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Consider the relational schema given below, where eId of the relation dependent is a foreign key referring to empId of the relation employee. Assume that every employee has at least one associated dependent in the dependent relation.

employee (empId, empName, empAge)

dependent(depId, eId, depName, depAge)

Consider the following relational algebra query:

$\Pi_{empId}\:(employee) - \Pi_{empId}\:(employee \bowtie_{(empId=eID) \wedge (empAge \leq depAge)} dependent)$

The above query evaluates to the set of empIds of employees whose age is greater than that of

1. some dependent.
2. all dependents.
3. some of his/her dependents.
4. all of his/her dependents.

(D) all of his/her dependents.

The inner query selects the employees whose age is less than or equal to at least one of his dependents. So, subtracting from the set of employees, gives employees whose age is greater than all of his dependents.
What's wrong in option B.Its also selecting all dependents value.
if $empId=eID$ this condition won't be there in question, then answer will be B. But in this case if any employee E1 is selected that means E1 age is greater than all of E1's dependent, E1 age might be smaller/greater than E2's dependent.

The below subquery after the subtraction sign produces id's of those employees who have at least one dependent with age greater than or equal the employee's age.

When the result of above subquery is subtracted from all employees, we get the employees whose age is greater than all dependents.

(C) some of his/her dependents

consider the qN as A-B

B select all the empID with age which is less than all his dependends

so subtracting above from the universal set give the opposite of which is "empID with age greater than atleast one of his dependents"

Procedure is correct. But
"B select all the empID with age which is less than all his dependends"

is wrong. Select statement returns empID even if it is true for at least one of the dependent. So, in B it is "some" and in A-B it will be "All"