1. Chapter 7: Relational Database Design

2. ©Silberschatz, Korth and Sudarshan7.2Database System Concepts Chapter 7: Relational Database Design First Normal Form Pitfalls in Relational Database Design Functional Dependencies Decomposition Boyce-Codd Normal Form Third Normal Form Multivalued Dependencies and Fourth Normal Form Overall Database Design Process

3. ©Silberschatz, Korth and Sudarshan7.3Database System Concepts First Normal Form Domain is atomic if its elements are considered to be indivisible units  Examples of non-atomic domains:  Set of names, composite attributes  Identification numbers like CS101 that can be broken up into parts (leads to encoding of information in application program rather than in the database) A relational schema R is in first normal form if the domains of all attributes of R are atomic Non-atomic values complicate storage and encourage redundant (repeated) storage of data  E.g. Set of accounts stored with each customer, and set of owners stored with each account (instead of relation depositor)  We assume all relations are in first normal form

4. ©Silberschatz, Korth and Sudarshan7.4Database System Concepts Pitfalls in Relational Database Design Relational database design requires that we find a “good” collection of relation schemas. A bad design may lead to  Repetition of Information.  Inability to represent certain information. Design Goals:  Avoid redundant data  Ensure that relationships among attributes are represented  Facilitate the checking of updates for violation of database integrity constraints.

5. ©Silberschatz, Korth and Sudarshan7.5Database System ConceptsExample Consider the relation schema: Lending-schema = (branch-name, branch-city, assets, customer-name, loan-number, amount) Redundancy:  Data for branch-name, branch-city, assets are repeated for each loan that a branch makes (functional dependency: branch-name  branch-city assets)  Wastes space  Complicates updating, introducing possibility of inconsistency of assets value Null values  Cannot store information about a branch if no loans exist  Can use null values, but they are difficult to handle.

6. ©Silberschatz, Korth and Sudarshan7.6Database System ConceptsDecomposition Decompose the relation schema Lending-schema into: Branch-schema = (branch-name, branch-city,assets) Loan-info-schema = (customer-name, loan-number, branch-name, amount) All attributes of an original schema (R) must appear in the decomposition (R 1, R 2 ): R = R 1  R 2 Lossless-join decomposition. For all possible relations r on schema R r =  R1 (r)  R2 (r)

7. ©Silberschatz, Korth and Sudarshan7.7Database System Concepts Example of Non Lossless-Join Decomposition Decomposition of R = (A, B) R 1 = (A)R 2 = (B) AB  121121 A  B 1212 r A(r)A(r)  B(r)  A (r)  B (r) AB  12121212

8. ©Silberschatz, Korth and Sudarshan7.8Database System Concepts Goal — Devise a Theory for the Following Decide whether a particular relation R is in “good” form. In the case that a relation R is not in “good” form, decompose it into a set of relations {R 1, R 2,..., R n } such that  each relation is in good form  the decomposition is a lossless-join decomposition Our theory is based on:  functional dependencies  multivalued dependencies

9. ©Silberschatz, Korth and Sudarshan7.9Database System Concepts Functional Dependencies Constraints on the set of legal relations. Require that the value for a certain set of attributes determines uniquely the value for another set of attributes. A functional dependency is a generalization of the notion of a key.

10. ©Silberschatz, Korth and Sudarshan7.10Database System Concepts Functional Dependencies (Cont.) Let R be a relation schema   R and   R The functional dependency    holds on R if in any legal relations r(R), for all pairs of tuples t 1 and t 2 in r, such that t 1 [  ] = t 2 [  ]  t 1 [  ] = t 2 [  ] Example: Consider r(A,B) with the following instance of r. On this instance, A  B does NOT hold, but B  A does hold. 14 1 5 37

11. ©Silberschatz, Korth and Sudarshan7.11Database System Concepts Functional Dependencies (Cont.) K is a superkey for relation schema R if and only if K  R K is a candidate key for R if and only if  K  R, and  for no   K,   R Functional dependencies allow us to express constraints that cannot be expressed using superkeys. Consider the schema: Loan-info-schema = (customer-name, loan-number, branch-name, amount). We expect this set of functional dependencies to hold: loan-number  amount loan-number  branch-name but would not expect the following to hold: loan-number  customer-name

