Basics of Relational Databases and Relational Algebra

1. Chapter 2: Relational Model • Structure of Relational Databases • Relational Algebra

2. Basic Structure • Formally, given sets D1, D2, …. Dn a relation r is a subset of D1 x D2 x … x Dn • Thus, a relation is a set of tuples (a1, a2, …, an) where each ai Di • Example: cust-name = {Jones, Smith, Curry, Lindsay}cust-street = {Main, North, Park}cust-city = {Harrison, Rye, Pittsfield} r = {(Jones, Main, Harrison), (Smith, North, Rye), (Curry, North, Rye), (Lindsay, Park, Pittsfield)} r is a relation over cust-name x cust-street x cust-city

3. Relations are Unordered • In this (mathematical) context, since a relation is a set, the order of tuples is irrelevant and may be thought of as arbitrary. • In a real DBMS, tuple order is typically very important and not arbitrary.

4. Table vs. Relation • The Table: • The Relation: • { (A-101,Downtown,500), (A-102,Perryridge,400), (A-201,Brighton,900), : • (A-305,Round Hill,350) }

5. Attribute Types • Each attribute of a relation has a name. • The set of allowed values for each attribute is called the domain of the attribute. • Attribute values are required to be atomic, that is, indivisible (note how this differs from ER modeling). • Multi-valued attribute values are not atomic. • Composite attribute values are not atomic.

6. The Evil Value “Null” • The special value null is an implicit member of every domain. • Tuples can have a null value for some of their attributes. • A null value can be interpreted in several ways: • Value is unknown • Value does not exist • Value is known and exists, but just hasn’t been entered yet • The null value causes complications in the definition of many operations. • We shall consider their effect later.

7. Relation Schema • Let A1, A2, …, An be attributes. Then R = (A1, A2, …, An ) is a relation schema. Customer-schema = (customer-name, customer-street, customer-city) • Sometimes referred to as a relational schema or relational scheme. • Let R be a relation schema. Then r(R) is a relation variable. customer (Customer-schema)

8. Database • A database consists of multiple relations. account - account informationdepositor - depositor information, i.e., who deposits into which accounts customer - customer information • Storing all information as a single relation is possible, as in: bank(account-number, balance, customer-name, ..) • This results in: • Repetition of information (e.g. two customers own an account) • The need for null values (e.g. represent a customer without an account). • Normalization (Chapter 7) deals with how to design relational schemas.

9. E-R Diagramfor the Banking Enterprise • The (modified) E-R diagram from the banking enterprise:

10. Relational Schemesfor the Banking Enterprise • The following relational schemes result: branch (branch-name, branch-city, assets) customer (customer-name, customer-street, customer-city) account (account-number, branch-name, balance) loan (loan-number, branch-name, amount) depositor (customer-name, account-number) borrower (customer-name, loan-number)

11. Schema Diagramfor the Banking Enterprise

12. Query Languages • Language in which user requests information from the database. • Categories of languages • procedural • non-procedural • “Pure” languages: • Relational Algebra (primarily procedural) • Tuple Relational Calculus (non-procedural) • Domain Relational Calculus (non-procedural) • Pure languages form underlying basis of “real” query languages.

13. Relational Algebra • Procedural language (primarily) • Six basic operators: • select • project • union • set difference • cartesian product • rename

14. Relational Algebra • Each operator takes one or more relations as input and gives a new relation as a result. • Each operation defines: • Requirements or constraints on its’ parameters. • Attributes in the resulting relation, including their types and names. • Which tuples will be included in the result.

15. Select Operation – Example • Relation r A B C D         1 5 12 23 7 7 3 10 • A=B ^ D > 5 (r) A B C D     1 23 7 10

16. Select Operation • Notation: p(r) where p is a selection predicateand r is a relation (or more generally, a relational algebra expression). • Defined as: p(r) = {t | t  rand p(t)} where p is a formula in propositional logic consisting of termsconnected by:  (and),  (or),  (not), and where each term can involve the comparison operators: =, , >, , <, 

17. Select Operation, Cont. • Example of selection:branch-name=“Perryridge”(account) • Logically, one can think of selection as performing a table scan, but technically this may or may not be the case, i.e., an index may be used.

