1. Chapter 13 Normalization

2. 2 Chapter 13 - Objectives u Purpose of normalization. u Problems associated with redundant data. u Identification of various types of update anomalies such as insertion, deletion, and modification anomalies. u How to recognize appropriateness or quality of the design of relations.

3. 3 Chapter 13 - Objectives u How functional dependencies can be used to group attributes into relations that are in a known normal form. u How to undertake process of normalization. u How to identify most commonly used normal forms, namely 1NF, 2NF, 3NF, and Boyce– Codd normal form (BCNF). u How to identify fourth (4NF) and fifth (5NF) normal forms.

4. 4 Recommended Reading u Chapter 13 of the text book u Module 9 of the Study Book –Available from the Course Website

5. 5 Normalization u Main objective in developing a logical data model for relational database systems is to create an accurate representation of the data, its relationships, and constraints. u To achieve this objective, must identify a suitable set of relations.

6. 6 From Conceptual Model to Logical Model 1. Analyse requirements 2. Draw up ERM diagram –See chapters 11 & 12 3. Map ERM diagram to Relational Model –See separate slides + Chapter 15 4. Normalize relations –See chapter 13 5. Optimize relations –‘de-’normalize to gain performance –Not covered in this course

7. 7 Normalization u Four most commonly used normal forms are first (1NF), second (2NF) and third (3NF) normal forms, and Boyce–Codd normal form (BCNF). u Based on functional dependencies among the attributes of a relation. u A relation can be normalized to a specific form to prevent possible occurrence of update anomalies.

8. 8 Data Redundancy u Major aim of relational database design is to group attributes into relations to minimize data redundancy and reduce file storage space required by base relations. u Problems associated with data redundancy are illustrated by comparing the following Staff and Branch relations with the StaffBranch relation.

9. 9 Data Redundancy

10. 10 Data Redundancy u StaffBranch relation has redundant data: details of a branch are repeated for every member of staff. u In contrast, branch information appears only once for each branch in Branch relation and only branchNo is repeated in Staff relation, to represent where each member of staff works.

11. 11 Update Anomalies u Relations that contain redundant information may potentially suffer from update anomalies. u Types of update anomalies include: –Insertion »E.g. Adding branch without staff –Deletion »E.g. Deleting Mary Howe –Modification »E.g. Changing address of B003

12. 12 Lossless-join and Dependency Preservation Properties u Two important properties of decomposition: - Lossless-join property enables us to find any instance of original relation from corresponding instances in the smaller relations. - Dependency preservation property enables us to enforce a constraint on original relation by enforcing some constraint on each of the smaller relations.

13. 13 Section 1 Functional Dependencies

14. 14 Functional Dependency (FD) u Main concept associated with normalization u Formally related to concept of keys u Functional Dependency –Describes relationship between attributes in a relation. –If A and B are attributes of relation R, B is functionally dependent on A (denoted A  B), if each value of A in R is associated with exactly one value of B in R.

15. 15 Functional Dependency u Property of the meaning (or semantics) of the attributes in a relation. u Diagrammatic representation: u Determinant of a functional dependency refers to attribute or group of attributes on left-hand side of the arrow.

16. 16 Example - Functional Dependency

17. 17 Functional Dependency u Main characteristics of functional dependencies used in normalization: –have a 1:1 relationship between attribute(s) on left and right-hand side of a dependency; –hold for all time; –are nontrivial »Impossible not to be satisfied »RHS is a subset of LHS

18. 18 Closure of FDs u Complete set of functional dependencies for a given relation can be very large. u Important to find an approach that can reduce set to a manageable size. u Need to identify set of functional dependencies (X) for a relation that is smaller than complete set of functional dependencies (Y) for that relation and has property that every functional dependency in Y is implied by functional dependencies in X.

19. 19 Closure of FDs u Set of all functional dependencies implied by a given set of functional dependencies X called closure of X (written X + ). u Set of inference rules, called Armstrong’s axioms, specifies how new functional dependencies can be inferred from given ones. u These rules are both sound and complete u Computing closure of a set of FDs is expensive –Size of closure may be exponential in the number of attributes!

20. 20 Armstrong’s Axioms u Let A, B, and C be subsets of the attributes of relation R. Armstrong’s axioms are as follows: 1. Reflexivity If B is a subset of A, then A  B 2. Augmentation If A  B, then AC  C 3. Transitivity If A  B and B  C, then A  C Four other rules mentioned in Study Book can be derived from Amstrong’s Axioms.

