L9 design2

Relational Database Design (II)
1

Roadmap of This Lecture
 Algorithms for Functional Dependencies (cont’d)
 Decomposition Using Multi-valued Dependencies
 More Normal Form
 Database-Design Process
 Modeling Temporal Data
2

Boyce-Codd Normal Form (Reminder)
    is trivial (i.e.,   )
  is a superkey for R
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:
Example schema not in BCNF:
instr_dept (ID, name, salary, dept_name, building, budget )
because dept_name building, budget
holds on instr_dept, but dept_name is not a superkey
3

Testing for BCNF
 To check if a non-trivial dependency   causes a violation of BCNF
1. compute + (the attribute closure of ), and
2. verify that it includes all attributes of R, that is, it is a superkey of R.
 Simplified test: To check if a relation schema R is in BCNF, it suffices
to check only the dependencies in the given set F for violation of BCNF,
rather than checking all dependencies in F+.
 If none of the dependencies in F causes a violation of BCNF, then
none of the dependencies in F+ will cause a violation of BCNF
either.
 However, simplified test using only F is incorrect when testing a
relation in a decomposition of R
 Consider R = (A, B, C, D, E), with F = { A  B, BC  D}
 Decompose R into R1 = (A,B) and R2 = (A,C,D, E)
 Neither of the dependencies in F contain only attributes from
(A,C,D,E) so we might be mislead into thinking R2 satisfies
BCNF.
 In fact, dependency AC  D in F+ shows R2 is not in BCNF.
4

Testing Decomposition for BCNF
 To check if a relation Ri in a decomposition of R is in BCNF,
 Either test Ri for BCNF with respect to the restriction of F to Ri
(that is, all FDs in F+ that contain only attributes from Ri)
 or use the original set of dependencies F that hold on R, but with
the following test:
– for every set of attributes   Ri, check that + (the
attribute closure of ) either includes no attribute of Ri- ,
or includes all attributes of Ri.
 If the condition is violated by some    in F, the
dependency
  (+ -  )  Ri
can be shown to hold on Ri, and Ri violates BCNF.
 We use above dependency to decompose Ri
5

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.
6

Example of BCNF Decomposition
 R = (A, B, C )
F = {A  B
B  C}
Key = {A}
 R is not in BCNF (B  C but B is not superkey)
 Decomposition
 R1 = (B, C)
 R2 = (A, B)
7

Example of BCNF Decomposition
 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)
8

BCNF Decomposition (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.
9

BCNF and Dependency Preservation
 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
It is not always possible to get a BCNF decomposition that is
dependency preserving
10

Third Normal Form: Motivation
 There are some situations where
 BCNF is not dependency preserving, and
 efficient checking for FD violation on updates is important
 Solution: define a weaker normal form, called Third
Normal Form (3NF)
 Allows some redundancy (with resultant problems; we will
see examples later)
 But functional dependencies can be checked on individual
relations without computing a join.
 There is always a lossless-join, dependency-preserving
decomposition into 3NF.
11

Third Normal Form (Reminder)
 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)
12

3NF Example
 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
13

 There is some redundancy in this schema
 Example of problems due to redundancy in 3NF
 R = (J, K, L)
F = {JK  L, L  K }
Redundancy in 3NF
J
j1
j2
j3
null
L
l1
l1
l1
l2
K
k1
k1
k1
k2
 repetition of information (e.g., the relationship l1, k1)
 (i_ID, dept_name)
 need to use null values (e.g., to represent the relationship
l2, k2 where there is no corresponding value for J).
 (i_ID, dept_nameI) if there is no separate relation mapping
instructors to departments
14

Testing for 3NF
 Optimization: Need to check only FDs in F, need not check all FDs in
F+.
 Use attribute closure to check for each dependency   , if  is a
superkey.
 If  is not a superkey, we have to verify if each attribute in  is
contained in a candidate key of R
 this test is rather more expensive, since it involves finding
candidate keys
 testing for 3NF has been shown to be NP-hard
 Interestingly, decomposition into third normal form (described
shortly) can be done in polynomial time
15

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 begin /* delete Rj */
Rj = Rk;
i= i -1;
end
return (R1, R2, ..., Ri)
16

3NF Decomposition Algorithm (Cont.)
 Above algorithm ensures:
 each relation schema Ri is in 3NF
 decomposition is dependency preserving and lossless-join
 Proof of correctness? (See textbook)
17

3NF Decomposition: An Example
 Relation schema:
cust_banker_branch = (customer_id, employee_id, branch_name, type )
 The functional dependencies for this relation schema are:
1. customer_id, employee_id  branch_name, type
2. employee_id  branch_name
3. 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
18

