Tag: dbms

ER Diagrams to Tables

Database Management System

Converting ER Diagrams to Tables-

 

After designing an ER Diagram,

  • ER diagram is converted into the tables in relational model.
  • This is because relational models can be easily implemented by RDBMS like MySQL , Oracle etc.

 

Following rules are used for converting an ER diagram into the tables-

 

Rule-01: For Strong Entity Set With Only Simple Attributes-

 

A strong entity set with only simple attributes will require only one table in relational model.

  • Attributes of the table will be the attributes of the entity set.
  • The primary key of the table will be the key attribute of the entity set.

 

Example-

 

 

Roll_no Name Sex
 

 

 

Schema : Student ( Roll_no , Name , Sex )

 

Also Read- Entity Sets in DBMS

 

Rule-02: For Strong Entity Set With Composite Attributes-

 

  • A strong entity set with any number of composite attributes will require only one table in relational model.
  • While conversion, simple attributes of the composite attributes are taken into account and not the composite attribute itself.

 

Example-

 

 

Roll_no First_name Last_name House_no Street City
 

 

 

 

Schema : Student ( Roll_no , First_name , Last_name , House_no , Street , City )

 

Also Read- Types of Attributes in DBMS

 

Rule-03: For Strong Entity Set With Multi Valued Attributes-

 

A strong entity set with any number of multi valued attributes will require two tables in relational model.

  • One table will contain all the simple attributes with the primary key.
  • Other table will contain the primary key and all the multi valued attributes.

 

Example-

 

 

Roll_no City
 

 

 

Roll_no Mobile_no
 

 

 

Rule-04: Translating Relationship Set into a Table-

 

A relationship set will require one table in the relational model.

Attributes of the table are-

  • Primary key attributes of the participating entity sets
  • Its own descriptive attributes if any.

Set of non-descriptive attributes will be the primary key.

 

Example-

 

 

Emp_no Dept_id since
 

 

Schema : Works in ( Emp_no , Dept_id , since )

 

NOTE-

 

If we consider the overall ER diagram, three tables will be required in relational model-

  • One table for the entity set “Employee”
  • One table for the entity set “Department”
  • One table for the relationship set “Works in”

 

Rule-05: For Binary Relationships With Cardinality Ratios-

 

The following four cases are possible-

 

Case-01: Binary relationship with cardinality ratio m:n

Case-02: Binary relationship with cardinality ratio 1:n

Case-03: Binary relationship with cardinality ratio m:1

Case-04: Binary relationship with cardinality ratio 1:1

 

Also read- Cardinality Ratios in DBMS

 

Case-01: For Binary Relationship With Cardinality Ratio m:n

 

 

Here, three tables will be required-

  1. A ( a1 , a2 )
  2. R ( a1 , b1 )
  3. B ( b1 , b2 )

 

Case-02: For Binary Relationship With Cardinality Ratio 1:n

 

 

Here, two tables will be required-

  1. A ( a1 , a2 )
  2. BR ( a1 , b1 , b2 )

 

NOTE- Here, combined table will be drawn for the entity set B and relationship set R.

 

Case-03: For Binary Relationship With Cardinality Ratio m:1

 

 

Here, two tables will be required-

  1. AR ( a1 , a2 , b1 )
  2. B ( b1 , b2 )

 

NOTE- Here, combined table will be drawn for the entity set A and relationship set R.

 

Case-04: For Binary Relationship With Cardinality Ratio 1:1

 

 

Here, two tables will be required. Either combine ‘R’ with ‘A’ or ‘B’

 

Way-01:

  1. AR ( a1 , a2 , b1 )
  2. B ( b1 , b2 )

 

Way-02:

  1. A ( a1 , a2 )
  2. BR ( a1 , b1 , b2 )

 

Thumb Rules to Remember

 

While determining the minimum number of tables required for binary relationships with given cardinality ratios, following thumb rules must be kept in mind-

  • For binary relationship with cardinality ration m : n , separate and individual tables will be drawn for each entity set and relationship.
  • For binary relationship with cardinality ratio either m : 1 or 1 : n , always remember “many side will consume the relationship” i.e. a combined table will be drawn for many side entity set and relationship set.
  • For binary relationship with cardinality ratio 1 : 1 , two tables will be required. You can combine the relationship set with any one of the entity sets.

