Chapter 7: Relational Database Design

Chapter 7: Relational Database Design

اسلاید 1: Chapter 7: Relational Database Design

اسلاید 2: Chapter 7: Relational Database DesignFeatures of Good Relational DesignAtomic Domains and First Normal FormDecomposition Using Functional DependenciesFunctional Dependency TheoryAlgorithms for Functional DependenciesDecomposition Using Multivalued Dependencies More Normal FormDatabase-Design ProcessModeling Temporal Data

اسلاید 3: The Banking Schemabranch = (branch_name, branch_city, assets)customer = (customer_id, customer_name, customer_street, customer_city)loan = (loan_number, amount)account = (account_number, balance)employee = (employee_id. employee_name, telephone_number, start_date)dependent_name = (employee_id, dname)account_branch = (account_number, branch_name)loan_branch = (loan_number, branch_name)borrower = (customer_id, loan_number)depositor = (customer_id, account_number)cust_banker = (customer_id, employee_id, type)works_for = (worker_employee_id, manager_employee_id)payment = (loan_number, payment_number, payment_date, payment_amount)savings_account = (account_number, interest_rate)checking_account = (account_number, overdraft_amount)

اسلاید 4: Combine Schemas?Suppose we combine borrow and loan to get bor_loan = (customer_id, loan_number, amount )Result is possible repetition of information (L-100 in example below)

اسلاید 5: A Combined Schema Without RepetitionConsider combining loan_branch and loanloan_amt_br = (loan_number, amount, branch_name)No repetition (as suggested by example below)

اسلاید 6: What About Smaller Schemas?Suppose we had started with bor_loan. How would we know to split up (decompose) it into borrower and loan?Write a rule “if there were a schema (loan_number, amount), then loan_number would be a candidate key”Denote as a functional dependency: loan_number  amountIn bor_loan, because loan_number is not a candidate key, the amount of a loan may have to be repeated. This indicates the need to decompose bor_loan.Not all decompositions are good. Suppose we decompose employee intoemployee1 = (employee_id, employee_name)employee2 = (employee_name, telephone_number, start_date)The next slide shows how we lose information -- we cannot reconstruct the original employee relation -- and so, this is a lossy decomposition.

اسلاید 7: A Lossy Decomposition

اسلاید 8: First Normal FormDomain is atomic if its elements are considered to be indivisible unitsExamples of non-atomic domains:Set of names, composite attributesIdentification numbers like CS101 that can be broken up into partsA relational schema R is in first normal form if the domains of all attributes of R are atomicNon-atomic values complicate storage and encourage redundant (repeated) storage of dataExample: Set of accounts stored with each customer, and set of owners stored with each accountWe assume all relations are in first normal form (and revisit this in Chapter 9)

اسلاید 9: First Normal Form (Cont’d)Atomicity is actually a property of how the elements of the domain are used.Example: Strings would normally be considered indivisible Suppose that students are given roll numbers which are strings of the form CS0012 or EE1127If the first two characters are extracted to find the department, the domain of roll numbers is not atomic.Doing so is a bad idea: leads to encoding of information in application program rather than in the database.

اسلاید 10: Goal — Devise a Theory for the FollowingDecide 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 {R1, R2, ..., Rn} such that each relation is in good form the decomposition is a lossless-join decompositionOur theory is based on:functional dependenciesmultivalued dependencies

اسلاید 11: Functional DependenciesConstraints 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.

