CS1010S — Programming Methodology
School of Computing
National University of Singapore
Midterm Test
Time allowed: 1 hour
30 Sep 2020
Instructions (please read carefully):
1. This is a CLOSED book assessment, but you are allowed to bring ONE double-sided A4
sheet of notes for this assessment.
2. The assessment paper contains FOUR (4) questions and comprises EIGHT (8) pages
including this cover page.
3. The time allowed for solving this test is 1 hour.
4. Weightage of each question is given in square brackets. The maximum attainable score is
50.
5. Enter all your answers in the correct space provided on Examplify. No other mode of
submission will be accepted.
6. Use of physical calculators are not allowed in the test. You may however use the on-screen
calculator in Examplify.
7. Marks may be deducted for i) unrecognisable handwriting, and/or ii) excessively long
code. A general guide would be not more than twice the length of our model answers.
GOOD LUCK!
CS1010S Midterm Test — 30 Sep 2020
Question 1: Python Expressions [15 marks]
There are several parts to this problem. Answer each part independently and separately.
In each part, one or more Python expressions are entered into the interpreter (Python shell).
Determine the printed output and write the exact output in the answer box. If the interpreter produces an error message, or enters an infinite loop, explain why and clearly state the
responsible evaluation step.
A.
a = 1
b = 2
def fib(a):
print(b)
def fab(b):
print(b - a)
fab(b - a)
print(b)
print(fib(b))
[5 marks]
B.
a = (1, (2, (3,)))
b = (4, 5)
if 1 in a:
print(2)
if 3 == a[1][1]:
print(4)
elif 5 == b[-1]:
b += 6
else:
print(7)
print(b)
[5 marks]
C.
d = 4
while d >= 0:
print(d)
if d % 2 == 1:
d = d - 1
print("c")
continue
if d % 3 == 0:
print("b")
break
d = d // 2
[5 marks]
2
CS1010S Midterm Test — 30 Sep 2020
Question 2: Zombie Island Apocalypse [14 marks]
Just when you thought that the world was recovering from COVID-19, a new zombie apocalypse has broken out on a desert island. You have been asked to model the outcome of the
outbreak with your newfound skills in Pythonfu. The outbreak starts on day 0 with one infected
person (zombie). The rate of the zombie infection is modelled by a parameter r0 . On each new
day, the zombies will infect an additional r0 × the existing zombie population, rounded up (you
can use the python function ceil() ).
For example,
with r0 = 0.5, the sequence of zombie population is 1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, · · ·
with r0 = 1, the sequence of zombie population is 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, · · ·
with r0 = 2, the sequence of zombie population is 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, · · ·
The function infected(r0, j, k) takes three inputs: r0 the parameter described above, and
non-negative integers j and k. It returns a tuple, representing the sequence of zombie population
of the outbreak from day j to day k inclusive, i.e., a slice of the complete sequence. If j > k,
infected will return an empty tuple.
Sample execution:
>>> infected(0.5, 0, 10)
(1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93)
>>> infected(1, 5,
(32, 64, 128, 256,
>>> infected(2, 3,
(27, 81, 243, 729,
10)
512, 1024)
7)
2187)
A. Provide a recursive implementation of
infected . Partial marks may be given for incom-
plete solution, e.g., returning the complete sequence instead of the slice.
[5 marks]
B. What is the order of growth in time and in space for the function you wrote in Part A?
Briefly explain your answer.
[2 marks]
C. Provide an iterative implementation of infected . Partial marks may be given for incomplete solution, e.g., returning the complete sequence instead of the slice.
[5 marks]
D. What is the order of growth in time and in space for the function you wrote in Part C?
Briefly explain your answer.
[2 marks]
3
CS1010S Midterm Test — 30 Sep 2020
Question 3: Higher-Order Zombies [4 marks]
A. [HARD] Consider the higher-order function
fold which was taught in class.
def fold(op, f, n):
if n == 0:
return f(0)
else:
return op(f(n), fold(op, f, n-1))
The function infected defined in Question 2 can be defined in terms of fold , by first obtaining the entire sequence and then slicing the tuple to the required section:
def infected(r0, j, k):
<PRE>
return fold(<T1>, <T2>, <T3>)[j:]
Please provide possible implementations for the terms T1, T2, and T3. You may optionally
define other functions in PRE if needed.
