1. ©Silberschatz, Korth and Sudarshan2.1Database System Concepts Chapter 2: Relational Model Structure of Relational Databases Relational Algebra Tuple Relational Calculus Domain Relational Calculus

2. ©Silberschatz, Korth and Sudarshan2.2Database System Concepts 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 tuples (a 1, a 2, …, a n ) where each a i  D i Example: cust-name= {Jones, Smith, Curry, Lindsay} cust-street= {Main, North, Park} cust-city= {Harrison, Rye, Pittsfield} r = {(Jones, Main, Harrison), (Smith, North, Rye), (Curry, North, Rye), (Lindsay, Park, Pittsfield)} r is a relation over cust-name x cust-street x cust-city

3. ©Silberschatz, Korth and Sudarshan2.3Database System Concepts Table vs. Relation The Table: The Relation: { (A-101,Downtown,500), (A-102,Perryridge,400), (A-201,Brighton,900), : (A-305,Round Hill,350) }

4. ©Silberschatz, Korth and Sudarshan2.4Database System Concepts 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 how this differs from ER modeling).  Multi-valued attribute values are not atomic.  Composite attribute values are not atomic.

5. ©Silberschatz, Korth and Sudarshan2.5Database System Concepts The Evil Value “Null” The special value null is an implicit member of every domain. Tuples can have a null value for some of their attributes. A null value can be interpreted in several ways:  Value is unknown  Value does not exist  Value is known and exists, but just hasn’t been entered yet 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. ©Silberschatz, Korth and Sudarshan2.6Database System Concepts Relation Schema Let A 1, A 2, …, A n be attributes. Then R = (A 1, A 2, …, A n ) is a relation schema. Customer-schema = (customer-name, customer-street, customer-city) A relation schema is also sometimes referred to as a relational schema or relational scheme. Let R be a relation schema. Then r(R) is a relation variable. customer (Customer-schema)

7. ©Silberschatz, Korth and Sudarshan2.7Database System Concepts Revisiting Keys Let R be a relation scheme, and let K  R K is a superkey of R if values for K are sufficient to identify a unique tuple for any possible relation r(R) Example: {customer-name, customer-street} {customer-name} are both superkeys of Customer. K is a candidate key if K is minimal Example: {customer-name} is a candidate key for Customer, since it is a superkey, and no subset of it is a superkey.

8. ©Silberschatz, Korth and Sudarshan2.8Database System Concepts Relations are Unordered In this (mathematical) context, since a relation is a set, the order of tuples is irrelevant and may be thought of as arbitrary. In a real DBMS, tuple order is typically very important and not arbitrary.

9. ©Silberschatz, Korth and Sudarshan2.9Database System Concepts Database A database consists of multiple relations. account-account information depositor-depositor information, i.e., who deposits into which accounts customer- customer information Storing all information as a single relation is possible, as in: bank(account-number, balance, customer-name,..) This 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. ©Silberschatz, Korth and Sudarshan2.10Database System Concepts E-R Diagram for the Banking Enterprise The (modified) E-R diagram from the banking enterprise:

11. ©Silberschatz, Korth and Sudarshan2.11Database System Concepts Relational Schemes for the Banking Enterprise The following relational schemes follow: 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)

12. ©Silberschatz, Korth and Sudarshan2.12Database System Concepts Schema Diagram for the Banking Enterprise

13. ©Silberschatz, Korth and Sudarshan2.13Database System Concepts Query Languages Language in which user requests information from the database. Categories of languages  procedural  non-procedural “Pure” languages:  Relational Algebra (primarily procedural)  Tuple Relational Calculus (non-procedural)  Domain Relational Calculus (non-procedural) Pure languages form underlying basis of “real” query languages.

14. ©Silberschatz, Korth and Sudarshan2.14Database System Concepts Relational Algebra Procedural language (primarily) Six basic operators:  select  project  union  set difference  Cartesian product  rename

15. ©Silberschatz, Korth and Sudarshan2.15Database System Concepts Relational Algebra Each operator takes one or more relations as input and gives a new relation as a result. Each operation defines:  Requirements or constraints on its’ parameters.  Attributes in the resulting relation, including their types and names.  Which tuples will be included in the result.

