6 relational schema_design

Prof P Sreenivasa Kumar
Department of CS&E, IITM
1
Database Design and Normal Forms
Database Design
coming up with a ‘good’ schema is very important
How do we characterize the “goodness” of a schema ?
If two or more alternative schemas are available
how do we compare them ?
What are the problems with “bad” schema designs ?
Normal Forms:
Each normal form specifies certain conditions
If the conditions are satisfied by the schema
certain kind of problems are avoided
Details follow….

2
An Example
student relation with attributes: studName, rollNo, sex, studDept
department relation with attributes: deptName, officePhone, hod
Several students belong to a department.
studDept gives the name of the student’s department.
Correct schema:
What are the problems that arise ?
studName
studName
rollNo
rollNo
sex
sex
studDept deptName
deptName
officePhone
officePhone
HOD
HOD
Incorrect schema:
Student Department
Student Dept

3
Problems with bad schema
Redundant storage of data:
Office Phone & HOD info - stored redundantly
once with each student that belongs to the department
wastage of disk space
A program that updates Office Phone of a department
must change it at several places
• more running time
• error - prone
Transactions running on a database
must take as short time as possible to increase transaction
throughput

4
Update Anomalies
Another kind of problem with bad schema
Insertion anomaly:
No way of inserting info about a new department unless
we also enter details of a (dummy) student in the department
Deletion anomaly:
If all students of a certain department leave
and we delete their tuples,
information about the department itself is lost
Update Anomaly:
Updating officePhone of a department
• value in several tuples needs to be changed
• if a tuple is missed - inconsistency in data

5
Normal Forms
First Normal Form (1NF) - included in the definition of a relation
Second Normal Form (2NF)
defined in terms of
Third Normal Form (3NF) functional dependencies
Boyce-Codd Normal Form (BCNF)
Fourth Normal Form (4NF) - defined using multivalued
dependencies
Fifth Normal Form (5NF) or Project Join Normal Form (PJNF)
defined using join dependencies

6
Functional Dependencies
A functional dependency (FD) X → Y
(read as X determines Y) (X ⊆ R, Y ⊆ R)
is said to hold on a schema R if
in any instance r on R,
if two tuples t1, t2 (t1 ≠ t2, t1 ∈ r, t2 ∈ r)
agree on X i.e. t1 [X] = t2 [X]
then they also agree on Y i.e. t1 [Y] = t2 [Y]
Note: If K ⊂ R is a key for R then for any A ∈ R,
K → A
holds because the above if …..then condition is
vacuously true

7
Functional Dependencies – Examples
Consider the schema:
Student ( studName, rollNo, sex, dept, hostelName, roomNo)
Since rollNo is a key, rollNo → {studName, sex, dept,
hostelName, roomNo}
Suppose that each student is given a hostel room exclusively, then
hostelName, roomNo → rollNo
Suppose boys and girls are accommodated in separate hostels, then
hostelName → sex
FDs are additional constraints that can be specified by designers

8
Trivial / Non-Trivial FDs
An FD X → Y where Y ⊆ X
- called a trivial FD, it always holds good
An FD X → Y where Y ⊈ X
- non-trivial FD
An FD X → Y where X ∩ Y = f
- completely non-trivial FD

9
Deriving new FDs
Given that a set of FDs F holds on R
we can infer that a certain new FD must also hold on R
For instance,
given that X → Y, Y → Z hold on R
we can infer that X → Z must also hold
How to systematically obtain all such new FDs ?
Unless all FDs are known, a relation schema is not fully specified

10
Entailment relation
We say that a set of FDs F ⊨{ X → Y}
(read as F entails X → Y or
F logically implies X → Y)
if in every instance r of R on which FDs F hold,
FD X → Y also holds.
Armstrong came up with several inference rules
for deriving new FDs from a given set of FDs
We define F+
= {X → Y | F ⊨X → Y}
F
+
: Closure of F

11
Armstrong’s Inference Rules (1/2)
1. Reflexive rule
F ⊨ {X → Y | Y ⊆ X} for any X. Trivial FDs
2. Augmentation rule
{X → Y} ⊨ {XZ → YZ}, Z ⊆ R. Here XZ denotes X ⋃ Z
3. Transitive rule
{X → Y, Y → Z} ⊨ {X → Z}
4. Decomposition or Projective rule
{X → YZ} ⊨ {X → Y}
5. Union or Additive rule
{X → Y, X → Z} ⊨ {X → YZ}
6. Pseudo transitive rule
{X → Y, WY → Z} ⊨ {WX → Z}

