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GATE and NET on Lossless Join Decomposition

Q: UGC NET: 2008 (D2)

Suppose R is a relation schema and F is a set of functional dependencies on R. Further, suppose R1 and R2 form a decomposition of R. Then the decomposition is a lossless join decomposition of R provided that:

Before you see the answer please follow the previous chapters: Functional Dependency Closer,  Concept of Lossless and Lossy Join Decomposition

Solution:

Explanation: 

The decomposition of relation R into R1 and R2 is said to be lossless join decomposition if

UGC-NET | UGC NET CS 2016 July – III | Question 10

Consider the table R with attributes A, B, and C. The functional dependencies that hold on R are: A → B, C → AB. Which of the following statements is/are true?

I. The decomposition of R into R1(C, A) and R2(A, B) is lossless.

II. The decomposition of R into R1(A, B) and R2(B, C) is lossy.

(a) Only I                 

(b) Only II

(c) Both I and II       

(d) Neither I nor II 

Before you see the answer please follow the previous chapters: Functional Dependency Closer,  Concept of Lossless and Lossy Join Decomposition

Solution:

Option (c) is correct.

Explanation: 

I.

Condition for Lossless Join Decomposition

Here, 

II.

GATE CS 2021 | Set 1 | Question 43

Before you see the answer please follow the previous chapters: Functional Dependency Closer, Concept of Lossless and Lossy Join Decomposition

Which one of the following options is correct? 

(A) D1 is lossless decomposition, but D2 is a lossy decomposition.

(B) D1 is a lossy decomposition, but D2 is a lossless decomposition.

(C) Both D1 and D2 are lossless decomposition.

(D) Both D1 and D2 are lossy decompositions.

Answers:

We know that the decomposition of R into R1 and R2 is lossless if

Now,

Now, 

So, it is lossless, we can join them.

So, it is lossless, we can merge them.

So, it is lossless, we can merge them.

So, it is also lossless, we will join them.

So, we can join them

Option (a) is the correct answer.