Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Chapter 8: Relational Database Design
©Silberschatz, Korth and Sudarshan8.2Database System Concepts - 6th Edition
Chapter 8: Relational Database Design
 Features of Good Relational Design
 Atomic Domains and First Normal Form
 Decomposition Using Functional Dependencies
 Functional Dependency Theory
 Algorithms for Functional Dependencies
 Decomposition Using Multivalued Dependencies
 More Normal Form
 Database-Design Process
 Modeling Temporal Data
©Silberschatz, Korth and Sudarshan8.3Database System Concepts - 6th Edition
University ERD for Reference in Examples
©Silberschatz, Korth and Sudarshan8.4Database System Concepts - 6th Edition
Combine Schemas?
 Suppose we combine instructor and department into inst_dept
 (No connection to relationship set inst_dept)
 Result is possible repetition of information
inst_dept
©Silberschatz, Korth and Sudarshan8.5Database System Concepts - 6th Edition
A Combined Schema Without Repetition
 Consider combining relations
 sec_class(sec_id, building, room_number) and
 section(course_id, sec_id, semester, year)
into one relation
 section(course_id, sec_id, semester, year,
building, room_number)
 No repetition in this case
©Silberschatz, Korth and Sudarshan8.6Database System Concepts - 6th Edition
What About Smaller Schemas?
 Suppose we had started with inst_dept. How would we know to split up
(decompose) it into instructor and department?
 Write a rule “if there were a schema (dept_name, building, budget), then
dept_name would be a candidate key”
 Denote as a functional dependency:
dept_name  building, budget
 In inst_dept, because dept_name is not a candidate key, the building
and budget of a department may have to be repeated.
 This indicates the need to decompose inst_dept
 Not all decompositions are good. Suppose we decompose
employee(ID, name, street, city, salary) into
employee1 (ID, name)
employee2 (name, street, city, salary)
 The next slide shows how we lose information -- we cannot reconstruct
the original employee relation -- and so, this is a lossy decomposition.
©Silberschatz, Korth and Sudarshan8.7Database System Concepts - 6th Edition
A Lossy Decomposition
©Silberschatz, Korth and Sudarshan8.8Database System Concepts - 6th Edition
Example of Lossless-Join Decomposition
 Lossless join decomposition
 Decomposition of R = (A, B, C)
R1 = (A, B) R2 = (B, C)
A B


1
2
A


B
1
2
r B,C(r)
A,B (r) B,C (r)
A B


1
2
C
A
B
B
1
2
C
A
B
C
A
B
A,B(r)
©Silberschatz, Korth and Sudarshan8.9Database System Concepts - 6th Edition
First Normal Form
 Domain is atomic if its elements are considered to be indivisible units
 Examples of non-atomic domains:
 Set of names, composite attributes
 Identification numbers like CS101 that can be broken up into
parts
 A relational schema R is in first normal form if the domains of all
attributes of R are atomic
 Non-atomic values complicate storage and encourage redundant
(repeated) storage of data
 Example: Set of accounts stored with each customer, and set of
owners stored with each account
 We assume all relations are in first normal form (and revisit this in
Chapter 22: Object Based Databases)
©Silberschatz, Korth and Sudarshan8.10Database System Concepts - 6th Edition
First Normal Form (Cont’d)
 Atomicity is actually a property of how the elements of the domain are
used.
 Example:
 Strings would normally be considered indivisible
 Suppose that students are given roll numbers which are strings of
the form CS0012 or EE1127
 If the first two characters are extracted to find the department,
the domain of roll numbers is not atomic.
 Doing so is a bad idea: leads to encoding of information in
application program rather than in the database.
©Silberschatz, Korth and Sudarshan8.11Database System Concepts - 6th Edition
Goal — Devise a Theory for the Following
 Decide whether a particular relation R is in “good” form.
 In the case that a relation R is not in “good” form, decompose it into a
set of relations {R1, R2, ..., Rn} such that
 each relation is in good form
 the decomposition is a lossless-join decomposition
 Our theory is based on:
 functional dependencies
 multivalued dependencies
©Silberschatz, Korth and Sudarshan8.12Database System Concepts - 6th Edition
Functional Dependencies
 Constraints on the set of legal relations.
 Require that the value for a certain set of attributes determines
uniquely the value for another set of attributes.
 A functional dependency is a generalization of the notion of a key.
©Silberschatz, Korth and Sudarshan8.13Database System Concepts - 6th Edition
Functional Dependencies (Cont.)
 Let R be a relation schema
  R and   R
 The functional dependency
  
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 , they also agree
on the attributes . That is,
t1[] = t2 []  t1[ ] = t2 [ ]
 Example: Consider r (A,B) with the following instance of r.
 On this instance, A  B does NOT hold, but B  A does
hold.
1 4
1 5
3 7
A B
©Silberschatz, Korth and Sudarshan8.14Database System Concepts - 6th Edition
Functional Dependencies (Cont.)
 K is a superkey 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   K,   R
 Functional dependencies allow us to express constraints that cannot be
expressed using superkeys. Consider the schema:
inst_dept (ID, name, salary, dept_name, building, budget ).
We expect these functional dependencies to hold:
dept_name  building
and ID  building
but would not expect the following to hold:
dept_name  salary
©Silberschatz, Korth and Sudarshan8.15Database System Concepts - 6th Edition
Use of Functional Dependencies
 We use functional dependencies to:
 test relations to see if they are legal under a given set of functional
dependencies.