12. ©Silberschatz, Korth and Sudarshan7.12Database System Concepts Use of Functional Dependencies We use functional dependencies to:  test relations to see if they are legal under a given set of functional dependencies.  If a relation r is legal under a set F of functional dependencies, we say that r satisfies F.  specify constraints on the set of legal relations  We say that F holds on R if all legal relations on R satisfy the set of functional dependencies F. Note: A specific instance of a relation schema may satisfy a functional dependency even if the functional dependency does not hold on all legal instances. For example, a specific instance of Loan-schema may, by chance, satisfy loan-number  customer-name.

13. ©Silberschatz, Korth and Sudarshan7.13Database System Concepts Functional Dependencies A  C C  A (satisfy)

14. ©Silberschatz, Korth and Sudarshan7.14Database System Concepts Functional Dependencies (Cont.) A functional dependency is trivial if it is satisfied by all instances of a relation  E.g.  customer-name, loan-number  customer-name  customer-name  customer-name  In general,    is trivial if    When we design a relational database, we first list those functional dependencies that must always hold The list of functional dependencies in banking database (next page)

15. ©Silberschatz, Korth and Sudarshan7.15Database System Concepts Functional dependencies in banking database On Branch-schema branch-name  branch-city branch-name  assets On Customer-schema customer-name  branch-city customer-name  customer-street On Loan-schema loan-number  amount loan-number  branch-name On Account-schema account-number  branch-name account-number  balance

16. ©Silberschatz, Korth and Sudarshan7.16Database System Concepts Closure of a Set of Functional Dependencies Given a set F of functional dependencies, there are certain other functional dependencies that are logically implied by F.  E.g. If A  B and B  C, then we can infer that A  C The set of all functional dependencies logically implied by F is the closure of F. We denote the closure of F by F +. We can find all of F + by applying Armstrong’s Axioms:  if   , then    (reflexivity)  if   , then      (augmentation)  if   , and   , then    (transitivity) These rules are  sound (generate only functional dependencies that actually hold) and  complete (generate all functional dependencies that hold).

17. ©Silberschatz, Korth and Sudarshan7.17Database System ConceptsExample R = (A, B, C, G, H, I) F = { A  B A  C CG  H CG  I B  H} some members of F +  A  H  by transitivity from A  B and B  H  AG  I  by augmenting A  C with G, to get AG  CG and then transitivity with CG  I  CG  HI  from CG  H and CG  I : “union rule” can be inferred from –definition of functional dependencies, or –Augmentation of CG  I to infer CG  CGI, augmentation of CG  H to infer CGI  HI, and then transitivity

18. ©Silberschatz, Korth and Sudarshan7.18Database System Concepts Procedure for Computing F + To compute the closure of a set of functional dependencies F: F + = F repeat for each functional dependency f in F + apply reflexivity and augmentation rules on f add the resulting functional dependencies to F + for each pair of functional dependencies f 1 and f 2 in F + if f 1 and f 2 can be combined using transitivity then add the resulting functional dependency to F + until F + does not change any further NOTE: We will see an alternative procedure for this task later

19. ©Silberschatz, Korth and Sudarshan7.19Database System Concepts Closure of Functional Dependencies (Cont.) We can further simplify manual computation of F + by using the following additional rules.  If    holds and    holds, then     holds (union)  If     holds, then    holds and    holds (decomposition)  If    holds and     holds, then     holds (pseudotransitivity) The above rules can be inferred from Armstrong’s axioms.

20. ©Silberschatz, Korth and Sudarshan7.20Database System Concepts Closure of Attribute Sets Given a set of attributes  define the closure of  under F (denoted by  + ) as the set of attributes that are functionally determined by  under F:    is in F +     + Algorithm to compute  +, the closure of  under F result :=  ; while (changes to result) do for each    in F do begin if   result then result := result   end

21. ©Silberschatz, Korth and Sudarshan7.21Database System Concepts Example of Attribute Set Closure R = (A, B, C, G, H, I) F = {A  B A  C CG  H CG  I B  H} (AG) + 1.result = AG 2.result = ABCG(A  C and A  B) 3.result = ABCGH(CG  H and CG  AGBC) 4.result = ABCGHI(CG  I and CG  AGBCH) Is AG a candidate key? 1. Is AG a super key? 1. Does AG  R? 2. Is any subset of AG a superkey? 1. Does A +  R? 2. Does G +  R?

