1. Module 5: Overview of Database Design -- Normalization

2. Outline Decomposition Functional Dependency
Decomposition Using Functional Dependency

3. Bad Schema
Consider a relation inst_dept, which represents the result of a natural join on the relations corresponding to instructor and department
This relation is in “bad form” since there is a possibility of repetition of information and we may need to use null values to represent information (department with no instructors).

4. How to Deal with a Bad Schema?
The only way to deal with a bad schema is to split the scheme (decompose) into a number of smaller schemas.
Hopefully, each of the smaller schemes is in a “good form”; that is, there is no repetition of information and there is no need to use null values.
In the case of the inst_dept schema , we would decompose it into two relation schemes – instructor and department .
We know that both instructor and department are in good form (how do we know this)?

5. Lossless-Join Decomposition
When we decompose a relation r into two relations r1 and r2 we must make sure that we can reconstruct the information that r originally contained. Reconstruction is achieved by using the natural-join operation
Decomposition of r = (A, B, C) into: r1 = (A, B) and r2 = (B, C)
 r1 (r)  r2 (r) B 1 2
B,C(r) C A   r
A,B(r)

6. Decomposition does not always work
Not all compositions are lossless-join decomposition
Suppose we decompose
employee(ID, name, street, city, salary) into:
employee1 (ID, name)
employee2 (name, street, city, salary)
The next slide shows how we lose information -- we cannot reconstruct the original employee relation using employee1 and employee2 -- and so, this is a lossy decomposition.

7. A Lossy Decomposition

8. Lossless-Join Decomposition Definition
Let R be a relation schema, and let R1 and R2 form a decomposition of R.
We say that the decomposition is a lossless decomposition if there is no loss of information by replacing R with two relation schemas R1 and R2 .
More precisely, we say the decomposition is lossless if, for all legal database instances (this is, all possible relations r on schema R)
r = R1 (r ) R2 (r )
We will define “legal database instances” later on.
We will show shortly how one can test to see if a decomposition is lossless.

9. Decomposition Theory Goal — Devise a Theory for the Following:
Decide whether a particular relation R is in “good” form.
In the case that a relation R is not in “good” form, decompose it into a set of relations {R1, R2, ..., Rn} such that
each relation is in good form
the decomposition is a lossless-join decomposition
Our theory is based on:
functional dependencies
multivalued dependencies

10. Functional Dependencies

11. Functional Dependencies
Constraints on the set of legal relations.
Require that the value for a certain set of attributes determines uniquely the value for another set of attributes.
A functional dependency is a generalization of the notion of a key.

12. Functional Dependencies (Cont.)
Let R be a relation schema, and let:
  R and   R
The functional dependency
  
holds on R if and only if for any legal relations r(R), whenever any two tuples t1 and t2 of r agree on the attributes , they also agree on the attributes . That is,
t1[] = t2 []  t1[ ] = t2 [ ]
Example: Consider r(A,B ) with the following instance of r.
On this instance, A  B does NOT hold, but B  A does hold. A 1 3 B 4 5 7

13. Functional Dependencies (Cont.)
K is a superkey for relation schema R if and only if K  R
K is a candidate key for R if and only if
K  R, and
for no   K,   R
Functional dependencies allow us to express constraints that cannot be expressed using superkeys. Consider the schema:
inst_dept (ID, name, salary, dept_name, building, budget ).
We expect these two functional dependencies to hold:
dept_name building
ID  building
but would not expect the following to hold:
dept_name  salary

14. Use of Functional Dependencies
We use functional dependencies to:
Test relations to see if they are legal under a given set of functional dependencies.
If a relation r is legal under a set F of functional dependencies, we say that r satisfies F.
specify constraints on the set of legal relations
We say that F holds on R if all legal relations on R satisfy the set of functional dependencies F.
Note: A specific instance of a relation schema may satisfy a functional dependency even if the functional dependency does not hold on all legal instances.
For example, a specific instance of instructor may, by chance, satisfy
name  ID.

15. Trivial Functional Dependencies
A functional dependency is trivial if it is satisfied by all instances of a relation
Example:
ID, name  ID
name  name
In general,    is trivial if   

16. Closure of a Set of Functional Dependencies
Given a set F of functional dependencies, there are certain other functional dependencies that are logically implied by F.
For example: If A  B and B  C, then we can infer that A  C
The set of all functional dependencies logically implied by F is the closure of F.
We denote the closure of F by F+.
F+ is a superset of F.

