Download presentation

Presentation is loading. Please wait.

Published byCandace Jennings Modified about 1 year ago

1
Chapter 15 Functional Dependencies and Normalization for Relational Databases Chapter 16 Relational Database Design Algorithms and Further Dependencies 1

2
2 Functional Dependency Armstrong’s Axioms Closure of FDs Closure of attribute(X) – Find Keys Minimal set of FDs Normalization Lossless Join Decomposition

3
Outline 3 1 Informal Design Guidelines for Relational Databases 1.1Semantics of the Relation Attributes 1.2 Redundant Information in Tuples and Update Anomalies 1.3 Null Values in Tuples 1.4 Spurious Tuples 2 Functional Dependencies (FDs) 2.1 Definition of FD 2.2 Inference Rules for FDs 2.3 Equivalence of Sets of FDs 2.4 Minimal Sets of FDs

4
Outline 4 3 Normal Forms Based on Primary Keys 3.1 Normalization of Relations 3.2 Practical Use of Normal Forms 3.3 Definitions of Keys and Attributes Participating in Keys 3.4 First Normal Form 3.5 Second Normal Form 3.6 Third Normal Form 4 General Normal Form Definitions (For Multiple Keys) 5 BCNF (Boyce-Codd Normal Form)

5
5 What is relational database design? The grouping of attributes to form "good" relation schemas Two levels of relation schemas The logical "user view" level The storage "base relation" level Design is concerned mainly with base relations What are the criteria for "good" base relations?

6
6 A simplified COMPANY relational database schema

7
1.2 Redundant Information in Tuples and Update Anomalies 7 Information is stored redundantly Wastes storage Causes problems with update anomalies Insertion anomalies Deletion anomalies Modification anomalies

8
8

9
9

10
10

11
Functional Dependency: 11 Attribute B is functionally dependent on attribute A if at each point in time, each value of A has only one value of B associated with it.

12
Examples: 12 a. SSN ENAME b. PNUMBER {PNAME, PLOCATION} c. {SSN, PNUMBER} HOURS

13
functional dependency 13 A functional dependency (X Y) between two sets of attributes X and Y that are subsets of R specifies a constraint on the possible tuples that can form a relation state r of R. The constraint is that, for any two tuples t 1 and t 2 in r that have t 1 [X] = t 2 [X], we must also have t 1 [Y] = t 2 [Y].

14
14 --X is a candidate key of R—this implies that X Y for any subset of attributes Y of R --If X Y in R, this does not say whether or not Y X in R.

15
Inference Rules: 15 F = {SSN {ENAME, BDATE, ADDRESS, DNUMBER}, DNUMBER {DNAME,DMGRSSN}} Can infer additional FDs such as: SSN {DNAME, DMGRSSN}, SSN SSN, DNUMBER DNAME

16
Inference rules ( Armstrong’s Axioms) 16 Given a set of FDs F, we can infer additional FDs that hold whenever the FDs in F hold Armstrong's inference rules: IR1. (Reflexive) If Y subset-of X, then X -> Y IR2. (Augmentation) If X -> Y, then XZ -> YZ (Notation: XZ stands for X U Z) IR3. (Transitive) If X -> Y and Y -> Z, then X -> Z IR1, IR2, IR3 form a sound and complete set of inference rules These are rules hold and all other rules that hold can be deduced from these

17
Inference rules ( Armstrong’s Axioms) - continued 17 IR4 (decomposition, or projective rule): {X YZ} |= X Y. IR5 (union, or additive rule): {X Y, X Z} |= X YZ. IR6 (pseudotransitive rule) {X Y, WY Z } |= WX Z.

18
Armstrong’s Axioms 18 IR1 – Reflexive rule X Y, then X Y ssn, name ssn IR2 – Augmentation rule X Y |= XZ YZ ssn name |= ssn, dob name,dob

19
19 IR3 – Transitive rule X Y, Y Z |= X Z Ssn phone# Phone# zip Ssn zip

20
PROOF OF IR4 (USING IR1 THROUGH IR3) 20 IR4 (decomposition, or projective, rule): {X YZ} |= X Y. 1. X YZ (given). 2. YZ Y (using IR1 and knowing that YZ Y). 3. X Y (using IR3 on 1 and 2).

21
PROOF OF IR5 (USING IR1 THROUGH IR3) 21 IR5 (union, or additive, rule) {X Y, X Z} |= X YZ. X Y (given). X Z (given). X XY (using IR2 on 1 by augmenting with X; notice that XX = X). XY YZ (using IR2 on 2 by augmenting with Y). X YZ (using IR3 on 3 and 4).