18. Project Operation – Example • Relation r: A B C     10 20 30 40 1 1 1 2 • A,C (r) A C A C     1 1 1 2    1 1 2 =

19. Project Operation • Notation:A1, A2, …, Ak (r) where A1, A2 are attribute names and r is a relation. • The result is defined as the relation of k columns obtained by erasing the columns that are not listed. • Duplicate rows are removed from result, since relations are sets. • Example: To eliminate the branch-name attribute of account:account-number, balance (account) • Note, however, that account is not actually modified.

20. Union Operation – Example • Relations r, s: A B A B    1 2 1   2 3 s r r  s A B     1 2 1 3

21. Union Operation • Notation: r s • Defined as: r s = {t | t  r or t  s} • For r s to be valid: • r,s must have the same arity (same number of attributes) • The attribute domains must be compatible (e.g., 2nd attribute of r deals with “the same type of values” as does the 2nd attribute of s) • Example: find all customers with either an account or a loan customer-name (depositor)  customer-name (borrower)

22. Set Difference Operation • Relations r, s: A B A B    1 2 1   2 3 s r r – s A B   1 1

23. Set Difference Operation, Cont. • Notation r – s • Defined as: r – s = {t | t rand t  s} • Set differences must be taken between compatible relations. • r and s must have the same arity • attribute domains of r and s must be compatible • Note that there is no requirement that the attribute names be the same. • So what about attributes names in the result? • Similarly for union.

24. Cartesian-Product Operation A B C D E • Relations r, s:   1 2     10 10 20 10 a a b b r s r x s: A B C D E         1 1 1 1 2 2 2 2         10 10 20 10 10 10 20 10 a a b b a a b b

25. Cartesian-Product Operation, Cont. • Notation r x s • Defined as: r x s = {tq | t  r and q  s} • The attributes of r and s are assumed to be disjoint, i.e., that R  S = . • If the attributes of r and s are not disjoint, then renaming must be used. • Each attributes name has its originating relations name as a prefix. • If r and s are the same relation, then the rename operation can be used.

26. A B C D E         1 1 1 1 2 2 2 2         10 10 20 10 10 10 20 10 a a b b a a b b A B C D E       10 20 20 a a b 1 2 2 Composition of Operations • Expressions can be built using multiple operations r x s A=C(r x s)

27. Rename Operation • The rename operator allows the results of an expression to be renamed. • Example: x (E) returns the expression E under the name X • If a relational-algebra expression E has arity n, then x(A1, A2, …, An)(E) returns the result of expression E under the name X, and with the attributes renamed to A1, A2, …., An.

28. Formal Definition • A basic expression in relational algebra consists of one of the following: • A relation in the database • A constant relation • Let E1 and E2 be relational-algebra expressions. Then the following are all also relational-algebra expressions: • E1 E2 • E1 - E2 • E1 x E2 • p (E1), P is a predicate on attributes in E1 • s(E1), S is a list consisting of attributes in E1 •  x(E1), x is the new name for the result of E1

29. Banking Example • Recall the relational schemes from the banking enterprise: branch (branch-name, branch-city, assets) customer (customer-name, customer-street, customer-city) account (account-number, branch-name, balance) loan (loan-number, branch-name, amount) depositor (customer-name, account-number) borrower (customer-name, loan-number)

30. Example Queries • Find all loans of over $1200 (a bit ambiguous). • amount > 1200 (loan) • Find the loan number for each loan with an amount greater than $1200. • loan-number (amount> 1200 (loan))

31. Example Queries • Find the names of all customers who have a loan, an account, or both. • customer-name (borrower)  customer-name (depositor) • Find the names of all customers who have a loan and an account. • customer-name (borrower)  customer-name (depositor)

32. Example Queries • Find the names of all customers who have a loan at the Perryridge branch. customer-name (branch-name=“Perryridge” (borrower.loan-number = loan.loan-number(borrower x loan))) • Notes: • The two selections could have been combined into one. • The selection on branch-name could have been applied to loan first, as shown on the next slide…

33. Example Queries • Alternative - Find the names of all customers who have a loan at the Perryridge branch. • customer-name(loan.loan-number = borrower.loan-number( borrower x branch-name = “Perryridge”(loan))) • Notes: • What are the implications of doing the selection first? • How does a non-Perryridge borrower get eliminated? • Couldn’t the amount and the branch-name attributes be eliminated from loan after the selection? • What would be the implications?

