This document discusses database normalization and functional dependencies. It provides examples of 1st, 2nd, and 3rd normal forms. It defines key concepts like functional dependencies, candidate keys, closure of attribute sets, minimal covers, and extraneous attributes. An example of a supplier-parts database is used to illustrate 2nd normal form. Functional dependencies indicate that city and status are not fully functionally dependent on the primary key, so the relation is not in 2nd normal form.
4th year paper presentation on the GPUVerify paper
http://www.doc.ic.ac.uk/~afd/papers/BettsCDQT_OOPSLA2012.html
http://multicore.doc.ic.ac.uk/tools/GPUVerify/
Sentient Arithmetic and Godel's Incompleteness TheoremsKannan Nambiar
For me, there is only one logic that we rational human beings are able to accept and appreciate, and that is the mathematical logic of ZF theory. But in the last century we found that ZF theory is not in a position to provide all that we want, and went in search of a new mode of thinking and got one which we called meta mathematics. My question is: if we can put the unambiguous logic of ZF theory on paper, why can't we do the same with meta mathematics. This paper is my feeble attempt in that direction.
4th year paper presentation on the GPUVerify paper
http://www.doc.ic.ac.uk/~afd/papers/BettsCDQT_OOPSLA2012.html
http://multicore.doc.ic.ac.uk/tools/GPUVerify/
Sentient Arithmetic and Godel's Incompleteness TheoremsKannan Nambiar
For me, there is only one logic that we rational human beings are able to accept and appreciate, and that is the mathematical logic of ZF theory. But in the last century we found that ZF theory is not in a position to provide all that we want, and went in search of a new mode of thinking and got one which we called meta mathematics. My question is: if we can put the unambiguous logic of ZF theory on paper, why can't we do the same with meta mathematics. This paper is my feeble attempt in that direction.
Functional dependencies in Database Management SystemKevin Jadiya
Slides attached here describes mainly Functional dependencies in database management system, how to find closure set of functional dependencies and in last how decomposition is done in any database tables
It is very helpful in learning all the basics concepts of DBMS starting from Introduction: An Overview of Database Management System to Data Modeling using the Entity-Relationship Model, PL/SQL, Transaction Processing Concept, and Concurrency Control Techniques plus important numerals in exam point of view can be learned.
Lecture notes on Functional dependencies, normal forms, first, second, third normal forms, BCNF, inclusion dependence, loss less join decompositions, normalization using FD, MVD, and JDs, alternative approaches to database design.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
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Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
2. SCHEMA FOR UNIVERSITY DATABASE
Let’s consider the relations
department <deptName, building, budget>
instructor <ID,name, deptName, salary>
S bi i d d Suppose we combine instructor and department
into inst_dept
2
3. Result is possible repetition of information
3
4. 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 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 budgetdept_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 deptThis 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)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 areconstruct the original employee relation and so, this is a
lossy decomposition 4
6. EXAMPLE OF LOSSLESS JOIN DECOMPOSITION
Lossless join decomposition
Decomposition of R = (A, B, C)
R = (A B) R = (B C)R1 = (A, B) R2 = (B, C)
A B A B CBC
1
2
1
2
r (r)
A
B
1
2
A
B
(r)r B,C(r)
A,B (r) B,C (r)
A B C
A
A,B(r)
1
2
A
B
6
7. 1ST NORMAL FORM
Domain is atomic if its elements are considered
to be indivisible units
Examples of non-atomic domains:
Set of names composite attributes 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 atomicf f
7
8. 1ST NORMAL FORM (CONTD)
Atomicity is actually a property of how they y p p y
elements of the domain are used.
Example: Strings would normally be considered
indivisibleindivisible
Suppose that students are given roll numbers which
are strings of the form CS0012 or EE1127
f f f 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.
8
9. GOAL- DEVISE A THEORY FOR THE
FOLLOWING
Decide whether a particular relation r is in “good”p g
form.
