R(a,b,c,d) with Fd\'s a->b, b->c , c->d , and d->a A) indicate all the BCNF viol

Question

R(a,b,c,d) with Fd's a->b, b->c , c->d , and d->a
A) indicate all the BCNF violationd. Do not forget to consider Fd's that are not in the given set, but follow from them. However it is necessary to give violations that have more than one attribute on the right side.
B) Decompse the relations, as necessary , into collections of relations that are in BCNF
R(a,b,c,d) with Fd's a->b, b->c , c->d , and d->a
A) indicate all the BCNF violationd. Do not forget to consider Fd's that are not in the given set, but follow from them. However it is necessary to give violations that have more than one attribute on the right side.
B) Decompse the relations, as necessary , into collections of relations that are in BCNF
R(a,b,c,d) with Fd's a->b, b->c , c->d , and d->a
A) indicate all the BCNF violationd. Do not forget to consider Fd's that are not in the given set, but follow from them. However it is necessary to give violations that have more than one attribute on the right side.
B) Decompse the relations, as necessary , into collections of relations that are in BCNF

Explanation / Answer

C->D and C->A: The set closure of C+ is { C, D, A } therefore the left hand side is not a superkey. BCNF Violation

D-> A: C->D: The set closure of D+is { D, A } therefore the left hand side is not a superkey. BCNF Violation

DB->C: The set closure of DB+is { D, B, C, A } therefore the left hand side of this FD is a superkey.

DECOMPOSE

A+ = { A }                  AB+ = { A,B,C,D }     BC+ = { B,C,D,A }     CD+ = { C,D,A }

B+ = { B }                   AC+ = { A,C,D }        BD+ = { B,D,A,C }

C+ = { C,D,A }           AD+ = { A,D }

D+ = { D,A }

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