Find minimal cover from given functional dependencies

Question

A -> BC

CD-> E

B->D

E->A

Comments

To find Minimum Cover , the following steps must be done in this order (#1 and then #2), or first #2 then #1:

1) Get rid of redundant attributes (reduce left sides)

2) Get rid of redundant dependencies

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Sir even if we do another way around, I don't think there will be incorrect results. See this: https://www.youtube.com/watch?v=o0GQQFu-5C0&index=5&list=PLmXKhU9FNesR1rSES7oLdJaNFgmuj0SYV

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@ smsubham  

yes it can change . You are correct.

first we can do  Get rid of redundant attributes (reduce left sides)

Then we can do  Get rid of redundant dependencies or vice versa..

 take few examples and check .

In best answer see there are 4 rules mention  we should follow.. here we first did get rid of redundants then reduce left side and it also gives correct answer.

Answer 1

there are three step to  calculate minimal cover of fds

step1: we decompose each fd with single attribute in RHS side {A->B,A->C,CD->E,B->D,E->A}

step2: now check for each dependencies it should or shouldn't present in set  , for this we hide that particular FD for which i m cecking and try to take closure of LHS side of that FD if we r able to derive the RHS side of that fd without using that FD then  that particular fd is not required in the set

for example for FD A->B we hide A->B and find closure of A+=A cant find B so we cant remove A->B from set thus we check for all the FDS

step3: For the remaining functional dependencies having more than one attribute on LHS, ex AB->C, check whether closure(A) contains B, if yes than further simplify the functional dependency as A->C. If viceversa is also true then we will have 2 sets of minimal cover, one having A->C and other containing B->C.

for this question we cannot remove any dependencies so minimal set is {A->BC,CD->E,B->D,E->A}

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4 rules to find Minimal cover :

1. Break down the RHS of each functional dependency into a single attribute .
2.  Minimize the LHS .
3. Minimize the set of functional dependencies .
4. Group the functional dependencies that have common LHS together into a Single FD .

Reference :

 http://www.mathcs.emory.edu/~cheung/Courses/377/Syllabus/9-NormalForms/FD-equi.html

Comments

@thanks sir , to give conclusion after long discussion and proof

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@saket @bikram

so can we say minimal cover and canonical cover are same.

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Here Minimal cover is {A->B, A->C,CD->E,B->D,E->A}

Canonical cover is {A->BC,CD->E,B->D,E->A}

Answer 2

A ---> B, A----> C , B ----> D, E ----> A, CD ----> E

minimal cover need  not to be unique but they are logically equivalent..

Comments

@Digvijay sir logically equivalent means if F ⊆ G and G ⊆F if both conditions satisfied?

Answer 3

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