In the case that a relation r is not in “good” form,
d it i t t f l ti { }decompose it into a set of relations {r1, r2, ..., rn}
such that
each relation is in good formg
the decomposition is a lossless-join decomposition
Our theory is based on:
f i l d d i functional dependencies
multivalued dependencies
9
10. FUNCTIONAL DEPENDENCYFUNCTIONAL DEPENDENCY
Constraints on the set of legal relationsg
Require that the value for a certain set of
attributes determines uniquely the value for
th t f tt ib tanother set of attributes
A functional dependency is a generalization of
the notion of a keythe notion of a key
10
11. FUNCTIONAL DEPENDENCYFUNCTIONAL DEPENDENCY
Let R be a relation schema
R and R R and R
The functional dependency
h ld R if d l if f l l l iholds 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
ib Th iattributes . That is,
t1,t2 r (t1[] = t2 [] t1[ ] = t2 [ ] )
11
12. EXAMPLE
Consider r(A,B ) with the following instance of r.( ) g
1 4
A B
On this instance, A B does NOT hold, but B
1 4
1 5
3 7
On this instance, A B does NOT hold, but B
A does hold
12
13. FUNCTIONAL DEPENDENCYFUNCTIONAL DEPENDENCY
K is a superkey for relation schema R if and only if K R
K is a candidate key for R if and only ify y
K R, and
for no K, R
Functional dependencies allow us to express constraints 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 buildingdept_name building
and ID building
but would not expect the following to hold:
dept_name salary 13
14. USE OF FUNCTIONAL DEPENDENCYUSE OF FUNCTIONAL DEPENDENCY
We use functional dependencies to:
test relations to see if they are legal under a given setes e a o s o see ey a e ega e a g ve se
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.
N t A ifi i t f l ti h Note: A specific instance of a relation schema
may satisfy a functional dependency even if the
functional dependency does not hold on all legal
i tinstances.
For example, a specific instance of instructor may, by
chance, satisfy
IDname ID. 14
15. FUNCTIONAL DEPENDENCY (CONTD)
A functional dependency is trivial if it isp y
satisfied by all instances of a relation
Example:
ID name ID ID, name ID
name name
In general, is trivial if
15
16. CLOSURE OF A SET OF FUNCTIONAL
DEPENDENCY
Given a set F of functional dependencies, therep ,
are certain other functional dependencies that
are logically implied by F.
F l If A B d B C th 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+ i t f F F+ is a superset of F.
16
17. CLOSURE OF A SET OF FUNCTIONAL
DEPENDENCY
We can find F+, the closure of F, by repeatedly, y p y
applying
Armstrong’s Axioms:
if th ( fl i it ) if , then (reflexivity)
if , then (augmentation)
if , and , then (transitivity)
These rules are
sound (generate only functional dependencies that
actually hold) andactually hold), and
complete (generate all functional dependencies that
hold).
17
18. EXAMPLE
R = (A, B, C, G, H, I)
F = { A B
A CA C
CG H
CG I
B H}B H}
some members of F+
A H
by transitivity from A B and B 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 transitivityand then transitivity
18
19. 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 +
f h i f f ti l d d i f d f i F +for each pair of functional dependencies f1and 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 furtheruntil F does not change any further
19
20. CLOSURE OF FDS
Additional rules:
If holds and holds, then holds
(union)
If holds then holds and holds If holds, then holds and holds
(decomposition)
If holds and holds, then holds
( d t iti it )(pseudotransitivity)
The above rules can be inferred from Armstrong’s
axioms.