12
Rules 4, 5, 6 are not really necessary.
For instance, Rule 5: {X → Y, X → Z} ⊨ {X → YZ} can be
proved using 1, 2, 3 alone
1) X → Y
2) X → Z
3) X → XY Augmentation rule on 1
4) XY → ZY Augmentation rule on 2
5) X → ZY Transitive rule on 3, 4.
Similarly, 4, 6 can be shown to be unnecessary.
But it is useful to have 4, 5, 6 as short-cut rules
given
Armstrong's Inference Rules (2/2)

13
Sound and Complete Inference Rules
Armstrong showed that
Rules (1), (2) and (3) are sound and complete.
These are called Armstrong’s Axioms (AA)
Soundness:
Every new FD X → Y derived from a given set of FDs F
using Armstrong's Axioms is such that F ⊨{X → Y}
Completeness:
Any FD X → Y logically implied by F (i.e. F ⊨ {X → Y})
can be derived from F using Armstrong’s Axioms

14
Proving Soundness
Suppose X → Y is derived from F using AA in some n steps.
If each step is correct then overall deduction would be correct.
Single step: Apply Rule (1) or (2) or (3)
Rule (1) – obviously results in correct FDs
Rule (2) – {X → Y}⊨ {XZ → YZ}, Z ⊆ R
Suppose t1, t2 ∈ r agree on XZ
⇒ t1, t2 agree on X
⇒ t1, t2 agree on Y (since X → Y holds on r)
⇒ t1, t2 agree as YZ
Hence Rule (2) gives rise to correct FDs
Rule (3) – {X → Y, Y → Z} ⊨ X → Z
Suppose t1, t2 ∈ r agree on X
⇒ t1, t2 agree on Y (since X → Y holds)
⇒ t1, t2 agree on Z (since Y → Z holds)

15
Proving Completeness of Armstrong’s Axioms (1/4)
Define X
+
F (closure of X wrt F)
= {A | X → A can be derived from F using AA}, A ∈ R
Claim1:
X → Y can be derived from F using AA iff Y ⊆ X
+
(If) Let Y = {A1, A2,…, An}. Y ⊆ X
+
⇒ X → Ai can be derived from F using AA (1 ≤ i ≤ n)
By union rule, it follows that X → Y can be derived from F.
(Only If) X → Y can be derived from F using AA
By projective rule X → Ai (1 ≤ i ≤ n)
Thus by definition of X+
, Ai ∈ X
+
⇒ Y ⊆ X
+

16
Completeness of Armstrong’s Axioms (2/4)
Completeness:
(F ⊨ {X → Y}) ⇒ X → Y follows from F using AA
We will prove the contrapositive:
X →Y can’t be derived from F using AA
⇒ F ⊭ {X → Y}
⇒ ∃ a relation instance r on R st all the FDs of
F hold on r but X → Y doesn’t hold.
Consider the relation instance r with just two tuples:
X
+
attributes Other attributes
r: 1 1 1 …1 1 1 1 …1
1 1 1 …1 0 0 0 …0

17
Claim 2: All FDs of F are satisfied by r
Suppose not. Let W → Z in F be an FD not satisfied by r
Then W ⊆ X+
and Z ⊈ X
+
Let A ∈ Z – X
+
Now, X → W follows from F using AA as W ⊆ X
+
(claim 1)
X → Z follows from F using AA by transitive rule
Z → A follows from F using AA by reflexive rule as A ∈ Z
X → A follows from F using AA by transitive rule
By definition of closures, A must belong to X
+
- a contradiction. r: 1 1 1 …1 1 1 1 …1
Hence the claim. 1 1 1 …1 0 0 0 …0
X+
R - X+
Completeness Proof (3/4)

18
Completeness Proof (4/4)
Claim 3: X → Y is not satisfied by r
Suppose not
Because of the structure of r, Y ⊆ X+
⇒ X → Y can be derived from F using AA
contradicting the assumption about X → Y
Hence the claim
Thus, whenever X → Y doesn’t follow from F using AA,
F doesn’t logically imply X → Y
Armstrong’s Axioms are complete.

