Canonical Cover

In the case of updating the database, the responsibility of the system is to check whether the existing functional dependencies are getting violated during the process of updating. In case of a violation of functional dependencies in the new database state, the rollback of the system must take place.

A canonical cover or irreducible set of functional dependencies FD is a simplified set of FD that has a similar closure as the original set FD.

Extraneous attributes

An attribute of an FD is said to be extraneous if we can remove it without changing the closure of the set of FD.

Example: Given a relational Schema R( A, B, C, D) and set of Function Dependency FD = { B → A, AD → BC, C → ABD }. Find the canonical cover?

Solution: Given FD = { B → A, AD → BC, C → ABD }, now decompose the FD using decomposition rule( Armstrong Axiom ).

1. B → A
2. AD → B ( using decomposition inference rule on AD → BC)
3. AD → C ( using decomposition inference rule on AD → BC)
4. C → A ( using decomposition inference rule on C → ABD)
5. C → B ( using decomposition inference rule on C → ABD)
6. C → D ( using decomposition inference rule on C → ABD)

Now set of FD = { B → A, AD → B, AD → C, C → A, C → B, C → D }

The next step is to find closure of the left side of each of the given FD by including that FD and excluding that FD, if closure in both cases are same then that FD is redundant and we remove that FD from the given set, otherwise if both the closures are different then we do not exclude that FD.

Calculating closure of all FD { B → A, AD → B, AD → C, C → A, C → B, C → D }

1a. Closure B+ = BA using FD = { B → A, AD → B, AD → C, C → A, C → B, C → D }

1b. Closure B+ = B using FD = { AD → B, AD → C, C → A, C → B, C → D }

From 1 a and 1 b, we found that both the Closure( by including B → A and excluding B → A ) are not equivalent, hence FD B → A is important and cannot be removed from the set of FD.

2 a. Closure AD+ = ADBC using FD = { B →A, AD → B, AD → C, C → A, C → B, C → D }

2 b. Closure AD+ = ADCB using FD = { B → A, AD → C, C → A, C → B, C → D }

From 2 a and 2 b, we found that both the Closure (by including AD → B and excluding AD → B) are equivalent, hence FD AD → B is not important and can be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → A, C → B, C → D }

3 a. Closure AD+ = ADCB using FD = { B →A, AD → C, C → A, C → B, C → D }

3 b. Closure AD+ = AD using FD = { B → A, C → A, C → B, C → D }

From 3 a and 3 b, we found that both the Closure (by including AD → C and excluding AD → C ) are not equivalent, hence FD AD → C is important and cannot be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → A, C → B, C → D }

4 a. Closure C+ = CABD using FD = { B →A, AD → C, C → A, C → B, C → D }

4 b. Closure C+ = CBDA using FD = { B → A, AD → C, C → B, C → D }

From 4 a and 4 b, we found that both the Closure (by including C → A and excluding C → A) are equivalent, hence FD C → A is not important and can be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → B, C → D }

5 a. Closure C+ = CBDA using FD = { B →A, AD → C, C → B, C → D }

5 b. Closure C+ = CD using FD = { B → A, AD → C, C → D }

From 5 a and 5 b, we found that both the Closure (by including C → B and excluding C → B) are not equivalent, hence FD C → B is important and cannot be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → B, C → D }

6 a. Closure C+ = CDBA using FD = { B →A, AD → C, C → B, C → D }

6 b. Closure C+ = CBA using FD = { B → A, AD → C, C → B }

From 6 a and 6 b, we found that both the Closure( by including C → D and excluding C → D) are not equivalent, hence FD C → D is important and cannot be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → B, C → D }

• Since FD = { B → A, AD → C, C → B, C → D } is the resultant FD, now we have checked the redundancy of the attribute, since the left side of FD AD → C has two attributes, let's check their importance, i.e. whether they both are important or only one.

Closure AD+ = ADCB using FD = { B →A, AD → C, C → B, C → D }

Closure A+ = A using FD = { B →A, AD → C, C → B, C → D }

Closure D+ = D using FD = { B →A, AD → C, C → B, C → D }

Since the closure of AD+, A+, D+ that we found are not all equivalent, hence in FD AD → C, both A and D are important attributes and cannot be removed.

