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for all the above questions answer the following :

a ) how many minimum relation tables are required which satisfy 1NF

b) how many minimum relation tables are required which satisfy 3NF

c) how many minimum relation tables are required which satisfy BCNF

d) minimum tables required

edited | 1.1k views
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make 5 diffrent problem ...)

and give proper entity then only noemal form concept clear

Here since nothing is mentioned about the attributes and keys of entities , so I will talk about INF form only as without attribute and key set we cannot formulate FDs and hence we cannot conclude about how many tables required for 2NF , 3NF etc which requires knowledge of keys and hence prime and non prime attributes.

So we talk about 1NF form only.

Now to decide minimum tables we should keep into account the cardinality of the entity sets that are participating into the relation and mode of participation i.e. total or partial.Both factors have an important role in deciding no of tables required.

For 1) :

Since it is many to one relation and participation is partial both sides of the relation , so only "many" side of the relation can be merged.Hence the tables will be : {E2R} , {E1} .Hence two tables required minimum for INF.

For 2) :

Now it is similar like previous one but the only difference is that E1 is participating totally , hence although it is in the "one " side of the relation , it can be merged.Hence in this case only 1 table required minimum.

For 3) :

Here total participation both sides hence both the entities can be merged with the relationship so only 1 table required here as well

For 4) :

Here we have total participation in "many" side of the relationship but "one" side of the relationship is participating partially .Hence the "one" side entity cannot be merged.Hence here we require 2 tables minimum. {E2R} , {E1}

For 5) :

Now here although it is a many to many relationship but total participation is here both sides of the relationship.Hence here also we require only 1 table minimum if no redundancy is considered (i.e. for 1NF only).

Note that in each of the above instances , we are merging the relationship set with either of the entity set or both of the entity sets , so no separate table is required for relationship set.

by Veteran (102k points)
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Thnaks habib..:)
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Most welcome..:)
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Sir, if we are considering 1nf only are (E1,R2) and (E2) also valid for the first case?
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@Habibkhan , 1NF just says that we should only have atomic values. So in all cases we can merge the tables I guess because even though redundancy will be there but all attributes will be atomic and hence in 1NF.
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Dumb people just know how to downvote without giving a appropriate reason. No use of giving answer to such people.
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Ya whoever has downvoted is wrong..They can discuss properly..
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@Habibkhan, can you check my comment please ?
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I have not downvoted ur comment.:).N as far as merging is concerned u have to merge as per the conventions mentioned in the answer. Else consistency of records cannot be ensured i.e. some new records may come out as a result of merging which is wrong..
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@Habibkhan, please check this question https://gateoverflow.in/143186/minimum-number-of-tables-to-represent-er-diagram

Bikram sir also said that 1 Table is enough for that question.

Also, how new records will come by merging ? Only some records will be redundant. I am not telling to take cross product. Just merge the elements in both tables which are actually related in one table.

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In case there is partial participation can't we insert null value after merging.That will mean we only need 1 in all cases.Is introducing null values violation of 1NF?
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in case 4, why can't we merge the entity with partial participation? one table should be fine, since we are concerned with finding the minimum number of table.
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if M:N mapping with total participation on M side only then how many tables req?

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same is my doubt...please any one clear it.
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1NF states that an attribute can have only one value from its domain. So, if we take just one table (in case of 5th diagram) then the tuples in the relation can have multiple values for the attribute that is reffering to the the other entity. Isn' it?
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so duplicate the row on the left side as many times as the row on right is repeating.

Example

a1 b1 related to ((c1, d1), (c2, d2), (c3, d3))

then in table u put

a1 b1 c1 d1

a1 b1 c2 d2

a1 b1 c3 d3

although redundancy will be there but table will be in 1nf. so i thnik only one table is required.

its late to answer but answering for anyone else who might end up here.