16. ©Silberschatz, Korth and Sudarshan2.16Database System Concepts Select Operation – Example Relation r ABCD   1 5 12 23 7 3 10  A=B ^ D > 5 (r) ABCD   1 23 7 10

17. ©Silberschatz, Korth and Sudarshan2.17Database System Concepts Select Operation Notation:  p (r) p is called the selection predicate r is a relation (or more generally a relational algebra expression). Defined as:  p ( r) = {t | t  r and p(t)} Where p is a formula in propositional logic consisting of terms connected by:  (and),  (or),  (not), and where each term can involve the comparison operators: =, , >, , <, 

18. ©Silberschatz, Korth and Sudarshan2.18Database System Concepts Select Operation, Cont. Example of selection:  branch-name=“Perryridge ” (account) Logically, one can think of selection as performing a table scan, but technically this may or may not be the case, i.e., an index may be used.

19. ©Silberschatz, Korth and Sudarshan2.19Database System Concepts Project Operation – Example Relation r: ABC  10 20 30 40 11121112 AC  11121112 = AC  112112  A,C (r)

20. ©Silberschatz, Korth and Sudarshan2.20Database System Concepts Project Operation Notation:  A1, A2, …, Ak (r) where A 1, A 2 are attribute names and r is a relation. The result is defined as the relation of k columns obtained by erasing the columns that are not listed. Duplicate rows are removed from result, since relations are sets. E.g. To eliminate the branch-name attribute of account:  account-number, balance (account)  Note, however, that account is not actually modified.

21. ©Silberschatz, Korth and Sudarshan2.21Database System Concepts Union Operation – Example Relations r, s: r  s: AB  121121 AB  2323 r s AB  12131213

22. ©Silberschatz, Korth and Sudarshan2.22Database 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 attribute of r deals with “the same type of values” as does the 2nd attribute of s) Example: find all customers with either an account or a loan  customer-name (depositor)   customer-name (borrower)

23. ©Silberschatz, Korth and Sudarshan2.23Database System Concepts Set Difference Operation Relations r, s: r – s : AB  121121 AB  2323 r s AB  1111

24. ©Silberschatz, Korth and Sudarshan2.24Database System Concepts Set Difference Operation, Cont. 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 What about attributes names in r, s and the result? Similarly for union.

25. ©Silberschatz, Korth and Sudarshan2.25Database System Concepts Cartesian-Product Operation 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

26. ©Silberschatz, Korth and Sudarshan2.26Database System Concepts Cartesian-Product Operation, Cont. Notation r x s Defined as: r x s = {tq | t  r and q  s} Assume that attributes of r and s are disjoint. (That is, R  S =  ). If the attributes of r and s are not disjoint, then renaming must be used.  Each attribute’s name has its’ originating relation name as a prefix.  If r and s are the same relation, then the rename operation can be used.

27. ©Silberschatz, Korth and Sudarshan2.27Database 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  1111222211112222 CD  10 20 10 20 10 E aabbaabbaabbaabb ABCDE  122122  20 aabaab

28. ©Silberschatz, Korth and Sudarshan2.28Database System Concepts Rename Operation Allows the results of a relational-algebra expression to be renamed. 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.

29. ©Silberschatz, Korth and Sudarshan2.29Database System Concepts Banking Example Recall the relational schemes from the banking enterprise: 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)

30. ©Silberschatz, Korth and Sudarshan2.30Database System Concepts Example Queries Find all loans of over $1200 (a bit ambiguous). Find the loan number for each loan with an amount greater than $1200.  amount > 1200 (loan)  loan-number (  amount > 1200 (loan))

31. ©Silberschatz, Korth and Sudarshan2.31Database System Concepts Example Queries Find the names of all customers who have a loan, an account, or both.  customer-name (borrower)   customer-name (depositor)  customer-name (borrower)   customer-name (depositor) Find the names of all customers who have a loan and an account.

32. ©Silberschatz, Korth and Sudarshan2.32Database System Concepts 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))) Notes:  The two selections could have been combined into one.  The selection on branch-name could have been applied to loan first (what would be the implications?)

33. ©Silberschatz, Korth and Sudarshan2.33Database System Concepts Example Queries Alternative - Find the names of all customers who have a loan at the Perryridge branch.  customer-name (  loan.loan-number = borrower.loan-number ( borrower x  branch-name = “Perryridge” (loan))) Notes:  How does a non-Perryridge borrower get eliminated?  Couldn’t the amount and the branch-name attributes be eliminated from loan after the selection?  What would be the implications?

