1. Prof. Sin-Min Lee Department of Computer Science
CS157A Lecture 20
Midterm 3 Revision 2
Prof. Sin-Min Lee
Department of Computer Science

2. Normal Forms 1NF 2NF 3NF BCNF 4NF 5NF Functional dependencies
Multivalued
dependencies
Join
dependencies

3. F+: dependencies induced by Armstrong’s Axioms
Axioms for reasoning about FD’s
(i), (ii) and (iii) are Armstrong’s axioms
F+ is the set of dependencies which can be deduced from F by applying three inference rules:
(i) reflexivity if Y  X then X ® Y
(ii) augmentation if X ® Y then XZ ® YZ
(iii) transitivity if X ® Y and Y ® Z then X ® Z

4. Completeness of Armstrong Axioms
Idea
If X ® Y cannot be deduced using Armstrong’s axioms then there is a relational instance for R in which all the dependencies in F are true,
but X ® Y does not hold
R=LMNO
X=L
F={L ® M , M ® N, O ® N}
then X+ = LMN
L ® O cannot
be deduced in F* L M N
Introduction
Presentation O
L ® O cannot
be deduced in F+
Counterexample:

5. Closure of a Set of Attributes
Let U be a set of attributes and F be a set of functional dependencies on U.
Suppose that X  U is a set of attributes.
Definition: X+ = { A | F X  A}
We would like to compute X+ |=

6. AB  AB, BC  BC, AC  AC, ABC  ABC, AB  A, AB  B, BC  B, BC  C,
Note that A, B, C, are attributes
We refer to the set {A,B} simply as AB
R = ( A, B, C )
F = { A  B, B  C }
F+ = { A  A, B  B, C  C,
AB  AB, BC  BC, AC  AC, ABC  ABC,
AB  A, AB  B,
BC  B, BC  C,
AC  A, AC  C,
ABC  AB, ABC  BC, ABC  AC,
ABC  A, ABC  B, ABC  C,
A  B, … (1) ( given )
B  C, … (2) ( given )
A  C, … (3) ( transitivity on (1) and (2) )
AC  BC, … (4) ( augmentation on (1) )
AC  B, … (5) ( decomposition on (4) )
A  AB, … (6) ( augmentation on (1) )
AB  AC, AB  C, B  BC,
A  AC, AB  BC, AB  ABC, AC  ABC, A  BC, A  ABC }
Using reflexivity, we
can generate all
trivial dependencies

7. Algorithm to compute closure of attributes X+ under F
closure := X ;
Repeat
for each U  V in F do
begin
if U  closure
then closure := closure  V ;
end
Until (there is no change in closure)

9. F = { A  B, A  C, CG  H, CG  I, B  H } To compute AG+
R( A, B, C, G, H, I )
F = { A  B, A  C, CG  H, CG  I, B  H }
To compute AG+
closure = AG
closure = ABG ( A  B )
closure = ABCG ( A  C )
closure = ABCGH ( CG  H )
closure = ABCGHI ( CG  I )
Is AG a candidate key?
AG  R
A+  R ?
G+  R ?

10. Example
R(ABCDE)
F={ABC, CEB, DA, BCE}
{A}+ =
{A,B}+ =
{B,D}+=

11. Theorem:
R - a relation schema
F - set of functional dependencies on R
The decomposition of R into relations with attribute sets
R1, R2 is a lossless-join decomposition iff
( R1  R2 )  R1  F + OR
( R1  R2 )  R2  F +
i.e., R1  R2 is a superkey for R1 or R2.
(the attributes common to R1 and R2 must contain a key for
either R1 or R2 ).

12. Example
R(A,B,C,D,E)
 = {R1(A,D), R2(A,B),R3(B,E), R4(C,D,E), R5(A,E)}
FD1. AC FD2. BC FD3. CD
FD4. DEC FD5. CE A.
Decide whether the decomposition is lossless.

13. BCNF Decomposition It is a lossless join decomposition.
Suppose R is not in BCNF, A is an attribute, and X  A is a FD where X  A =  that violates the condition.
Remove A from R
Create a new relational schema XA
Repeat this process until all the relations are in BCNF
It is a lossless join decomposition.
But not necessary dependency preserving

14. Key is C
CSJDPQV
SDP
JS
SDP
CSJDQV
SDP
JS
CJDQV JS

15. Does not preserve JPC, we can add a schema:
Key is C
SDP
CSJDPQV
JS
SDP
JP C
SDP
CSJDQV
JS JS
CJDQV
The result is in BCNF
Does not preserve JPC, we can add a schema:
CJP
Each of SDP, JS, CJDQV, CJP is in BCNF, but there is
redundancy in CJP.

