Determine Decomposition Is Lossless Or Lossy

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

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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Summary Article Name
Determine Decomposition Is Lossless Or Lossy
Description
To determine whether decomposition is lossless or lossy, we check 3 conditions. Lossless Decomposition occurs if all conditions satisfy. Lossy decomposition occurs if any one condition fails.
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Gate Vidyalay
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