22. ©Silberschatz, Korth and Sudarshan7.22Database System Concepts Uses of Attribute Closure There are several uses of the attribute closure algorithm: Testing for superkey:  To test if  is a superkey, we compute  +, and check if  + contains all attributes of R. Testing functional dependencies  To check if a functional dependency    holds (or, in other words, is in F + ), just check if    +.  That is, we compute  + by using attribute closure, and then check if it contains .  Is a simple and cheap test, and very useful Computing closure of F  For each   R, we find the closure  +, and for each S   +, we output a functional dependency   S.

23. ©Silberschatz, Korth and Sudarshan7.23Database System Concepts Canonical Cover Sets of functional dependencies may have redundant dependencies that can be inferred from the others  Eg: A  C is redundant in: {A  B, B  C, A  C}  Parts of a functional dependency may be redundant  E.g. on RHS: {A  B, B  C, A  CD} can be simplified to {A  B, B  C, A  D}  E.g. on LHS: {A  B, B  C, AC  D} can be simplified to {A  B, B  C, A  D} Intuitively, a canonical cover of F is a “minimal” set of functional dependencies equivalent to F, with no redundant dependencies or having redundant parts of dependencies

24. ©Silberschatz, Korth and Sudarshan7.24Database System Concepts Extraneous Attributes Consider a set F of functional dependencies and the functional dependency    in F.  Attribute A is extraneous in  if A   and F logically implies (F – {    })  {(  – A)   }.  Attribute A is extraneous in  if A   and the set of functional dependencies (F – {    })  {   (  – A)} logically implies F. Example: Given F = {A  C, AB  C }  B is extraneous in AB  C because A  C logically implies AB  C. Example: Given F = {A  C, AB  CD}  C is extraneous in AB  CD since A  C can be inferred even after deleting C

25. ©Silberschatz, Korth and Sudarshan7.25Database System Concepts Testing if an Attribute is Extraneous Consider a set F of functional dependencies and the functional dependency    in F.  To test if attribute A   is extraneous in  (check if (  – {A})   ) 1. compute (  – {A}) + using the dependencies in F 2. check that (  – {A}) + contains  ; if it does, A is extraneous  To test if attribute A   is extraneous in  1. compute  + using only the dependencies in F’ = (F – {    })  {   (  – A)}, (check if   A can be inferred from F’) 2. check that  + contains A; if it does, A is extraneous Example:  R = (A, B, C, D, E) F = { AB  CD A  E E  C } To check if C is extraneous in AB  CD

26. ©Silberschatz, Korth and Sudarshan7.26Database System Concepts Canonical Cover A canonical cover for F is a set of dependencies F c such that  F logically implies all dependencies in F c, and  F c logically implies all dependencies in F, and  No functional dependency in F c contains an extraneous attribute, and  Each left side of functional dependency in F c is unique. To compute a canonical cover for F: F c = F repeat Use the union rule to replace any dependencies in F c of the form  1   1 and  1   2 with  1   1  2 Find a functional dependency    in F c with an extraneous attribute either in  or in  If an extraneous attribute is found, delete it from    until F c does not change Note: Union rule may become applicable after some extraneous attributes have been deleted, so it has to be re-applied

27. ©Silberschatz, Korth and Sudarshan7.27Database System Concepts Example of Computing a Canonical Cover R = (A, B, C) F = {A  BC B  C A  B AB  C} Combine A  BC and A  B into A  BC  Set is now {A  BC, B  C, AB  C} A is extraneous in AB  C because B  C logically implies AB  C.  Set is now {A  BC, B  C} C is extraneous in A  BC since A  BC is logically implied by A  B and B  C. The canonical cover is: A  B B  C

