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Normalization. ISYS 464. Database Design Based on ERD. Strong entity: Create a table that includes all simple attributes Composite Weak entity: add owner primary key Multi-valued attribute: Create a table for each multi-valued attribute Key + attribute Relationship: 1:1, 1:M - PowerPoint PPT Presentation
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Normalization
ISYS 464
Database Design Based on ERD • Strong entity: Create a table that includes all simple attributes
– Composite• Weak entity: add owner primary key• Multi-valued attribute: Create a table for each multi-valued
attribute– Key + attribute
• Relationship: – 1:1, 1:M
• Relationship table: for partial participation to avoid null• Foreign key
– M:M: relationship table– N-ary relationship: relationship table– Recursive relationship
• Attribute of relationship• Superclass and subclass• Note: The database designed according to these
rules will meet the 3NF requirements.
Top-Down vs Bottom-Up
• Top-Down database design:– Users’ requirements: forms,reports,views, etc.– Data model: ERD– Relational database design
• Bottom-Up database design:– Users’ requirements: forms,reports,views, etc.– Universal relation: A relation contains all attributes
required by users and relationships between attributes.• View aggregation
– Relational database design
ExampleEmployee/Dependent report:
EmpID: E101 Ename: Peter
Address: 123 XYZ St
DependentName Relationship DOB
Nancy Daughter 1/1/95
Alan Son 12/25/03
EmpDependent Table:
EmpID EmpName Address DepName Relation DepDOBE101 Peter 123 XYZ St Nancy D 1/1/95
E101 Peter 123 XYZ St Alan S 12/25/03
Note: This database is able to produce the report, but has duplicated data.
Update Anomalies Due To Duplication
• Modification anomaly:– Inconsistent data
• Insertion Anomalies:– Enter an employee with no dependent– Null
• Deletion Anomaly:– If Nancy and Alan become independent.
Example
• Employee Table:– SSN, Ename, Sex, DOB, Phone– Employee may have more than 1 phone.
• If Peter has two phones, 7890 and 7892:– SSN, Ename, Sex, DOB, Phone– 1234 Peter M 7/4/75 7890– 1234 Peter M 7/4/75 7892
Normalization• Decompose unsatisfactory relation into smaller
relations with desirable properties.– No duplication
• The original relation can be recovered by applying natural join to the smaller relations.– So that no information is lost in the process.
• Keys and function dependency:– Which field is the key field of the EMpDependent
Table?• EmpID + DepName
Function Dependency
• Relationship between attributes
• X -> Y– The value of X uniquely determines the value
of Y.– Y is functionally dependent on X.– A value of X is associated with only one value
of Y.
Example• Employee table:
– SSN Ename Sex DOB– S1 Peter M 1/1/75– S2 Paul M 12/25/80– S3 Mary F 7/4/72
• Function Dependencies:– SSN -> Ename, SSN ->Sex, SSN -> DOB– SSN -> Ename, Sex, DOB
• Any other FD:– Ename -> SSN ?– Ename -> Sex ?– DOB -> SSN ?
• What is the key of Employee table:– SSN
• Observations:– All non-key fields are functionally dependent on SSN.
– There is no other FD.
– The only FD is the key dependency.
– There is no data duplication in the Employee table.
If we mix multivalue attribute with regular attributes in one table
• Employee Table:– SSN, Ename, Sex, DOB, Phone– Employee may have more than 1 phone.
• FD:– SSN -> Ename, Sex, DOB– SSN -> Phone ?
• Key: • Duplication ?
Example 2
• EmpDependent table:– EmpID, Ename, Address, Depname, Relation,
DepDOB
• FD:– EmpID ->Ename, Address
• Key: EmpID + Depname
If we mix two entities with 1:M relationship in one table
• FacultyStudent table:– Faculty Advise Student: 1:M relationship– FID, Fname, SID, Sname, SAddress
• FD:– FID -> Fname– SID -> Sname, SAddress, FID, Fname
• Key: SID
If we mix two entities with M:M relationship in one table
• BankAccount table:– Acct#, AcctType, CID, Cname, Address, OpenDate,
Balance• Function Dependencies:
– Acct# -> AcctType Y– Acct# -> CID N– Acct# -> Cname N– Acct# -> OpenDate Y– Acct# ->Balance Y– CID -> Cname, Address Y
• Key: Acct# + CID• Duplication ? Y
Normalization Process
• Inputs: – A “universal relation”
– Function dependencies
• Output: Normalized tables• Process:
– Decompose the unnormalized relation into smaller relations such that in each relation the non key fields are functionally dependent on the key, the whole key, and nothing but the key. So help me Codd!
First Normal Form
• The fields of a relation are all simple attribute.– All relational database tables meet this
requirement.
• EmpDependent table:– EmpID, Ename, Address, Depname, Relation, DepDOB
– First normal form? Yes
– Second normal form?
Second Normal Form
• The non-key fields are functionally dependent on the key, and the whole key.– FD:
• EmpID ->Ename, Address
– Key: EmpID + Depname – Ename and Address depend on part of the key.
• Every non-key field is fully functionally dependent on the key.
• Decompose the EMpDependent table into two tables:– EmpID, Ename, Address– EmpID, Depname, Relation, DepDOB
• Employee Table:– SSN, Ename, Sex, DOB, Phone– Employee may have more than 1 phone.
• FD:– SSN -> Ename, Sex, DOB,– SSN -> Phone ?