22
PROOF OF IR6 (USING IR1 THROUGH IR3) 22 IR6 (pseudotransitive rule) {X Y, WY Z } |= WX Z. X Y (given). WY Z (given). WX WY (using IR2 on 1 by augmenting with W). WX Z (using IR3 on 3 and 2).

23
Exercise 23 a) {w y, x z} |= {wx y} {w y, x z} |= {wx y} w y Given wx xyAug(X) xy y Reflexive wx yTransitive STUID DOB C# C_TITLE STUID, C# DOB

24
Closure of FDs (F + ) 24 Set of FDs that could be derived from FDs using Armstrong’s Axioms ssn name ssn dob => Ssn name, dob

25
closure sets with respect to F (Example) 25 F = { SSN ENAME, NUMBER { NAME,PLOCATION}, {SSN, PNUMBER} HOURS } { SSN } + = { SSN, ENAME } { PNUMBER } + = { PNUMBER, PNAME, PLOCATION } { SSN, PNUMBER } + = { SSN, PNUMBER, ENAME, PNAME, PLOCATION, HOURS }

26
Inference Rules for FDs (3) 26 Closure of a set F of FDs is the set F + of all FDs that can be inferred from F Closure of a set of attributes X with respect to F is the set X + of all attributes that are functionally determined by X X + can be calculated by repeatedly applying IR1, IR2, IR3 using the FDs in F

27
Closure of attribute X (X + ) 27 Closure of a set of attributes X with respect to F is the set X + of all attributes that are functionally determined by X Algorithm 16.1 Determining X +, the closure of X under F X + := X Repeat old X + := X + ; for each functional dependency Y Z in F do if X + Y then X + := X + U Z; Until ( X + = old X + ); p 310

28
Example R(A,B,C,D,M,F,G) 28 FDs: A B B CD C M D FG A+ = A = AB (A B) = ABCD (B CD) = ABCDM (C M) = ABCDFGM (D FG) A IS A KEY OF THIS RELATION

29
R(STUID,F,M,L,STREET,CITY,STATE,ZIP,C#,CNAME,GRADE) 29 STUID F,M,L,STREET, CITY,STATE,ZIP C# CNAME ZIP CITY, STATE STUID,C# GRADE (STUID)+ = STUID = STUID, F,M,L,STREET, CITY,STATE,ZIP (C#)+ = C# = C#, CNAME (STUID,C#)+ = STUID,C# =STUID,C#, F,M,L,STREET, CITY,STATE,ZIP = STUID,C#, F,M,L,STREET, CITY,STATE,ZIP,CNAME = STUID,C#, F,M,L,STREET, CITY,STATE,ZIP,CNAME,GRADE

30
Algorithm 16.2 (a): Finding a Key K for R Given a set F of Functional Dependencies 30 Input: A universal relation R and a set of functional dependencies F on the attributes of R. 1. Set K := R; 2. For each attribute A in K { Compute (K - A)+ with respect to F; If (K - A)+ contains all the attributes in R, then set K := K - {A}; }

31
Find key of R(ABCDFGMN) 31 FDs: AC B B CD C MN D FG Is (AC) a candidate key? (AC)+ = ABCDFGMN A+ = A C+ = CMN (AC)+ = AC = ABC (AC B) = ABCD (B CD) = ABCDMN (C MN) = ABCDMNFG (D FG)

32
Find key of R(ABCDFGMN) 32 FDs: AC B B CD C MN D FG A C Is (AC) a candidate key? (AC)+ = ABCDFGMN A+ = AC = ABC (AC B) = ABCD (B CD) = ABCDMN (C MN) = ABCDMNFG (D FG)

33
Find key of R(ABCDFGMN) 33 FDs: A B B CD C MN D FG ABCDFGMN ACDFGMN A B AFGMN A B, B CD AFG A B, B C, C MN A A B, B D, D FG

34
Find key of R(ABCDFGMN) 34 FDs: AC B B CD C MN D FG A C Is (AC) a candidate key? (AC)+ = ABCDFGMN A+ = C+ =

35
Example: 35 Patient(Patient#, Pat_Name, Pat_Adr, Med_Exam_Description, Doctor_Number, Med_Exam#, Date_Test, Test_Result, Doctor_Name) Key:

36
Example: 36 Hotel_Bill(Guest_Id, Room#, Room_Rate, Date_Checkin, Total_Amount, Guest_Fname, Guest_Lname, InvoiceNumber, Guest_Address) Key:

37
16.1.3 Minimal Sets of Functional Dependencies 37 F is minimal if: 1. Every dependency in F has a single attribute for its right- hand side. 2. We cannot replace any dependency X A in F with a dependency Y A, where Y is a proper subset of X, and still have a set of dependencies that is equivalent to F. 3. We cannot remove any dependency from F and still have a set of dependencies that is equivalent to F.