Note: You are to use the higher-order function and not solve it recursively or iteratively or use
a formula.
[4 marks]
4
CS1010S Midterm Test — 30 Sep 2020
Question 4: Hotel Reservations [17 marks]
INSTRUCTIONS: Please read the entire question clearly before you attempt this problem!! You are also not to use any Python data types which have not yet been taught in
class.
Breaking News [OK to skip] The Singapore Tourism Board (STB) said on Wednesday (Sep
16) that every Singapore citizen aged 18 and above this year will soon receive S$100 worth of
SingapoRediscovers vouchers, which can be used on staycations, attraction tickets and tours.
The vouchers are expected to support the local tourism business, which has been badly hit by
the COVID-19 pandemic and border restrictions.
Source: CNA
You have been hired by The Waffles Hotel to help develop a new software booking system
to deal with the expected surge in hotel reservations as dutiful Singaporeans do their part to
stimulate the local economy by taking government-paid staycations to work from hotel.
Your first task is to design a booking data type to capture a room booking. A guest makes a
booking by stating the size of the room required, and the check-in and check-out dates. To
simplify the handling of dates, the hotel’s existing system automatically converts dates into
days since the hotel’s founding, i.e., dates are simply running integers.
Though guests will specify a check-out day in their booking, it should not be counted in the
reservation system as hotels are only interested in the nights that are stayed. For example, a
guest who checks-in on day 20 and checks-out on day 22 is considered to be staying on day 20
and 21.
The booking data type should be supported by the following functions:
• make_booking(room_size, check_in, check_out) takes in a room size (which is
the number of guests in the room), and the check-in and check-out day.
• size_of(booking) takes in a booking, and returns the size of the room that is booked.
• booked_days(booking) takes in an booking, and returns a tuple containing the days
that the room is being booked.
Sample Execution:
>>> booking = make_booking(2, 27, 30)
>>> size_of(booking)
2
>>> booked_days(booking)
(27, 28, 29)
A. Decide on a data structure using tuples to represent a booking, and implement the functions
make_booking , size_of and booked_days .
5
[3 marks]
CS1010S Midterm Test — 30 Sep 2020
[Important!] For the remaining parts of this question, you should not break the abstraction
of a booking in your code. A sample execution is given on the next page for reference.
B. The hotel will need to keep track of all the bookings made, using a reservation schedule.
Decide on a data structure using tuples to represent a reservation schedule and support it with
the following functions:
• make_empty_reservations() takes no inputs and returns an empty reservation schedule.
• add_booking(booking, reservations) takes in a booking and the existing reservation schedule, and returns a new reservation schedule with the booking added to it.
[2 marks]
C. The hotel staff needs to know what are the current bookings for a given room size, for a particular day. Implement the function current_bookings(reservations, room_size, day)
that takes in the reservation schedule, the room size and the day, and returns a tuple of bookings
that fulfils the criteria.
[4 marks]
D. The hotel of course has a limited number of rooms, which is denoted using a tuple, where
the index of the elements denote the room size and the value denotes the number of rooms
it has. For example, a hotel with 10 single rooms (for one person), 30 double rooms (for
two persons) and 5 quad rooms (for 4 persons) who have its capacity represented as a tuple
(0, 10, 30, 0, 5) .
Assume a global variable hotel_capacity defines the Waffles Hotel capacity.
The function can_accept(booking, reservations) takes in the current reservation schedule, and a new booking, and returns True if the hotel has an available room, of the matching
size, to service the entire booking, and False if there is one day that the hotel has no rooms,
of the matching size, available for the booking.
Note that the same room does not have to be used for the entire booking. We force our guest to
change rooms if needed.
Provide an implementation for the function can_accept .
[4 marks]
E. Now if a booking was rejected because there are no available rooms of the given size, the
hotel might have a larger room that can accommodate the booking if the guest is willing to
upgrade. For example, a guest requesting a single room can be put into a quad room if there are
no single or double rooms available. But a booking cannot be “downgraded”, i.e. into smaller
rooms from what was requested.
Implement the function accept_upgrade(booking, reservations) that takes in the same
inputs as can_accept , and returns a booking with the next available room size if one is available, and False if the hotel is completely unable to service the booking.