 

Rule-06: For Binary Relationship With Both Cardinality Constraints and Participation Constraints-

 

  • Cardinality constraints will be implemented as discussed in Rule-05.
  • Because of the total participation constraint, foreign key acquires NOT NULL constraint i.e. now foreign key can not be null.

 

Case-01: For Binary Relationship With Cardinality Constraint and Total Participation Constraint From One Side-

 

 

Because cardinality ratio = 1 : n , so we will combine the entity set B and relationship set R.

Then, two tables will be required-

  1. A ( a1 , a2 )
  2. BR ( a1 , b1 , b2 )

Because of total participation, foreign key a1 has acquired NOT NULL constraint, so it can’t be null now.

 

Case-02: For Binary Relationship With Cardinality Constraint and Total Participation Constraint From Both Sides-

 

If there is a key constraint from both the sides of an entity set with total participation, then that binary relationship is represented using only single table.

 

 

Here, Only one table is required.

  • ARB ( a1 , a2 , b1 , b2 )

 

Rule-07: For Binary Relationship With Weak Entity Set-

 

Weak entity set always appears in association with identifying relationship with total participation constraint.

 

 

Here, two tables will be required-

  1. A ( a1 , a2 )
  2. BR ( a1 , b1 , b2 )

 

Next Article- Practice Problems On Converting ER Diagrams to Tables

 

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Determine Decomposition Is Lossless Or Lossy

Database Management System

Decomposition in DBMS-

 

Before you go through this article, make sure that you have gone through the previous article on Decomposition in DBMS.

 

We have discussed-

  • Decomposition is a process of dividing a single relation into two or more sub relations.
  • Decomposition of a relation can be completed in the following two ways-

 

 

  1. Lossless Join Decomposition
  2. Lossy Join Decomposition

 

In this article, we will learn how to determine whether the decomposition is lossless or lossy.

 

Determining Whether Decomposition Is Lossless Or Lossy-

 

Consider a relation R is decomposed into two sub relations R1 and R2.

Then,

  • If all the following conditions satisfy, then the decomposition is lossless.
  • If any of these conditions fail, then the decomposition is lossy.

 

Condition-01:

 

Union of both the sub relations must contain all the attributes that are present in the original relation R.

Thus,

R1 ∪ R2 = R

 

Condition-02:

 

  • Intersection of both the sub relations must not be null.
  • In other words, there must be some common attribute which is present in both the sub relations.

Thus,

R1 ∩ R2 ≠ ∅

 

Condition-03:

 

Intersection of both the sub relations must be a super key of either R1 or R2 or both.

Thus,

R1 ∩ R2 = Super key of R1 or R2

 

PRACTICE PROBLEMS BASED ON DETERMINING WHETHER DECOMPOSITION IS LOSSLESS OR LOSSY-

 

Problem-01:

 

Consider a relation schema R ( A , B , C , D ) with the functional dependencies A → B and C → D. Determine whether the decomposition of R into R1 ( A , B ) and R2 ( C , D ) is lossless or lossy.

 

Solution-

 

To determine whether the decomposition is lossless or lossy,

  • We will check all the conditions one by one.
  • If any of the conditions fail, then the decomposition is lossy otherwise lossless.

 

Condition-01:

 

According to condition-01, union of both the sub relations must contain all the attributes of relation R.

So, we have-

 R1 ( A , B ) ∪ R2 ( C , D )

= R ( A , B , C , D )

Clearly, union of the sub relations contain all the attributes of relation R.

Thus, condition-01 satisfies.

 

Condition-02:

 

According to condition-02, intersection of both the sub relations must not be null.

So, we have-

R1 ( A , B ) ∩ R2 ( C , D )

= Φ

Clearly, intersection of the sub relations is null.

So, condition-02 fails.

Thus, we conclude that the decomposition is lossy.