19
Consequence of Completeness of AA
X
+
= {A | X → A follows from F using AA}
= {A | F ⊨ X → A}
Similarly
F
+
= {X → Y | F ⊨ X → Y}
= {X → Y | X → Y follows from F using AA}

20
Computing closures
The size of F
+
can sometimes be exponential in the size of F.
For instance, F = {A → B1, A → B2,….., A → Bn}
F
+
= {A → X} where X ⊆ {B1, B2,…,Bn}.
Thus |F
+
| = 2
n
Computing F
+
: computationally expensive
Fortunately, checking if X → Y ∈ F+
can be done by checking if Y ⊆ X
+
F
Computing attribute closure (X
+
F) is easier

21
Computing X+
F
We compute a sequence of sets X0, X1,… as follows:
X0:= X; // X is the given set of attributes
Xi+1:= Xi ∪ {A | there is a FD Y → Z in F
and A ∈ Z and Y ⊆ Xi}
Since X0 ⊆ X1 ⊆ X2 ⊆ ... ⊆ Xi ⊆ Xi+1 ⊆ ...⊆ R
and R is finite,
There is an integer i st Xi = Xi+1 = Xi+2 =…
and X+
F is equal to Xi.

22
Normal Forms – 2NF
Full functional dependency:
An FD X → A for which there is no proper subset Y of X
such that Y → A
(A is said to be fully functionally dependent on X)
2NF: A relation schema R is in 2NF if
every non-prime attribute is fully functionally dependent on any
key of R
prime attribute: A attribute that is part of some key
non-prime attribute: An attribute that is not part of any key

23
Example
1) Book (authorName, title, authorAffiliation, ISBN, publisher,
pubYear )
Keys: (authorName, title), ISBN
Not in 2NF as authorName Æ authorAffiliation
(authorAffiliation is not fully functionally dependent on the
first key)
2) Student (rollNo, name, dept, sex, hostelName, roomNo,
admitYear)
Keys: rollNo, (hostelName, roomNo)
Not in 2NF as hostelName → sex
student (rollNo, name, dept, hostelName, roomNo, admitYear)
hostelDetail (hostelName, sex)
- There are both in 2NF

24
Transitive Dependencies
Transitive dependency:
An FD X → Y in a relation schema R for which there is a set of
attributes Z ⊆ R such that
X → Z and Z → Y and Z is not a subset of any key of R
Ex: student (rollNo, name, dept, hostelName, roomNo, headDept)
Keys: rollNo, (hostelName, roomNo)
rollNo → dept; dept → headDept hold
So, rollNo → headDept a transitive dependency
Head of the dept of dept D is stored redundantly in every tuple
where D appears.
Relation is in 2NF but redundancy still exists.

25
Normal Forms – 3NF
Relation schema R is in 3NF if it is in 2NF and no non-prime
attribute of R is transitively dependent on any key of R
student (rollNo, name, dept, hostelname, roomNo, headDept)
is not in 3NF
Decompose: student (rollNo, name, dept, hostelName, roomNo)
deptInfo (dept, headDept)
both in 3NF
Redundancy in data storage - removed

26
Another definition of 3NF
Relation schema R is in 3NF if for any nontrivial FD X → A
either (i) X is a superkey or (ii) A is prime.
Suppose some R violates the above definition
⇒ There is an FD X → A for which both (i) and (ii) are false
⇒ X is not a superkey and A is non-prime attribute
Two cases arise:
1) X is contained in a key – A is not fully functionally dependent
on this key
- violation of 2NF condition
2) X is not contained in a key
K → X, X → A is a case of transitive dependency
(K – any key of R)

27
Motivating example for BCNF
gradeInfo (rollNo, studName, course, grade)
Suppose the following FDs hold:
1) rollNo, course → grade Keys:
2) studName, course → grade (rollNo, course)
3) rollNo → studName (studName, course)
4) studName → rollNo
For 1,2 lhs is a key. For 3,4 rhs is prime
So gradeInfo is in 3NF
But studName is stored redundantly along with every course
being done by the student

