ER model

Question

Consider the following entity relationship diagram and three possible relationship sets (I, II, and III) for this E-R diagram.

If different symbols stand for different values (eg. t1 is definately not equal to t2) then which of the following could NOT be the relationship set for the ER diagram?

(A) I only
(B) I and II only
(C) II only
(D) I, II, and III

Answer 1

When a relationship's degree is more than binary (ternary and 4-ary....) and there is more than one arraw then in literature it is interpreted in two ways.

1st interpretation: 

Since S has arrow it is interpreted as

for every PQT there is atmost one S  [ PQT  S]

Since T has arrow it is interpreted as 

for every PQS there is atmost one T  [PQS T]

2nd interpretation:

Since S, T have arrows, in this interpretation it is interpreted as PS and PT and QS, QT

Since there are two interpretations if you follow first interpretation then Answer is (A) and if you follow 2nd interpretation then it is (B).

Hence , both of the options are true.

Comments

very nice..post it on techtud also mam has just solved for option A

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and thanks a lot

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Mention not plz bro..:)

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@habibkhan you are duing excellent job bro... really... hats of to you.. :)

Answer 2

In the above question P and Q are  combinely  act as the primary key ,since all the attributes can be derived by them.

In 1 relationship table , p and q are not uniquely deriving t. So it is wrong.

In 2 relationship table, pand q are not uniquely deriving s and t.

In 3 rd relationship table p  and q are uniquely deriving  s and t . so it is right.

Hence answer should be (B).

Comments

how u found P and Q are primary key?

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Since P to S ,P to T  have 1 to many relationship. similarly Q to S and Q to  t have 1 to many relationship, so they can combinely act as primary key in relation.

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actually correct option is A
http://www.techtud.com/comment/7076#comment-7076

see this and please explain in detail m nt getting this solution