1. Relational Database Schema Designer Using Bernstein’s Algorithm
Project Team: P10
adinda | daryl | xiaojie | woan ni

2. Introduction
Non- 3NF
3NF
Bernstein’s Algorithm

3. Key Features FD VALIDATION 3NF COMPLIANCE GUI
User graphic interface (GUI) is provided for user, which will allow easy adding, editing and deletion of function dependencies for user.
The software also checks if the user-provided functional dependencies are valid before proceeding.
Base on the user-provided functional dependencies, the software will derive the 3NF database base on Bernstein’s algorithm and result will be displayed on our software user interface.
Our software provides the results of intermediate steps of the Bernstein’s algorithm till the final result.
Our software provides the closure for each of the functional dependencies based on the user-provided functional dependencies.
As Bernstein’s algorithm is not perfect, our software provides the solution for 2 Bernstein’s shortcomings.

4. Key Features MANAGE SHORTCOMINGS STEP-BY-STEP CLOSURE
User graphic interface (GUI) is provided for user, which will allow easy adding, editing and deletion of function dependencies for user.
The software also checks if the user-provided functional dependencies are valid before proceeding.
Base on the user-provided functional dependencies, the software will derive the 3NF database base on Bernstein’s algorithm and result will be displayed on our software user interface.
Our software provides the results of intermediate steps of the Bernstein’s algorithm till the final result.
Our software provides the closure for each of the functional dependencies based on the user-provided functional dependencies.
As Bernstein’s algorithm is not perfect, our software provides the solution for 2 Bernstein’s shortcomings.

5. Software Demonstration

6. Code Implementation PRE-STEP FUNCTION DEPENDENCY FUNCTION DEPENDENCY
FUNCTIONAL
DEPENDENCY
FUNCTIONAL
DEPENDENCY
3NF RELATION
FUNCTIONAL
DEPENDENCY :
STEP 6
INPUT
BERNSTEIN’S
OUTPUT

7. Pre-step Merged dependencies FUNCTION Validated dependencies
same LHS
Derive closures
FUNCTION
DEPENDENCY
Validated dependencies
FUNCTION
DEPENDENCY
FUNCTIONAL
DEPENDENCY
Functional Dependency
FUNCTION
DEPENDENCY
FUNCTION
DEPENDENCY
FUNCTION
DEPENDENCY
FUNCTIONAL
DEPENDENCY
Functional Dependency
Closures
FUNCTION
DEPENDENCY
FUNCTION
DEPENDENCY
FUNCTIONAL
DEPENDENCY
Closure

8. STEP 1 LHS1 is subset of LHS2 LHS1 = A no RHS1 = E STEP 1 A → E yes
Remove extraneous attributes
A → E
yes
A subset of RHS1, x, is subset of LHS2
LHS2 = AE
RHS2 = B no
AE → B
yes
LHS2 = LHS2 - x

9. Remove extraneous attributes
STEP 1
A is subset of AE
LHS1 = A
RHS1 = E no
STEP 1
Remove extraneous attributes
A → E
yes
A subset of
E, E, is subset of AE
LHS2 = AE
RHS2 = B no
AE → B
yes
LHS2 = AE - E = A

10. Remove extraneous attributes
STEP 1
LHS1 = A
RHS1 = E
STEP 1
Remove extraneous attributes
A → E
A → E
LHS2 = AE
RHS2 = B
AE → B
A → B

11. STEP 1 LHS1 = A no RHS1 = E STEP 1 A → E yes
LHS1 is a subset of LHS2 and RHS1 is a subset of RHS2
LHS1 = A
RHS1 = E no
STEP 1
Remove extraneous attributes
A → E
yes
LHS2 - LHS1 =
RHS2 - RHS1
LHS2 = AB
RHS2 = BE no
AB → BE
yes
Remove FD2

12. STEP 1 A is a subset of AB and E is a subset of BE LHS1 = A no
RHS1 = E no
STEP 1
Remove extraneous attributes
A → E
yes
AB - A =
BE - E
LHS2 = AB
RHS2 = BE no
AB → BE
yes
Remove AB → BE

13. Remove extraneous attributes
STEP 1
LHS1 = A
RHS1 = E
STEP 1
Remove extraneous attributes
A → E
A → E
LHS2 = AB
RHS2 = BE
AB → BE

14. STEP 2 RHS1 is a subset of LHS2 LHS1 = A no RHS1 = BE A → BE STEP 2
Find minimal cover
A → BE
yes
Append RHS2 to
closure of LHS1
LHS2 = E
RHS2 = C
E → C
Remove FD2

15. STEP 2 E is a subset of BE LHS1 = A no RHS1 = BE A → BE STEP 2 yes
Find minimal cover
A → BE
yes
Append C to closure of A
LHS2 = E
RHS2 = C
E → C
Remove E → C

16. STEP 2 LHS1 = A RHS1 = BE A → BE STEP 2 Find minimal cover A+ → ABCE
LHS2 = E
RHS2 = C
E → C

17. STEP 3 A → B STEP 3 LHS1 = LHS2 Partition into relations no A → C yes
Put FD1 and FD2 in one relation, H
B → C

18. Partition into relations
STEP 3 H1
A → B
STEP 3
Partition into relations
A → B
A → C
A → C H2
B → C
B → C

19. STEP 4 Group FDs with the same closure Group has more than 1 elements
Merge equivalent keys
FD2
Find equivalent FDs and put them in J
Do nothing to corresponding H :
Find union of FDs in group
FDN
Subtract J from union as H

