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Relational Algebra Ch. 7.4 – 7.6 John Ortiz

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Lecture 4Relational Algebra2 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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Lecture 4Relational Algebra3 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.

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Lecture 4Relational Algebra4 Preliminaries A query is applied to relation instances, and the result of a query is also a relation instance. Schemas of input & result relations are fixed (determined by relations & query language constructs). A query is specified against schemas (regardless of instances). Attributes may be referenced either by names or by positions (two notation systems).

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Lecture 4Relational Algebra5 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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Lecture 4Relational Algebra6 Selection Format: selection-condition (R). Choose tuples that satisfy the selection condition. Result has identical schema as the input. Major = ‘CS’ (Students) Students Result Selection condition is a Boolean expression including =, ,, , and, or, not.

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Lecture 4Relational Algebra7 Projection Format: attribute-list (R). Retain only those columns in the attribute-list. Result must eliminate duplicates. Major (Students) StudentsResult Operations can be composed. Name, GPA ( Major = ‘CS’ (Students))

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Lecture 4Relational Algebra8 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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Lecture 4Relational Algebra9 Example of Cross Product Students Awards Students Awards

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Lecture 4Relational Algebra10 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)) Students CS_Students

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Lecture 4Relational Algebra11 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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Lecture 4Relational Algebra12 Examples of Set Operations TAs RAs TAs RAs TAs RAs TAs RAs

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Lecture 4Relational Algebra13 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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Lecture 4Relational Algebra14 Examples of Joins Theta Join. Students Students.Age<=Profs.Age Profs Students Profs Result

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Lecture 4Relational Algebra15 Examples of Joins (cont.) Equijoin. Students Prof=PID AND Name=Pname Profs Result Natural Join. Students Profs Result

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Lecture 4Relational Algebra16 Some Questions About Joins * What is the result of R1 R2 if they do not have a common attribute? What is the result of R R? Consider relations Students(SSN, Name, GPA, Major, Age, PSSN) Profs(PSSN, Name, Office, Age, Dept) Which type of join should be used to find pairs of names of students and their advisors? Can a natural join be used? How?

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Lecture 4Relational Algebra17 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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Lecture 4Relational Algebra18 An Example of Division Takes CS_Req Takes Result CS_Req What is the meaning of this expression?

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Lecture 4Relational Algebra19 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) Students Result

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Lecture 4Relational Algebra20 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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Lecture 4Relational Algebra21 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. Can also have Theta Outer Joins.

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Lecture 4Relational Algebra22 Examples of Outer Joins Left Outer Join. Students Awards Students Awards Result

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Lecture 4Relational Algebra23 Relational Algebra Exercises Find the result of these expressions. R S R R.C=S.C S B,E (( B,C R) ( E<7 S)) ( A,B R) - S(A,B) ( D,C S) RS

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Lecture 4Relational Algebra24 Queries In Relational Algebra Consider the following database schema: Students(SSN, Name, GPA, Age, MajorDept) Enrollment(SSN, CourseNo, Grade) Courses(CourseNo, Title, DName) Departments(DName, Location, Phone) Two methods: Use temporary relations. One expression per query.

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Lecture 4Relational Algebra25 Queries In Relational Algebra List student name and course title such that the student has an A in the course and the course is not offered by the student’s major department. Find those students who got an A in any course. Find the department of the students and the courses. Find the final answer.

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Lecture 4Relational Algebra26 Queries In Relational Algebra Find names of students who have the highest GPA. Find students whose GPA are lower than some other students. Find students not in the first group. Find the names of those in the second group.

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Lecture 4Relational Algebra27 Queries In Relational Algebra Find names of students who took every course taken by John Green. Find names of students who never had a grade lower than C. Find names of departments that offer courses taken by both Mary Smith and John Green. Find SSNs and names of students who major in CS and take CS544.

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Lecture 4Relational Algebra28 Queries In Relational Algebra List name and id of CS students along with titles of courses that they take. Students.SSN, Name, Title ( Students.MajorDept = ‘CS’ and Students.SSN = Enrollment.SSN and Enrollment.CourseNo = Courses.CourseNo (Students Enrollment Courses)) = Students.SSN,Name,Title (( MajorDept=‘CS’ (Students) Enrollment) Enrollment.CourseNo=Courses.CourseNo Courses) = Students.SSN,Name,Title ( MajorDept=‘CS’ (Students) Students.SSN=Enrollment.SSN (Enrollment Courses))

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Lecture 4Relational Algebra29 Summary Relational model provides simple yet powerful formal query languages. Relational algebra is procedural and used for internal representation of queries. Several ways to express a given query. DBMS should choose the most efficient plan. Any language able to express all relational algebra queries is relational complete.

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Lecture 4Relational Algebra30 Summary (cont.) Lots useful properties. C1 ( C2 (R)) = C2 ( C1 (R)) = C1 and C2 (R) L1 ( L2 (R)) = L1 (R), if L1 L2 R1 R2 = R2 R1 R1 (R2 R3) = (R1 R2) R3 R1 R2 = R2 R1 R1 (R2 R3) = (R1 R2) R3

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Lecture 4Relational Algebra31 Look Ahead Next topic: Translation form ER/EER to relational model Read from the textbook: Chapter 14.1 – 14.2

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