1. Functional Dependencies and Normalization 1 Instructor: Mohamed Eltabakh meltabakh@cs.wpi.edu

2. What to Cover Functional Dependencies (FDs) Closure of Functional Dependencies Lossy & Lossless Decomposition Normalization 2

3. Decomposing Relations Greg Dave sName p2 p1 pNumber MMs2 MMs1 pNamesNumber StudentProf FDs: pNumber  pName Greg Dave sName p2 p1 pNumber s2 s1 sNumber Student p2 p1 pNumber MM pName Professor Greg Dave sName MM pName S2 S1 sNumber Student p2 p1 pNumber MM pName Professor 3 Lossless Lossy

4. Lossless vs. Lossy Decomposition Assume R is divided into R1 and R2 Lossless Decomposition R1 natural join R2 should create exactly R Lossy Decomposition R1 natural join R2 adds more records (or deletes records) from R 4

5. Lossless Decomposition 5 Greg Dave sName p2 p1 pNumber MMs2 MMs1 pNamesNumber StudentProf FDs: pNumber  pName Greg Dave sName p2 p1 pNumber s2 s1 sNumber Student p2 p1 pNumber MM pName Professor Lossless Student & Professor are lossless decomposition of StudentProf (Student ⋈ Professor = StudentProf)

6. Lossy Decomposition 6 Greg Dave sName p2 p1 pNumber MMs2 MMs1 pNamesNumber StudentProf FDs: pNumber  pName Greg Dave sName MM pName S2 S1 sNumber Student p2 p1 pNumber MM pName Professor Lossy Student & Professor are lossy decomposition of StudentProf (Student ⋈ Professor != StudentProf)

7. Goal: Ensure Lossless Decomposition How to ensure lossless decomposition? Answer: The common columns must be candidate key in one of the two relations 7

8. Back to our example Greg Dave sName p2 p1 pNumber MMs2 MMs1 pNamesNumber StudentProf FDs: pNumber  pName Greg Dave sName p2 p1 pNumber s2 s1 sNumber Student p2 p1 pNumber MM pName Professor Greg Dave sName MM pName S2 S1 sNumber Student p2 p1 pNumber MM pName Professor 8 Lossless Lossy pNumber is candidate key pName is not candidate key

9. What to Cover Functional Dependencies (FDs) Closure of Functional Dependencies Lossy & Lossless Decomposition Normalization 9

10. 10

11. Normalization First Normal Form (1NF) Boyce-Codd Normal Form (BCNF) Third Normal Form (3NF) Canonical Cover of FDs 11

12. Normalization Set of rules to avoid “bad” schema design Decide whether a particular relation R is in “good” form If not, decompose R to be in a “good” form Several levels of normalization First Normal Form (1NF) BCNF Third Normal Form (3NF) Fourth Normal Form (4NF) If a relation is in a certain normal form, then it is known that certain kinds of problems are avoided or minimized 12

13. First Normal Form (1NF) Attribute domain is atomic if its elements are considered to be indivisible units (primitive attributes) Examples of non-atomic domains are multi-valued and composite attributes A relational schema R is in first normal form (1NF) if the domains of all attributes of R are atomic 13 We assume all relations are in 1NF

14. First Normal Form (1NF): Example 14 Since all attributes are primitive  It is in 1NF

15. Boyce-Codd Normal Form (BCNF): Definition A relation schema R is in BCNF with respect to a set F of functional dependencies if for all functional dependencies in F + of the form α → β where α ⊆ R and β ⊆ R, then at least one of the following holds:  α → β is trivial (i.e.,β ⊆ α)  α is a superkey for R 15 Remember: Candidate keys are also superkeys Remember: Candidate keys are also superkeys

16. BCNF: Example 16 sNumbersNamepNumberpName s1Davep1MM s2Gregp2ER s3Mikep1MM Student Student InfoProfessor Info Is relation Student in BCNF given pNumber  pName It is not trivial FD pNumber is not a key in Student relation How to fix it and make it in BCNF??? NO

17. Decomposing a Schema into BCNF If R is not in BCNF because of non-trivial dependency α → β, then decompose R R is decomposed into two relations R1 = (α U β ) -- α is super key in R1 R2 = (R- (β - α)) -- R2.α is foreign keys to R1.α 17

18. Example of BCNF Decomposition sNumbersNamepNumberpName s1Davep1MM s2Gregp2MM StudentProf FDs: pNumber  pName sNumbersNamepNumber s1Davep1 s2Gregp2 Student pNumberpName p1MM p2MM Professor FOREIGN KEY: Student (PNum) references Professor (PNum) 18

19. What is Nice about this Decomposing ??? R is decomposed into two relations R1 = (α U β ) -- α is super key in R1 R2 = (R- (β - α)) -- R2.α is foreign keys to R1.α 19 This decomposition is lossless (Because R1 and R2 can be joined based on α, and α is unique in R1) This decomposition is lossless (Because R1 and R2 can be joined based on α, and α is unique in R1) When you join R1 and R2 on α, you get R back without lose of information

