Question

Consider a relation with schema R(C,T,S,N,G) and FDs C->T, CS->G, S->N.

a) Find a key for the relation using closure test.
b) Is relation R in BCNF? If not, decompose R and make the resulting relations in BCNF.

Answer 1

X= {C,S}
C->T: X+ {C,T,S}
CS-> G: X+ {C,T,S,G)
S-> N: X+ {C,T,S,N,G}

Neither C or S is a superkey.

But I'm not sure how to do the res.t