16. Possible refinement Key is C SDP SDP SDQ SDQ
CSJDPQV
Key is C
SDP
SDP
SDP
CSJDQV
SDQ
SDQ
SDQ
CSJDV
SD is a key in SDP and SDQ,
There is no dependency between P and Q
we can combine SDP and SDQ into one schema
Resulting in SDPQ, CSJDV

17. Overview
It is possible to decompose any relational schema into a set of relational schemas with the following properties:
1) Attribute Preserving
2) FDs preserving
3) Lossless Join

18. The Decomposition Algorithm
Step 1. Find a minimal cover G for F
Step 2. For each left-hand side X that appears in G, create a relation schema with attributes
{X  A1  A2, ... ,  Am}
where X  A1, X  A2, ... , X  Am are all dependencies in G with X as left-hand side.
Step 3. If none of the relation schemas contains a key of R, create one more relation schema that contains attributes that form a key for R.

19. Example Consider R(A, B, C, D, E) and F={AB  C, A  BE, C E}
Step 1. minimal cover
Fmin={AC, AB, CE}
Step 2. R1(A,B,C), R2(C,E)
Step 3. Key: {A,D}
we have R3(A,D)
 Final Result:
R1(A, B,C), R2(C,E), and R3(A,D)

20. BCNF Decomposition Property LJ1:
A decomposition D={R1, R2} of R has the lossless join property with respect to a set of functional dependencies F on R if and only if either
the FD ((R1  R2)  (R1 - R2)) is in F+, or
the FD ((R1  R2)  (R2 - R1)) is in F+

21. BCNF Decomposition Property LJ2:
If a decomposition D={R1, R2, ..., Rm} of R has the lossless join property with respect to a set of functional dependencies F on R, and if a decomposition D1={Q1, Q2, ... ,Qk} of Ri has the lossless join property with respect to the projection of F on Ri, then the decomposition
D2={R1, R2, ... Ri-1, Q1, Q2, ..., Qk, Ri+1, ..., Rm} of R has the lossless join property with respect to F

22. BCNF Decomposition - Algorithm
1. Set D  {R}
2. 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 (X  Y)
end;

23. BCNF Decomposition - Example
Consider R(A,B,C,D) and
F={A  B, B  C, D  B}
Decompose R into BCNF relations.
Step 1.D={R(A,B,C,D)}
Step 2. Loop 1.R is not in BCNF because A  B and A is not a superkey
decompose R into R1(A, C, D), and R2(A, B)
Loop 2. R1 is not in BCNF because A  C and A is not a superkey
decompose R1 into R11(A, D) and R12(A, C)
Result:D={R11(A,D), R12(A,C), R2(A,B)}

24. Overview Given a relation schema R(A1, A2, ... , An).
If R is not in the third normal form (3NF), we wan to decompose it into a set of relation schema D= { R1, R2, ... ,Rm }, where each Ri is in 3NF, such that the following conditions are held:
Attribute preservation condition.
Dependency preservation condition
Lossless join condition

25. Attribute Preservation Condition
Attribute preservation condition states that the union of attributes of Ri equal to the set of attributes of R.
 For example: Given R(A, B, C, D) and the decomposition
D1={ R1(A,B), R2(B,C) and R3(A,C,D)}. D1 satisfies the attribute preservation condition.

26. Attribute Preservation Condition
Given R(A, B, C, D) and the decomposition
D2={R1(A, B), R2(B,C), R3(A, C)},
The attribute preservation condition is violated because D is missing (not preserved in the decomposition).

27. Dependency Preservation Condition
We say that a decomposition D={R1, R2, ... , Rm} of R is dependency preserving with respect to F if the union of the projections of F on each Ri in D is equivalent to F. That is:
((F(R1)  ...  F(Rm))+ = F+
Given a set of dependencies F on R, the projection of F on Ri, denoted by F(Ri) where Ri is a subset of R, is the set of dependencies X  Y in F+ such that the attributes in X  Y are all contained in Ri.

28. Dependency Preservation Condition
Given R(A, B, C, D) and F = { A  B, B  C, C  D}
 Let D1={R1(A,B), R2(B,C), R3(C,D)}
F(R1)={A  B}
F(R2)={B  C}
F(R3)={C  D}
FDs are preserved.