34. ©Silberschatz, Korth and Sudarshan2.34Database System Concepts Example Queries Find the names of all customers who have a loan at the Perryridge branch but no account at any branch of the bank.  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number (borrower x loan))) –  customer-name (depositor)

35. ©Silberschatz, Korth and Sudarshan2.35Database System Concepts Example Queries Find the largest account balance. Requires a cartesian product between account and itself, resulting in ambiguity of attribute names. Resolved by renaming one instance of the account relation as d.  balance (account) -  account.balance (  account.balance < d.balance (account x  d (account)))

36. ©Silberschatz, Korth and Sudarshan2.36Database System Concepts Formal Definition A basic expression in relational algebra consists of one of the following:  A relation in the database  A constant relation Let E 1 and E 2 be relational-algebra expressions. Then the following are all also 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 attributes in E 1   x (E 1 ), x is the new name for the result of E 1

37. ©Silberschatz, Korth and Sudarshan2.37Database System Concepts Additional Operations The following operations do not add any power to relational algebra, but simplify common queries.  Set intersection  Natural join  Theta join  Division  Assignment

38. ©Silberschatz, Korth and Sudarshan2.38Database 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)

39. ©Silberschatz, Korth and Sudarshan2.39Database System Concepts Set-Intersection Operation, Cont. Relation r, s: r  s A B  121121  2323 r s  2

40. ©Silberschatz, Korth and Sudarshan2.40Database System Concepts Natural-Join Operation Notation: r s Let r and s be relations on schemas R and S respectively. Then, r s is a relation that:  Has all attributes in R  S  Contains the following tuples: For each pair of tuples t r and t s from r and s, respectively: –If t r and t s have the same value on all attributes in R  S, add a tuple t to the result, where: »t has the same value as t r on attributes in R »t has the same value as t s on attributes in S

41. ©Silberschatz, Korth and Sudarshan2.41Database System Concepts Natural-Join Example Relational schemes for relations r and s, respectively: R = (A, B, C, D) S = (E, B, D) Resulting schema for r s : (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))

42. ©Silberschatz, Korth and Sudarshan2.42Database System Concepts Natural Join Operation Relations r, s: Contents of r s: AB  1241212412 CD  aababaabab B 1312313123 D aaabbaaabb E  r AB  1111211112 CD  aaaabaaaab E  s

43. ©Silberschatz, Korth and Sudarshan2.43Database System Concepts Natural Join – Another Example Find the names of all customers who have a loan at the Perryridge branch. Original Expression: Using the Natural Join Operator: Note that the selection still could have been applied to loan first.  customer-name (  branch-name=“Perryridge” (  borrower.loan-number = loan.loan-number (borrower x loan)))  customer-name (  branch-name = “Perryridge” (borrower loan))

44. ©Silberschatz, Korth and Sudarshan2.44Database System Concepts Theta-Join Operation Notation: r θ s Let r and s be relations on schemas R and S respectively. Then, r θ s is a relation that:  Has all attributes in R  S including duplicates.  Contains the following tuples: For each pair of tuples t r and t s from r and s, respectively: –If θ evaluates to true for t r and t s, add a tuple t to the result, where: »t has the same value as t r on attributes in R »t has the same value as t s on attributes in S

45. ©Silberschatz, Korth and Sudarshan2.45Database System Concepts Theta-Join Example Example: R = (A, B, C, D) S = (E, B, D) Resulting schema: (r.A, r.B, r.C, r.D, s.E, s.B, s.D) r θ s is defined as:  θ (r x s))

46. ©Silberschatz, Korth and Sudarshan2.46Database System Concepts Theta Join – Another Example Consider the following relational schemes: Exam = (Exam#, Average) Score = (SS#, Exam#, Grade) Find the SS#s for those students who scored less than average on some exam. Along with the SS# list the exam#, exam average, and the students’ grade.  Score.SS#, Exam.Exam#, Exam.Average, Score.Grade ( Exam Exam.Exam# = Score.Exam#  Exam.Average > Score.Grade Score)

47. ©Silberschatz, Korth and Sudarshan2.47Database System Concepts 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 S  R, and let the attributes of R and S be: 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 ) where: r  s = { t | t   R-S (r)   u  s ( tu  r ) } r  s

48. ©Silberschatz, Korth and Sudarshan2.48Database System Concepts Division Operation – Example Relations r, s: r  s:r  s: A B  1212 AB  1231113461212311134612 r s

49. ©Silberschatz, Korth and Sudarshan2.49Database System Concepts 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

50. ©Silberschatz, Korth and Sudarshan2.50Database System Concepts Division Operation (Cont.) Definition in terms of the basic algebra operation Let r(R) and s(S) be relations, and let S  R. Then: r  s =  R-S (r) –  R-S ( (  R-S (r) x s) –  R-S,S (r)) To see why:   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. Property:  Let q = r  s  Then q is the largest relation satisfying q x s  r

51. ©Silberschatz, Korth and Sudarshan2.51Database System Concepts Assignment Operation The assignment operation (  ) provides a convenient way to express complex queries.  A query is written as a sequence of assignments.  Assignment is always 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

