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vishal_19_official
Asked: December 14, 20192019-12-14T20:57:00+05:30 2019-12-14T20:57:00+05:30In: Computer Science

Justify “Any relation which is in BCNF is in 3NF but converse is not true”.

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Justify “Any relation which is in BCNF is in 3NF but converse is not true”.
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    2019-12-15T16:02:20+05:30Added an answer on December 15, 2019 at 4:02 pm

    Consider the schema and functional dependency set of Empdept given below:

    Empdept (emp# ,Dept#, Manager#, Dept_Name , Dept_Loc)

    Emp#Dept#  manager# manager#  Dept #

    Normalisation:

    1st normal form: the relion is already in 1nf as each cell is single valued.

    2nd Normal form: A relation is in second normal form if it is 1NF and every non key attribute is fully dependent on each candidate key of the relation. Thus R is also in 2NF.

    3rd Normal form: A relation is in second normal form if it is 2NF and every non key attribute is non-transitively dependent on each candidate key of the relation. This above relation is in 3NF as every non key attribute is non-transitively dependent on each candidate key of the relation.

    Here primary key is emp# and candidate key is (emp# ,Dept#, Manager#)

    BCNF Normal form: A relation is in second normal form if it is 3NF and if X->A in R and A is not in X, then X is a candidate key. The above relations satisfy this condition.

    The above relation is not in BCNF as in Fd manager#  Dept # , dept# is dependent on manager# but manager#is not in candidate key.

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