Describe the relations that would be produced by the following relational algebra operations:
a) ?hotelNo (?price ? 50 (Room) )
b) ?Hotel.hotelNo ? Room.hotelNo(Hotel ? Room)
c) ?hotelName (Hotel Hotel.hotelNo ? Room.hotelNo (?price ? 50 (Room)) )
d) Guest (?dateTo ? ‘1-Jan-2007’ (Booking))
e) Hotel Hotel.hotelNo ? Room.hotelNo (?price ? 50 (Room)) )
f) ?guestName, hotelNo (Booking Booking.guestNo ? Guest.guestNo Guest) ?
A. This will produce a relation with a single attribute (hotelNo) giving the number of those
hotels with a room price greater than £50.
B. This will produce a join of the Hotel and Room relations containing all the attributes of
both Hotel and Room (there will be two copies of the hotelNo attribute). Essentially this
will produce a relation containing all rooms at all hotels.
D. This will produce a (left outer) join of Guest and those tuples of Booking with an end date
(dateTo) greater than or equal to 1-Jan-2007. All guests who don’t have a booking with
such a date will still be included in the join. Essentially this will produce a relation
containing all guests and show the details of any bookings they have beyond 1-Jan-2002.
E. This will produce a (semi) join of Hotel and those tuples of Room with a price greater
than £50. Only those Hotel attributes will be listed. Essentially this will produce a relation
containing all the details of all hotels with a room price above £50.
F. This will produce a relation containing the names of all guests who have booked all hotels in London.
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(Simulation: The Tortoise and the Hare) In this exercise, you’ll re-create the classic race of the tortoise and the hare. You’ll use random number generation to develop a simulation of this memorable event.
Our contenders begin the race at “square 1” of 70 squares. Each square represents a possible position along the race course. The finish line is at square 70. The first contender to reach or pass square 70 is rewarded with a pail of fresh carrots and lettuce. The course weaves its way up the side of a slippery mountain, so occasionally the contenders lose ground. There is a clock that ticks once per second. With each tick of the clock, your program should use function moveTortoise and moveHare to adjust the position of the animals according to the rules in Fig. 7.21. These functions should use pointer-based pass-by-reference to modify the position of the tortoise and the hare.
Use variables to keep track of the positions of the animals (i.e., position numbers are 1–70). Start each animal at position 1 (i.e., the “starting gate”). If an animal slips left before square 1, move the animal back to square 1. Generate the percentages in the preceding table by producing a random integer i in the range
1 ? i ? 10. For the tortoise, perform a “fast plod” when 1 ? i ? 5, a “slip” when 6 ? i ? 7 or a “slow plod” when 8 ? i ? 10. Use a similar technique to move the hare. Begin the race by printing
BANG !!!!!
AND THEY'RE OFF !!!!!
For each tick of the clock (i.e., each repetition of a loop), print a 70-position line showing the letter T in the tortoise’s position and the letter H in the hare’s position. Occasionally, the contenders land on the same square. In this case, the tortoise bites the hare and your program should print OUCH!!! beginning at that position. All print positions other than the T, the H or the OUCH!!! (in
case of a tie) should be blank. After printing each line, test if either animal has reached or passed square 70. If so, print the winner and terminate the simulation. If the tortoise wins, print TORTOISE WINS!!! YAY!!! If the are wins, print Hare wins. Yuch. If both animals win on the same clock tick, you may want to favor the tortoise (the “underdog”), or you may want to print It's a tie. If neither animal wins, perform the loop again to simulate the next tick of the clock.