34. Example Queries • Find the names of all customers who have a loan at the Perryridge branch but no account at any branch of the bank. customer-name (branch-name = “Perryridge” (borrower.loan-number = loan.loan-number (borrower x loan))) – customer-name(depositor)

35. Example Queries • Find the largest account balance. • Requires a cartesian product between account and itself, resulting in ambiguity of attribute names. • Resolved by renaming one instance of the account relation as d. balance(account)– account.balance(account.balance < d.balance(account x rd(account)))

36. Additional Operations • The following operations do not add any power to relational algebra, but simplify common queries. • Set intersection • Natural join • Theta join • Division • Assignment • All of the above can be defined in terms of the six basic operators.

37. Set-Intersection Operation • Notation: r s • Defined as: rs = { t | trandts } • Assume: • r, s have the same arity • attributes of r and s are compatible • Note: rs = r - (r - s)

38. A B A B    1 2 1   2 3 A B  2 Set-Intersection Operation, Cont. • Relation r, s: • r  s r s

39. Natural-Join Operation • Notation: r s • Let r and s be relations on schemas R and S respectively. • r s is a relation that: • Has all attributes in R S • For each pair of tuples tr and ts from r and s, respectively, if tr and ts have the same value on all attributes in RS, add a tuple t to the result, where: • t has the same value as tr on attributes in R • t has the same value as ts on attributes in S

40. Natural-Join Example • Relational schemes for relations r and s, respectively: R = (A, B, C, D) S = (E, B, D) • Resulting schema for rs : (A, B, C, D, E) • rs is defined as:r.A, r.B, r.C, r.D, s.E (r.B = s.B  r.D = s.D (r x s))

41. Natural Join Example • Relations r, s: • Contents of r s: B D E A B C D 1 3 1 2 3 a a a b b           1 2 4 1 2      a a b a b r s A B C D E      1 1 1 1 2      a a a a b     

42. Natural Join – Another Example • Find the names of all customers who have a loan at the Perryridge branch. Original Expression: Using the Natural Join Operator: • Specifying the join explicitly makes it look nicer, plus it helps the query optimizer. customer-name (branch-name=“Perryridge” (borrower.loan-number = loan.loan-number(borrower x loan))) customer-name(branch-name = “Perryridge”(borrower loan))

43. Theta-Join Operation • Notation: r θ s • Let r and s be relations on schemas R and S respectively. • Then, r θ s is a relation that: • Has all attributes in R S including duplicates. • For each pair of tuples tr and ts from r and s, respectively, if θ evaluates to true for tr and ts, then add a tuple t to the result, where: • t has the same value as tr on attributes in R • t has the same value as ts on attributes in S • rθ s is defined as:θ (r x s)

44. Theta-Join Example • Example: R = (A, B, C, D) S = (E, B, D) • Resulting schema: (r.A, r.B, r.C, r.D, s.E, s.B, s.D)

45. Theta Join – Another Example • Consider the following relational schemes: Exam = (Exam#, Average) Score = (SS#, Exam#, Grade) • Find the SS#s for those students who scored less than average on some exam. Along with the SS# list the exam#, exam average, and the students’ grade. • Score.SS#, Exam.Exam#, Exam.Average, Score.Grade ( • Exam Exam.Exam# = Score.Exam#  Exam.Average > Score.Grade Score)

46. Division Operation • Notation: • Suited to queries that include the phrase “for all” • Let r and s be relations on schemas R and S respectively where S  R, and let the attributes of R and S be: R = (A1, …, Am, B1, …, Bn) S = (B1, …, Bn) The result of r  s is a relation on schema R – S = (A1, …, Am) where: r  s = { t | t   R-S(r)   u  s ( tu  r ) } r  s

47. Division – Example #1 Relations r, s: r  s: A B B A   1 2            1 2 3 1 1 1 3 4 6 1 2 s r

48. Division – Example #2 Relations r, s: r  s: D E A B C A B C D E   a a   a b 1 1         a a a a a a a a         a a b a b a b b 1 1 1 1 3 1 1 1 A B C s r

49. Division – Example #3 Relations r, s: r  s: A B C D E B D A C E         a a a a a a a a         a a b a b a b b 1 1 1 1 3 1 1 1 a a a b     1 1 s r

50. Division Operation (Cont.) • Let r(R) and s(S) be relations, and let S  R. Then: r  s = R-S (r) – R-S ( (R-S(r) x s) – R-S,S(r)) To see why: • R-S,S(r) simply reorders attributes of r • R-S(R-S(r) x s) – R-S,S(r)) gives those tuples t in R-S(r) such that for some tuple u  s, tu  r. • Property: • Let q = r  s • Then q is the largest relation satisfying q x s r