1. Assignment 2 Relational Algebra Which tables? What operations?
Common attributes?
What result (attributes)?
Syntax (Standard Notations and Symbols)
Product: X
Join:

2. I a) R (A1, A2) R = {(1, 2), (4, 3), (2, 9), (3, 8)} S (B1, B2, A1)
S = {(x, a, 1), (y, b, 1), (z, a, 4)}
A1, B2 (A1 < 4 and B1  x (S))
= A1, B2 ( { (y, b, 1) } )
= { (1, b) }

3. I b) R (A1, A2) R = {(1, 2), (4, 3), (2, 9), (3, 8)} S (B1, B2, A1)
S = {(x, a, 1), (y, b, 1), (z, a, 4)}
A2, B1 ( R.A1 > S.A1 ( R  S))
= { (3, x), (3, y), (8, x), (8, y), (9, x), (9, y)}
R  S
(1, 2, x, a, 1)
(1, 2, y, b, 1)
(1, 2, z, a, 4)
(4, 3, x, a, 1)
(4, 3, y, b, 1)
(4, 3, z, a, 4)
(2, 9, x, a, 1)
(2, 9, y, b, 1)
(2, 9, z, a, 4)
(3, 8, x, a, 1)
(3, 8, y, b, 1)
(3, 8, z, a, 4)
R.A1 > S.A1 ( R  S)
(4, 3, x, a, 1)
(4, 3, y, b, 1)
(2, 9, x, a, 1)
(2, 9, y, b, 1)
(3, 8, x, a, 1)
(3, 8, y, b, 1)

4. I c) R (A1, A2) R = {(1, 2), (4, 3), (2, 9), (3, 8)} S (B1, B2, A1)
S = {(x, a, 1), (y, b, 1), (z, a, 4)}
R S
= { (1, 2, x, a), (1, 2, y, b), (4, 3, z, a)}
R  S
(1, 2, x, a, 1)
(1, 2, y, b, 1)
(1, 2, z, a, 4)
(4, 3, x, a, 1)
(4, 3, y, b, 1)
(4, 3, z, a, 4)
(2, 9, x, a, 1)
(2, 9, y, b, 1)
(2, 9, z, a, 4)
(3, 8, x, a, 1)
(3, 8, y, b, 1)
(3, 8, z, a, 4)
R.A1 = S.A1 ( R  S)
(1, 2, x, a, 1)
(1, 2, y, b, 1)
(4, 3, z, a, 4)

5. I d) R (A1, A2) R = {(1, 2), (4, 3), (2, 9), (3, 8)} S (B1, B2, A1)
S = {(x, a, 1), (y, b, 1), (z, a, 4)}
A2, B1, B2, S.A1 (R S)
= { (2, x, a, 1), (2, y, b, 1), (3, z, a, 4)}
R  S
(1, 2, x, a, 1)
(4, 3, x, a, 1)
(2, 9, x, a, 1)
(3, 8, x, a, 1)
(1, 2, y, b, 1)
(4, 3, y, b, 1)
(2, 9, y, b, 1)
(3, 8, y, b, 1)
(1, 2, z, a, 4)
(4, 3, z, a, 4)
(2, 9, z, a, 4)
(3, 8, z, a, 4)
R S
(1, 2, x, a, 1)
(1, 2, y, b, 1)
(4, 3, z, a, 4)

6. I e) R (A1, A2) R = {(1, 2), (4, 3), (2, 9), (3, 8)} S (B1, B2, A1)
S = {(x, a, 1), (y, b, 1), (z, a, 4)}
R S
= { (1, 2, x, a), (1, 2, y, b), (4, 3, z, a), (2, 9, null, null), (3, 8, null, null)}
R  S
(1, 2, x, a, 1)
(1, 2, y, b, 1)
(1, 2, z, a, 4)
(4, 3, x, a, 1)
(4, 3, y, b, 1)
(4, 3, z, a, 4)
(2, 9, x, a, 1)
(2, 9, y, b, 1)
(2, 9, z, a, 4)
(3, 8, x, a, 1)
(3, 8, y, b, 1)
(3, 8, z, a, 4)
R S
(1, 2, x, a, 1)
(1, 2, y, b, 1)
(4, 3, z, a, 4)
(2, 9, null, null, null)
(3, 8, null, null, null)

