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Consider $R(A,B,C,D,E,F,G)$ having $FDโ€™s=\left\{AC\rightarrow G,D\rightarrow EG,BC\rightarrow D,CG\rightarrow BD,ACD\rightarrow B,CE\rightarrow AG\right\}$ The number of different minimal cover possible$?$

i m getting this only $\left\{AC\rightarrow G,D\rightarrow EG,BC\rightarrow D,CG\rightarrow B,CE\rightarrow A\right\}$
  • ๐Ÿšฉ Duplicate | ๐Ÿ‘ฎ Hira Thakur | ๐Ÿ’ฌ โ€œhttps://gateoverflow.in/242802/made-easy-test-series?show=243059โ€
in Databases by Loyal (5.2k points) 1 flag
reopened by | 184 views
0

I'm also getting only one
see this https://gateoverflow.in/271091/gatebook-2019-dbms2-10
what is the answer?

0
is the ans 2?
0
can you share another one?
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{ACD->B,AC->G,BC->D,D->EG,CE->A}

@Lakshman Patel RJIT

the link u mentioned is not opening!

0
0

yes answer is 2.

0
I'm not getting the procedure?

and I'm try to do some other way
0
same,i also doing in 3 phases in first phase CG->D,ACD->B,CE->G removed

we cant CG->B as B is only in RHS in this production only

We cant CE->A for same reason so i m not noticing any multiplicity/prallalism any.
+1
please wait some time i will explain and add the answer,ok?
0

plz explain this soln?

2 Answers

+1 vote

Given $FDโ€™s=\left\{AC\rightarrow G,D\rightarrow EG,BC\rightarrow D,CG\rightarrow BD,ACD\rightarrow B,CE\rightarrow AG\right\}$

Finding the minimal cover:-

$(1)$ Split the FD's such a way that $RHS$ contain only single attribute.

Here,

         $AC\rightarrow G$

          $D\rightarrow E$

          $D\rightarrow G$

          $BC\rightarrow D$

          $CG\rightarrow B$

          $CG\rightarrow D$

,       $ACD\rightarrow B$

         $CE\rightarrow A$

         $CE\rightarrow G$

$(2)$ find the redundant FD's and delete them from the set.

         $AC\rightarrow G$ $[$Take this FD and find the attribute closure without this FD.If we are able to find the 'G' without this FD.then we can say that this FD is redundant and remove from the set$]$

Now, $(AC)^{+}=AC$ we are not able to find the $'G'$. So we can say that this FD is not redundant.

Similarly for all those FD's

          $D\rightarrow E$         $[(D)^{+}=DG]$

           $D\rightarrow G$      $[(D)^{+}=DE]$

          $BC\rightarrow D$        $[(BC)^{+}=BC]$

          $CG\rightarrow B$           $[(CG)^{+}=CGDEAB]$ It remove from the set. It is not considered are FD for finding the remaining FD attribute closure.

          $CG\rightarrow D$          $[(CG)^{+}=CG]$

         $ACD\rightarrow B$         $[(ACD)^{+}=ACDEG]$

         $CE\rightarrow A$             $[(CE)^{+}=CEGD]$

         $CE\rightarrow G$             $[(CE)^{+}=CEAGD]$ It remove from the set. It is not considered are FD for finding the remaining FD attribute closure.

Now, we got 

           $AC\rightarrow G$

            $D\rightarrow E$       

             $D\rightarrow G$      

             $BC\rightarrow D$

             $CG\rightarrow D$         

,            $ACD\rightarrow B$        

              $CE\rightarrow A$     

$(3)$ Find the redundant attributes on $LHS$ and delete them. 

            $AC\rightarrow G$  $[(C)^{+}=C$ ,$(A)^{+}=A]$  So, here nothing is deleted.

             $D\rightarrow E$ Here only one attribute on $LHS$ so nothing is redundant      

             $D\rightarrow G$ Here only one attribute on $LHS$ so nothing is redundant     

             $BC\rightarrow D$  Here We can delete $'B'$,if $(C)^{+}$ contain $B$  $(or)$   We can delete $'C'$,if $(B)^{+}$ contain $'C'$

            lets find $(B)^{+}=B$

                          $(C)^{+}=C$

            So, here nothing is deleted.

