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Normalization

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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

Examples to follow next.

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What is Normalization?

Normalization uses concept of dependencies Functional Dependencies

Technique used: Decomposition Break R (A, B, C, D) into R1 (A, B) and R2 (B, C, D)

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Insert Anomaly

sNumber sName pNumber pName

s1 Dave p1 MM

s2 Greg p2 ER

Student

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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Delete Anomaly

sNumber sName pNumber pName

s1 Dave p1 MM

s2 Greg p2 ER

Student

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.

Note: In both cases, minimum cardinality of Professor in the correspondingER schema is 0

Student

sNumber

sName

Professor

pNumber

pName

HasAdvisor

(1,1) (0,1)

years

Delete Anomaly: We are not able to perform a delete without losing some “valid” information.

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Update Anomaly

sNumber sName pNumber pName

s1 Dave p1 MM

s2 Greg p1 MM

Student

Question: Can we simply update a professor’s name ?Note: To update the name of a professor, we have to update in multiple tuples.

Student

sNumber

sName

Professor

pNumber

pName

HasAdvisor

(1,1) (0,*)

years

Note the maximum cardinality of Professor in the corresponding ER schema is *

Update Anomaly: To update a value, we have to update multiple rows. Update anomalies are due to redundancy.

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Normalization

Need a method to find such “dependencies” exist between attributes Functional dependencies

Need a method to remove such harmful dependencies, when they exist Relational decomposition

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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 need not be minimal

A prime attribute: an attribute that is part of a key

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Functional Dependencies (FDs)

sNumber sName address

1 Dave 144FL

2 Greg 320FL

Student

Suppose we have the FD: sName address

That is, there is a function from sName to address

Meaning: For any two rows in the Student relation with the same value for sName, the value for address must be same.

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FD and Keys

sNumber sName address

1 Dave 144FL

2 Greg 320FL

Student

Questions : • Does a key implies functional dependencies? Which ones ?• Does a functional dependency imply keys ? Which ones ?

Observation : Any key (primary or candidate) or superkey of a relation R functionally determines all attributes of R.

Primary Key : <sNumber>FD : sName address

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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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Inferring FDs

Suppose we have : 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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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 is the set of all FDs that

are implied by F.

Computing closure of a set of FDs can be expensive. Size of closure is exponential in # attrs!

F

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Reasoning About FDs

Instead of computing closure F+ of a set of FDs Too expensive

Typically, we just need to know if a given FD X Y is in closure of a set of FDs F.

Algorithm for efficient check: Compute attribute closure of X (denoted ) wrt F:

Set of all attributes A such that X A is in There is a linear time algorithm to compute this.

Check if Y is in X+ . If yes, then X A in F+.

X

F

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Reasoning About FDs (Contd.)

Does F = {A B, B C, C D E } imply A E?

Question : i.e, is A E in the functional set closure ?

Equivalent Question : Is E in the attribute closure ?

A

F

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Algorithm for Inference of FDs

Computing the closure of set of attributes {A1, A2, …, An}, denoted {A1, A2, …, An}+

1. Let X = {A1, A2, …, An}

2. If there exists a FD : B1, B2, …, Bm C, such that every Bi X, then X = X C

3. Repeat step 2 until no more attributes can be added.

4. {A1, A2, …, An}+ = X

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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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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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Decomposing Relations

sNumber sName pNumber pName

s1 Dave p1 MM

s2 Greg p2 MM

StudentProf

FDs: pNumber pName

sNumber sName pNumber

s1 Dave p1

s2 Greg p2

Student

pNumber pName

p1 MM

p2 MM

Professor

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Decomposition

Decomposition:

Must be Lossless (no spurious tuples)

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Decomposition: Lossless Join Property

sNumber sName pName

S1 Dave MM

S2 Greg MM

Student

pNumber pName

p1 MM

p2 MM

Professor

sNumber sName pNumber pName

s1 Dave p1 MM

s1 Dave p2 MM

s2 Greg p1 MM

s2 Greg p2 MM

StudentProf

SpuriousTuples

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Normalization

How do I decide if I need to further decompose?

Once decided, what is the algorithm for decomposing?