38
Algorithm 16.2 Finding a minimal cover F for a set of functional dependencies E (p550) 38 Input: a set of functional dependencies E. 1. Set F := E. 2. Replace each functional dependency X {A 1, A 2,…, A n } in F by the n functional dependencies X A 1, X A 2,..., X A n. 3. For each functional dependency X A in F for each attribute B that is an element of X if ((F - {X A}) D {(X - {B}) A}) is equivalent to F, then replace X A with (X - {B}) A in F. 4. For each remaining functional dependency X A in F if (F - {X A}) is equivalent to F, then remove X A from F.

39
Example 39 1) The set of functional dependencies F of relation R(ABCDEFGHIJKLM) is A A,B,C,D,E B C,D,E,F C E,H D,E F D F,G K L L K a) Find a minimal cover of the set F. b) Find a candidate key for the relation R.

40
40 Normalization: The process of decomposing complex data structures into simple relations according to a set of dependency rules. – reduce anomalies (update, insert, delete) first proposed by Codd (1972) Assumption: all non-key fields will be updated frequently - tend to penalize retrieval

41
Problems with update anomalies 41 Insertion anomalies Deletion anomalies Modification anomalies

42
EXAMPLE OF AN UPDATE ANOMALY 42 Consider the relation: EMP_PROJ(Emp#, Proj#, Ename, Pname, No_hours) Update Anomaly: Changing the name of project number P1 from “Billing” to “Customer-Accounting” may cause this update to be made for all 100 employees working on project P1.

43
EXAMPLE OF AN INSERT ANOMALY 43 Consider the relation: EMP_PROJ(Emp#, Proj#, Ename, Pname, No_hours) Insert Anomaly: Cannot insert a project unless an employee is assigned to it. Conversely Cannot insert an employee unless a he/she is assigned to a project.

44
EXAMPLE OF AN DELETE ANOMALY 44 Consider the relation: EMP_PROJ(Emp#, Proj#, Ename, Pname, No_hours) Delete Anomaly: When a project is deleted, it will result in deleting all the employees who work on that project. Alternately, if an employee is the sole employee on a project, deleting that employee would result in deleting the corresponding project.

45
15.3 Normalization of Relations (1) 45 Normalization: The process of decomposing unsatisfactory "bad" relations by breaking up their attributes into smaller relations Normal form: Condition using keys and FDs of a relation to certify whether a relation schema is in a particular normal form

46
Normalization of Relations (2) 46 2NF, 3NF, BCNF based on keys and FDs of a relation schema 4NF based on keys, multi-valued dependencies : MVDs; 5NF based on keys, join dependencies : JDs (Chapter 11) Additional properties may be needed to ensure a good relational design (lossless join, dependency preservation; Chapter 11)

47
Definitions of Keys and Attributes Participating in Keys (2) 47 If a relation schema has more than one key, each is called a candidate key. One of the candidate keys is arbitrarily designated to be the primary key, and the others are called secondary keys. A Prime attribute must be a member of some candidate key A Nonprime attribute is not a prime attribute— that is, it is not a member of any candidate key.

48
Problems: 48

49
49

50
Decomposition by Analysis and Synthesis techniques 50 Decomposition by analysisDecomposition by synthesis

51
51

52
Normalization into 1NF 52

53
Normalization of nested relations into 1NF 53

54
54 Unnormalized & 1NF relation

55
55

56
First Normal Form(1NF): 56 A relation is in first normal form if and only if every attribute is single-valued for each tuple. -- 1NF disallows having a set of values, a tuple of values, or a combination of both as an attribute value for a single tuple. In other words, 1NF disallows "relations within relations" or "relations as attributes of tuples." The only attribute values permitted by 1NF are single atomic (or indivisible) values.

