Database Slides on Normalization

Chapter 11 proportional Database Design algorithmic rules and Further Dependencies Chapter Outline ? ? ? ? ? ? ? 0. plan a inflexible of Relations 1. Properties of comparative rots 2. Algorithms for relative Database Schema 3. Multivalued Dependencies and one-fourth general approach pattern 4. kernel Dependencies and Fifth Normal Form 5. inclusion body Dependencies 6. Other Dependencies and Normal Forms DESIGNING A SET OF RELATIONS ? Goals ? lossless touch base property (a must) ? Algorithm 11. 1 tests for general losslessness. Algorithm 11. decomposes a comparison into BCNF components by sacrificing the dependency preservation. 4NF (based on multi-valued dependencies) 5NF (based on wedlock dependencies) ? Dependency preservation property ? ? Additional normal course of instructions ? ? 1. Properties of Relational Decompositions ? Relation Decomposition and Insufficiency of Normal Forms ? Universal Relation Schema ? A relative outline R = A1, A2, , An that includes entirely the attributes of the database. Every attribute let on is unique. ? Universal copulation assumption ? (Cont) ? Decomposition ? ? Attribute preservation put in ?The process of decomposing the universal relation lineation R into a sort out of relation dodgings D = R1,R2, , Rm that go forth become the relational database outline by using the usable dependencies. Each attribute in R will appear in at least one relation schema Ri in the radioactive decay so that no attributes are lost. (Cont) ? ? another(prenominal) goal of annihilation is to hold in separately(prenominal)(prenominal) individual relation Ri in the guff D be in BCNF or 3NF. Additional properties of decomposition are needed to prevent from generating spurious tuples (Cont) ? Dependency Preservation blank space of a Decomposition ? definition Given a garment of dependencies F on R, the projection of F on Ri, denoted by pRi(F) where Ri is a subset of R, is the set of dependencies X Y in F+ such that the attributes in X U Y are each(prenominal) contained in Ri. Hence, the projection of F on each relation schema Ri in the decomposition D is the set of in operation(p) dependencies in F+, the closure of F, such that every(prenominal) their left- and right-hand-side attributes are in Ri. (Cont. ) ? Dependency Preservation holding of a Decomposition (cont. ) ? Dependency Preservation blank space ? ? A decomposition D = R1, R2, Rm of R is dependency-preserving with compliments to F if the union of the projections of F on each Ri in D is equivalent to F that is ((? R1(F)) U . . . U (? Rm(F)))+ = F+ (See examples in Fig 10. 12a and Fig 10. 11) ? Claim 1 ? It is always possible to happen a dependency-preserving decomposition D with notice 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 attribut es 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) pairing 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 disk operating system r of R that satisfies F, the pursuit holds, where * is the natural join of all the transaction in D (? R1(r), , ? Rm(r)) = r ? Note The word loss in lossless refers to loss of tuition, not to loss of tuples. In fact, for loss of information a better term is addition of spurious information physical exercise 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 Join Decomposition ?? NO (Cont. ) Lossless (Non-additive) Join Property of a Decomposition (cont. ) Algorithm 11. 1 Testing for Lossless Join Property excitant A universal relation R, a decomposition D = R1, R2, , Rm of R,and a set F of functional dependencies. 1.Create an initial ground substance S with one row i for each relation Ri in D, and one column j for each attribute Aj in R. 2. Set S(i,j)=bij for all matrix entries. (/* each bij is a perspicuous symbol 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) then set S(i,j)= aj ? (/* each aj is a explicit symbol associated with index (j) */) ? CONTINUED on NEXT SLIDE (Cont. ) 4. double over the following loop 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 homogeneous symbols in the columns corresponding to attributes in X make the symbols in each column that correspond to an attribute in Y be the aforementioned(prenomin al) in all these rows as follows If any of the rows has an a symbol for the column, set the other rows to that uniform 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 make up entirely of a symbols, then the decomposition has the lossless join property otherwise it does not. (Cont. ) Lossless (nonadditive) join test for n-ary decompositions. (a) Case 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 two 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 wholly 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 Input A universal relation R and a set of functional dependencies F on the attributes of R. 1. Set D = R 2.While 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 invent a functional dependency X Y in Q that violates BCNF replace Q in D by two relation schemas (Q Y) and (X U Y) Assumption No cipher values 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 stripped 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 key 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. Eliminate 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. Multivalued 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 SUPPLY with no MVDs is in 4NF but not in 5NF if it has the JD(R1, R2, R3). (d) Decomposing the relation SUPPLY into the 5NF relations R1, R2, and R3. (Cont. ) Definition ? A multivalued dependency (MVD) X >> Y specified on relation schema R, where X and Y are both subsets of R, specifies the following simplicity 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 use 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 4NF 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 divergent 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 univer sal 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 choose 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 legal 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 Dependencies and Fourth Normal Form Join Dependencies and Fifth Normal FormTutorial/Quiz 4 Q1) trade 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