1. Schema Refinement and Normal Forms
Chapter 19
The slides for this text are organized into chapters. This lecture covers Chapter 15.
Chapter 1: Introduction to Database Systems
Chapter 2: The Entity-Relationship Model
Chapter 3: The Relational Model
Chapter 4 (Part A): Relational Algebra
Chapter 4 (Part B): Relational Calculus
Chapter 5: SQL: Queries, Programming, Triggers
Chapter 6: Query-by-Example (QBE)
Chapter 7: Storing Data: Disks and Files
Chapter 8: File Organizations and Indexing
Chapter 9: Tree-Structured Indexing
Chapter 10: Hash-Based Indexing
Chapter 11: External Sorting
Chapter 12 (Part A): Evaluation of Relational Operators
Chapter 12 (Part B): Evaluation of Relational Operators: Other Techniques
Chapter 13: Introduction to Query Optimization
Chapter 14: A Typical Relational Optimizer
Chapter 15: Schema Refinement and Normal Forms
Chapter 16 (Part A): Physical Database Design
Chapter 16 (Part B): Database Tuning
Chapter 17: Security
Chapter 18: Transaction Management Overview
Chapter 19: Concurrency Control
Chapter 20: Crash Recovery
Chapter 21: Parallel and Distributed Databases
Chapter 22: Internet Databases
Chapter 23: Decision Support
Chapter 24: Data Mining
Chapter 25: Object-Database Systems
Chapter 26: Spatial Data Management
Chapter 27: Deductive Databases
Chapter 28: Additional Topics 1

2. The Evils of Redundancy
Redundancy is at the root of several problems associated with relational schemas:
redundant storage, insert/delete/update anomalies
Integrity constraints, in particular functional dependencies, can be used to identify schemas with such problems and to suggest refinements.
Main refinement technique: decomposition (replacing ABCD with, say, AB and BCD, or ACD and ABD).
Decomposition should be used judiciously:
Is there a reason to decompose a relation?
What problems (if any) does the decomposition cause? 14

3. Example in page 607:
Hourly_Emps(ssn, name, lot, rating, hourly_wages, hours_worked)
suppose hourly_wages attribute is determined by the rating attribute.
For a given rating value, there is only one permissible hourly_wages
value.
This integrity constraint (IC) is an example of functional dependency.
See problems in next page

4. Figure 19.1 Problems due to R W : Redundant storage
Update anomaly: Can we change W in just the 1st tuple of SNLRWH?
Insertion anomaly: What if we want to insert an employee and don’t know the hourly wage for his rating?
Deletion anomaly: If we delete all employees with rating 5, we lose the information about the wage for rating 5! 17

5. Can null values help?
Null values cannot help eliminate redundant storage or update anomalies
The insertion anomaly in the previous page may be solved by using null values, null values cannot address all insertion anomalies: we cannot record the hourly wages for a rating unless there is an employee with that rating (primary key field cannot store null)
Null values can not solve deletion anomalies either

6. Use of Decompositions Decomposing Hourly_Emps into two relations:
page 609
Are all problems solved?

7. Problems related to decomposition
Queries over the original relation may require us to join the decomposed relations. If such queries are common, the performance penalty of decomposing the relation may not be acceptable.
In this case we may choose to live with some of the problems of redundancy and not decompose the relation.
It is important to be aware of the potential problems caused by such residual redundancy in the design and to take steps to avoid them (e.g., by adding some checks to application code).
A good DB designer should have a firm grasp of normal forms and what problems they do (or do not) alleviate, the technique of decomposition, and potential problems with decompositions.

8. Functional Dependencies (FDs)
A functional dependency X Y holds over relation R if, for every allowable instance r of R:
t1 r, t2 r, (t1) = (t2) implies (t1) = (t2)
i.e., given two tuples in r, if the X values agree, then the Y values must also agree. (X and Y are sets of attributes.)
An FD is a statement about all allowable relations.
Must be identified based on semantics of application.
Given some allowable instance r1 of R, we can check if it violates some FD f, but we cannot tell if f holds over R!
K is a candidate key for R means that K R
However, K R does not require K to be minimal! 15

9. Examples motivating schema refinement
ER design CAN generate some schemas with redundancy problems, because it is a complex, subjective process, and certain constraints are NOT expressible in terms of ER diagrams.
Constraints on an entity set
Constraints on a relationship set
Identifying attributes of entities
Identifying entity sets

