ER diagram

Question

Minimum number of tables required for above diagram ________ 

Comments

ans is 4

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E3R4

E1R1E2R2

R3

E4R5

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why E1 and E2 are merged both are strong entities ryt?? one table for each ....?

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Because of total participation

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@anjan just follow these rules , you will be easily solve these questions.

CASE --->ONE-ONE

A. Both sides partial participation { E1<---R-->E2 }

     In this case you can merge R either side . so two table will be required.

A. Both sides total participation { E1<==R==>E2 }

     In this case you can merge all in a single table . so only 1 table will be required.

A. one sides total participation { E1<==R--->E2 }

     In this case also you can merge all in a single table . so only 1 table will be required.

read this and then try to answer the above ques .

THE ans is 4 . { E3R4 , E1R1E2R2 ,R3, E4R5 }

CASE --->1-m

A. { E1<---R------E2 }

In this case relation will be merged with the entity having ONE mapping  , So we got two table ( E1 ,E2R)

CASE --->n-m

A. { E1-----R------E2 }

In this case relation can not  be merged with the any entity ,and  we got 3 tables ( E1 ,E2, R) where R stores the foreign key to both E1 and E2.

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Thanks a lot :)

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Thanks bro :) Very helpful

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why ans is not 5?

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i got 5

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@raviyogi Isn't there any effect of Total Participation in One to Many and Many to Many relationships?

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no effect@vishal

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@raviyogi tell me what will be the minimum of relation required for M:1 relationship if the 1 side is fully participated. its going to one na.

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@raviyogi Dear, can you provide a reference where are the rules for merging the two "Entity" Sets given (in cases of total participation)? I did not find any such rules in Korth's book (Sixth Edition).

Thank You.

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@raviyogi does this rules you explained above applicable for weak entity depending on a strong entity?