Support easy, efficient access to large data sets.
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Formal Relational Query Languages
Two mathematical query languages form the basis for “real” languages (e.g., SQL), and for implementation:
1. Relational algebra: more operational, very useful for understanding meanings of queries and for representing execution plans.
2. Relational calculus: lets users describe what they want, rather than how to compute it.
Difference
The relational algebra might suggest these steps to retrieve the phone numbers and names of book stores that supply Some Sample Book: Join books and titles over the BookstoreID. Restrict the result of that join to tuples for the book Some Sample Book. Project the result of that restriction over StoreName and StorePhone.
The relational calculus would formulate a descriptive, declarative way: Get StoreName and StorePhone for supplies such that there exists a title
BK with the same BookstoreID value and with a BookTitle value of Some Sample Book.
The relational algebra and the relational calculus are essentially logically equivalent: for any algebraic expression, there is an equivalent expression in the calculus, and vice versa
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Relational Algebra Basic Operations:
Selection (): choose a subset of rows. Projection (): choose a subset of columns. Cross Product (): Combine two tables. Union (): unique tuples from either table. Set difference (): tuples in R1 not in R2. Renaming (): change names of tables & columns
Additional Operations (for convenience): Intersection, joins (very useful), division, outer joins,
aggregate functions, etc.
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Selection Format: selection-condition(R). Choose tuples that
satisfy the selection condition. Result has identical schema as the input.
Major = ‘CS’ (Students)
SID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS 678 Ken 3.5 Math
StudentsSID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS
Result
Selection condition is a Boolean expression including =, , <, , >, , and, or, not.
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Projection Format: attribute-list(R). Retain only those
columns in the attribute-list. Result must eliminate duplicates.
Major(Students)
SID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS 678 Ken 3.5 Math
StudentsMajor CS Math
Result
Operations can be composed.
Name, GPA(Major = ‘CS’ (Students))
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Cross Product
Format: R1 R2. Each row of R1 is paired with each row of R2.
Result schema consists of all attributes of R1 followed by all attributes of R2.
Problem: Columns may have identical names. Use notation R.A, or renaming attributes.
Only some rows make sense. Often need a selection to follow.
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Example of Cross Product
SID Name GPA Major SID Amount Year 456 John 3.4 CS 456 1500 1998 456 John 3.4 CS 678 3000 2000 457 Carl 3.2 CS 456 1500 1998 457 Carl 3.2 CS 678 3000 2000 678 Ken 3.5 Math 456 1500 1998 678 Ken 3.5 Math 678 3000 2000
Students Awards
SID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS 678 Ken 3.5 Math
StudentsSID Amount Year 456 1500 1998 678 3000 2000
Awards
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Renaming
Format: S(R) or S(A1, A2, …)(R): change the name of relation R, and names of attributes of R
CS_Students(Major = ‘CS’ (Students))
SID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS 678 Ken 3.5 Math
StudentsSID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS
CS_Students
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Union, Intersection, Set Difference Format: R1 R2 (R1 R2, R1 R2). Return all
tuples that belong to either R1 or R2 (to both R1 and R2; to R1 but not to R2).
Requirement: R1 and R2 are union compatible. With same number of attributes. Corresponding attributes have same domains.
Schema of result is identical to that of R1. May need renaming.
Duplicates are eliminated.
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Examples of Set Operations
SID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS 678 Ken 3.5 Math
TAsSID Name GPA Major 456 John 3.4 CS 223 Bob 2.95 Ed
RAs
SID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS 678 Ken 3.5 Math 223 Bob 2.95 Ed
TAs RAsSID Name GPA Major 456 John 3.4 CS
TAs RAs
SID Name GPA Major 457 Carl 3.2 CS 678 Ken 3.5 Math
TAs RAs
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Joins Theta Join.
Format: R1 join-condition R2.
Returns tuples in join-condition(R1 R2)
Equijoin. Same as Theta Join except the join-condition
contains only equalities. Natural Join.
Same as Equijoin except that equality conditions are on common attributes and duplicate columns are eliminated.
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Examples of Joins
Theta Join.