28. ©Silberschatz, Korth and Sudarshan7.28Database System Concepts Goals of Normalization Decide whether a particular relation R is in “good” form. In the case that a relation R is not in “good” form, decompose it into a set of relations {R 1, R 2,..., R n } such that  each relation is in good form  the decomposition is a lossless-join decomposition Our theory is based on:  functional dependencies  multivalued dependencies

29. ©Silberschatz, Korth and Sudarshan7.29Database System ConceptsDecomposition Figure 7.1: (a bad database design) Lending -schema = (branch-name, branch-city, assets, customer-name, loan-number, amount)  Repetition of information (add a new loan)  Inability to represent certain information (a branch no loan) Observation: branch-name  branch-city branch-name  assets but not branch-name  loan-number From above example, we should decompose a relation schema into several schema with fewer attributes

30. ©Silberschatz, Korth and Sudarshan7.30Database System ConceptsDecomposition Lending-schema is decomposed into two schemas: Branch-customer-schema = (branch-name, branch-city, assets, customer-name) Customer-loan-schema = (customer-name, loan-number, amount) branch-customer =  branch-name, branch-city, assets, customer-name (Lending) customer-loan =  customer-name, loan-number, amount  (Lending) Figure7.9 and Figure 7.10 We wish to find all branches that have loans with amount less than $1000: Branch-customer  Customer-loan (Figure 7.11) We are no longer able to represent in the database information about which customers are borrowers from which branch

31. ©Silberschatz, Korth and Sudarshan7.31Database System Concepts Lending

32. ©Silberschatz, Korth and Sudarshan7.32Database System Concepts Branch-customer Customer-loan

33. ©Silberschatz, Korth and Sudarshan7.33Database System Concepts Branch-customer Customer-loan

34. ©Silberschatz, Korth and Sudarshan7.34Database System ConceptsDecomposition Because of this loss of information, we call the decomposition of Lending-schema into Branch-customer-schema and Customer- loan-schema a lossy decomposition, or a lossy-join decomposition A decomposition that is not a lossy-join decomposition is a lossless-join decomposition Consider another alternative design, in which Lending-schema is decomposed into two schemas:  Branch-schema = (branch-name, branch-city, assets)  Loan-info-schema = (branch-name, customer-name, loan-number, amount) This decomposition is a lossless-join decomposition, because branch-name  branch-city assets

35. ©Silberschatz, Korth and Sudarshan7.35Database System ConceptsDecomposition All attributes of an original schema (R) must appear in the decomposition (R 1, R 2 ): R = R 1  R 2 Lossless-join decomposition. For all possible relations r on schema R r =  R1 (r)  R2 (r) A decomposition of R into R 1 and R 2 is lossless join if and only if at least one of the following dependencies is in F + :  R 1  R 2  R 1  R 1  R 2  R 2

36. ©Silberschatz, Korth and Sudarshan7.36Database System ConceptsDecomposition Example: Lending-schema = (branch-name, branch-city, assets, customer-name, loan-number, amount). The set F of functional dependencies that we require to hold on Lending-schema are branch-name  branch-city assets loan-number  amount branch-name We decompose Lending-schema into two schemas Branch-schema = (branch-name, branch-city, assets) Loan-info-schema = (branch-name, customer-name, loan-number, amount) Since branch-name  branch-name branch-city assets This decomposition is a lossless-join decomposition Next, we decompose Loan-info-schema into Loan-schema = (branch-name, loan-number, amount) borrower-schema = (customer-name, loan-number) Since loan-number  amount branch-name This decomposition is a lossless-join decomposition

37. ©Silberschatz, Korth and Sudarshan7.37Database System Concepts Dependency Preservation When an update is made to the database, the system should be able to check that the update will satisfy all the given functional dependencies To check updates efficiently, we should design relational database schemas that allow update validation without the computation of joins F : a set of functional dependencies on a schema R R 1, R 2, …, R n : a decomposition of R F i : a subset of all functional dependencies in F + that include only attributes of R i The set of restrictions F 1, F 2, …, F n is the set of dependencies that can be checked efficiently Let F’ = F 1  F 2  …  F n Dependency preserving decomposition: A decomposition has the property F’ + = F +

38. ©Silberschatz, Korth and Sudarshan7.38Database System Concepts Normalization Using Functional Dependencies When we decompose a relation schema R with a set of functional dependencies F into R 1, R 2,.., R n we want  Lossless-join decomposition: Otherwise decomposition would result in information loss.  No redundancy: The relations R i preferably should be in either Boyce- Codd Normal Form or Third Normal Form.  Dependency preservation: Let F i be the set of dependencies F + that include only attributes in R i.  Preferably the decomposition should be dependency preserving, that is, (F 1  F 2  …  F n ) + = F +  Otherwise, checking updates for violation of functional dependencies may require computing joins, which is expensive.