17. Computing F+
We can compute F+, the closure of F, by repeatedly applying Armstrong’s Axioms:
if   , then    (reflexivity)
if   , then      (augmentation)
if   , and   , then    (transitivity)
These rules are
sound (generate only functional dependencies that actually hold), and
complete (generate all functional dependencies that hold).

18. Example of Armstrong’s Axioms
R = (A, B, C, G, H, I) F = { A  B A  C CG  H CG  I B  H}
some members of F+
A  H
by transitivity from A  B and B  H
AG  I
by augmenting A  C with G, to get AG  CG and then transitivity with CG  I
CG  HI
by augmenting CG  I with CG, to infer CG  CGI,
and augmenting of CG  H with I, to infer CGI  HI,
and then transitivity

19. Procedure for Computing F+
F + = F repeat for each functional dependency f in F apply reflexivity and augmentation rules on f add the resulting functional dependencies to F + for each pair of functional dependencies f1and f2 in F if f1 and f2 can be combined using transitivity then add the resulting functional dependency to F + until F + does not change any further
NOTE: We shall see an alternative procedure for this task later

20. Canonical Cover
Sets of functional dependencies may have redundant dependencies that can be inferred from the others
For example: A  C is redundant in: {A  B, B  C, A C}
Parts of a functional dependency may be redundant
E.g.: on RHS: {A  B, B  C, A  CD} can be simplified to {A  B, B  C, A  D}
A  C, A  D
E.g.: on LHS: {A  B, B  C, AC  D} can be simplified to {A  B, B  C, A  D}
AC  CD, CD D
Intuitively, a canonical cover of F, denoted by Fc is a “minimal” set of functional dependencies equivalent to F, having no redundant dependencies or redundant parts of dependencies.
We show how to compute Fc later on in Module 8.

21. Decomposition Using Functional Dependency

22. Lossless-join Decomposition
We can now formally define lossless-join decomposition.
Let R be a relation schema, and let R1 and R2 form a decomposition of R. For the case of R = (R1, R2), we require that for all possible relations r on schema R
r = R1 (r ) R2 (r )
A decomposition of R into R1 and R2 is lossless join if at least one of the following dependencies is in F+:
R1  R2  R1
R1  R2  R2

23. Example R = (A, B, C) F = {A  B, B  C) R1 = (A, B), R2 = (B, C)
Lossless-join decomposition:
R1  R2 = {B} and B  BC
R1 = (A, B), R2 = (A, C)
R1  R2 = {A} and A  AB

24. Dependency Persevering Decomposition
Constraints, including functional dependencies, are costly to check (when one updates or inserts new tuples) unless they pertain to only one relation
It is desirable to be able to test only those dependencies on each individual relation of a decomposition in order to ensure that all functional dependencies hold.
A decomposition is dependency preserving If it is sufficient to test only those dependencies on each individual relation of a decomposition in order to ensure that all functional dependencies hold.

25. Example R = (A, B, C) F = {A  B, B  C) R1 = (A, B), R2 = (B, C)
Lossless-join decomposition:
R1  R2 = {B} and B  BC
Dependency preserving
R1 = (A, B), R2 = (A, C)
R1  R2 = {A} and A  AB
Not dependency preserving (cannot check B  C without computing R1 R2)

26. What do Know Thus Far?
We know that any acceptable decomposition must be a lossless-join decomposition
We know how to test to see if the decompose of a relation into a set of relations a lossless-join decomposition.
We know that is desirable to have a lossless-join decomposition that is dependency preserving.
We do NOT know (yet) is -- how to decide if a given relation is in good form.

27. Boyce-Codd Normal Form
A relation schema R is in BCNF with respect to a set F of functional dependencies if for all functional dependencies in F+ of the form
 
where   R and   R, at least one of the following holds:
   is trivial (i.e.,   )
 is a superkey for R
Example schema not in BCNF:
instr_dept (ID, name, salary, dept_name, building, budget )
dept_name building, budget holds on instr_dept,
but dept_name is not a superkey

28. Decomposing a Schema into BCNF
Suppose we have a schema R and a non-trivial dependency  in F+ that causes a violation of BCNF.
We decompose R into:
(U  )
( R - (  -  ) )
In our example,
R = dept_name, building, budget, ID, name, salary
 = dept_name
 = building, budget
and inst_dept is replaced by
R1 = (U  ) = ( dept_name, building, budget )
R2 = ( R - (  -  ) ) = ( ID, name, salary, dept_name )
Are R1 and R2 in BCNF?