52. ©Silberschatz, Korth and Sudarshan2.52Database System Concepts 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)) Query 2  customer-name, branch-name (depositor account)   temp(branch-name ) ({(“Downtown”), (“Uptown”)})

53. ©Silberschatz, Korth and Sudarshan2.53Database System Concepts 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))

54. ©Silberschatz, Korth and Sudarshan2.54Database System Concepts Extended Relational Algebra Operations Generalized Projection Outer Join Aggregate Functions

55. ©Silberschatz, Korth and Sudarshan2.55Database System Concepts Generalized Projection Extends the projection operation by allowing arithmetic functions to be used in the projection list.  F1, F2, …, Fn (E) E is any relational-algebra expression Each of F 1, F 2, …, F n are arithmetic expressions involving constants and attributes in the schema of E. Given relation credit-info(customer-name, limit, credit-balance), find how much more each person can spend:  customer-name, limit – credit-balance (credit-info)

56. ©Silberschatz, Korth and Sudarshan2.56Database System Concepts Aggregate Functions An aggregation function takes a collection of values and returns a single value as a result. avg- average value min- minimum value max- maximum value sum- sum of values count- number of values

57. ©Silberschatz, Korth and Sudarshan2.57Database System Concepts The Aggregate Operator Aggregation functions are specified in relational algebra using the aggregate operator: G1, G2, …, Gn g F1( A1), F2( A2),…, Fn( An) (E)  E is any relational-algebra expression.  G 1, G 2 …, G n is a list of attributes on which to group (can be empty) Similar to sorting.  Each F i is an aggregate function.  Each A i is an attribute name.

58. ©Silberschatz, Korth and Sudarshan2.58Database System Concepts Aggregate Operation – Example Relation r: AB   C 7 3 10 g sum(c) (r) sum-C 27 Could add min, max, and other aggregates to the above expression as well.

59. ©Silberschatz, Korth and Sudarshan2.59Database System Concepts Aggregate Operation – Example Grouping is somewhat like sorting, although not identical. Relation account grouped by branch-name: branch-name g sum(balance) (account) branch-nameaccount-numberbalance Perryridge Brighton Redwood A-102 A-201 A-217 A-215 A-222 400 900 750 700 branch-namebalance Perryridge Brighton Redwood 1300 1500 700

60. ©Silberschatz, Korth and Sudarshan2.60Database System Concepts Aggregate Functions (Cont.) Consider the following relational scheme: History = (Student-Name, Department, Course-Number, Grade) Give a list of student names, and for each name list the average course grade for each department in which the student has taken classes. SmithCSE87 SmithMTH93 JonesCHM88 JonesCSE75 BrownPSY97 : Adding count(Course-Number) would tell how many courses the student had in each department. Similarly min and could be added. student-name, department g avg(grade) (History)

61. ©Silberschatz, Korth and Sudarshan2.61Database System Concepts Aggregate Functions (Cont.) Result of an aggregate function does not have a name:  The rename operation can be used to give it a name.  For convenience, we permit renaming as part of aggregate operation. branch-name g sum(balance) as sum-balance (account)

62. ©Silberschatz, Korth and Sudarshan2.62Database System Concepts Aggregate Functions and Null Values Null values are controversial. Various proposals exist throughout the research literature on whether null values should be allowed and, if so, how they should affect operations.  See www.dbazine.com/ofinterest/oi-articles/pascal27 for example.www.dbazine.com/ofinterest/oi-articles/pascal27 Null values can frequently be eliminated through normalization and decomposition.

63. ©Silberschatz, Korth and Sudarshan2.63Database System Concepts Aggregate Functions and Null Values For duplicate elimination and grouping, null is treated like any other value, i.e., two nulls are assumed to be the same.  This is arbitrary, but consistent with SQL.  Alternative - assume each null is different from every other. Aggregate functions (except for count) simply ignore null values.  This behavior is also arbitrary (i.e., could return null as a result instead), but consistent with the semantics of SQL. Note how this can be misleading:  Is avg(grade) actually a class average?

64. ©Silberschatz, Korth and Sudarshan2.64Database System Concepts Null Values and Expression Evaluation The result of any arithmetic expression involving null is null. Comparisons with null returns the special truth value unknown. Three-valued logic using the truth value unknown:  OR: (unknown or true)= true (unknown or false) = unknown (unknown or unknown) = unknown  AND: (true and unknown) = unknown (false and unknown) = false (unknown and unknown) = unknown  NOT: (not unknown) = unknown Value of a predicate is treated as false if it evaluates to unknown.