39. ©Silberschatz, Korth and Sudarshan7.39Database System ConceptsExample R = (A, B, C) F = {A  B, B  C) R 1 = (A, B), R 2 = (B, C)  Lossless-join decomposition: R 1  R 2 = {B} and B  BC  Dependency preserving R 1 = (A, B), R 2 = (A, C)  Lossless-join decomposition: R 1  R 2 = {A} and A  AB  Not dependency preserving (cannot check B  C without computing R 1 R 2 )

40. ©Silberschatz, Korth and Sudarshan7.40Database System Concepts Testing for Dependency Preservation Algorithm for testing if a decomposition D is dependency preservation: compute F + for each schema R i in D do begin Fi := the restriction of F + to R i ; end F’ :=  for each restriction F i do begin F’ = F’  Fi end compute F’ + ; if (F’ + = F+) then return (true) else return (false);

41. ©Silberschatz, Korth and Sudarshan7.41Database System Concepts Testing for Dependency Preservation To check if a dependency  is preserved in a decomposition of R into R 1, R 2, …, R n we apply the following simplified test (with attribute closure done w.r.t. F)  result =  while (changes to result) do for each R i in the decomposition t = (result  R i ) +  R i result = result  t  If result contains all attributes in , then the functional dependency    is preserved. We apply the test on all dependencies in F to check if a decomposition is dependency preserving This procedure takes polynomial time, instead of the exponential time required to compute F + and (F 1  F 2  …  F n ) + (A, B, C, D) A  B AB  CD * B  C C  D (A, B, C) (A, D)

42. ©Silberschatz, Korth and Sudarshan7.42Database System Concepts Boyce-Codd Normal Form     is trivial (i.e.,    )  is a superkey for R 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    , where   R and   R, at least one of the following holds: A database design is in BCNF if each relation schema in the database is in BCNF

43. ©Silberschatz, Korth and Sudarshan7.43Database System ConceptsExample Customer-schema = (customer-name, customer-street, customer-city) customer-name  customer-street customer-city Customer-schema is in BCNF Branch-schema = (branch-name, assets, branch-city) branch-name  assets branch-city Branch-schema is in BCNF Loan -info-schema = (branch-name, customer-name, loan- number, amount) loan-number  branch-name amount Loan-info-schema is not in BCNF Loan-info-schema suffers from the problem of information repetition: (Downtown, Mr. Bell, L-44, 1000) (Downtown, Ms. Bell, L-44, 1000) Decompose Loan-info-schema into two schemas: Loan-schema = (branch-name, loan-number, amount) Borrower-schema = (customer-name, loan-number) This decomposition is a lossless-join decomposition

44. ©Silberschatz, Korth and Sudarshan7.44Database System Concepts Testing for BCNF To check if a non-trivial dependency     causes a violation of BCNF 1. compute  + (the attribute closure of  ), and 2. verify that it includes all attributes of R, that is, it is a superkey of R. Simplified test: To check if a relation schema R with a given set of functional dependencies F is in BCNF, it suffices to check only the dependencies in the given set F for violation of BCNF, rather than checking all dependencies in F +.  We can show that if none of the dependencies in F causes a violation of BCNF, then none of the dependencies in F + will cause a violation of BCNF either. However, using only F is incorrect when testing a relation in a decomposition of R  E.g. Consider R (A, B, C, D), with F = { A  B, B  C}  Decompose R into R 1 (A,B) and R 2 (A,C,D)  Neither of the dependencies in F contain only attributes from (A,C,D) so we might be mislead into thinking R 2 satisfies BCNF.  In fact, dependency A  C in F + shows R 2 is not in BCNF.

