1. CSC 453 Database Systems Lecture
Tanu Malik
College of CDM
DePaul University

2. Today Quiz Errata Normalization Functional Dependencies Review Closure
Minimal Cover
Equivalence

3. Why Normalization? Databases are published in bulk.
City of Chicago datasets
Property Tax Bill data
Scientific data

4. Transaction Data TransID Amount # Shares Investment Risk Category
YTD Yield 1
100 10
Gold
Low
3.0% 2
200
Pirates+
High
50.2% 3
202 20 4
505 25 5
900 40
US Bonds
0.1% 6
650
Treasure Hunt
Medium
12.17% 7
350 14 8
420

5. Redundancy in Data TransID Amount # Shares Investment Risk Category
YTD Yield 1
100 10
Gold
Low
3.0% 2
200
Pirates+
High
50.2% 3
202 20 4
505 25 5
900 40
US Bonds
0.1% 6
650
Treasure Hunt
Medium
12.17% 7
350 14 8
420

6. Repeating Combinations
TransID
Amount
# Shares
Investment
Risk Category
YTD Yield 1
100 10
Gold
Low
3.0% 2
200
Pirates+
High
50.2% 3
202 20 4
505 25 5
900 40
US Bonds
0.1% 6
650
Treasure Hunt
Medium
12.17% 7
350 14 8
420
Leads to:
Wasted space
Insert/Update/Delete Anomalies

7. Anomalies
Update Anomaly: The hourly_wages in the first tuple could be updated without making a similar change in the second tuple.
Insertion Anomaly: Cannot insert a tuple for the hourly_wages unless have an employee with that wage.
Deletion Anomaly: If we delete all tuples with a given rating value (e.g., we delete the tuples for Charlie and Don) we lose the association between that rating value and its hourly_wage value.

8. Functional Dependencies

9. Functional Dependencies
Set of attributes Y is functionally dependent on set of attributes X if and only if the values of X uniquely determine the values of Y
Also “X determines Y”, “X  Y”
X is called the determinant
A functional dependency must hold for all possible relation states to be valid

10. Do not Trust the Data Cannot derive FDs from data
It appears that City  State
AddressID
City
State
ZipCode 1
Chicago IL
60604 2
Portland ME
04104 3
60611 4
Boston MA
02468
Move lower?

11. False Function Dependency
New data may violate this “FD”
Need other ways to validate
AddressID
City
State
ZipCode 1
Chicago IL
60604 2
Portland ME
04104 3
60611 4
Boston MA
02468 5 OR
97232

12. Where do FDs Come From? Given by domain experts
Deduced by the database designer
Laws of Physics
X, Y, Z  Location in 3D space
Not based on data

13. Inference Rules Reflexivity: If X includes Y, then X  Y
Augmentation: If X  Y, then XZ  YZ
Transitivity: If X  Y and Y  Z, then X  Z
[Decomposition: If X  YZ, then X  Y]
[Union: If X  Y and X  Z, then X  YZ]
[Pseudo-transitivity: If X  Y and WY  Z,
then WX  Z]

14. Search notion of a f.d Accessing ‘x’ requires a column name and rowidB C 1 a x 2 b y 3 4 c
A BC
BC A

15. Closures
For a set of functional dependencies F, the closure of F (F+) is the set of all functional dependencies that can be derived from F
F+ can be constructed from the closures under F of all possible sets of attributes X
For F and a set of attributes X, the closure of X under F (X+) is the set of all attributes that can be determined from X

16. Finding a Closure of X under F
To find the closure of X under F:
set X+ = X
repeat
set oldX+ = X+
for each Y  Z in F do
if X+ includes Y, then set X+ = X+ U Z
until oldX+ = X+

17. Types of Keys Super Key: Candidate Key Primary Key
For each set X, find the closure of X under F
If X+ contains all attributes, then X is a superkey
Candidate Key
If X is a superkey, but no subset Y of X is a superkey, then X is a candidate key
Primary Key
Choose a candidate key to be the primary key

18. Find Candidate Keys R(ABCDEFGH) F = {AB →C, A → DE, B → F, F → GH}
F = {AB →C, BD → EF, AD → G, A → H}
R(ABCDE)
F = {BC → ACE, D →B}
F = {AB → CD, D →A, BC →DE}

19. More examples
R(WXYZ)
Z → W Y → XZ WX → Y

20. Equivalent Set of F.Ds
Given a relation R and two set of functional dependencies F and G on R, the equivalence question is if
F ⊆ G,
G ⊆ F,
F = G, or
F ̸= G.

