1. 2.1 Chapter 2: Relational Model

2. 2.2 Chapter 2: Relational Model Structure of Relational Databases Fundamental Relational-Algebra-Operations Additional Relational-Algebra-Operations Extended Relational-Algebra-Operations Null Values Modification of the Database

3. 2.3 Example of a Relation

4. 2.4 Basic Structure Formally, given sets D 1, D 2, …. D n a relation r is a subset of D 1 x D 2 x … x D n (D i is the domain for attribute A i ) Thus, a relation is a set of n-tuples (a 1, a 2, …, a n ) where each a i  D i Example: If customer_name = {Jones, Smith, Curry, Lindsay} customer_street = {Main, North, Park} customer_city = {Harrison, Rye, Pittsfield} Then r = { (Jones, Main, Harrison), (Smith, North, Rye), (Curry, North, Rye), (Lindsay, Park, Pittsfield) } is a relation over customer_name x customer_street x customer_city

5. 2.5 Attribute Types Each attribute of a relation has a name The set of allowed values for each attribute is called the domain of the attribute Attribute values are (normally) required to be atomic; that is, indivisible Note: multivalued attribute values are not atomic Note: composite attribute values are not atomic The special value null is a member of every domain The null value causes complications in the definition of many operations We shall ignore the effect of null values in our main presentation and consider their effect later

6. 2.6 Relation Schema A 1, A 2, …, A n are attributes R = (A 1, A 2, …, A n ) is a relation schema Example: Customer_schema = (customer_name, customer_street, customer_city) r(R) is a relation on the relation schema R Example: customer (Customer_schema)

7. 2.7 Relation Instance The current values (relation instance) of a relation are specified by a table An element t of r is a tuple, represented by a row in a table Jones Smith Curry Lindsay customer_name Main North Park customer_street Harrison Rye Pittsfield customer_city customer attributes (or columns) tuples (or rows)

8. 2.8 Relations are Unordered Order of tuples is irrelevant (tuples may be stored in an arbitrary order) Example: account relation with unordered tuples

9. 2.9 Database A database consists of multiple relations Information about an enterprise is broken up into parts, with each relation storing one part of the information account : stores information about accounts depositor : stores information about which customer owns which account customer : stores information about customers Storing all information as a single relation such as bank(account_number, balance, customer_name,..) results in repetition of information (e.g., two customers own an account) the need for null values (e.g., represent a customer without an account) Normalization theory (Chapter 7) deals with how to design relational schemas

10. 2.10 Keys Let K  R K is a superkey of R if values for K are sufficient to identify a unique tuple of each possible relation r(R) If t 1 and t 2 are in r and t 1  t 2, then t 1 [K]  t 2 [K] Example: {customer_name, customer_street} and {customer_name} are both superkeys of Customer, if no two customers can possibly have the same name. K is a candidate key if K is minimal Example: {customer_name} is a candidate key for Customer, since it is a superkey (assuming no two customers can possibly have the same name), and no subset of it is a superkey. A relation schema R 1 derived from an E-R schema may include the primary key of another relation schema R 2. All attributes in the primary key are called a foreign key of R 1

11. 2.11 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)

12. 2.12 E-R Diagram for the Banking Enterprise

13. 2.13 Schema Diagram for the Banking Enterprise

14. 2.14 The customer Relation

15. 2.15 The depositor Relation

16. 2.16 Query Languages Language in which user requests information from the database. Categories of languages Procedural Non-procedural, or declarative “Pure” languages: Relational algebra Tuple relational calculus Domain relational calculus Pure languages form underlying basis of query languages that people use.

17. 2.17 Relational Algebra Procedural language Six basic operators select:  project:  union:  set difference: – Cartesian product: x rename:  The operators take one or two relations as inputs and produce a new relation as a result.

