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3nf Synthesis Example Essays

Algorithms for Relational Database Schema Design

We now give three algorithms for creating a relational decomposition from a uni-versal relation. Each algorithm has specific properties, as we discuss next.

1. Dependency-Preserving Decomposition into 3NF Schemas

Algorithm 16.4 creates a dependency-preserving decomposition D = {R1, R2, ..., Rm} of a universal relation R based on a set of functional dependencies F, such thateach Ri in D is in 3NF. It guarantees only the dependency-preserving property; it does not guarantee the nonadditive join property. The first step of Algorithm 16.4 is to find a minimal cover G for F; Algorithm 16.2 can be used for this step. Note that multiple minimal covers may exist for a given set F (as we illustrate later in the example after Algorithm 16.4). In such cases the algorithms can potentially yield multiple alternative designs.

Algorithm 16.4. Relational Synthesis into 3NF with Dependency Preservation

Input: A universal relationRand a set of functional dependenciesFon theattributes of R.

        Find a minimal cover G for F (use Algorithm 16.2);

        For each left-hand-side X of a functional dependency that appears in G, cre-ate a relation schema in D with attributes {X∪ {A1} ∪ {A2} ... ∪ {Ak} }, where XA1, XA2, ..., XAk are the only dependencies in G with X as the left-hand-side (X is the key of this relation);

        Place any remaining attributes (that have not been placed in any relation) in a single relation schema to ensure the attribute preservation property.

Example of Algorithm 16.4. Consider the following universal relation:

U(Emp_ssn,Pno,Esal,Ephone,Dno,Pname,Plocation)

Emp_ssn, Esal, Ephone refer to the Social Security number, salary, and phone numberof the employee. Pno, Pname, and Plocation refer to the number, name, and location of the project. Dno is department number.

The following dependencies are present:

FD1: Emp_ssn→ {Esal, Ephone, Dno}

FD2: Pno→ { Pname, Plocation}

FD3: Emp_ssn, Pno→ {Esal, Ephone, Dno, Pname, Plocation}

By virtue of FD3, the attribute set {Emp_ssn, Pno} represents a key of the universal relation. Hence F, the set of given FDs includes {Emp_ssn→Esal, Ephone, Dno;

Pno → Pname, Plocation; Emp_ssn, Pno → Esal, Ephone, Dno, Pname, Plocation}.

By applying the minimal cover Algorithm 16.2, in step 3 we see that Pno is a redun-dant attribute in Emp_ssn, Pno→Esal, Ephone, Dno. Moreover, Emp_ssn is redun-dant in Emp_ssn, Pno→Pname, Plocation. Hence the minimal cover consists of FD1 and FD2 only (FD3 being completely redundant) as follows (if we group attributes with the same left-hand side into one FD):

Minimal cover G: {Emp_ssn→Esal, Ephone, Dno; Pno→Pname, Plocation}

By applying Algorithm 16.4 to the above Minimal cover G, we get a 3NF design con-sisting of two relations with keys Emp_ssn and Pno as follows:

R1(Emp_ssn,Esal,Ephone,Dno)

R2(Pno,Pname,Plocation)

An observant reader would notice easily that these two relations have lost the original information contained in the key of the universal relation U (namely, that there are certain employees working on certain projects in a many-to-many relationship). Thus, while the algorithm does preserve the original dependencies, it makes no guarantee of preserving all of the information. Hence, the resulting design is a lossy design.

Claim 3. Every relation schema created by Algorithm 16.4 is in 3NF. (We willnot provide a formal proof here;6 the proof depends on G being a minimal set of dependencies.)

It is obvious that all the dependencies in G are preserved by the algorithm because each dependency appears in one of the relations Ri in the decomposition D. Since G is equivalent to F, all the dependencies in F are either preserved directly in the decomposition or are derivable using the inference rules from Section 16.1.1 from those in the resulting relations, thus ensuring the dependency preservation prop-erty. Algorithm 16.4 is called a relational synthesis algorithm, because each relation schema Ri in the decomposition is synthesized (constructed) from the set of functional dependencies in G with the same left-hand-side X.

2. Nonadditive Join Decomposition into BCNF Schemas

The next algorithm decomposes a universal relation schema R = {A1, A2, ..., An} into a decomposition D = {R1, R2, ..., Rm} such that each Ri is in BCNF and the decomposition D has the lossless join property with respect to F. Algorithm 16.5 utilizes Property NJB and Claim 2 (preservation of nonadditivity in successive decompositions) to create a nonadditive join decomposition D = {R1, R2, ..., Rm} of a universal relation R based on a set of functional dependencies F, such that each Ri in D is in BCNF.