65. ©Silberschatz, Korth and Sudarshan2.65Database System Concepts Null Values and Expression Evaluation, Cont. Why doesn’t a comparison with null simply result in false?  If false was used instead of unknown, then: not (A < 5) would not be equivalent to: A >= 5 Why would this be a problem? How does a comparison with null resulting in unknown help?  Consider not (A = 5?

66. ©Silberschatz, Korth and Sudarshan2.66Database System Concepts Outer Join An extension of the join operation that avoids loss of information. Computes the join and then adds tuples from one relation that do not match tuples in the other relation. Typically introduces null values.

67. ©Silberschatz, Korth and Sudarshan2.67Database System Concepts 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

68. ©Silberschatz, Korth and Sudarshan2.68Database System Concepts Outer Join – Example Inner Join loan Borrower loan-numberamount L-170 L-230 3000 4000 customer-name Jones Smith branch-name Downtown Redwood Left Outer Join loan Borrower Jones Smith null loan-numberamount L-170 L-230 L-260 3000 4000 1700 customer-namebranch-name Downtown Redwood Perryridge

69. ©Silberschatz, Korth and Sudarshan2.69Database System Concepts 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

70. ©Silberschatz, Korth and Sudarshan2.70Database System Concepts Example Left-Outer Join Consider the following relational schemes: Student = (SS#, Address, Date-of-Birth) Grade-Point-Average = (SS#, GPA) Consider the following query: “Create a list of all student SS#’s and their GPAs. Be sure to include all students, including first semester freshman, who do not have a GPA.” Solution: Student Grade-Point-Average

71. ©Silberschatz, Korth and Sudarshan2.71Database System Concepts Modification of the Database The database contents can be modified with operations:  Deletion  Insertion  Updating All these operations are expressed using the assignment operator.

72. ©Silberschatz, Korth and Sudarshan2.72Database System Concepts Deletion A deletion is expressed in relational algebra by: r  r – E where r is a relation and E is a relational algebra query. Only whole tuples can be deleted, not specific attribute values.

73. ©Silberschatz, Korth and Sudarshan2.73Database System Concepts Deletion Examples Delete all account records with a branch name equal to Perryridge. Delete all accounts at branches located in Needham. r 1    branch-city = “Needham” (account branch) r 2   account-number, branch-name, balance (r 1 ) r 3   customer-name, account-number (depositor r 2 ) account  account – r 2 depositor  depositor – r 3 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)

74. ©Silberschatz, Korth and Sudarshan2.74Database System Concepts An Alternative Version Delete all accounts at branches located in Needham. Which version is preferable? r 1   branch-name (   branch-city = “Needham” (branch)) r 2   account-number (  account-number, branch-name (account) r 1 ) account  account – (account r 2 ) depositor  depositor – (depositor r 3 )

75. ©Silberschatz, Korth and Sudarshan2.75Database System Concepts Insertion To insert data into a relation, we specify either:  A single tuple, or  An expression whose result is a set of tuples. In relational algebra, an insertion is expressed by: r  r  E where r is a relation and E is a relational algebra expression. The insertion of a single tuple is expressed by letting E be a constant relation containing one tuple.

76. ©Silberschatz, Korth and Sudarshan2.76Database System Concepts 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 at the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account. Is r 1 overly complex? account  account  {(A-973, “Perryridge”, 1200)} depositor  depositor  {(“Smith”, A-973)} r 1  (  branch-name = “Perryridge” (borrower loan)) account  account   loan-number, branch-name,200 (r 1 ) depositor  depositor   customer-name, loan-number (r 1 )

77. ©Silberschatz, Korth and Sudarshan2.77Database System Concepts Updating Generalized projection can be used to change a value in a tuple without changing all values in the tuple. r   F1, F2, …, FI, (r) Each F i is either:  The ith attribute of r, if the ith attribute is not updated, or,  If the attribute is to be updated F i is an expression, involving only constants and attributes of r, which gives the new value for the attribute.

78. ©Silberschatz, Korth and Sudarshan2.78Database System Concepts Update Examples Make interest payments by increasing all balances by 5 percent. Pay 6 percent interest to all accounts with balances over $10,000 and pay 5 percent interest to all others. 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.

79. ©Silberschatz, Korth and Sudarshan2.79Database System Concepts Views Views are very important, but we will not consider them until chapter 3.

