1. PMIT-6102 Advanced Database Systems By- Jesmin Akhter Assistant Professor, IIT, Jahangirnagar University

2. Lecture 02 Overview of Relational DBMS

3. Outline Overview of Relational DBMS  Structure of Relational Databases  Relational Algebra

4. Why Relational DBMS Most of the distributed database technology has been developed using the relational model  Very simple model.  Often a good match for the way we think about our data. Example of a Relation: account (account-number, branch-name, balance)

5. Relational Design Simplest approach (not always best): convert each Entity Set to a relation and each relationship to a relation. Entity Set  Relation Entity Set attributes become relational attributes. Becomes : account (account-number, branch-name, balance) Slide 5 account account-number balance branch-name

6. Relational Model Table = relation. Column headers = attributes. Row = tuple Relation schema = name(attributes) + other structure info., e.g., keys, other constraints. Example: Account (account-number, branch-name, balance)  Order of attributes is arbitrary, but in practice we need to assume the order given in the relation schema. Relation instance is current set of rows for a relation schema. Database schema = collection of relation schemas. Slide 6 Account

7. 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 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, customer-street, customer-city Slide 7

8. A1 A2 A3... An a1 a2 a3 an b1 b2 a3 cn a1 c3 b3 bn. x1 v2 d3 wn Relational Data Model Set theoretic Domain — set of values like a data type n-tuples (V1,V2,...,Vn) s.t., V1  D1, V2  D2,...,Vn  Dn Tuples = members of a relation inst. Arity = number of domains Components = values in a tuple Domains — corresp. with attributes Cardinality = number of tuples Relation as table Rows = tuples Columns = components Names of columns = attributes Set of attribute names = schema REL (A1,A2,...,An) Arity CardinalityCardinality Attributes Component Tuple Slide 8

9. Name address tel # 5 3 7 Cardinality of domain Domains N A T N1 A1 T1 N2 A2 T2 N3 A3 T3 N4 T4 N5 T5 T6 T7 Relation: Example Domain of Relation N A T N1 A1 T1 N1 A1 T2 N1 A1 T3. N1 A1 T7 N1 A2 T1 N1 A3 T1 N2 A1 T1 Arity3 Cardinality<=5x3x7 of relation Tuple Domain Component Attribute Slide 9

10. 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  E.g. multivalued attribute values are not atomic  E.g. composite attribute values are not atomic The special value null is a member of every domain Slide 10

11. Relation Schema A 1, A 2, …, A n are attributes R = (A 1, A 2, …, A n ) is a relation schema E.g. Customer-schema = (customer-name, customer-street, customer-city) r(R) is a relation on the relation schema R E.g.customer (Customer-schema) Slide 11

12. 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) Slide 12

13. Relations are Unordered Order of tuples is irrelevant (tuples may be stored in an arbitrary order) E.g. account relation with unordered tuples 13

14. 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 E.g.: 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. customer own two account)  the need for null values (e.g. represent a customer without an account) Normalization theory deals with how to design relational schemas Slide 14

15. The customer Relation 15 The branch Relation The depositor Relation Account Relation

16. borrower Relation 16 The Loan Relation Loan-numberBranch-nameamount L-11Round Hill900 L-14Downtown1500 L-15Perryridge1500 L-16Perryridge1300 L-17Downtown1000 L-23Redwood2000 L-93Mianus500

17. E-R Diagram for the Banking Enterprise 17 Total Participation mean Every account must be related via account-branch to some branch Arrow from account-branch to branch mean Each account is for a single branch

18. Keys A super key of an entity set is a set of one or more attributes whose values uniquely determine each entity. A candidate key of an entity set is a minimal super key  Customer-id is candidate key of customer  account-number is candidate key of account Although several candidate keys may exist, one of the candidate keys is selected to be the primary key.