 If a relation r is legal under a set F of functional dependencies, we
say that r satisfies F.
 specify constraints on the set of legal relations
 We say that F holds on R if all legal relations on R satisfy the set
of functional dependencies F.
 Note: A specific instance of a relation schema may satisfy a functional
dependency even if the functional dependency does not hold on all legal
instances.
 For example, a specific instance of instructor may, by chance, satisfy
name  ID.
©Silberschatz, Korth and Sudarshan8.16Database System Concepts - 6th Edition
Functional Dependencies (Cont.)
 A functional dependency is trivial if it is satisfied by all instances of a
relation
 Example:
 ID, name  ID
 name  name
 In general,    is trivial if   
©Silberschatz, Korth and Sudarshan8.17Database System Concepts - 6th Edition
Closure of a Set of Functional
Dependencies
 Given a set F of functional dependencies, there are certain other
functional dependencies that are logically implied by F.
 For example: If A  B and B  C, then we can infer that A 
C
 The set of all functional dependencies logically implied by F is the
closure of F.
 We denote the closure of F by F+.
 F+ is a superset of F.
©Silberschatz, Korth and Sudarshan8.18Database System Concepts - 6th Edition
Boyce-Codd Normal Form
    is trivial (i.e.,   )
  is a superkey for R
A relation schema R is in BCNF with respect to a set F of
functional dependencies if for all functional dependencies in F+ of
the form
  
where   R and   R, at least one of the following holds:
Example schema not in BCNF:
instr_dept (ID, name, salary, dept_name, building, budget )
because dept_name  building, budget
holds on instr_dept, but dept_name is not a superkey
©Silberschatz, Korth and Sudarshan8.19Database System Concepts - 6th Edition
Decomposing a Schema into BCNF
 Suppose we have a schema R and a non-trivial dependency  
causes a violation of BCNF.
We decompose R into:
• R1 = (   )
• R2 = ( R - (  -  ) )
 In our example,
  = dept_name
  = building, budget
and inst_dept is replaced by
 dept = (   ) = ( dept_name, building, budget )
 inst = ( R - (  -  ) ) = ( ID, name, salary, dept_name )
©Silberschatz, Korth and Sudarshan8.20Database System Concepts - 6th Edition
BCNF and Dependency Preservation
 Constraints, including functional dependencies, are costly to check in
practice unless they pertain to only one relation
 If it is sufficient to test only those dependencies on each individual
relation of a decomposition in order to ensure that all functional
dependencies hold, then that decomposition is dependency
preserving.
 Because it is not always possible to achieve both BCNF and
dependency preservation, we consider a weaker normal form, known
as third normal form.
©Silberschatz, Korth and Sudarshan8.21Database System Concepts - 6th Edition
Third Normal Form
 A relation schema R is in third normal form (3NF) if for all:
   in F+
at least one of the following holds:
    is trivial (i.e.,   )
  is a superkey for R
 Each attribute A in  –  is contained in a candidate key for R.
(NOTE: each attribute may be in a different candidate key)
 If a relation is in BCNF it is in 3NF (since in BCNF one of the first two
conditions above must hold).
 Third condition is a minimal relaxation of BCNF to ensure dependency
preservation (will see why later).
©Silberschatz, Korth and Sudarshan8.22Database System Concepts - 6th Edition
Goals of Normalization
 Let R be a relation scheme with a set F of functional dependencies.
 Decide whether a relation scheme R is in “good” form.
 In the case that a relation scheme R is not in “good” form,
decompose it into a set of relation scheme {R1, R2, ..., Rn} such that
 each relation scheme is in good form
 the decomposition is a lossless-join decomposition
 Preferably, the decomposition should be dependency preserving.
©Silberschatz, Korth and Sudarshan8.23Database System Concepts - 6th Edition
How good is BCNF?
 There are database schemas in BCNF that do not seem to be
sufficiently normalized
 Consider a relation
inst_info (ID, child_name, phone)
 where an instructor may have more than one phone and can have
multiple children
ID child_name phone
99999
99999
99999
99999
David
David
William
William
512-555-1234
512-555-4321
512-555-1234
512-555-4321
inst_info
©Silberschatz, Korth and Sudarshan8.24Database System Concepts - 6th Edition
 There are no non-trivial functional dependencies and therefore the
relation is in BCNF
 Insertion anomalies – i.e., if we add a phone 981-992-3443 to 99999,
we need to add two tuples
(99999, David, 981-992-3443)
(99999, William, 981-992-3443)
How good is BCNF? (Cont.)
©Silberschatz, Korth and Sudarshan8.25Database System Concepts - 6th Edition
 Therefore, it is better to decompose inst_info into:
This suggests the need for higher normal forms, such as
Fourth Normal Form (4NF), which we shall see later.
How good is BCNF? (Cont.)
ID child_name
99999
99999
99999
99999
David
David
William
William
inst_child
ID phone
99999
99999
99999
99999
512-555-1234
512-555-4321
512-555-1234
512-555-4321
inst_phone
©Silberschatz, Korth and Sudarshan8.26Database System Concepts - 6th Edition
Functional-Dependency Theory
 We now consider the formal theory that tells us which functional
dependencies are implied logically by a given set of functional
dependencies.