21. 21 Attribute Closure  Typically want to check if given FD X  Y is in F + –To check if X is a candidate key for relation R –(Also need to check if X is minimal) u X + is called closure of attribs X under F –Defined as the set of attribs functionally determined by X under F –X  Y is in F + iff Y  X + u Example –Is AG a candidate key for R(A,B,C,G,H,I) under set of FDs F = {A  B, A  C, CG  H, CG  I, B  H} ? –Compute AG + using algorithm in StudyBook –Check to see if AG is minimal (compute A + and G + )

22. 22 Minimal Covers u Minimal Sets of FDs –Set of all FDs for a relation can be large –Reduce number of FDs by eliminating redundancies –Textbook gives different definition than StudyBook »Both amount to the same concept u Unfortunately there can be several minimal covers

23. 23 Minimal Covers u Textbook definition: –A set of FDs Y is covered by a set of FDs X if every FD in Y is in X + –X is minimal set of FDs if: »Every FD in X has a single attribute on its RHS »A  B cannot be replaced by C  B with C  A without losing equivalence »Not possible to remove a FD in X without losing equivalence –A minimal cover for X is a minimal set of FDs X min that is equivalent to X

24. 24 Summary u Relation R yields set of FDs F –Possible to compute closure F + »Includes all FDs derivable from F, needed for BCNF –Possible to compute attribute closure of X under F »Useful to identify candidate keys –Possible to compute minimal cover F M »Excludes redundancies, needed for 3NF u Relationship between F, F +, and F M : –F  F + and F  F M +, but not F M  F u Examples on next 7 slides are from Study Book

25. 25 Examples – (a) functional dependencies u staffNo  sName, position, salary, branchNo, bAddress {PK} u branchNo  bAddress u bAddress  branchNo u branchNo, position  salary u bAddress, position  salary

26. 26 Functional Dependencies u Three methods to identify FDs for a relation: –Create adequate test data »populate the relation with several tuples (rows), inserting all possible data combinations, while identifying which combinations should not be valid –Reason about the relation itself »look for rules (also called predicates) –Translate any given assumptions about the data

27. 27 Functional Dependencies u Diagrammatic notation:

28. 28 Examples – (b) closure of set of FDs  Given R(A,B,C,G,H,I) and F = {A  B, A  C, CG  H, CG  I, B  H} u Compute F + using Amstrong’s Axioms: –A  H(since A  B and B  H)(Transitivity) –CG  CG  H and CG  I  (Union) –AG  AG  CG and CG  I  (Aug+trans) –… u Can be very large  Needed for BCNF definition

29. 29 Examples – (c) attribute closure  Given R(A,B,C,G,H,I) and F = {A  B, A  C, CG  H, CG  I, B  H} u Is AG a candidate key for R? u Calculate AG + :( ref. algorithm in StudyBook) –Result := AG –Result := AGB (A  B and A  result(AG) ) –Result := AGBC(A  C and A  AGB ) –Result := AGBCH(B  H and B  result(AGBC) ) –Result := AGBCHI(CG  I and CG  result(AGBCH) ) u Thus, AG is at least a super key; if minimal, also a candidate key u Calculate A + and G + separately in the same manner

30. 30 Examples – (d) minimal set of FDs  Given R(A,B,C) and F = {A  BC, B  C, A  B, AB  C} u Compute F M (ref. algorithm in StudyBook) 1. Merge all FDs with the same LHS in F: A  BC(A  B is merged) 2. AB  C contains extra attrib A in LHS since B  C F – {AB  C}  {B  C} = {A  BC, B  C} 3. C is extra attrib in RHS of A  BC since B  C F – {A  BC}  {A  B} = {A  B, B  C}  Thus, F M = {A  B, B  C}

31. 31 Examples – (e) finding candidate keys (1/2)  Given R(A,B,C,D,E) and F = {A  B, BC  E, ED  A} u Step 1: find super keys –A) check whether determinants are super keys »A+ = ABBC+ = BCEED+ = ABED(No) –B) create super keys using F »(R – A+)  A = ACDE is a super key for R »(R – BC+)  BC = ABCD is a super key for R »(R – ED+)  ED = CDE is a super key for R u Step 2: find candidate keys –For each super key, try minimalizing –Compute attrib. closure of all subsets, test equality with R –Example: super key ACDE, take all subsets

32. 32 Examples – (e) finding candidate keys (2/2) u Super key ACDE, check subsets: »AC+ = ABCE not R »AD+ = ABDnot R »AE+ = ABEnot R »CD+ = CDnot R »CE+ = CEnot R »DE+ = ABDEnot R »ACD+ = ABCDEcandidate key »ACE+ = ABCEnot R »CDE+ = ABCDEcandidate key »ADE+ = ABDEnot R u Need to do the same for the 2 other super keys u All candidate keys for R: ACD, CDE, and BCD

33. 33 Section 2 Normalization

34. 34 The Process of Normalization u Formal technique for analyzing (decomposing) a relation based on its primary key and functional dependencies between its attributes. u Often executed as a series of steps. Each step corresponds to a specific normal form, which has known properties. u As normalization proceeds, relations become progressively more restricted (stronger) in format and also less vulnerable to update anomalies.