80. ©Silberschatz, Korth and Sudarshan2.80Database System Concepts Tuple Relational Calculus A nonprocedural query language, where each query is of the form { t | P (t) }  read as “the set of all tuples t such that predicate P is true for t”  P is a formula similar to that of the predicate calculus

81. ©Silberschatz, Korth and Sudarshan2.81Database System Concepts Predicate Calculus Formula The predicate P(t) will contain:  Attribute, tuple and relation variables and constants If t is a tuple variable, t[A] denotes the value of t on attribute A t  r denotes that tuple t is in relation r  Comparison operators: (e.g., , , , , ,  )  Connectives: and (  ), or (v)‚ not (  )  Implication (  ): x  y, if x if true, then y is true x  y  x v y  Quantifiers:  t  r (Q(t))  ”there exists” a tuple t in relation r such that Q(t) is true  t  r (Q(t))  Q(t) is true “for all” tuples t in relation r => See the book for a more complete and precise definition (pages 166 & 167)

82. ©Silberschatz, Korth and Sudarshan2.82Database System Concepts 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)

83. ©Silberschatz, Korth and Sudarshan2.83Database System Concepts Example Queries Find the loan-number, branch-name, and amount for loans of over $1200 How about the following? {t |  s  loan (t = s  s [amount]  1200)} {t | t  loan  t [amount]  1200} {t | t [amount]  1200}

84. ©Silberschatz, Korth and Sudarshan2.84Database System Concepts Example Queries Find the loan number for each loan having an amount greater than $1200 Why not the following? Notice that a relation on schema [loan-number] is implicitly defined by the query {t |  s  loan (t[loan-number] = s[loan-number]  s[amount]  1200)} {t | t  loan  t [amount]  1200}

85. ©Silberschatz, Korth and Sudarshan2.85Database System Concepts Example Queries Find the names of all customers having a loan, an account, or both at the bank {t |  s  borrower( t[customer-name] = s[customer-name])   u  depositor( t[customer-name] = u[customer-name])} Find the names of all customers who have a loan and an account at the bank {t |  s  borrower( t[customer-name] = s[customer-name])   u  depositor( t[customer-name] = u[customer-name])}

86. ©Silberschatz, Korth and Sudarshan2.86Database System Concepts Example Queries Find the names of all customers having a loan at the Perryridge branch {t |  s  borrower( t[customer-name] = s[customer-name]   u  loan(u[branch-name] = “Perryridge”  u[loan-number] = s[loan-number]))  not  v  depositor (v[customer-name] = t[customer-name]) } Find the names of all customers who have a loan at the Perryridge branch, but no account at any branch of the bank {t |  s  borrower(t[customer-name] = s[customer-name]   u  loan(u[branch-name] = “Perryridge”  u[loan-number] = s[loan-number]))}

87. ©Silberschatz, Korth and Sudarshan2.87Database System Concepts Example Queries Find the names of customers and their cities of residence for those customer having a loan from the Perryridge branch. Note that the above contains a parenthetical mistake. {t |  s  loan(s[branch-name] = “Perryridge”   u  borrower (u[loan-number] = s[loan-number]  t [customer-name] = u[customer-name])   v  customer (u[customer-name] = v[customer-name]  t[customer-city] = v[customer-city])))}

88. ©Silberschatz, Korth and Sudarshan2.88Database System Concepts Safety of Expressions It is possible to write tuple calculus expressions that generate infinite relations.  For example, {t |  t  r} results in an infinite relation if the domain of any attribute of relation r is infinite To guard against the problem, we restrict the set of allowable expressions to what are called “safe” expressions. An expression {t | P(t)} in the tuple relational calculus is safe if every component of t (i.e., the result) appears in one of the relations, tuples, or constants that appear in P  NOTE: this is more than just a syntax condition.  Example: { t | t[A]=5  true } is not safe --- it defines an infinite set with attribute values that do not appear in any relation or tuples or constants in P.

89. ©Silberschatz, Korth and Sudarshan2.89Database System Concepts Example Queries Find the names of all customers who have an account at all branches located in Brooklyn: Is the first term in the predicate necessary?  Consider a branch relation that consists of no Brooklyn branches. {t |  c  customer (t[customer.name] = c[customer-name])   s  branch(s[branch-city] = “Brooklyn”   u  account(s[branch-name] = u[branch-name]   v  depositor(v[account-number] = u[account-number]  t[customer-name] = v[customer-name])))}

90. ©Silberschatz, Korth and Sudarshan2.90Database System Concepts Domain Relational Calculus A nonprocedural query language equivalent in power to the tuple relational calculus Each query is an expression of the form: {  x 1, x 2, …, x n  | P(x 1, x 2, …, x n )}  x 1, x 2, …, x n represent domain variables  P represents a formula similar to that of the predicate calculus

