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Lecture No 16 Asma Ahmad Relational Algebra I

Database Systems. Lecture No 16 Asma Ahmad Relational Algebra I. March 10, 2011. Relational Query Languages Formal Relational Query Languages Operations in Relational Algebra Selection Projection Cross Product Renaming Union, Intersection, Set Difference Join and its types Division

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Lecture No 16 Asma Ahmad Relational Algebra I

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  1. Database Systems Lecture No 16Asma AhmadRelational Algebra I March 10, 2011

  2. Relational Query Languages Formal Relational Query Languages Operations in Relational Algebra Selection Projection Cross Product Renaming Union, Intersection, Set Difference Join and its types Division Grouping & Aggregate Functions Examples Summary Relational Algebra

  3. Relational Query Languages • Query languages: allow manipulation and retrievalof 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.

  4. 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 understanding meanings of queries and for representing execution plans. • Relational calculus: lets users describe what they want, rather than how to compute it.

  5. 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

  6. 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.

  7. Result Students Selection • Format: selection-condition(R). Choose tuples that satisfy the selection condition. • Result has identical schema as the input. Major = ‘CS’(Students) • Selection condition is a Boolean expression including =, , <, , >, , and, or, not.

  8. Students Result Projection • Format: attribute-list(R). Retain only those columns in the attribute-list. • Result must eliminate duplicates. Major(Students) • Operations can be composed. Name, GPA(Major = ‘CS’(Students))

  9. 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.

  10. Students  Awards Students Awards Example of Cross Product

  11. Students CS_Students 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))

  12. 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.

  13. TAs  RAs TAs  RAs TAs RAs TAs  RAs Examples of Set Operations

  14. 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.

  15. Students Profs Result Examples of Joins • Theta Join. Students Students.Age<=Profs.Age Profs

  16. Result Result Examples of Joins (cont.) • Equijoin. Students Prof=PID AND Name=Pname Profs • Natural Join. Students Profs

  17. Relational Algebra Defined:Where is it in DBMS? Optimized Relational algebra expression Relational algebra expression Query execution plan Executable code SQL Code generator parser Query optimizer DBMS

  18. 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.

  19. 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.

  20. Students Awards Result Examples of Outer Joins • Left Outer Join. Students Awards

  21. Employee Department Department Brown B Jones C Smith A Head B Black B White Employee Department Head Jones B Black Smith B Black Join: An Observation Some tuples don’t contribute to the result, they get lost.

  22. 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 ...

  23. Employee Department Employee Department Department Brown B Jones C Smith Employee Department Left A Head B Employee Department Head Black B Brown A null White Jones B Black Smith B Black Left Outer Join

  24. Employee Department Employee Department Department Brown B Jones C Smith Employee Department Right A Head B Employee Department Head Black B Jones B Black White Smith B Black null C White Right Outer Join

  25. Employee Department Employee Department Department Brown B Jones C Smith Employee Department Full A Head B Black B White Smith null C B Black White Full Outer Join Employee Department Head Brown A null Jones B Black

  26. 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

  27. Takes CS_Req Result An Example of Division Takes  CS_Req • What is the meaning of this expression?

  28. 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

  29. Example - Division • Identify all clients who have viewed all properties with three rooms. (clientNo, propertyNo(Viewing))  (propertyNo(rooms = 3 (PropertyForRent)))

  30. DIVISION () : QUERIES THAT INCLUDE THEPHRASE “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 APPEAR IN S WITH EVERY BRANCH NAME IN R. CUSTOMER-NAME,BRANCH-NAME (DEPOSIT) BRANCH-NAME(BRANCH-CITY=“BROOKLYN” (BRANCH))

  31. Students Result 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)

  32. Sample Schema for Exercises Student(ID, name, address, GPA, SAT) Campus(location, enrollment, rank) Apply(ID, location, date, major, decision)

  33. name, address (GPA>3.7  decision=‘No’ major=‘CS’ (Student Student.ID=Apply.ID Apply)) name, address ((Students) Students.ID= Not_Apply.ID (Not_Apply(ID (Students) - ID (Apply)))) Sample Queries Find 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.

  34. Another Database Schema 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)

  35. 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 )

  36. 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-NAMECUSTOMER-NAME = BANKER-NAME (CLIENT).

  37. CARTESIAN PRODUCT () : Q. FIND ALL CLIENTS OF BANKER JOHNSON AND THE CITY IN WHICH THEY LIVE. BANKER-NAME=“JOHNSON” (CLIENTCUSTOMER) NOW CLIENT.CUTOMER-NAME COLUMN CONTAINS ONLY CUSTOMERS of Banker JOHNSON. TO FIND ALL THE CLIENTS OF BANKER JOHNSON, WE WRITE CLIENT.CUSTOMER-NAME = CUSTOMER.CUSTOMER- NAME (BANKER-NAME=“JOHNSON” (CLIENTCUSTOMER))

  38. NOW WE WANT ONLY CUSTOMER-NAME AND CUSTOMER-CITY, WE WRITE CLIENT.CUSTOMER-NAME,CUSTOMER-CITY(CLIENT.CUSTOMER-NAME = CUSTOMER.CUSTOMER-NAME (BANKER-NAME = “JOHNSON” (CLIENT*CUSTOMER)))

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