19. Determining Keys from E-R Sets Strong entity set. The primary key of the entity set becomes the primary key of the relation. Weak entity set. The primary key of the relation consists of the union of the primary key of the strong entity set and the discriminator of the weak entity set. Relationship set. The union of the primary keys of the related entity sets becomes a super key of the relation.  For binary many-to-one relationship sets, the primary key of the “many” entity set becomes the relation’s primary key.  For one-to-one relationship sets, the relation’s primary key can be that of either entity set.  For many-to-many relationship sets, the union of the primary keys becomes the relation’s primary key Slide 19

20. Schema Diagram for the Banking Enterprise 20

21. Query Languages Language in which user requests information from the database. Categories of languages  Procedural o User instructs the system to perform a sequence of operations on the database to compute the desired result.  non-procedural o User describes the desired information without giving a specific procedure for obtaining that information. “Pure” languages:  Relational Algebra  Tuple Relational Calculus  Domain Relational Calculus Pure languages form underlying basis of query languages that people use. Slide 21

22. Relational Algebra Procedural language Six basic operators  select  project  union  set difference  Cartesian product  rename The operators take two or more relations as inputs and give a new relation as a result. Slide 22

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

24. 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: =, , >, . <.  Slide 24

25. Example of selection:  branch-name = “Perryridge” (loan) 25  branch-name=“Perryridge” (loan)

26. Project Operation – Example Relation r: ABC  10 20 30 40 11121112 AC  11121112 = AC  112112  A,C (r) Slide 26 Duplicate rows removed

27. 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) Slide 27

28. Union Operation – Example Relations r, s: r  s: AB  121121 AB  2323 r s AB  12131213 Slide 28

29. 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) Slide 29

30. Names of All Customers Who Have Either a Loan or an Account 30  customer-name (depositor)   customer-name (borrower) Union Operation

31. 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 Slide 31

32. Set Difference Operation – Example Relations r, s: r – s : AB  121121 AB  2323 r s AB  1111 Slide 32

33. 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. Slide 33

34. 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 34

35. 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 Slide 35

36. Rename Operation 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. Slide 36

37. 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) Slide 37

38. 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)) Slide 38 Loan-numberBranch-nameamount L-14Downtown1500 L-15Perryridge1500 L-16Perryridge1300 L-23Redwood2000 Loan-number L-14 L-15 L-16 L-23 loan

39. 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) Slide 39

40. 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))) Slide 40

41. Result of borrower  loan 41

42. Result of  branch-name = “Perryridge” (borrower  loan) 42  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number (borrower x loan)))  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number (borrower x loan))) –  customer-name (depositor) Customer-name Adams

43. Customers With An Account But No Loan 43  customer-name (depositor)-  customer-name (borrower)

44. 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))) A.accA.bal ac11000 ac23000 ac32000 d.accd.bal ac11000 ac23000 ac32000 A.accA.bald.accd.bal ac11000ac11000 ac11000ac23000 ac11000ac32000 ac23000ac11000 ac23000ac23000 ac23000ac32000 ac3 2000ac11000 ac3 2000ac23000 ac3 2000ac32000 Slide 44 A.bal 1000 2000 A.bal 1000 3000 2000 A.bal 1000 2000 = A.bal 3000 - (account x  d (account)  account.balance (  account.balance < d.balance (account x  d (account)) Result Duplicate removed

45. Formal Definition 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 Slide 45

46. 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 Slide 46

47. 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) Slide 47

48. Set-Intersection Operation - Example Relation r, s: r  s A B  121121  2323 r s  2 Slide 48

49. 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 o t has the same value as t r on r o 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)) Slide 49

50. 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 Slide 50

51. Result of  customer-name, branch-name (depositor account) 51 Result of  customer-name, loan-number, amount (borrower loan)

52. Division Operation 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 ) } r  s Slide 52

53. Division Operation – Example Relations r, s: r  s:r  s: A B  1212 AB  1231113461212311134612 r s 53

54. Another Division Example AB  aaaaaaaaaaaaaaaa CD  aabababbaabababb E 1111311111113111 Relations r, s: r  s:r  s: D abab E 1111 AB  aaaa C  r s 54