اسلاید 12: Functional Dependencies (Cont.)Let R be a relation schema  R and   RThe functional dependency    holds on R if and only if for any legal relations r(R), whenever any two tuples t1 and t2 of r agree on the attributes , they also agree on the attributes . That is, t1[] = t2 []  t1[ ] = t2 [ ] Example: Consider r(A,B ) with the following instance of r.On this instance, A  B does NOT hold, but B  A does hold. 41 537

اسلاید 13: Functional Dependencies (Cont.)K is a superkey for relation schema R if and only if K  RK is a candidate key for R if and only if K  R, andfor no   K,   RFunctional dependencies allow us to express constraints that cannot be expressed using superkeys. Consider the schema:bor_loan = (customer_id, loan_number, amount ).We expect this functional dependency to hold:loan_number  amountbut would not expect the following to hold: amount  customer_name

اسلاید 14: Use of Functional DependenciesWe 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 relationsWe 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 may, by chance, satisfy amount  customer_name.

اسلاید 15: Functional Dependencies (Cont.)A functional dependency is trivial if it is satisfied by all instances of a relationExample: customer_name, loan_number  customer_name customer_name  customer_nameIn general,    is trivial if   

اسلاید 16: Closure of a Set of Functional DependenciesGiven a set F set of functional dependencies, there are certain other functional dependencies that are logically implied by F.For example: If A  B and B  C, then we can infer that A  CThe set of all functional dependencies logically implied by F is the closure of F.We denote the closure of F by F+.F+ is a superset of F.

اسلاید 17: Boyce-Codd Normal Form   is trivial (i.e.,   ) is a superkey for RA 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:Example schema not in BCNF:bor_loan = ( customer_id, loan_number, amount )because loan_number  amount holds on bor_loan but loan_number is not a superkey

اسلاید 18: Decomposing a Schema into BCNFSuppose we have a schema R and a non-trivial dependency  causes a violation of BCNF.We decompose R into:(U  )( R - (  -  ) )In our example,  = loan_number = amountand bor_loan is replaced by (U  ) = ( loan_number, amount )( R - (  -  ) ) = ( customer_id, loan_number )

اسلاید 19: BCNF and Dependency PreservationConstraints, including functional dependencies, are costly to check in practice unless they pertain to only one relationIf it is sufficient to test only those dependencies on each individual relation of a decomposition in order to ensure that all functional dependencies hold, then that decomposition is dependency preserving.Because it is not always possible to achieve both BCNF and dependency preservation, we consider a weaker normal form, known as third normal form.

اسلاید 20: Third Normal FormA 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 REach 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 (will see why later).

اسلاید 21: Goals of NormalizationLet R be a relation scheme with a set F of functional dependencies.Decide whether a relation scheme R is in “good” form.In the case that a relation scheme R is not in “good” form, decompose it into a set of relation scheme {R1, R2, ..., Rn} such that each relation scheme is in good form the decomposition is a lossless-join decompositionPreferably, the decomposition should be dependency preserving.

اسلاید 22: How good is BCNF?There are database schemas in BCNF that do not seem to be sufficiently normalized Consider a database classes (course, teacher, book ) such that (c, t, b)  classes means that t is qualified to teach c, and b is a required textbook for cThe database is supposed to list for each course the set of teachers any one of which can be the course’s instructor, and the set of books, all of which are required for the course (no matter who teaches it).

اسلاید 23: There are no non-trivial functional dependencies and therefore the relation is in BCNF Insertion anomalies – i.e., if Marilyn is a new teacher that can teach database, two tuples need to be inserted(database, Marilyn, DB Concepts) (database, Marilyn, Ullman)courseteacherbookdatabasedatabasedatabasedatabasedatabasedatabaseoperating systemsoperating systemsoperating systemsoperating systemsAviAviHankHankSudarshanSudarshanAviAvi PetePeteDB ConceptsUllmanDB ConceptsUllmanDB ConceptsUllmanOS ConceptsStallingsOS ConceptsStallingsclassesHow good is BCNF? (Cont.)

اسلاید 24: Therefore, it is better to decompose classes into:courseteacherdatabasedatabasedatabaseoperating systemsoperating systemsAviHankSudarshanAvi Jimteachescoursebookdatabasedatabaseoperating systemsoperating systemsDB ConceptsUllmanOS ConceptsShawtextThis suggests the need for higher normal forms, such as Fourth Normal Form (4NF), which we shall see later.How good is BCNF? (Cont.)