28
Boyce - Codd Normal Form (BCNF)
Relation schema R is in BCNF if for every nontrivial
FD X → A, X is a superkey of R.
In gradeInfo, FDs 3, 4 are nontrivial but lhs is not a superkey
So, gradeInfo is not in BCNF
Decompose:
gradeInfo (rollNo, course, grade)
studInfo (rollNo, studName)
Redundancy allowed by 3NF is disallowed by BCNF
BCNF is stricter than 3NF
3NF is stricter than 2NF

29
Decomposition of a relation schema
If R doesn’t satisfy a particular normal form,
we decompose R into smaller schemas
What’s a decomposition?
R = (A1, A2,…, An)
D = (R1, R2,…, Rk) st Ri ⊆ R and R = R1 ∪ R2 ∪ … ∪ Rk
(Ri’s need not be disjoint)
Replacing R by R1, R2,…, Rk – process of decomposing R
Ex: gradeInfo (rollNo, studName, course, grade)
R1: gradeInfo (rollNo, course, grade)
R2: studInfo (rollNo, studName)

30
Desirable Properties of Decompositions
Not all decomposition of a schema are useful
We require two properties to be satisfied
(i) Lossless join property
- the information in an instance r of R must be preserved in the
instances r1, r2,…,rk where ri = pRi
(r)
(ii) Dependency preserving property
- if a set F of dependencies hold on R it should be possible to
enforce F by enforcing appropriate dependencies on each ri

31
Lossless join property
F – set of FDs that hold on R
R – decomposed into R1, R2,…,Rk
Decomposition is lossless wrt F if
for every relation instance r on R satisfying F,
r = pR1
(r) * pR2
(r) *…* pRk
(r)
R = (A, B, C); R1 = (A, B); R2 = (B, C)
r: A B C r1: A B r2: B C r1 * r2: A B C
a1 b1 c1 a1 b1 b1 c1 a1 b1 c1
a2 b2 c2 a2 b2 b2 c2 a1 b1 c3
a3 b1 c3 a3 b1 b1 c3 a2 b2 c2
a3 b1 c1
a3 b1 c3
Spurious tuples
Original info
is distorted
Lossy join
Lossless joins
are also called
non-additive joins

32
Dependency Preserving Decompositions
Decomposition D = (R1, R2,…,Rk) of schema R preserves a set
of dependencies F if
(pR1
(F) ∪ pR2
(F) ∪… ∪ pRk
(F))
+
= F
+
Here, pRi
(F) = { (X Æ Y) ∈ F
+
| X ⊆ Ri, Y ⊆ Ri}
(called projection of F onto Ri)
Informally, any FD that logically follows from F must also
logically follow from the union of projections of F onto Ri’s
Then, D is called dependency preserving.

33
An example
Schema R = (A, B, C)
FDs F = {A → B, B → C, C → A}
Decomposition D = (R1 = {A, B}, R2 = {B, C})
pR1
(F) = {A → B, B → A}
pR2
(F) = {B → C, C → B}
(pR1
(F) ∪ pR2
(F))+
= {A → B, B → A,
B → C, C → B,
A → C, C → A} = F+
Hence Dependency preserving

34
Testing for lossless decomposition property(1/6)
R – given schema with attributes A1,A2, …, An
F – given set of FDs
D – {R1,R2, …, Rm} given decomposition of R
Is D a lossless decomposition?
Create an m × n matrix S with columns labeled as A1,A2, …, An
and rows labeled as R1,R2, …, Rm
Initialize the matrix as follows:
set S(i,j) as symbol bij for all i,j.
if Aj is in the scheme Ri, then set S(i,j) as symbol aj , for all i,j

35
After S is initialized, we carry out the following process on it:
repeat
for each functional dependency U → V in F do
for all rows in S which agree on U-attributes do
make the symbols in each V- attribute column
the same in all the rows as follows:
if any of the rows has an “a” symbol for the column
set the other rows to the same “a” symbol in the column
else // if no “a” symbol exists in any of the rows
choose one of the “b” symbols that appears
in one of the rows for the V-attribute and
set the other rows to that “b” symbol in the column
until no changes to S
At the end, if there exists a row with all “a” symbols then D is
lossless otherwise D is a lossy decomposition