10. Constraints on Entity Set
Consider relation obtained from Hourly_Emps:
Hourly_Emps (ssn, name, lot, rating, hrly_wages, hrs_worked)
Notation: We will denote this relation schema by listing the attributes: SNLRWH
This is really the set of attributes {S,N,L,R,W,H}.
Sometimes, we will refer to all attributes of a relation by using the relation name. (e.g., Hourly_Emps for SNLRWH)
Some FDs on Hourly_Emps:
ssn is the key: S SNLRWH
rating determines hrly_wages: R W 16

11. Example (Contd.) Problems due to R W : Redundant storage
Update anomaly: Can we change W in just the 1st tuple of SNLRWH?
Insertion anomaly: What if we want to insert an employee and don’t know the hourly wage for his rating?
Deletion anomaly: If we delete all employees with rating 5, we lose the information about the wage for rating 5!
This FD (RW) can NOT
be expressed in terms of the
ER model. Only FDs that
determine all attributes of a
relation (i.e., key constraints)
can be expressed in the ER
model. Therefore we can NOT
detect it when we considered
Hourly_Emps as an entity set
during ER modeling. 17

12. Example (Contd.)
Wages
Hourly_Emps2 17

13. Refining an ER Diagram Before:
1st diagram translated: Workers(S,N,L,D,S) Departments(D,M,B)
Lots associated with workers.
Suppose all workers in a dept are assigned the same lot: D L
Redundancy; fixed by: Workers2(S,N,D,S) Dept_Lots(D,L)
Can fine-tune this: Workers2(S,N,D,S) Departments(D,M,B,L)
lot
dname
budget
did
since
name
Works_In
Departments
Employees
ssn
After:
lot
dname
budget
did
since
name
Works_In
Departments
Employees
ssn 18

14. Reasoning About FDs
Given some FDs, we can usually infer additional FDs:
ssn did, did lot implies ssn lot
An FD f is implied by a set of FDs F if f holds whenever all FDs in F hold.
= closure of F is the set of all FDs that are implied by F.
Armstrong’s Axioms (X, Y, Z are sets of attributes):
Reflexivity: If X  Y, then X Y
Augmentation: If X Y, then XZ YZ for any Z
Transitivity: If X Y and Y Z, then X Z
These are sound and complete inference rules for FDs! 19

15. Reasoning About FDs (Contd.)
Couple of additional rules (that follow from AA):
Union: If X  Y and X  Z, then X  YZ (XX XY and XYYZ; XXYZ)
Decomposition: If X  YZ, then X  Y and X  Z (YZY and YZ Z)
Example: Contracts(cid,sid,jid,did,pid,qty,value), and:
C is the key: C  CSJDPQV
A project purchases a given part using a single contract: JP  C
A dept purchases at most one part from a supplier: SD  P
JP  C, C  CSJDPQV imply JP  CSJDPQV
SD  P implies SDJ  JP
SDJ  JP, JP  CSJDPQV imply SDJ  CSJDPQV 20

16. Reasoning About FDs (Contd.)
Computing the closure of a set of FDs can be expensive. (Size of closure is exponential in # attrs!)
Typically, we just want to check if a given FD X Y is in the closure of a set of FDs F. An efficient check:
Compute attribute closure of X (denoted ) wrt F:
Set of all attributes A such that X A is in
There is a linear time algorithm to compute this.
Check if Y is in
Does F = {A B, B C, C D E } imply A E?
i.e, is A E in the closure ? Equivalently, is E in ? 21

17. Apply algorithm in Figure 19.4 to compute attribute closure for A+
closure = A;
repeat until there is no change: {
if there is an FD U  V in F s.t. U  closure,
then set closure = closure  V }
U V
closure = A FD: A B
closure = AB FD: B C
closure = ABC
A+ = {A,B,C} and E  A+ , hence A E  F+

18. Another Example ― 習題 19.2 第一小題
Given a relation R with 5 attributes ABCDE and the following FDs: A  B, BC  E, and ED  A. We want to find all keys for R. Any key K for R must satisfy K  ABCDE. There are = 31 different possibilities.
Let’s first try K = A; we need to show A  ABCDE
closure = A;
U V
closure = A FD: A B
closure = AB
A+ = {A,B} and ABCDE  A+ , hence A ABCDE  F+. In other words, A is not a key for R.

