Keys and Functional Dependency

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Keys and Functional Dependency
CS157B
Lecture 7

• Prof. Sin-Min Lee
• Department of Computer Science
• San Jose State University

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Data Normalization

• Primarily a tool to validate and improve a
  logical design so that it satisfies certain
  constraints that avoid unnecessary duplication of
  data.
• The process of decomposing relations with
  anomalies to produce smaller, well-structured
  relations.
• Primary Objective Reduce Redundancy,Reduce
  nulls,
• Improve modify activities
• insert,
• update,
• delete,
• but not read
• Price degraded query, display, reporting

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Functional Dependency and Keys

• Functional Dependency The value of one attribute
  (the determinant) determines the value of another
  attribute.
• Candidate Key Each non-key field is functionally
  dependent on every candidate key.

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Functional dependency

• a constraint between two attributes (columns) or
  two sets of columns
• A ? B if for every valid instance of A, that
  value of A uniquely determines the value of B
• or A ?B if there exists at most one value of B
  for every value of A

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Functional Dependencies
R
X Y Z

• FDs defined over two sets of attributes X, Y
  Ì R
• Notation X à Y reads as X determines Y
• If X à Y, then all tuples that agree on X must
  also agree on Y

1 2 3 2 4 5 1 2 4 1 2 7 2 4 8 3 7 9
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Functional Dependencies (example)
X Y Z
X Y Z
1 2 3 2 4 5 1 2 4 1 2 7 2 4 8 3 7 9
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functional dependency

• some examples
• SSN ? Name, Address, Birthdate
• VIN ? Make, Model, Color
• note the LHS is the determinant
• so functional dependency is the technical term
  for determines

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Candidate Keys

• an attribute (or set of attributes) that uniquely
  identifies a row
• primary key is a special candidate key
• values cannot be null
• e.g.
• ENROLL (Student_ID, Name, Address, )
• PK Student_ID
• candidate key Name, Address

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Armstrongs Axioms

• For computing the set of FDs that follow a given
  FD, the
• following rules called Armstrongs axioms are
  useful
• Reflexivity If B ? A, then A ? B
• Augmentation If A ? B, then A ? C ? B ? C
  Note also that if A ? B, then A ? C ? B for any
  set of attributes C.
• Transitivity If A ? B and B ? C then A ? C

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Example of 3NF but not BCNF

• R(A B C D)
• 1 2 1 2
• 1 3 2 1
• 2 1 1 1
• 1 3 2 2
• 2 1 3 2
• Is this table BCNF?

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Projecting FDs

• Given a relation R (A,B,C,D) and F(R) A?B,
  B?C, C?D.
• Suppose S is projected from R as S(A,C,D). What
  is F(S).
• To compute F(S), start by computing the closures
  of all attributes
• in S.
• In R, A A?B, A?C, A?D
• In S, A A?C, A?D
• C C?D and
• D D
• Since A contains all attributes of S, it is not
  required to compute
• (AC), (AD) or (ACD).

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Inference Rules for FDs
A1, A2, , An ? B1, B2, , Bm
Splitting rule and Combining rule
Is equivalent to
A1, A2, , An ? B1 A1, A2, , An ? B2 . . . .
. A1, A2, , An ? Bm
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Inference Rules for FDs(continued)
Trivial Rule
A1, A2, , An ? Ai
where i 1, 2, ..., n
Why ?
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Inference Rules for FDs(continued)
Transitive Closure Rule
A1, A2, , An ? B1, B2, , Bm
If
and
B1, B2, , Bm ? C1, C2, , Cp
A1, A2, , An ? C1, C2, , Cp
then
Why ?
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Example (continued)
1. name ? color 2. category ? department 3.
color, category ? price
Start from the following FDs
Infer the following FDs
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Another Rule
Augmentation
A1, A2, , An ? B
If
then
A1, A2, , An , C1, C2, , Cp ? B
Augmentation follows from trivial rules and
transitivityHow ?
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Problem infer ALL FDs

• Given a set of FDs, infer all possible FDs
• How to proceed ?
• Try all possible FDs, apply all 3 rules
• E.g. R(A, B, C, D) how many FDs are possible ?
• Drop trivial FDs, drop augmented FDs
• Still way too many
• Better use the Closure Algorithm (next)

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Closure of a set of Attributes
Given a set of attributes A1, , An The
closure, A1, , An , is the set of attributes
Bs.t. A1, , An ? B
name ? color category ? department color,
category ? price
Example
Closures name name, color
name, category name, category, color,
department, price color color
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Closure Algorithm
Start with XA1, , An. Repeat until X doesnt
change do if B1, , Bn ? C is a FD
and B1, , Bn are all in X then
add C to X.
Example
name ? color category ? department color,
category ? price
name, category name, category,
color, department, price
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Example
A, B ? C A, D ? E B ? D A, F ? B
R(A,B,C,D,E,F)
Compute A,B X A, B,
Compute A, F X A, F,