29. Dependency Preservation Condition
Given R(A, B, C, D) and F = { A  B, B  C, C  D}
 Let D2={R1(A,B}, R2(B,C), R3(A, D)}, then FDs are not preserved.

30. Dependency Preservation Condition
Given R(A, B, C, D) and F = { A  B, B  C, C  D}
 Let D2={R1(A,B}, R2(B,C), R3(A, D)}, then FDs are not preserved.

31. Dependency Preservation Condition
We want to preserve the dependencies because each dependency in F represents a constraint on a database.

32. Dependency Preservation Condition
If one of the dependencies is not represented by the dependencies on some individual relation Ri of the decomposition, we will not be able to enforce this constraint by looking only at an individual relation, instead, to enforce the constraint, we will have to join two or more of the relations in the decomposition and then check that functional dependency hold in the result of the join operation. This is very inefficient.

33. Multivalued Dependencies and Fourth Normal Form
Formal Definition of Multivalued Dependency

34. Multi-Valued Dependency
Problem: multi-valued (or binary join) dependency
Definition: If every instance of schema R can be (losslessly) decomposed using attribute sets (X, Y) such that:
r =  X (r)
 Y (r)
then a multi-valued dependency exists
Ex: Person= SSN,PhoneN(Person)
 SSN,ChildSSN(Person)

35. Fourth Normal Form
A schema is in fourth normal form if for every non-trivial multi-valued dependency:
R = X Y
either:
- X  Y or Y  X (trivial case) or
- X  Y is a superkey of R
(i.e., X  Y R )

36. 4th Normal Form No multivalued dependencies and BCNF
Create separate tables for each separate functional dependency

37. Example
SalesForce (State, SalesPerson)
Delivery (State, Delivery)

38. Beyond 4th Normal Form 5th Normal Form Domain Key Normal Form (DKNF)
Project-Join Normal Form
Domain Key Normal Form (DKNF)

39. Assume the relation R contains the following two tuples R(A B C D)
( ) …
( ) …
  What other tuples must R contain so that A ->-> B and A ->-> C hold for R ?
Answer: The tuples that must be included due to the two multi-valued dependency are:
( )
( )
( )
( )
( ) second round
( ) second round

40. Example Consider the following relation and determinants.
R(a,b,c,d) a,c -> b,d a,d -> b
To be in BCNF, all valid determinants must be a candidate key. In the relation R, a,c->b,d is the determinate used, so the first determinate is fine.
a,d->b suggests that a,d can be the primary key, which would determine b. However this would not determine c. This is not a candidate key, and thus R is not in BCNF.

41. Tuple Relational Calculus
based on specifying a number of tuple variables
a tuple variable refers to any tuple

42. Simple example 1
To find all employees whose salary is greater than $50,000
{t| EMPLOYEE(t) and t.Salary>5000}
where
EMPLOYEE(t) specifies the range of tuple variable t
The above operation selects all the attributes

43. Simple example 2
To find only the names of employees whose salary is greater than $50,000
{t.FNAME, t.NAME| EMPLOYEE(t) and t.Salary>5000}
The above is equivalent to
SELECT T.FNAME, T.LNAME
FROM EMPLOYEE T
WHERE T.SALARY > 5000

44. Elements of a tuple calculus
In general, we need to specify the following in a tuple calculus expression:
Range Relation (I.e, R(t)) = FROM
Selected combination= WHERE
Requested attributes= SELECT

45. Elements of formula A formula is made of Predicate Calculus atoms:
an atom of the from R(ti)
ti.A op tj.B op{=, <,>,..}
F1 And F2 where F1 and F2 are formulas
F1 OR F2
Not (F1)
F’=(t) (F) or F’= (t) (F)
 Y friends (Y, John)
X likes(X, ICE_CREAM)

46. More Example
For every project located in ‘Stafford’, retrieve the project number, the controlling department number, and the last name, birthrate, and address of the manger of that department.

