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I think the question is incorrect… as there are only few FD’s given.

Assuming 1NF as there is single partialFunctional Dependency. 

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Bikram asked Aug 26, 2017
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Decompose the following table into BCNF:$R(ABCD)$$A \rightarrow C$$C \rightarrow A$$AB \rightarrow D$The result is:(AC) (ABD)(AC) (CBD)(AB)(BAD)Both (A) & (B)