19. Another Example ― 習題 19.2 第一小題
Now you try K = BCD; you need to show BCD  ABCDE

20. Another Example ― 習題 19.2 第一小題
Show BCD  ABCDE
closure = BCD;
U V
closure = BCD FD: BC  E
closure = BCDE FD: ED  A
Closure=ABCDE
BCD+ = {A,B,C,D,E} and ABCDE  BCD+ , hence BCD ABCDE  F+ . In other words, BCD is a key for R.

21. Another Example ― 習題 19.2 第一小題
You try K = ACD; you need to show ACD  ABCDE
You try K = CDE; you need to show CDE  ABCDE

22. Another Example ― 習題 19.2 第一小題
Hence, there are three keys for R:
CDE, ACD, BCD

23. Normal Forms
Returning to the issue of schema refinement, the first question to ask is whether any refinement is needed!
If a relation is in a certain normal form (BCNF, 3NF etc.), it is known that certain kinds of problems are avoided/minimized. This can be used to help us decide whether decomposing the relation will help.
Role of FDs in detecting redundancy:
Consider a relation R with 3 attributes, ABC.
No FDs hold: There is no redundancy here.
Given A B: Several tuples could have the same A value, and if so, they’ll all have the same B value! 22

24. Normal Forms Normal forms based on FDs:
First normal form (1NF)
second normal form (2NF)
third normal form (3NF)
Boyce-Codd normal form (BCNF)
1NF: every field contains only atomic values, that is, no lists or sets.

25. Boyce-Codd Normal Form (BCNF)
Reln R with FDs F is in BCNF if, for every X A (single attribute) in F
A X (called a trivial FD), or
X contains a key for R (i.e., X is a superkey).
In other words, R is in BCNF if the only non-trivial FDs that hold over R are key constraints (see Figure 19.5).
Given a Reln R with a set F of FDs, to show whether R is in BCNF, it is sufficient to check whether the left side of each FD in F is a superkey 23

26. Another Example ― 習題 19.2 第三小題
Given a relation R with 5 attributes ABCDE and the following FDs: A  B, BC  E, and ED  A. Is R in BCNF?
We have shown that there are 3 keys for R: CDE, ACD, BCD.
R is not in BCNF because none of A, BC, and ED contain a key.

27. Boyce-Codd Normal Form (BCNF)
BCNF ensures that no redundancy can be detected using FD information alone (because key is unique).
It is thus the most desirable normal form (from the point of view of redundancy) if we take into account only FD information. 23

28. Third Normal Form (3NF)
Reln R with FDs F is in 3NF if, for every X A in F
A X (called a trivial FD), or
X contains a key for R (i.e., X is a superkey), or
A is part of some key for R.
Minimality of a key is crucial in third condition above!
If R is in BCNF, obviously in 3NF.
If R is in 3NF, some redundancy is possible. It is a compromise, used when BCNF not achievable (e.g., no ``good’’ decomp, or performance considerations).
Lossless-join, dependency-preserving decomposition of R into a collection of 3NF relations always possible. 24

29. Another Example ― 習題 19.2 第二小題
Given a relation R with 5 attributes ABCDE and the following FDs: A  B, BC  E, and ED  A. Is R in 3NF?
We have shown that there are 3 keys for R: CDE, ACD, BCD.
R is in 3NF because the attribute on the right hand side of each FD, i.e., B, E, and A, are all parts of keys.

30. Suppose a dependency X  A causes a violation of 3NF (2 cases to consider)
X is a proper subset of some key K (see Fig. 19.7): this is called partial dependency.
In this case, we store (X,A) pairs redundantly.
As an example, consider the Reserves relation with attribute SBDC where S is sid, B is bid, D is day, and C denotes the credit card to which the reservation is charged. The only key is SBD and we have S  C (sid determines credit card). In this case, we store the credit card number for a sailor as many times as there are reservations for that sailor.

31. Suppose a dependency X  A causes a violation of 3NF (the second case)
X is not a proper subset of any key (see Fig. 19.8): such a dependency is called a transitive dependency.
As an example, consider the Hourly_Emps relation with attributes SNLRWH. The only key is S, but there is a FD R  W, which gives rise to the chain S  R  W.
The consequence is that we cannot record the fact that employee S has rating R without knowing the hourly wage for that rating.
This condition leads to insertion, deletion, and update anomalies.

