maximum and minimum number of tuples in natural join

ⅰ亾dé卋堺 提交于 2021-02-02 09:17:55

问题


I came across a question that states

Consider the following relation schema pertaining to a students

  • database: Student (rollno, name, address)
  • Enroll (rollno, courseno, coursename)

where the primary keys are shown underlined. The number of tuples in the Student and Enroll tables are 120 and 8 respectively. What are the maximum and minimum number of tuples that can be present in (Student * Enroll), where '*' denotes natural join ?

I have seen several solutions on Internet like this or this

As per my understanding. maximum tuples should be 8 and minimum should be 8 as well, since for each (rollnum,course) there should be a roll num in Students. Anyone who can help in this regard


回答1:


If there was a referential constraint in place ensuring that every rollno in Enroll must also appear in Student then your answer of 8 for both minimum and maximum would be correct. The question doesn't actually mention any such constraint however. There's no need to assume that the RI constraint exists just because the rollno attribute appears in both tables. So the best answer is 0 minimum and 8 maximum. If it's a multiple-choice question and 0,8 isn't one of the given answers then answer 8,8 instead - and tell your teacher that the question is unclear.




回答2:


I hope, you understood what Natural Join exactly is. You can review here.

If the tables R and S contains common attributes and value of that attribute in each tuple in both tables are same, then the natural join will result n*m tuples as it will return all combinations of tuples.

Consider following two tables

Table R (With attributes A and C)

 A  |  C
----+----
 1  |  2
 3  |  2

Table S (With attributes B and C)

 B  |  C
----+----
 4  |  2
 5  |  2
 6  |  2

Result of natural join R * S (If domain of attribute C in the two tables are same )

 A | B |  C
---+---+----
 1 | 4 |  2
 1 | 5 |  2
 1 | 6 |  2
 3 | 4 |  2
 3 | 5 |  2
 3 | 6 |  2  

You can see both R and S contain the attribute C whose value is 2 in each and every tuple. Table R contains 2 tuples, Table S contains 3 tuples, where Result table contains 2*3=6 tuples.

Moreover, while performing a natural join, if there were no common attributes between the two relations, Natural join will behave as Cartesian Product. In that case, you'll obviously have m x n as maximum number of tuples.

Consider following two tables

Table R (With attributes A and B)

 A  |  B
----+----
 1  |  2
 3  |  2

Table S (With attributes C and D)

 C  |  D
----+----
 4  |  2
 5  |  2

Result of natural join R * S

 A | B |  C |  D
---+---+----+----
 1 | 2 |  4 |  2
 1 | 2 |  5 |  2
 3 | 2 |  4 |  2
 3 | 2 |  5 |  2

Hope this helps.




回答3:


If you are asking about the maximum number of tuple that could appear in the natural join of R and S the its the Cartesian product of both the tuples




回答4:


Yes answer should be 8,8 . Because Rollno is key in Student table and rollno,courseno are compound key . Relationships between Student and enrol table is 1:M . So maximum number of tuples is same as many side ie. 8 And minimum number of tuples is 8 if Foreign key exist other wise 0.

So answer is 8,8 .



来源:https://stackoverflow.com/questions/22673235/maximum-and-minimum-number-of-tuples-in-natural-join

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