DATABASE DESIGN -2 part1

DATABASE DESIGN -2

DESIGNING A SET OF RELATIONS

The Approach of Relational Synthesis (Bottom-up Design): Assumes that all possible functional dependencies are known. First constructs a minimal set of FDs

Then applies algorithms that construct a target set of 3NF or BCNF relations.

Additional criteria may be needed to ensure the the set of relations in a relational database are satisfactory (see Algorithms 11.2 and 11.4).

Goals:

Lossless join property (a must)

Algorithm 11.1 tests for general losslessness.

Dependency preservation property

Algorithm 11.3 decomposes a relation into BCNF components by sacrificing the dependency preservation.

Additional normal forms

4NF (based on multi-valued dependencies) 5NF (based on join dependencies)

Properties of Relational Decompositions

Relation Decomposition and Insufficiency of Normal Forms:

Universal Relation Schema:

A relation schema R = {A1, A2, …, An} that includes all the attributes of the database.

Universal relation assumption: Every attribute name is unique

Decomposition:

The process of decomposing the universal relation schema R into a set of relation schemas D = {R1,R2, …, Rm} that will become the relational database schema by using the functional dependencies.

Attribute preservation condition:

Each attribute in R will appear in at least one relation schema Ri in the decomposition so that no attributes are “lost”.

Another goal of decomposition is to have each individual relation Ri in the decomposition D be in BCNF or 3NF.

Additional properties of decomposition are needed to prevent from generating spurious tuples.

Dependency Preservation Property of a Decomposition:

Definition: Given a set 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 υ Y are all contained in Ri.

Hence, the projection of F on each relation schema Ri in the decomposition D is the set of functional dependencies in F+, the closure of F, such that all their left- and right-hand- side attributes are in Ri.

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 equivalent to F; that is

((πR1(F)) υ . . . υ (πRm(F)))+ = F+

(See examples in Fig 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.

Lossless (Non-additive) Join Property of 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:

* (π R1(r), ..., πRm(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 better term is “addition of spurious information”.

Algorithm 11.1: Testing for Lossless Join Property

Input: A universal relation R, a decomposition D = {R1, R2, ..., Rm} of R, and a set F of functional dependencies.

1. Create an initial matrix 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 distinct 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 distinct symbol associated with index (j) *)

4. Repeat 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 same symbols in the columns corresponding to attributes in X

{make the symbols in each column that correspond 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 entirely of “a” symbols, then the decomposition has the lossless join property; otherwise it does not.

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.

 

Lossless (nonadditive) join test for n-ary decompositions.

(c) Case 2: Decomposition of EMP_PROJ into EMP, PROJECT, and WORKS_ON satisfies test

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 only if either

The f.d. ((R1 ∩ R2) → (R1- R2)) is in F+, or

The f.d. ((R1 ∩ R2) →(R2 - R1)) is in F+.

Successive Lossless Join Decomposition:

Claim 2 (Preservation of non-additivity in successive decompositions):

If a decomposition D = {R1, R2, ..., Rm} of R has the lossless (non-additive) join property with respect to a set of functional dependencies F on R, and if a decomposition Di = {Q1, Q2, ..., Qk} of Ri has the lossless (non-additive) join property with respect to the projection of F on Ri, then the decomposition D2 = {R1, R2, ..., Ri-1, Q1, Q2, ..., Qk, Ri+1, ..., Rm} of R has the non-additive join property with respect to F

2. Algorithms for Relational Database Schema Design

Algorithm 11.2: Relational Synthesis into 3NF with Dependency

Preservation (Relational Synthesis Algorithm)

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);

2. For each left-hand-side X of a functional dependency that appears in G, create a relation schema in D with attributes {X υ {A1} υ {A2} ...

υ {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. Place any remaining attributes (that have not been placed in any

relation) in a single relation schema to ensure the attribute preservation property. Claim 3: Every relation schema created by Algorithm 11.2 is in 3NF.

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;

find a functional dependency X → Y in Q that violates BCNF;

replace Q in D by two relation schemas (Q - Y) and (X υ Y);

}; Assumption: No null values are allowed for the join attributes.

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).

2. For each left-hand-side X of a functional dependency that appears in G, create a relation schema in D with attributes {X υ {A1} υ {A2} ...

υ {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)

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};

Discussion of Normalization Algorithms:

Problems:

The database designer must first specify all the relevant functional dependencies among the database attributes.

These algorithms are not deterministic in general.

It is not always possible to find a decomposition into relation schemas that preserves dependencies and allows each relation schema in the decomposition to be in BCNF (instead of 3NF as in Algorithm 11.4).