[4 marks]
6
CS1010S Midterm Test — 30 Sep 2020
Sample execution:
>>> hotel_capacity = (0, 1, 1)
# hotel only has 1 single and 1 double room
>>> b1 = make_booking(1, 0, 2)
>>> b2 = make_booking(1, 1, 3)
>>> b3 = make_booking(2, 2, 3)
# check-in day 0, check-out day 2
>>> schedule = make_empty_reservations()
>>> can_accept(b1, schedule)
True
>>> schedule = add_booking(b1, schedule)
>>> b1 in current_bookings(schedule, 1, 0)
True
>>> b1 in current_bookings(schedule, 1, 2)
False
>>> can_accept(b2, schedule)
False
>>> can_accept(b3, schedule)
True
# b1 is booked on day 0
# b1 is not booked on day 2
# single room is booked on day 1
# double room is available
>>> accept_upgrade(b2, schedule) != False
True
# upgrade to double room is available
>>> accept_upgrade(b3, schedule) != False
True
# double room available
>>> b2 = accept_upgrade(b2, schedule)
>>> schedule = add_booking(b2, schedule)
# booking is upgraded
# double room booked day 1 to day 2
>>> accept_upgrade(b3, schedule)
False
— END
OF
# only single room available on day 2
QUESTIONS —
7
CS1010S Midterm Test — 30 Sep 2020
Appendix
The following are some functions that were introduced in class. For your reference, they are
reproduced here.
def sum(term, a, next, b):
if a > b:
return 0
else:
return term(a) + sum(term, next(a), next, b)
def product(term, a, next, b):
if a > b:
return 1
else:
return term(a) * product(term, next(a), next, b)
def fold(op, f, n):
if n == 0:
return f(0)
else:
return op(f(n), fold(op, f, n-1))
def enumerate_interval(low, high):
return tuple(range(low,high+1))
def map(fn, seq):
if seq == ():
return ()
else:
return (fn(seq[0]),) + map(fn, seq[1:])
def filter(pred, seq):
if seq == ():
return ()
elif pred(seq[0]):
return (seq[0],) + filter(pred, seq[1:])
else:
return filter(pred, seq[1:])
def accumulate(fn, initial, seq):
if seq == ():
return initial
else:
return fn(seq[0], accumulate(fn, initial, seq[1:]))
— END
OF
8
PAPER —
CS1010S — Programming Methodology
School of Computing
National University of Singapore
Midterm Test — Answer Sheet
Time allowed: 1 hour
30 Sep 2020
Student No:
S
O
L
U
T
I
O
N
S
Instructions (please read carefully):
1. Write down your student number on this answer sheet. DO NOT WRITE YOUR NAME!
2. This answer sheet comprises NINE (9) pages.
3. All questions must be answered in the space provided; no extra sheets will be accepted as
answers. You may use the extra page provided at the back if you need more space for your
answers.
4. You must submit only the ANSWER SHEET and no other documents. The question set
may be used as scratch paper.
5. An excerpt of the question may be provided to aid you in answering in the correct box. It
is not the exact question. You should refer to the original question in the question booklet.
6. You are allowed to use pencils, ball-pens or fountain pens, as you like as long as it is
legible (no red color, please).
7. Marks may be deducted for i) unrecognisable handwriting, and/or ii) excessively long
code. A general guide would be not more than twice the length of our model answers.
GOOD LUCK!
For Examiner’s Use Only
Question
Marks
Remarks
Q1
/ 15
Q2
/ 14
Q3
/
Q4
/ 17
Total
/ 50
4
Answer Sheet
CS1010S Midterm Test — 30 Sep 2020
This page is intentionally left blank.
Use it ONLY if you need extra space for your answers, in which case indicate the question
number clearly as well as in the original answer box. DO NOT use this for your rough work.
2
Answer Sheet
CS1010S Midterm Test — 30 Sep 2020
Question 1A
a = 1
b = 2
def fib(a):
print(b)
def fab(b):
print(b - a)
fab(b - a)
print(b)
print(fib(b))
2
-2
2
None
[5 marks]
Tests the understanding of scoping and whether
students are careful with parameter substitution.
Checks if students understand what the difference
between print and return.
+1 2
+1 -2
+1 2
+1 None
+1 Correct order of output
Note Additional marks may be deducted if the
answer does not demonstrate understanding of
scoping, even if some of the output match.