21. Example
F={A→B,AB→C,D→AC,D→E}
G={A→BC,D→AE}

22. Algorithm Must bring F and G on the same starting point. How?
Compute the closure set of left hand attributes in F, but by using the f.ds in G.
Compute the closure set of left hand attributes in G but using the f.ds in F.
Compare F’s search power as stated by its f.d set is equivalent to the search power of G.

23. Minimal Cover
F is a minimal cover if F no dependency can be removed from F
In other words, F is an irreducible funcitonal dependency set

24. Finding Minimal Cover
Find a minimal cover by eliminating the redundant/extraneous set of attributes in the set F .
Given a f.d α → β, 3 kinds of redundancies:
the redundancy is on the r.h.s,
the redundancy is on l.h.s, or
the entire f.d is redundant.

25. Eliminate redundancy on R.H.S
Use decomposition rule to decompose all f.d’s in F.
α → βγ to α → β, and α → γ
We have only one attribute on r.h.s, either the entire f.d will be redundant or the f.d will not be redundant.

26. Eliminate entire f.d
Eliminate a f.d if you see it exactly repeated again or can be inferred transitively.
Given a α → β in F,
Find α+ first using F,
compute α+ but this time using F − {α → β}.
If α+ does not change then surely the f.d is redundant.

27. Eliminate redundancy on l.h.s
Given a f.d such αγ → β, determine if α → γ holds in F′ = F −{αγ → β}.
If γ can be searched by just α in F’ without the given f.d, then clearly γ is redundant in the given f.d to search for β.

28. Finding Minimal Cover Minimal/Canonical Cover Algorithm (CCA)
Given FD set, F, CCA finds minimal FD set equivalent to F
minimal: can’t find another equivalent FD set with fewer FD’s

29. Minimal Cover Algorithm
ALGORITHM minimal-cover (X: FD Set)
BEGIN
REPEAT UNTIL STABLE
1. Remove all trivial FDs.
2. Where possible, apply DECOMPOSITION rule (A’s Axioms)
3. Remove all extraneous attributes:
a. Test if B extraneous in A → BC (B extraneous if (A → B)  (F – {A → BC} U {A → C})+)
b. Test if B extraneous in AB → C
(B extraneous if (A → C)  F+)
END

30. Extraneous Attributes
1. Extraneous in RHS?
e.g.: Can we replace A → BC with A → C?
(i.e.: Is B extraneous in A → BC?)
2. Extraneous in LHS?
e.g.: Can we replace AB → C with A → C?
(i.e.: Is B extraneous in AB → C?)
Simple (but expensive) test:
1. Replace A → BC (or AB → C) with A → C in F
Define F2 = F – {A → BC}  {A → C} OR
F2 = F – {AB → C}  {A → C}
2. Test: Is F2+ = F+? If yes, then B was extraneous

31. Extraneous Attributes
A. RHS: Is B extraneous in A → BC?
Step 1: F2 = F – {A → BC}  {A → C}
Step 2: F+ = F2+?
To simplify step 2, observe that F2+  F+
(i.e.: no new FD’s in F2+)
Why?
Have effectively removed A → B from F
When is F+ = F2+?
A: When (A → B)  F2+
Idea: If F2+ includes: A → B and A → C,
then it includes A → BC

32. Extraneous Attributes
A. RHS:
Given F = {A → BC, B → C},
is C extraneous in A → BC?
Why or why not?
A: Yes, because
(A → C)  {A → B, B → C}+
Proof: 1. A → B Given
2. B → C Given
3. A → C transitivity, (1) and (2)
Use Armstrong’s axioms in proof

