The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
+2 votes
Answer is 2 minimal and 2 canonical covers.

Please give full explanation of how to solve.
asked in Databases by Junior (821 points) | 642 views

3 Answers

+4 votes
Miinimal (canonical) cover of a set of FDs is the minimal set of FDs such that all other FDs can be derived.

So, {X -> Y, Y -> Z, Z-> X}, {X->Z, Z-> Y, Y->X}
answered by Veteran (407k points)
+1 vote

minimal cover is known as canonical cover
here two canonical cover are possible
we get canonical cover by eliminating redundant FDs
1st cv :
becoz z->x so we can replace x by z in y->xz ( becoz if child derivates sumthing dan child will surely do dat) nd get y->z (1)
by spliting we cn write x->yz as x->y (2) nd x->z(3)
if we combine (2) nd (1) by union rule we get x->z
which is nothing bt (3) FD so we cn elliminate (1) nd (2) becoz de r redundant
nd we got 1st canonical cover as {x->z}
similarly we got 2nd minimal cover as {y->z}.

see this:

answered by Active (2k points)
But with these two covers x -> z and y -> z, we lost the relation between x and y.

Neither do we have x->y nor y->x
If those two are minimal dependency, thnt there that is lossy .. Llss in FDs
+1 vote
x→y and z→x are minimal cover
answered by Boss (38.2k points)
edited by
x -> z?

Sir if remove x -> z  then by taking (x)+=xyz So is it n,t redudant .

Related questions

+1 vote
3 answers
+1 vote
3 answers
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
49,532 questions
54,136 answers
71,060 users