Question 1B
a = (1, (2, (3,)))
b = (4, 5)
if 1 in a:
print(2)
if 3 == a[1][1]:
print(4)
elif 5 == b[-1]:
b += 6
else:
print(7)
print(b)
[5 marks]
Tests tuple indexing and concatenation, managing
nested tuples, and differences in tuples vs integers. Tests understanding of independence of each
if statement.
+1
+2
+1
+1
2
State error occurs
Does not print more output after error occurs
Specifies TypeError or pinpoints error line
a += (6) or states cannot add tuple and int
Sympathy Marks
2m if printed 2, 4, b . Demonstrates if/elif/else
understanding
2m if printed 2, 7, b . Demonstrates if/elif/else
understanding
2
TypeError: cannot add tuple and int 3m if printed 2, (4, 5, 6,) . Only mistake is
failing to recognize (6) causes an error when
added to tuple.
3
Answer Sheet
CS1010S Midterm Test — 30 Sep 2020
Question 1C
d = 4
while d >= 0:
print(d)
if d % 2 == 1:
d = d - 1
print("c")
continue
if d % 3 == 0:
print("b")
break
d = d // 2
4
2
1
c
0
b
[5 marks]
General guidelines: 1 mark for demonstration of
understanding how a while-loops works. 2 marks
each for correct understanding of continue and
break .
5m
4m
3m
2m
1m
1m
0m
4 2 1 c 0 b
4 2 1 c b
4 2 1 c 0 0 ...
4 2 1 c
4 2 1 0 0 ...
4 2 1
States infinite loop without
demonstrating 4 2 1
4
Answer Sheet
CS1010S Midterm Test — 30 Sep 2020
Question 2A Recursive implementation.
[5 marks]
def infected(r0, j, k):
def helper(k):
if k == 0:
return (1,)
else:
return helper(k-1) + (ceil(helper(k-1)[-1] * (1+r0)),)
return helper(k)[j:]
Question 2B Order of growth for Part A.
[2 marks]
Time: O(2k ), there are two recursive calls to helper(k-1) for helper(k) , which is a tree
recursion with branching factor of 2.
O(k2 ), if the result of helper(k-1) is stored in a variable and reused, there is only one
recursive call. Thus, there is a total of k recursive calls, with each call creating a tuple of
increasing length.
Space:O(k2 ), the first recursive call to helper(k-1) will return a tuple of length k. This will
be stored in memory as the second call is executed. In the final leg, there will be k recursive
calls, with each call stack storing a tuple of decreasing size.
O(k), if the result of helper(k-1) is stored in a variable and reused, there is only one
recursive call. The tuple returned by each call is immediately replaced by a new tuple. So
only space for the final tuple is needed.
Question 2C Iterative implementation.
[5 marks]
def infected(r0, j, k):
result = ()
if j == 0:
result = (1,)
num = 1
for i in range(1, k + 1):
num = ceil(num * (1 + r0))
if j <= i:
result += (num,)
return result
5
Answer Sheet
CS1010S Midterm Test — 30 Sep 2020
Question 2D Order of growth for Part C.
[2 marks]
Time: O(k2 ), there is a total of k steps in the loop, and each iteration constructs a new tuple
of increasing size.
Space:O(k), each new tuple created is stored over the previous one, and finally a tuple of k
length is created.
Question 3A
[4 marks]
*optional
<PRE>:
<T1>: lambda a, b: b + (ceil(b[-1] * (1+r0)),)
<T2>: lambda a: (1,)
<T3>: k
Question 4A
[3 marks]
def make_booking(room_size, start_day, end_day):
return (room_size, tuple(range(start_day, end_day)))
def size_of(booking):
return booking[0]
def booked_days(booking):
return booking[1]
6
Answer Sheet
CS1010S Midterm Test — 30 Sep 2020
Question 4B
[2 marks]
def make_empty_reservations():
return ()
def add_booking(booking, reservations):
return reservations + (booking,)
Question 4C
[4 marks]
def current_bookings(reservations, room_size, day):
bookings = ()
for booking in reservations:
if size_of(booking) == room_size and day in booked_days(booking):
bookings += (booking,)
return bookings
Question 4D Provide an implementation.