33. Extraneous Attributes
B. LHS: Is B extraneous in AB → C?
Step 1: F2 = F – {AB → C} U {A → C}
Step 2: F+ = F2+?
To Simplify step 2, observe that F2+  F+
(i.e.: there may be new FD’s in F2+)
Why?
A → C “implies” AB → C.
Thus, all FD’s in F+ also in F2+.
But AB → C does not “imply” A → C.
Thus, all FD’s in F2+, not necessarily in F+.
When is F+ = F2+?
A: When (A → C)  F+
Idea: If (A → C) F+ , then it will include all FD’s of F2+

34. Minimal Cover Algorithm
Example: Determine the minimal cover of
F = {A → BC, B → CE, A → E}
Iteration 1:
a. F = {A → B A → C, A → E, B → E, B → C}
Must check for up to 5 extraneous attributes
Is A → B? extraneous No
Is A → C? extraneous
Yes: (A → C)  {A → BE, B → CE}+
1. A → BE Given
2. A → B Decomposition (1)
3. B → CE Given
4. B → C Decomposition (3)
5. A  C Trans (2,4)
Is B → E? extraneous …

35. Minimal Cover Algorithm
Example (cont.): F = {A → BC, B → CE, A → E}
Iteration 1:
Iteration 1:
a. F = {A → B A → C, A → E, B → E, B → C}
Extraneous atts:
B extraneous in A → BCE? No
C extraneous in A → BCE? Yes…
F = {A → B, A → E, B → E, B → C}
Is A → E? extraneous
Yes: (A → E)  {A → B, B → E, B → C}+
1. A → B Given
2. B → CE Given
3. B → E Decomposition (2)
4. A → E Trans (1,3)
F = {A → B, B → E, B → C}
E extraneous in B → E? No
C extraneous in B → C? No

36. Minimal Cover Algorithm
Example (cont.): F = {A → BC, B → CE, A → E}
Iteration 1:
a. F = {A → BCE, B → CE}
b. Extraneous atts:
B extraneous in A → BCE? No
C extraneous in A → BCE? Yes…
E extraneous in A → BE? Yes…
E extraneous in B → CE? No
C extraneous in B → CE? No
Iteration 2:
a. F = {A → B, B → CE}
b. Extraneous atts:
E extraneous in B → CE? No
C extraneous in B → CE? No
DONE!

37. Minimal Cover of a set of FDs
F = {A → BC,
B → CE,
A → E,
AC → H,
D → B}
F = {A → BC,
B → C,
A → B,
AB → C,
AC → D}
A-> B, B -> CE, AC-> H, D->B

38. Minimal Cover Algorithm
Given:
Determine canonical cover of F:
F = {A → BC,
B → CE,
A → E,
AC → H,
D → B}
Fc = {A → BH,
B → CE,
D → B}
Another Example:
F = {A → BC,
B → C,
A → B,
AB → C,
AC → D}
cc algorithm
Fc = {A → BD,
B → C}

39. Minimal Cover Exercise
Relation R (A, B, C, D)
Set F
ABCD
BC
AB
ABC
BD
Compute canonical cover

40. Normalization

41. Relational Table A B C D E H I TransID Amount # Shares Investment
Risk Category
YTD Yield
Transaction
Date 1
100 10
Gold
Low
3.0%
10/01/2016 2
200
Pirates+
High
50.2% 3
202 20
10/05/2016 4
505 25
10/06/2017 5
900 40
US Bonds
0.1%
10/08/2017 6
650
Treasure
Medium
12.17% 7
350 14 8
420
A-> B, C, D
D-> E,H,
A,D->I
A->B,C,D
D->E,H
A->I
TransID → Amount, # of Shares, Investment
Investment → Risk Category, YTD Yield
TransID, Investment → Transaction Date

42. Option 1 TransID Amount # Shares Transaction Date 1 100 10 10/01/2016
200 3
202 20
10/05/2016 4
505 25
10/06/2017 5
900 40
10/08/2017 6
650 7
350 14 8
420
Investment
Risk Category
YTD Yield
Gold
Low
3.0%
Pirates+
High
50.2%
US Bonds
0.1%
Treasure
Medium
12.17%
A->B,C,I
D->E,H
TransID → Amount, # of Shares, Investment
Investment → Risk Category, YTD Yield
TransID, Investment → Transaction Date