36
R = (rollNo, name, advisor, advisorDept, course, grade)
FD’s = { rollNo → name; rollNo → advisor; advisor → advisorDept
rollNo, course → grade}
D : { R1 = (rollNo, name, advisor), R2 = (advisor, advisorDept),
R3 = (rollNo, course, grade) }
Matrix S : (Initial values)
rollNo name advisor advisor
Dept
course grade
R1 a1 a2 a3 b14 b15 b16
R2 b21 b22 a3 a4 b25 b26
R3 a1 b32 b33 b34 a5 a6

37
Matrix S : (After enforcing rollNo → name & rollNo → advisor)
Dept
course grade
R1 a1 a2 a3 b14 b15 b16
R2 b21 b22 a3 a4 b25 b26
R3 a1 b32a2 b33a3 b34 a5 a6

38
Matrix S : (After enforcing advisor → advisorDept )
No more changes. Third row with all a symbols. So a lossless join.
Dept
course grade
R1 a1 a2 a3 b14a4 b15 b16
R2 b21 b22 a3 a4 b25 b26
R3 a1 b32a2 b33a3 b34a4 a5 a6

39
R – given schema. F – given set of FDs
The decomposition of R into R1, R2 is lossless wrt F if and only if
either R1 ∩ R2 → (R1 – R2) belongs to F
+
or
R1 ∩ R2 → (R2 – R1) belongs to F
+
Eg. gradeInfo (rollNo, studName, course, grade)
with FDs = {rollNo, course → grade; studName, course → grade;
rollNo → studName; studName → rollNo}
decomposed into
grades (rollNo, course, grade) and studInfo (rollNo, studName)
is lossless because
rollNo → studName

40
A property of lossless joins
D1: (R1, R2,…, RK) lossless decomposition of R wrt F
D2: (Ri1, Ri2,…, Rip) lossless decomposition of Ri wrt Fi = pRi
(F)
Then
D = (R1, R2, … , Ri-1, Ri1, Ri2, …, Rip, Ri+1,…, Rk) is a
lossless decomposition of R wrt F
This property is useful in the algorithm for BCNF decomposition

41
Algorithm for BCNF decomposition
R – given schema. F – given set of FDs
D = {R} // initial decomposition
while there is a relation schema Ri in D that is not in BCNF do
{ let X → A be the FD in Ri violating BCNF;
Replace Ri by Ri1 = Ri – {A} and Ri2 = X ∪ {A} in D;
}
Decomposition of Ri is lossless as
Ri1 ∩ Ri2 = X, Ri2 – Ri1 = A and X → A
Result: a lossless decomposition of R into BCNF relations

42
Dependencies may not be preserved (1/2)
Consider the schema: townInfo (stateName, townName, distName)
with the FDs F: ST → D (town names are unique within a state)
D → S
Keys: ST, DT. – all attributes are prime
– relation in 3NF
Relation is not in BCNF as D → S and D is not a key
Decomposition given by algorithm: R1: TD R2: DS
Not dependency preserving as pR1
(F) = trivial dependencies
pR2
(F) = {D → S}
Union of these doesn’t imply ST → D
ST → D can’t be enforced unless we perform a join.
S T D

43
Dependencies may not be preserved (2/2)
Consider the schema: R (A, B, C)
with the FDs F: AB → C and C → B
Keys: AB, AC – relation in 3NF (all attributes are prime)
– Relation is not in BCNF as C → B and C is not a key
Decomposition given by algorithm: R1: CB R2: AC
Not dependency preserving as pR1
(F) = trivial dependencies
pR2
(F) = {C → B}
Union of these doesn’t imply AB → C
All possible decompositions: {AB, BC}, {BA, AC}, {AC, CB}
Only the last one is lossless!
Lossless and dependency-preserving decomposition doesn't exist.