• Key: SSN + Phone• 2NF? No• Decompose into two tables:
– SSN, Ename, Sex, DOB– SSN, Phone
• FacultyStudent table:– Faculty Advise Student: 1:M relationship– FID, Fname, Office, SID, Sname, SAddress
• FD:– FID -> Fname, Office– SID -> Sname, SAddress, FID, Fname, Office
• Key: SID• 2NF ? Yes• Duplication? Yes• Why?
– All non-key fields depend on the whole key, but not Nothing But the Key!
• SID -> FID, Fname, Office• FID -> Fname, Office
Transitive Dependency
• If X -> Y, and Y->Z then X -> Z.
• Z if transitively dependent on the key.
• SID -> FID, FID -> Fname, Office– SID -> Fname, Office– Fname and Office are transitively dependent on
SID.
Third Normal Form
• Every non-key field is:– Fully functionally dependent on the key, and– Non-transitively dependent on the key.
• Decompose:– FID, Fname, Office– SID, FID, Sname, SAddress
ExampleCustomer/Orders report:
CID: C101 Cname: Peter
Address: 123 XYZ St
OID Odate SalesPerson Amount
O25 1/1/04 John 125
O30 2/25/04 Alan 500
CustomerOrders Table:CID CName Address OID Odate SalesPerson Amount
C101 Peter 123 XYZ St O25 1/1/04 John 125
C101 Peter 123 XYZ St O30 2/25/04 Alan 500
Example
• Key: OID• FD:
– OID -> CID, Cname, Address, Odate, SalesPerson, Amount– CID -> Cname, Address
• 2NF? Yes• 3 NF? No• Decompose:
– CID, Cname, Address– OID, CID, Odate, SalesPerson, Amount
Example with 1:M Relationship
• FacultyStudent table:– Faculty Advise Student: 1:M relationship– FID, Fname, SID, Sname, SAddress
• FD: – FID -> Fname– SID -> Sname, Saddress
• Key: SID• 2NF? Yes• 3NF? No, because SID ->FID, FID -> Fname• Decompose:
– Table 1: FID, Fname– Tablw 2: SID, FID, Sname, SAddress
Example with M:M Relationship
• BankAccount table:– Acct#, AcctType, CID, Cname, Address, OpenDate,
Balance• Function Dependencies:
– Acct# -> AcctType, OpenDate, Balance– CID -> Cname, Address
• Key: Acct# + CID• 2NF? No
– Decompose: • Table 1: Acct# -> AcctType, OpenDate, Balance• Table 2: CID -> Cname, Address• Table 3: Acct# , CID
• 3NF? Yes
Is It Really A Ternary Relationship?
Supplier Project
Part
Qty
If each part is supplied by only one supplier?
• Relationship Table:– SID, PrtID, PjID, Qty
• Key: PrtID + PjID -> SID, Qty• FD: PrtID SID• 2ND NF? No• Decompose:
– Table1: PrtID, SID– Table2: PrtID, PjID, Qty
Database Design Based on ERD • Strong entity: Create a table that includes all simple attributes
– Composite• Weak entity: add owner primary key• Multi-valued attribute: Create a table for each multi-valued
attribute– Key + attribute
• Relationship: – 1:1, 1:M
• Relationship table: for partial participation to avoid null• Foreign key
– M:M: relationship table– N-ary relationship: relationship table– Recursive relationship
• Attribute of relationship• Superclass and subclass• Note: The database designed according to these
rules will meet the 3NF requirements.
Boyce-Codd Normal Form
• 3NF: All non-key attributes are dependent on the key and nothing else.
• Boy-Codd Normal Form: – A relation is in 3NF, and– Key attributes cannot be dependent on non-key
attributes.• Whenever X -> A holds, then X is a key.
Example
• Relation: A, B, C
• FD:– A, B C– C B
• Key: A, B
• It is in 3 NF, but not in BC NF.
Employee Part
Warehouse
Stock
M
MM1
We also know each part in a warehouse is managed by one employee, and an employee may manage many parts.
• Table:– WID, PID, EID, Stock– LA P1 E5 40– LA P5 E6 50– LA P7 E5 45
• Key: WID + PID EID, Stock• FD: EID WID• 3 NF? Yes• BC NF? No• Decompose:
– Table 1: EID, WID– Table 2: EID, PID, Stock
Alternative Approach
Key: EID + PID WID, Stock
FD: EID WID
2NF? No
Table 1: EID, WID
Table 2: EID, PID, Stock
Potential violation of BCNF may occur when:1. The relation has more than 1 composite candidate keys;2. The candidate keys have at least one attribute in common.
1. WID + PID2. EID + PID
Multi-Valued Dependency
• A -- >> B
• A -- >> C
• For each value of A there is a set of values for B and a set of values for C. However, the set of values for B and C are independent of each other.
Fourth Normal Form
• A relation is in Boyce-Codd normal form and does not contain multi-valued dependencies.
• Student: A student may have many majors and many phones.– SID, Sname, Major, Phone– S1 Peter CIS 1234– S1 Peter CIS 3456– S1 Peter Acct 1234– S1 Peter Acct 3456
• Key: SID + Major + Phone• FD: SID Sname• 2NF? No• Decompose:
– Table 1: SID, Sname– Table 2: SID, Major, Phone
• Table 2 in 3NF? Yes; in 4 NF? No– SID, Major– SID, Phone
Denormalization
• The refinement to the relational schema such that the degree of normalization for a modified relation is less than the degree of at least one of the original relations.
• Objective:– Speed up processing
