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Here directly dependency not given. Which normal form it doesnot hold , how to eliminate it? I got here 1:m, or 1:1 dependency. Is it any value to ans this question?

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I AM ALSO GETTING RELATIONSHIP AS 1:M 

AS SPECIALIZATION AS RELATIONSHIP AND EACH EMPLOYEE HAS MANY SPECALIZATION.

 

NOW ABOUT NORMAL FORM I AM THINKING IT IS 3NF NOT BCNF 

 

JUST CREATE TABLE AS

EMPID SPECIALITY  MANGERID
A 1 C

2 B
C 1 C
D 2 M

(EMPID+ SPECIALITY) UNIQUE AS STATEMENT EMPLOYEE CAN HAVE MANY SPECIALITY. 

NOW MANAGER ID IS UNIQUE WITH RESPECT TO SPECIALITY AS EACH MANAGER ONE SPECIALITY.

SO THINK OD EMPID AS A AND SPECIAL AS B , MNGID AS C

SO AB------>C

 AND C-------------> B

SO BY THESE IT IS 3NF BUT NOT BCNF AS C NOT CANDIDATE KEY.


IF ANYTHING WRONG LET ME KNOW :)

edited by
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The given relation is:

$emp(empid, speciality, managerid)$

The following functional dependencies hold:

$managerid \to speciality $

$empid,speciality \to managerid$

Thus, we see that (empid, speciality) is the candidate key.

Thus, the relation is in 3NF as the RHS of first dependency is prime attribute while the LHS of the second dependency is the superkey.

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