18. 2.18 Select Operation – Example Relation r ABCD   1 5 12 23 7 3 10   A=B ^ D > 5 (r) ABCD   1 23 7 10

19. 2.19 Select Operation Notation:  p (r) p is called the selection predicate Defined as:  p (r) = {t | t  r and p(t)} Where p is a formula in propositional calculus consisting of terms connected by :  (and),  (or),  (not) Each term is one of: op or where op is one of: =, , >, . <.  Example of selection:  branch_name=“Perryridge” (account)  amount >1200 (loan)  customer-name=banker-name (loan-officer) loan-officer(customer-name, banker-name, loan-number)

20. 2.20 Figure 2.9 Result of Figure 2.9 Result of  branch_name = “Perryridge” (loan)

21. 2.21 Project Operation – Example Relation r: ABC  10 20 30 40 11121112 AC  11121112 = AC  112112  A,C (r)

22. 2.22 Project Operation Notation: 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 Example: To eliminate the branch_name attribute of account  account_number, balance (account) Composition of Relational operations E.g., Find the names of those customers who live in Harrison:  customer (  customer-city=“Harrison ” (customer) )

23. 2.23 Union Operation – Example Relations r, s: r  s: AB  121121 AB  2323 r s AB  12131213

24. 2.24 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 (example: 2 nd column of r deals with the same type of values as does the 2 nd column of s) Example: Find all customers with either an account or a loan  customer_name (depositor)   customer_name (borrower)

25. 2.25 Set Difference Operation – Example Relations r, s: r – s: AB  121121 AB  2323 r s AB  1111

26. 2.26 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 E.g., Find all customers of the bank who have an account but not a loan  customer-name (depositor) –  customer-name (borrower)

27. 2.27 Cartesian-Product Operation – Example Relations r, s: r x s: AB  1212 AB  1111222211112222 CD  10 20 10 20 10 E aabbaabbaabbaabb CD  20 10 E aabbaabb r s

28. 2.28 Cartesian-Product Operation Notation r x s Defined as: r x s = {t q | t  r and q  s} The same attribute name may appear in both r and s. Attach to the attribute the name of the relation from which the attribute originally came. For example, the relation schema for r = borrower x loan is (borrower.customer-name, borrower.loan-number, loan.loan- number, loan.branch-name, loan.amount) We can write the relation schema for r as (customer-name, borrower.loan-number, loan.loan-number, branch-name, amount)

29. 2.29 Composition of Operations Can build expressions using multiple operations Example:  A=C (r x s) r x s  A=C (r x s) AB  1111222211112222 CD  10 20 10 20 10 E aabbaabbaabbaabb ABCDE  122122  20 aabaab

30. 2.30 Composition of Operations Find the names of all customers who have a loan at the Perryridge branch  branch-name = “Perryridge” (borrower x loan)  customer-name (  borrower.loan-number=loan.loan-number ^ branch-name = “Perryridge” (borrower x loan) )

31. 2.31 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.

32. 2.32 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))

33. 2.33 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 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)

34. 2.34 Example Queries Find the names of all customers who have a loan at the Perryridge branch. 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)))

35. 2.35 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)))

36. 2.36 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

37. 2.37 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

38. 2.38 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)

39. 2.39 Set-Intersection Operation – Example Relation r, s: r  s A B  121121  2323 rs  2 Find the names of all customers who have a loan and an account at bank.  customer-name (borrower)   customer-name (depositor)

40. 2.40 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))

41. 2.41 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

42. 2.42 Natural Join Operation – Example Find the names of all customers who have a loan at the bank  customer-name (  borrower.loan-number = loan.loan-number (borrower x loan)))  customer-name (borrower loan) Find the names of all branches with customers who have an account in the bank and who live in “Harrison”  branch-name (  customer-city = “Harrison” (customer account depositor)) Find all customers who have both a loan and an account at the bank  customer-name (borrower depositor))

43. 2.43 Bank Example Queries Find the names of all customers who have a loan and an account at bank.  customer_name (borrower)   customer_name (depositor) Find the name of all customers who have a loan at the bank and the loan amount

44. 2.44 Bank Example Queries Find all customers who have an account from at least the “Downtown” and the Uptown” branches.  customer_name (  branch_name = “Downtown ” (depositor account ))   customer_name (  branch_name = “Uptown ” (depositor account))

45. Database System Concepts, 5 th Ed. ©Silberschatz, Korth and Sudarshan See www.db-book.com for conditions on re-usewww.db-book.com End of Chapter 2