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ER- no of tables?

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4? E1,E2,E3,R2
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isn't R1 is m:n relationship as well which needs an another table? @Abhisek Tiwari 4

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then also i think we can do it in 4 if no case for normalization is mentioned.

is there cardinality info?
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no..but i think the given anwer is 3
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i think 3 will be the answer
+2
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If R1 is 1:1 with total participation on either side, then number of relations should be only 2.

E1E2R1(P,Q,R,S)  and E3R2(T,U,R)
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but i don't think it's total participation on both sides..
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Either doesn't mean both.
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according to me it is 1:1 relationship with partial on either end(total+partial)
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thanks @adarsh_1997 for pic. i was not sure about these rep and treated R2 as many to many.

i think u must be sure about correctness of this snap?

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I think it can be minimized ---- >  2 table

right ??

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