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Using Closure to Infer ALL FDs
Example
A, B ? CA, D ? B B ? D
Step 1 Compute X, for every X
A A, B BD, C C, D D AB ABCD,
AC AC, AD ABCD ABC ABD ACD ABCD
(no need to compute why ?) BCD BCD, ABCD
ABCD
Step 2 Enumerate all FDs X ? Y, s.t. Y ? X and
X?Y ?
AB ? CD, AD?BC, ABC ? D, ABD ? C, ACD ? B
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Problem Finding FDs

• Approach 1 During Database Design
• Designer derives them from real-world knowledge
  of users
• Problem knowledge might not be available
• Approach 2 From a Database Instance
• Analyze given database instance and find all FDs
  satisfied by that instance
• Useful if designers dont get enough information
  from users
• Problem FDs might be artifical for the given
  instance

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Find All FDs
Do all FDsmake sensein practice ?
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Answer
Course ? Dept, Room Dept, Room ? Course Student,
Dept ? Course, Room Student, Course ? Dept,
Room Student, Room ? Dept, Course
Do all FDsmake sensein practice ?
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Keys

• A key is a set of attributes A1, ..., An s.t. for
  any other attribute B, we have A1, ..., An ? B
• A minimal key is a set of attributes which is a
  key and for which no subset is a key
• Note book calls them superkey and key

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Computing Keys

• Compute X for all sets X
• If X all attributes, then X is a key
• List only the minimal keys
• Note there can be many minimal keys !
• Example R(A,B,C), AB?C, BC?AMinimal keys AB
  and BC

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Examples of Keys

• Product(name, price, category, color)
• name, category ? price
• category ? color
• Keys are name, category and all supersets
• Enrollment(student, address, course, room, time)
• student ? address
• room, time ? course
• student, course ? room, time
• Keys are

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candidate key

• a candidate key must satisfy
• unique identification.
• implies that each nonkey attribute is
  functionally dependent on the key (for not(A ? B)
  to be true, A must occur more than once (with a
  different B), or A must map to more than one B in
  a given row)
• nonredundancy
• no attribute in the key can be deleted and still
  be unique
• minimal set of columns (Simsion)

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keys and dependencies
EMPLOYEE1 (Emp_ID, Name, Dept_Name, Salary)
determinant
functional dependency
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EMPLOYEE2 (Emp_ID, Course_Title, Name,
Dept_Name, Salary, Date_Completed)
not fully functionally dependant on the primary
key
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Keys

• The difference between a key and a superkey is
  that a key has to be minimal.
• Example
• SSN is a key for EMPLOYEE, whereas SSN,
  SSN,ENAME, SSN, ENAME, BDATE are all
  superkeys.

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Keys

• one of the candidate keys is designated to be the
  primary key.
• Each relation schema must have a primary key.
• For example, SSN is the only candidate key for
  EMPLOYEE, so it is also the primary key.

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R(A B C D E)

• FD1. A -gt C
• FD2. BC -gtD
• FD3. E -gtAB
• result A
• By FD1. A -gt C A? result
• result A, C
• By FD2. BC -gt D BC? result
• ?result A, C
• By FD3. E -gtAB E? result
• ?result A, C
• ? A A, C

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• Similarly B B
• C C
• D D
• E E,A,B,C,D
• E is a candidate key
• Now, we see
• AB ABCD AC AC AD
  ACD
• BC BCD BD BD CD
  CD
• ABC ABCD ABD ABCD BCD
  BCD
• ACD ACD

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What is the largest normal form of this table?
R(A B C D E)
FD1. A -gtC FD2. BC -gtD FD3. E -gtAB Answer E
is the only candidate key of R The non-prime
attributes are A, B, C, D As FD!. A-gtC, we have
transitive dependency. Thus R(ABCD) is 2NF but
not 3NF
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What is Normalization?

• The purpose of normalization is to produce a
  stable set of relations that is a faithful model
  of the operations of the enterprise. By following
  the principles of normalization, we can achieve a
  design that is highly flexible, allowing the
  model to be extended when needed to account for
  new attributes, entity sets, and relationships.

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Normal Forms

• A relation is in specific normal form if it
  satisfies the set of requirements or constraints
  for that form. All of the normal forms are nested
  in that each satisfies the constraints of the
  previous one but is a "better" form because each
  eliminates flaws found in the previous

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Steps in Normalization
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2NF

• a relation is in second normal form if it is in
  first normal form AND every nonkey attribute is
  fully functionally dependant on the primary key
• i.e. remove partial functional dependencies, so
  no nonkey attribute depends on just part of the
  key

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EMPLOYEE2 (Emp_ID, Course_Title, Name,
Dept_Name, Salary, Date_Completed)
not fully functionally dependant on the primary
key
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Second Normal Form ( 2NF )

• it is based on the concept of full functional
  dependency.
• A functional dependency X?Y is a full functional
  dependency , for any attribute A ? X, X - A
  ? Y.