29. BCNF Decomposition Algorithm
result := {R }; done := false; compute F +; while (not done) do if (there is a schema Ri in result that is not in BCNF) then begin let    be a nontrivial functional dependency that holds on Ri such that   Ri is not in F + and    = ; result := (result – Ri )  (Ri – )  (,  ); end else done := true;
Note: each Ri is in BCNF, and decomposition is lossless-join.

30. BCNF Decomposition Algorithm Example
R = (A, B, C )
Functional dependencies
{A  B, B  C}
BCNF Decomposition:
B  C
but B is not a superkey.
We replace R by:
R1 = (B, C)
R2 = (A, B)
R1 and R2 in BCNF

31. BCNF Decomposition Algorithm Example
class (course_id, title, dept_name, credits, sec_id, semester, year, building, room_number, capacity, time_slot_id)
Functional dependencies:
course_id→ title, dept_name, credits
building, room_number→capacity
course_id, sec_id, semester, year→building, room_number, time_slot_id
A candidate key {course_id, sec_id, semester, year}.
BCNF Decomposition:
course_id→ title, dept_name, credits holds
but course_id is not a superkey.
We replace class by:
course(course_id, title, dept_name, credits)
class-1 (course_id, sec_id, semester, year, building, room_number, capacity, time_slot_id)

32. BCNF Decomposition Algorithm Example (Cont.)
course is in BCNF
How do we know this?
building, room_number→capacity holds on class-1
But {building, room_number} is not a superkey for class-1.
We replace class-1 by:
classroom (building, room_number, capacity)
section (course_id, sec_id, semester, year, building, room_number, time_slot_id)
classroom and section are in BCNF.

33. BCNF and Dependency Preservation
It is not always possible to get a BCNF decomposition that is
dependency preserving
R = (J, K, L ) F = {JK  L L  K } Two candidate keys = JK and JL
R is not in BCNF
Any decomposition of R will fail to preserve
JK  L
This implies that testing for JK  L requires a join
Because it is not always possible to achieve both BCNF and dependency preservation, we consider a weaker normal form, known as third normal form.

34. Third Normal Form
A relation schema R is in third normal form (3NF) if for all:
   in F+
at least one of the following holds:
   is trivial (i.e.,   )
 is a superkey for R
Each attribute A in {  –  } is contained in a candidate key for R.
(NOTE: each attribute may be in a different candidate key)
If a relation is in BCNF it is in 3NF (since in BCNF one of the first two conditions above must hold).
Third condition is a minimal relaxation of BCNF to ensure dependency preservation (we will see why later).

35. 3NF Example
R = (J, K, L ) F = {JK  L L  K } Two candidate keys = JK and JL
We have seen before that R is not in BCNF
R, however, is in 3NF
JK is a superkey
L  K and L is NOT a superkey, but:
{ K } – { L } = { K } and
K is contained in the candidate key JK

36. 3NF Example Using University schema
Relation dept_advisor:
dept_advisor (s_ID, i_ID, dept_name) F = {s_ID, dept_name  i_ID, i_ID  dept_name}
Two candidate keys:
{s_ID, dept_name} and {i_ID, s_ID}
R is in 3NF
s_ID, dept_name  i_ID
{s_ID,dept_name} is a superkey
i_ID  dept_name
dept_name is contained in a candidate key

37. 3NF Decomposition Algorithm
Let Fc be a canonical cover for F; i := 0; for each functional dependency    in Fc do if none of the schemas Rj, 1  j  i contains   then begin i := i + 1; Ri :=   end if none of the schemas Rj, 1  j  i contains a candidate key for R then begin i := i + 1; Ri := any candidate key for R; end /* Optionally, remove redundant relations */
repeat if any schema Rj is contained in another schema Rk then /* delete Rj */ Rj = R;; i=i-1; return (R1, R2, ..., Ri)