57
First Normal Form 57 Disallows composite attributes multivalued attributes nested relations; attributes whose values for an individual tuple are non-atomic

58
58

59
Second Normal Form (1) 59 Uses the concepts of FDs, primary key Definitions Prime attribute: An attribute that is member of the primary key K Full functional dependency: a FD Y -> Z where removal of any attribute from Y means the FD does not hold any more Examples: {SSN, PNUMBER} -> HOURS is a full FD since neither SSN -> HOURS nor PNUMBER -> HOURS hold {SSN, PNUMBER} -> ENAME is not a full FD (it is called a partial dependency ) since SSN -> ENAME also holds

60
Second Normal Form (2) 60 A relation schema R is in second normal form (2NF) if every non-prime attribute A in R is fully functionally dependent on the primary key R can be decomposed into 2NF relations via the process of 2NF normalization

61
Second Normal Form(2NF): 61 A relation is in second normal form if and only if it is in first normal form and all the nonkey attributes are fully functionally dependent on the key. Grade_report( SID, C#, stu_name, course_title) not in 2NF ( there are partial dependencies )

62
Full functional dependency 62 A functional dependency X Y is a full functional dependency if removal of any attribute A from X means that the dependency does not hold any more; that is, for any attribute A X, (X - {A}) does not functionally determine Y.

63
(definition from the book) 63 Second normal form (2NF) is based on the concept of full functional dependency. A functional dependency X Y is a full functional dependency if removal of any attribute A from X means that the dependency does not hold any more; that is, for any attribute A X, (X - {A}) does not functionally determine Y. A functional dependency X Y is a partial dependency if some attribute A X can be removed from X and the dependency still holds; that is, for some A X, (X - {A}) Y.

64
Second normal form is violated when a non-key field is a fact about a subset of a key. It is only relevant when the key is composite, i.e., consists of several fields 64 Function Dependencies (F): Warehouse Warehouse_address Part, Warehouse quantity William Kent (CACM1982)

65
65 Normalizing into 2NF and 3NF

66
3.4 Third Normal Form (1) 66 Definition: Transitive functional dependency: a FD X -> Z that can be derived from two FDs X -> Y and Y -> Z Examples: SSN -> DMGRSSN is a transitive FD Since SSN -> DNUMBER and DNUMBER -> DMGRSSN hold SSN -> ENAME is non-transitive Since there is no set of attributes X where SSN -> X and X -> ENAME

67
Third Normal Form (2) 67 A relation schema R is in third normal form (3NF) if it is in 2NF and no non-prime attribute A in R is transitively dependent on the primary key R can be decomposed into 3NF relations via the process of 3NF normalization NOTE: In X -> Y and Y -> Z, with X as the primary key, we consider this a problem only if Y is not a candidate key. When Y is a candidate key, there is no problem with the transitive dependency. E.g., Consider EMP (SSN, Emp#, Salary ). Here, SSN -> Emp# -> Salary and Emp# is a candidate key.

68
68 Successive Normalization of LOTS into 2NF and 3NF

69
Normal Forms Defined Informally 69 1 st normal form All attributes depend on the key 2 nd normal form All attributes depend on the whole key 3 rd normal form All attributes depend on nothing but the key

70
70 SUMMARY OF NORMAL FORMS based on Primary Keys

71
General Normal Form Definitions (For Multiple Keys) (1) 71 The above definitions consider the primary key only The following more general definitions take into account relations with multiple candidate keys A relation schema R is in second normal form (2NF) if every non-prime attribute A in R is fully functionally dependent on every key of R

72
General Normal Form Definitions (2) 72 Definition: Superkey of relation schema R - a set of attributes S of R that contains a key of R A relation schema R is in third normal form (3NF) if whenever a FD X -> A holds in R, then either: (a) X is a superkey of R, or (b) A is a prime attribute of R NOTE: Boyce-Codd normal form disallows condition (b) above

73
73 Third Normal Form(3NF): A relation is in third normal form if it is in second normal form and no nonkey attribute is transitively dependent on the key. Course_info(C#, course_title, Instructure_name, Inst_phone) Not in 3NF – there is a transitive dependency C# course_title, Instructure_name Instructure_name Inst_phone

74
From the book 74 Third normal form (3NF) is based on the concept of transitive dependency. A functional dependency X Y in a relation schema R is a transitive dependency if there is a set of attributes Z that is neither a candidate key nor a subset of any key of R, and both X Z and Z Y hold. Codd’s original definition, a relation schema R is in 3NF if it satisfies 2NF and no nonprime attribute of R is transitively dependent on the primary key

75
Third normal form is violated when a non- key field is a fact about another non-key field 75 Function Dependencies (F): Employee Department Department Location William Kent ( CACM 1982)

