1. Normalization
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2. Why Normalization? To remove potential redundancy in design
Redundancy causes several anomalies: insert, delete and update
Redundancy wastes storage, and often slows down query processing
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3. Insert Anomaly Student sNumber sName pNumber pName s1 Dave p1 MM s2
Greg p2 ER
Question: Could we insert any professor ?
Note: We cannot insert a professor who has no students.
Insert Anomaly: We are not able to insert “valid” value/(s)
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4. Delete Anomaly Student sNumber sName pNumber pName s1 Dave p1 MM s2
Greg p2 ER
Question: Can we delete a student and keep a professor info ?
Note: We cannot delete a student that is the only student of a professor.
Delete Anomaly: We are not able to perform a delete without losing some
“valid” information.
Note: In both cases, minimum cardinality of Professor in the corresponding
ER schema is 0
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5. Update Anomaly Student sNumber sName pNumber pName s1 Dave p1 MM s2
Greg
Question: Can we simply update a professor’s name ?
Note: To update the name of a professor, we have to update in multiple tuples.
Update Anomaly: To update a value, we have to update multiple rows.
Update anomalies are due to redundancy.
Note the maximum cardinality of Professor in the corresponding ER schema is *
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6. Normalization Need a method to find “dependencies” between attributes
Functional dependencies
Need a method to remove such harmful dependencies, when they exist
Relational decomposition
Break R (A,B,C,D) into R1 (A, B) and R2 (B, C, D)
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7. Keys : Revisited
A key for a relation R (a1, a2, …, an) is a set of attributes, K, that together uniquely determine the values for all attributes of R.
A key is minimal: no subset of K is a key.
A superkey may not be minimal
A prime attribute: an attribute that is part of a key
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8. Functional Dependencies (FDs)
Student
sNumber
group
address 1 DB
144FL 2 AI
320FL
Suppose we have the FD: group address
That is, there is a function from group to address
Meaning: For any two rows in the Student relation with the same value for group, the value for address must be same.
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9. FD and Keys
Student
sNumber
group
address 1 DB
144FL 2 AI
320FL
Primary Key : <sNumber>
FD : group  address
Questions :
Does a key implies functional dependencies? Which ones ?
Does a functional dependency imply keys ? Which ones ?
We assume NO NULL values here.
Observation :
Any key (primary or candidate) or superkey of a relation R functionally determines all attributes of R.
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10. Properties of FDs Consider A, B, C, Z are sets of attributes
Reflexive (trivial FD): if A  B, then A  B
Transitive: if A  B, and B  C, then A  C
Augmentation: if A  B, then AZ  BZ
Union: if A  B, A  C, then A  BC
Decomposition: if A  BC, then A  B, A  C
Note: Sound and complete inference rules for FDs
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11. Inferring FDs Suppose we have : Questions :
a relation R (A, B, C) and
functional dependencies A  B, B  C, C  A
Questions :
What is a key for R?
Should we split R into multiple relations?
We can infer A  ABC, B  ABC, C  ABC.
Hence A, B, C are all keys.
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12. Reasoning About FDs
An FD f is implied by a set of FDs F if f holds whenever all FDs in F hold.
Closure of F, denoted by F+, is the set of all FDs that are implied by F.
Computing closure F+ of a set of FDs can be expensive.
Size of closure is exponential in # attrs!
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13. Reasoning About FDs But given question :
Is X  Y in closure of a set of FDs F?
Fortunately, computing just attribute closure is sufficient (and linear time complexity)
Compute attribute closure of X, denoted X+, wrt F:
Set of all attributes A such that X  A is in F+
Check if Y is in X+ . If yes, then X  Y in F+.
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14. Reasoning About FDs (Contd.)
Does F = {A B, B C, C D E } imply A E?
Question :
i.e, is A E in the closure F+ ?
Equivalent Question :
Is E in the attribute closure ?
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15. Algorithm for Inference of FDs
Computing the closure of set of attributes {A1, A2, …, An}:
Let X = {A1, A2, …, An}
If there exists a FD : B1, B2, …, Bm  C, such that every Bi  X, then X = X  C
Repeat step 2 until no more attributes can be added.
{A1, A2, …, An}+ = X
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16. Inferring FDs: Example 1
Consider R (A, B, C, D, E)
with FDs A  B, B  C, CD  E
Does A  E? (Is A  E in F+ ?)
Rephrase as : Is E in A+ ?
Let us compute {A}+
{A}+ = {A, B, C}
Therefore, A  E is false
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17. Inferring FDs: Example 2
Given R (A, B, C), and
FDs : A  B, B  C, C  A
What are possible keys for R ?
Compute the closure of attributes:
{A}+ = {A, B, C}
{B}+ = {A, B, C}
{C}+ = {A, B, C}
So keys for R are <A>, <B>, <C>
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18. Decomposing Relations
StudentProf
Greg
Dave
sName p2 p1
pNumber MM s2 s1
pName
sNumber
FDs: pNumber  pName
Greg
Dave
sName p2 p1
pNumber s2 s1
sNumber
Student MM
pName
Professor
Greg
Dave
sName MM
pName S2 S1
sNumber
Student p2 p1
pNumber
Professor
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19. Decomposition Decomposition: Must be Lossless (no spurious tuples)
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20. Decomposition: Lossless Join
Greg
Dave
sName p2 p1
pNumber MM s2 s1
pName
sNumber
Greg
Dave
sName MM
pName S2 S1
sNumber
Student p2 p1
pNumber
Professor
StudentProf
sNumber
sName
pNumber
pName s1
Dave p1 MM p2 s2
Greg
Spurious
Tuples
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21. Normalization
Once decided, what is the algorithm for (lossless) decomposing?
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22. Normalization Step : Decompose
Consider relation R with set of attributes AR. Consider a FD : A  B (such that no other attribute in (AR – A – B) is functionally determined by A).
If A is not a superkey for R, we may decompose R as:
Create R’ with attributes (AR – B)
Create R’’ with attributes A  B
Key for R’’ = A
Foreign key : R’ (A) references R’’ (A)
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23. Example Decomposition Revisited
StudentProf
sNumber
sName
pNumber
pName s1
Dave p1 MM s2
Greg p2
FDs: pNumber  pName
Student
Professor
sNumber
sName
pNumber s1
Dave p1 s2
Greg p2
pNumber
pName p1 MM p2
FOREIGN KEY: Student (PNum) references Professor (PNum)
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24. Normalization How do I decide if I need to further decompose?
Once decided, what is the algorithm for decomposing?
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