 We then develop algorithms to generate lossless decompositions into
BCNF and 3NF
 We then develop algorithms to test if a decomposition is dependency-
preserving
©Silberschatz, Korth and Sudarshan8.27Database System Concepts - 6th Edition
Closure of a Set of Functional
Dependencies
 Given a set F set of functional dependencies, there are certain other
functional dependencies that are logically implied by F.
 For e.g.: If A  B and B  C, then we can infer that A  C
 The set of all functional dependencies logically implied by F is the
closure of F.
 We denote the closure of F by F+
.
©Silberschatz, Korth and Sudarshan8.28Database System Concepts - 6th Edition
Closure of a Set of Functional
Dependencies
 We can find F+, the closure of F, by repeatedly applying
Armstrong’s Axioms:
 if   , then    (reflexivity)
 if   , then      (augmentation)
 if   , and   , then    (transitivity)
 These rules are
 sound (generate only functional dependencies that actually hold),
and
 complete (generate all functional dependencies that hold).
©Silberschatz, Korth and Sudarshan8.29Database System Concepts - 6th Edition
Example
 R = (A, B, C, G, H, I)
F = { A  B
A  C
CG  H
CG  I
B  H}
 Some members of F+
 A  H
 by transitivity from A  B and B  H
 AG  I
 by augmenting A  C with G, to get AG  CG
and then transitivity with CG  I
 CG  HI
 by augmenting CG  I to infer CG  CGI,
and augmenting of CG  H to infer CGI  HI,
and then transitivity
©Silberschatz, Korth and Sudarshan8.30Database System Concepts - 6th Edition
Procedure for Computing F+
 To compute the closure of a set of functional dependencies F:
F+ = F
repeat
for each functional dependency f in F+
apply reflexivity and augmentation rules on f
add the resulting functional dependencies to F+
for each pair of functional dependencies f1 and f2 in F+
if f1 and f2 can be combined using transitivity
then add the resulting functional dependency to F+
until F+ does not change any further
NOTE: We shall see an alternative procedure for this task later
©Silberschatz, Korth and Sudarshan8.31Database System Concepts - 6th Edition
Closure of Functional Dependencies
(Cont.)
 Additional rules:
 If    holds and    holds, then     holds (union)
 If     holds, then    holds and    holds
(decomposition)
 If    holds and     holds, then     holds
(pseudotransitivity)
The above rules can be inferred from Armstrong’s axioms.
©Silberschatz, Korth and Sudarshan8.32Database System Concepts - 6th Edition
Closure of Attribute Sets
 Given a set of attributes , define the closure of  under F (denoted
by +) as the set of attributes that are functionally determined by 
under F
 Algorithm to compute +, the closure of  under F
result := ;
while (changes to result) do
for each    in F do
begin
if   result then result := result  
end
©Silberschatz, Korth and Sudarshan8.33Database System Concepts - 6th Edition
Example of Attribute Set Closure
 R = (A, B, C, G, H, I)
 F = {A  B
A  C
CG  H
CG  I
B  H}
 (AG)+
1. result = AG
2. result = ABCG (A  C and A  B)
3. result = ABCGH (CG  H and CG  AGBC)
4. result = ABCGHI (CG  I and CG  AGBCH)
 Is AG a candidate key?
1. Is AG a super key?
1. Does AG  R? == Is (AG)+  R
2. Is any subset of AG a superkey?
1. Does A  R? == Is (A)+  R
2. Does G  R? == Is (G)+  R
©Silberschatz, Korth and Sudarshan8.34Database System Concepts - 6th Edition
Uses of Attribute Closure
There are several uses of the attribute closure algorithm:
 Testing for superkey:
 To test if  is a superkey, we compute +, and check if + contains
all attributes of R.
 Testing functional dependencies
 To check if a functional dependency    holds (or, in other
words, is in F+), just check if   +.
 That is, we compute + by using attribute closure, and then check
if it contains .
 Is a simple and cheap test, and very useful
 Computing closure of F
 For each   R, we find the closure +, and for each S  +, we
output a functional dependency   S.
©Silberschatz, Korth and Sudarshan8.35Database System Concepts - 6th Edition
Canonical Cover Fc
 Sets of functional dependencies may have redundant dependencies
that can be inferred from the others
 For example: A  C is redundant in: {A  B, B  C, A  C}
 Parts of a functional dependency may be redundant
 E.g.: on RHS: {A  B, B  C, A  CD} can be simplified
to
{A  B, B  C, A  D}
 E.g.: on LHS: {A  B, B  C, AC  D} can be simplified
to
{A  B, B  C, A  D}
 Intuitively, a canonical cover of F is a “minimal” set of functional
dependencies equivalent to F, having no redundant dependencies or
redundant parts of dependencies
©Silberschatz, Korth and Sudarshan8.36Database System Concepts - 6th Edition
Extraneous Attributes
 Consider a set F of functional dependencies and the functional
dependency    in F.
 Attribute A is extraneous in  if A  
and F logically implies (F – {  })  {( – A)  }.
 Attribute A is extraneous in  if A  
and the set of functional dependencies
(F – {  })  {  ( – A)} logically implies F.
 Note: implication in the opposite direction is trivial in each of the
cases above, since a “stronger” functional dependency always
implies a weaker one
 Example: Given F = {A  C, AB  C }
 B is extraneous in AB  C because {A  C, AB  C} logically
implies A  C (i.e. the result of dropping B from AB  C).