76
Summary of Normal Forms Based on Primary Keys and corresponding Normalization. 76 First (1NF) Relation should have no nonatomic attributes or nested relations Form new relations for each nonatomic attribute or nested relation Second (2NF) For relations where primary key contains multiple attributes, no nonkey attribute should be functionally dependent on a part of the primary key Decompose and set up a new relation for each partial key with its dependent attribute(s). Make sure to keep a relation with the original primary key and any attributes that are fully functionally dependent on it. Third (3NF) Relation should not have a nonkey attribute functionally determined by another nonkey attribute (or by a set of nonkey attributes.) That is, there should be no transitive dependency of a nonkey attribute on the primary key. Decompose and set up a relation that includes the nonkey attribute(s) that functionally determine(s) other nonkey attribute(s).

77
77 Unnormalized & 1NF relation

78
78 University Database (Third Normal Form relations)

79
79 Philip Bernstein, “Synthesizing Third Normal Form Relations from Functional Dependencies,” ACM Transactions on Database Systems, 1976, pp 277-298. (University of Toronto) To find 3NF relations - Find Minimal Cover of FDs - Each FD— convert to 3NF relation - If key of the Universal DB is not in any relations then create a relation from the key

80
80 A) INVOICE(INVOICE#, CUST_ID, CUST_NAME, BILL_DATE, TOT_AMOUNT, SALES_REP#, SALES_REP_NAME) Key: Normal Form: 3NF relations:

81
81 B) PRJ_EMP(PROJECT#, MANAGER_ID, PROJECT_TITLE, EMP_ID, EMP_LNAME, EMP_FNAME, MANAGER_NAME, EXPECT_END_PROJECT_DATE, EMP_HOURLY_RATE,HOURS_WORK_PER_PRJ) Key: Normal Form: 3NF relations:

82
82

83
Boyce-Codd Normal Form(BCNF): 83 A relation is in Boyce-Codd normal form if and only if every determinant is a candidate key (1974). ( every relation in BCNF is also in 3NF) Determinant– attribute(s) on the LHS of FD A BC (A is determinant)

84
BCNF (Boyce-Codd Normal Form) 84 A relation schema R is in Boyce-Codd Normal Form (BCNF) if whenever an FD X -> A holds in R, then X is a superkey of R Each normal form is strictly stronger than the previous one Every 2NF relation is in 1NF Every 3NF relation is in 2NF Every BCNF relation is in 3NF There exist relations that are in 3NF but not in BCNF The goal is to have each relation in BCNF (or 3NF)

85
85

86
86

87
Boyce-Codd Normal Form 87

88
A relation TEACH that is in 3NF but not in BCNF 88

89
Achieving the BCNF by Decomposition (1) 89 Two FDs exist in the relation TEACH: fd1: { student, course} -> instructor fd2: instructor -> course {student, course} is a candidate key for this relation and that the dependencies shown follow the pattern in Figure 10.12 (b). So this relation is in 3NF but not in BCNF A relation NOT in BCNF should be decomposed so as to meet this property, while possibly forgoing the preservation of all functional dependencies in the decomposed relations. (See Algorithm 11.3)

90
Multivalued Dependency 90 Let R be a relation having attributes or sets of attributes A,B,C. There is a multivalued dependence of attribute B on attribute A if and only if the set of B values associated with a given A value is independent of the C values. A > B,C B -/-> C or C -/-> B

91
Fourth Normal Form(4NF): 91 A relation is in fourth normal form if and only if it is in Boyce-Codd normal form and there are no nontrivial multivalued dependencies.

92
92

93
93 Employee > Skill, Language Language --/ skill EmployeeSkillLanguage EmployeeSkillEmployeeLanguage William Kent (1982)

94
Fifth Normal Form(5NF): 94 A relation is in fifth normal form if no remaining nonloss projections are possible, except the trivial one in which the key appears in each projection.

95
95 Fifth normal form when its information content cannot be reconstructed from several smaller record types William Kent (CACM 1982)

96
Domain-Key Normal Form(DKNF): 96 A relation is in domain-key normal form if every constraint is a logical consequence of domain constraints or key constraints.

