1. Designing Relational Databases
B.Ramamurthy
4/9/2019
B.ramamurthy

2. Problems with Simple Relations
Repetition of information
Inability to represent certain information
Loss of information from decomposing relations
4/9/2019
B.ramamurthy

3. Example Consider a lending schema
lending-schema (bname, bcity, assets, cname,loanNo, amt)
when you want to insert a new loan do you have insert assets too?
Ex: insert (Perry, Horseneck, , Adams, l-13, 1500)
4/9/2019
B.ramamurthy

4. Example (contd.)
How will you represent branch details if there was no loan in the branch.
We will need null values.
Null values are difficult to handle.
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B.ramamurthy

5. Solution : alternative design
Decomposition
Decompose lending-schema into branch-customer-schema(bcs) and customer-loan-schema(cls)
bcs(bname, bcity, assets, cname)
cls(cname, loanNo, amt)
Is this decomposition correct? Lossless?
4/9/2019
B.ramamurthy

6. Decomposition
bcs natural-join cls should result in our original lending but does it?
For example, we cannot answer the question “Find all branches that have made loan < 100” on “lending-schema” will result in Mianus and Round Hill.
But the same query on bcs joined with cls results in Mianus, Round Hill and Downtown.
Some information was lost in decomposition that leads us to the incorrect answer.
4/9/2019
B.ramamurthy

7. How can we make decomposition lossless?
Impose set of constraints (including functional dependencies).
For example
bname --> {bcity, assets} would have taken care of the problem above.
4/9/2019
B.ramamurthy

8. Formally Stating.. Let C be a set of constraints.
A decomposition {R1, R2, .. Rn} of a relation scheme R is a lossless join for if all relations r on schema R that are legal under C,
r = R1 ( r) join R2 ( r ) … join Rn ( r )
4/9/2019
B.ramamurthy

9. Normalization using Functional Dependencies
Using functional dependencies we can define several “normal forms” that represent good database deign.
There are several normal forms : 1NF, 2NF, etc.
We will cover 3NF and Boyce-Codd Normal Form (BCNF)
4/9/2019
B.ramamurthy

10. Desirable Properties Lossless-join decomposition
Dependency preservation
Lack of redundancy
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B.ramamurthy

11. BCNF
A relation schema R is in BCNF with respect to a set of functional dependencies if for all functional dependencies in F+ of the form a-->b at least one of the following holds,
a-->b is a trivial functional dependency (that is b is subset of a)
a is superkey for schema R
4/9/2019
B.ramamurthy

12. Example customer-scheme(cname,cstreet,ccity) cname-->cstreet,ccity
branch-scheme(banme, assets, bcity)
bname-->asset,bcity
loan-info-schema(bname, cname, loanNo, amt)
loanNo-->amt,bname
Is not
BCNF
4/9/2019
B.ramamurthy

13. Solution Decompose loan-info-scheme into
loan-schema (bname, loanNo, amt)
Borrower-schema (cnmae, loanNo)
4/9/2019
B.ramamurthy

14. BCNF Decomposition result = {R} done = false Compute F+ while not done
if there is a schema Ri that is not BCNF
let a --> b be the non-trivial fn.dep in Ri
s.t. a --> Ri is not in F+ and ab =0
result = (result -Ri)  (Ri-b)  (a,b)
else done=true;
4/9/2019
B.ramamurthy

15. Lossless Decomposition
If R is split into R1 and R2, for the decomposition to be loss less the at least one of the two should hold:
R1  R2 --> R1
R1  R2 --> R2
4/9/2019
B.ramamurthy

16. Dependency Preservation
When a relational schema D defined by functional dependency F is decomposed into Ri {i = 1,2,3 ..n}, each functional dependency should be testable by at least one of Ri.
Formally, let F+ be the closure F and let F’+ be the closure of dependencies covered by Ri. F+ == F+’ for dependency preservation.
4/9/2019
B.ramamurthy

17. 3NF : Third Normal Form
BCNF is quite stringent that sometimes it may not be possible to cover all the functional dependencies after decomposition.
3NF solves this problem by relaxing the BCNF requirements as follows:
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B.ramamurthy

18. 3NF : Definition
A relation schema R is in BCNF with respect to a set of functional dependencies if for all functional dependencies in F+ of the form a-->b at least one of the following holds,
a-->b is a trivial functional dependency (that is b is subset of a)
a is superkey for schema R
Each attribute A in b-->a is contained in a candidate key for R.
4/9/2019
B.ramamurthy

19. 3NF : Algorithm Let Fc be the canonical cover of F. j = 0;
for each dependency a-->b if none of schemes in Ri contains ab then
j = j+1;
Rj = ab;
if none of the schemas contains a candidate key for R then
j = j+ 1;
Rj = any candidate key for R;
return (Rj)
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B.ramamurthy

20. Examples : 7.2, 7.3, 7.8
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B.ramamurthy

21. Functional Dependencies (review)
A functional dependency is a relationship between attributes
Examples:
Student ID determines Major: StuID => Major
StuID => (Name, Major)
(StuID, Course) => Grade
Attribute(s) to the left of arrow called determinant(s)
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B.ramamurthy 3 3

22. Anomalies in a Relation (review)
An anomaly is a weakness in the way a relation is set up
Assume a single table representing student-courses (!)
Consider the following table:
Id Name Major Course-Id Course-Desc
45 Meyer CS CSE Operating Systems
56 Beaver CE CSE Database Concepts
23 Teller CE CSE Architecture
This table has all three anomalies: insert, update, and delete
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B.ramamurthy 4 4

23. Insert, Update and Delete Anomalies (review)
Insert anomaly is caused when a new course needs to be inserted that is not registered by a student
Update anomaly caused when Major is updated in one row only for Id 45.
Delete anomaly caused when Id 56 removed from the table; we lose a CSE course description
Anomalies are avoided by splitting the offending relation into multiple relations (decomposition)
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24. Fourth Normal Form (4NF)
A multi-valued dependency exists when
there are at least three attributes A, B, and C in a relation and
for each value of A there is a well-defined set of values for B, and a well-defined set of values for C,
but the set of values of B is independent of set C
A relation is in 4NF if it is already in 3NF and has no multi-valued dependencies
Every possible combination of the two multi-valued attributes have to be stored in the database thus leading to redundancy and consequent anomalies
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B.ramamurthy 8 10

25. Fourth Normal Form (4NF) - Example
Course-Id Instructor Textbook MGS Clay Hansen MGS Clay Kroenke MGS Drake Hansen MGS Drake Kroenke
By placing the multi-valued attributes in tables by themselves, we can convert the above to 4NF
Change to: COURSE-INST (Course-Id, Instructor) COURSE-TEXT (Course-Id, Textbook)
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26. Normal Forms
5NF
4NF
BCNF
3NF
2NF
1NF
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B.ramamurthy

27. Homework #3 :Due date 4/11 in class : Hardcopy
1. Consider a scheme R (A,B,C,D,E) and the two set of functional dependencies. Are the two sets F1 and F2 equivalent?
F1 : A-->B, AB--> C, D--> AC, D --> E
F2 : A--> BC , D--> AE
2. Consider the schema R (A,B,C,D,E,F).
AB--> C, C --> A, BC -->D, AC D--> B
BE-->C, CE-->FA, CF-->BD, D-->EF
a. Find the closure.
b. Find the set candidate keys.
Show all the steps.
4/9/2019
B.ramamurthy