 Example: Given F = {A  C, AB  CD}
 C is extraneous in AB  CD since AB  C can be inferred even
after deleting C
©Silberschatz, Korth and Sudarshan8.37Database System Concepts - 6th Edition
Testing if an Attribute is Extraneous
 Consider a set F of functional dependencies and the functional
dependency    in F.
 To test if attribute A   is extraneous in 
1. compute ({} – A)+ using the dependencies in F
2. check that ({} – A)+ contains ; if it does, A is extraneous in 
 To test if attribute A   is extraneous in 
1. compute + using only the dependencies in
F’ = (F – {  })  { ( – A)},
2. check that + contains A; if it does, A is extraneous in 
©Silberschatz, Korth and Sudarshan8.38Database System Concepts - 6th Edition
Canonical Cover
 A canonical cover for F is a set of dependencies Fc such that
 F logically implies all dependencies in Fc, and
 Fc logically implies all dependencies in F, and
 No functional dependency in Fc contains an extraneous attribute, and
 Each left side of functional dependency in Fc is unique.
 To compute a canonical cover for F:
repeat
Use the union rule to replace any dependencies in F
1  1 and 1  2 with 1  1 2
Find a functional dependency    with an
extraneous attribute either in  or in 
/* Note: test for extraneous attributes done using Fc, not F*/
If an extraneous attribute is found, delete it from   
until F does not change
 Note: Union rule may become applicable after some extraneous attributes
have been deleted, so it has to be re-applied
©Silberschatz, Korth and Sudarshan8.39Database System Concepts - 6th Edition
Computing a Canonical Cover
 R = (A, B, C)
F = {A  BC
B  C
A  B
AB  C}
 Combine A  BC and A  B into A  BC
 Set is now {A  BC, B  C, AB  C}
 A is extraneous in AB  C
 Check if the result of deleting A from AB  C is implied by the other dependencies
 Yes: in fact, B  C is already present!
 Set is now {A  BC, B  C}
 C is extraneous in A  BC
 Check if A  BC is logically implied by A  B and the other dependencies
 Yes: using transitivity on A  B and B  C we get A  C and then using union
– Can use attribute closure of A in more complex cases
 The canonical cover is: A  B
B  C
©Silberschatz, Korth and Sudarshan8.40Database System Concepts - 6th Edition
Lossless-join Decomposition
 For the case of R = (R1, R2), we require that for all possible relations r
on schema R
r = R1 (r ) R2 (r )
 A decomposition of R into R1 and R2 is lossless join if at least one of
the following dependencies is in F+:
 R1  R2  R1
 R1  R2  R2
 The above functional dependencies are a sufficient condition for
lossless-join decomposition; the dependencies are a necessary
condition only if all constraints are functional dependencies
©Silberschatz, Korth and Sudarshan8.41Database System Concepts - 6th Edition
Example
 R = (A, B, C)
F = {A  B, B  C)
 Can be decomposed in two different ways
 R1 = (A, B), R2 = (B, C)
 Lossless-join decomposition:
R1  R2 = {B} and B  BC
 Dependency preserving
 R1 = (A, B), R2 = (A, C)
 Lossless-join decomposition:
R1  R2 = {A} and A  AB
 Not dependency preserving
(cannot check B  C without computing R1  R2)
©Silberschatz, Korth and Sudarshan8.42Database System Concepts - 6th Edition
Dependency Preservation
 Let Fi be the set of dependencies F+ that include only attributes in Ri
 A decomposition is dependency preserving, if
(F1  F2  …  Fn )+ = F+
 If it is not, then checking updates for violation of functional
dependencies may require computing joins, which is expensive.
©Silberschatz, Korth and Sudarshan8.43Database System Concepts - 6th Edition
Testing for Dependency Preservation
 To check if a dependency    is preserved in a decomposition
of R into R1, R2, …, Rn we apply the following test
(with attribute closure done with respect to F)
 result = 
while (changes to result) do
for each Ri in the decomposition
t = (result  Ri)+  Ri
result = result  t
 If result contains all attributes in , then
the functional dependency    is preserved.
 We apply the test on all dependencies in F to check if a
decomposition is dependency preserving
 This procedure takes polynomial time, instead of the exponential
time required to compute F+ and (F1  F2  …  Fn)+
©Silberschatz, Korth and Sudarshan8.44Database System Concepts - 6th Edition
Example
 R = (A, B, C )
F = {A  B
B  C}
Key = {A}
 R is not in BCNF
 Decomposition R1 = (A, B), R2 = (B, C)
 R1 and R2 in BCNF
 Lossless-join decomposition
 Dependency preserving
©Silberschatz, Korth and Sudarshan8.45Database System Concepts - 6th Edition
Testing for BCNF
 To check if a non-trivial dependency   causes a violation of BCNF
1. compute + (the attribute closure of ), and
2. verify that it includes all attributes of R, that is, it is a superkey of R.
 Simplified test: To check if a relation schema R is in BCNF, it suffices
to check only the dependencies in the given set F for violation of BCNF,
rather than checking all dependencies in F+.
 If none of the dependencies in F causes a violation of BCNF, then
none of the dependencies in F+ will cause a violation of BCNF
either.
 However, simplified test using only F is incorrect when testing a
relation in a decomposition of R
 Consider R = (A, B, C, D, E), with F = { A  B, BC  D}
 Decompose R into R1 = (A,B) and R2 = (A,C,D, E)
 Neither of the dependencies in F contain only attributes from
(A,C,D,E) so we might be mislead into thinking R2 satisfies
BCNF.