35. 35 Relationship Between Normal Forms

36. 36 Three methods, increasing difficulty –Text Book approach »Up to 3NF, based on primary key only »BCNF based on all candidate keys for a relation »Will be assessed –Text Book General Definitions for 2NF, 3NF »Based on all candidate keys for a relation »Catches some extra update anomalies »May be assessed –Study Book extra information »Based on all candidate keys for a relation »Formal algorithms using F + and F M »Will not be assessed

37. 37 Unnormalized Form (UNF) u A table that contains one or more repeating groups. –Multiple values for a group of attribs per tuple, given a primary key u To create an unnormalized table: –transform data from information source (e.g. form) into table format with columns and rows.

38. 38 First Normal Form (1NF) u A relation in which the intersection of each row and column contains one and only one value. –Cfr. ‘multi-valued’ attributes in E-ER Model

39. 39 UNF to 1NF u Nominate an attribute or group of attributes to act as the key for the unnormalized table. u Identify repeating group(s) in unnormalized table which repeats for the key attribute(s).

40. 40 UNF to 1NF u Remove repeating group by: –entering appropriate data into the empty columns of rows containing repeating data (‘flattening’ the table). Or by –placing repeating data along with copy of the original key attribute(s) into a separate relation u Both approaches induce 1NF, but update anomalies are still very likely

41. 41 UNF to 1NF: example u First approach CR76John KayPG46 LawrenceCO40T. Murphy PG165 Novar DrCO93T. Shaw CR56A. StewartPG46 LawrenceCO40T. Murphy PG362 Manor DCO93T. Shaw CR76John KayPG46 LawrenceCO40T. Murphy CR76John KayPG165 Novar DrCO93T. Shaw CR56A. StewartPG46 LawrenceCO40T. Murphy CR56A. StewartPG362 Manor DCO93T. Shaw UNF 1NF

42. 42 UNF to 1NF: example u Second approach CR76John Kay CR56A. Stewart CR76John KayPG46 LawrenceCO40T. Murphy CR76John KayPG165 Novar DrCO93T. Shaw CR56A. StewartPG46 LawrenceCO40T. Murphy CR56A. StewartPG362 Manor DCO93T. Shaw 1NF

43. 43 Second Normal Form (2NF)  Based on concept of full functional dependency: –A and B are (groups of) attributes of a relation, –B is fully dependent on A if B is functionally dependent on A but not on any proper subset of A. u 2NF - A relation that is in 1NF and every non- primary-key attribute is fully functionally dependent on the primary key.

44. 44 1NF to 2NF  Identify primary key for the 1NF relation.  Identify functional dependencies in the relation.  If partial dependencies exist on the primary key remove them by placing them in a new relation along with copy of their determinant.

45. 45 1NF to 2NF: example CR76John KayPG46 LawrenceCO40T. Murphy CR76John KayPG165 Novar DrCO93T. Shaw CR56A. StewartPG46 LawrenceCO40T. Murphy CR56A. StewartPG362 Manor DCO93T. Shaw 1NF Fd1 {PK} Fd2 Partial dependency on PK Fd3Partial dependency on PK 2NF Client(clientNo, cName) Prop(propNo, pAdd, pOwnNo, pOwnName) PropRent(clientNo, propNo)

46. 46 Third Normal Form (3NF)  Based on concept of transitive dependency: –A, B and C are (groups of) attributes of a relation such that if A  B and B  C, –then C is transitively dependent on A through B. (Provided that A is not functionally dependent on B or C). u 3NF - A relation that is in 1NF and 2NF and in which no non-primary-key attribute is transitively dependent on the primary key.

47. 47 2NF to 3NF  Identify the primary key in the 2NF relation.  Identify functional dependencies in the relation.  If transitive dependencies exist on the primary key remove them by placing them in a new relation along with copy of their determinant.

48. 48 2NF to 3NF: example PG46 LawrenceCO40T. Murphy PG165 Novar DrCO93T. Shaw PG46 LawrenceCO40T. Murphy PG362 Manor DCO93T. Shaw 2NF Fd1 {PK} Fd2 transitive dependency on PK 3NF Client(clientNo, cName) Prop(propNo, pAdd, pOwnNo) Owner(pOwnNo, pOwnName) PropRent(clientNo, propNo)

49. 49 General Definitions of 2NF and 3NF u Previous definitions are only an approximation u Second normal form (2NF) –A relation that is in 1NF and every non- primary-key attribute is fully functionally dependent on any candidate key. u Third normal form (3NF) –A relation that is in 1NF and 2NF and in which no non-primary-key attribute is transitively dependent on any candidate key.