43. Option 2 TransID Amount # Shares Transaction Date 1 100 10 10/01/2016
200 3
202 20
10/05/2016 4
505 25
10/06/2017 5
900 40
10/08/2017 6
650 7
350 14 8
420
Investment
Risk Category
YTD Yield
Transaction
Date
Gold
Low
3.0%
10/01/2016
Pirates+
High
50.2%
10/05/2016
10/06/2017
US Bonds
0.1%
10/08/2017
Treasure
Medium
12.17%
A->B,C,I
D->E,H
TransID → Amount, # of Shares, Investment
Investment → Risk Category, YTD Yield
TransID, Investment → Transaction Date

44. Issues with Options
Option1: This option does not generate the original table as there is no common attribute
Option2: This option has a common attribute but generates spurious tuples

45. Goals of Decomposition
Redundancy avoidance
Avoid any unnecessary data duplication
Want to be able to reconstruct what you started with (lossless)
Avoid Option1 and Option2
Dependency preservation
Minimize the cost of integrity constrains

46. Nonadditive OR Lossless Join
If we decompose a relation state and combine the resulting restrictions with a natural join, we would expect to get the original relation state
There should not be any spurious tuples that were not present in the original relation state…
For the nonadditive join property to hold, this must be the case for every relation state

47. Ensuring Lossless Joins
A Decomposition of R, R = R1  R2 is Lossless iff
R1 ∩ R2 → R1 or
R1 ∩ R2 → R2 or both
(i.e.: Intersecting atts must form a super key for one of the relations)
Intuition: Original relation R has n tuples A A R1 ● …  ● R2  
▪ A a key  |R2|  n, & Relationship with R1 is n:1
▪ A not a key  |R1| = n
 n tuples in result

48. Dependency Preservation
The union over all i in {1,…,m} of the projection of F on Ri is equivalent to F
We want the set of all projections to be equivalent to F – the decomposition neither destroys any functional dependencies in R nor introduces any new ones…
We can always ensure that this property is preserved, but it is not automatic…

49. Projections Suppose we have a set of functional dependencies F over R
For each Ri, consider all XY in F such that both X and Y are subsets of Ri
This set is called the projection of F on Ri
The projection represents the set of all constraints that F puts on the attributes of Ri

50. Projections: Intuition
R = (A, B, C, D), set of FDs - F
R1 = (A, B, C), R2 = (C, D)
Projection of F on R1 is all “local” FDs for R1
All FDs in F that use attributes (A, B, C)
Projection of F on R2 is all “local” FDs for R2
All FDs in F that use attributes (C, D)

51. Dependency Preservation Test
The union over all i in {1,…,m} of the projection of F on Ri is equivalent to F
We want the set of all projections to be equivalent to F – the decomposition does not destroy any functional dependencies in R.
We can always ensure that this property is preserved, but it is not automatic…

52. Dependency Preservation
The projection of F on Ri is the set of all XY in F+ such that both X and Y are subsets of Ri
The projection represents the full set of constraints that F puts on the attributes of Ri
For the dependency preservation property to hold, the union of the projections must be equivalent to F

53. Projections and Dep. Preservation
R = {A, B, C, D}
F = {A  BD, B  DC, C  D}
Let R = R1, R2 where R1 = {A,B,C}, R2 = {C,D}
Compute the projection of F on R1
Compute the projection of F on R2
Is the decomposition dependency preserving? 

54. Normalization
Database normalization, or simply normalization, is the process of organizing the columns (attributes) and tables (relations) of a relational database to reduce data redundancy and improve data integrity.

55. 4 kinds of Normal Forms First Normal Form Second Normal Form
Third Normal Form
Boyce-Codd Normal Form
Fourth Normal Form

56. Prime Vs Non-Prime Attributes
Prime Attribute: Any attribute that is specified in the candidate key
Non-Prime Attribute: Any attribute not part of the candidate key

57. First Normal Form (1NF)
First Normal Form: Each domain of a column must contain only atomic values, and each column in a record must have at most one value from its domain
To get 1NF, remove composite attributes and multi-valued attributes
Composite attributes are easy – just subdivide them into multiple attributes…

58. Removing Multi-Valued Attributes
Remove the multi-valued attribute from the relation
Create a new relation with the primary key of the original relation and the multi-valued attribute
For each tuple in original relation with k values, add k tuples to the new relation
Primary key of new relation contains all attributes; primary key of original relation becomes foreign key in new relation referencing original relation