 In fact, dependency AC  D in F+ shows R2 is not in BCNF.
©Silberschatz, Korth and Sudarshan8.46Database System Concepts - 6th Edition
Testing Decomposition for BCNF
 To check if a relation Ri in a decomposition of R is in BCNF,
 Either test Ri for BCNF with respect to the restriction of F to Ri
(that is, all FDs in F+ that contain only attributes from Ri)
 or use the original set of dependencies F that hold on R, but with
the following test:
– for every set of attributes   Ri, check that + (the
attribute closure of ) either includes no attribute of Ri- ,
or includes all attributes of Ri.
 If the condition is violated by some    in F, the
dependency
  (+ -  )  Ri
can be shown to hold on Ri, and Ri violates BCNF.
 We use above dependency to decompose Ri
So it is a trivial FD.
So  is a superkey.
©Silberschatz, Korth and Sudarshan8.47Database System Concepts - 6th Edition
BCNF Decomposition Algorithm
result := {R };
done := false;
compute F+;
while (not done) do
if (there is a schema Ri in result that is not in BCNF)
then begin
let    be a nontrivial functional dependency that
holds on Ri such that   Ri is not in F+,
and    = ;
result := (result – Ri )  (Ri – )  (, );
end
else done := true;
Note: each Ri is in BCNF, and decomposition is lossless-join.
©Silberschatz, Korth and Sudarshan8.48Database System Concepts - 6th Edition
Example of BCNF Decomposition
 R = (A, B, C )
F = {A  B
B  C}
Key = {A}
 R is not in BCNF (B  C but B is not a superkey)
 Decomposition
 R1 = (B,C)
 R2 = (A,B)
©Silberschatz, Korth and Sudarshan8.49Database System Concepts - 6th Edition
Example of BCNF Decomposition
 class (course_id, title, dept_name, credits, sec_id, semester, year,
building, room_number, capacity, time_slot_id)
 Functional dependencies:
 course_id  title, dept_name, credits
 building, room_number  capacity
 course_id, sec_id, semester, year  building, room_number,
time_slot_id
 A candidate key {course_id, sec_id, semester, year}.
 BCNF Decomposition:
 course_id  title, dept_name, credits holds
 but course_id is not a superkey.
 We replace class by:
 course(course_id, title, dept_name, credits)
 class-1 (course_id, sec_id, semester, year, building,
room_number, capacity, time_slot_id)
©Silberschatz, Korth and Sudarshan8.50Database System Concepts - 6th Edition
BCNF Decomposition (Cont.)
 course(course_id, title, dept_name, credits) is in BCNF
 How do we know this?
 class-1 (course_id, sec_id, semester, year, building,
room_number, capacity, time_slot_id)
 building, room_number  capacity holds on class-1
 but {building, room_number} is not a superkey for class-1.
 We replace class-1 by:
 classroom (building, room_number, capacity)
 section (course_id, sec_id, semester, year, building,
room_number, time_slot_id)
 classroom and section are in BCNF.
©Silberschatz, Korth and Sudarshan8.51Database System Concepts - 6th Edition
BCNF and Dependency Preservation
 Relation dept_advisor (s_ID, i_ID, dept_name)
F = { s_ID, dept_name  i_ID,
i_ID  dept_name }
Two candidate keys = s_ID, dept_name and
s_ID, i_ID
 dept_advisor is not in BCNF
 Any decomposition of dept_advisor will fail to preserve
s_ID, dept_name  i_ID
This implies that testing for s_ID, dept_name  i_ID
requires a join
It is not always possible to get a BCNF decomposition that is
dependency preserving
©Silberschatz, Korth and Sudarshan8.52Database System Concepts - 6th Edition
Third Normal Form: Motivation
 There are some situations where
 BCNF is not dependency preserving, and
 efficient checking for FD violation on updates is important
 Solution: define a weaker normal form
 called Third Normal Form (3NF)
 Allows some redundancy (with resultant problems; we will
see examples later)
 But functional dependencies can be checked on individual
relations without computing a join.
 There is always a lossless-join, dependency-preserving
decomposition into 3NF.
©Silberschatz, Korth and Sudarshan8.53Database System Concepts - 6th Edition
3NF Example
 Relation dept_advisor (s_ID, i_ID, dept_name)
 F = { s_ID, dept_name  i_ID,
i_ID  dept_name }
 Two candidate keys: s_ID, dept_name, and i_ID, s_ID
 dept_advisor is in 3NF
 s_ID, dept_name  i_ID
 s_ID, dept_name is a candidate key
 i_ID  dept_name
 dept_name is contained in a candidate key
©Silberschatz, Korth and Sudarshan8.54Database System Concepts - 6th Edition
Redundancy in 3NF
J
j1
j2
j3
null
L
l1
l1
l1
l2
K
k1
k1
k1
k2
 repetition of information (e.g., the relationship l1, k1)
 (i_ID, dept_name)
 need to use null values (e.g., to represent the relationship
l2, k2 where there is no corresponding value for J).
 (i_ID, dept_name) if there is no separate relation mapping
instructors to departments
 There is some redundancy in this schema
 Example of problems due to redundancy in 3NF
 R = (J, K, L)
F = {JK  L, L  K }
©Silberschatz, Korth and Sudarshan8.55Database System Concepts - 6th Edition
Testing for 3NF
 Optimization: Need to check only FDs in F, need not check all FDs in
F+.