Algorithm 16.5. Relational Decomposition into BCNF with NonadditiveJoin Property

Input: A universal relationRand a set of functional dependenciesFon theattributes of R.

        Set D := {R} ;

        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;

find a functional dependency XY in Q that violates BCNF; replace Q in D by two relation schemas (QY) and (XY);

} ;

Each time through the loop in Algorithm 16.5, we decompose one relation schema Q that is not in BCNF into two relation schemas. According to Property NJB forbinary decompositions and Claim 2, the decomposition D has the nonadditive join property. At the end of the algorithm, all relation schemas in D will be in BCNF. The reader can check that the normalization example in Figures 15.12 and 15.13 basi-cally follows this algorithm. The functional dependencies FD3, FD4, and later FD5 violate BCNF, so the LOTS relation is decomposed appropriately into BCNF rela-tions, and the decomposition then satisfies the nonadditive join property. Similarly, if we apply the algorithm to the TEACH relation schema from Figure 15.14, it is decomposed into TEACH1(Instructor, Student) and TEACH2(Instructor, Course) because the dependency FD2 Instructor→Course violates BCNF.

In step 2 of Algorithm 16.5, it is necessary to determine whether a relation schema Q is in BCNF or not. One method for doing this is to test, for each functionaldependency XY in Q, whether X+ fails to include all the attributes in Q, thereby determining whether or not X is a (super)key in Q. Another technique is based on an observation that whenever a relation schema Q has a BCNF violation, there exists a pair of attributes A and B in Q such that {Q – {A, B} } →A; by computing the closure {Q – {A, B} }+ for each pair of attributes {A, B} of Q, and checking whether the closure includes A (or B), we can determine whether Q is in BCNF.

3. Dependency-Preserving and Nonadditive (Lossless) Join Decomposition into 3NF Schemas

So far, in Algorithm 16.4 we showed how to achieve a 3NF design with the potential for loss of information and in Algorithm 16.5 we showed how to achieve BCNF design with the potential loss of certain functional dependencies. By now we know that it is not possible to have all three of the following: (1) guaranteed nonlossy design, (2) guaranteed dependency preservation, and (3) all relations in BCNF. As we have said before, the first condition is a must and cannot be compromised. The second condition is desirable, but not a must, and may have to be relaxed if we insist on achieving BCNF. Now we give an alternative algorithm where we achieve conditions 1 and 2 and only guarantee 3NF. A simple modification to Algorithm 16.4, shown as Algorithm 16.6, yields a decomposition D of R that does the following:

        Preserves dependencies

        Has the nonadditive join property

        Is such that each resulting relation schema in the decomposition is in 3NF

Because the Algorithm 16.6 achieves both the desirable properties, rather than only functional dependency preservation as guaranteed by Algorithm 16.4, it is preferred over Algorithm 16.4.

Algorithm 16.6. Relational Synthesis into 3NF with Dependency Preservationand Nonadditive Join Property

Input: A universal relationRand a set of functional dependenciesFon theattributes of R.

        Find a minimal cover G for F (use Algorithm 16.2).

        For each left-hand-side X of a functional dependency that appears in G, cre-ate a relation schema in D with attributes {X∪ {A1} ∪ {A2} ... ∪ {Ak} }, where XA1, XA2, ..., XAk are the only dependencies in G with X as left-hand-side (X is the key of this relation).

        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.7 (Algorithm 16.2(a) may be used to find a key.)

        Eliminate redundant relations from the resulting set of relations in the relational database schema. A relation R is considered redundant if R is a projection of another relation S in the schema; alternately, R is subsumed by S.

Step 3 of Algorithm 16.6 involves identifying a key K of R. Algorithm 16.2(a) can be used to identify a key K of R based on the set of given functional dependencies F. Notice that the set of functional dependencies used to determine a key in Algorithm 16.2(a) could be either F or G, since they are equivalent.

Example 1 of Algorithm 16.6. Let us revisit the example given earlier at the endof Algorithm 16.4. The minimal cover G holds as before. The second step produces relations R1 and R2 as before. However, now in step 3, we will generate a relation corresponding to the key {Emp_ssn, Pno}. Hence, the resulting design contains:

R1(Emp_ssn,Esal,Ephone,Dno)

R2(Pno,Pname,Plocation)

R3(Emp_ssn,

From G.A.R.D. Realms


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