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Answer is 2 minimal and 2 canonical covers.

Please give full explanation of how to solve.

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}
+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}.

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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
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If those two are minimal dependency, thnt there that is lossy .. Llss in FDs
+1 vote
x→y and z→x are minimal cover
edited by
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x -> z?
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Sir if remove x -> z  then by taking (x)+=xyz So is it n,t redudant .

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