اسلاید 25: Functional-Dependency TheoryWe now consider the formal theory that tells us which functional dependencies are implied logically by a given set of functional dependencies.We then develop algorithms to generate lossless decompositions into BCNF and 3NFWe then develop algorithms to test if a decomposition is dependency-preserving

اسلاید 26: Closure of a Set of Functional DependenciesGiven a set F set of functional dependencies, there are certain other functional dependencies that are logically implied by F.For example: If A  B and B  C, then we can infer that A  CThe 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).

اسلاید 27: ExampleR = (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  HAG  I by augmenting A  C with G, to get AG  CG and then transitivity with CG  I CG  HI by augmenting CG  I to infer CG  CGI, and augmenting of CG  H to infer CGI  HI, and then transitivity

اسلاید 28: 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 f1and f2 in F + if f1 and f2 can be combined using transitivity then add the resulting functional dependency to F + until F + does not change any furtherNOTE: We shall see an alternative procedure for this task later

اسلاید 29: 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.

اسلاید 30: Closure of Attribute SetsGiven a set of attributes a, define the closure of a under F (denoted by a+) as the set of attributes that are functionally determined by a under F Algorithm to compute a+, the closure of a under F result := a; while (changes to result) do for each    in F do begin if   result then result := result   end

اسلاید 31: Example of Attribute Set ClosureR = (A, B, C, G, H, I)F = {A  B A  C CG  H CG  I B  H}(AG)+1.result = AG2.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? Is AG a super key?Does AG  R? == Is (AG)+  RIs any subset of AG a superkey?Does A  R? == Is (A)+  RDoes G  R? == Is (G)+  R

اسلاید 32: Uses of Attribute ClosureThere 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 dependenciesTo 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 usefulComputing closure of FFor each   R, we find the closure +, and for each S  +, we output a functional dependency   S.

اسلاید 33: Canonical CoverSets of functional dependencies may have redundant dependencies that can be inferred from the othersFor example: A  C is redundant in: {A  B, B  C}Parts of a functional dependency may be redundantE.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, having no redundant dependencies or redundant parts of dependencies

اسلاید 34: Extraneous AttributesConsider 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.Note: implication in the opposite direction is trivial in each of the cases above, since a “stronger” functional dependency always implies a weaker oneExample: Given F = {A  C, AB  C }B is extraneous in AB  C because {A  C, AB  C} logically implies A  C (I.e. the result of dropping B from AB  C).Example: Given F = {A  C, AB  CD}C is extraneous in AB  CD since AB  C can be inferred even after deleting C

اسلاید 35: Testing if an Attribute is ExtraneousConsider a set F of functional dependencies and the functional dependency    in F.To test if attribute A   is extraneous in  compute ({} – A)+ using the dependencies in F check that ({} – A)+ contains A; if it does, A is extraneousTo test if attribute A   is extraneous in  compute + using only the dependencies in F’ = (F – {  })  { ( – A)}, check that + contains A; if it does, A is extraneous

اسلاید 36: Canonical CoverA canonical cover for F is a set of dependencies Fc such that F logically implies all dependencies in Fc, and Fc logically implies all dependencies in F, andNo functional dependency in Fc contains an extraneous attribute, andEach left side of functional dependency in Fc is unique.To compute a canonical cover for F: repeat Use the union rule to replace any dependencies in F 1  1 and 1  2 with 1  1 2 Find a functional dependency    with an extraneous attribute either in  or in  If an extraneous attribute is found, delete it from    until F does not changeNote: Union rule may become applicable after some extraneous attributes have been deleted, so it has to be re-applied

