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Relational Algebra and SQL

CS 157 Lecture 13. Relational Algebra and SQL. Prof. Sin-Min Lee Department of Computer Science San Jose State University. Functional Dependencies. A form of constraint hence, part of the schema Finding them is part of the database design Also used in normalizing the relations

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Relational Algebra and SQL

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  1. CS 157 Lecture 13 Relational Algebra and SQL Prof. Sin-Min Lee Department of Computer Science San Jose State University

  2. Functional Dependencies • A form of constraint • hence, part of the schema • Finding them is part of the database design • Also used in normalizing the relations • Warning: this is the most abstract, and “hardest” part of the course.

  3. Functional Dependencies Definition: If two tuples agree on the attributes A1, A2, …, An then they must also agree on the attributes B1, B2, …, Bm Formally: A1, A2, …, An B1, B2, …, Bm

  4. Examples • EmpID  Name, Phone, Position • Position  Phone • but Phone  Position EmpID Name Phone Position E0045 Smith 1234 Clerk E1847 John 9876 Salesrep E1111 Smith 9876 Salesrep E9999 Mary 1234 Lawyer

  5. In General • To check A  B, erase all other columns • check if the remaining relation is many-one (called functional in mathematics)

  6. Example EmpID Name Phone Position E0045 Smith 1234 Clerk E1847 John 9876 Salesrep E1111 Smith 9876 Salesrep E9999 Mary 1234 Lawyer Position  Phone

  7. Typical Examples of FDs Product: name  price, manufacturer Person: ssn  name, age Company: name  stockprice, president

  8. Example Product(name, category, color, department, price) Consider these FDs: namecolor categorydepartment color, categoryprice What do they say ?

  9. Example • FD’s are constraints on relations: • On some instances they hold • On others they don’t namecolor categorydepartment color, categoryprice Does this instance satisfy all the FDs ?

  10. Example namecolor categorydepartment color, categoryprice What about this one ?

  11. Example If some FDs are satisfied, thenothers are satisfied too namecolor categorydepartment color, categoryprice If all these FDs are true: name, categoryprice Then this FD also holds: Why ??

  12. Inference Rules for FD’s A1, A2, …, An B1, B2, …, Bm Splitting rule and Combining rule Is equivalent to A1, A2, …, An B1 A1, A2, …, An B2 . . . . . A1, A2, …, An Bm

  13. Inference Rules for FD’s(continued) Trivial Rule A1, A2, …, An Ai where i = 1, 2, ..., n Why ?

  14. Inference Rules for FD’s(continued) Transitive Closure Rule A1, A2, …, An B1, B2, …, Bm If and B1, B2, …, Bm  C1, C2, …, Cp A1, A2, …, An C1, C2, …, Cp then Why ?

  15. Example (continued) 1. namecolor 2. categorydepartment 3. color, categoryprice Start from the following FDs: Infer the following FDs:

  16. Example (continued) 1. namecolor 2. categorydepartment 3. color, categoryprice Answers:

  17. Another Example • Enrollment(student, major, course, room, time) student  major major, course  room course  time What else can we infer ?

  18. Another Rule Augmentation A1, A2, …, An B If then A1, A2, …, An , C1, C2, …, Cp  B Augmentation follows from trivial rules and transitivityHow ?

  19. R(A B C D) 1 2 2 1 Functional Dependency Graph 2 1 2 3 A B C D 1 2 4 2 3 1 2 1 3 1 1 4

  20. Problem: infer ALL FDs Given a set of FDs, infer all possible FDs How to proceed ? • Try all possible FDs, apply all 3 rules • E.g. R(A, B, C, D): how many FDs are possible ? • Drop trivial FDs, drop augmented FDs • Still way too many • Better: use the Closure Algorithm (next)

  21. Relational Algebra • Procedural language • Six basic operators • select • project • union • set difference • Cartesian product • rename • The operators take two or more relations as inputs and give a new relation as a result.

