Module 4- Database Management System by Navathe

Copyright © 2007 Ramez Elmasri and Shamkant B. Navathe
Module4
Relational Database Design
Algorithms and Further
Dependencies

Inference Rules for FDs (1)
 Given a set of FDs F, we can infer additional FDs that
hold whenever the FDs in F hold
 Armstrong's inference rules:
 IR1. (Reflexive) If Y subset-of X, then X -> Y
 IR2. (Augmentation) If X -> Y, then XZ -> YZ
 (Notation: XZ stands for X U Z)
 IR3. (Transitive) If X -> Y and Y -> Z, then X -> Z
 IR1, IR2, IR3 form a sound and complete set of
inference rules
 These are rules hold and all other rules that hold can be
deduced from these

 Some additional inference rules that are useful:
 Decomposition: If X -> YZ, then X -> Y and X ->
Z
 Union: If X -> Y and X -> Z, then X -> YZ
 Psuedotransitivity: If X -> Y and WY -> Z, then
WX -> Z
 The last three inference rules, as well as any
other inference rules, can be deduced from IR1,
IR2, and IR3 (completeness property)

 Closure of a set F of FDs is the set F+ of all FDs
that can be inferred from F
 Closure of a set of attributes X with respect to F
is the set X+ of all attributes that are functionally
determined by X
 X+ can be calculated by repeatedly applying IR1,
IR2, IR3 using the FDs in F

ALGORITHM - CLOSURE
Slide 11- 6

Algorithm for 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};
}
Slide 11- 7

Equivalence of Sets of FDs
 Two sets of FDs F and G are equivalent if:
 Every FD in F can be inferred from G, and
 Every FD in G can be inferred from F
 Hence, F and G are equivalent if F+ =G+
 Definition (Covers):
 F covers G if every FD in G can be inferred from F
 (i.e., if G+ subset-of F+)
 F and G are equivalent if F covers G and G covers F
 There is an algorithm for checking equivalence of sets of
FDs

Minimal Sets of FDs (1)
 A set of FDs is minimal if it satisfies the
following conditions:
1. Every dependency in F has a single attribute for
its RHS.
2. We cannot remove any dependency from F and
have a set of dependencies that is equivalent to
F.
3. We cannot replace any dependency X -> A in F
with a dependency Y -> A, where Y proper-
subset-of X ( Y subset-of X) and still have a set
of dependencies that is equivalent to F.

Minimal Sets of FDs (2)
 Every set of FDs has an equivalent minimal set
 There can be several equivalent minimal sets
 There is no simple algorithm for computing a
minimal set of FDs that is equivalent to a set F of
FDs
 To synthesize a set of relations, we assume that
we start with a set of dependencies that is a
minimal set

Properties of Relational
Decompositions (1)
 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.

Decompositions (2)
 Relation Decomposition and
Insufficiency of Normal Forms (cont.):
 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”.

Decompositions (2)
 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

Decompositions (3)
 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.

Decompositions (4)
 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
equivalent to F; that is
((R1(F)) υ . . . υ (Rm(F)))+ = F+
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.

Decompositions (5)
 Lossless (Non-additive) Join 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 better term is
“addition of spurious information”

Algorithm: 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) *)

Algorithm: Testing for Lossless Join
Property
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.

Properties of Relational Decompositions
(8)
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.

Properties of Relational Decompositions (8)
Lossless (nonadditive) join
test for n-ary
decompositions.
(c) Case 2: Decomposition
of EMP_PROJ into EMP,
PROJECT, and
WORKS_ON satisfies test.

Decompositions (9)
 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+.

Decompositions (10)
 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.

Algorithm: Relational Synthesis into 3NF
with Dependency Preservation (Relational
Synthesis Algorithm)

Algorithms for Relational Database
Schema Design (2)
 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.

Algorithms for Relational Database
Schema Design (4)

Module 4- Database Management System by Navathe

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