20. StudentProf = Student ⋈ Professor sNumbersNamepNumberpName s1Davep1MM s2Gregp2MM StudentProf FDs: pNumber  pName sNumbersNamepNumber s1Davep1 s2Gregp2 Student pNumberpName p1MM p2MM Professor BCNF decomposition rule create lossless decomposition 20

21. Multi-Step Decomposition Relation R and functional dependency F R = (customer_name, loan_number, branch_name, branch_city, assets, amount ) F = {branch_name  assets branch_city, loan_number  amount branch_name} Is R in BCNF ?? Based on branch_name  assets branch_city R1 = (branch_name, assets, branch_city) R2 = (customer_name, loan_number, branch_name, amount) Are R1 and R2 in BCNF ? Divide R2 based on loan_number  amount branch_name R3 = (loan_number, amount, branch_name) R4 = (customer_name, loan_number) 21 NO R2 is not Final Schema has R1, R3, R4

22. What is NOT Nice about BCNF Before decomposition, we had set of functional dependencies FDs (Say F) 22 After decomposition, do we still have the same set of FDs or we lost something ??

23. What is NOT Nice about BCNF Dependency Preservation After the decomposition, all FDs in F + should be preserved BCNF does not guarantee dependency preservation Can we always find a decomposition that is both BCNF and preserving dependencies? No…This decomposition may not exist That is why we study a weaker normal form called (third normal form –3NF) 23

24. Dependency Preserving Assume R is decomposed to R1 and R2 Dependencies of R1 and R2 include: Local dependencies α → β All columns of α and β must be in a single relation Global Dependencies Use transitivity property to form more FDs across R1 and R2 relations 24 Does these dependencies match the ones in R ? Yes  Dependency preserving No  Not dependency preserving

25. Example of Lost FD Assume relation R(C, S, J, D, T, Q, V) C is key, JT  C and SD  T C  CSJDTQV (C is key) -- Good for BCNF JT  CSJDTQV (JT is key) -- Good for BCNF SD  T (SD is not a key) –Bad for BCNF Decomposition: R1(C, S, J, D, Q, V) and R2(S, D, T) Does C  CSJDTQV still exist? Yes: C  CSJDQV (local), SD  T (local), C  CSJDQVT (global) 25 Lossless & in BCNF

26. Example of Lost FD (Cont’d) Assume relation R(C, S, J, D, T, Q, V) C is key, JT  C and SD  T C  CSJDTQV (C is key) -- Good for BCNF JT  CSJDTQV (JT is key) -- Good for BCNF SD  T (SD is not a key) –Bad for BCNF Decomposition: R1(C, S, J, D, Q, V) and R2(S, D, T) Does SD  T still exist? Yes: SD  T (local) 26 Lossless & in BCNF

27. Example of Lost FD (Cont’d) Assume relation R(C, S, J, D, T, Q, V) C is key, JT  C and SD  T C  CSJDTQV (C is key) -- Good for BCNF JT  CSJDTQV (JT is key) -- Good for BCNF SD  T (SD is not a key) –Bad for BCNF Decomposition: R1(C, S, J, D, Q, V) and R2(S, D, T) Does JT  CSJDTQV still exist? No this one is lost (no way from the local FDs to get this one) 27 Lossless & in BCNF

28. Dependency Preservation Test Assume R is decomposed into R1 and R2 The closure of FDs in R is F + The FDs in R1 and R2 are F R1 and F R2, respectively Then dependencies are preserved if: F + = (F R1 union F R2 ) + 28 local dependencies in R1 local dependencies in R2

29. Back to Our Example Assume relation R(C, S, J, D, T, Q, V) C is key, JT  C and SD  T C  CSJDTQV (C is key) -- Good for BCNF JT  CSJDTQV (JT is key) -- Good for BCNF SD  T (SD is not a key) –Bad for BCNF Decomposition: R1(C, S, J, D, Q, V) and R2(S, D, T) F + = {C  CSJDTQV, JT  CSJDTQV, SD  T} F R1 = {C  CSJDQV}  local for R1 F R2 = {SD  T}  local for R2 F R1 U F R2 = {C  CSJDQV, SD  T} (F R1 U F R2 ) + = {C  CSJDQV, SD  T, C  T} 29 JT  C is still missing

30. Dependency Preservation BCNF does not necessarily preserve FDs. But 3NF is guaranteed to be able to preserve FDs. 30

31. Normalization First Normal Form (1NF) Boyce-Codd Normal Form (BCNF) Third Normal Form (3NF) Canonical Cover of FDs 31

32. Third Normal Form: Motivation There are some situations where BCNF is not dependency preserving Solution: Define a weaker normal form, called Third Normal Form (3NF) Allows some redundancy (we will see examples later) But all FDs are preserved 32 There is always a lossless, dependency- preserving decomposition in 3NF

33. Normal Form : 3NF Relation R is in 3NF if, for every FD in F + α  β, where α ⊆ R and β ⊆ R, at least one of the following holds:  α → β is trivial (i.e.,β ⊆ α)  α is a superkey for R  Each attribute in β-α is part of a candidate key (prime attribute) 33 L.H.S is superkey OR R.H.S consists of prime attributes L.H.S is superkey OR R.H.S consists of prime attributes