 Use attribute closure to check for each dependency   , whether 
is a superkey.
 If  is not a superkey, we have to verify if each attribute in  is
contained in a candidate key of R
 This test is rather more expensive, since it involve finding all
candidate keys
 Testing for 3NF has been shown to be NP-hard
 Interestingly, decomposition into third normal form (described
shortly) can be done in polynomial time
©Silberschatz, Korth and Sudarshan8.56Database System Concepts - 6th Edition
3NF Decomposition Algorithm for schema R
Let Fc be a canonical cover for F;
i := 0;
for each functional dependency    in Fc do
if none of the schemas Rj, 1  j  i contains  
then begin
i := i + 1;
Ri :=  
end
if none of the schemas Rj, 1  j  i contains a candidate key for R
then begin
i := i + 1;
Ri := any candidate key for R;
end
-- Optionally, remove redundant relations
while any schema Rj is contained in another schema Rk
-- delete Rj
Rj = Ri ;
i=i-1;
return (R1, R2, ..., Ri )
©Silberschatz, Korth and Sudarshan8.57Database System Concepts - 6th Edition
3NF Decomposition Algorithm (Cont.)
 Above algorithm ensures:
 each relation schema Ri is in 3NF
 decomposition is dependency preserving and lossless-join
 Proof of correctness is at end of this presentation (click here)
©Silberschatz, Korth and Sudarshan8.58Database System Concepts - 6th Edition
3NF Decomposition: An Example
 Relation schema:
cust_banker_branch = (customer_id, employee_id, branch_name, type )
 The functional dependencies for this relation schema are:
1. customer_id, employee_id  branch_name, type
2. employee_id  branch_name
3. customer_id, branch_name  employee_id
 We first compute a canonical cover
 branch_name is extraneous in the r.h.s. of the 1st dependency
 No other attribute is extraneous, so we get
FC = { customer_id, employee_id  type
employee_id  branch_name
customer_id, branch_name  employee_id }
 What are candidate keys?
©Silberschatz, Korth and Sudarshan8.59Database System Concepts - 6th Edition
3NF Decompsition Example (Cont.)
 The for loop generates following 3NF schema:
(customer_id, employee_id, type )
(employee_id, branch_name)
(customer_id, branch_name, employee_id)
 Observe that (customer_id, employee_id, type ) contains a
candidate key of the original schema, so no further relation schema
needs be added
 At end of for loop, detect and delete schemas, such as (employee_id,
branch_name), which are subsets of other schemas
 result will not depend on the order in which FDs are considered
 The resultant simplified 3NF schema is:
(customer_id, employee_id, type)
(customer_id, branch_name, employee_id)
©Silberschatz, Korth and Sudarshan8.60Database System Concepts - 6th Edition
Comparison of BCNF and 3NF
 It is always possible to decompose a relation into a set of relations
that are in 3NF such that:
 the decomposition is lossless
 the dependencies are preserved
 It is always possible to decompose a relation into a set of relations
that are in BCNF such that:
 the decomposition is lossless
 it may not be possible to preserve dependencies.
©Silberschatz, Korth and Sudarshan8.61Database System Concepts - 6th Edition
Design Goals
 Goal for a relational database design is:
 BCNF.
 Lossless join.
 Dependency preservation.
 If we cannot achieve this, we accept one of
 Lack of dependency preservation
 Redundancy due to use of 3NF
 Interestingly, SQL does not provide a direct way of specifying functional
dependencies other than superkeys.
Can specify FDs using assertions, but they are expensive to test, (and
currently not supported by any of the widely used databases!)
 Even if we had a dependency preserving decomposition, using SQL we
would not be able to efficiently test a functional dependency whose left
hand side is not a key.
©Silberschatz, Korth and Sudarshan8.62Database System Concepts - 6th Edition
PB168, Vlastislav Dohnal, FI MUNI, 2014 62
Second Normal Form
 A functional dependency    is called a partial dependency
 if there is a subset  of , i.e.   , such that   .
 We say that  is partially dependent on .
 A relation R is in second normal form (2NF) if it is in 1NF and
each attribute A in R meets one of the following:
 A appears in a candidate key;
 A is not partially dependent on a candidate key.
 Every 3NF is in 2NF.
 Every partial dependency is a transitive dependency.
©Silberschatz, Korth and Sudarshan8.63Database System Concepts - 6th Edition
Multivalued Dependencies
 Suppose we record names of children, and phone numbers for
instructors:
 inst_child(ID, child_name)
 inst_phone(ID, phone_number)
 If we were to combine these schemas to get
 inst_info(ID, child_name, phone_number)
 Example data:
(99999, David, 512-555-1234)
(99999, David, 512-555-4321)
(99999, William, 512-555-1234)
(99999, William, 512-555-4321)
 This relation is in BCNF
 Why?
©Silberschatz, Korth and Sudarshan8.64Database System Concepts - 6th Edition
Multivalued Dependencies (MVDs)
 Let R be a relation schema and let   R and   R. The
multivalued dependency
  
holds on R if in any legal relation r(R), for all pairs for tuples t1 and t2
in r such that t1[] = t2 [], there exist tuples t3 and t4 in r such that:
t1[] = t2 [] = t3 [] = t4 []
t3[] = t1 []
t3[R – ] = t2[R – ]
t4 [] = t2[]
t4[R – ] = t1[R – ]
©Silberschatz, Korth and Sudarshan8.65Database System Concepts - 6th Edition
MVD (Cont.)