Students Students.Age<=Profs.Age Profs
SID Name GPA Age Prof 456 John 3.4 29 123 457 Carl 3.2 35 123 678 Ken 3.5 25 154
StudentsPID Pname Age Dept 123 John 35 CS 154 Scott 28 Math
Profs
SID Name GPA Age Prof PID Pname Age Dept 456 John 3.4 29 123 123 John 35 CS 457 Carl 3.2 35 123 123 John 35 CS 678 Ken 3.5 25 154 123 John 35 CS 678 Ken 3.5 25 154 154 Scott 28 Math
Result
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Examples of Joins (cont.) Equijoin.
Students Prof=PID AND Name=Pname Profs
SID Name GPA Age Prof PID Pname Age Dept 456 John 3.4 29 123 123 John 35 CS
Result
Natural Join. Students Profs
SID Name GPA Age Prof PID Pname Dept 457 Carl 3.2 35 123 123 John CS
Result
Relational Algebra Defined:Where is it in DBMS?
parser
SQLRelationalalgebraexpression
OptimizedRelationalalgebraexpression
Query optimizer
Codegenerator
Queryexecutionplan
Executablecode
DBMS
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Dangling Tuples in Join
Usually, only a subset of tuples of each relation will actually participate in a join.
Tuples of a relation not participating in a join are dangling tuples.
How do we keep dangling tuples in the result of a join? (Why do we want to do that?) Use null values to indicate a “no-join” situation.
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Outer Joins Left Outer Join.
Format: R1 R2. Similar to a natural join but keep all dangling tuples of R1.
Right Outer Join. Format: R1 R2. Similar to a natural join but keep
all dangling tuples of R2. (Full) Outer Join.
Format: R1 R2. Similar to a natural join but keep all dangling tuples of both R1 & R2.
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Examples of Outer Joins
Left Outer Join.
Students Awards
SID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS 678 Ken 3.5 Math
StudentsSID Amount Year 456 1500 1998 678 3000 2000
Awards
SID Name GPA Major Amount Year 456 John 3.4 CS 1500 1998 457 Carl 3.2 CS Null Null 678 Ken 3.5 Math 3000 2000
Result
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Join: An Observation
Some tuples don’t contribute to the result, they get lost.
Employee DepartmentBrownJonesSmith
ABB
DepartmentBC
HeadBlackWhite
Employee Department HeadJonesSmith
BB
BlackBlack
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Outer Join
An outer join extends those tuples with null values that would get lost by an (inner) join.
The outer join comes in three versions left: keeps the tuples of the left argument, extending them
with nulls if necessary right: ... of the right argument ... full: ... of both arguments ...
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Left Outer Join
Employee DepartmentBrownJonesSmith
ABB
EmployeeDepartment
BC
HeadBlackWhite
Department
Employee Department HeadBrownJones
AB
nullBlack
Smith B Black
Employee DepartmentLeft
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Right Outer Join
Employee DepartmentBrownJonesSmith
ABB
EmployeeDepartment
BC
HeadBlackWhite
Department
Employee Department HeadJonesSmith
BB
BlackBlack
null C White
Employee DepartmentRight
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Full Outer Join
Employee DepartmentBrownJonesSmith
ABB
EmployeeDepartment
BC
HeadBlackWhite
Department
Employee Department HeadBrownJones
AB
nullBlack
Smith B Black
Employee DepartmentFull
null C White
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Division
Format: R1 R2.
Restriction: Every attribute in R2 is in R1. For R1(A1, ..., An, B1, ..., Bm) R2(B1, ...,
Bm) and T = A1, ..., An (R1), Return the subset of T, say W, such that every tuple in W R2 is in R1.
W is the largest subset of T, such that,
(W R2) R1
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An Example of Division
Takes CS_Req
SID CNO 456 CS210 456 CS321 456 CS135 457 CS210 457 CS321 532 CS210 678 CS321
TakesSID 456 457
ResultCNO CS210 CS321
CS_Req
What is the meaning of this expression?
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Division (Definition) R S
Defines a relation over the attributes C that consists of set of tuples from R that match combination of every tuple in S.
Expressed using basic operations:T1 C(R)
T2 C((S X T1) – R)
T T1 – T2
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Example - Division Identify all clients who have viewed all
properties with three rooms.
(clientNo, propertyNo(Viewing)) (propertyNo(rooms = 3 (PropertyForRent)))
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DIVISION () : QUERIES THAT INCLUDE THE PHRASE “FOR ALL”.
