DBMS Korth Edition 6 Exercise 6 Question 14 (Page No. 253)

Question

Consider the following relational schema for a library:

member(memb_no, name, dob)
books(isbn, title, authors, publisher)
borrowed(memb_no, isbn, date)

Write the following queries in relational algebra.
a. Find the names of members who have borrowed any book published
by “McGraw-Hill”.
b. Find the name of members who have borrowed all books published
by “McGraw-Hill”.
c. Find the name and membership number of members who have borrowed
more than five different books published by “McGraw-Hill”.
d. For each publisher, find the name and membership number of members
who have borrowed more than five books of that publisher.
e. Find the average number of books borrowed per member. Take into
account that if a member does not borrow any books, then that
member does not appear in the borrowed relation at all.

Comments

1. $\large \Pi _{name}(member \Join (Borrowed \Join (\Pi _{isbn}(\sigma _{publisher = "McGraw-Hill"} (Books) ) )))$

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i think 2nd query should output members who have borrowed all the books published by McGraw-Hill like if there are 2 books "a" and "b" then the members who have borrowed both the books should be present in resulting relation.

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yes you are  right ,is my query solution wrong for second?

(all name of member)-(name of member who borrowed book which is not published by Mcgraw Hill )

=name of member who  have borrowed all books published by "McGraw-Hill"

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a member who borrowed all books  published by "McGraw-Hill" can also borrow some other book by some other publication and will still be included in resulting relation.

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2. Πname(member)Πname(member) −−  Πname(member⋈borrowed⋈(σpublisher≠McGraw−HillBooks)

@Prateek, Since Member Name is not a key in the Member table.. So, it is never a good idea to Solve the Queries in terms of it (Non-key). For example, Let All the members have same name "Ram" then, Your Query will return Empty table if even a single person fulfills the condition. 

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Yup got it,thank u both of u

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For Second query  Π_name ( Member ⋈ Π_{M_Num}( Borrow / Π_isbn ( σ_{publisher="McGraw−Hill"}(Books))))

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is it possible to write the queries in relation algebra for 3,4,5 questions?

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is it possible to write the queries in relation algebra for 3,4,5 questions?

I am doubtful for 4,5. But 3 can be written in RA.  

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can u please provide me the solution

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$(Borrow / \Pi _{isbn}(\sigma _{publisher = "McGraw-Hill"}(Books)))$

Shaik Masthan what will be the result of this sub-query?

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here / is Division operator 

therefore (Borrow / Π_isbn(σ_{publisher="McGraw−Hill"}(Books)))

gives the all the tuples in Borrow which has linked with every isbn of (σ_{publisher="McGraw−Hill"}(Books))

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oh yes. thanks

Answer 1

Please give the answers to me