The document discusses functional dependencies and database normalization. It provides examples of functional dependencies and explains key concepts like: - Functional dependencies define relationships between attributes in a relation. - Armstrong's axioms are properties used to derive functional dependencies. - Decomposition aims to eliminate redundancy and anomalies by breaking relations into smaller, normalized relations while preserving information and dependencies. - A decomposition is lossless if it does not lose any information, and dependency preserving if the original dependencies can be maintained on the decomposed relations.

1. Prepared by:
RAJ NAIK
SY CE-1
150410107053

2.  Fuctional dependencies play a key role in differentiating
good database designs from bad database designs.
 Let R be a relation schema,
α  R and β  R
 The functional dependency
α  β
holds on R if and only if for any legal relations
r(R), whenever any two tuples t1 and t2 of R agree on the
attributes α, they also agree on the attributes β. That
is,
 t1 [α] = t2 [α]  t1 [β] = t2 [β ]

3.  Example:
Consider r(A,B) with the following instance of r.
 On this instance, A  B does NOT hold, but B  A does
hold.
 A functional dependency is a generalization of the notion
of a key
1 4
1 7
4 6

4.  K is a superkey or for relation schema R if and only if K
 R
 K is a candidate key for R if and only if
K  R, and
for no α  K, α  R
 Functional dependencies allow us to express
constraints that cannot be expressed using superkeys.

5. Consider the schema:
inst_dept (ID, name, salary, dept_name, building,
budget )
 We expect these functional dependencies to hold:
dept_name  building
ID  building
 but would not expect the following to hold:
dept_name  salary

6.  A functional dependency is trivial if it is satisfied by
all instances of a relation
 Example:
ID, name  ID
name  name
In general, α  β is trivial if β  α

7.  The most important properties are Armstrong's
axioms, which are used in database normalization:
 Subset Property
If Y is a subset of X, then X → Y
 Augmentation
If X → Y, then XZ → YZ
 Transitivity
If X → Y and Y → Z, then X → Z

8.  From these rules, we can derive these secondary rules:
 Union:
If X → Y and X → Z, then X → YZ
 Decomposition:
If X → YZ, then X → Y and X → Z
 Pseudotransitivity:
f X → Y and WY → Z, then XW → Z

9.  We can avoid update anomalies by decomposition of
the original relation.
 The relation scema R is decomposed into relation
schemas:
 {R1,R2,R3,…….,Rn} in such a way that
R1  R2  …  Rn = R

10. Lending_schema =
(Branch_name, Branch_city, Assets,
Cusomer_name, Loan_number, Amount)
 The Lending schema is decomposed into following two
relations:
1) Branch_customer_schema =
(Branch_name, Branch_city, Assets,
Cusomer_name)
2) Customer_loan_schema =
(Cusomer_name, Loan_number, Amount)

11. Branch_Name Branch_City Assets Customer_Name
Downtown Brooklyn 90000 Jones
Redwood Palo Alto 21000 Smith
Perryridge Horseneck 90000 Jackson
Mianus Horseneck 80000 Jones
Northtown Rye 37000 Hayes
Customer_Name Loan_No Amount
Jones L-17 10000
Smith L-23 2000
Jackson L-14 1500
Jones L-11 900
Hayes L-16 1300
The Branch_customer Relation
The Customer_loan Relation

12. Branch_Na
me
Branch_Ci
ty
Assets Customer_
Name
Loan_No Amount
Downtown Brooklyn 90000 Jones L-17 10000
Downtown Brooklyn 90000 Jones L-11 9000
Redwood Palo Alto 21000 Smith L-23 2000
Perryridge Horseneck 90000 Jackson L-14 1500
Mianus Horseneck 80000 Jones L-11 900
Mianus Horseneck 80000 Jones L-17 1000
Northtown Rye 37000 Hayes L-16 1300
The result of this join is shown below:
The relation Branch_customer Customer_loan

13. 1. If the relation Branch_customer and Customer_loan
is compared with the original lending relation, we
observe that:
 Every tuple in lending relation is in Branch_customer
and Customer_loan relation.
 The relation Branch_customer and Customer_loan
contain following additional tuples:
 (Downtown,Brooklyn,90000,Jones,L-11,9000)
 (Mianus,Horseneck,80000,Jones,L-17,1000)

14. 2. Consider the query:
 “Find all the branches that have made a loan in an
amount less than Rs. 1000.”
 The result of this query using the lending relation is :
Mianus.
 The result of the same query using the relation
Branch_customer and Customer_loan is: Downtown
and Mianus.

15. 3. From the above two cases we can say that
the decomposition of lending relation into
Branch_customer and Customer_loan
results in bad database design. It results in
loss of information and repetition of
information.
4. Because of loss of information, we call the
decomposition of lending relation into
Branch_customer and Customer_loan a
lossy decomposition or a lossy-join
decomposition.

16.  Decomposition help us eliminate redundancy, root of
data anomalies.
 There are two important properties associated with
decomposition:
 1. Lossless join
 2. Dependency Preservation

17.  This property ensures that any instance of the original relation
can be identifies from the corresponding instances in the smaller
relation.
 Let R be a relational schema, and let F be a set of functional
dependencies on R. Let R1 and R2 form a decomposition of R.
This decomposition is a lossless-join decomposition of R if at
least one of the following functional dependencies is in F+
R1  R2  R1
R1  R2  R2
 In other word, if R1  R2 forms a superkey of either R1 or R2, the
decomposition of R is a lossless-join decomposition.

18. Question: Let R(A,B,C) and F=(AB).Check the
decomposition of R into R1(A,B) and R2(A,C).
 R1  R2 is A which is comman attribute.
A B is the FD in F. (augmentation rule)
A AB which is relation R1.
Thus,
 R1  R2 is satisfies.
 Hence, above decomposition is lossless-join
decomposition.

19. Question: Let R(A,B,C) and F=(A B).Check the
decomposition of R into R1(A,B) and R2(B,C).
 R1  R2 is B which is comman attribute.
 But B is not a superkey of either R1 or R2.
 Hence this decomposition is not a lossless-join
decomposition.

20.  This property ensures that a constraint on the original
relation can be maintained by simply enforcing some
constraint on each of the smaller relation.
 Given a relational schema R and set of fuctional
dependencies associated with it is F.
 R is decomposed into the relation schema R1,R2,…Rn
with functional dependencies F1,F2,…Fn.

21.  Then the decomposition is dependency preserving
if the closure of F’ (where F’=F1,F2,….Fn) is
identical to F+.
F’+ =F+
 Getting lossless decomposition is necessary.

22. Question: R=(A, B, C), F={A  B, B  C} Decomposition
of R: R1=(A, C) R2=(B, C) Does this decomposition preserve
the given dependencies?
 In R1 the following dependencies hold:
F1’={A  A, C  C, A  C, AC  AC}
 In R2 the following dependencies hold:
F2’= {B  B, C  C, B  C, BC  BC}
 The set of nontrivial dependencies hold on R1 and R2:
F‘ = {B  C, A  C}
 A  B can not be derived from F’, so this decomposition is
NOT dependency preserving.

23. Question: R=(A, B, C), F={A  B, B  C}
Decomposition of R: R1=(A, B) R2=(B, C) Does this
decomposition preserve the given dependencies?
 In R1 the following dependencies hold:
F1={A  B, A  A, B  B, AB  AB}
 In R2 the following dependencies hold:
F2= {B  B, C  C, B  C, BC  BC}
 F’= F1’ U F2’ = {A  B, B  C, trivial dependencies}
 In F’ all the original dependencies occur, so this
decomposition preserves dependencies.