1. Sridhar Narayan narayans@uncw.edu
Normalization
Sridhar Narayan

2. EMP_PROJ Something feels wrong about this design
SSN
PNUMBER
HOURS
ENAME
PNAME
PLOC E1 P1 20
Joe
CIS Roof
UNCW P2
Restaurant
Mayfaire E2 40
Something feels wrong about this design
Try adding a row – Insertion anomaly
Try deleting a row – Deletion anomaly
Try updating a row – Update anomaly
Need a formal way to reason about what is wrong with it and how to fix it

3. Functional Dependency
Constraints between attribute sets in a relation
If X and Y are sets of attributes of a relation R, and whenever two tuples in R have the same X-values they also have the same Y-values, we say that X functionally determines Y.

4. Functional Dependency
Written as X -> Y
X functionally determines Y
Y is functionally determined by X
X is the determinant, Y is the dependent
Examples
SSN -> SSN (trivial dependency)
PNUMBER -> PNAME
SSN -> ENAME
SSN, PNUMBER -> HOURS

5. Functional Dependency
Between sets of attributes, not just single attributes
Holds for all time, not just for a particular instance (snapshot) of a relation
Formally states constraints that exist for the relation
These constraints are in addition to those imposed by primary keys and foreign keys

6. Functional dependencies and keys
If X functionally determines all attributes of R, then X is a super key
If X is irreducible, i.e. every member of X is essential for the functional dependencies to hold, then X is a candidate key.
Attributes that are a part of a candidate key are key attributes

7. Examples Super key: Candidate key:
SSN
PNUMBER
HOURS
ENAME
PNAME
PLOC E1 P1 20
Joe
CIS Roof
UNCW P2
Restaurant
Mayfaire E2 40
Super key:
SSN, PNUMBER, PNAME -> SSN, PNUMBER, HOURS, ENAME, PNAME, PLOC
Candidate key:
SSN, PNUMBER -> SSN, PNUMBER, HOURS, ENAME, PNAME, PLOC

8. Redundancy
If in a relation R, A -> B and A is not a candidate key for R, then R will involve some redundancy.
SSN
PNUMBER
HOURS
ENAME
PNAME
PLOC
Intuitively, all functional dependencies in a relation should involve candidate keys to eliminate redundancy

9. Normalization
A process that utilizes functional dependencies to identify relation schemas that have an undesirable form (redundancy) and decomposes them into smaller schema in which the redundancy has been eliminated.

10. Decomposition Decomposition should be Lossless join
Allow exact recovery of the original schema (without spurious tuples)
Dependency preserving
Allow dependencies to be checked without requiring a join

11. Lossy decomposition SSN PNUMBER HOURS ENAME PNAME PLOC E1 P1 20 Joe
CIS Roof
UNCW P2
Restaurant
Mayfaire E2 40
SSN
PNUMBER
HOURS
ENAME E1 P1 20
Joe P2 E2 40
ENAME
PNAME
PLOC
Joe
CIS Roof
UNCW
Restaurant
Mayfaire

12. Natural join to recover original
SSN
PNUMBER
HOURS
ENAME
PNAME
PLOC E1 P1 20
Joe
CIS Roof
UNCW P2
Restaurant
Mayfaire E2 40 E2 P1 40
Joe
Restaurant
Mayfaire

13. Heath’s Theorem If relation R = {A,B,C} where A,B,C are attribute sets
and A -> B
then R1= {A, B} and R2 = {A, C} represents a lossless join decomposition

14. Levels of normalization
First normal form – 1NF
Second normal form – 2NF
Third normal form – 3NF
Boyce-Codd Normal Form - BCNF
Increasingly stringent requirements

15. Normal Forms
1NF
2NF
BCNF
3NF

16. First normal form
Relation is in 1NF if all attribute values are atomic (By definition, all relations are in 1NF)
D_NAME
D_NUM
MGR_SSN
D_LOCATIONS
RESEARCH 5
{Lumberton, Red Springs, Raeford}
Assume that a department can have multiple locations, like {Lumberton, Red Springs, Raeford}
Relation not in 1NF