اسلاید 37: Computing a Canonical CoverR = (A, B, C) F = {A  BC B  C A  B AB  C}Combine A  BC and A  B into A  BCSet is now {A  BC, B  C, AB  C}A is extraneous in AB  CCheck if the result of deleting A from AB  C is implied by the other dependenciesYes: in fact, B  C is already present!Set is now {A  BC, B  C}C is extraneous in A  BC Check if A  C is logically implied by A  B and the other dependenciesYes: using transitivity on A  B and B  C. Can use attribute closure of A in more complex casesThe canonical cover is: A  B B  C

اسلاید 38: Lossless-join DecompositionFor the case of R = (R1, R2), we require that for all possible relations r on schema Rr = R1 (r ) R2 (r ) A decomposition of R into R1 and R2 is lossless join if and only if at least one of the following dependencies is in F+:R1  R2  R1R1  R2  R2

اسلاید 39: ExampleR = (A, B, C) F = {A  B, B  C)Can be decomposed in two different waysR1 = (A, B), R2 = (B, C)Lossless-join decomposition: R1  R2 = {B} and B  BCDependency preservingR1 = (A, B), R2 = (A, C)Lossless-join decomposition: R1  R2 = {A} and A  ABNot dependency preserving (cannot check B  C without computing R1 R2)

اسلاید 40: Dependency Preservation Let Fi be the set of dependencies F + that include only attributes in Ri. A decomposition is dependency preserving, if (F1  F2  …  Fn )+ = F +If it is not, then checking updates for violation of functional dependencies may require computing joins, which is expensive.

اسلاید 41: Testing for Dependency PreservationTo check if a dependency    is preserved in a decomposition of R into R1, R2, …, Rn we apply the following test (with attribute closure done with respect to F)result =  while (changes to result) do for each Ri in the decomposition t = (result  Ri)+  Ri result = result  tIf 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 preservingThis procedure takes polynomial time, instead of the exponential time required to compute F+ and (F1  F2  …  Fn)+

اسلاید 42: ExampleR = (A, B, C ) F = {A  B B  C} Key = {A}R is not in BCNFDecomposition R1 = (A, B), R2 = (B, C)R1 and R2 in BCNFLossless-join decompositionDependency preserving

اسلاید 43: Testing for BCNFTo check if a non-trivial dependency  causes a violation of BCNF1. 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 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+.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 RConsider R = (A, B, C, D, E), with F = { A  B, BC  D}Decompose R into R1 = (A,B) and R2 = (A,C,D, E) Neither of the dependencies in F contain only attributes from (A,C,D,E) so we might be mislead into thinking R2 satisfies BCNF. In fact, dependency AC  D in F+ shows R2 is not in BCNF.

اسلاید 44: Testing Decomposition for BCNFTo check if a relation Ri in a decomposition of R is in BCNF, Either test Ri for BCNF with respect to the restriction of F to Ri (that is, all FDs in F+ that contain only attributes from Ri)or use the original set of dependencies F that hold on R, but with the following test:for every set of attributes   Ri, check that + (the attribute closure of ) either includes no attribute of Ri- , or includes all attributes of Ri.If the condition is violated by some   in F, the dependency  (+ - )  Ri can be shown to hold on Ri, and Ri violates BCNF.We use above dependency to decompose Ri

اسلاید 45: BCNF Decomposition Algorithmresult := {R }; done := false; compute F +; while (not done) do if (there is a schema Ri in result that is not in BCNF) then begin let    be a nontrivial functional dependency that holds on Ri such that   Ri is not in F +, and    = ; result := (result – Ri )  (Ri – )  (,  ); end else done := true; Note: each Ri is in BCNF, and decomposition is lossless-join.

