1. Consider a relation R with six attributes A, B, C, D, E, F. You are given the

Question

1. Consider a relation R with six attributes A, B, C, D, E, F. You are given the following dependences: C F, E BC D, D B, and ED C. Is{(ABDE)} a candidate key of this relation? If not, is {(ADE)}? Justify your answer. Answer:

2. Consider the relation F with six attributes F,R,I,D,A,Y with the following dependences: R I, RY F, FY A, and FA R.

a) List all keys for F and justify your answer. Answer:

b) Identify prime and non-prime attributes and justify your answer. Answer:

c) Classify each functional dependency and justify your answer. Answer:

d) Determine normal form for the relation F and justify your answer. Answer:

Explanation / Answer

Solution:

So our dependensies are:

C=> F

EBC=> D

D=> B

ED=> C

Let's see if ABDE will be able to drive all the atributes then we can say it's a super key, after getting the super key we will decide the candidate key, because minimal super key is called candidate key.

so, (ABDE)^+ (Closure), ED toghether will get C now it will become ABCDE and by using C=> F

So (ABDE)^+= ABCDEF

we will get ABCDEF, So we can say that ABDE is super key for given dependencies. But is it a candiate key as well, No it is not. Please check the closure of ADE to understand.

Now (ADE)^+, D=> B will give ABDE, and just like above further tuples can be acheived so ADE is also a super key. But is it a candidate key,

to know that we have to compute closure of it's subsets,

AD^+= ADB (It's not)

and

DE^+= DBE (It's not)

AE^+= AE (It's not)

So now we can say that ADE is a candidate key for given dependencies and ABDE is a super key (But not a candidate key).

I hope this helps. Don't forget to give a thumbs up if you like this.

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