55. Division Operation (Cont.) r  s =  R-S (r) –  R-S ( (  R-S (r) x s) –  R-S,S (r))   R-S,S (r) simply reorders attributes of r   R-S (  R-S (r) x s) –  R-S,S (r)) gives those tuples t in  R-S (r) such that for some tuple u  s, tu  r. Slide 55

56. Assignment Operation The assignment operation (  ) provides a convenient way to express complex queries.  Write query as a sequential program consisting of o a series of assignments o 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: Write r  s as temp1   R-S (r) temp2   R-S ((temp1 x s) –  R-S,S (r)) result = temp1 – temp2  The result to the right of the  is assigned to the relation variable on the left of the .  May use variable in subsequent expressions. Slide 56

57. Example Queries Find all customers who have an account from at least the “Downtown” and the Uptown” branches. where CN denotes customer-name and BN denotes branch-name. Query 1  CN (  BN=“Downtown ” (depositor account))   CN (  BN=“Uptown ” (depositor account)) Slide 57

58. Find all customers who have an account at all branches located in Brooklyn city. Example Queries  customer-name, branch-name (depositor account)   branch-name (  branch-city = “Brooklyn” (branch)) Slide 58 depositor account branch

59. Result of  branch-name (  branch-city = “Brooklyn” (branch)) 59

60. Outer Join An extension of the join operation that avoids loss of information. Computes the join and then adds tuples form one relation that does not match tuples in the other relation to the result of the join. Slide 60

61. Outer Join – Example Relation loan Relation borrower customer-nameloan-number Jones Smith Hayes L-170 L-230 L-155 3000 4000 1700 loan-numberamount L-170 L-230 L-260 branch-name Downtown Redwood Perryridge Slide 61

62. Outer Join – Example Inner Join loan Borrower loan-numberamount L-170 L-230 3000 4000 customer-name Jones Smith branch-name Downtown Redwood Jones Smith null loan-numberamount L-170 L-230 L-260 3000 4000 1700 customer-namebranch-name Downtown Redwood Perryridge Left Outer Join loan Borrower Slide 62

63. Outer Join – Example Right Outer Join loan borrower Full Outer Join loan-numberamount L-170 L-230 L-155 3000 4000 null customer-name Jones Smith Hayes branch-name Downtown Redwood null loan-numberamount L-170 L-230 L-260 L-155 3000 4000 1700 null customer-name Jones Smith null Hayes branch-name Downtown Redwood Perryridge null Slide 63

64. Null Values It is possible for tuples to have a null value, denoted by null, for some of their attributes null signifies an unknown value or that a value does not exist. The result of any arithmetic expression involving null is null. Slide 64

65. Modification of the Database The content of the database may be modified using the following operations:  Deletion  Insertion  Updating All these operations are expressed using the assignment operator. Slide 65

66. Deletion Examples Delete all account records in the Perryridge branch. Delete all loan records with amount in the range of 0 to 50 loan  loan –   amount  0  and amount  50 (loan) account  account –  branch-name = “Perryridge” (account) Slide 66

67. Insertion Examples Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account. account  account  {(“Perryridge”, A-973, 1200)} depositor  depositor  {(“Smith”, A-973)} r 1  (  branch-name = “Perryridge” (borrower loan)) account  account   branch-name, account-number,200 (r 1 ) depositor  depositor   customer-name, loan-number (r 1 ) Slide 67

68. Update Examples Make interest payments by increasing all balances by 5 percent. Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent account   AN, BN, BAL * 1.06 (  BAL  10000 (account))   AN, BN, BAL * 1.05 (  BAL  10000 (account)) account   AN, BN, BAL * 1.05 (account) where AN, BN and BAL stand for account-number, branch-name and balance, respectively. Slide 68

69. Thank You