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ER model

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4 4 votes

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

2 Answers

8 8 votes

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 \rightarrow S]

Since T has arrow it is interpreted as 

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

2nd interpretation:

Since S, T have arrows, in this interpretation it is interpreted as P\rightarrowS and P\rightarrowT and Q\rightarrowS, Q\rightarrowT

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.

2 2 votes
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).
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