ternary relationship

Question

Consider the following ER diagram:

If A has 100 entities, B has 1000 entities, and C has 10 entities, what is the maximum number of triples of entities that could be in the relationship set for R?

Comments

Sir,
I only have a query
In a relation table we are having three attributes (attribute from A,B,C)
out of 3 attributes which attribute from(a,b,c) we  consider as a primary key of Relation Table.

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A will be the primary key for R

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@Arjun sir pls explain

Answer 1

1000 tuples will be in relationship set!

Comments

My interpretation :

Create a relation table for R. This relation table will contain A as its primary key attribute. (Please refer Korth book for creating relation tables for n-ary relationship sets).

Now since there are at max 100 different values of A's primary key, so only 100 different entities should be present in R.

 

Please point out any possible errors in my interpretation. Thank you.

Answer 2

In relation R participation of A is many while participation of B and C is one . That means there is 1 to 1 relation between B and C,many-one between A and B, many to one between A and C.

A to B is many to one that means in every relationship,  only one entity of B can participate and many entity of A is participate in one relationship. So, total number of triples =  1000 triples .

As B have 1000 triples and each entity of B must take participation (maximum participation) .....(i)

B and C is one to one. So, only 10 triples are possible.  ...(ii)

A to C also many to one relation, C will also form 10 triples. ...(iii)

among (i) , (ii) and (iii) 1000 is maximum number of triples of entities that could be possible in the relationship set for R.

So maximum triples can be 1000 .

Comments

<AB, AC, BC > forms a triples.

We have to find a maximum entity in this triples.

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Rule 1: for unique pair of a&b there can only be unique value of c.

Rule 2: for a unique pair of a&c there can only be unique value of b.

a&b together can make 100*1000=100,000 unique combinations but c has only 10 values. hence R can't have 100,000 enteries.

Since number of unique combinations of a&c are 1000 and b has 1000 entities.

hence we can only have max 1000 records in Relationship set as said by @Bikram

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@Arjun @Manu Thakur and sir if they asked about minimum number of triples...then it would be 10??

As A and B can have combination of 100000 entities and with C it can 10 entities...So on overall combination 10 entities are possible??

Answer 3

edited####=>>

I strongly suggest that all answers here are wrong and interpreted the question very wrongfully..

@Arjun@Bikram@Manu Thakur

in a relationship in ER model with n relations participating together we NEVER TAKE 2 RELATIONS AT A TIME AND SAY 1:m or 1:1 etc..

THE RULE IS TAKE (n-1) at a time with the other one..

so here AB and C; BC and A;AC and B are only 3 categories of saying whether they are x:y (x,y ∈ {1,m} ) .

YOU CAN'T JUST INVENT A 1:M OR WHATEVER RELATIONSHIP BETWEEN ONLY 2 OF THEM...THIS INTERPRETATION IS NOT CORRECT.

TO DO THAT FIRST CONVERT IT TO SOME DUAL RELATIONSHIP MODELS WHICH IS OF NO NEED HERE.

for A,B,C its m,1,1 given in diagram

meaning..

stmt_1)for one pair of A,B there can be at most 1 C

stmt_2)for one pair of A,C there can be at most 1 B

stmt_3)for one pair of B,C there can be at most m A's

so,  A,B->C ; A,C->B

******we can never know if A->B , A->C, B->C , C->B holds or not.*******

here Candidate keys are both A,B and A,C

and A,B,C is a strict superkey.

now, this two fd together never indicates B->C or C->B

A1,B1,C1

A2,B1,C1

A3,B1,C2

all can be in R

 

EVEN IF THEY DID, THAT DOES NOT MEAN THAT YOU STATE A RELATIONSHIP ONLY BETWEEN 2 OF THEM DIRECTLY.

IN A STATEMENT DICTATING RELATIONSHIP, 3 RELATIONS MUST BE THERE IF YOU'RE INTERCEPTING TERNARY RELATIONSHIP.

now the very fact that both AB and AC are candidate keys in turn makes each of A,B & C a prime attribute.

so none of them can contain NULL value.

now it asks for maximum number of triples in R,

so answer must be    100*min(1000,10)=1000