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ICOM 5016 – Introduction to Database Systems

ICOM 5016 – Introduction to Database Systems. Lecture 6 Dr. Manuel Rodriguez Department of Electrical and Computer Engineering University of Puerto Rico, Mayagüez. Chapter 3: Relational Model. Structure of Relational Databases Relational Algebra Tuple Relational Calculus

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ICOM 5016 – Introduction to Database Systems

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  1. ICOM 5016 – Introduction to Database Systems Lecture 6 Dr. Manuel Rodriguez Department of Electrical and Computer Engineering University of Puerto Rico, Mayagüez

  2. Chapter 3: Relational Model • Structure of Relational Databases • Relational Algebra • Tuple Relational Calculus • Domain Relational Calculus • Extended Relational-Algebra-Operations • Modification of the Database • Views

  3. Projection Operation • Given a relation R, the projection operation is used to create a new relation S, such that each tuple ts is formed by taking a tuple tR and removing one or more columns. • Formally, the projection of R over columns A1, A2, …,An is defined as:

  4. 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 =

  5. 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)

  6. 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

  7. 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)

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

  9. 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

  10. 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 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

  11. 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. • A tuple is r x s is made by concatenating the columns from the first tuple, with the those of the second tuple.

  12. 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 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

  13. 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.

  14. 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)

  15. 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))

  16. 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)

  17. 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)

  18. Example Queries • Find the names of all customers who have a loan at the Perryridge branch. Two possible solutions follow: • 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))

  19. 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)))

  20. Formal Definition • A basic expression in the relational algebra consists of either one of the following: • A relation in the database • A constant relation • Let E1 and E2 be relational-algebra expressions; the following are all 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 some of the attributes in E1 • x(E1), x is the new name for the result of E1

  21. 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

  22. 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)

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

  24. Natural-Join Operation • Notation: r s • Let r and s be relations on schemas R and S respectively. Then, r s is a relation on schema R S obtained as follows: • Consider 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, add a tuple t 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), and R S = (B,D) • 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))

  25. 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     

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