Hence resultant FD = { B → A, AD → C, C → B, C → D } and we can rewrite it as

FD = { B → A, AD → C, C → BD } is Canonical Cover of FD = { B → A, AD → BC, C → ABD }.

Example 2: Given a relational Schema R( W, X, Y, Z) and set of Function Dependency FD = { W → X, Y → X, Z → WXY, WY → Z }. Find the canonical cover?

Solution: Given FD = { W → X, Y → X, Z → WXY, WY → Z }, now decompose the FD using decomposition rule( Armstrong Axiom ).

1. W → X
2. Y → X
3. Z → W ( using decomposition inference rule on Z → WXY )
4. Z → X ( using decomposition inference rule on Z → WXY )
5. Z → Y ( using decomposition inference rule on Z → WXY )
6. WY → Z

Now set of FD = { W → X, Y → X, WY → Z, Z → W, Z → X, Z → Y }

The next step is to find closure of the left side of each of the given FD by including that FD and excluding that FD, if closure in both cases are same then that FD is redundant and we remove that FD from the given set, otherwise if both the closures are different then we do not exclude that FD.

Calculating closure of all FD { W → X, Y → X, Z → W, Z → X, Z → Y, WY → Z }

1 a. Closure W+ = WX using FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY → Z }

1 b. Closure W+ = W using FD = { Y → X, Z → W, Z → X, Z → Y, WY → Z }

From 1 a and 1 b, we found that both the Closure (by including W → X and excluding W → X ) are not equivalent, hence FD W → X is important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY → Z }

2 a. Closure Y+ = YX using FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY → Z }

2 b. Closure Y+ = Y using FD = { W → X, Z → W, Z → X, Z → Y, WY → Z }

From 2 a and 2 b we found that both the Closure (by including Y → X and excluding Y → X ) are not equivalent, hence FD Y → X is important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY → Z }

3 a. Closure Z+ = ZWXY using FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY → Z }

3 b. Closure Z+ = ZYX using FD = { W → X, Y → X, Z → X, Z → Y, WY → Z }

From 3 a and 3 b, we found that both the Closure (by including Z → W and excluding Z → W ) are not equivalent, hence FD Z → W is important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY → Z }

4 a. Closure Z+ = ZXWY using FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY → Z }

4 b. Closure Z+ = ZWYX using FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

From 4 a and 4 b, we found that both the Closure (by including Z → X and excluding Z → X ) are equivalent, hence FD Z → X is not important and can be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

5 a. Closure Z+ = ZYWX using FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

5 b. Closure Z+ = ZWX using FD = { W → X, Y → X, Z → W, WY → Z }

From 5 a and 5 b, we found that both the Closure (by including Z → Y and excluding Z → Y ) are not equivalent, hence FD Z → X is important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

6 a. Closure WY+ = WYZX using FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

6 b. Closure WY+ = WYX using FD = { W → X, Y → X, Z → W, Z → Y }

From 6 a and 6 b, we found that both the Closure (by including WY → Z and excluding WY → Z) are not equivalent, hence FD WY → Z is important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

Since FD = { W → X, Y → X, Z → W, Z → Y, WY → Z } is the resultant FD now, we have checked the redundancy of the attribute, since the left side of FD WY → Z has two attributes at its left, let's check their importance, i.e. whether they both are important or only one.

Closure WY+ = WYZX using FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

Closure W+ = WX using FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

Closure Y+ = YX using FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

Since the closure of WY+, W+, Y+ that we found are not all equivalent, hence in FD WY → Z, both W and Y are important attributes and cannot be removed.

Hence resultant FD = { W → X, Y → X, Z → W, Z → Y, WY → Z } and we can rewrite it as:

FD = { W → X, Y → X, Z → WY, WY → Z } is Canonical Cover of FD = { W → X, Y → X, Z → WXY, WY → Z }.

Example 3: Given a relational Schema R( V, W, X, Y, Z) and set of Function Dependency FD = { V → W, VW → X, Y → VXZ }. Find the canonical cover?

Solution: Given FD = { V → W, VW → X, Y → VXZ }. now decompose the FD using the decomposition rule (Armstrong Axiom).