اسلاید 46: Example of BCNF DecompositionR = (A, B, C ) F = {A  B B  C} Key = {A}R is not in BCNF (B  C but B is not superkey)DecompositionR1 = (B, C)R2 = (A,B)

اسلاید 47: Example of BCNF DecompositionOriginal relation R and functional dependency F R = (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}DecompositionR1 = (branch_name, branch_city, assets )R2 = (branch_name, customer_name, loan_number, amount )R3 = (branch_name, loan_number, amount )R4 = (customer_name, loan_number )Final decomposition R1, R3, R4

اسلاید 48: BCNF and Dependency PreservationR = (J, K, L ) F = {JK  L L  K } Two candidate keys = JK and JLR is not in BCNFAny decomposition of R will fail to preserveJK  L This implies that testing for JK  L requires a joinIt is not always possible to get a BCNF decomposition that is dependency preserving

اسلاید 49: Third Normal Form: MotivationThere are some situations where BCNF is not dependency preserving, and efficient checking for FD violation on updates is importantSolution: define a weaker normal form, called Third Normal Form (3NF)Allows some redundancy (with resultant problems; we will see examples later)But functional dependencies can be checked on individual relations without computing a join.There is always a lossless-join, dependency-preserving decomposition into 3NF.

اسلاید 50: 3NF ExampleRelation R:R = (J, K, L ) F = {JK  L, L  K }Two candidate keys: JK and JLR is in 3NFJK  LJK is a superkey L  KK is contained in a candidate key

اسلاید 51: Redundancy in 3NFJj1j2j3nullLl1l1l1l2Kk1k1k1k2repetition of information (e.g., the relationship l1, k1) need to use null values (e.g., to represent the relationship l2, k2 where there is no corresponding value for J).There is some redundancy in this schemaExample of problems due to redundancy in 3NFR = (J, K, L) F = {JK  L, L  K }

اسلاید 52: Testing for 3NFOptimization: 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 Rthis test is rather more expensive, since it involve finding candidate keystesting for 3NF has been shown to be NP-hardInterestingly, decomposition into third normal form (described shortly) can be done in polynomial time

اسلاید 53: 3NF Decomposition AlgorithmLet Fc be a canonical cover for F; i := 0; for each functional dependency    in Fc do if none of the schemas Rj, 1  j  i contains   then begin i := i + 1; Ri :=   end if none of the schemas Rj, 1  j  i contains a candidate key for R then begin i := i + 1; Ri := any candidate key for R; end return (R1, R2, ..., Ri)

اسلاید 54: 3NF Decomposition Algorithm (Cont.)Above algorithm ensures:each relation schema Ri is in 3NFdecomposition is dependency preserving and lossless-joinProof of correctness is at end of this file (click here)

اسلاید 55: ExampleRelation schema:cust_banker_branch = (customer_id, employee_id, branch_name, type )The functional dependencies for this relation schema are: customer_id, employee_id  branch_name, type employee_id  branch_nameThe for loop generates:(customer_id, employee_id, branch_name, type )It then generates (employee_id, branch_name)but does not include it in the decomposition because it is a subset of the first schema.

اسلاید 56: Comparison of BCNF and 3NFIt is always possible to decompose a relation into a set of relations that are in 3NF such that:the decomposition is losslessthe dependencies are preservedIt is always possible to decompose a relation into a set of relations that are in BCNF such that:the decomposition is losslessit may not be possible to preserve dependencies.

اسلاید 57: Design GoalsGoal for a relational database design is:BCNF.Lossless join.Dependency preservation.If we cannot achieve this, we accept one ofLack of dependency preservation Redundancy due to use of 3NFInterestingly, SQL does not provide a direct way of specifying functional dependencies other than superkeys.Can specify FDs using assertions, but they are expensive to testEven if we had a dependency preserving decomposition, using SQL we would not be able to efficiently test a functional dependency whose left hand side is not a key.