47. Domain Relational Calculus (DRC)
Another type of formal predicate calculus-based language
QBE is based on DRC
The language shares a lot of similarities with the tuple calculus

48. DRC The only difference is the type of variables:
variables range over singles values from domains of attributes
An expression of DRC is:
{x1, x2,…,xn|COND(x1,x2,…,xn, xn+2,…,xn+m)}
where x1,x2,…,xn+m are domain var range over attributers
COND is a condition (or formula)

49. Examples
Retrieve the birthdates and address of the employee whose name is ‘John B. Smith’
{uv| (q)(r)(s) (EMPLOYEE(qrstuvwxyz) and q=‘John’ and r=‘B’ and s=‘Smith’

50. Alternative notation
Ssign the constants ‘John’, ‘B’, and ‘Smith’ directly
{uv|EMPLOYEE (‘John’, ’B’, ’Smith’ ,t ,u ,v ,x ,y ,z)}

51. More example
Retrieve the name and address of all employees who work for the ‘Reseach’ department
{qsv | ( z) EMPLOYEE(qrstuvwxyz) and ( l) ( m) (DEPARTMENT (lmno) and l=‘Research’ and m=z))}

52. More example
List the names of managers who have at least on e dependent
{sq| ( t) EMPLOYEE(qrstuvwxyz) and (( j)( DEPARTMENT (hijk) and (( l) | (DEPENTENT (lmnop) and t=j and t=l))))}

53. Characteristics of a Decomposition
Two important characteristics of a decomposition:
lossless join: necessary, otherwise original relation cannot be recreated, even if tables are not modified
dependency preserving: allows us to check that inserts/updates are correct without joining the sub-relations

54. Lossless Join T C S Smith DB Cohen Jones OS Levy C S DB Cohen OS Levy

55. Checking
Check for a lossless join using the algorithm from class (with the a-s and b-s)
Check for dependency preserving using an algorithm shown today

56. Dependency Preservation
R=ABC
Decomposition {AB, AC}
Dependencies {AB, BC}.
Is it lossless?
Does this decomposition preserve BC?

57. Dependency Preservation (cont’d)B A
100 10 1 2
300 20 3 B A 10 1 2 30 3 4 C A
100 1 2
300 3
400 4

58. Definitions
We define S (F) to be the set of dependencies XY in F+ such that X and Y are in S.
We say that a decomposition R1...Rn of R is dependency preserving if for all instances r of R that satisfy the FDs of R:
(R1 (F) U ... U Rn (F))+ = F+
Note that one inclusion clearly holds always.
This definition implies an exponential algorithm to check if a decomposition is dependency preserving
We give a polynomial algorithm

59. Algorithm Let R be a relation, decomposed into R1, R2,…,Rn
Let F be a set of functional dependencies
To check whether R1,…,Rn preserves all the functional dependencies in F, run the algorithm on the next slide for each X -> Y in F
If the answer is “Yes” for all FDs, then the decomposition preserves F
If the answer is “No” for at least one FD, then the decomposition does not preserve F

60. Testing Dependency Preservation
To check if the decomposition preserves XY:
Z:=X
while changes to Z occur do
for i=1 to n do
Z:= Z  ((Z  Ri)+  Ri)
if YZ
return “yes”
else
return “no”

61. Example (1) R=ABCD F = {A -> B, B -> C, C -> D, D -> A}
R1=AB, R2=BC, R3=CD
Is this decomposition dependency preserving?

62. Example (2) R = ABCDE F = {A -> ABCDE, BC -> A, DE -> C}
Suppose we decompose R into ABDE and DEC.
Is the decomposition dependency preserving?

63. Normal Forms
The basic idea: if a relation is in one of these forms, then it avoids certain problems (e.g., redundancy)
Normal Forms:
BCNF: Every dependency X->A in F+ must be (1) trivial or (2) X is a super-key
3NF: Every dependency X->A in F+ must be (1) trivial, (2) X is a super-key or (3) A is an attribute of a key

64. Example
Reminder F+ = {X -> X+ | exist Y->Z in F st Y in X and Z not in X}
Suppose that R = ABC. For each of the following values of F, decide whether R is in BCNF/3NF:
F = {}
F = {A -> B}
F = {A -> B, A -> C}
F = {A -> B, B -> C}
F = {A -> B, BC -> A}

65. Decomposition into 3NF
Given a relation R with functional dependencies F
Step 1: Find a non-redundant cover G of F
Step 2: For each FD XA in G, create a schema XA
Step 3: If no schema created so far contains a key, add a key as a schema
Step 4: Remove schemas that are contained in other schemas
The result is a decomposition into 3NF that is dependency preserving and has a lossless join

66. Example
Find a decomposition into 3NF for the relation R = ABCDEFGH, with the functional dependencies F = {AB, ABCDE, EFGH, ACDFEG}

67. Example Non-redundant cover G = {AB, ACDE, EFG, EFH} Key ACDF
Schema: AB, ACDE, EFG, EFH, ACDF