50. 50 Boyce–Codd Normal Form (BCNF) u Based on functional dependencies that take into account all candidate keys in a relation, however BCNF also has additional constraints compared with general definition of 3NF. u BCNF - A relation is in BCNF if and only if every determinant is a candidate key.

51. 51 Boyce–Codd normal form (BCNF) u Difference between 3NF and BCNF is that for a functional dependency A  B, 3NF allows this dependency in a relation if B is a primary- key attribute and A is not a candidate key. u Whereas, BCNF insists that for this dependency to remain in a relation, A must be a candidate key. u Every relation in BCNF is also in 3NF. However, relation in 3NF may not be in BCNF.

52. 52 Boyce–Codd normal form (BCNF) u Violation of BCNF is quite rare. u Potential to violate BCNF may occur in a relation that: –contains two (or more) composite candidate keys; –the candidate keys overlap (ie. have at least one attribute in common).

53. 53 Example BCNF Violation clientNointerviewDateinterviewTimestaffNoroomNo CR7613 May 0210.30SG5G101 CR5613 May 0212.00SG5G101 CR7413 May 0212.00SG37G102 CR561 July 0210.30SG5G103 fd1PK fd2CK fd3CK fd4 Relation is in ‘simple 3NF’ (not in ‘general 3NF’) But not in BCNF since LHS of fd4 is not a candidate key

54. 54 Problems with BCNF u 3NF properties: –Always possible to decompose to 3NF »Where decomposition is lossless, and »Dependencies are preserved u BCNF properties: –When decomposing to BCNF, dependencies may be lost (e.g. fd3 in previous slide) u Stay at 3NF or go to BCNF? –3NF: more redundancy –BCNF: loss of constraint

55. 55 Review of Normalization (UNF to BCNF)

56. 56 Review of Normalization (UNF to BCNF) UNF 1NF

57. 57 Review of Normalization (UNF to BCNF) 1NF

58. 58 Review of Normalization (UNF to BCNF)

59. 59 Review of Normalization (UNF to BCNF) u Resulting relational schema: –Property(propertyNo, pAddress) –Staff(staffNo, sName) –Inspection(propertyNo, iDate, iTime, comments, staffNo) –StaffCar(staffNo, iDate, carReg) u Loss of fd5, since fd4 causes redundancy u PK, FK mechanism can be used to compute lossless join

60. 60 Fourth Normal Form (4NF) u Although BCNF removes anomalies due to functional dependencies, another type of dependency called a multi-valued dependency (MVD) can also cause data redundancy. u Possible existence of MVDs in a relation is due to 1NF and can result in data redundancy. branchNosNameoName B03{Beech, Ford}{Farrel, Murphy}

61. 61 Additional Normal Forms u 4NF and 5NF u These are optional w.r.t. the process of normalization u Assessment: –Concepts may be assessed –Normalization to 4NF and/or 5NF will not be assessed

62. 62 Fourth Normal Form (4NF) - MVD u MVD –Dependency between attributes (for example, A, B, and C) in a relation, –such that for each value of A there is a set of values for B and a set of values for C. –However, set of values for B and C are independent of each other. branchNosNameoName B03BeechFarrel B03BeechMurphy B03FordFarrel B03FordMurphy

63. 63 Fourth Normal Form (4NF) u MVD between attributes A, B, and C in a relation using the following notation: A   B A   C

64. 64 Fourth Normal Form (4NF) u MVD can be further defined as being trivial or nontrivial. – MVD A   B in relation R is defined as being trivial if (a) B is a subset of A or (b) A  B = R. – MVD is defined as being nontrivial if neither (a) nor (b) are satisfied. – Trivial MVD does not specify a constraint on a relation, while a nontrivial MVD does specify a constraint.

65. 65 Fourth Normal Form (4NF)  Defined as a relation that is in BCNF and contains no nontrivial MVDs. u Note that we must allow non-trivial MVDs that are in fact FDs

66. 66 4NF - Example BCNF 4NF

67. 67 Fifth Normal Form (5NF) u A relation decomposed into two relations must have lossless-join property, which ensures that no spurious tuples are generated when relations are reunited through a natural join. u However, there are requirements to decompose a relation into more than two relations. u Although rare, these cases are managed by join dependency and fifth normal form (5NF).

68. 68 Fifth Normal Form (5NF) u A relation that has no (non-trivial) join dependency. u Example of a JD

69. 69 5NF - Example PropItemSup(A,B,C) satisfies JD (R1(A,B), R2(B,C), R3(A,C)) Any join between two relations produces spurious tuples Join between the three produces legal state of PropItemSup