self doubt dbms

Question

Is Minimal cover of FD'S and a minimal set of FD is different things?

Comments

As per my knowledge,

Minimal set means it is either canonical cover or minimal cover of the original FD set.

Minimal cover means, every FD has only one attribute at RIGHT side of FD.

Canonical cover means, every FD has unique attributes at LEFT side of FD to compare with another FD's in the set.

 

ex:-

Let R(A,B,C,D,X) have following FD set: {AX -> BD,B-> C,AX --> C}

Minimal cover is :- {AX -> B,AX -> D,B-> C}

Canonical cover is :- {AX -> BD,B-> C}

Note that Both are Minimal Sets only and those are equivalent also.

 

OBSERVATION :-

There exist exactly one Canonical cover which is equivalent to Minimal cover of FD's (sometimes those are equal also.)

 ===> No.of Minimal covers for original FD set = No.of Canonical covers for original FD set

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better to ask direct question, not just theory

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Sir (either canonical cover or minimal cover of the original FD set.) means if 2 fd's are F AND G then F is covering G and G covering F? and because of this, both are equivalence?

In some questions they directly asked which fd  is minimal cover for given fd so we can use the F ⊆ G AND G ⊆ F if this condition satisfied then yes given fd is minimal cover of given fd. (from the options we can assume G one by one)

but in some questions, they asked for no of such fd's which are minimal cover of given FD here I am not able to answer them.

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Didn't get your doubts... Post with the questions, then may i help to you

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consider (A,B,C,D,E,F) be a relational schema with following dependencies AC->G , D->EG ,B->CD, CG->BD        ACD->B ,CE->AG the no of diffferent minimal cover possible?

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Sir (either canonical cover or minimal cover of the original FD set.) means if 2 fd's are F AND G then F is covering G and G covering F? and because of this, both are equivalence?

Did you mean F is minimal cover and G is Canonical cover?

if yes, then your reason is correct.

 

In some questions they directly asked which fd  is minimal cover for given fd so we can use the F ⊆ G AND G ⊆ F if this condition satisfied then yes given fd is minimal cover of given fd. (from the options we can assume G one by one)

Did you mean, multiple choice question ?

if yes, then "yes, we can easily check one by one from the options and easily get correct answer."

 

but in some questions, they asked for no of such fd's which are minimal cover of given FD here I am not able to answer them.

consider (A,B,C,D,E,F) be a relational schema with following dependencies AC->G , D->EG ,B->CD, CG->BD        ACD->B ,CE->AG the no of diffferent minimal cover possible?

yes, for fill in the blanks, we have to think, every possibility, it is some what difficult. But note that in GATE they will give the small questions which can be solve atmost 5 min, the example which you provided is a very lengthy one, i hope it asked in Madeeasy test series. check this link https://gateoverflow.in/242802/made-easy-test-series

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If  F ⊆ G it means G Covers F i.e all  fds of F are implied in G also and G has more expressive power and both F &G are not equivalent.

 If F ⊆ G AND G ⊆ F when this condition holds true ,means both are equivalent i.e F=G

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yes, sir that was made easy test series que. thats why I was not showing question because of the rules in GO. thanks, sir you have cleared my all doubt.

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brother, there is no rule in GO that, you can't ask some other test series questions.... If you want to add, you can add them., but i want to give One suggestion that most of the aspirants doesn't follow is " with your question, add your approach and your doubt then someone will easily help you "