اسلاید 58: 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 t1 and t2 in r such that t1[] = t2 [], there exist tuples t3 and t4 in r such that: t1[] = t2 [] = t3 [] = t4 [] t3[] = t1 [] t3[R – ] = t2[R – ] t4 [] = t2[] t4[R – ] = t1[R – ]

اسلاید 59: MVD (Cont.)Tabular representation of   

اسلاید 60: ExampleLet R be a relation schema with a set of attributes that are partitioned into 3 nonempty subsets.Y, Z, WWe say that Y  Z (Y multidetermines Z ) if and only if for all possible relations r (R )< y1, z1, w1 >  r and < y2, z2, w2 >  rthen< y1, z1, w2 >  r and < y2, z2, w1 >  rNote that since the behavior of Z and W are identical it follows that Y  Z if Y  W

اسلاید 61: Example (Cont.)In our example:course  teacher course  bookThe above formal definition is supposed to formalize the notion that given a particular value of Y (course) it has associated with it a set of values of Z (teacher) and a set of values of W (book), and these two sets are in some sense independent of each other.Note: If Y  Z then Y  ZIndeed we have (in above notation) Z1 = Z2 The claim follows.

اسلاید 62: Use of Multivalued DependenciesWe use multivalued dependencies in two ways: 1.To test relations to determine whether they are legal under a given set of functional and multivalued dependencies2.To specify constraints on the set of legal relations. We shall thus concern ourselves only with relations that satisfy a given set of functional and 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.

اسلاید 63: Theory of MVDsFrom the definition of multivalued dependency, we can derive the following rule:If   , then   That is, every functional dependency is also a multivalued dependencyThe 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.We can manage with such reasoning for very simple multivalued dependencies, which seem to be most common in practiceFor complex dependencies, it is better to reason about sets of dependencies using a system of inference rules (see Appendix C).

اسلاید 64: Fourth Normal FormA 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 RIf a relation is in 4NF it is in BCNF

اسلاید 65: Restriction of Multivalued DependenciesThe restriction of D to Ri is the set Di consisting ofAll functional dependencies in D+ that include only attributes of RiAll multivalued dependencies of the form   (  Ri) where   Ri and    is in D+

اسلاید 66: 4NF Decomposition Algorithm result: = {R}; done := false; compute D+; Let Di denote the restriction of D+ to Ri while (not done) if (there is a schema Ri in result that is not in 4NF) then begin let    be a nontrivial multivalued dependency that holds on Ri such that   Ri is not in Di, and ; result := (result - Ri)  (Ri - )  (, ); end else done:= true;Note: each Ri is in 4NF, and decomposition is lossless-join

اسلاید 67: ExampleR =(A, B, C, G, H, I)F ={ A  BB  HICG  H }R is not in 4NF since A  B and A is not a superkey for RDecompositiona) R1 = (A, B) (R1 is in 4NF)b) R2 = (A, C, G, H, I) (R2 is not in 4NF)c) R3 = (C, G, H) (R3 is in 4NF)d) R4 = (A, C, G, I) (R4 is not in 4NF)Since A  B and B  HI, A  HI, A  Ie) R5 = (A, I) (R5 is in 4NF)f)R6 = (A, C, G) (R6 is in 4NF)

اسلاید 68: Further Normal FormsJoin dependencies generalize multivalued dependencieslead to project-join normal form (PJNF) (also called fifth normal form)A class of even more general constraints, leads to a normal form called domain-key normal form.Problem with these generalized constraints: are hard to reason with, and no set of sound and complete set of inference rules exists.Hence rarely used

اسلاید 69: Overall Database Design ProcessWe have assumed schema R is givenR could have been generated when converting E-R diagram to a set of tables.R could have been a single relation containing all attributes that are of interest (called universal relation).Normalization breaks R into smaller relations.R could have been the result of some ad hoc design of relations, which we then test/convert to normal form.