38. 3NF Decomposition Algorithm (Cont.)
Above algorithm ensures:
each relation schema Ri is in 3NF
decomposition is dependency preserving and lossless-join
Proof of correctness is at end of this presentation

39. Redundancy in 3NF Consider the relation R R = (J, K, L)
F = {JK  L, L  K }
And an instance table: J j1 j2 j3
null L K I1 I2 k1 k2
What is wrong with the table?
Repetition of information
Need to use null values (e.g., to represent the relationship l2, k2 where there is no corresponding value for J)

40. Design Goals Goal for a relational database design is: BCNF.
Lossless join.
Dependency preservation.
If we cannot achieve this and dependency preservation is important, we use 3NF.
Interestingly, SQL does not provide a direct way of specifying functional dependencies other than superkeys.
Can specify FDs using assertions, but they are expensive to test, (and currently not supported by any of the widely used databases!)
Even if we had a dependency preserving decomposition, using SQL we would not be able to efficiently test a functional dependency whose left hand side is not a key.

41. How Good is BCNF?
There are database schemas in BCNF that do not seem to be sufficiently normalized
Consider a relation
inst_info (ID, child_name, phone)
where an instructor may have more than one phone and can have multiple children
99999
David
William
ID child_name phone
inst_info

42. How good is BCNF? (Cont.)
There are no non-trivial functional dependencies and therefore the relation is in BCNF
Insertion anomalies – i.e., if we add a phone to 99999, we need to add two tuples
(99999, David, ) (99999, William, )

43. Fourth Normal Form (4NF)
It is better to decompose inst_info into:
99999
David
William
ID child_name
ID phone
inst_child
inst_phone
This suggests the need for higher normal forms, such as Fourth Normal Form (4NF), which we shall see later.

44. End of Module 5

45. Lossless-Join Decomposition
A decomposition that does not result in loss of information is termed a Lossless join decomposition.
Whenever we decompose a relation into two relations, we will insist that the decomposition will be a lossless-join.
Decomposition of R = (A, B, C) R1 = (A, B) R2 = (B, C) A B   1 2 r C A   B 1 2
A,B(r) B 1 2
B,C(r) C A A B   1 2 C
A (r) B (r)

46. Lossless-Join Decomposition
When we decompose a relation scheme R into two relation schemes R1 and R2 we must make sure that we can always reconstruct the information that any database instance of R contained. Reconstruction is achieved by using the natural-join operation
Decomposition of R = (A, B, C) R1 = (A, B) R2 = (B, C A B   1 2 r C A   B 1 2
A,B(r) B 1 2
B,C(r) C A A B   1 2 C
A (r) B (r)

47. Chapter E: Overview of Database Design
Features of Good Relational Design
Normalization

48. 3NF Decomposition Algorithm Example
Relation schema:
cust_banker_branch = (customer_id, employee_id, branch_name, type )
The functional dependencies for this relation schema are:
customer_id, employee_id  branch_name, type
employee_id  branch_name
customer_id, branch_name  employee_id
We first compute a canonical cover
branch_name is extraneous in the r.h.s. of the 1st dependency
No other attribute is extraneous, so we get FC =
customer_id, employee_id  type employee_id  branch_name customer_id, branch_name  employee_id

49. 3NF Decomposition Algorithm Example (Cont.)
The for loop generates following 3NF schema:
(customer_id, employee_id, type )
(employee_id, branch_name)
(customer_id, branch_name, employee_id)
Observe that (customer_id, employee_id, type ) contains a candidate key of the original schema, so no further relation schema needs be added
At end of for loop, detect and delete schemas, such as (employee_id, branch_name), which are subsets of other schemas
result will not depend on the order in which FDs are considered
The resultant simplified 3NF schema is:
(customer_id, employee_id, type)

50. Redundancy in 3NF (Cont.)
The above example corresponds to:
dept_advisor (s_ID, i_ID, dept_name)
F = {s_ID, dept_name  i_ID, i_ID  dept_name}
Repetition of information (e.g., the relationship l1, k1)
(i_ID, dept_name)
Need to use null values (e.g., to represent the relationship
i_ID, dept_name where there is no corresponding value for s_ID).
(i_ID, dept_name) if there is no separate relation mapping instructors to departments