Consider a relation geq which represents "greater than or equal to", that is, $(x,y) \in $ geq only if $y \geq x$.

create table geq ( ib integer not null, ub integer not null, primary key ib, foreign key (ub) references geq on delete cascade );

Which of the following is possible if tuple (x,y) is deleted?

- A tuple (z,w) with z > y is deleted
- A tuple (z,w) with z > x is deleted
- A tuple (z,w) with w < x is deleted
- The deletion of (x,y) is prohibited

### 13 Comments

https://www.geeksforgeeks.org/gate-gate-cs-2001-question-50/

Solution becomes easier by taking examples

"foreign key (ub) references geq" Means referring to primary key of table geq(if foreign key references to same table then referenced key must be primary key of same table with same data type). In this case Parent table(Foreign key column) and child table (referenced column) both denoting same table.

**Referenced column by Foreign key in same table or in child table must be a primary key.**

NOTE-If more than 1 table involved in foreign key then you can replace the word primary key with candidate key above.

## 5 Answers

Answer: C

The table can be depicted as:

$$\begin{array}{|c|c|c|} \hline \textbf{ib(PK)} & \textbf {ub(FK)} \\\hline \text {z} & \text{w = u}\\\hline \text{u} & \text{v = x} \\\hline \text{x} & \text{y} \\\hline \end{array}$$

If $(x,y)$ is deleted then from the above table:

- $v\leq y$ $($as $v=x)$
- $u<v\leq y, u!=v$ $($as $v=x$ and ib is the Primary Key$)$
- $w<v\leq y$ $($as $w=u)$
- $z<w<v\leq y, z!=w$ $($as $w=u$ and ib is the Primary Key$)$

As, it can be seen that $w<v$ or $w<x$ $($as $v=x)$ so C is the answer.

### 16 Comments

@ankyAS when you solve by taking an example in any example, form example such that it will be worst case. I mean here it is given that on delete cascade, form example when there will be many cascading delete due to deletion of one tuple. In second cascade delete you will get the answer. :)

w <= x (and also z<=x) would be the correct option.

Say suppose (2,3) is deleted then we additionally delete (2,2). That is the reason we need to have <= and not just <.

Ex:

lb |
ub |

0 | 0 |

0 | 1 |

0 | 2 |

0 | 3 |

1 | 1 |

1 | 2 |

1 | 3 |

2 | 2 |

2 | 3 |

3 | 3 |

If in this table (2,3) is deleted, then additionally we need to delete the below records:

lb |
ub |

0 | 0 |

0 | 1 |

0 | 2 |

1 | 1 |

1 | 2 |

2 | 2 |

** But this is wrong, because we can't have duplicates in the primary key (ib) in the first place.**

Hence Option C is correct.

See, if we write three relation $\left ( x,y \right ),\left ( y,z \right ),\left ( z,w \right )$, then see there is a on delete cascade relationship in it. But in question they have only given $\left ( x,y \right ),\left ( z,w \right )$and asked to find inner one, that is $\left ( y,z \right )$

tuple (x,y) is to be deleted so let's take it into relation.

Now we have to observe the effect on tuple (z,w) so we will take it also in our table.

So till now, we are having two tuples,** (x,y)** and **(z,w)**. Now let's consider each option:

a) z>y means (y,z) tuple is there. So deletion of (x,y) doesn't give any cascade relation between (x,y),(z,w) and (y,z)

b) z>x means (x,z) exists in relation. But since 'ib' is primary key, we can't repeat 'x' .

**c) w<x means (w,x) exists. On deleting (x,y), (w,x) should be deleted and then (z,w) to maintain DELETE CASCADE. **

Hence option C is correct choice

### 9 Comments

Consider the below example

lb | ub |
---|---|

5 | 8 |

3 | 5 |

8 | 8 |

2 | 3 |

Deleting first tuple will force us to delete 2 nd tuple

And deleting 2 nd tuple will force us to delete 4 th tuple.

Now because of deletion of first tuple (5,3), 4 th tuple (2,3) should be deleted ... Here z = 2 and x = 5 and z<x

So option C) is the answer ...

Hence the table geq is both the master ( which has the referenced key ) as well as the child table (which has the referencing key).

The table has two constraint, one is that if there is a tuple ( x, y ), then y is greater than or equal to x, And the other is referential integrity constraint, which is on-cascade-delete on the foreign key.

On-cascade-delete says, that “When the referenced row is deleted from the other table (master table), then delete also from the child table”.

Suppose the instance in the given relation is the following:

x y

-----

5 6

4 5

3 4

6 6

Now if we delete tuple (5,6) then tuple ( 4,5 ) should also be deleted ( as 5 in the tuple (4, 5) was referencing to 5 in the tuple(5,6) which no longer exist, hence the referencing tuple should also be deleted), and as (4,5) got deleted hence tuple (3,4) should also be deleted for the same reason.

Therefore in total 3 rows have to be deleted if tuple ( 5,6 ) is deleted.

Now from the above instance we can say that if (x,y), i.e. ( 5,6 ) gets deleted then a tuple ( z, w) i.e, ( 3, 4) is also deleted. And we can see here that w < x. Hence option C.