32. Redundancy is possible with 3NF
Let us revisit the Reserves relation with attributes SBDC and the FD S  C, which states a sailor uses a unique credit card to pay for reservations.
S is not a key, and C is not part of a key (the only key is SBD). Hence, this relation is not in 3NF; (S,C) pairs stored redundantly.

33. Redundancy is possible with 3NF
If we also know that credit cards uniquely identify the owner, we have the FD C S, which means that CBD is also a key for Reserves.
Therefore, the dependency S  C does not violate 3NF, and Reserves is in 3NF.
Nonetheless, in all tuples containing the same S value, the same (S,C) pair is redundantly recorded.

34. Second Normal Form (2NF)
2NF are relations in which partial dependencies are not allowed.
Thus, if a relation is in 3NF (which precludes both partial and transitive dependencies), it is also in 2NF.

35. 習題 19.1 第3小題
Given a set of FDs for the relation schema R(A,B,C,D) with
Primary key AB under which R is in 1NF but not in 2NF.

36. 習題 19.1 第3小題
Given a set of FDs for the relation schema R(A,B,C,D) with
Primary key AB under which R is in 1NF but not in 2NF.
Answer: A  C or A  D or B  C or B  D.

37. 習題 19.1 第4小題
Given a set of FDs for the relation schema R(A,B,C,D) with
Primary key AB under which R is in 2NF but not in 3NF.

38. 習題 19.1 第4小題
Given a set of FDs for the relation schema R(A,B,C,D) with
Primary key AB under which R is in 2NF but not in 3NF.
Answer: D  C or C  D or AC  D or AD  C or BC  D or BD  C

39. 3NF but not BCNF
Given a set of FDs for the relation schema R(A,B,C,D) with
Primary key AB under which R is in 3NF but not in BCNF.

40. 3NF but not BCNF
Given a set of FDs for the relation schema R(A,B,C,D) with
primary key AB under which R is in 3NF but not in BCNF.
Answer: C  A or C  B or D  A or D  B or CD  A or CD  B or CDA  B or CDB  A

41. BCNF Given a set of FDs for the relation schema R(A,B,C,D) with
Primary key AB under which R is in BCNF.

42. BCNF Given a set of FDs for the relation schema R(A,B,C,D) with
Primary key AB under which R is in BCNF.
Answer: AB  ABCD

43. 習題 19.1 第5小題
Consider the relation schema R(A,B,C), which has the FD B  C. If A is a candidate key for R, is it possible for R to be in BCNF? If so, under what conditions? If not, explain why not.

44. 習題 19.1 第5小題
Consider the relation schema R(A,B,C), which has the FD B  C. If A is a candidate key for R, is it possible for R to be in BCNF? If so, under what conditions? If not, explain why not.
Answer: If B is also a key, then R is in BCNF.

45. Decomposition of a Relation Scheme
Suppose that relation R contains attributes A1 ... An. A decomposition of R consists of replacing R by two or more relations such that:
Each new relation scheme contains a subset of the attributes of R (and no attributes that do not appear in R), and
Every attribute of R appears as an attribute of one of the new relations.
Intuitively, decomposing R means we will store instances of the relation schemes produced by the decomposition, instead of instances of R. 26

46. Example Decomposition
Decompositions should be used only when needed.
SNLRWH has FDs S SNLRWH and R W
Second FD causes violation of 3NF; W values repeatedly associated with R values. Easiest way to fix this is to create a relation RW to store these associations, and to remove W from the main schema:
i.e., we decompose SNLRWH into SNLRH and RW 27

47. Example (Contd.)
Wages
Hourly_Emps2 17

48. Example Decomposition
The information to be stored consists of SNLRWH tuples. If we just store the projections of these tuples onto SNLRH and RW, are there any potential problems that we should be aware of? 27

49. Problems with Decompositions
There are three potential problems to consider:
Some queries become more expensive.
e.g., How much did employee Guldu earn? (salary = W*H)
Given instances of the decomposed relations, we may not be able to reconstruct the corresponding instance of the original relation!
Fortunately, not in the SNLRWH example.
Checking some dependencies may require joining the instances of the decomposed relations.
Tradeoff: Must consider these issues vs. redundancy. 28

50. Lossless Join Decompositions
Decomposition of R into X and Y is lossless-join w.r.t. a set of FDs F if, for every instance r that satisfies F:
(r) (r) = r
It is always true that r (r) (r)
In general, the other direction does not hold! If it does, the decomposition is lossless-join.
Definition extended to decomposition into 3 or more relations in a straightforward way.
It is essential that all decompositions used to deal with redundancy be lossless! (Avoids Problem (2).) 29

51. More on Lossless Join
Lossy
Decomp.
The decomposition of R into X and Y is lossless-join wrt F if and only if the closure of F contains:
X Y X, or
X Y Y
In particular, the decomposition of R into UV and R - V is lossless-join if U V holds over R. 30

52. An example of lossless decomposition
We decompose Hourly_Emps with attributes SNLRWH into SNLRH and RW. R is common to both decomposed relations, and RW holds, this decomposition is lossless.