20
21. FUNCTIONAL DEPENDENCY EXAMPLE
Flight <flight no, c arr, c dept, pl type>g f g _ , _ , _ p , p _ yp
Seats_free <flight_no, date, seats_avl>
The following FDs hold
flight_no → c_arr
flight_no → c_dept
flight no → pl type flight_no → pl_type
flight_no, date → seats_avl
21
22. FUNCTIONAL DEPENDENCY EXAMPLE
Stud addr <name, address>_ ,
Stud_grade <name, subject, grade>
FDs that hold are
name → address
name, subject → grade
22
23. Which FDs hold here?
X Y Z W
x1 y1 z1 w11 y1 1 1
x1 y2 z1 w2
x2 y2 z2 w2
x2 y3 z2 w3
x3 y3 z2 w4
23
25. FULL FUNCTIONAL DEPENDENCY
When the functional dependency is ‘minimal’ inp y
size (i.e., containing non redundant terms)
FD X →A for which there is no proper subset Y of
X h th t Y A (A i id t b f llX such that Y →A (A is said to be fully
functionally dependent on X)
25
26. CLOSURE OF ATTRIBUTE SETS
The set of all attributes functionally determinedy
by α under a set F of FDs
It is denoted by α+
Let’s consider the following example
A → BC
AC → D AC → D
D → B
AB → D
So
A+={A,B,C,D}, B+={B}, …
26
27. COVER OF A SET OF FDS
Let f and g be two FDs on a relation scheme RLet f and g be two FDs on a relation scheme R.
Then f is a cover of g if f+=g+
This is also known as f is equivalent to g
f
A→BC
g
A→BC
B →C
A →B
AB →C
A→BC
B →C
AB →C
Here f+=g+
So g covers fSo g covers f
27
28. MINIMAL COVER OR CANONICAL COVER
A cover is said to be minimal if it has no
redundant terms
Denoted by Fc
Example:
Fc
F
A → BC
A → CD
D → B
A → BC
AC → D
D → B
AB → D
28
29. EXTRANEOUS ATTRIBUTE
An attribute of a FD is said to be extraneous if
we can remove it without changing the closure of
the set of FDs
F ll Formally,
Consider a set F of FDs and α→β in F
Attribute A is extraneous in α if A α and F logically Attribute A is extraneous in α if A α, and F logically
implies (F-{α → β}) U{(α –A) → β}
Attribute A is extraneous in β if A β, and F logically
implies (F {α → β}) U{α → (β A)}implies (F-{α → β}) U{α → (β - A)}
29
30. EXAMPLE
Suppose F:{AB →C and A →C}pp { }
Then B is extraneous in AB →C
Again F:{AB →CD and A →C}
Then C is extraneous in AB →CD
30
31. Included in the
definition of
NORMAL FORMS
First Normal Form (1NF)
definition of
relation
( )
Second Normal Form (2NF)
Third Normal Form (3NF)
Defined in terms
of FDs
Boyce-Codd Normal Form (BCNF)
Fourth Normal Form (4NF) Defined using
MVDs
Fifth Normal Form (5NF)
Also known as Project Join Normal Form (PJNF)
MVDs
Defined usingDefined using
join dependency
31
32. 2ND NORMAL FORM
2NF: A relation schema R is in 2NF if
it is 1NF and every non-key attribute is fully
functionally dependent on the primary key of R
key attribute: An attribute that is part of some key key attribute: An attribute that is part of some key
non-key attribute: An attribute that is not part of any
key
32
33. EXAMPLE
Let’s consider the following supplier-partsg pp p
database system
first <sno, status, city, pno, qty>
Here a possible primary key is (sno, pno)
FDs for relation first
citysno
status
qty
pno
33
34. Instance of relation first
sno status city pno qty
s1 20 mumbai p1 300
s1 20 mumbai p2 200
s1 20 mumbai p3 400s1 20 mumbai p3 400
s1 20 mumbai p4 200
s1 20 mumbai p5 100
1 20 b i 6 700s1 20 mumbai p6 700
s2 10 chennai p1 200
s2 10 chennai p2 120
s3 10 chennai p2 340
s4 20 mumbai p2 230
s4 20 mumbai p4 432s4 20 mumbai p4 432
s4 20 mumbai p5 120
34
35. ANOMALIES
Insert:
Insertion not possible until a supplier supplied some
items
Ex s5 located in Delhi in cannot be inserted Ex. s5 located in Delhi in cannot be inserted
Delete:
May loose some additional informationy
Ex. if s3, p2 is deleted then we loose the information
that s3 is located in Chennai
Update: Update:
Same city value appears in many places
Ex. if s1 moves from Mumbai to Ahmedabad then
update is to be done in many places 35