45. ©Silberschatz, Korth and Sudarshan7.45Database System Concepts Testing for BCNF An alternative BCNF test is sometimes easier than computing every dependency in F + To check if a relation R i in a decomposition of R is in BCNF, we apply this test  For every subset  of attributes in R i, check that  + (the attribute closure of  under F) either includes no attribute of R i - , or includes all attributes of R i If the condition is violated by some set of attributes  in R i, the following functional dependency must be in F +     (  + -  )  R i

46. ©Silberschatz, Korth and Sudarshan7.46Database System Concepts BCNF Decomposition Algorithm result := {R}; done := false; compute F + ; while (not done) do if (there is a schema R i in result that is not in BCNF) then begin let     be a nontrivial functional dependency that holds on R i such that    R i is not in F +, and    =  ; result := (result – R i )  (R i –  )  ( ,  ); end else done := true; Note: each R i is in BCNF, and decomposition is lossless-join. Since the dependency   holds on R i, schema R i is decomposed into (R i -  ) and ( ,  ), and (R i -  )  ( ,  ) = 

47. ©Silberschatz, Korth and Sudarshan7.47Database System Concepts Example of BCNF Decomposition Lending-schema = (branch-name, branch-city, assets, customer-name, loan-number, amount) F = {branch-name  assets branch-city loan-number  amount branch-name} Key = {loan-number, customer-name} For the dependency branch-name  assets branch-city, Lending-schema is not in BCNF: Branch = (branch-name, branch-city, assets) Loan-info = (branch-name, customer-name, loan-number, amount) For the dependency loan-number  branch-name amount, Loan-info is not in BCNF: Loan = (branch-name, loan-number, amount) is in BCNF Borrower = (customer-name, loan-number) is in BCNF Not every BCNF decomposition is dependency preserving

48. ©Silberschatz, Korth and Sudarshan7.48Database System Concepts Example of BCNF Decomposition Banker-schema = (branch-name, customer-name, banker-name) banker-name  branch-name branch-name customer-name  banker-name Since banker-name is not a superkey, Banker-schema is not in BCNF The BCNF decomposition: Banker-branch = (banker-name, branch-name) Customer-banker = (customer-name, banker-name) This decomposition is not dependency preserving: F 1 = {banker-name  branch-name} F 2 =  F’ = {banker-name  branch-name} F’ +  F + Hence, this decomposition is not dependency preserving

49. ©Silberschatz, Korth and Sudarshan7.49Database System Concepts Third Normal Form: Motivation There are some situations where  BCNF is not dependency preserving, and  efficient checking for FD violation on updates is important Solution: define a weaker normal form, called Third Normal Form.  Allows some redundancy (with resultant problems; we will see examples later)  But FDs can be checked on individual relations without computing a join.  There is always a lossless-join, dependency-preserving decomposition into 3NF.

50. ©Silberschatz, Korth and Sudarshan7.50Database System Concepts Third Normal Form A relation schema R is in third normal form (3NF) if for all:    in F + at least one of the following holds:     is trivial (i.e.,    )   is a superkey for R  Each attribute A in  –  is contained in a candidate key for R. (NOTE: each attribute may be in a different candidate key) If a relation is in BCNF it is in 3NF (since in BCNF one of the first two conditions above must hold). Third condition is a minimal relaxation of BCNF to ensure dependency preservation.

51. ©Silberschatz, Korth and Sudarshan7.51Database System Concepts 3NF (Cont.) Example  R = (J, K, L) F = {JK  L, L  K}  Two candidate keys: JK and JL  R is in 3NF JK  LJK is a superkey L  KK is contained in a candidate key BCNF decomposition has (JL) and (LK) Testing for JK  L requires a join There is some redundancy in this schema Equivalent to example in book: Banker-schema = (branch-name, customer-name, banker-name) banker-name  branch name branch name customer-name  banker-name

52. ©Silberschatz, Korth and Sudarshan7.52Database System Concepts Testing for 3NF Optimization: Need to check only FDs in F, need not check all FDs in F +. Use attribute closure to check, for each dependency   , if  is a superkey. If  is not a superkey, we have to verify if each attribute in  is contained in a candidate key of R  this test is rather more expensive, since it involve finding candidate keys  testing for 3NF has been shown to be NP-hard  Interestingly, decomposition into third normal form (described shortly) can be done in polynomial time