34. Testing for 3NF Use attribute closure to check for each dependency α → β, if α is a superkey If α is not a superkey, we have to verify if each attribute in (β- α) is contained in a candidate key of R 34

35. 3NF: Example Lot (ID, county, lotNum, area, price, taxRate) Primary key: ID Candidate key: FDs: county  taxRate area  price Decomposition based on county  taxRate Lot (ID, county, lotNum, area, price) County (county, taxRate) 35 Is relation Lot in 3NF ? NO Are relations Lot and County in 3NF ? Lot is not

36. 3NF: Example (Cont’d) Lot (ID, county, lotNum, area, price) County (county, taxRate) Candidate key for Lot: FDs: county  taxRate area  price Decompose Lot based on area  price Lot (ID, county, lotNum, area) County (county, taxRate) Area (area, price) 36 Is every relation in 3NF ? YES

37. Comparison between 3NF & BCNF ? If R is in BCNF, obviously R is in 3NF If R is in 3NF, R may not be in BCNF 3NF allows some redundancy and is weaker than BCNF 3NF is a compromise to use when BCNF with good constraint enforcement is not achievable Important: Lossless, dependency-preserving decomposition of R into a collection of 3NF relations always possible ! 37

38. Normalization First Normal Form (1NF) Boyce-Codd Normal Form (BCNF) Third Normal Form (3NF) Canonical Cover of FDs 38

39. Canonical Cover of FDs 39

40. Canonical Cover of FDs Canonical Cover (Minimal Cover) = G Is the smallest set of FDs that produce the same F + There are no extra attributes in the L.H.S or R.H.S of and dependency in G Given set of FDs (F) with functional closure F + Canonical cover of F is the minimal subset of FDs (G), where G + = F + 40 Every FD in the canonical cover is needed, otherwise some dependencies are lost

41. Example : Canonical Cover Given F: A  B, ABCD  E, EF  GH, ACDF  EG Then the canonical cover G: A  B, ACD  E, EF  GH 41 The smallest set (minimal) of FDs that can generate F +

42. Computing the Canonical Cover Given a set of functional dependencies F, how to compute the canonical cover G 42

43. Example : Canonical Cover (Lets Check L.H.S) Given F = {A  B, ABCD  E, EF  G, EF  H, ACDF  EG} Union Step: {A  B, ABCD  E, EF  GH, ACDF  EG} Test ABCD  E Check A: {BCD}+ = {BCD}  A cannot be deleted Check B: {ACD} + = {A B C D E}  Then B can be deleted Now the set is: {A  B, ACD  E, EF  GH, ACDF  EG} Test ACD  E Check C: {AD}+ = {ABD}  C cannot be deleted Check D: {AC}+ = {ABC}  D cannot be deleted 43

44. Example: Canonical Cover (Lets Check L.H.S-Cont’d) Now the set is: {A  B, ACD  E, EF  GH, ACDF  EG} Test EF  GH Check E: {F}+ = {F}  E cannot be deleted Check F: {E}+ = {E}  F cannot be deleted Test ACDF  EG None of the H.L.S can be deleted 44

45. Example: Canonical Cover (Lets Check R.H.S) Now the set is: {A  B, ACD  E, EF  GH, ACDF  EG} Test EF  GH Check G: {EF}+ = {E F H}  G cannot be deleted Check H: {EF}+ = {E F G}  H cannot be deleted Test ACDF  EG Check E: {ACDF}+ = {A B C D F E G}  E can be deleted Now the set is: {A  B, ACD  E, EF  GH, ACDF  G} 45

46. Example: Canonical Cover (Lets Check R.H.S-Cont’d) Now the set is: {A  B, ACD  E, EF  GH, ACDF  G} Test ACDF  G Check G: {ACDF}+ = {A B C D F E G}  G can be deleted Now the set is: {A  B, ACD  E, EF  GH} 46 The canonical cover is: {A  B, ACD  E, EF  GH} The canonical cover is: {A  B, ACD  E, EF  GH}

47. Canonical Cover Used to find the smallest (minimal) set of FDs that have the same closure as the original set. Used in the decomposition of relations to be in 3NF The resulting decomposition is lossless and dependency preserving 47

48. Done with Normalization First Normal Form (1NF) Boyce-Codd Normal Form (BCNF) Third Normal Form (3NF) Canonical Cover of FDs 48

49. Questions ? 49

50. What You Learned Data Models Entity-Relationship Model & ERD Relational Model Conversion between the data models Relational Algebra & Operators Structured Query Language SQL DML: Data Manipulation Language DDL: Data Definition Language 50

51. What You Learned (Cont’d) Advanced SQL Triggers, Views, Cursors, Stored Procedures and Functions PL/SQL Functional Dependencies Normalization Rules 51

52. In Advanced Courses Things get more interesting Indexing Techniques Transaction Management Query Optimization Handling of Big Data And many more … 52