20. STEP 4 J J
FD1
STEP 4
Merge equivalent keys J J
FD2 H H : H H H
FDN

21. STEP 5 compare H(x) with H(y) where x != y
Get closure of FD of H(x), C1
C1 =
LHS of FD of H(y)
STEP 5
Remove transitive dependencies H1 H H H FD
compare H(x) with H(y+1)
Get closure of FD of H(y), C2
C2 =
RHS of FD of H(x) H2 H H FD
Remove FD of H(x)
Add Js to Hs

22. Combine LHS and RHS as attributes
STEP 6
STEP 6
Generate
relations
Relation H H H FD
Combine LHS and RHS as attributes
Relation
Set LHS of FD as key H H FD

23. Bernstein’s Shortcoming 1
Problem
Cannot guarantee losslessness
Solution
Combine all attributes from step 6
Find relations that are subset of others
Find missing attributes
Obtain keys to form new relation
To create the new relation, the first step is to combine all attributes in every relation by intersection from step 6’s result and in the process identifying which relation is a subset of the other relations (R ⊆ R’). The second step is to find the missing attribute. We cross check the user’s attribute inputs against our new combined attribute set. Finally, using the relations (where, R ⊄ any R’) found in the second step, we obtain a key of the relation (if multiple keys exist, only use the first key) and by combining the keys with the missing attribute we form our new relation.

24. Bernstein’s Shortcoming 1
R = {A, B, C, D}
F = {A, B → C, A → C, D}
R1 = {A, C, D} [after Bernstein’s]
R2 = {A, B} [new relation added]

25. Bernstein’s Shortcoming 2
Problem
Does not include all keys
Solution
For two arbitrary keys (X,Y) of all keys that can be formed among relations
If the closure X+ shares some attributes with Y,
A new key Z formed, where Z = (Y - (X+ ∩ Y)) ∪ X
If Z ⊆ R (where Y is a key in R) and (all the keys in R)⊄Z
Z is the missing key.
Repeat above steps until no more new key found.
To find all the keys, we compare all the keys among all the relations that have been computed after step 6. For two arbitrary keys (X, Y), first we compute the closure of one key (X+). If the intersection of the closure and the other key exists (if X+ ∩ Y ≠ ∅), we form a new key (Z) by replacing the intersection of Y with X and eliminate duplicate attributes (Z = (Y - (X+ ∩ Y)) ∪ X). If the relation (Rx) which has the key Y contains all the attributes in Z and all the keys of the relation are not a subset of Z ((Z ∈ Rx) && (for all keys (Rx) ⊄Z), we conclude that the new key (Z) should be added to remove the shortcoming of Bernstein’s algorithm.

26. Bernstein’s Shortcoming 2
R = {A, B, C, D}
F = {A, B → C, C → B}
R1 = {A, B, C, D} [after Bernstein’s]
R2 = {C, B} [after Bernstein’s]
given a relation R = {A, B, C, D} and respective functional dependencies F = {A, B → C, C → B}. After applying Bernstein’s algorithm, the result will be: R1 (A, B, C, D) where the key for R1 is A, B and R2 (C, B) where the key for R2 is C.

27. Bernstein’s Shortcoming 3
Problem
Superfluous attributes
Solution
Get all functional dependencies that cannot be retrieved from J set. For two arbitrary functional dependencies from above set (X → Y and X’ → Y)
If X’+ ⊆ X+
All attributes in X’ are all redundant attributes
Remove all these attributes from all the relations that have been affected by any J set
To find the superfluous attributes, first we find all the functional dependencies that do not share the same closure with any functional dependencies in the J set. Second, for any two functional dependencies in these functional dependencies (X → Y and X’ → Y), if the closure of one functional dependency is a subset of the closure of the other functional dependency (if X’+ ⊆ X+), then all the attributes on the left hand side of the first functional dependency (X’) are redundant attributes. Finally, we remove these redundant attributes from all the relations that have been affected by any J set that has been computed in step 4.

28. Bernstein’s Shortcoming 3
R = {A, B, C, D, E, F}
F = {A, D → B, B → C, C → D, A, B → E, A, C → F}
R1 = {A, B, C, D, E, F}
R2 = {B, C}
R3 = {C, D}
For instance, given a relation R = {A, B, C, D, E, F} and respective functional dependencies F = {A, D → B, B → C, C → D, A, B → E, A, C → F}. After applying Bernstein’s algorithm, the result will be : R1 (A, B, C, D, E, F) R2 (B, C) R3 (C,D) where the key for R1 is A, B, A, C, A, D and key for R2 is B and key for R3 is C. For the above example, C is a superfluous attribute.

29. Bernstein’s Shortcoming 5
Problem
Superfluous redundant attributes
Solution
For all functional dependencies that have been computed after step 5
Take two arbitrary functional dependencies (X → Y and X’ → Y’)
If Y ∩ Y’ ≠ ∅
If Y ∩ Y’ ⊄keys(R)
Remove all the attributes in Y ∩ Y’ from Y
Repeat above steps until no more attribute can be removed
To remove superfluous attributes, we compare all the functional dependencies that have computed after step 5. For two arbitrary functional dependencies (X → Y and X’ → Y’), if the right hand side of one functional dependency shares a certain common attributes of the right hand side of the other functional dependency (if Y ∩ Y’ ≠ ∅), then for all the attributes in this intersection, if the attribute is not a part of any key among all the relations (Y ∩ Y’ ⊄keys(R)), we remove this attribute from Y. We recursively perform the above step until no more attribute can be removed.

30. Bernstein’s Shortcoming 5
R = {Model#, Serial#, Price, Color, Name, Year}
F = {{Model#, Serial#} → {Price, Color}, {Model# → Name}, {Serial# → Year}, {Name, Year → Price}}
R1 (Model#, Serial#, Price, Color)
R2 (Model#, Name)
R3 (Serial#, Year)
R4(Name, Year, Price)

31. Questions and Answers