What is the difference between having the JOIN(⋈) inside the relation instance and before the other selection operator (like SELECTσ)?

Question

I'm having trouble understanding how to provide proper relational algebra expressions to some of my assignment problems. I realized instead of having the JOIN selection operator outside the relation instance, maybe I can put it also inside it. I'm sorry if I don't make much sense but here's what I am thinking about:

The question is

Find the name of all employees whose salary is more than 15,000.00 and managed by Paul

WORKS (employee-name, company-name, salary)

MANAGES (employee-name, manager-name, position)

My answer is Π(employee-name) ((MANAGES⋈(σ(salary > 15000 ∧ manager-name="Paul")(WORKS))) https://i.stack.imgur.com/ZQ6ZM.png

But should I just instead do this Π(employee-name) (( σ(salary > 15000 ∧ manager-name="Paul")(WORKS⋈MANAGES) https://i.stack.imgur.com/6nwIy.png

[Re relational querying.](https://stackoverflow.com/a/24425914/3404097) — philipxy, Jun 29 '21 at 22:45
Why would you do either? PS Please read the edit help on formatting code inline & in blocks. Please indent code reasonably. — philipxy, Jun 29 '21 at 22:46

score 0 · Answer 1 · answered Jun 29 '21 at 16:29

0

The problem with your solution is the right operand of the JOIN:

σ(salary > 15000 ∧ manager-name="Paul")(WORKS)

the attribute manager-name is inside MANAGES, not WORKS, so you cannot use it in the restriction. The correct solution is either the second one, or alternatively, the following:

Π(employee-name) ((σ(manager-name="Paul")(MANAGES))⋈(σ(salary > 15000)(WORKS)))

answered Jun 29 '21 at 16:29

Renzo

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Ohhhh I get the difference now....I'll be applying that knowledge in more of my assignments. Thank you so much! – Nick Jun 29 '21 at 16:45

What is the difference between having the JOIN(⋈) inside the relation instance and before the other selection operator (like SELECTσ)?

1 Answers1