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

©Silberschatz, Korth and Sudarshan3.2Database System Concepts 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

©Silberschatz, Korth and Sudarshan3.3Database System Concepts Projection Operation Given a relation R, the projection operation is used to create a new relation S, such that each tuple t s is formed by taking a tuple t R and removing one or more columns. Formally, the projection of R over columns A 1, A 2, …,A n is defined as:

©Silberschatz, Korth and Sudarshan3.4Database System Concepts Project Operation – Example Relation r: ABC  AC  = AC   A,C (r)

©Silberschatz, Korth and Sudarshan3.5Database System Concepts Project Operation Notation:  A1, A2, …, Ak (r) where A 1, A 2 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)

©Silberschatz, Korth and Sudarshan3.6Database System Concepts Union Operation – Example Relations r, s: r  s: AB  AB  2323 r s AB 

©Silberschatz, Korth and Sudarshan3.7Database System Concepts 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)

©Silberschatz, Korth and Sudarshan3.8Database System Concepts Set Difference Operation – Example Relations r, s: r – s : AB  AB  2323 r s AB  1111

©Silberschatz, Korth and Sudarshan3.9Database System Concepts 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

©Silberschatz, Korth and Sudarshan3.10Database System Concepts Cartesian-Product Operation-Example Relations r, s: r x s: AB  1212 AB  CD  E aabbaabbaabbaabb CD  E aabbaabb r s

©Silberschatz, Korth and Sudarshan3.11Database System Concepts 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.

©Silberschatz, Korth and Sudarshan3.12Database System Concepts Composition of Operations Can build expressions using multiple operations Example:  A=C (r x s) r x s  A=C (r x s) AB  CD  E aabbaabbaabbaabb ABCDE   20 aabaab

©Silberschatz, Korth and Sudarshan3.13Database System Concepts 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 A 1, A2, …., An.

©Silberschatz, Korth and Sudarshan3.14Database System Concepts 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)

©Silberschatz, Korth and Sudarshan3.15Database System Concepts Example Queries Find all loans of over $1200 Find the loan number for each loan of an amount greater than $1200  amount > 1200 (loan)  loan-number (  amount > 1200 (loan))

©Silberschatz, Korth and Sudarshan3.16Database System Concepts Example Queries Find the names of all customers who have a loan, an account, or both, from the bank Find the names of all customers who have a loan and an account at bank.  customer-name (borrower)   customer-name (depositor)  customer-name (borrower)   customer-name (depositor)

©Silberschatz, Korth and Sudarshan3.17Database System Concepts Example Queries Find the names of all customers who have a loan at the Perryridge branch. 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 )  customer-name (  branch-name=“Perryridge ” (  borrower.loan-number = loan.loan-number (borrower x loan)))

©Silberschatz, Korth and Sudarshan3.18Database System Concepts Example Queries Find the names of all customers who have a loan at the Perryridge branch. Two possible solutions follow:  Query 2  customer-name (  loan.loan-number = borrower.loan-number ( (  branch-name = “Perryridge” (loan)) x borrower))  Query 1  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number (borrower x loan)))

©Silberschatz, Korth and Sudarshan3.19Database System Concepts 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)))

©Silberschatz, Korth and Sudarshan3.20Database System Concepts 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 E 1 and E 2 be relational-algebra expressions; the following are all relational-algebra expressions:  E 1  E 2  E 1 - E 2  E 1 x E 2   p (E 1 ), P is a predicate on attributes in E 1   s (E 1 ), S is a list consisting of some of the attributes in E 1   x (E 1 ), x is the new name for the result of E 1

©Silberschatz, Korth and Sudarshan3.21Database System Concepts 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

©Silberschatz, Korth and Sudarshan3.22Database System Concepts 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)

©Silberschatz, Korth and Sudarshan3.23Database System Concepts Set-Intersection Operation - Example Relation r, s: r  s A B   2323 r s  2

©Silberschatz, Korth and Sudarshan3.24Database System Concepts Notation: r s Natural-Join Operation 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 t r from r and t s from s.  If t r and t s 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 r on 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))

©Silberschatz, Korth and Sudarshan3.25Database System Concepts Natural Join Operation – Example Relations r, s: AB  CD  aababaabab B D aaabbaaabb E  r AB  CD  aaaabaaaab E  s r s