not every bcnf  decomposition is dependancy preserving with example?

    consider the relation schema.
            the set of fds is following.

    1. banker namep--> bname
    2. bname,cname -->banker_name

    the schema is not in bcnf as banker name isn''t super key
*         if we apply our algorithem we nmay obtain the decomposition.
           * cust_banker=(cname,banker_name)
              the closure of this dependancy does not include second one.
              -- thus the vialation of bname,cname-->bnker-name cannot be detected unless a join is computed .
              This shows that not every bcnf decomposition is dependancy perserving.      

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by suraj   dube  in Gujarati Literature  on 3/29/2015 11:19:35 PM  

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