 

Problem-02:

 

Consider a relation schema R ( A , B , C , D ) with the following functional dependencies-

A → B

B → C

C → D

D → B

Determine whether the decomposition of R into R1 ( A , B ) , R2 ( B , C ) and R3 ( B , D ) is lossless or lossy.

 

Solution-

 

Strategy to Solve

 

When a given relation is decomposed into more than two sub relations, then-

  • Consider any one possible ways in which the relation might have been decomposed into those sub relations.
  • First, divide the given relation into two sub relations.
  • Then, divide the sub relations according to the sub relations given in the question.

As a thumb rule, remember-

Any relation can be decomposed only into two sub relations at a time.

 

Consider the original relation R was decomposed into the given sub relations as shown-

 

 

Decomposition of R(A, B, C, D) into R'(A, B, C) and R3(B, D)-

 

To determine whether the decomposition is lossless or lossy,

  • We will check all the conditions one by one.
  • If any of the conditions fail, then the decomposition is lossy otherwise lossless.

 

Condition-01:

 

According to condition-01, union of both the sub relations must contain all the attributes of relation R.

So, we have-

 R‘ ( A , B , C ) ∪ R3 ( B , D )

= R ( A , B , C , D )

Clearly, union of the sub relations contain all the attributes of relation R.

Thus, condition-01 satisfies.

 

Condition-02:

 

According to condition-02, intersection of both the sub relations must not be null.

So, we have-

 R‘ ( A , B , C ) ∩ R3 ( B , D )

= B

Clearly, intersection of the sub relations is not null.

Thus, condition-02 satisfies.

 

Condition-03:

 

According to condition-03, intersection of both the sub relations must be the super key of one of the two sub relations or both.

So, we have-

 R‘ ( A , B , C ) ∩ R3 ( B , D )

= B

Now, the closure of attribute B is-

B+ = { B , C , D }

Now, we see-

  • Attribute ‘B’ can not determine attribute ‘A’ of sub relation R’.
  • Thus, it is not a super key of the sub relation R’.
  • Attribute ‘B’ can determine all the attributes of sub relation R3.
  • Thus, it is a super key of the sub relation R3.

 

Clearly, intersection of the sub relations is a super key of one of the sub relations.

So, condition-03 satisfies.

Thus, we conclude that the decomposition is lossless.

 

Decomposition of R'(A, B, C) into R1(A, B) and R2(B, C)-

 

To determine whether the decomposition is lossless or lossy,

  • We will check all the conditions one by one.
  • If any of the conditions fail, then the decomposition is lossy otherwise lossless.

 

Condition-01:

 

According to condition-01, union of both the sub relations must contain all the attributes of relation R’.

So, we have-

 R1 ( A , B ) ∪ R2 ( B , C )

= R’ ( A , B , C )

Clearly, union of the sub relations contain all the attributes of relation R’.

Thus, condition-01 satisfies.

 

Condition-02:

 

According to condition-02, intersection of both the sub relations must not be null.

So, we have-

 R1 ( A , B ) ∩ R2 ( B , C )

= B

Clearly, intersection of the sub relations is not null.

Thus, condition-02 satisfies.

 

Condition-03:

 

According to condition-03, intersection of both the sub relations must be the super key of one of the two sub relations or both.

So, we have-

 R1 ( A , B ) ∩ R2 ( B , C )

= B

Now, the closure of attribute B is-

B+ = { B , C , D }

Now, we see-

  • Attribute ‘B’ can not determine attribute ‘A’ of sub relation R1.
  • Thus, it is not a super key of the sub relation R1.
  • Attribute ‘B’ can determine all the attributes of sub relation R2.
  • Thus, it is a super key of the sub relation R2.

 

Clearly, intersection of the sub relations is a super key of one of the sub relations.

So, condition-03 satisfies.

Thus, we conclude that the decomposition is lossless.

 

Conclusion-

Overall decomposition of relation R into sub relations R1, R2 and R3 is lossless.

 

Next Article- Introduction to Normal Forms

 

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Closure in DBMS | Steps to Find Closure

Database Management System

Closure of an Attribute Set-

 

  • The set of all those attributes which can be functionally determined from an attribute set is called as a closure of that attribute set.
  • Closure of attribute set {X} is denoted as {X}+.

