1. THIRD NORMAL FORM (3NF)
A relation R is in BCNF if whenever a FD XA holds in R, one of the following statements is true:
XA is a trivial FD, or
X is a superkey, or
A is part of some key

2. Why 3NF?
A relation R is in 3NF if whenever a FD XA holds in R, one of the following statements is true:
XA is a trivial FD, or
X is a superkey, or
A is part of some key
By making an exception for certain dependencies involving some key attributes, we can ensure that every relation schema can be decomposed into a collection of 3NF with two desirable properties: lossless-join and dependency-preserving.

3. Cases of 3NF Violation
If XY causes a violation of 3NF, there are two cases:
X is a proper subset of some key (partial dependency)
Partial dependency causes data redundancy: Since XY and X is not a key, X could be redundant, so Y.
X is not a proper subset of any key (transitive dependency)
KEY
Attributes X
Attribute Y
KEY
Attributes X
Attribute Y
Attributes X
KEY
Attribute Y
Transitive dependency KXY makes it impossible to record the values of (K, X, Y) unless all of them are known: Since KXY, we cannot associate an X value with a K value unless we also associate an X value with Y value

4. Dependency-Preserving Decomposition
Inputs:
A relation R with a set Fmin of FDs that is minimum cover
D(R1, R2, …, Rn) is a lossless-join decomposition of R
Find all dependencies in Fmin that are not preserved
For each such dependency XA, create a relation schema XA and add it to the decomposition of R
Every dependency in Fmin is now preserved
Proof: XA is in 3rd NF
X must be a key for XA, Since XA is in a minimal cover, YA does not hold for any Y that is a subset of X
For any other dependencies hold over XA, say YZ, in Fmin, it must satisfy 3rd NF conditions
If Z is A, Y must be X
If Z is not A, Z must be part of X
R and F
For each X->Y in F, check if it violates BCNF. If yes, decompose R into R-Y and XY
Let R1, …, Rn be the resulted decomposition. Compute F+, and for each X->Y in F+, find its corresponding Ri (if X and Y are both in Ri). This generates F1, …, Fn for R1, …, Rn, respectively.
Find Fmin. For each X->Y in Fmin, check if it is included in the union of F1, …, Fn. If not, add XY in R’s decomposition.

5. A. Lossless-Join Decomposition
Top-Down Approach: Lossless-Join and Dependency Preserving Decomposition into 3NF
A. Lossless-Join Decomposition
Set D{R}
While there is a relation schema Q in D that is not in BCNF do
begin
Choose a relation schema Q in D that is not in BCNF;
Find a functional dependency XY in Q that violates BCNF;
Replace Q in D by two schemas (Q-Y) and (XUY)
end;
B. Dependency-Preserving Decomposition
Assume the decomposition is D(R1, R2, …, Rn) and the FD sets are accordingly F1, F2, …, and Fn (let their union be F’)
For each dependency XA in the original F (needs to be a minimum cover), check if it can be inferred from F’
If not, create a relation schema XA and add it to the decomposition of R

6. Exercise Top-down approach R(ABCDE) F={ABCDE,ED,AB,ACD}
Loss-less join decomposition: R(ACBDE) is not in BCNF
3. Dependency-preserving decomposition:
{ABCDE, ED, ABB, ACD}+ == {ACE, AB, ED}+ ??
/* Find a minimum cover first */
/* If XY is not preserved, add (XY) into the decomposition */
ED
R1(ACBE)
R2(ED)
AB
R1(ACE)
R2(AB)

7. Bottom-up Approach: Lossless-Join and Dependency Preserving Decomposition into 3NF
Inputs:
A relation R and a set of functional dependencies F on the attributes of R.
Find a minimal cover G of F.
For each left-hand side X of a FD G, create a relation schema in D with attributes
where XA1, XA2, …, XAm are the only dependencies in G with X as the left-hand side.
Prove that this relation is in 3rd NF
If none of the relation schemas in D contains a key of R, then create one more relation schema in D that contains attributes that form a key of R.
Prove that this decomposition is lossless-join
X->Ai is in Fmin. Let Y->Z
If Z is Ai, Y must be X, then Y->Z obey BNCF.
If Z is not Ai, then Z must be part of X. Since X is a key, Z is part of some key. So Y->Z obey 3NF.
Since K determines all other attributes, so

8. Exercise R(ABCDE) F={ABCDE,ED,AB,ACD}
Bottom up approach (Synthesis):
Step 1: Find a minimum cover, G={ACE,ED,AB}
Step 2: R1(ACE), R2(ED), R3(AB)
Step 3: Is this a lossless-join decomposition?

9. Normalization Review Functional Dependency Normal Forms Decomposition
Amstrong’s axioms
Attribute closure (A+)
Dependency closure (F+)
Minimum cover (Fmin)
Normal Forms
BCNF
3NF
Decomposition
Lossless join
Dependency preserving
Conceptual design
Schemas
ICs

10. Determine Normal Forms
BCNF
For each XA, is it a trivial dependency?
Is X a superkey?
3NF
Suppose XA violate BCNF
Is A part of some key?

11. Exercise 1 For each of the following relation schemas and sets of FDs
R(ABCD) with FDs ABC, CD, and DA
R(ABCD) with FDs BC and BD.
R(ABCD) with FDs ABC, BCD, CDA, and ADB
Check if they are in BCNF or 3NF, if not, perform a lossless join and dependency preserving decomposition
BCNF
For each XA, is it a trivial dependency?
Is X a superkey?
3NF
Suppose XA violate BCNF
Is A part of some key?

12. Exercise 2
Prove that, if R is in 3NF and every key is simple (i.e, a single attribute), then R is in BCNF
Prove that, if R has only one key, it is in BCNF if and only if it is in 3NF.

13. Quiz
For each of the following relation schemas Indicate the strongest normal form of each of the following relations
R1(ABCDE) F1={AB, CD, ACEABCDE}
R2(ABCEF) F2={ABC, BF, FE}
Consider a relation R with five attributes: ABCDE. F={AB, BCE, EDA}
Are {ECD}, {ACD}, {BCD} keys for R?
Is R in BCNF? Why?
Is R in 3NF? Why?

14. Dependency preservation
Conceptual Schema ( ER diagram )
DBMS independent
Data Model Mapping
DBMS specific
Conceptual Schema ( Relations )
Normalization
BCNF/3NF?
Decomposition
Lossless join
Dependency preservation