91. ©Silberschatz, Korth and Sudarshan2.91Database System Concepts Example Queries Find the loan-number, branch-name, and amount for loans of over $1200 {  c, a  |  l (  c, l   borrower   b(  l, b, a   loan  b = “Perryridge”))} or {  c, a  |  l (  c, l   borrower   l, “Perryridge”, a   loan)} Find the names of all customers who have a loan from the Perryridge branch and the loan amount: {  c  |  l, b, a (  c, l   borrower   l, b, a   loan  a > 1200)} Find the names of all customers who have a loan of over $1200 {  l, b, a  |  l, b, a   loan  a > 1200}

92. ©Silberschatz, Korth and Sudarshan2.92Database System Concepts Example Queries Find the names of all customers having a loan, an account, or both at the Perryridge branch: {  c  |  s, n (  c, s, n   customer)   x,y,z(  x, y, z   branch  y = “Brooklyn”)   a,b(  x, y, z   account   c,a   depositor)} Find the names of all customers who have an account at all branches located in Brooklyn: {  c  |  l ({  c, l   borrower   b,a(  l, b, a   loan  b = “Perryridge”))   a(  c, a   depositor   b,n(  a, b, n   account  b = “Perryridge”))}

93. ©Silberschatz, Korth and Sudarshan2.93Database System Concepts Safety of Expressions {  x 1, x 2, …, x n  | P(x 1, x 2, …, x n )} is safe if all of the following hold: 1.All values that appear in tuples of the expression are values from dom(P) (that is, the values appear either in P or in a tuple of a relation mentioned in P). 2.For every “there exists” subformula of the form  x (P 1 (x)), the subformula is true if and only if there is a value of x in dom(P 1 ) such that P 1 (x) is true. 3. For every “for all” subformula of the form  x (P 1 (x)), the subformula is true if and only if P 1 (x) is true for all values x from dom (P 1 ).

94. End of Chapter 3

95. ©Silberschatz, Korth and Sudarshan2.95Database System Concepts Views Views are very important, but we will not consider them until chapter 4, so goto slide 88. In some cases, it is not desirable for all users to see the entire logical model (i.e., all the actual relations stored in the database.) Consider a person who needs to know a customer’s loan number but has no need to see the loan amount. This person should see a relation described, in the relational algebra, by  customer-name, loan-number (borrower loan) Any relation that is not of the conceptual model but is made visible to a user as a “virtual relation” is called a view.

96. ©Silberschatz, Korth and Sudarshan2.96Database System Concepts View Definition A view is defined using the create view statement which has the form create view v as <query expression where is any legal relational algebra query expression. The view name is represented by v. Once a view is defined, the view name can be used to refer to the virtual relation that the view generates. View definition is not the same as creating a new relation by evaluating the query expression  Rather, a view definition causes the saving of an expression; the expression is substituted into queries using the view.

97. ©Silberschatz, Korth and Sudarshan2.97Database System Concepts View Examples Consider the view (named all-customer) consisting of branches and their customers. We can find all customers of the Perryridge branch by writing: create view all-customer as  branch-name, customer-name (depositor account)   branch-name, customer-name (borrower loan)  branch-name (  branch-name = “Perryridge” (all-customer))

98. ©Silberschatz, Korth and Sudarshan2.98Database System Concepts Updates Through View Database modifications expressed as views must be translated to modifications of the actual relations in the database. Consider the person who needs to see all loan data in the loan relation except amount. The view given to the person, branch- loan, is defined as: create view branch-loan as  branch-name, loan-number (loan) Since we allow a view name to appear wherever a relation name is allowed, the person may write: branch-loan  branch-loan  {(“Perryridge”, L-37)}

99. ©Silberschatz, Korth and Sudarshan2.99Database System Concepts Updates Through Views (Cont.) The previous insertion must be represented by an insertion into the actual relation loan from which the view branch-loan is constructed. An insertion into loan requires a value for amount. The insertion can be dealt with by either.  rejecting the insertion and returning an error message to the user.  inserting a tuple (“L-37”, “Perryridge”, null) into the loan relation Some updates through views are impossible to translate into database relation updates  create view v as  branch-name = “Perryridge” (account)) v  v  (L-99, Downtown, 23) Others cannot be translated uniquely  all-customer  all-customer  {(“Perryridge”, “John”)} Have to choose loan or account, and create a new loan/account number!