 

Steps to Find Closure of an Attribute Set-

 

Following steps are followed to find the closure of an attribute set-

 

Step-01:

 

Add the attributes contained in the attribute set for which closure is being calculated to the result set.

 

Step-02:

 

Recursively add the attributes to the result set which can be functionally determined from the attributes already contained in the result set.

 

Example-

 

Consider a relation R ( A , B , C , D , E , F , G ) with the functional dependencies-

A → BC

BC → DE

D → F

CF → G

 

Now, let us find the closure of some attributes and attribute sets-

 

Closure of attribute A-

 

A+   = { A }

= { A , B , C }                          ( Using A → BC )

= { A , B , C , D , E }               ( Using BC → DE )

= { A , B , C , D , E , F }          ( Using D → F )

= { A , B , C , D , E , F , G }    ( Using CF → G )

Thus,

A+ = { A , B , C , D , E , F , G }

 

Closure of attribute D-

 

D+   = { D }

= { D , F }   ( Using D → F )

We can not determine any other attribute using attributes D and F contained in the result set.

Thus,

D+ = { D , F }

 

Closure of attribute set {B, C}-

 

{ B , C }+= { B , C }

= { B , C , D , E }               ( Using BC → DE )

= { B , C , D , E , F }          ( Using D → F )

= { B , C , D , E , F , G }    ( Using CF → G )

Thus,

{ B , C }+ = { B , C , D , E , F , G }

 

Finding the Keys Using Closure-

 

Super Key-

 

  • If the closure result of an attribute set contains all the attributes of the relation, then that attribute set is called as a super key of that relation.
  • Thus, we can say-

“The closure of a super key is the entire relation schema.”

 

Example-

 

In the above example,

  • The closure of attribute A is the entire relation schema.
  • Thus, attribute A is a super key for that relation.

 

Candidate Key-

 

  • If there exists no subset of an attribute set whose closure contains all the attributes of the relation, then that attribute set is called as a candidate key of that relation.

 

Example-

 

In the above example,

  • No subset of attribute A contains all the attributes of the relation.
  • Thus, attribute A is also a candidate key for that relation.

 

Also Read- How To Find Candidate Keys?

 

PRACTICE PROBLEM BASED ON FINDING CLOSURE OF AN ATTRIBUTE SET-

 

Problem-

 

Consider the given functional dependencies-

AB → CD

AF → D

DE → F

C → G

F → E

G → A

 

Which of the following options is false?

(A) { CF }+ = { A , C , D , E , F , G }

(B) { BG }+ = { A , B , C , D , G }

(C) { AF }+ = { A , C , D , E , F , G }

(D) { AB }+ = { A , C , D , F ,G }

 

Solution-

 

Let us check each option one by one-

 

Option-(A):

 

{ CF }+  = { C , F }

      = { C , F , G }                     ( Using C → G )

      = { C , E , F , G }               ( Using F → E )

      = { A , C , E , E , F }          ( Using G → A )

      = { A , C , D , E , F , G }    ( Using AF → D )

 

Since, our obtained result set is same as the given result set, so, it means it is correctly given.

 

Option-(B):

 

{ BG }+  = { B , G }

      = { A , B , G }                   ( Using G → A )

      = { A , B , C , D , G }        ( Using AB → CD )

 

Since, our obtained result set is same as the given result set, so, it means it is correctly given.

 

Option-(C):

 

{ AF }+  = { A , F }

      = { A , D , F }               ( Using AF → D )

      = { A , D , E , F }          ( Using F → E )

 

Since, our obtained result set is different from the given result set, so,it means it is not correctly given.

 

Option-(D):

 

{ AB }+  = { A , B }

      = { A , B , C , D }            ( Using AB → CD )

      = { A , B , C , D , G }      ( Using C → G )

 

Since, our obtained result set is different from the given result set, so,it means it is not correctly given.

Thus,

Option (C) and Option (D) are correct.

 

Next Article- 10 Different Kinds of Keys in DBMS

 

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Decomposition in DBMS | Lossless | Lossy

Database Management System

Decomposition of a Relation-

 

The process of breaking up or dividing a single relation into two or more sub relations is called as decomposition of a relation.