36. DECOMPOSITION
The relation first must be decomposed in such af p
way so that the decomposed relations satisfy 2NF
second <sno, status, city> and
sp <sno, pno, qty>
FDs for the above relations
sno
city
sno
qty
status
pno
qty
36
37. EXAMPLE OF 2NF RELATIONS
second sno pno qty
sp
sno status city
s1 20 mumbai
2 10 h i
s1 p1 300
s1 p2 200
s1 p3 400
s2 10 chennai
s3 10 chennai
s4 20 mumbai
s1 p3 400
s1 p4 200
s1 p5 100
1 6 700
s5 30 delhi
s1 p6 700
s2 p1 200
s2 p2 120
s3 p2 340
s4 p2 230
s4 p4 432
37
s4 p4 432
s4 p5 120
38. Thus in r(A,B,C,D) if (A,B) is a primary key and( , , , ) ( , ) p y y
A →D holds
Then by 2NF r can be replaced by r1 and r2 as
f llfollows
r1(A,D) primary key {A}
r2(A,B,C) primary key {A,B} and foreign key A( , , ) p y y { , } g y
references r1(A)
38
39. 3NF
A relation is in 3NF iff
it is in 2NF and
every non-key attribute is non-transitively dependent
on the primary keyon the primary key
No transitive dependency means no mutual
dependency
Now consider relation second
city
sno
status
39
40. ANOMALIES
Insert:
A particular city has a particular status
Ex: any supplier in city Kanpur has 10 status
C t b i t d til th i t ll li Cannot be inserted until there is actually a supplier
located in that city
Delete:
If we delete S5 then we lose information that Delhi
has status 30
Update: Update:
The status of a given city appears in many places
So updating the status value may be problematic
40
41. Now if we decompose the relation second into twop
relations such that they satisfy 3NF
sc <sno, city>
cs <city, status>
The FDs of the above relations are
sno city statuscity
41
42. Thus if r(A,B,C) and A is a primary key and B →( , , ) p y y
C holds
Then by 3NF r can be replaced by
r1 (B,C) and B is a primary key
r2(A,B) and A is a primary key and foreign key B
references r1(A)
42
43. PROPERTIES OF DECOMPOSITION
Decomposition1: Relation second is decomposedp p
into
sc <sno, city>
< it t t > cs <city,status>
Decomposition2: Relation second is decomposed
into
sc <sno,city>
ss <sno,status>
Whi h f h b d i i i Which of the above decomposition is
lossless and dependency preserving?
43
44. DESIRABLE PROPERTIES OF
DECOMPOSITION
Lossless joinj
When decomposing a relation into number of smaller
ones then it is crucial that the decomposition be
losslesslossless
Dependency preservation
The system must not create relation that does not
satisfy all the given functional dependencies
44
45. LOSSLESS JOIN
Let R be a relation schema and F be a set of
functional dependencies
Let R1 and R2 form a decomposition of R
The decomposition will be lossless if atleast one
of the following functional dependencies is in F+
R1∩R2→R1R1∩R2→R1
R1∩R2→R2
In other words, R1 ∩ R2 forms a super key of
ith R Reither R1 or R2
45
46. DEPENDENCY PRESERVATION
Create legal relations preserving theg p g
dependencies
Let F be a set of functional dependencies on a
h R d l t R R R bschema R and let R1, R2, …, Rn be a
decomposition of R
The restriction of F to Ri is the set of all The restriction of F to Ri is the set of all
functional dependencies in F+ that include only
attributes of Ri
Th f i i F F F i h f The set of restrictions F1, F2, …, Fn is the set of
dependencies that can be checked efficiently
Now we check whether testing only the Now we check whether testing only the
restrictions is sufficient?
46
47. Let F′=F1 U F2 U … U Fn1 2 n
F′ is the set of all functional dependencies on
schema R but in general F′≠F
But if F′+=F+ is satisfied then we say that it is a
dependency preserving decomposition
47
48. DEPENDENCY PRESERVING: EXAMPLE
Example:p
Suppose F={A →B, B →C} and the original relation is
r<A,B,C>
And the decompositions are r <A B> and r <A C> And the decompositions are r1<A,B> and r2<A,C>
Is it dependency preserving?
