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Given R with n tuples S with m tuples n<m  then How many minimum and maximum tuples in follwing relations . Please Justify with Reason / Examples

  • R-S
  • S-R
  • R Left Join S
  • R  Natural Join S
  • R/S
  • S/R
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4 Comments

Confused ! What is correct one ? :(
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@pc,if no tuple is common then we go for cartesian product.am i right??
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Exactly what I have told above. Even shared few References :)
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1 Answer

4 votes
4 votes

[Someone should verify . This is what I think]

1. R UNION S

  • max : n+m
    Reason : union we add all the tuples from both relations. ie When R and S have no common tuple.

  • min: n
    Reason : The minimum is n (the greatest of the two sizes, m and n). When all the tuples of R also exist in S.

2. R INTERSECTION S

  • max : m ( m<n )
    Reason : both relation contains same tuples then we may get maximum m keys

  • min: 0
    Reason : taking m=n=null if no common tuples in both relations

3. R - S

  • max : m 
    Reason : if they are disjoint then in R-S we will get all tuples of R

  • min: 0
    Reason : if all tuples in R is also present in S
     

4. S - R

  • max : n 
    Reason : as explained above
     

  • min: n-m
    Reason : m<n  there will be some tuples in S after deleting the common tuples

5. R natural join S

  • max : n*m
    Reason :  if no matching key constraints natural join will produce Cartesian product )
     

  • min:  0
    Reason : Identical with case 2 (INTERSECTION).

6. R LEFT OUTER JOIN S

  • max : m*n
    Reason : if all rows in left tables matches with all rows in right table

  • min: m  (could be 0 when m= 0)
    Reason : The minimum is 1 when m=1 , minimum is 2 when m=2, minimum is 0 when m=0

7. R / S

  • max : m 
    Reason :   when n=0

  • min: 0
    Reason :  Consider that relational division is similar to integer division. 3 / 7 gives 0 in integer division for example. Try to convert this into relational division

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3 Comments

For 7th point, if m=0 then R / S  has zero tuples. Isn't it?
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6.max of left natural join

How will all rows of left table match with all of right?
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For natural join

mn is max.

Case 1: if there is a common attribute between and , and every row of R matches with the each row of S - i.e., the join attribute has the same value in all the rows of both and ,
Case 2: If there is no common attribute between R and S .

Min is 0.

If There is a common attribute between R and S and nothing matches.

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