53. 習題 19.8 第2小題的 (b)
Is the following decomposition of R = ABCDEG, with the FD
set F = {AB  C, AC  B, AD  E, B  D, BC  A, E 
G}, lossless-join?
{ABC, ACDE, ADG}
首先 ABC join ACDE = ABCDE. 共同attributes是
AC. 這個decomposition是否為lossless, 我們必須檢查
AC  B or AC  DE 是否  F+ ?
答案是肯定, 因為 AC  B  F.
接下來 ABCDE join ADG = ABCDEG.共同attributes是
AD. 這個decomposition是否為lossless, 我們必須檢查
AD  BCE or AD  G 是否  F+ ?
答案是肯定, 因為 AD  E 與 E  G  F, 故 AD  G  F+.
故此decomposition是lossless-join

54. 習題 19.8 第2小題的 (a)
Is the following decomposition of R = ABCDEG, with the FD
set F = {AB  C, AC  B, AD  E, B  D, BC  A, E 
G}, lossless-join?
{AB, BC, ABDE, EG}
Answer: this decomposition is not lossless-join.

55. Dependency Preserving Decomposition
Consider CSJDPQV, C is key, JP C and SD P.
BCNF decomposition: CSJDQV and SDP (lossless)
Problem: Checking JP C requires a join!
Dependency preserving decomposition (Intuitive):
If R is decomposed into X, Y and Z, and we enforce the FDs that hold on X, on Y and on Z, then all FDs that were given to hold on R must also hold. (Avoids Problem (3).)
Projection of set of FDs F: If R is decomposed into X, ... projection of F onto X (denoted FX ) is the set of FDs U V in F+ (closure of F ) such that U, V are in X. 3

56. Dependency Preserving Decompositions (Contd.)
Decomposition of R into X and Y is dependency preserving if (FX  FY ) + = F +
i.e., if we consider only dependencies in the closure F + that can be checked in X without considering Y, and in Y without considering X, these imply all dependencies in F +.
Important to consider F +, not F, in this definition:
ABC, A  B, B  C, C  A, decomposed into AB and BC.
Is this dependency preserving? (next page) Is C  A preserved????? 4

57. Dependency Preserving Decompositions (Contd.)
F = {AB, B C, C A}
F+ = {A  C, B  A, C  B, AB, B C, C A}
F{A,B} = {A B, B A}
F{B,C} = {B C, C B}
(F{A,B} F{B,C} )+ = {A B, B A, B C, C B}+ = {C A, A C, A B, B A, B C, C B} = F+ 4

58. Dependency Preserving Decompositions (Contd.)
Dependency preserving does not imply lossless join:
ABC, A  B, decomposed into AB and BC.
And vice-versa! (Example?) 4

59. 習題 19.8 第2小題的 (b)
Is the following decomposition of R = ABCDEG, with the FD
set F = {AB  C, AC  B, AD  E, B  D, BC  A, E 
G}, is dependency-preserving?
{ABC, ACDE, ADG}

60. 習題 19.8 第2小題的 (b)
Is the following decomposition of R = ABCDEG, with the FD
set F = {AB  C, AC  B, AD  E, B  D, BC  A, E 
G}, is dependency-preserving?
{ABC, ACDE, ADG}
F = {AB  C, AC  B, AD  E, B  D, BC  A, E  G}
F+ = {AC  D, AD  G, AB  C, AC  B, AD  E, B  D, BC  A, E  G}
F{ABC} = {AB  C, AC  B, BC  A}
F{ACDE} = {AC  D, AD  E}
F{ADG} = {AD  G}
(F{ABC} F{ACDE} F{ADG} )+ = {AB  C, AC  B, BC  A,
AC  D, AD  E, AD  G}  F+ (少了B  D, E  G)
故此decomposition非dependency-preserving.