 Tabular representation of   
©Silberschatz, Korth and Sudarshan8.66Database System Concepts - 6th Edition
Example
 Let R be a relation schema with a set of attributes that are partitioned
into 3 nonempty subsets.
Y, Z, W
 We say that Y  Z (Y multidetermines Z )
if and only if for all possible relations r (R )
< y1, z1, w1 >  r and < y1, z2, w2 >  r
then
< y1, z1, w2 >  r and < y1, z2, w1 >  r
 Note that since the behavior of Z and W are identical it follows that
Y  Z if Y  W
©Silberschatz, Korth and Sudarshan8.67Database System Concepts - 6th Edition
Example (Cont.)
 In our example:
ID  child_name
ID  phone_number
 The above formal definition is supposed to formalize the notion that given
a particular value of Y (ID) it has associated with it a set of values of Z
(child_name) and a set of values of W (phone_number), and these two
sets are in some sense independent of each other.
 Note:
 If Y  Z then Y  Z
 Indeed we have (in above notation) Z1 = Z2
The claim follows.
©Silberschatz, Korth and Sudarshan8.68Database System Concepts - 6th Edition
Use of Multivalued Dependencies
 We use multivalued dependencies in two ways:
1. To test relations to determine whether they are legal under a
given set of functional and multivalued dependencies
2. To specify constraints on the set of legal relations. We shall
thus concern ourselves only with relations that satisfy a given
set of functional and multivalued dependencies.
 If a relation r fails to satisfy a given multivalued dependency, we can
construct a relations r that does satisfy the multivalued
dependency by adding tuples to r.
©Silberschatz, Korth and Sudarshan8.69Database System Concepts - 6th Edition
Theory of MVDs
 From the definition of multivalued dependency, we can derive the
following rule:
 If   , then   
That is, every functional dependency is also a multivalued dependency
 The closure D+ of D is the set of all functional and multivalued
dependencies logically implied by D.
 We can compute D+ from D, using the formal definitions of
functional dependencies and multivalued dependencies.
 We can manage with such reasoning for very simple multivalued
dependencies, which seem to be most common in practice
 For complex dependencies, it is better to reason about sets of
dependencies using a system of inference rules (see Appendix C).
©Silberschatz, Korth and Sudarshan8.70Database System Concepts - 6th Edition
Fourth Normal Form
 A relation schema R is in 4NF with respect to a set D of functional and
multivalued dependencies if for all multivalued dependencies in D+ of
the form   , where   R and   R, at least one of the following
hold:
    is trivial (i.e.,    or    = R)
  is a superkey for schema R
 If a relation is in 4NF it is in BCNF
©Silberschatz, Korth and Sudarshan8.71Database System Concepts - 6th Edition
Restriction of Multivalued Dependencies
 The restriction of D to Ri is the set Di consisting of
 All functional dependencies in D+ that include only attributes of Ri
 All multivalued dependencies of the form
  (  Ri)
where   Ri and    is in D+
©Silberschatz, Korth and Sudarshan8.72Database System Concepts - 6th Edition
4NF Decomposition Algorithm
result: = {R};
done := false;
compute D+;
Let Di denote the restriction of D+ to Ri
while (not done)
if (there is a schema Ri in result that is not in 4NF) then
begin
let    be a nontrivial multivalued dependency that holds
on Ri such that   Ri is not in Di, and ;
result := (result - Ri)  (Ri - )  (, );
end
else done:= true;
Note: each Ri is in 4NF, and decomposition is lossless-join
©Silberschatz, Korth and Sudarshan8.73Database System Concepts - 6th Edition
Example
 R =(A, B, C, G, H, I)
F ={ A  B
B  HI
CG  H }
 R is not in 4NF since A  B and A is not a superkey for R
 Decomposition
a) R1 = (A, B) (R1 is in 4NF)
b) R2 = (A, C, G, H, I) (R2 is not in 4NF, decompose into R3 and R4)
c) R3 = (C, G, H) (R3 is in 4NF)
d) R4 = (A, C, G, I) (R4 is not in 4NF, decompose into R5 and R6)
 A  B and B  HI  A  HI, (MVD transitivity), and
 and hence A  I (MVD restriction to R4)
e) R5 = (A, I) (R5 is in 4NF)
f)R6 = (A, C, G) (R6 is in 4NF)
©Silberschatz, Korth and Sudarshan8.74Database System Concepts - 6th Edition
Further Normal Forms
 Join dependencies generalize multivalued dependencies
 lead to project-join normal form (PJNF) (also called fifth normal
form)
 A class of even more general constraints, leads to a normal form
called domain-key normal form.
 Problem with these generalized constraints: are hard to reason with,
and no set of sound and complete set of inference rules exists.
 Hence rarely used
©Silberschatz, Korth and Sudarshan8.75Database System Concepts - 6th Edition
Overall Database Design Process
 We have assumed schema R is given
 R could have been generated when converting E-R diagram to a set
of tables.
 R could have been a single relation containing all attributes that are
of interest (called universal relation).
 R could have been the result of some ad hoc design of relations,
which we then test/convert to normal form.