 

Properties of Decomposition-

 

The following two properties must be followed when decomposing a given relation-

 

1. Lossless decomposition-

 

Lossless decomposition ensures-

  • No information is lost from the original relation during decomposition.
  • When the sub relations are joined back, the same relation is obtained that was decomposed.

Every decomposition must always be lossless.

 

2. Dependency Preservation-

 

Dependency preservation ensures-

  • None of the functional dependencies that holds on the original relation are lost.
  • The sub relations still hold or satisfy the functional dependencies of the original relation.

 

Types of Decomposition-

 

Decomposition of a relation can be completed in the following two ways-

 

 

1. Lossless Join Decomposition-

 

  • Consider there is a relation R which is decomposed into sub relations R1 , R2 , …. , Rn.
  • This decomposition is called lossless join decomposition when the join of the sub relations results in the same relation R that was decomposed.
  • For lossless join decomposition, we always have-

 

R1 ⋈ R2 ⋈ R3 ……. ⋈ Rn = R 

where ⋈ is a natural join operator

 

Example-

 

Consider the following relation R( A , B , C )-

 

A B C
1 2 1
2 5 3
3 3 3

R( A , B , C )

 

Consider this relation is decomposed into two sub relations R1( A , B ) and R2( B , C )-

 

 

The two sub relations are-

 

A B
1 2
2 5
3 3

R1( A , B )

 

B C
2 1
5 3
3 3

R2( B , C )

 

Now, let us check whether this decomposition is lossless or not.

For lossless decomposition, we must have-

R1 ⋈ R2 = R

 

Now, if we perform the natural join ( ⋈ ) of the sub relations R1 and R2 , we get-

 

A B C
1 2 1
2 5 3
3 3 3

 

This relation is same as the original relation R.

Thus, we conclude that the above decomposition is lossless join decomposition.

 

NOTE-

 

  • Lossless join decomposition is also known as non-additive join decomposition.
  • This is because the resultant relation after joining the sub relations is same as the decomposed relation.
  • No extraneous tuples appear after joining of the sub-relations.

 

2. Lossy Join Decomposition-

 

  • Consider there is a relation R which is decomposed into sub relations R1 , R2 , …. , Rn.
  • This decomposition is called lossy join decomposition when the join of the sub relations does not result in the same relation R that was decomposed.
  • The natural join of the sub relations is always found to have some extraneous tuples.
  • For lossy join decomposition, we always have-

 

R1 ⋈ R2 ⋈ R3 ……. ⋈ Rn ⊃ R 

where ⋈ is a natural join operator

 

Example-

 

Consider the following relation R( A , B , C )-

 

A B C
1 2 1
2 5 3
3 3 3

R( A , B , C )

 

Consider this relation is decomposed into two sub relations as R1( A , C ) and R2( B , C )-

 

 

The two sub relations are-

 

A C
1 1
2 3
3 3

R1( A , B )

 

B C
2 1
5 3
3 3

R2( B , C )

 

Now, let us check whether this decomposition is lossy or not.

For lossy decomposition, we must have-

R1 ⋈ R2 ⊃ R

 

Now, if we perform the natural join ( ⋈ ) of the sub relations R1 and R2 we get-

 

A B C
1 2 1
2 5 3
2 3 3
3 5 3
3 3 3

 

This relation is not same as the original relation R and contains some extraneous tuples.

Clearly, R1 ⋈ R2 ⊃ R.

Thus, we conclude that the above decomposition is lossy join decomposition.

 

NOTE-

 

  • Lossy join decomposition is also known as careless decomposition.
  • This is because extraneous tuples get introduced in the natural join of the sub-relations.
  • Extraneous tuples make the identification of the original tuples difficult.

 

Next Article- Rules to Determine Lossless and Lossy Decomposition

 

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Equivalence of Two Sets of Functional Dependencies

Database Management System

Equivalence of Two Sets of Functional Dependencies-

 

Before you go through this article, make sure that you have gone through the previous article on Functional Dependency.