اسلاید 70: ER Model and NormalizationWhen an E-R diagram is carefully designed, identifying all entities correctly, the tables generated from the E-R diagram should not need further normalization.However, in a real (imperfect) design, there can be functional dependencies from non-key attributes of an entity to other attributes of the entityExample: an employee entity with attributes department_number and department_address, and a functional dependency department_number  department_addressGood design would have made department an entityFunctional dependencies from non-key attributes of a relationship set possible, but rare --- most relationships are binary

اسلاید 71: Denormalization for PerformanceMay want to use non-normalized schema for performanceFor example, displaying customer_name along with account_number and balance requires join of account with depositorAlternative 1: Use denormalized relation containing attributes of account as well as depositor with all above attributesfaster lookupextra space and extra execution time for updatesextra coding work for programmer and possibility of error in extra codeAlternative 2: use a materialized view defined as account depositorBenefits and drawbacks same as above, except no extra coding work for programmer and avoids possible errors

اسلاید 72: Other Design IssuesSome aspects of database design are not caught by normalizationExamples of bad database design, to be avoided: Instead of earnings (company_id, year, amount ), use earnings_2000, earnings_2001, earnings_2002, etc., all on the schema (company_id, earnings).Above are in BCNF, but make querying across years difficult and needs new table each yearcompany_year(company_id, earnings_2000, earnings_2001, earnings_2002)Also in BCNF, but also makes querying across years difficult and requires new attribute each year.Is an example of a crosstab, where values for one attribute become column namesUsed in spreadsheets, and in data analysis tools

اسلاید 73: Modeling Temporal DataTemporal data have an association time interval during which the data are valid.A snapshot is the value of the data at a particular point in time.Adding a temporal component results in functional dependencies likecustomer_id  customer_street, customer_citynot to hold, because the address varies over timeA temporal functional dependency holds on schema R if the corresponding functional dependency holds on all snapshots for all legal instances r (R )

اسلاید 74: End of Chapter

اسلاید 75: Proof of Correctness of 3NF Decomposition Algorithm

اسلاید 76: Correctness of 3NF Decomposition Algorithm3NF decomposition algorithm is dependency preserving (since there is a relation for every FD in Fc)Decomposition is losslessA candidate key (C ) is in one of the relations Ri in decompositionClosure of candidate key under Fc must contain all attributes in R. Follow the steps of attribute closure algorithm to show there is only one tuple in the join result for each tuple in Ri

اسلاید 77: Correctness of 3NF Decomposition Algorithm (Cont’d.)Claim: if a relation Ri is in the decomposition generated by the above algorithm, then Ri satisfies 3NF.Let Ri be generated from the dependency   Let   B be any non-trivial functional dependency on Ri. (We need only consider FDs whose right-hand side is a single attribute.)Now, B can be in either  or  but not in both. Consider each case separately.

اسلاید 78: Correctness of 3NF Decomposition (Cont’d.)Case 1: If B in :If  is a superkey, the 2nd condition of 3NF is satisfiedOtherwise  must contain some attribute not in Since   B is in F+ it must be derivable from Fc, by using attribute closure on .Attribute closure not have used  . If it had been used,  must be contained in the attribute closure of , which is not possible, since we assumed  is not a superkey.Now, using  (- {B}) and   B, we can derive  B(since    , and B   since   B is non-trivial)Then, B is extraneous in the right-hand side of  ; which is not possible since   is in Fc.Thus, if B is in  then  must be a superkey, and the second condition of 3NF must be satisfied.

اسلاید 79: Correctness of 3NF Decomposition (Cont’d.)Case 2: B is in .Since  is a candidate key, the third alternative in the definition of 3NF is trivially satisfied.In fact, we cannot show that  is a superkey.This shows exactly why the third alternative is present in the definition of 3NF.Q.E.D.

اسلاید 80: Figure 7.5: Sample Relation r

اسلاید 81: Figure 7.6

اسلاید 82: Figure 7.7

اسلاید 83: Figure 7.15: An Example of Redundancy in a BCNF Relation

اسلاید 84: Figure 7.16: An Illegal R2 Relation

اسلاید 85: Figure 7.18: Relation of Practice Exercise 7.2