ME Workbook

Question

1. How many minimum tables required which satisfy 1NF.

2. How many minimum tables required which satisfy BCNF.

3. How many minimum tables required which satisfy 4NF.

Answer 1

Minimum tables required which satisfy 1NF     ====   4

( because of multivalued attributes )

 

 Minimum tables required which satisfy BCNF  ===== 4

(beacause each attribute is primary key )

Comments

I think number of tables from 1NF to BCNF will be 2 only,

because in one table we will keep all the multivalued attr along with key attr and in the second table we will keep all single valued attribute along with key attr.

and in 4NF we will keep each multivalued attribute in diff table along with key attr

hence answer should be

1 > 2

2 > 2

3 > 4

correct me if i am wrong ?????

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Given Answer is wrong
For 1 NF -> 1 table is required
MV attr can still be placed in 1 NF table
like if EID is 1 and there are two Phone No. for EID 1 you can still make a table like this where Primary Key is Now EID Phone No.
1 - A
1 - B
For BCNF -> 2 table required
For 4 NF -> 4 table required

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@aehkn for 2NF and 3NF we need 2 tables right?

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I think yes

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BCNF satisfies 1NF, 2NF, 3NF but not 4 NF

right?

that is why  u telling 2 table for BCNF.

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Relation1(EID, Ename, Age) primary Key : EID
Relation2(EID, PolicyNo., PhNo. , Email )   primary key : EID PolicyNo. PhNo. Email

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@aekhn

for some EID, it might be case that it has no phone no., in that case phone no. = NULL, but prime attributes should never be NULL according to entitiy intigrity constraint. so EID PolicyNo. PhNo. Email can not be primary key..

for 3 NF, 4 tables are required, one for orginal entitiy and 3 each for multivalued attribute.

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@ joshi_nitish

we use multivalued attribute for 4 NF

not for 3 NF

rt?

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In R(Eid,Ph.No)    Ph.No still can be null so EIC is violated here too so will this Relation hold?

Answer 2

For 1NF We can make a single table by making Eid, policyNumber, email, and phone number a single candidate key.