1. V → W
2. VW → X
3. Y → V ( using decomposition inference rule on Y → VXZ )
4. Y → X ( using decomposition inference rule on Y → VXZ )
5. Y → Z ( using decomposition inference rule on Y → VXZ )

Now set of FD = { V → W, VW → X, Y → V, Y → X, Y → Z }.

The next step is to find closure of the left side of each of the given FD by including that FD and excluding that FD, if closure in both cases are same then that FD is redundant and we remove that FD from the given set, otherwise if both the closures are different then we do not exclude that FD.

Calculating closure of all FD { V → W, VW → X, Y → V, Y → X, Y → Z }.

1 a. Closure V+ = VWX using FD = {V → W, VW → X, Y → V, Y → X, Y → Z}

1 b. Closure V+ = V using FD = {VW → X, Y → V, Y → X, Y → Z }

From 1 a and 1 b, we found that both the Closure( by including V → W and excluding V → W ) are not equivalent, hence FD V → W is important and cannot be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → X, Y → Z }.

2 a. Closure VW+ = VWX using FD = { V → W, VW → X, Y → V, Y → X, Y → Z }

2 b. Closure VW+ = VW using FD = { V → W, Y → V, Y → X, Y → Z }

From 2 a and 2 b, we found that both the Closure( by including VW → X and excluding VW → X ) are not equivalent, hence FD VW → X is important and cannot be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → X, Y → Z }.

3 a. Closure Y+ = YVXZW using FD = { V → W, VW → X, Y → V, Y → X, Y → Z }

3 b. Closure Y+ = YXZ using FD = { V → W, VW → X, Y → X, Y → Z }

From 3 a and 3 b, we found that both the Closure( by including Y → V and excluding Y → V ) are not equivalent, hence FD Y → V is important and cannot be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → X, Y → Z }.

4 a. Closure Y+ = YXVZW using FD = { V → W, VW → X, Y → V, Y → X, Y → Z }

4 b. Closure Y+ = YVZWX using FD = { V → W, VW → X, Y → V, Y → Z }

From 4 a and 4 b, we found that both the Closure( by including Y → X and excluding Y → X ) are equivalent, hence FD Y → X is not important and can be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → Z }.

5 a. Closure Y+ = YZVWX using FD = { V → W, VW → X, Y → V, Y → Z }

5 b. Closure Y+ = YVWX using FD = { V → W, VW → X, Y → V }

From 5 a and 5 b, we found that both the Closure( by including Y → Z and excluding Y → Z ) are not equivalent, hence FD Y → Z is important and cannot be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → Z }.

Since FD = { V → W, VW → X, Y → V, Y → Z } is the resultant FD now, we have checked the redundancy of the attribute, since the left side of FD VW → X has two attributes at its left, let's check their importance, i.e. whether they both are important or only one.

Closure VW+ = VWX using FD = { V → W, VW → X, Y → V, Y → Z }

Closure V+ = VWX using FD = { V → W, VW → X, Y → V, Y → Z }

Closure W+ = W using FD = { V → W, VW → X, Y → V, Y → Z }

Since the closure of VW+, V+, W+ we found that all the Closures of VW and V are equivalent, hence in FD VW → X, W is not at all an important attribute and can be removed.

Hence resultant FD = { V → W, V → X, Y → V, Y → Z } and we can rewrite asit 

FD = { V → WX, Y → VZ } is Canonical Cover of FD = { V → W, VW → X, Y → VXZ }.

CONCLUSION: From the above three examples we conclude that canonical cover / irreducible set of functional dependency follows the following steps, which we need to follow while calculating Canonical Cover.

STEP 1: For a given set of FD, decompose each FD using the decomposition rule (Armstrong Axiom) if the right side of any FD has more than one attribute.

STEP 2: Now make a new set of FD having all decomposed FD.

STEP 3: Find closure of the left side of each of the given FD by including that FD and excluding that FD, if closure in both cases are same then that FD is redundant and we remove that FD from the given set, otherwise if both the closures are different then we do not exclude that FD.

STEP 4: Repeat step 4 till all the FDs in FD set are complete.