53. ©Silberschatz, Korth and Sudarshan7.53Database System Concepts 3NF Decomposition Algorithm Let F c be a canonical cover for F; i := 0; for each functional dependency    in F c do if none of the schemas R j, 1  j  i contains   then begin i := i + 1; R i :=   end if none of the schemas R j, 1  j  i contains a candidate key for R then begin i := i + 1; R i := any candidate key for R; end return (R 1, R 2,..., R i )

54. ©Silberschatz, Korth and Sudarshan7.54Database System Concepts 3NF Decomposition Algorithm (Cont.) Above algorithm ensures:  each relation schema R i is in 3NF  decomposition is dependency preserving and lossless-join

55. ©Silberschatz, Korth and Sudarshan7.55Database System ConceptsExample Relation schema: Banker-info-schema = (branch-name, customer-name, banker-name, office-number) The functional dependencies for this relation schema are: banker-name  branch-name office-number customer-name branch-name  banker-name The key is: {customer-name, branch-name}

56. ©Silberschatz, Korth and Sudarshan7.56Database System Concepts Applying 3NF to Banker-info-schema The for loop in the algorithm causes us to include the following schemas in our decomposition: Banker-office-schema = (banker-name, branch-name, office-number) Banker-schema = (customer-name, branch-name, banker-name) Since Banker-schema contains a candidate key for Banker-info-schema, we are done with the decomposition process.

57. ©Silberschatz, Korth and Sudarshan7.57Database System Concepts Comparison of BCNF and 3NF It is always possible to decompose a relation into relations in 3NF and  the decomposition is lossless  the dependencies are preserved It is always possible to decompose a relation into relations in BCNF and  the decomposition is lossless  it may not be possible to preserve dependencies.

58. ©Silberschatz, Korth and Sudarshan7.58Database System Concepts Comparison of BCNF and 3NF (Cont.) J j 1 j 2 j 3 null L l1l1l1l2l1l1l1l2 K k1k1k1k2k1k1k1k2 A schema that is in 3NF but not in BCNF has the problems of repetition of information (e.g., the relationship l 1, k 1 ) need to use null values (e.g., to represent the relationship l 2, k 2 where there is no corresponding value for J). Example of problems due to redundancy in 3NF  R = (J, K, L) F = {JK  L, L  K} Equivalent to example in book: Banker-schema = (customer-name, branch-name, banker-name) banker-name  branch name branch name customer-name  banker-name

59. ©Silberschatz, Korth and Sudarshan7.59Database System Concepts Design Goals Goal for a relational database design is:  BCNF.  Lossless join.  Dependency preservation. If we cannot achieve this, we accept one of  Lack of dependency preservation  Redundancy due to use of 3NF

60. ©Silberschatz, Korth and Sudarshan7.60Database System Concepts Multivalued Dependencies Example: BC-schema = (loan-number, customer-name, customer-street, customer-city) customer-name  customer-street customer-city  This schema is not in BCNF  Assume that our bank is attracting wealthy customers who have several addresses, we no longer wish to enforce the dependency customer-name  customer-street customer-city  If the functional dependency is removed, then BC-schema is in BCNF. Hence, BC-schema still have the problem of repetition of information To deal with this problem, a new form of constraint, called a multivalued dependency, is defined Multivalued dependencies are used to define a normal form This normal form, called fourth normal form (4NF), is more restrictive than BCNF

61. ©Silberschatz, Korth and Sudarshan7.61Database System Concepts Since there are non-trivial dependencies, (course, teacher, book) is the only key, and therefore the relation is in BCNF Insertion anomalies – i.e., if Sara is a new teacher that can teach database, two tuples need to be inserted (database, Sara, DB Concepts) (database, Sara, Ullman) courseteacherbook database operating systems Avi Hank Sudarshan Avi Jim DB Concepts Ullman DB Concepts Ullman DB Concepts Ullman OS Concepts Shaw OS Concepts Shaw classes