Q. FIND ALL CUSTOMERS WHO HAVE AN ACCOUNT AT ALL BRANCHES LOCATED IN BROOKLYN. TO OBTAIN ALL BRANCHES IN BROOKLYN :R = BRANCH-NAME(BRANCH-CITY=“BROOKLYN” (BRANCH))
TO OBTAIN ALL CUSTOMER-NAME, BRANCH-NAME PAIRS FOR WHICH THE CUSTOMER HAS AN ACCOUNT AT A BRANCH :S = CUSTOMER-NAME,BRANCH-NAME (DEPOSIT)
NOW TO FIND CUSTOMERS WHO APPEARIN S WITH EVERY BRANCH NAME IN R. CUSTOMER-NAME,BRANCH-NAME (DEPOSIT) BRANCH-NAME(BRANCH-
CITY=“BROOKLYN” (BRANCH))
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Grouping & Aggregate Functions Format: group_attributes F aggregate_functions ( r ) Partition a relation into groups Apply aggregate function to each group Output grouping and aggregation values, one
tuple per group Ex: Major F count(SID), avg(GPA) (Students)
SID Name GPA Major 456 John 3.4 CS 457 Carl 3.2 CS 678 Ken 3.5 Math
StudentsMajor count(SID) avg(GPA) CS 2 3.3 Math 1 3.5
Result
Sample Schema for Exercises
Student(ID, name, address, GPA, SAT)
Campus(location, enrollment, rank)
Apply(ID, location, date, major, decision)
Sample QueriesFind Names and addresses of all students with GPA > 3.7 who applied to CS major and were rejected.
List name and address of all students who didn’t apply anywhere.
name, address ((Students) Students.ID= Not_Apply.ID
(Not_Apply(ID (Students) - ID (Apply))))
name, address (GPA>3.7 decision=‘No’ major=‘CS’
(Student Student.ID=Apply.ID Apply))
BRANCH-SCHEME (BRANCH-NAME, ASSETS, BRANCH-CITY) CUSTOMER (CUSTOMER-NAME, STREET, CUSTOMER-CITY). DEPOSIT (BRANCH-NAME, ACCOUNT-NUMBER, CUSTOMER-NAME, BALANCE) BORROW-SCHEME = (BRANCH-NAME, LOAN-NUMBER, CUSTOMER-NAME, AMOUNT) CLIENT (CLIENT-NAME, BANKER-NAME)
Another Database Schema
SELECT () : Q. SELECT THOSE TUPLES OF THE BORROW RELATION WHERE BRANCH IS “PERRYRIDGE”.BRANCH-NAME = “PERRYRIDGE”(BORROW)
Q. FIND ALL TUPLES IN WHICH AMOUNT BORROWED IS LESS THAN $1200. AMOUNT < 1200 (BORROW)
WE CAN USE THE FOLLOWING OPERATIONS :=, , , , , . ALSO, WE DENOTE AND BY , OR BY .
Q. FIND THOSE TUPLES PERTAINING TO LOANS OF MORE THAN $1200 MADE BY THE PERRYRIDGE BRANCH. BRANCH-NAME = “PERRYRIDGE” AMOUNT > 1200 (BORROW )
PROJECT () : Q. SHOW CUSTOMERS AND THEIR BORROWING BRANCH-NAME.BRANCH-NAME, CUSTOMER-NAME(BORROW)
Q. FIND ALL THOSE CUSTOMERS WHO HAVE THE SAME NAME AS THEIR PERSONAL BANKER. CUSTOMER-NAMECUSTOMER-NAME = BANKER-NAME (CLIENT).
CARTESIAN PRODUCT () :
Q. FIND ALL CLIENTS OF BANKER JOHNSON AND THE CITY IN WHICH THEY LIVE. BANKER-NAME=“JOHNSON” (CLIENTCUSTOMER)
NOW CLIENT.CUTOMER-NAME COLUMN CONTAINS ONLY CUSTOMERS of Banker JOHNSON.
TO FIND ALL THE CLIENTS OF BANKER JOHNSON, WE WRITECLIENT.CUSTOMER-NAME = CUSTOMER.CUSTOMER-
NAME ( BANKER-NAME=“JOHNSON”
(CLIENTCUSTOMER))