17. Is this a resolution? D_NAME D_NUM MGR_SSN D_LOCATIONS RESEARCH 5
Lumberton
Red Springs
Raeford

18. Decomposition D_NAME D_NUM MGR_SSN D_LOCATIONS D_NAME D_NUM MGR_SSN

19. Second Normal Form: 2NF A relation is in 2NF if It is in 1NF, and
If the non-key attributes are fully (irreducibly) dependent on the primary key

20. Example: EMP_PROJ SSN -> ENAME PNUMBER -> PNAME, PLOC
HOURS
ENAME
PNAME
PLOC
Functional Dependencies?
SSN -> ENAME
PNUMBER -> PNAME, PLOC
{SSN, PNUMBER} -> HOURS
Relation not in 2NF
Non-key attributes ENAME, and PLOC and PNAME, are not fully dependent on the primary key

21. 2NF 1a 1b 2nF ? Solution? Decompose SSN PNUMBER HOURS SSN PNUMBER
ENAME
PNAME
PLOC
2nF ?

22. Decompose further…
SSN
ENAME 2a
2NF 2b
SSN
PNUMBER
PNAME
PLOC
2nF ?

23. 3a 2NF 3b And a little more… 3b is a part of 1a, so drop it. PNUMBER
PNAME
PLOC 3a
2NF 3b
3b is a part of 1a, so drop it.
SSN
PNUMBER

24. 1a 2NF 2a 2NF 3a 2NF 2NF Normalization SSN PNUMBER HOURS SSN ENAME
PNAME
PLOC 3a
2NF

25. More than one way to get here
SSN
PNUMBER
HOURS
ENAME
PNAME
PLOC 1a
2NF
PNUMBER
PNAME
PLOC
Not2NF
SSN
PNUMBER
HOURS
ENAME 1b

26. 2a 2NF Not2NF 2b Decompose further… SSN PNUMBER HOURS SSN PNUMBER
ENAME 2b

27. And a little bit more 3a
2NF
SSN
ENAME
Redundant 3b
SSN
PNUMBER

28. 3NF Normalization A relation is in 3NF if It is in 2NF, and
If the non-key attributes are mutually independent. That is, no functional dependencies exist between non-key attributes.

29. Example: EMP_DEPT SSN -> {ENAME, DOB, ADDRESS, DNUM}
DNAME
DMGRSSN
Functional Dependencies?
SSN -> {ENAME, DOB, ADDRESS, DNUM}
DNUM -> {DNAME, DMGRSSN}
Redundancy?
Relation in 1NF ?
2NF ?
3NF ?

30. 3NF Normalization
SSN
ENAME
DOB
ADDRESS
DNUM 1a 1b
DNUM
DNAME
DMGRSSN

31. BCNF Normalization S#
SNAME P#
QTY S1
Acme Supply P1
100 S2
Gem Mfg
200 P2
400
S# and SNAME – Supplier# and Supplier Name are unique
FDs
S# -> SNAME
SNAME -> S#
S#,P# -> QTY
SNAME, P# -> QTY
Candidate keys
S#, P# and SNAME, P#

32. BCNF Normalization Redundancy? 1NF? 2NF? 3NF? S# SNAME P# QTY S1
Acme Supply P1
100 S2
Gem Mfg
200 P2
400
Redundancy?
1NF?
2NF?
3NF?

33. BCNF
Relation is in BCNF if and only if the only determinants are candidate keys
FDs
S# -> SNAME
SNAME -> S#
S#,P# -> QTY
SNAME, P# -> QTY

34. BCNF Normalization S# P# QTY S1 P1 100 S2 200 P2 400 S# SNAME S1
Acme Supply S2
Gem Mfg
Two candidate keys: S#
SNAME