44
Equivalent Dependency Sets
F, G – two sets of FDs on schema R
F is said to cover G if G ⊆ F+
(equivalently G+
⊆ F+
)
F is equivalent to G if F+
= G+
(or, F covers G and G covers F)
Note: To check if F covers G,
it’s enough to show that for each FD X → Y in G, Y ⊆ X
+
F

45
Canonical covers or Minimal covers
It is of interest to reduce a set of FDs F into a “standard” form
F′ such that F′ is equivalent to F.
We define that a set of FDs F is in ‘minimal form’ if
(i) the rhs of any FD of F is a single attribute
(ii) there are no redundant FDs in F
that is, there is no FD X → A in F
s.t (F – {X → A}) is equivalent to F
(iii) there are no redundant attributes on the lhs of any FD in F
that is, there is no FD X → A in F s.t there is Z ⊂ X for which
F – {X → A} ∪ {Z → A} is equivalent to F
Minimal Covers
useful in obtaining a lossless, dependency-preserving
decomposition of a scheme R into 3NF relation schemas

46
Algorithm for computing a minimal cover
R – given Schema or set of attributes; F – given set of fd’s on R
Step 1: G := F
Step 2: Replace every fd of the form X → A1A2A3…Ak in G
by X → A1; X → A2; X → A3; … ; X → Ak
Step 3: For each fd X → A in G do
for each B in X do
if A ∈ (X – B)+ wrt G then
replace X → A by (X – B) → A
Step 4: For each fd X → A in G do
if (G – { X → A})+ = G+ then
replace G by G – { X → A}

47
3NF decomposition algorithm
R – given Schema; F – given set of fd’s on R in minimal form
Use BCNF algorithm to get a lossless decomposition D = (R1, R2,…,Rk)
Note: each Ri is already in 3NF (it is in BCNF in fact!)
Algorithm: Let G be the set of fd’s not preserved in D
For each fd Z → A that is in G
Add relation scheme S = (B1,B2, …, Bs,A) to D. // Z = {B1,B2, …, Bs}
As Z → A is in F which is a minimal cover,
there is no proper subset X of Z s.t X → A. So Z is a key for S!
Any other fd X → C on S is such that C is in {B1,B2, …, Bs}.
Such fd’s do not violate 3NF because each Bj’s is prime a attribute!
Thus any scheme S added to D as above is in 3NF.
D continues to be lossless even when we add new schemas to it!

48
Multi-valued Dependencies (MVDs)
studCourseEmail(rollNo,courseNo,emailAddr)
a student enrolls for several courses and has several email addresses
rollNo →→ courseNo ( read as rollNo multi-determines courseNo )
If (CS05B007, CS370, shyam@gmail.com)
(CS05B007, CS376, shyam@yahoo.com) appear in the data then
(CS05B007, CS376, shyam@gmail.com)
(CS05B007, CS370, shyam@yahoo.com)
should also appear for, otherwise, it implies that having gmail
address has something to with doing course CS370 !!
By symmetry, rollNo →→ emailAddr

49
More about MVDs
Consider studCourseGrade(rollNo,courseNo,grade)
Note that rollNo →→ courseNo does not hold here even though
courseNo is a multi-valued attribute of student
If (CS05B007, CS370, A)
(CS05B007, CS376, B) appear in the data then
(CS05B007, CS376, A)
(CS05B007, CS370, B) will not appear !!
Attribute ‘grade’ depends on (rollNo,courseNo)
MVD’s arise when two unrelated multi-valued attributes of an
entity are sought to be represented together.

50
More about MVDs
Consider
studCourseAdvisor(rollNo,courseNo,advisor)
Note that rollNo →→ courseNo holds here
If (CS05B007, CS370, Dr Ravi)
(CS05B007, CS376, Dr Ravi)
appear in the data then swapping courseNo values
gives rise to existing tuples only.
But, since rollNo → advisor and (rollNo, courseNo) is the key,
this gets caught in checking for 2NF itself.

51
Alternative definition of MVDs
Consider R(X,Y,Z)
Suppose that X →→ Y and by symmetry X →→ Z
Then, decomposition D = (XY, XZ) should be lossless
That is, for any instance r on R, r = π XY(r) * π XZ(r)

52
MVDs and 4NF
An MVD X →→ Y on scheme R is called trivial if either
Y ⊆ X or R = X ∪ Y. Otherwise, it is called nontrivial.
4NF: A relation R is in 4NF if it is in BCNF and for every
nontrivial MVD X →→ A, X must be a superkey of R.
studCourseEmail(rollNo,courseNo,emailAddr)
is not in 4NF as
rollNo →→ courseNo and
rollNo →→ emailAddr
are both nontrivial and rollNo is not a superkey for the
relation

6 relational schema_design

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