             $CG\rightarrow D$  $[(C)^{+}=C$ ,$(G)^{+}=G]$  So, here nothing is deleted.        

,            $ACD\rightarrow B$ $[(AC)^{+}=ACGD$  $,$   $(D)^{+}=DEG]$ we can say that $'D'$ is deleted.       

              $CE\rightarrow A$  $[(C)^{+}=C$, $(E)^{+}=E]$  So, here nothing is deleted.

Now we got

           $AC\rightarrow G$

            $D\rightarrow E$       

             $D\rightarrow G$      

             $BC\rightarrow D$

             $CG\rightarrow D$         

,            $AC\rightarrow B$        

              $CE\rightarrow A$ 

Now again find the redundant $FD's$

            $AC\rightarrow G$   $[(AC)^{+}=ACBDEG$  It remove from the set. It is not considered are FD for finding the remaining FD attribute closure.

            $D\rightarrow E$  $[(D)^{+}=DG]$  

             $D\rightarrow G$   $[(D)^{+}=DE]$     

             $BC\rightarrow D$$[(BC)^{+}=BC]$  

             $CG\rightarrow D$   $[(CG)^{+}=CG]$        

,            $AC\rightarrow B$   $[(AC)^{+}=AC]$       

              $CE\rightarrow A$ $[(CE)^{+}=CE]$  

Now we got

            $D\rightarrow E$       

             $D\rightarrow G$      

             $BC\rightarrow D$

             $CG\rightarrow D$         

,            $AC\rightarrow B$        

              $CE\rightarrow A$ 

We can write like this 

So, minimal cover are

             $D\rightarrow EG$      

             $BC\rightarrow D$

             $CG\rightarrow D$         

,            $AC\rightarrow B$        

              $CE\rightarrow A$ 

 

$(1)$ For the given FD's when we split I take to change the order $($I take FD's from the point $(1)$ $)$

        $AC\rightarrow G$

          $D\rightarrow E$

          $D\rightarrow G$

          $BC\rightarrow D$

         $CG\rightarrow D$

         $CG\rightarrow B$

,       $ACD\rightarrow B$

         $CE\rightarrow A$

         $CE\rightarrow G$

 

$(2)$ find the redundant FD's and delete them from the set.

     $AC\rightarrow G$ $[$Take this FD and find the attribute closure without this FD.If we are able to find the 'G' without this FD.then we can say that this FD is redundant and remove from the set$]$

Now, $(AC)^{+}=AC$ we are not able to find the $'G'$. So we can say that this FD is not redundant.

Similarly for all those FD's

          $D\rightarrow E$         $[(D)^{+}=DG]$

           $D\rightarrow G$      $[(D)^{+}=DE]$

          $BC\rightarrow D$        $[(BC)^{+}=BC]$

          $CG\rightarrow D$           $[(CG)^{+}=CGBDEA]$ It remove from the set. It is not considered are FD for finding the remaining FD attribute closure.

          $CG\rightarrow B$          $[(CG)^{+}=CG]$

         $ACD\rightarrow B$         $[(ACD)^{+}=ACDBG]$ It remove from the set. It is not considered are FD for finding the remaining FD attribute closure.

         $CE\rightarrow A$             $[(CE)^{+}=CEGBD]$

         $CE\rightarrow G$             $[(CE)^{+}=CEAG]$ It remove from the set. It is not considered are FD for finding the remaining FD attribute closure.

 

Now we got

     $AC\rightarrow G$

       $D\rightarrow E$ 

       $D\rightarrow G$  

        $BC\rightarrow D$     

        $CG\rightarrow B$   

        $CE\rightarrow A$ 

 

$(3)$ Find the redundant attributes on $LHS$ and delete them. 

            $AC\rightarrow G$  $[(C)^{+}=C$ ,$(A)^{+}=A]$  So, here nothing is deleted.

             $D\rightarrow E$ Here only one attribute on $LHS$ so nothing is redundant      

             $D\rightarrow G$ Here only one attribute on $LHS$ so nothing is redundant     

             $BC\rightarrow D$  Here We can delete $'B'$,if $(C)^{+}$ contain $B$  $(or)$   We can delete $'C'$,if $(B)^{+}$ contain $'C'$

            lets find $(B)^{+}=B$

                          $(C)^{+}=C$

                 So, here nothing is deleted.