59. Second Normal Form (2NF)
Second Normal Form: 1NF, plus every non-prime attribute in the relation is determined by the entire primary key (but not by any subset)
To get 2NF, eliminate partial dependencies on the primary key
XY is a partial dependency if ZY for some subset Z of X

60. Removing Partial Dependencies
Find all dependencies where a subset of the primary key determines some non-prime attribute(s)
Starting with the smallest subset, do the following:
1. Remove all attributes on the right-hand side from the relation and put them in a new relation
2. Add the attributes in the determinant(l.h.s) to the new relation; make them the primary key, and make them a foreign key in the original relation referencing the new table
3. Remove from any remaining partial dependencies any attributes removed from the original relation

61. Converting 1NF to 2NF
(EmpID, EmpLName, EmpFName, Dept, ProjCode, Hours)
Functional Dependencies
EmpID  EmpLName, EmpFName, Dept
EmpID, ProjCode  Hours

62. 1NF vs 2NF Data EmpID EmpLName EmpFName Dept ProjCode Hours 550 Smith
Winston
Accounting
101 20
252 10
601
Barney
Finance 5
390
Hammond
Evey
Personnel
995 25
001
Preston
Bill
Special Events
100
Logan
Ted
007
Bond
James
505
Lane
Lois
Media Relations

63. Converting 1NF to 2NF
Decompose (EmpID, EmpLName, EmpFName, Dept, ProjCode, Hours)
Remove the offending FD
EmpID EmpLName, EmpFName, Dept
Two tables
(EmpID, EmpLName, EmpFName, Dept)
(EmpID, ProjCode, Hours)

64. Towards 2NF EmpID EmpLName EmpFName Dept 550 Smith Winston Accounting
601
Barney
Finance
390
Hammond
Evey
Personnel
001
Preston
Bill
Special Events
100
Logan
Ted
007
Bond
James
505
Lane
Lois
Media Relations
EmpID
ProjCode
Hours
550
101 20
252 10
601 5
390
995 25
001
100
007
505

65. Towards 2NF EmpID  EmpLName, EmpFName, Dept EmpID, ProjCode  Hours
550
Smith
Winston
Accounting
601
Barney
Finance
390
Hammond
Evey
Personnel
001
Preston
Bill
Special Events
100
Logan
Ted
007
Bond
James
505
Lane
Lois
Media Relations
EmpID
ProjCode
Hours
550
101 20
252 10
601 5
390
995 25
001
100
007
505
EmpID 
EmpLName, EmpFName, Dept
EmpID, ProjCode 
Hours

66. Third Normal Form (3NF)
Third Normal Form: 2NF, plus every non-prime attribute in the relation is determined only by the primary key of the relation.
To get 3NF, eliminate transitive dependencies on the primary key
XY is a transitive dependency if XZ and ZY for some Z that is disjoint from X

67. Removing Transitive Dependencies
Find all dependencies where a set of attributes disjoint from the primary key determines some non-prime attribute(s)
Starting with the smallest such set, do the following:
1. Remove all attributes on the right-hand side from the relation and put them in a new relation
2. Add the attributes in the determinant to the new relation; make them the primary key, and make them a foreign key in the original relation referencing the new relation
3. Remove from any remaining transitive dependencies any attributes removed from the original relation

68. Converting 2NF to 3NF
Schema: (First , Last, Address, City, State, Zip)
First, Last  Address, City, State, Zip
Transitive functional dependency:
Zip  City, State

69. 2NF Data Table First Last Address City State ZipCode Henry Bienen
2145 Sheridan Rd
Evanston IL
60202
Helmut
Epp
55 E. Jackson St
Chicago
60604
Denis
Stein
55 E. Jackson St.
David
Miller
243 S. Wabash Av.
Gary
5000 Forbes Av.
Pittsburgh PA
15213
Leighton
77 Beacon St
Cambridge MA
02139

70. Decompose the Tables City State ZipCode Evanston IL 60202 Chicago
First
Last
Address
ZipCode
Henry
Bienen
2145 Sheridan Rd
60202
Helmut
Epp
55 E. Jackson St
60604
Denis
Stein
55 E. Jackson St.
David
Miller
243 S. Wabash Av.
Gary
5000 Forbes Av.
15213
Leighton
77 Beacon St
02139
City
State
ZipCode
Evanston IL
60202
Chicago
60604
Pittsburgh PA
15213
Cambridge MA
02139

