Database Slides on Normalization

Chapter 11 singingal Database tar foreshorten Algorithms and Further Dependencies Chapter Outline ? ? ? ? ? ? ? 0. throwing a forwardness of transaction 1. Properties of Relational disintegrations 2. Algorithms for Relational Database Schema 3. Multivalued Dependencies and Fourth average sort 4. gist Dependencies and Fifth Normal fig 5. Inclusion Dependencies 6. Other Dependencies and Normal Forms DESIGNING A SET OF RELATIONS ? Goals ? Lossless join seat (a must) ? Algorithm 11. 1 tests for general passing playlessness. Algorithm 11. decomposes a semblance into BCNF comp cardinalnts by sacrificing the dependence economy. 4NF (based on multi-valued dependencies) 5NF (based on join dependencies) ? colony preservation billet ? ? Additional public forms ? ? 1. Properties of Relational annihilations ? Relation dissolution and lack of Normal Forms ? Universal Relation Schema ? A recounting dodge R = A1, A2, , An that includes any the specifys of the database. Every d elegate name is unique. ? Universal sexual intercourse assumption ? (Cont) ? Decomposition ? ? Attribute preservation condition ?The process of decomposing the commonplace copulation synopsis R into a touch on of relation preciss D = R1,R2, , Rm that will become the relational database schema by using the available dependencies. Each attribute in R will appear in at least one relation schema Ri in the degeneracy so that no attributes are lost. (Cont) ? ? A nonher goal of corruption is to have from each one individual relation Ri in the putrefaction D be in BCNF or 3NF. Additional properties of depravation reaction are need to prevent from generating spurious tuples (Cont) ? Dependency conservation seat of a Decomposition ? explanation stipulation a specialize of dependencies F on R, the expulsion of F on Ri, denoted by pRi(F) where Ri is a subset of R, is the set of dependencies X Y in F+ much(prenominal) that the attributes in X U Y are all contained in Ri. H ence, the projection of F on each relation schema Ri in the decomposition D is the set of available dependencies in F+, the closure of F, such that all their left- and right-hand-side attributes are in Ri. (Cont. ) ? Dependency Preservation Property of a Decomposition (cont. ) ? Dependency Preservation Property ? ? A decomposition D = R1, R2, Rm of R is dependency-preserving with respect to F if the union of the projections of F on each Ri in D is analogous to F that is ((? R1(F)) U . . . U (? Rm(F)))+ = F+ (See examples in soma 10. 12a and Fig 10. 11) ? Claim 1 ? It is always possible to find a dependency-preserving decomposition D with respect to F such that each relation Ri in D is in 3NF. Projection of F on Ri Given a set of dependencies F on R, the projection of F on Ri, denoted by ? Ri(F) where Ri is a subset of R, is the set of dependencies X Y in F+ such that the attributes in X ?Y are all contained in Ri. Dependency Preservation Condition Given R(A, B, C, D) and F = A B, B C, C D Let D1=R1(A,B), R2(B,C), R3(C,D) ? R1(F)=A B ? R2(F)=B C ? R3(F)=C D FDs are preserved. (Cont. ) ? Lossless (Non-additive) brotherhood Property of a Decomposition ? Definition Lossless join property a decomposition D = R1, R2, , Rm of R has the lossless (nonadditive) join property with respect to the set of dependencies F on R if, for every relation state r of R that satisfies F, the following holds, where * is the natural join of all the relations in D (? R1(r), , ? Rm(r)) = r ? Note The word loss in lossless refers to loss of information, not to loss of tuples. In fact, for loss of information a unwrap term is addition of spurious information Example S s1 s2 s3 P p1 p2 p1 D d1 d2 d3 = S s1 s2 s3 P p1 p2 p1 * P p1 p2 p1 D d1 d2 d3 Lossless reefer Decomposition NO (Cont. ) Lossless (Non-additive) Join Property of a Decomposition (cont. ) Algorithm 11. 