 Normalization breaks R into smaller relations.
©Silberschatz, Korth and Sudarshan8.76Database System Concepts - 6th Edition
E-R Model and Normalization
 When an E-R diagram is carefully designed, identifying all entities
correctly, the tables generated from the E-R diagram should not need
further normalization.
 However, in a real (imperfect) design, there can be functional
dependencies from non-key attributes of an entity to other attributes of
the entity
 Example: an employee entity with attributes
department_name and building,
and a functional dependency
department_name building
 Good design would have made department an entity set
 Functional dependencies from non-key attributes of a relationship set
possible, but rare --- most relationships are binary
©Silberschatz, Korth and Sudarshan8.77Database System Concepts - 6th Edition
Denormalization for Performance
 May want to use non-normalized schema for performance
 For example, displaying prereqs(course_id,prereq_course_id) along with
course title requires join of course with prereq
 Alternative 1:
Use denormalized relation containing attributes of course as well as
prereq with all above attributes
 faster lookup
 extra space and extra execution time for updates
 extra coding work for programmer and possibility of error in extra code
 Alternative 2:
Use a materialized view defined as course  prereq
 Benefits and drawbacks same as above, except no extra coding work
for programmer and avoids possible errors
©Silberschatz, Korth and Sudarshan8.78Database System Concepts - 6th Edition
Other Design Issues
 Some aspects of database design are not caught by normalization
 Examples of bad database design, to be avoided:
Instead of earnings (company_id, year, amount ), use
 earnings_2004, earnings_2005, earnings_2006, etc., all on the same
schema (company_id, earnings).
 Above are in BCNF, but make querying across years difficult and
needs new table each year
 company_year (company_id, earnings_2004, earnings_2005,
earnings_2006)
 Also in BCNF, but also makes querying across years difficult and
requires new attribute each year.
 Is an example of a crosstab, where values for one attribute
become column names
 Used in spreadsheets, and in data analysis tools
©Silberschatz, Korth and Sudarshan8.79Database System Concepts - 6th Edition
Modeling Temporal Data
 Temporal data have an association time interval during which the
data are valid.
 A snapshot is the value of the data at a particular point in time
 Several proposals to extend ER model by adding valid time to
 attributes, e.g., address of an instructor at different points in time
 entities, e.g., time duration when a student entity exists
 relationships, e.g., time during which an instructor was associated
with a student as an advisor.
 But no accepted standard
 Adding a temporal component results in functional dependencies like
ID  street, city
not to hold, because the address varies over time
 A temporal functional dependency X  Y holds on schema R if the
functional dependency X  Y holds on all snapshots for all legal
instances r(R).
©Silberschatz, Korth and Sudarshan8.80Database System Concepts - 6th Edition
Modeling Temporal Data (Cont.)
 In practice, database designers may add start and end time attributes
to relations
 E.g., course(course_id, course_title) is replaced by
course(course_id, course_title, start, end)
 Constraint: no two tuples can have overlapping valid times
– Hard to enforce efficiently
 Foreign key references may be to current version of data, or to data at
a point in time
 E.g., student transcript should refer to course information at the
time the course was taken
Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
End of Chapter
Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Proof of Correctness of 3NF
Decomposition Algorithm
©Silberschatz, Korth and Sudarshan8.83Database System Concepts - 6th Edition
Correctness of 3NF Decomposition
Algorithm
 3NF decomposition algorithm is dependency preserving (since there
is a relation for every FD in Fc)
 Decomposition is lossless
 A candidate key (C) is in one of the relations Ri in decomposition
 Closure of candidate key under Fc must contain all attributes in
R.
 Follow the steps of attribute closure algorithm to show there is
only one tuple in the join result for each tuple in Ri
©Silberschatz, Korth and Sudarshan8.84Database System Concepts - 6th Edition
Correctness of 3NF Decomposition
Algorithm (Cont’d.)
Claim: if a relation Ri is in the decomposition generated by the
above algorithm, then Ri satisfies 3NF.
 Let Ri be generated from the dependency   
 Let   B be any non-trivial functional dependency on Ri. (We need only
consider FDs whose right-hand side is a single attribute.)
 Now, B can be in either  or  but not in both. Consider each case
separately.
©Silberschatz, Korth and Sudarshan8.85Database System Concepts - 6th Edition
Correctness of 3NF Decomposition
(Cont’d.)
 Case 1: If B in :
 If  is a superkey, the 2nd condition of 3NF is satisfied
 Otherwise  must contain some attribute not in 
 Since   B is in F+ it must be derivable from Fc, by using attribute
closure on .
 Attribute closure not have used  . If it had been used,  must
be contained in the attribute closure of , which is not possible, since
we assumed  is not a superkey.
 Now, using  (- {B}) and   B, we can derive  B
(since    , and B   since   B is non-trivial)
 Then, B is extraneous in the right-hand side of  ; which is not
possible since   is in Fc.
 Thus, if B is in  then  must be a superkey, and the second
condition of 3NF must be satisfied.
©Silberschatz, Korth and Sudarshan8.86Database System Concepts - 6th Edition
Correctness of 3NF Decomposition
(Cont’d.)
 Case 2: B is in .
 Since  is a candidate key, the third alternative in the definition of
3NF is trivially satisfied.
 In fact, we cannot show that  is a superkey.
 This shows exactly why the third alternative is present in the
definition of 3NF.
Q.E.D.