61. Decomposition into BCNF
Consider relation R with FDs F. If X Y violates BCNF, decompose R into R - Y and XY.
Repeated application of this idea will give us a collection of relations that are in BCNF; lossless join decomposition, and guaranteed to terminate.
e.g., CSJDPQV, key C, JP C, SD P, J S
To deal with SD P, decompose into SDP, CSJDQV.
To deal with J S, decompose CSJDQV into JS and CJDQV
In general, several dependencies may cause violation of BCNF. The order in which we ``deal with’’ them could lead to very different sets of relations! 5

62. BCNF and Dependency Preservation
In general, there may not be a dependency preserving decomposition into BCNF.
e.g., CSZ, CS  Z, Z  C (CS is key)
Can’t decompose while preserving 1st FD; not in BCNF.
Similarly, decomposition of CSJDQV into SDP, JS and CJDQV is not dependency preserving (w.r.t. the FDs JP  C, SD  P and J  S).
However, it is a lossless join decomposition.
In this case, adding JPC to the collection of relations gives us a dependency preserving decomposition.
JPC tuples stored only for checking FD! (Redundancy!)
Hence, redundancy can still occur across relations, even though there is no redundancy within a relation. 6

63. Decomposition into 3NF
Obviously, the algorithm for lossless join decomp into BCNF can be used to obtain a lossless join decomp into 3NF (typically, can stop earlier).
To ensure dependency preservation, one idea:
If X Y is not preserved, add relation XY.
Problem is that XY may violate 3NF! e.g., consider the addition of CJP to `preserve’ JP C. What if we also have J C ?
Refinement: Instead of the given set of FDs F, use a minimal cover for F. 7

64. Minimal Cover for a Set of FDs
Minimal cover G for a set of FDs F:
Closure of F = closure of G.
Right hand side of each FD in G is a single attribute.
If we modify G by deleting an FD or by deleting attributes from an FD in G, the closure changes.
Intuitively, every FD in G is needed, and ``as small as possible’’ in order to get the same closure as F. 8

65. Minimal Cover for a Set of FDs
e.g., AB, ABCD  E, EF  G, EF  H, ACDF  EG has the following minimal cover:
A  B, ACD  E, EF  G and EF  H
Steps:
Put FDs in a standard form: AB, ABCD  E, EF  G, EF  H, ACDF  E, ACDF  G
Minimize the left side of each FD: ABCD E can be minimized to become ACD E (use algorithm in Fig. 15.6)
Delete redundant FDs: ACDF G can be deleted because we have A B, ACD E, EF G (again use algorithm in Fig. 15.6); Similarly ACDF E can be deleted as well 8

66. Minimal Cover for a Set of FDs
AB, ABCD  E, EF  G, EF  H, ACDF  EG
Step 1
AB, ABCD  E, EF  G, EF  H, ACDF  E, ACDF  G
Step 2
AB, ACD  E, EF  G, EF  H, ACDF  E, ACDF  G
Step 3
AB, ACD  E, EF  G, EF  H 8

67. Algorithm in Figure 15.6
XY  F+ is equivalent to Y  X+
closure = X;
do until no change {
if there is U V in F and U closure then
closure = closure  V }

68. Minimal Cover for a Set of FDs
Step 2:
ABCD E (?) F+ is equivalent to E (?) {ABCD}+
closure UV U  closure
{ABCD} ACD E {ACD}  {ABCD}
{ABCDE}
E  {ABCDE}  {ABCD}+
Hence, ABCD E  F+ 8

69. Dependency-preserving decomposition into 3NF
Let R be a relation with a set F of FDs that is a minimal cover, and let R1, R2, …, Rn be a lossless-join decomposition of R. For 1  i  n, suppose that each Ri is in 3NF and let Fi denote the projection of F onto the attributes of Ri. Do the following:
Identify the set N of dependencies in F that are not preserved, that is, not included in the closure of the union of Fis
For each FD XA in N, create a relation schema XA and add it to the decomposition of R.

70. Summary of Schema Refinement
If a relation is in BCNF, it is free of redundancies that can be detected using FDs. Thus, trying to ensure that all relations are in BCNF is a good heuristic.
If a relation is not in BCNF, we can try to decompose it into a collection of BCNF relations.
Must consider whether all FDs are preserved. If a lossless-join, dependency preserving decomposition into BCNF is not possible (or unsuitable, given typical queries), should consider decomposition into 3NF.
Decompositions should be carried out and/or re-examined while keeping performance requirements in mind. 9