1. Database System Concepts, 5 th Ed. ©Silberschatz, Korth and Sudarshan See www.db-book.com for conditions on re-usewww.db-book.com Chapter 2: Relational Model

2. ©Silberschatz, Korth and Sudarshan2.2Database System Concepts - 5 th Edition, Oct 5, 2006 Examples Consider the following relations: Student(roll, name, address, phone) Course(code, title) Registered(roll, code) Find out the course code in which at least one student is registered π code ( Registered) Find out the titles of registered courses π title (  Course.code = Registered.code (Course X Registered )) Find out the course code in which no student is registered π code ( Course ) - π code ( Registered )

3. ©Silberschatz, Korth and Sudarshan2.3Database System Concepts - 5 th Edition, Oct 5, 2006 Examples Find out the student names and course titles they registered to π name, title (  Student.roll = Registered.roll^ Registered.code=Course.code (Student X Registered X Course)) Consider the following relations: Student(roll, name, address, phone) Course(code, title) Registered(roll, code) Name of student who are registered to ‘Database’ or ‘Algorithms’ course π name (  Student.roll = Registered.roll^ Registered.code=Course.code (Student X Registered X (σ title=’Database’ Course)) ) ∪ π name (  Student.roll = Registered.roll^ Registered.code=Course.code (Student X Registered X (σ title=’Algorithm’ Course)) )

4. ©Silberschatz, Korth and Sudarshan2.4Database System Concepts - 5 th Edition, Oct 5, 2006 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 returns the result of expression E under the name X, and with the attributes renamed to A 1, A 2, …., A n.

5. ©Silberschatz, Korth and Sudarshan2.5Database System Concepts - 5 th Edition, Oct 5, 2006 Banking Example 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)

6. ©Silberschatz, Korth and Sudarshan2.6Database System Concepts - 5 th Edition, Oct 5, 2006 Example Queries Find the largest account balance Strategy:  Find those balances that are not the largest –Rename account relation as d so that we can compare each account balance with all others  Use set difference to find those account balances that were not found in the earlier step. The query is:  balance (account) -  account.balance (  account.balance < d.balance (account x  d (account)))

7. ©Silberschatz, Korth and Sudarshan2.7Database System Concepts - 5 th Edition, Oct 5, 2006 Example Queries Find the names of all customers who have a loan at the Dhanmondi branch. Query 2  customer_name (  loan.loan_number = borrower.loan_number ( (  branch_name = “Dhanmondi ” (loan)) x borrower)) Query 1  customer_name (  branch_name = “Dhanmondi” (  borrower.loan_number = loan.loan_number (borrower x loan)))

8. ©Silberschatz, Korth and Sudarshan2.8Database System Concepts - 5 th Edition, Oct 5, 2006 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

9. ©Silberschatz, Korth and Sudarshan2.9Database System Concepts - 5 th Edition, Oct 5, 2006 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

10. ©Silberschatz, Korth and Sudarshan2.10Database System Concepts - 5 th Edition, Oct 5, 2006 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)

11. ©Silberschatz, Korth and Sudarshan2.11Database System Concepts - 5 th Edition, Oct 5, 2006 Set-Intersection Operation – Example Relation r, s: r  s A B  121121  2323 rs  2

12. ©Silberschatz, Korth and Sudarshan2.12Database System Concepts - 5 th Edition, Oct 5, 2006 Example Query Find the names of all customers who have a loan and an account at the bank.  customer_name (borrower)   customer_name (depositor)

13. ©Silberschatz, Korth and Sudarshan2.13Database System Concepts - 5 th Edition, Oct 5, 2006 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) 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))

14. ©Silberschatz, Korth and Sudarshan2.14Database System Concepts - 5 th Edition, Oct 5, 2006 Natural Join Operation – Example Relations r, s: AB  1241212412 CD  aababaabab B 1312313123 D aaabbaaabb E  r AB  1111211112 CD  aaaabaaaab E  s r s

15. ©Silberschatz, Korth and Sudarshan2.15Database System Concepts - 5 th Edition, Oct 5, 2006 Example Query Find the name and loan amount of all customers who have a loan at the bank.  customer_name, amount (borrower loan)  customer. customer_name (  customer_city = “Jessore ” (customer depositor account )) Find the names of all customers who have an account in the bank and who live in Jessore Natural Join is associative

16. ©Silberschatz, Korth and Sudarshan2.16Database System Concepts - 5 th Edition, Oct 5, 2006 Example Query Find the names of all customers who have a loan and an account at the bank.  customer_name (borrower)   customer_name (depositor)  customer_name (borrower depositor) When R S =R X S ? If R  S= φ

17. ©Silberschatz, Korth and Sudarshan2.17Database System Concepts - 5 th Edition, Oct 5, 2006 Summary

18. ©Silberschatz, Korth and Sudarshan2.18Database System Concepts - 5 th Edition, Oct 5, 2006 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 R = (A 1, …, A m, B 1, …, B n ) S = (B 1, …, B n ) The result of r  s is a relation on schema R – S = (A 1, …, A m ) r  s = { t | t   R-S (r)   u  s ( tu  r ) } Where tu means the concatenation of tuples t and u to produce a single tuple r  s

19. ©Silberschatz, Korth and Sudarshan2.19Database System Concepts - 5 th Edition, Oct 5, 2006 Division Operation – Example Relations r, s: r  s: B A  1212 AB  1231113461212311134612 r s

20. ©Silberschatz, Korth and Sudarshan2.20Database System Concepts - 5 th Edition, Oct 5, 2006 Another Division Example AB  aaaaaaaaaaaaaaaa CD  aabababbaabababb E 1111311111113111 Relations r, s: r  s: D abab E 1111 AB  aaaa C  r s

21. ©Silberschatz, Korth and Sudarshan2.21Database System Concepts - 5 th Edition, Oct 5, 2006 Find all customers who have an account at all branches located in Brooklyn city. Bank Example Queries  customer_name, branch_name (depositor account)   branch_name (  branch_city = “Brooklyn” (branch))

22. ©Silberschatz, Korth and Sudarshan2.22Database System Concepts - 5 th Edition, Oct 5, 2006 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: The result to the right of the  is assigned to the relation variable on the left of the . May use variable in subsequent expressions. r 1    branch_city = “dhaka” (account branch ) r 2   account_number (r 1 )