Question

Exercise #4
Using the Functional Dependencies,
F = {A → BC ; CD → E ; B→D ; E→A}
a) Compute the closure of F (F+).
b) Is true / false : F ⊨  E → BC?
c) Provide the minimal cover Fc (min(F)) using steps shown in the class.
d) List of the candidate keys for R

Answer 1

Hey, let's solve each of your question one by one.

We have given some Functional dependecies as -

F = {A → BC ; CD → E ; B→D ; E→A}

a) Closure of F: For calculating the closure of F, we need to calculate the attribute closure of F as -

(A)⁺ = {A, B, C, D, E}

(B)⁺ = {B, D}

(C)⁺ = {C}

(D)⁺ = {D}

(E)⁺ = {A, B, C, D, E}

(CD)⁺ = {A, B, C, D, E}

From the above closures, we can see the closure of A, E, CD can derive all the attributes.

b) F: E -> BC is True/False

So we have F: E -> BC

Let's get in to the relation to see if this is true or not.

We have

E - > A , and also,

A -> BC

So, As we study in the transitive rule,

if A determines B and B determine C, then A must also determine C.

Hence by applying the transitive rule here, we can say that

E → BC is true

C)

Steps to find canonical cover -

1. The left side of each functional dependency in F is unique which means the attributes on the left shouldn't be same.
2. No attributes in the left side or right side of any functional dependency is extraneous which means there shoud not be any extra dependency which doesn't in need.
3. Hence, In the our question, the canonical cover F_c is equal to F.

d) So Here we need to find the candidate key:

• So candidate keys are those mimimal super keys that have capability to derive all the attributes.
• As in the First Question we saw that A, E, CD can derive all the attribute -

(A)⁺ = {A, B, C, D, E}

(B)⁺ = {B, D}

(C)⁺ = {C}

(D)⁺ = {D}

(E)⁺ = {A, B, C, D, E}

(CD)⁺ = {A, B, C, D, E}

Hence, candidate key for relation R are - {A, E, CD}.

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