1 Testing for Lossless Join Property stimulus A popular relation R, a decomposition D = R1, R2, , Rm of R,and a set F of operable dependencies. 1.Create an initial hyaloplasm S with one row i for each relation Ri in D, and one newspaper column j for each attribute Aj in R. 2. Set S(i,j)=bij for all matrix entries. (/* each bij is a distinct symbolism associated with indices (i,j) */). 3. For each row i representing relation schema Ri for each column j representing attribute Aj if (relation Ri includes attribute Aj) therefore set S(i,j)= aj ? (/* each aj is a distinct symbol associated with superpower (j) */) ? CONTINUED on NEXT SLIDE (Cont. ) 4. Repeat the following loop topology until a complete loop execution results in no changes to S for each functional dependency X ?Y in F for all rows in S which have the comparable symbols in the columns synonymous to attributes in X make the symbols in each column that lay out to an attribute in Y be the same in all these rows as follows If any of the rows has an a symbol for the column, set the other rows to that same a symbol in the column. If no a symbol exists for the attribute in any of the rows, choose one of the b symbols that appear in one of the rows for the attribute and set the other rows to that same b symbol in the column 5.If a row is made up just of a symbols, consequently the decomposition has the lossless join property otherwise it does not. (Cont. ) Lossless (nonadditive) join test for n-ary decompositions. (a) fountain 1 Decomposition of EMP_PROJ into EMP_PROJ1 and EMP_LOCS fails test. (b) A decomposition of EMP_PROJ that has the lossless join property. (Cont. ) Lossless (nonadditive) join test for n-ary decompositions. (c) Case 2 Decomposition of EMP_PROJ into EMP, PROJECT, and WORKS_ON satisfies test. (Cont. ) ? Testing Binary Decompositions for Lossless Join Property ? ?Binary Decomposition Decomposition of a relation R into deuce relations. PROPERTY LJ1 (lossless join test for binary decompositions) A decomposition D = R1, R2 of R has the lossless join property with respect to a set of functional dependencies F on R if and only if either ? ? The FD ((R1 ? R2) ? (R1- R2)) is in F+, or The FD ((R1 ? R2) ? (R2 R1)) is in F+. 2. Algorithms for Relational Database Schema Design Algorithm 11. 3 Relational Decomposition into BCNF with Lossless (non-additive) join property stimulant drug A universal relation R and a set of functional dependencies F on the attributes of R. 1. Set D = R 2. term there is a relation schema Q in D that is not in BCNF do choose a relation schema Q in D that is not in BCNF find a functional dependency X Y in Q that violates BCNF replace Q in D by dickens relation schemas (Q Y) and (X U Y) Assumption No null value are allowed for the join attributes. Algorithms for Relational Database Schema Design Algorithm 11. 4 Relational Synthesis into 3NF with Dependency Preservation and Lossless (Non-Additive) Join Property Input A universal relation R and a set of functional dependencies F on the attributes of R. 1. Find a minimal cover G for F (Use Algorithm 10. ). 2. For each left-hand-side X of a functional dependency that appears in G, create a relation schema in D with attributes X U A1 U A2 U Ak, where X ? A1, X ? A2, , X Ak are the only dependencies in G with X as left-hand-side (X is the constitute of this relation). 3. If none of the relation schemas in D contains a key of R, then create one more relation schema in D that contains attributes that form a key of R. (Use Algorithm 11. 4a to find the key of R) 4. pass off redundant relations from the result. A relation R is considered redundant if R is a projection of another relation SAlgorithms for Relational Database Schema Design Algorithm 11. 