 

In DBMS,

  • Two different sets of functional dependencies for a given relation may or may not be equivalent.
  • If F and G are the two sets of functional dependencies, then following 3 cases are possible-

 

Case-01: F covers G (F ⊇ G)

Case-02: G covers F (G ⊇ F)

Case-03: Both F and G cover each other (F = G)

 

Case-01: Determining Whether F Covers G-

 

Following steps are followed to determine whether F covers G or not-

 

Step-01:

 

  • Take the functional dependencies of set G into consideration.
  • For each functional dependency X → Y, find the closure of X using the functional dependencies of set G.

 

Step-02:

 

  • Take the functional dependencies of set G into consideration.
  • For each functional dependency X → Y, find the closure of X using the functional dependencies of set F.

 

Step-03:

 

  • Compare the results of Step-01 and Step-02.
  • If the functional dependencies of set F has determined all those attributes that were determined by the functional dependencies of set G, then it means F covers G.
  • Thus, we conclude F covers G (F ⊇ G) otherwise not.

 

Case-02: Determining Whether G Covers F-

 

Following steps are followed to determine whether G covers F or not-

 

Step-01:

 

  • Take the functional dependencies of set F into consideration.
  • For each functional dependency X → Y, find the closure of X using the functional dependencies of set F.

 

Step-02:

 

  • Take the functional dependencies of set F into consideration.
  • For each functional dependency X → Y, find the closure of X using the functional dependencies of set G.

 

Step-03:

 

  • Compare the results of Step-01 and Step-02.
  • If the functional dependencies of set G has determined all those attributes that were determined by the functional dependencies of set F, then it means G covers F.
  • Thus, we conclude G covers F (G ⊇ F) otherwise not.

 

Case-03: Determining Whether Both F and G Cover Each Other-

 

  • If F covers G and G covers F, then both F and G cover each other.
  • Thus, if both the above cases hold true, we conclude both F and G cover each other (F = G).

 

PRACTICE PROBLEM BASED ON EQUIVALENCE OF FUNCTIONAL DEPENDENCIES-

 

Problem-

 

A relation R (A , C , D , E , H) is having two functional dependencies sets F and G as shown-

 

Set F-

A → C

AC → D

E → AD

E → H

 

Set G-

A → CD

E → AH

 

Which of the following holds true?

(A) G ⊇ F

(B) F ⊇ G

(C) F = G

(D) All of the above

 

Solution-

 

Determining whether F covers G-

 

Step-01:

 

  • (A)+ = { A , C , D }               // closure of left side of A → CD using set G
  • (E)+ = { A , C , D , E , H }    // closure of left side of E → AH using set G

 

Step-02:

 

  • (A)+ = { A , C , D }               // closure of left side of A → CD using set F
  • (E)+ = { A , C , D , E , H }    // closure of left side of E → AH using set F

 

Step-03:

 

Comparing the results of Step-01 and Step-02, we find-

  • Functional dependencies of set F can determine all the attributes which have been determined by the functional dependencies of set G.
  • Thus, we conclude F covers G i.e. F ⊇ G.

 

Determining whether G covers F-

 

Step-01:

 

  • (A)+ = { A , C , D }               // closure of left side of A → C using set F
  • (AC)+ = { A , C , D }            // closure of left side of AC → D using set F
  • (E)+ = { A , C , D , E , H }    // closure of left side of E → AD and E → H using set F

 

Step-02:

 

  • (A)+ = { A , C , D }                // closure of left side of A → C using set G
  • (AC)+ = { A , C , D }             // closure of left side of AC → D using set G
  • (E)+ = { A , C , D , E , H }    // closure of left side of E → AD and E → H using set G

 

Step-03:

 

Comparing the results of Step-01 and Step-02, we find-

  • Functional dependencies of set G can determine all the attributes which have been determined by the functional dependencies of set F.
  • Thus, we conclude G covers F i.e. G ⊇ F.

 

Determining whether both F and G cover each other-

 

  • From Step-01, we conclude F covers G.
  • From Step-02, we conclude G covers F.
  • Thus, we conclude both F and G cover each other i.e. F = G.

 

Thus, Option (D) is correct.

 

Next Article- Canonical Cover in DBMS

 

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