71. Decompose the Tables ZipCode  City, State
First
Last
Address
ZipCode
Henry
Bienen
2145 Sheridan Rd
60202
Helmut
Epp
55 E. Jackson St
60604
Denis
Stein
55 E. Jackson St.
David
Miller
243 S. Wabash Av.
Gary
5000 Forbes Av.
15213
Leighton
77 Beacon St
02139
City
State
ZipCode
Evanston IL
60202
Chicago
60604
Pittsburgh PA
15213
Cambridge MA
02139
ZipCode  City, State
First, Last  First, Last, Address, ZipCode

72. Third Normal Form
Third Normal Form: Every non-prime attribute is fully functionally dependent on every candidate key, and no non-prime attribute is transitively dependent on any candidate key
“For every non-trivial functional dependency XA, either X is a superkey or A is a prime attribute.”

73. Third Normal Form
A relation schema is in third nomal form (3NF) if for all:
X → Y in F+
at least one of the following holds:
X → Y is trivial (i.e. X ∈ Y)
X is a superkey for R
each attribute A in Y – X is contained in a candidate key for R, i.e. A is a prime attribute.

74. Example R = (A,B,C,D) A→ C D BA→ C The candiate key is ?
3NF decomposition is ?
R1 = (C, A, D)
R2 = (B, A, C)

75. 3NF Decomposition R(A, B, C, D) R(A,B,C,D,E) R(A, B, C, D, E, F)
FDs: A B, C D
R(A,B,C,D,E)
FDs: CE, BC
R(A, B, C, D, E, F)
F = { AB  CD, D  A, C  EF}
(AB)
(AC)
(AB) (DB)

76. 3NF-Checking Order What is the candidate key? Which FDs violate 3NF?
Decompose to 3NF.
Is the decomposition lossless?
Is it dependency preserving?

77. 3NF Decomposition
Input: A universal relation R and a set of functional dependencies F on R
Output: A decomposition D of R into 3NF schemas with dependency preservation and nonadditive join

78. 3NF-Normalization Algorithm (3NF Normalization):
Input: Relation R with FDs F c
Output: 3NF decomposition D of R
D = {}
For every XY in F add sub-relation Q =(XY) to D, unless
some sub-relation in D already contains all of XY: don’t add Q
some sub-relation(S) in D is contained in XY: replace S with Q(XY)
If no relation in D contains a key of R, then add
new relation Q(X) on some key X of R

79. The End Result A collection of relations, each in 3NF
Each relation has a primary key
(We are assuming that there is only one candidate key…)
Every non-prime attribute in a relation is determined by its entire primary key
No non-prime attribute in a relation is determined by any attributes other than its entire primary key
Information can reconstructed using joins, and stored in views if desired

80. 3NF Decomposition R={A, B, C, D, E, F} F = { AB  CD, C  EF, D  A }
Non-prime determines a prime attribute 

81. Boyce-Codd Normal Form
Boyce-Codd Normal Form (BCNF): For every non-trivial functional dependency XA, it must be the case that X is a superkey
“Every determinant must contain a candidate key”
X must be a superkey even if A is a prime attribute

82. BCNF Decomposition
Input: A universal relation R and a set of functional dependencies F on R
Output: A decomposition D of R into BCNF schemas with nonadditive join
Algorithm on next page
Algorithm does not guarantee dependency preservation

83. BCNF Decomposition Algorithm ALGORITHM BCNF (R: Relation, F: FD set)
BEGIN
1. D  {R}
3. While some X → Y holds in some Ri(A1,…,An) in D and
(X → Y) is not trivial, X is not a superkey of Ri
Ri1  X+ ∩({A1,…,An})
Ri2  X  ({A1,…,An} - X+ )
Result  Result – {Ri}  {Ri1,Ri2}
4. Return result
END

84. BCNF Example: R = (A, B, C) F = {A → B, B → C} Is R in BCNF?
A: Consider the nontrivial dependencies in F:
1. A → B, A → R (A is a key)
2. B → C, B → A (B is not a key)
Therefore, R not in BCNF

85. BCNF Example: Q: Is the decomposition lossless?
R = R1  R2 R1 = (A, B); R2 = (B, C)
F = {A → B, B → C}
Are R1, R2 in BCNF?
A: 1. Test R1:
A → B covered, A → R1 (all other FD’s covered trivial)
2. Test R2:
B → C covered, B → R2 (all other FD’s covered trivial)
 R1, R2 in BCNF
Q: Is the decomposition lossless?