7. II (a) List all hotels Hotel (hotelNo, hotelName, city) * (Hotel)
Use H for Hotel
* (H)
 hotelNo, hotelName, city (Hotel)
hotelNo (Hotel)
Incorrect!
All attributes when not specified.

8. b) List all single rooms with a price below £20 per night
Room (roomNo, hotelNo, type, price)
(type = ‘Single’ and price < 20) (Room)

9. b) List all single rooms with a price below £20 per night
* (  (type = ‘Single’ and price < 20) (Room) )
Projection is NOT Needed!
(type = Single and price < 20) (Room)
 (type = “Single” and price < 20) (Room)
 (type = ‘Single’ and price < £20) (Room)
 (type = ‘Single’ and price < 20 pounds) (Room)
(type = ‘Single’  price < 20) (Room)
Incorrect notation!

10. c) List the name and addresses of all guests
Guest (guestNo, guestName, guestAddress)
guestName, guestAddress (Guest)
One table only!
guestAddress: (street, city, state, zipcode)
composite attribute

11. c) List the name and addresses of all guests
Guest (guestNo, guestName, guestAddress)
guestName, guestAddress (Guest)
guestName, guestAddress (Guest Booking Hotel )
Don’t need Joins!

12. d) List the price and type of all rooms at the Grosvenor Hotel
Room (RoomNo, HotelNo, Type, Price)
Hotel (HotelNo, hotelName)
price, type ( hotelName = ‘Grosvenor’ (Room Hotel))
Common attribute: HotelNo
price, type (Room (hotelName = ‘Grosvenor’) (Hotel))
Incorrect! It’s the same as the following:
 price, type ((hotelName = ‘Grosvenor’) (Room  Hotel))

13. e) List all guests currently staying at the Grosvenor Hotel
Guest (guestNo, guestName, guestAddress)
Hotel (hotelNo, hotelName)
Booking (hotelNo, guestNo, dateFrom, dateTo, roomNo)
 Guest.* (
(fromDate <= CurrentDate and toDate >= CurrentDate) and
hotelName = ‘Grosvenor’ (Guest Booking Hotel))

14. e) List all guests currently staying at the Grosvenor Hotel
Guest (guestNo, guestName, guestAddress)
Hotel (hotelNo, hotelName)
Booking (hotelNo, guestNo, dateFrom, dateTo, roomNo)
What is the common attribute between Guest and Hotel?
 Guest.* (
(fromDate <= CurrentDate and toDate >= CurrentDate) and
hotelName = ‘Grosvenor’ (Guest Hotel Booking) )
hotelName = ‘Grosvenor’ (Guest (Hotel Booking)))

15. f) List the details of all rooms at the Grosvenor Hotel, including the name of the guest staying in the room if the room is occupied
Room Guest
 Room.*, guestName
 (hotelName = ‘Grosvenor’ and fromDate <= CurrentDate and CurrentDate <= toDate)
((Room Hotel) (Booking Guest))
Not correct!
 Room.*, guestName (
( hotelName = ‘Grosvenor’ (Room Hotel) )
( (fromDate <= CurrentDate and CurrentDate <= toDate) (Booking Guest)) )
A room without current booking will join a guest who booked the room before
The room will be removed by the selection outside the outer join

16. Ctrl+G: 8. 5/10 Ctrl+I: Incorrect answer: -0
Ctrl+G: 8.5/10 Ctrl+I: Incorrect answer: -0.5 Ctrl+S: Incorrect style: -0.5 Ctrl+T: Missing table(s): -0.5 Ctrl+N: Missing name: -1 Ctrl+N: Incorrect file name: -1