Is my answer correct because here i am not sure if the decomposition lossy or lo

Question

Is my answer correct because here i am not sure if the decomposition lossy or lossless

the question: Consider R(A,B,C,D,E) with {BC->A, D->AE, B->C}

It is decomposed into R1(C,D), R2(B,D) and R3(A,D,E).

Is the decomposition lossy? You must use the chase matrix algorithm (EN Algorithm 16.3) to show your reasoning.

The decompoistion of R into R'(A,B,D,E) and R1(C,D) is lossless since the common attribute is D and D->AE , which is a key in R1. The decompoistion of R' into R2(B,D) and R3(A,D,E) is lossless since the common attribute is D and D->AE in R3. Thus, the overall decomposition is lossless.

Using the chase matrix algorithm:

We use the canonical form: {BC->A, D->AE, B->C}

Step 1. Create a table of 5 columns (number of columns and 4 rows (number of relations). Populate it with b(i,j).

Relation

A

B

C

D

E

R1

b(1,1)

b(1,2)

b(1,3)

b(1,4)

b(1,5)

R2

b(2,1)

b(2,2)

b(2,3)

b(2,4)

b(2,5)

R3

b(3,1)

b(3,2)

b(3,3)

b(3,4)

b(3,5)

Step 2. For each relation Ri, set all attribute Aj that appears in Ri from b(i,j) to a(j).

Relation

A

B

C

D

E

R1

b(1,1)

b(1,2)

a(3)

a(4)

b(1,5)

R2

b(2,1)

a(2)

b(2,3)

a(4)

b(2,5)

R3

a(1)

b(3,2)

b(3,3)

a(4)

a(5)

Step 3. While changes can be made with a FD X-> Y, with two rows in the table having the common X values in the following manner:

for every attribute W in Y:

If one cell is an a and the other cell is an b, change the b to the a.

If both cells are b's, change them to the same b.

Applying BC->A: no change.

Applying D->AE:

Relation

A

B

C

D

E

R1

b(1,1)

b(1,2)

a(3)

a(4)

b(1,5)

R2

b(1,1)

a(2)

b(2,3)

a(4)

b(1,5)

R3

a(1)

b(3,2)

b(3,3)

a(4)

a(5)

Applying B->C: no change.

Applying BC->A: no change.

Applying D->AE:

Relation

A

B

C

D

E

R1

a(1)

b(1,2)

a(3)

a(4)

a(5)

R2

a(1)

a(2)

b(2,3)

a(4)

a(5)

R3

a(1)

b(3,2)

b(3,3)

a(4)

a(5)

Relation

A

B

C

D

E

R1

b(1,1)

b(1,2)

b(1,3)

b(1,4)

b(1,5)

R2

b(2,1)

b(2,2)

b(2,3)

b(2,4)

b(2,5)

R3

b(3,1)

b(3,2)

b(3,3)

b(3,4)

b(3,5)

Explanation / Answer

You are correct.

priciple of lossy and lossless say that common attribute is key of any relation then it is called lossless and D is common attribute and it is key R3so it is lossless and your matrix algo also seen right.

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