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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 Domain Relational Calculus Extended Relational-Algebra-Operations Modification of the Database Views Database System Concepts 3.2 ©Silberschatz, Korth and Sudarshan 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: S A1 , A2 ,..., An ( R) {ts ( A1 , A2 ,..., An ) | t R R, t R ( B1 , B2 ,..., Bk ), and { A1 , A2 ,..., An } {B1 , B2 ,..., Bk }} Database System Concepts 3.3 ©Silberschatz, Korth and Sudarshan Project Operation – Example Relation r: A,C (r) Database System Concepts A B C 10 1 20 1 30 1 40 2 A C A C 1 1 1 1 1 2 2 = 3.4 ©Silberschatz, Korth and Sudarshan 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) Database System Concepts 3.5 ©Silberschatz, Korth and Sudarshan Union Operation – Example Relations r, s: A B A B 1 2 2 3 1 s r r s: Database System Concepts A B 1 2 1 3 3.6 ©Silberschatz, Korth and Sudarshan 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) Database System Concepts 3.7 ©Silberschatz, Korth and Sudarshan Set Difference Operation – Example Relations r, s: A B A B 1 2 2 3 1 s r r – s: Database System Concepts A B 1 1 3.8 ©Silberschatz, Korth and Sudarshan Set Difference Operation Notation r – s Defined as: r – s = {t | t r and 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 Database System Concepts 3.9 ©Silberschatz, Korth and Sudarshan Cartesian-Product Operation-Example Relations r, s: A B C D E 1 2 10 10 20 10 a a b b r s r x s: Database System Concepts 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 3.10 ©Silberschatz, Korth and Sudarshan 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. Database System Concepts 3.11 ©Silberschatz, Korth and Sudarshan Composition of Operations Can build expressions using multiple operations Example: A=C(r x s) rxs A=C(r x s) Database System Concepts 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 1 2 2 10 20 20 a a b 3.12 ©Silberschatz, Korth and Sudarshan 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. Database System Concepts 3.13 ©Silberschatz, Korth and Sudarshan 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) Database System Concepts 3.14 ©Silberschatz, Korth and Sudarshan 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)) Database System Concepts 3.15 ©Silberschatz, Korth and Sudarshan 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) Database System Concepts 3.16 ©Silberschatz, Korth and Sudarshan 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) Database System Concepts 3.17 ©Silberschatz, Korth and Sudarshan 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)) Database System Concepts 3.18 ©Silberschatz, Korth and Sudarshan 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 d (account))) Database System Concepts 3.19 ©Silberschatz, Korth and Sudarshan 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 Database System Concepts 3.20 ©Silberschatz, Korth and Sudarshan 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 Database System Concepts 3.21 ©Silberschatz, Korth and Sudarshan Set-Intersection Operation Notation: r s Defined as: r s ={ t | t r and t s } Assume: r, s have the same arity attributes of r and s are compatible Note: r s = r - (r - s) Database System Concepts 3.22 ©Silberschatz, Korth and Sudarshan Set-Intersection Operation - Example Relation r, s: A B 1 2 1 A r rs Database System Concepts A B 2 B 2 3 s 3.23 ©Silberschatz, Korth and Sudarshan 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 t on r r t has the same value as t s on s Example: R = (A, B, C, D) S = (E, B, D) Result schema = (A, B, C, D, E), and R S = (B,D) r s is defined as: r.A, r.B, r.C, r.D, s.E (r.B = s.B r.D = s.D (r x s)) Database System Concepts 3.24 ©Silberschatz, Korth and Sudarshan Natural Join Operation – Example Relations r, s: A B C D B D E 1 2 4 1 2 a a b a b 1 3 1 2 3 a a a b b r r s Database System Concepts s A B C D E 1 1 1 1 2 a a a a b 3.25 ©Silberschatz, Korth and Sudarshan