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1 Lecture 07: Relational Algebra

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2 Outline Relational Algebra (Section 6.1)

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3 Relational Algebra Formalism for creating new relations from existing ones Its place in the big picture: Declarative query language Algebra Implementation SQL, relational calculus Relational algebra

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4 Relational Algebra Five operators: –Union: –Difference: - –Selection: –Projection: –Cartesian Product: Derived or auxiliary operators: –Intersection, complement –Joins (natural,equi-join, theta join, semi-join) –Renaming:

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5 1. Union and 2. Difference R1 R2 Example: –ActiveEmployees RetiredEmployees R1 – R2 Example: –AllEmployees − RetiredEmployees

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6 What about Intersection ? It is a derived operator R1 R2 = R1 – (R1 – R2) Also expressed as a join (will see later) Example –UnionizedEmployees RetiredEmployees

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7 3. Selection Returns all tuples which satisfy a condition Notation: c (R) Examples – Salary > (Employee) – name = “Smith” (Employee) The condition c can be =,, , <> [in SQL: SELECT * FROM Employee WHERE Salary > 40000]

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8 Find all employees with salary more than $40,000. Salary > (Employee)

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9 4. Projection Eliminates columns, then removes duplicates Notation: A1,…,An (R) Example: project to social-security number and names: – SSN, Name (Employee) – Output schema: Answer(SSN, Name) [In SQL: SELECT DISTINCT SSN, Name FROM Employee]

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10 SSN, Name (Employee)

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11 5. Cartesian Product Combine each tuple in R1 with each tuple in R2 Notation: R1 R2 Example: –Employee Dependents Very rare in practice; mainly used to express joins [In SQL: SELECT * FROM R1, R2]

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13 Relational Algebra Five operators: –Union: –Difference: - –Selection: –Projection: –Cartesian Product: Derived or auxiliary operators: –Intersection, complement –Joins (natural,equi-join, theta join, semi-join) –Renaming:

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14 Renaming Changes the schema, not the instance Schema: R(A 1, …, A n ) Notation: B1,…,Bn (R) Example: – LastName, SocSocNo (Employee) –Output schema: Answer(LastName, SocSocNo) [in SQL: SELECT Name AS LastName, SSN AS SocSocNo FROM Employee]

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15 Renaming Example Employee NameSSN John Tony LastNameSocSocNo John Tony LastName, SocSocNo (Employee)

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16 Natural Join Notation: R1 ⋈ R2 Meaning: R1 ⋈ R2 = A ( C (R1 R2)) Where: –The selection C checks equality of all common attributes –The projection eliminates the duplicate common attributes [in SQL: SELECT DISTINCT R1.A, R1. B, R2.C FROM R1, R2 WHERE R1.B = R2.B Schema: R1(A,B), R2(B,C)]

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17 Natural Join Example Employee NameSSN John Tony Dependents SSNDname Emily Joe NameSSNDname John Emily Tony Joe Employee Dependents = Name, SSN, Dname ( SSN=SSN2 (Employee x SSN2, Dname (Dependents))

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18 Natural Join R= S= R ⋈ S= AB XY XZ YZ ZV BC ZU VW ZV ABC XZU XZV YZU YZV ZVW

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19 Natural Join Given the schemas R(A, B, C, D), S(A, C, E), what is the schema of R ⋈ S ? Given R(A, B, C), S(D, E), what is R ⋈ S ? Given R(A, B), S(A, B), what is R ⋈ S ?

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20 Theta Join A join that involves a predicate R1 ⋈ R2 = (R1 R2) Here can be any condition

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21 Eq-join A theta join where is an equality R1 ⋈ A=B R2 = A=B (R1 R2) Example: –Employee ⋈ SSN=SSN Dependents Most useful join in practice (difference to natural join?)

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22 Semijoin R ⋉ S = A1,…,An (R ⋈ S) Where A 1, …, A n are the attributes in R Example: –Employee ⋉ Dependents

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23 Semijoins in Distributed Databases Semijoins are used in distributed databases SSNName... SSNDnameAge... Employee Dependents network Employee ⋈ ssn=ssn ( age>71 (Dependents)) T = SSN age>71 (Dependents) R = Employee ⋉ T Answer = R ⋈ Dependents

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24 Complex RA Expressions Person Purchase Person Product name=fred name=gizmo pid ssn seller-ssn=ssnpid=pidbuyer-ssn=ssn name

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25 Application: Query Rewriting for Optimization Reserves Sailors sid=sid bid=100 rating > 5 sname Reserves Sailors sid=sid bid=100 sname rating > 5 (Scan; write to temp T1) (Scan; write to temp T2) The earlier we process selections, less tuples we need to manipulate higher up in the tree (predicate pushdown) Disadvantages?

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26 Algebraic Laws (Examples) Commutative and Associative Laws –R ∩ S = S ∩ R, R ∩ (S ∩ T) = (R ∩ S) ∩ T –R S = S R, R (S T) = (R S) T Laws involving selection – C AND C’ (R) = C ( C’ (R)) = C (R) ∩ C’ (R) – C (R S) = C (R) S When C involves only attributes of R Laws involving projections – M ( N (R)) = M,N (R)

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27 Operations on Bags A bag = a set with repeated elements All operations need to be defined carefully on bags {a,b,b,c} {a,b,b,b,e,f,f}={a,a,b,b,b,b,b,c,e,f,f} {a,b,b,b,c,c} – {b,c,c,c,d} = {a,b,b,d} C (R): preserve the number of occurrences A (R): no duplicate elimination Cartesian product, join: no duplicate elimination Important ! Relational Engines work on bags, not sets !

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28 Finally: RA has Limitations ! Cannot compute “transitive closure” Find all direct and indirect relatives of Fred Cannot express in RA !!! Need to write C program Name1Name2Relationship FredMaryFather MaryJoeCousin MaryBillSpouse NancyLouSister

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29 Formulating queries in RA Consider a database for student enrollment for courses, and books used in the courses –STUDENT (SSN, Name, Major, Bdate) –COURSE (Course#, Cname, Dept) –ENROLL (SSN, Course#, Quarter, Grade) –BOOK_ADOPTION (Course#, Quarter, Book_ISBN) –TEXT (Book_ISBN, Book_Title, Publisher, Author)

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30 Formulating queries in RA Specify the following queries in relational algebra –List the number of courses (Course#) taken by all students named ‘John Smith’ in Winter 1999 (i.e., Quarter = W99) –List any department which has all its adopted books published by ‘BC Publishing’

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31 Formulating Queries in RA Course# ( Quarter=W99 (( Name= ‘John Smith’ (STUDENT) ⋈ ENROLL)) OtherDept = Dept (( Publisher <> ‘PS Publishers’ (BOOK_ADOPTION ⋈ TEXT)) ⋈ COURSE) AllDept = Dept (BOOK_ADOPTION ⋈ COURSE) Answer = AllDept - OtherDept And how will you express it in SQL? WHY?

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