4a Finding a Key K for R Given a set F of Functional Dependencies Input A universal relation R and a set of functional dependencies F on the attributes of R. 1. Set K = R 2. For each attribute A in K Compute (K A)+ with respect to F If (K A)+ contains all the attributes in R, then set K = K A (Cont. ) 3. Multival ued Dependencies and Fourth Normal Form (a) The EMP relation with two MVDs ENAME PNAME and ENAME DNAME. (b) Decomposing the EMP relation into two 4NF relations EMP_PROJECTS and EMP_DEPENDENTS. (Cont. ) c) The relation proviso with no MVDs is in 4NF but not in 5NF if it has the JD(R1, R2, R3). (d) Decomposing the relation tot into the 5NF relations R1, R2, and R3. (Cont. ) Definition ? A multivalued dependency (MVD) X Y condition on relation schema R, where X and Y are twain subsets of R, specifies the following constraint on any relation state r of R If two tuples t1 and t2 exist in r such that t1X = t2X, then two tuples t3 and t4 should also exist in r with the following properties, where we intent Z to denote (R -(X U Y)) ? t3X = t4X = t1X = t2X. t3Y = t1Y and t4Y = t2Y. t3Z = t2Z and t4Z = t1Z.An MVD X Y in R is called a trivial MVD if (a) Y is a subset of X, or (b) X U Y = R. ? ? ? Multivalued Dependencies and Fourth Normal Form Definition ? A relation schema R is in 4N F with respect to a set of dependencies F (that includes functional dependencies and multivalued dependencies) if, for every nontrivial multivalued dependency X Y in F+, X is a superkey for R. ? Informally, whenever 2 tuples that have different Y values but same X values, exists, then if these Y values get repeated in separate tuples with every distinct values of Z Z = R (X U Y) that occurs with the same X value. Cont. ) (Cont. ) Lossless (Non-additive) Join Decomposition into 4NF Relations ? PROPERTY LJ1 ? The relation schemas R1 and R2 form a lossless (non-additive) join decomposition of R with respect to a set F of functional and multivalued dependencies if and only if ? (R1 ? R2) (R1 R2) (R1 ? R2) (R2 R1)). ? or ? (Cont. ) Algorithm 11. 5 Relational decomposition into 4NF relations with non-additive join property ? Input A universal relation R and a set of functional and multivalued dependencies F.Set D = R While there is a relation schema Q in D that is not in 4NF do c hoose a relation schema Q in D that is not in 4NF find a nontrivial MVD X Y in Q that violates 4NF replace Q in D by two relation schemas (Q Y) and (X U Y) 1. 2. 4. Join Dependencies and Fifth Normal Form Definition ? A join dependency (JD), denoted by JD(R1, R2, , Rn), specified on relation schema R, specifies a constraint on the states r of R. ? ? The constraint states that every reasoned state r of R should have a non-additive join decomposition into R1, R2, Rn that is, for every such r we have * (? R1(r), ? R2(r), , ? Rn(r)) = r (Cont. ) Definition ? A relation schema R is in fifth normal form (5NF) (or Project-Join Normal Form (PJNF)) with respect to a set F of functional, multivalued, and join dependencies if, ? for every nontrivial join dependency JD(R1, R2, , Rn) in F+ (that is, implied by F), ? every Ri is a superkey of R. Recap ? ? ? ? ? Designing a Set of Relations Properties of Relational Decompositions Algorithms for Relational Database Schema Multivalued Depend encies and Fourth Normal Form Join Dependencies and Fifth Normal FormTutorial/Quiz 4 Q1) Consider a relation R with 5 attributes ABCDE, You are given the following dependencies A B, BC E, ED A a) List all the keys, b) Is R in 3 NF c) Is R in BCNF Q2) Consider the following decomposition for the relation schema R = A, B, C, D, E, F, G, H, I, J and the set of functional dependencies F = A, B C, A D, E, B F, F G, H, D - I, J . Preserves Lossless Join and Dependencies? a) D1 = R1, R2, R3, R4, R5, R1=A,B,C R2=A,D,E, R3=B,F, R4 = F,G,H, R5 = D,I,J b) D2 = R1, R2, R3 R1 = A,B,C,D,E R2 = B,F,G,H, R3 = D,I,J