Answer to Question #267117 in Databases | SQL | Oracle | MS Access for saurav kumar

Question

Q3.A. Consider the relation schema R = (E,F,G,H,I,J,K,L,M,N) And the set of functional dependencies {{E,F} -> {G}, {F} -> {I,J},{E,H} ->{K,L}->{M},{K}->{M},{L}->{N}} on R. Find the key for R?

Q3. B. Consider the schema R(ABCDEFGHIJ) and functional dependencies {FDs=(AB->C,A->D,B->F,F->GH,D->IJ)} and decompositions

a) {D=(ABCDE,BFGH,DIJ)}

b) {D=(ABCD,DE,BF,FGH,DIJ)} Check whether the decomposition is lossless or not? 

Answer 1

A.

the Given Relation-Schema is R (E, F, G, H, I, J, K, L, M, N) and the Given functional dependencies :

{E, F} → {G}  (i)  {K} → {M}  (iv)

{F} → {I, J}  (ii)  {L} → {N}  (v)

{E, H} → {K, L}  (iii)

Now we need to find function dependency whose right side is R.

using Transitivity on (i) and (ii) we get

{E, F} → {G, I, J} .......(vi) ·

⚈ using preudotransitivity on (iii) and (vi) we get

{E, F, H} → {G, I, J, K, L} ........(vii)

⚈ using Decomposition on (vii) we get

{E, F, H} → {K} and {E, F, H} → {L}

Combining above with (iv) and (v) respectively

{E, F, H} → {M} and {E, F, H} → {N}

Now, finally performing union of these with (vii) we get:

{E, F, H} → {G, I, J, K, L, M, N} ...........(viii)

Also {E, F, H} → {E, F, H} is trivial, combine this with (viii) using union, we get:

{E, F, H} → {E, F, G, H, I, J, K, L, M, N}

⇒ {E, F, H} → R

So, {E, F, H} is a key of R.