            $CG\rightarrow B$  $[(C)^{+}=C$ ,$(G)^{+}=G]$  So, here nothing is deleted.              

              $CE\rightarrow A$  $[(C)^{+}=C$, $(E)^{+}=E]$  So, here nothing is deleted.

We got 

      $AC\rightarrow G$

       $D\rightarrow E$ 

       $D\rightarrow G$  

        $BC\rightarrow D$     

        $CG\rightarrow B$   

        $CE\rightarrow A$ 

Now again find the redundant $FD's$,but no one $FDs'$ are redundant.

We can write like this 

So, minimal cover are:

     $AC\rightarrow G$

       $D\rightarrow EG$      

       $BC\rightarrow D$     

        $CG\rightarrow B$   

        $CE\rightarrow A$ 

 

 

$(1)$ For the given FD's when we split I take to change the order $($I take FD's from the point $(1)$ $)$

        $AC\rightarrow G$

          $D\rightarrow E$

          $D\rightarrow G$

          $BC\rightarrow D$

         $CG\rightarrow B$

         $CG\rightarrow D$

,       $ACD\rightarrow B$

         $CE\rightarrow G$

         $CE\rightarrow A$

In this case, we got

 minimal cover are

             $D\rightarrow EG$      

             $BC\rightarrow D$

             $CG\rightarrow D$         

,            $AC\rightarrow B$        

              $CE\rightarrow A$ 

                                  $(OR)$   

 

       $AC\rightarrow G$

          $D\rightarrow E$

          $D\rightarrow G$

          $BC\rightarrow D$

         $CG\rightarrow D$

         $CG\rightarrow B$

,       $ACD\rightarrow B$

         $CE\rightarrow G$

         $CE\rightarrow A$

In this case, we got

 minimal cover are:

     $AC\rightarrow G$

       $D\rightarrow EG$      

       $BC\rightarrow D$     

        $CG\rightarrow B$   

        $CE\rightarrow A$ 

So,number of minimal cover are $2$

by Veteran (58.8k points)
edited by
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@ ma'am

@

@

@

Please see my solution and verify where I'm wrong?

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@Lakshman Patel RJIT Awesome Brother!

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@

answer will be $2$ (or) $4?$

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I am also getting 2 But in madeEasy explanation they are saying 4 as answer
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you see   comment in top,they add madeeasy answer as well,

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In my solution portal this is the solution they have provided.

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How $CE\rightarrow AF$?
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in the first part of finding out the minimal cover, u claimed that D is redundant attribute and so we can remove D from ACD->B. Isnt A also a redundant attribute as if we take the closure of CD, we can derive A?

So we could have removed A also right in place of D?

 

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Yeah, but at a time we can remove $D$ or $A$
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yaa..thats fine..but if we consider A as redundant there, then in the next step while finding out redundant FDs again,AC->G doesnt become redundant anymore..So would this be considered as a minimal cover also if we consider A as redundant attribute instead of D?
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Yes, and we get more I think.
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u mean we will get more minimal covers by considering A as redundant attribute?
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i think we get more

i'm not sure
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ok..I have one more doubt:

You have considered interchanging the orders of CG->B and CG->D..why didnt u consider interchanging the orders of CE->A and CE->G and also D->E and D->G? Is it becoz that even after interchanging their orders they result in the same minimal covers?
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might be possible lets find out!
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yes they give same.
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ok..thanks!
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$AC\rightarrow G,D\rightarrow EG,BC\rightarrow D,CG\rightarrow D,CD\rightarrow B,CE\rightarrow A$ is also possible
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@Soumya29

please help in this question.

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@Lakshman Patel RJIT

I got one minimal cover

$A\rightarrow G$

$D\rightarrow EG$

$B\rightarrow D$

$C\rightarrow D$

$AC\rightarrow B$

$CE\rightarrow A$

$CE\rightarrow G$

Now chking any more remaining or not

https://www.geeksforgeeks.org/canonical-cover/

@Shaik Masthan @Soumya29 @minal

Check this

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please see my answer.
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sir while taking 2nd possible FD why you r not changing the order of CE-> AG part
0 votes
4 minimal cover
by Junior (849 points)
0
Will you please elaborate!
0
what's the answer?
0
4 is correct

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