62. ©Silberschatz, Korth and Sudarshan7.62Database System Concepts Therefore, it is better to decompose classes into: courseteacher database operating systems Avi Hank Sudarshan Avi Jim teaches coursebook database operating systems DB Concepts Ullman OS Concepts Shaw text We shall see that these two relations are in Fourth Normal Form (4NF)

63. ©Silberschatz, Korth and Sudarshan7.63Database System Concepts Multivalued Dependencies (MVDs) Let R be a relation schema and let   R and   R. The multivalued dependency    holds on R if in any legal relation r(R), for all pairs for tuples t 1 and t 2 in r such that t 1 [  ] = t 2 [  ], there exist tuples t 3 and t 4 in r such that: t 1 [  ] = t 2 [  ] = t 3 [  ] t 4 [  ] t 3 [  ] = t 1 [  ] t 3 [R –  ] = t 2 [R –  ] t 4  ] = t 2 [  ] t 4 [R –  ] = t 1 [R –  ]

64. ©Silberschatz, Korth and Sudarshan7.64Database System Concepts MVD (Cont.) Tabular representation of   

65. ©Silberschatz, Korth and Sudarshan7.65Database System Concepts Use of Multivalued Dependencies If a relation r fails to satisfy a given multivalued dependency, we can construct a relations r that does satisfy the multivalued dependency by adding tuples to r. customer-name  customer-street customer-city The closure D + of D is the set of all functional and multivalued dependencies logically implied by D.  We can compute D + from D, using the formal definitions of functional dependencies and multivalued dependencies.

66. ©Silberschatz, Korth and Sudarshan7.66Database System Concepts Fourth Normal Form From the definition of multivalued dependency, we can derive the following rule:  If   , then    That is, every functional dependency is also a multivalued dependency A relation schema R is in 4NF with respect to a set D of functional and multivalued dependencies if for all multivalued dependencies in D + of the form   , where   R and   R, at least one of the following hold:     is trivial (i.e.,    or    = R)   is a superkey for schema R If a relation is in 4NF it is in BCNF  If a schema R is not in BCNF, then there is a nontrivial functional dependency   holding on R, where  is not a superkey. Since   implies  , R cannot be in 4NF

67. ©Silberschatz, Korth and Sudarshan7.67Database System Concepts 4NF Decomposition Algorithm result: = {R}; done := false; compute D + ; Let D i denote the restriction of D + to R i while (not done) if (there is a schema R i in result that is not in 4NF) then begin let    be a nontrivial multivalued dependency that holds on R i such that   R i is not in D i, and  ; result := (result - R i )  (R i -  )  ( ,  ); end else done:= true; Note: each R i is in 4NF, and decomposition is lossless-join

68. ©Silberschatz, Korth and Sudarshan7.68Database System ConceptsExample R =(A, B, C, G, H, I) F ={ A  B B  HI CG  H } R is not in 4NF since A  B and A is not a superkey for R Decomposition a) R 1 = (A, B) (R 1 is in 4NF) b) R 2 = (A, C, G, H, I) (R 2 is not in 4NF) c) R 3 = (C, G, H) (R 3 is in 4NF) d) R 4 = (A, C, G, I) (R 4 is not in 4NF) Since A  B and B  HI, A  HI, A  I e) R 5 = (A, I) (R 5 is in 4NF) f ) R 6 = (A, C, G) (R 6 is in 4NF) 4NF decomposition: {(A, B), (C, G, H), (A, I), (A, C, G)} It fails to preserve the multivalued dependency B  HI

69. ©Silberschatz, Korth and Sudarshan7.69Database System ConceptsExample BC-schema = (loan-number, customer-name, customer-street, customer-city) customer-name  loan-number customer-name  customer-street customer-city Since customer-name  loan-number is a nontrivial multivalued dependency, and customer-name is not a superkey, BC-schema is decomposed into two relation schema: Borrower = (customer-name, loan-number) Customer = (customer-name, customer-street, customer-city)  These two relation schemas are in 4NF  Eliminate the problem of the redundancy of BC-schema Let R 1 and R 2 form a decomposition of R. This decomposition is a lossless-join decomposition of R if at least one of the following multivalued dependencies is in D + : R 1  R 2  R 1 R 1  R 2  R 2