3NF Decompsition 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)
(customer_id, branch_name, employee_id)
19

Comparison of BCNF and 3NF
 It is always possible to decompose a relation into a set of relations
that are in 3NF such that:
 the decomposition is lossless
 the dependencies are preserved
 It is always possible to decompose a relation into a set of relations
that are in BCNF such that:
 the decomposition is lossless
 it may not be possible to preserve dependencies.
20

Design Goals
 Goal for a relational database design is:
 BCNF.
 Lossless join.
 Dependency preservation.
 If we cannot achieve this, we accept one of
 Lack of dependency preservation
 Redundancy due to use of 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.
21

Multivalued Dependencies
 Suppose we record names of children, and phone numbers for
instructors:
 inst_child(ID, child_name)
 inst_phone(ID, phone_number)
 If we were to combine these schemas to get
 inst_info(ID, child_name, phone_number)
 Example data:
(99999, David, 512-555-1234)
(99999, David, 512-555-4321)
(99999, William, 512-555-1234)
(99999, William, 512-555-4321)
 This relation is in BCNF
 Why?
22

Multivalued Dependencies (MVDs)
 Let R be a relation schema and let   R and   R. The
multivalued dependency
  
holds on R if in any legal relation r(R), for all pairs of tuples t1 and t2 in
r such that t1[] = t2 [], there exist tuples t3 and t4 in r such that:
t1[] = t2 [] = t3 [] = t4 []
t3[] = t1 []
t3[R – ] = t2[R – ]
t4 [] = t2[]
t4[R – ] = t1[R – ]
23

MVD (Cont.)
 Tabular representation of   
24

Example
 Let R be a relation schema with a set of attributes that are partitioned
into 3 nonempty subsets.
Y, Z, W
 We say that Y  Z (Y multidetermines Z )
if and only if for all possible relations r (R )
< y1, z1, w1 >  r and < y1, z2, w2 >  r
then
< y1, z1, w2 >  r and < y1, z2, w1 >  r
 Note that since the behavior of Z and W are identical it follows that
Y  Z if Y  W
25

Example (Cont.)
 In our example:
ID  child_name
ID  phone_number
 The above formal definition is supposed to formalize the notion that given
a particular value of Y (ID) it has associated with it a set of values of Z
(child_name) and a set of values of W (phone_number), and these two
sets are in some sense independent of each other.
 Note:
 If Y  Z then Y  Z
 Indeed we have (in above notation) Z1 = Z2
The claim follows.
26

Use of Multivalued Dependencies
 We use multivalued dependencies in two ways:
1. To test relations to determine whether they are legal under a
given set of functional and multivalued dependencies
2. To specify constraints on the set of legal relations. We shall
thus concern ourselves only with relations that satisfy a given
set of functional and multivalued dependencies.
 If a relation r fails to satisfy a given multivalued dependency, we can
construct a relations r that does satisfy the multivalued
dependency by adding tuples to r.
27

Theory of MVDs
 From the definition of multivalued dependency, we can derive the
following rule:
 If   , then   
That is, every functional dependency is also a multivalued dependency
 The closure D+ of D is the set of all functional and multivalued
dependencies logically implied by D.
 We can compute D+ from D, using the formal definitions of
functional dependencies and multivalued dependencies.
 We can manage with such reasoning for very simple multivalued
dependencies, which seem to be most common in practice
 For complex dependencies, it is better to reason about sets of
dependencies using a system of inference rules.
28

Inference rules for MVDs
1. Reflexivity rule. If α is a set of attributes, and β ⊆ α, then α → β
holds.
2. Augmentation rule. If α → β holds, and γ is a set of attributes, then
γα → γβ holds.
3. Transitivity rule. If α → β holds, and β → γ holds, then α → γ holds.
4. Complementation rule. If α →→ β holds, then α →→ R − β − α
holds.
5. Multivalued augmentation rule. If α →→ β holds, and γ ⊆ R and δ ⊆
γ, then γα →→ δβ holds.
6. Multivalued transitivity rule. If α →→ β holds, and β →→ γ holds,
then α →→ γ − β holds.
7. Replication rule. If α → β holds, then α →→ β.
8. Coalescence rule. If α →→ β holds, and γ ⊆ β, and there is a δ such
that δ ⊆ R, and δ ∩ β = ∅, and δ → γ, then α → γ holds.
29

Fourth Normal Form
 A relation schema R is in 4NF with respect to a set D of functional and
multivalued dependencies if for all multivalued dependencies in D+ of
the form   , where   R and   R, at least one of the following
hold:
    is trivial (i.e.,    or    = R)
  is a superkey for schema R
 If a relation is in 4NF it is in BCNF
30