100. ©Silberschatz, Korth and Sudarshan2.100Database System Concepts Views Defined Using Other Views One view may be used in the expression defining another view A view relation v 1 is said to depend directly on a view relation v 2 if v 2 is used in the expression defining v 1 A view relation v 1 is said to depend on view relation v 2 if either v 1 depends directly to v 2 or there is a path of dependencies from v 1 to v 2 A view relation v is said to be recursive if it depends on itself.

101. ©Silberschatz, Korth and Sudarshan2.101Database System Concepts View Expansion A way to define the meaning of views defined in terms of other views. Let view v 1 be defined by an expression e 1 that may itself contain uses of view relations. View expansion of an expression repeats the following replacement step: repeat Find any view relation v i in e 1 Replace the view relation v i by the expression defining v i until no more view relations are present in e 1 As long as the view definitions are not recursive, this loop will terminate

102. ©Silberschatz, Korth and Sudarshan2.102Database System Concepts Result of  branch-name = “Perryridge” (loan)

103. ©Silberschatz, Korth and Sudarshan2.103Database System Concepts Loan Number and the Amount of the Loan

104. ©Silberschatz, Korth and Sudarshan2.104Database System Concepts Names of All Customers Who Have Either a Loan or an Account

105. ©Silberschatz, Korth and Sudarshan2.105Database System Concepts Customers With An Account But No Loan

106. ©Silberschatz, Korth and Sudarshan2.106Database System Concepts Result of borrower  loan

107. ©Silberschatz, Korth and Sudarshan2.107Database System Concepts Result of  branch-name = “Perryridge” (borrower  loan)

108. ©Silberschatz, Korth and Sudarshan2.108Database System Concepts Result of  customer-name

109. ©Silberschatz, Korth and Sudarshan2.109Database System Concepts Result of the Subexpression

110. ©Silberschatz, Korth and Sudarshan2.110Database System Concepts Largest Account Balance in the Bank

111. ©Silberschatz, Korth and Sudarshan2.111Database System Concepts Customers Who Live on the Same Street and In the Same City as Smith

112. ©Silberschatz, Korth and Sudarshan2.112Database System Concepts Customers With Both an Account and a Loan at the Bank

113. ©Silberschatz, Korth and Sudarshan2.113Database System Concepts Result of  customer-name, loan-number, amount (borrower loan)

114. ©Silberschatz, Korth and Sudarshan2.114Database System Concepts Result of  branch-name (  customer-city = “Harrison” ( customer account depositor))

115. ©Silberschatz, Korth and Sudarshan2.115Database System Concepts Result of  branch-name (  branch-city = “Brooklyn” (branch))

116. ©Silberschatz, Korth and Sudarshan2.116Database System Concepts Result of  customer-name, branch-name (depositor account)

117. ©Silberschatz, Korth and Sudarshan2.117Database System Concepts The credit-info Relation

118. ©Silberschatz, Korth and Sudarshan2.118Database System Concepts Result of  customer-name, (limit – credit-balance) as credit-available (credit-info).

119. ©Silberschatz, Korth and Sudarshan2.119Database System Concepts The pt-works Relation

120. ©Silberschatz, Korth and Sudarshan2.120Database System Concepts The pt-works Relation After Grouping

121. ©Silberschatz, Korth and Sudarshan2.121Database System Concepts Result of branch-name  sum(salary) (pt-works)

122. ©Silberschatz, Korth and Sudarshan2.122Database System Concepts Result of branch-name  sum salary, max(salary) as max- salary (pt-works)

123. ©Silberschatz, Korth and Sudarshan2.123Database System Concepts The employee and ft-works Relations

124. ©Silberschatz, Korth and Sudarshan2.124Database System Concepts The Result of employee ft-works

125. ©Silberschatz, Korth and Sudarshan2.125Database System Concepts The Result of employee ft-works

126. ©Silberschatz, Korth and Sudarshan2.126Database System Concepts Result of employee ft-works

127. ©Silberschatz, Korth and Sudarshan2.127Database System Concepts Result of employee ft-works

128. ©Silberschatz, Korth and Sudarshan2.128Database System Concepts Tuples Inserted Into loan and borrower

129. ©Silberschatz, Korth and Sudarshan2.129Database System Concepts Names of All Customers Who Have a Loan at the Perryridge Branch

130. ©Silberschatz, Korth and Sudarshan2.130Database System Concepts E-R Diagram

131. ©Silberschatz, Korth and Sudarshan2.131Database System Concepts The branch Relation

132. ©Silberschatz, Korth and Sudarshan2.132Database System Concepts The loan Relation

133. ©Silberschatz, Korth and Sudarshan2.133Database System Concepts The borrower Relation

134. ©Silberschatz, Korth and Sudarshan2.134Database System Concepts 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