UGC NET CSE | June 2014 | Part 2 | Question: 49

Question

Let $R =\{A, B, C, D, E, F\}$ be a relation schema with the following dependencies $C \rightarrow F$, $E \rightarrow A$, $EC \rightarrow D$, $A \rightarrow B$. Which of the following is a key for $R$ ? 

• A. CD 
• B. EC 
• C. AE 
• D. AC 

Answer 1

{CD}⁺={CDF}

{EC}⁺={ABCDEF}

{AE}⁺={ABE}

{AC}⁺={ABCF}

Hence,Option(B)EC is the correct choice.

Answer 2

ans is B

Find the closure set of all the options give. If any closure covers all the attributes of the relation R then that is the key.
Algorithm to find Closure Set
Step1: Equate an attribute or attributes to X for which closure needs to be identified.
Step2: Take each FD (functional dependency) one by one and check whether the left side of FD is available in X, if yes then add the right side attributes to X if it is not available.
Step3: Repeat step 2 as many times as possible to cover all FD's.
Step4: After no more attributes can be added to X declare it as the closure set.
FDs: C->F, E->A, EC->D, A->B Find closure set for CD.
X = CD = CDF {C->F}
No more attributes can be added to X. 
Hence closure set of CD = CDF 
Find closure set for EC.
X = EC = ECF {C->F} = ECFA {E->A} = ECFAD {EC->D} = ECFADB {A->B}
Closure set of EC covers all the attributes of the relation R.

Answer 3

ANSWER: OPTION B
You can directly find the candidate key using two rules or you can say necessary attributes
An Attribute A is said to be a necessary attribute if
(a)  A occurs only in L.H.S. (left hand side ) of the fd's(functional dependencies) in F  and/or
(b)  A is an Attribute in relation, But A doses not occur either in L.H.S. or R.H.S. of any fd in F.
In other words, Necessary attributes never occur in the RHS of any fd in F.
Here we can directly see, both attribute E and C not occur in RHS of any fd. So, both of are necessary part of a candidate key.
for reference : http://csc.lsu.edu/~jianhua/fd_slide2_09.pdf  (Page 3)

nF