STEP 5: After STEP 4, find resultant FD = { B → A, AD → C, C → B, C → D } which are not redundant.

STEP 6: Check redundancy of attribute, by selecting those FDs from FD sets which are having more than one attribute on its left, let's an FD AD → C has two attributes at its left, let's check their importance, i.e. whether they both are important or only one.

STEP 6 a: Find Closure AD+

STEP 6 b: Find Closure A+

STEP 6 c: Find Closure D+

Compare Closure of STEP (6a, 6b, 6c) if the closure of AD+, A+, D+ are not equivalent, hence in FD AD → C, both A and D are important attributes and cannot be removed, otherwise, we remove the redundant attribute.

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Inclusion Dependency

• Multivalued dependency and join dependency can be used to guide database design although they both are less common than functional dependencies.
• Inclusion dependencies are quite common. They typically show little influence on the design of the database.
• The inclusion dependency is a statement in which some columns of a relation are contained in other columns.
• The example of inclusion dependency is a foreign key. In one relation, the referring relation is contained in the primary key column(s) of the referenced relation.
• Suppose we have two relations R and S which were obtained by translating two entity sets such that every R entity is also an S entity.
• Inclusion dependency would happen if projecting R on its key attributes yields a relation that is contained in the relation obtained by projecting S on its key attributes.
• In inclusion dependency, we should not split groups of attributes that participate in an inclusion dependency.
• In practice, most inclusion dependencies are key-based that is involved only keys.

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Join Dependency

• Join decomposition is a further generalization of Multivalued dependencies.
• If the join of R1 and R2 over C is equal to relation R, then we can say that a join dependency (JD) exists.
• Where R1 and R2 are the decompositions R1(A, B, C) and R2(C, D) of a given relations R (A, B, C, D).
• Alternatively, R1 and R2 are lossless decompositions of R.
• A JD ⋈ {R1, R2,..., Rn} is said to hold over a relation R if R1, R2,....., Rn is a lossless-join decomposition.
• The *(A, B, C, D), (C, D) will be a JD of R if the join of join's attribute is equal to the relation R.
• Here, *(R1, R2, R3) is used to indicate that relation R1, R2, R3, and so on are a JD of R.

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Multivalued Dependency

• Multivalued dependency occurs when two attributes in a table are independent of each other but, both depend on a third attribute.
• A multivalued dependency consists of at least two attributes that are dependent on a third attribute that's why it always requires at least three attributes.

Example: Suppose there is a bike manufacturer company that produces two colors(white and black) of each model every year.

BIKE_MODEL	MANUF_YEAR	COLOR
M2011	2008	White
M2001	2008	Black
M3001	2013	White
M3001	2013	Black
M4006	2017	White
M4006	2017	Black

Here columns COLOR and MANUF_YEAR are dependent on BIKE_MODEL and independent of each other.

In this case, these two columns can be called multivalued dependent on BIKE_MODEL. The representation of these dependencies is shown below:

1. BIKE_MODEL   →  →  MANUF_YEAR  
2. BIKE_MODEL   →  →  COLOR  

This can be read as "BIKE_MODEL multidetermined MANUF_YEAR" and "BIKE_MODEL multidetermined COLOR".

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Relational Decomposition

• When a relation in the relational model is not relationship normal form then the decomposition of a relation is required.
• In a database, it breaks the table into multiple tablinappropriatees.
• If the relation has no proper decomposition, then it may lead to problems like loss of information.
• Decomposition is used to eliminate some of the problems of bad design like anomalies, inconsistencies, and redundancy.

Types of Decomposition

Lossless Decomposition

• If the information is not lost from the relation that is decomposed, then the decomposition will be lossless.
• The lossless decomposition guarantees that the joining of relations will result in the same relation as it was decomposed.
• The relation is said to be lossless decomposition if natural joins of all the decomposition give the original relation.