Restriction of Multivalued Dependencies
 The restriction of D to Ri is the set Di consisting of
 All functional dependencies in D+ that include only attributes of Ri
 All multivalued dependencies of the form
  (  Ri)
where   Ri and    is in D+
31

4NF Decomposition Algorithm
result: = {R};
done := false;
compute D+;
Let Di denote the restriction of D+ to Ri
while (not done)
if (there is a schema Ri in result that is not in 4NF) then
begin
let    be a nontrivial multivalued dependency that holds
on Ri such that   Ri is not in Di, and ;
result := (result - Ri)  (Ri - )  (, );
end
else done:= true;
Note: each Ri is in 4NF, and decomposition is lossless-join
32

Example
 R =(A, B, C, G, H, I)
D ={ A  B
B  HI
CG  H }
 R is not in 4NF since A  B and A is not a super-key for R
 Decomposition
a) R1 = (A, B) (R1 is in 4NF)
b) R2 = (A, C, G, H, I) (R2 is not in 4NF, decompose into R3 and R4)
c) R3 = (C, G, H) (R3 is in 4NF)
d) R4 = (A, C, G, I) (R4 is not in 4NF, decompose into R5 and R6)
 A  B and B  HI
gives A  HI, (MVD transitivity), and
 hence A  I (MVD restriction to R4)
e) R5 = (A, I) (R5 is in 4NF)
f) R6 = (A, C, G) (R6 is in 4NF)
33

Further Normal Forms
 Join dependencies generalize multivalued dependencies
 lead to project-join normal form (PJNF) (also called fifth normal
form)
 A class of even more general constraints, leads to a normal form
called domain-key normal form.
 Problem with these generalized constraints: are hard to reason with,
and no set of sound and complete set of inference rules exists.
 Hence rarely used
34

Overall Database Design Process
 We have assumed schema R is given
 R could have been generated when converting E-R diagram to a set
of tables.
 R could have been a single relation containing all attributes that are
of interest (called universal relation).
 Normalization breaks R into smaller relations.
 R could have been the result of some ad hoc design of relations,
which we then test/convert to normal form.
35

ER Model and Normalization
 When an E-R diagram is carefully designed, identifying all entities
correctly, the tables generated from the E-R diagram should not need
further normalization.
 However, in a real (imperfect) design, there can be functional
dependencies from non-key attributes of an entity to other attributes of
the entity
 Example: an employee entity with attributes
department_name and building,
and a functional dependency
department_name building
 Good design would have made department an entity
 Functional dependencies from non-key attributes of a relationship set
possible, but rare --- most relationships are binary
36

Denormalization for Performance
 May want to use non-normalized schema for performance
 For example, displaying prereqs along with course_id, and title requires
join of course with prereq
 Alternative 1: Use denormalized relation containing attributes of course
as well as prereq with all above attributes
 faster lookup
 extra space and extra execution time for updates
 extra coding work for programmer and possibility of error in extra code
 Alternative 2: use a materialized view defined as
course prereq
 Benefits and drawbacks same as above, except no extra coding work
for programmer and avoids possible errors
37

Other Design Issues
 Some aspects of database design are not caught by normalization
 Examples of bad database design, to be avoided:
Instead of earnings (company_id, year, amount ), use
 earnings_2004, earnings_2005, earnings_2006, etc., all on the
schema (company_id, earnings).
 Above are in BCNF, but make querying across years difficult and
needs new table each year
 company_year (company_id, earnings_2004, earnings_2005,
earnings_2006)
 Also in BCNF, but also makes querying across years difficult and
requires new attribute each year.
 Is an example of a crosstab, where values for one attribute
become column names
 Used in spreadsheets, and in data analysis tools
 Good for display only
38

Modeling Temporal Data
 Temporal data have an association time interval during which the
data are valid.
 A snapshot is the value of the data at a particular point in time
 Several proposals to extend ER model by adding valid time to
 attributes, e.g., address of an instructor at different points in time
 entities, e.g., time duration when a student entity exists
 relationships, e.g., time during which an instructor was associated
with a student as an advisor.
 But no accepted standard
 Adding a temporal component results in functional dependencies like
ID  street, city
not to hold, because the address varies over time
 A temporal functional dependency X  Y holds on schema R if the
functional dependency X  Y holds on all snapshots for all legal
instances r (R).
t
39

Modeling Temporal Data (Cont.)
 In practice, database designers may add start and end time attributes
to relations
 E.g., course(course_id, course_title) is replaced by
course(course_id, course_title, start, end)
 Constraint: no two tuples can have overlapping valid times
– Hard to enforce efficiently
 Foreign key references may be to current version of data, or to data at
a point in time
 E.g., student transcript should refer to course information at the
time the course was taken
40

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