Number of minimal candidate keys

Question

A relation R(A,B,C,D,E,F) holds following FDs
$AB\rightarrow C$
$C\rightarrow D$
$D\rightarrow EA$
$E\rightarrow F$
$F\rightarrow B$

The number of minimal candidate keys of R are _______.

Comments

5 candidate keys are present

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I i think we get only 5 CK

Answer 1

Only 2 candidate keys which are minimal means length of 1 which are C and D.

D⁺ ->EAFBCD               [minimal since length is 1 ]

C⁺ -> D⁺ -> EAFBCD     [minimal since length is 1 ]

(AB)⁺-> C⁺ -> D⁺ -> EAFBCD   [not minimal since length is 2]

(AF)⁺ -> (AFB)⁺-> C⁺ -> D⁺ -> EAFBCD [ not minimal since length is 2 ]

(AE)⁺ -> (AF)⁺-> C⁺ -> D⁺ -> EAFBCD [ not minimal since length is 2 ]

Comments

Why AB, AE, AF are not minimal candidate keys ? Please provide the source which distinguishes between candidate key and minimal candidate key.

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read conversation. there is no minimal candiate key term present . here minimal means minimal character.

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if they ask for no of super key then it should be 48.

ryt??

Answer 2

As per the given FDs

AB→C
C→D
D→EA
E→F
F→B

We find that, every attributes are present in RHS of the FD. So we'll first go through length one of candidate key.

A⁺ =A

B⁺ =B

C⁺= (C, D, E, A, F, B)  // C is a minimal Candidate Key.

D⁺= (C, D, E, A, F, B)  // D is a minimal Candidate Key.

E⁺= (E, F, B)

F⁺=(F, B)

Here we cannot got for candidate key of length two, as the question is asking about "minimal" Candidate key.

Hence the number of minimal Candidate key is 2.

Comments

I feel AB, AE and AF should also be considered "minimal". Usually minimal means "something from which we cannot remove anything" as against "smallest" or "minimum". From any of these AB, AE and AF, we cannot remove any single attribute without making it non-candidate key. Should we also be considering AB, AE and AF also as MINIMAL candidate keys?

Answer 3

here C can identify the attributes , D, AB also , and You cannot identify only with E  and F as E and F cannot derive AB or C or D...but they can derive B so if we have AE, AB they can derive all the attributes now ypu cannot add C,D to E,F as C,D are already candidate keys , if u add they are made into super keys.... so number of candidate keys are 5...

but they asked minimal candidate keys so 2 which are C,D

Comments

I feel AB, AE and AF should also be considered "minimal". Usually minimal means "something from which we cannot remove anything" as against "smallest" or "minimum". From any of these AB, AE and AF, we cannot remove any single attribute without making it non-candidate key. Should we also be considering AB, AE and AF also as MINIMAL candidate keys?