search
Log In
2 votes
295 views

in Databases
edited by
295 views
1

as per my knowledge,

in the FD CD -> E, D is not redundant....

why ?

due to CD -> E only, C -> D happens via E ->A  A -> B and B -> D.

 

what is the solution they provided?

0
Yes that's what I did wrong in this question.. I chose A as answer but answer is B
0

@Shaik Masthan @tusharforever @srestha I have a doubt here.

CD ---> E , D is extraneous attribute only if C closure can generate D on removal of D? 

I dont get this condition would you please elaborate?

0
i didn't get your doubt correctly
0

@Shaik Masthan

My doubt is why  in the FD CD -> E, D is not redundant....?

0
Then no comments
1
Thanks alot I appreciate your time.!!

1 Answer

1 vote
Answer is b because left side of each function dependencies  is unique.
0

@Jony Bhatt

What do you mean by left side of each function is unique??

0
The condition for which you talking about "left side of each function dependencies  is unique", may be necessary but not sufficient.

Related questions

0 votes
2 answers
1
1 vote
2 answers
2
558 views
Consider the attribute set R = ABCDEGH (where Functional Dependencies are F= {AB·--> C,AC --> B: AD ---> E, B -----> D, BE --> A, B -> G}.
asked Jun 19, 2015 in Databases gauravalgo 558 views
0 votes
3 answers
3
235 views
Consider the following relation R(A, B, C, D, E, F, G) and set of functional dependencies. F={BCD → A, BC → E, A → F, F → G, A→G, C → D} Which of the following is minimal cover of F? {BC → A, BC → E, A → F, C → D, A → G} {BC → A, B → E, A → F, F → G, C → D} {BC → A, BC → E, A → F, F → G, C → D} {BC → A, BC → E, A → F, C → D}
asked Jan 5, 2019 in Databases Shivam Kasat 235 views
...