48
49. TESTING FOR DEPENDENCY PRESERVATION
Compute F+
For each schema Ri in D do
Begin
Fi:=restrictions of F+ to Ri;
D is an input set
and
D={R1,R2,…,Rn} of
decomposed Fi: restrictions of F to Ri;
End
F′:=
F h t i ti F d
decomposed
relation schemas
For each restriction Fi do
Begin
F′=F′ U Fi
End
Compute F′+
If (F′+ =F+) then return true; If (F F ) then return true;
else return false;
49
50. BOYCE/CODD NORMAL FORM (BCNF)
Can handle relation with
two or more candidate keys
composite candidate keys
l d k overlapped keys
The above conditions might not occur very often
For a relation where the above does not hold For a relation where the above does not hold,
3NF and BCNF are equivalent
BCNF is strictly stronger than 3NF definition
50
51. DETERMINANT
In a FD, the left side is termed as determinant,
whereas the right side is termed as dependent
A relation is in BCNF iff every determinant
i did t kis a candidate key
Assumption
The determinants are not too big The determinants are not too big
All FDs are nontrivial
51
52. Let’s check whether the following relations are in
BCNF
Relation first <sno, status, city, pno, qty>
Three determinants – {sno}, {city}, {sno, pno}{ } { y} { p }
Only the last one was candidate key
So not in BCNF
Relation second <sno, status, city>, , y
Two determinants- {sno}, {city}
Only sno is candidate key
So not in BCNF
Relation sp <sno, pno, qty>
One determinant-{sno,pno}
That is also candidate keyThat is also candidate key
It is in BCNF
52
53. EXAMPLE 1
Now let us consider relation suppliers <sno,pp ,
sname, status, city>
Here both sno and sname are candidate keys
So FDs of this relation
sno status
itcitysname
So here suppliers is in BCNF 53
54. EXAMPLE 2
Now consider the relation ssp <sno, sname, pno,p , , p ,
qty>
Here the candidate keys are {sno, pno} and
{ }{sname, pno}
So here the candidate keys overlap
But is it BCNF? But is it BCNF?
No, as the relation contains two other determinants
{sno} and {sname}
And these are not candidate keys
54
55. EXAMPLE 2 (CONTD)
So a possible decomposition will bep p
ss <sno, sname>
sp<sno,pno,qty>
A d h lid d i i And another valid decomposition
ss <sno, sname>
sp<sname,pno,qty>sp sname,pno,qty
55
56. EXAMPLE 3
Let’s consider a relation sjt <s,j,t>j ,j,
Here attributes s: student, j: subject and t:
teacher
The meaning of each tuple
“student s is taught subject j by teacher t”
Now the following constraints apply Now the following constraints apply
1. For each subject, each student of the subject
is taught by only one teacher
2. Each teacher teaches only one subject
3. However, each subject is taught by several
teachers
56
57. FDs
{s,j}→t
t→j
j→t does not hold
jt
What is a possible instance of relation sjt?
s j t
Sarala Maths Prof Raj
sjt
Sarala Maths Prof. Raj
Sarala Physics Prof. Atul
Uma Maths Prof. Raj
57Uma Physics Prof. Pathak
58. So what are the candidate keys?y
{s,j} and {s,t}
But the relation is not in BCNF as the
d i i did kdeterminant t is not a candidate key
So how do we decompose sjt?
Relation sjt can be decomposed into Relation sjt can be decomposed into
st<s,t>
tj<t,j>
58
59. st tj
s t
Sarala Prof. Raj
st
t j
Prof. Raj Maths
tj
Sarala Prof. Atul
Uma Prof. Raj
Uma Prof Pathak
Prof. Atul Physics
Prof. Pathak Physics
Uma Prof. Pathak
There is a problem with this decomposition.
The decomposition is not independent.
Because of FD {s,j}→t
59
60. So the main problem with the last decompositionp p
is that the relations cannot be independently
updated
Wh l ti t b d d i t When a relation cannot be decomposed into
independent components then it is said to be
atomic
So sometime there may be conflicts between
BCNF components
D i i i d d Decomposing into independent components
Thus it may not always possible to satisfy both of
them at the same time
60