97
97

98
98

99
99

100
100

101
Exercise!! 101 Classify each of the following relations as 1NF, 2NF, 3NF, BCNF, 4NF, or 5NF. (State any assumptions you make) a) COURSE(COURSE#,C_TITLE,INS_NAME,TEXT_BOOK,PUBLISHER) b) ENROLLMENT(STUID,COURSE#,STUNAME,COURSE_TITLE, GRADE) c) PART(PART#,PART_NAME,DEPARTMENT,PRICE)

102
102 d) EMPLOYEE(EMPID,EMPNAME,DATE_HIRED,JOB_TITLE, JOB_LEVEL,PROJ#,PROJ_TITLE,DUEDATE) e) BOOK(ISBN,TITLE,AUTHOR,AUTHOR_INSTITUTE,PRICE, PUBLISHER_NAME,PUBLISHER_ADR) f) CUSTOMER(CUSTID,CUSTNAME, PHONE#, STREET,CITY,STATE,ZIP)

103
Lossless Join Decomposition 103 (R1, R2) is a decomposition of R F is a set of functional dependencies (R1, R2) is lossless join with to F if and only if R1 R2 R1 – R2 or R1 R2 R2 – R1 And R1 U R2 = R

104
104

105
Dependency preservation 105 Dependency preservation: Let F i be the set of dependencies F + that include only attributes in R i. dependency preserving (F 1 F 2 … F n ) + = F + Otherwise, checking updates for violation of functional dependencies may require computing joins, which is expensive.

106
106 Example -- dependency preservation: R = (A, B, C) F = {A B, B C) Two ways to decompose this relation: 1) R 1 = (A, B), R 2 = (B, C) Lossless-join decomposition: R 1 R 2 = {B} and R 2 - R 1 = C Dependency preserving 2) R 1 = (A, B), R 2 = (A, C) Lossless-join decomposition: R 1 R 2 = {A} and R 1 - R 2 = B Not dependency preserving (cannot check B C without computing R 1 x R 2 ) 3NF – always dependency preservation BCNF – not always

107
107

108
108

109
109

110
Exercise 110 1) The set of functional dependencies F of relation R(ABCDEFGHIJKLM) is A A,B,C,D,E B C,D,E,F C E,H D,E F D F,G K L L K a) Find a minimal cover of the set F. b) Find a candidate key for the relation R. c) Find a lossless join decomposition of R into 3NF (also give a proof of lossless join).

111
111 2) The set of functional dependencies F of relation R(ABCDEFGHIJKL) is A ---> B,C,D,E,K BC ---> D,E,F C ---> K,H D,B---> G D ---> F,G K ---> L a) Find a minimal cover of the set F. b) Find a candidate key for the relation R. c) Find a lossless join decomposition of R into 3NF (also give a proof of lossless join).

112
112 REFERENCES 1. E.F. Codd, "A Relational Model of Data for Large Shared Data Banks", Comm. ACM 13 (6), June 1970, pp. 377-387. The original paper introducing the relational data model. 2. E.F. Codd, "Normalized Data Base Structure: A Brief Tutorial", ACM SIGFIDET Workshop on Data Description, Access, and Control, Nov. 11-12, 1971, San Diego, California, E.F. Codd and A.L. Dean (eds.). An early tutorial on the relational model and normalization. 3. E.F. Codd, "Further Normalization of the Data Base Relational Model", R. Rustin (ed.), Data Base Systems (Courant Computer Science Symposia 6), Prentice-Hall, 1972. Also IBM Research Report RJ909. The first formal treatment of second and third normal forms. 4. C.J. Date, An Introduction to Database Systems (third edition), Addison-Wesley, 1981. An excellent introduction to database systems, with emphasis on the relational. 5. R. Fagin, "Multivalued Dependencies and a New Normal Form for Relational Databases", ACM Transactions on Database Systems 2 (3), Sept. 1977. Also IBM Research Report RJ1812. The introduction of fourth normal form. 6. R. Fagin, "Normal Forms and Relational Database Operators", ACM SIGMOD International Conference on Management of Data, May 31-June 1, 1979, Boston, Mass. Also IBM Research Report RJ2471, Feb. 1979. The introduction of fifth normal form. 7. W. Kent, "A Primer of Normal Forms", IBM Technical Report TR02.600, Dec. 1973. An early, formal tutorial on first, second, and third normal forms. 8. T.-W. Ling, F.W. Tompa, and T. Kameda, "An Improved Third Normal Form for Relational Databases", ACM Transactions on Database Systems, 6(2), June 1981, 329-346. One of the first treatments of inter-relational dependencies.

113
Exercise 14.18 (p498) 113 a) {w y, x z} |= {wx y} {w y, x z} |= {wx y} w y Given wx xyAug(X) xy y Reflexive wx yTransitive STUID DOB C# C_TITLE STUID, C# DOB

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google