86. BCNF Decompose R into BCNF: R = (A, B, C, D, E, H)
F = {A → BC, E → HA}
Decompose R into BCNF:

87. BCNF Decomposition Decomposition #1: R = R1  R3  R4 Q: Is this DP?
R = (A, B, C, D, E, H)
F = {A → BC, E → HA}
(Note: Fc = F)
Decomposition #1: R = R1  R3  R4
R = (A, B, C, D, E, H)
Decompose on A → BC
R1 = (A, B, C)
R2 = (A, D, E, H)
Decompose on E → HA
R3 = (A, E, H)
R4 = (D, E)
Q: Is this DP?
A: Yes. All Fc covered by R1, R3, R4. Therefore F+ covered

88. BCNF Decomposition (cont.)
R = (A, B, C, D, E, H)
F = {A → BC, E → HA}
(Note: Fc = F)
Decomposition #2: R = R1  R3  R5  R6
R = (A, B, C, D, E, H)
Decompose on A → B
R1 = (A, B)
R2 = (A, C, D, E, H)
Decompose on E → HA
R3 = (A, E, H)
R4 = (C, D, E)
Decompose on E → C
R5 = (C, E)
R6 = (E, D)
Q: Not DP. Why?
A: A → C not covered by R1, R3, R5 , R6.

89. More BCNF (cont.)
Q: Can we decompose on FD’s in Fc to get a DP BCNF decomposition?
A: Sometimes, BCNF + DP not possible
Example:
R = (J, K, L)
F = {JK → L, L → K}
(Fc = F)
Decompose on
Or:
JK → L
L → K
L → K
JK → L

90. Not DP: JK → L not covered
More BCNF (cont.)
Q: Can we decompose on FD’s in Fc to get a DP BCNF decomposition?
A: Sometimes, BCNF + DP not possible
R = (J, K, L)
F = {JK → L, L → K}
Decomposition #1:
Decomposition #2:
R = (J, K, L)
Decompose on L → K
R = (J, K, L)
Decompose on JK → L
R1 = (L, K)
R2 = (J, L)
R2 = (J, K, L)
R2 = (J, K)
Not DP: JK → L not covered

91. BCNF Decomposition R (S, P, Q, X, Y, N, C)
F = { S  NC, P  XY, SP  Q , QP }
Decompose to BCNF
Is it dependency preserving?

92. BCNF Decomposition R(A, B, C, D) R(A, B, C, D, E, F) FDs: A B, C D
FDs: AC B, C D
R(A, B, C, D, E, F)
F = { AB  CD, C  EF, D  A }
(AB)
(AC)
(AB) (DB) (CB)

93. Properties of Decompositions
When we work with BCNF, we must look at properties involving multiple relations:
Nonadditive (Lossless) Join: No tuples that are not in the original relation (spurious tuples) are generated when decomposed relations are joined
Dependency Preservation: Every functional dependency in the original relation is represented somewhere in the decomposition

94. BCNF vs. 3NF Every relation in BCNF is in 3NF
Not every relation in 3NF is in BCNF
3NF relations that are not in BCNF fail because some prime attribute is determined by something that is not a superkey – this is allowed by 3NF but not by BCNF
Decomposing tables into BCNF can be tricky – functional dependencies can be lost!

95. Example 3NF vs BCNF Client, Office  (Client, Office, Account)
Joe 1 B
Mary
John C 2
Client, Office  (Client, Office, Account)
Account  Office

96. Remarks on Algorithms
Different runs may yield different results, depending on the order in which attributes and functional dependencies are considered
We must know all functional dependencies
We can’t always guarantee dependency preservation for BCNF, but we can generate a 3NF decomposition and then consider the individual relations in the result