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

  23. Select Operation • Notation: p(r) • p is called the selection predicate • Defined as: p(r) = {t | t  rand p(t)} Where p is a formula in propositional calculus consisting of terms connected by :  (and),  (or),  (not)Each term is one of: <attribute> op <attribute> or <constant> where op is one of: =, , >, . <.  • Example of selection:branch-name=“Perryridge”(account)

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

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

  26. 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

  27. Union Operation • Notation: r s • Defined as: r s = {t | t  r or t  s} • For r s to be valid. 1. r,s must have the same arity (same number of attributes) 2. The attribute domains must be compatible (e.g., 2nd column of r deals with the same type of values as does the 2nd column of s) • E.g. to find all customers with either an account or a loancustomer-name (depositor)  customer-name (borrower)

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

  29. Set Difference Operation • 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

  30. Cartesian-Product Operation-Example A B C D E Relations r, s:   1 2     10 10 20 10 a a b b r s r xs: A B C D E         1 1 1 1 2 2 2 2         10 19 20 10 10 10 20 10 a a b b a a b b

  31. Cartesian-Product Operation • Notation r x s • Defined as: r x s = {t q | t  r and q  s} • Assume that attributes of r(R) and s(S) are disjoint. (That is, R  S = ). • If attributes of r(R) and s(S) are not disjoint, then renaming must be used.

  32. Composition of Operations • Can build expressions using multiple operations • Example: A=C(r x s) • r x s • A=C(r x s) A B C D E         1 1 1 1 2 2 2 2         10 19 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

  33. Rename Operation • Allows us to name, and therefore to refer to, the results of relational-algebra expressions. • Allows us to refer to a relation by more than one name. 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.

  34. Banking Example branch (branch-name, branch-city, assets) customer (customer-name, customer-street, customer-only) account (account-number, branch-name, balance) loan (loan-number, branch-name, amount) depositor (customer-name, account-number) borrower (customer-name, loan-number)

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

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

  37. 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))) • Find the names of all customers who have a loan at the Perryridge branch but do not have an account at any branch of the bank. customer-name (branch-name = “Perryridge” (borrower.loan-number = loan.loan-number(borrower x loan))) – customer-name(depositor)

  38. Example Queries • Find the names of all customers who have a loan at the Perryridge branch. • Query 1customer-name(branch-name = “Perryridge” (borrower.loan-number = loan.loan-number(borrower x loan)))  Query 2 customer-name(loan.loan-number = borrower.loan-number( (branch-name = “Perryridge”(loan)) x borrower) )

  39. Example Queries Find the largest account balance • Rename account relation as d • The query is: balance(account) - account.balance (account.balance < d.balance(account x rd (account)))

  40. Additional Operations We define additional operations that do not add any power to the relational algebra, but that simplify common queries. • Set intersection • Natural join • Division • Assignment

  41. 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)

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

  43. Natural-Join Operation • Notation: r s • Let r and s be relations on schemas R and S respectively.The result is a relation on schema R S which is obtained by considering each pair of tuples tr from r and ts from s. • If tr and ts have the same value on each of the attributes in RS, a tuple t is added to the result, where • t has the same value as tr on r • t has the same value as ts on s • Example: R = (A, B, C, D) S = (E, B, D) • Result schema = (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))

  44. r s Natural Join Operation – Example • Relations 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     

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

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

  47. Another Division Example Relations r, s: A B C D E D 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 b 1 1 s r A B C r s:   a a  

  48. Division Operation (Cont.) • Property • Let q – r  s • Then q is the largest relation satisfying q x s r • Definition in terms of the basic algebra operationLet r(R) and s(S) be relations, and let S  R 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.

  49. Assignment Operation • The assignment operation () provides a convenient way to express complex queries, write query as a sequential program consisting of a series of assignments followed by an expression whose value is displayed as a result of the query. • Assignment must always be made to a temporary relation variable. • Example: Write r  s as temp1 R-S (r)temp2  R-S ((temp1 x s) – R-S,S(r))result = temp1 – temp2 • The result to the right of the  is assigned to the relation variable on the left of the . • May use variable in subsequent expressions.

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