Let R(A,B,C,D,E,P,G) be a relational schema in which the following FDs are known to hold:AB->CDDE->PC->EP->CB->GThe relation schema R isin 3NF, but not in BCNFin BCNFin 2NF, but not in 3NFnot in 2NF

To determine the normal form of a relation, we need to examine its functional dependencies (FDs).

1. First, we need to find the candidate keys for the relation R. A candidate key is a minimal set of attributes that can uniquely identify a tuple. From the given FDs, we can see that AB is a candidate key because it determines all other attributes (AB->CD, DE->P, P->C, C->E, B->G).

2. Next, we check if the relation is in 1NF, 2NF, 3NF, or BCNF.

   • A relation is in 1NF if it has an atomic value for each attribute in a tuple. Since there is no information given that contradicts this, we can assume that the relation is in 1NF.

   • A relation is in 2NF if it is in 1NF and every non-prime attribute is fully functionally dependent on any candidate key. In this case, all non-prime attributes (C, D, E, P, G) are fully functionally dependent on the candidate key AB. So, the relation is in 2NF.

   • A relation is in 3NF if it is in 2NF and every non-prime attribute is non-transitively dependent on every candidate key. Here, all non-prime attributes are non-transitively dependent on the candidate key AB. So, the relation is in 3NF.

   • A relation is in BCNF if it is in 3NF and for every functional dependency X -> Y, X is a superkey. However, in this case, we have the FDs P->C and B->G where neither P nor B are superkeys. So, the relation is not in BCNF.

Therefore, the relation schema R is in 3NF, but not in BCNF.