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Second Normal Form

• A relation is in second normal form (2NF) if and
  only if it is in first normal form and all the
  nonkey attributes are fully functionally
  dependent on the key.

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Second Normal Form

• A table is in second normal form (2NF) if
• It is in 1NF
• It includes no partial dependencies. No
  attribute is dependent on only a portion of the
  primary key.

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2NF

• a relation is in 2NF if it is in 1NF and any one
  of these is true
• the PK consists of only 1 attribute
• all attributes are part of the PK (no nonkey
  attributes)
• every nonkey attribute is functionally dependant
  on the whole PK

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2NF (Example)
A B C D
2 Candidate Keys
R with keyAB is NOT 2NF
R with keyAC is NOT 2NF
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Second Normal Form ( 2NF )
fd1
fd2
fd3

• SSN, PNUMBER?HOURS is a fully dependency
  (neither SSN?HOURS nor PNUMBER?HOURS holds).

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Second Normal Form ( 2NF )
EMP_PROJ
fd1
fd2
fd3
2NF NORMALIZATION
EP2
EP3
EP1
fd2
fd1
fd3

• The functional dependencies fd1,fd2,fd3 lead to
  the decomposition of EMP_PROJ into the three
  relation schemas EP1,EP2,EP3, each of which is in
  2NF.

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1NF ?2NF

• EMPLOYEE2 (Emp_ID, Course_Title, Name,
  Dept_Name, Salary, Date_Completed)
• ?
• EMPLOYEE1 (Emp_ID, Name, Dept_Name, Salary)
• and
• EMP_COURSE (Emp_ID, Course_Title, Date_Completed)
• EMPLOYEE1 satisfies condition1
• EMP_COURSE satisfies condition3

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3NF

• a relation is in third normal form if it is in
  2NF, AND no transitive dependencies exist
• transitive dependency is a functional dependency
  between nonkey attributes

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transitive dependency
transitive dependency
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transitive dependency

• same problems
• insertion anomaly (no salesman without a
  customer)
• deletion anomaly (if a salesman is assigned to
  only 1 customer, and the customer is deleted, we
  lose the salesman!)
• modification (update) anomaly (reassign
  salesperson to region)

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Converting to 2NF

• To convert from 1NF to 2NF, list each key
  component and then the key itself.
• Each component will become the key in a new table.

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Our Example in 2NF
Table Name PROJECT
PROJ_NUM
PROJ_NAME
Table Name EMPLOYEE
CHG_HOUR
EMP_NUM
EMP_NAME
JOB_CLASS
Table Name ASSIGN
HOURS
PROJ_NUM
EMP_NUM
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Problems with 2NF

• Transitive Dependency
• An attribute that is dependent on a non-prime
  attribute exhibits transitive dependency.
• Still leads to data anomalies.

CHG_HOUR
EMP_NUM
EMP_NAME
JOB_CLASS
Our example contains the transitive
dependency JOB_CLASS -----gt CHG_HOUR
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Second Normal Form

• Second normal form
• Let R be a relation, and let F be the set of
  governing FDs. An attribute belongs to R is
  prime if a key of R contains A. In other words,
  A is prime in R if there exists KltR such that
  (1) K-gtR,
• (2) for all B belongs to K, (K-B)-gtR not
  belongs to F, and
• (3) A belongs to K

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Third Normal Form

• Third normal form
• Let R be a relation, a subset of the universal
  relation, in the context of a set of FDs F. R
  satisfies third normal form if for every
  nontrival X-gtA belong to F, either
• (1). X is superkey for R or
• (2). A is a prime attribute in R.

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Third Normal Form ( 3 NF )

• Third normal form is based on the concept of
  transitive dependency.
• A functional dependency X?Y in a relation schema
  R is a transitive dependency if there is a set of
  attributes Z that is not a subset of any key of
  R, and both X?Z and Z?Y hold.

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Third Normal Form ( 3 NF )
EMP_DEPT
ENAME SSN BDATE ADDRESS DNUMBER DNAME
DMGRSSN

• example
• the dependency SSN?DMGRSSN is transitive through
  DNUMBER in EMP_DEPT, because both the
  dependencies SSN?DNUMBER and DNUMBER?DMGRSSN hold
  and DNUMBER is not a subset of the key of
  EMP_DEPT.

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General Definitions of Second and Third Normal
Forms

• A relation schema R is in second normal form
  (2NF) if every nonprime attribute A in R is fully
  functionally dependent on every key of R.
• A relation schema R is in 3NF if, whenever a
  functional dependency X? A holds in R, either
• (a) X is a superkey of R.
• (b) A is a prime attribute of R.

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