Example:

EMPLOYEE_DEPARTMENT table:

EMP_ID	EMP_NAME	EMP_AGE	EMP_CITY	DEPT_ID	DEPT_NAME
22	Denim	28	Mumbai	827	Sales
33	Alina	25	Delhi	438	Marketing
46	Stephan	30	Bangalore	869	Finance
52	Katherine	36	Mumbai	575	Production
60	Jack	40	Noida	678	Testing

The above relation is decomposed into two relations EMPLOYEE and DEPARTMENT

EMPLOYEE table:

EMP_ID	EMP_NAME	EMP_AGE	EMP_CITY
22	Denim	28	Mumbai
33	Alina	25	Delhi
46	Stephan	30	Bangalore
52	Katherine	36	Mumbai
60	Jack	40	Noida

DEPARTMENT table

DEPT_ID	EMP_ID	DEPT_NAME
827	22	Sales
438	33	Marketing
869	46	Finance
575	52	Production
678	60	Testing

Now, when these two relations are joined on the common column "EMP_ID", then the resultant relation will look like this:

Employee ⋈ Department

EMP_ID	EMP_NAME	EMP_AGE	EMP_CITY	DEPT_ID	DEPT_NAME
22	Denim	28	Mumbai	827	Sales
33	Alina	25	Delhi	438	Marketing
46	Stephan	30	Bangalore	869	Finance
52	Katherine	36	Mumbai	575	Production
60	Jack	40	Noida	678	Testing

Hence, the decomposition is Lossless join decomposition.

Dependency Preserving

• It is an important constraint of the database.
• Independence preservation, at least one decomposed table must satisfy every dependency.
• If a relation R is decomposed into relations R1 and R2, then the dependencies of R either must be a part of R1 or R2 or must be derivable from the combination of functional dependencies of R1 and R2.
• For example, suppose there is a relation R (A, B, C, D) with a functional dependency set (A->BC). The relational R is decomposed into R1(ABC) and R2(AD) which is dependency preserving because FD A->BC is a part of relation R1(ABC).

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Fifth normal form (5NF)

• A relation is in 5NF if it is in 4NF and not contains any join dependency and joining should be lossless.
• 5NF is satisfied when all the tables are broken into as many tables as possible in order to avoid redundancy.
• 5NF is also known as Project-join normal form (PJ/NF).

Example

SUBJECT	LECTURER	SEMESTER
Computer	Anshika	Semester 1
Computer	John	Semester 1
Math	John	Semester 1
Math	Akash	Semester 2
Chemistry	Praveen	Semester 1

In the above table, John takes both Computer and Math class for Semester 1 but he doesn't take Math class for Semester 2. In this case, combination of all these fields required to identify a valid data.

Suppose we add a new Semester as Semester 3 but do not know about the subject and who will be taking that subject so we leave Lecturer and Subject as NULL. But all three columns together acts as a primary key, so we can't leave other two columns blank.

So to make the above table into 5NF, we can decompose it into three relations P1, P2 & P3:

P1

SEMESTER	SUBJECT
Semester 1	Computer
Semester 1	Math
Semester 1	Chemistry
Semester 2	Math

P2

SUBJECT	LECTURER
Computer	Anshika
Computer	John
Math	John
Math	Akash
Chemistry	Praveen

P3

SEMSTER	LECTURER
Semester 1	Anshika
Semester 1	John
Semester 1	John
Semester 2	Akash
Semester 1	Praveen

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Fourth normal form (4NF)

• A relation will be in 4NF if it is in Boyce Codd normal form and has no multi-valued dependency.
• For a dependency A → B, if for a single value of A, multiple values of B exists, then the relation will be a multi-valued dependency.

Example

STUDENT

STU_ID	COURSE	HOBBY
21	Computer	Dancing
21	Math	Singing
34	Chemistry	Dancing
74	Biology	Cricket
59	Physics	Hockey

The given STUDENT table is in 3NF, but the COURSE and HOBBY are two independent entity. Hence, there is no relationship between COURSE and HOBBY.

In the STUDENT relation, a student with STU_ID, 21 contains two courses, Computer and Math and two hobbies, Dancing and Singing. So there is a Multi-valued dependency on STU_ID, which leads to unnecessary repetition of data.

So to make the above table into 4NF, we can decompose it into two tables:

STUDENT_COURSE

STU_ID	COURSE
21	Computer
21	Math
34	Chemistry
74	Biology
59	Physics

STUDENT_HOBBY

STU_ID	HOBBY
21	Dancing
21	Singing
34	Dancing
74	Cricket
59	Hockey

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