[4 marks]
def can_accept(booking, reservations):
size = size_of(booking)
for day in booked_days(booking):
num_booked = len(current_bookings(reservations, size, day))
if hotel_capacity[size] <= num_booked:
return False
return True
7
Answer Sheet
CS1010S Midterm Test — 30 Sep 2020
Question 4E
[4 marks]
def accept_upgrade(booking, reservations):
for size in range(size_of(booking), len(hotel_capacity)):
if can_accept(booking, reservations):
return booking
else:
booking = make_booking(size+1, booked_days(booking)[0],
booked_days(booking)[-1] + 1)
return False
8
Answer Sheet
CS1010S Midterm Test — 30 Sep 2020
This page is intentionally left blank.
Use it ONLY if you need extra space for your answers, in which case indicate the question
number clearly. DO NOT use it for your rough work.
— END OF ANSWER SHEET —
9
Solutions
CS1010S Midterm Test — 30 Sep 2020
Question 1A
a = 1
b = 2
def fib(a):
print(b)
def fab(b):
print(b - a)
fab(b - a)
print(b)
print(fib(b))
2
-2
2
None
[5 marks]
Tests the understanding of scoping and whether
students are careful with parameter substitution.
Checks if students understand what the difference
between print and return.
+1 2
+1 -2
+1 2
+1 None
+1 Correct order of output
Note Additional marks may be deducted if the
answer does not demonstrate understanding of
scoping, even if some of the output match.
Question 1B
a = (1, (2, (3,)))
b = (4, 5)
if 1 in a:
print(2)
if 3 == a[1][1]:
print(4)
elif 5 == b[-1]:
b += 6
else:
print(7)
print(b)
[5 marks]
Tests tuple indexing and concatenation, managing
nested tuples, and differences in tuples vs integers. Tests understanding of independence of each
if statement.
+1
+2
+1
+1
2
State error occurs
Does not print more output after error occurs
Specifies TypeError or pinpoints error line
a += (6) or states cannot add tuple and int
Sympathy Marks
2m if printed 2, 4, b . Demonstrates if/elif/else
understanding
2m if printed 2, 7, b . Demonstrates if/elif/else
understanding
2
TypeError: cannot add tuple and int 3m if printed 2, (4, 5, 6,) . Only mistake is
failing to recognize (6) causes an error when
added to tuple.
3
Solutions
CS1010S Midterm Test — 30 Sep 2020
Question 1C
d = 4
while d >= 0:
print(d)
if d % 2 == 1:
d = d - 1
print("c")
continue
if d % 3 == 0:
print("b")
break
d = d // 2
4
2
1
c
0
b
[5 marks]
General guidelines: 1 mark for demonstration of
understanding how a while-loops works. 2 marks
each for correct understanding of continue and
break .
5m
4m
3m
2m
1m
1m
0m
4 2 1 c 0 b
4 2 1 c b
4 2 1 c 0 0 ...
4 2 1 c
4 2 1 0 0 ...
4 2 1
States infinite loop without
demonstrating 4 2 1
4
Solutions
CS1010S Midterm Test — 30 Sep 2020
Question 2A Recursive implementation.
[5 marks]
def infected(r0, j, k):
def helper(k):
if k == 0:
return (1,)
else:
return helper(k-1) + (ceil(helper(k-1)[-1] * (1+r0)),)
return helper(k)[j:]
Question 2B Order of growth for Part A.
[2 marks]
Time: O(2k ), there are two recursive calls to helper(k-1) for helper(k) , which is a tree
recursion with branching factor of 2.
O(k2 ), if the result of helper(k-1) is stored in a variable and reused, there is only one
recursive call. Thus, there is a total of k recursive calls, with each call creating a tuple of
increasing length.
Space:O(k2 ), the first recursive call to helper(k-1) will return a tuple of length k. This will
be stored in memory as the second call is executed. In the final leg, there will be k recursive
calls, with each call stack storing a tuple of decreasing size.
O(k), if the result of helper(k-1) is stored in a variable and reused, there is only one
recursive call. The tuple returned by each call is immediately replaced by a new tuple. So
only space for the final tuple is needed.
Question 2C Iterative implementation.
[5 marks]
def infected(r0, j, k):
result = ()
if j == 0:
result = (1,)
num = 1
for i in range(1, k + 1):
num = ceil(num * (1 + r0))
if j <= i:
result += (num,)
return result
5
Solutions
CS1010S Midterm Test — 30 Sep 2020
Question 2D Order of growth for Part C.
[2 marks]
Time: O(k2 ), there is a total of k steps in the loop, and each iteration constructs a new tuple
of increasing size.
Space:O(k), each new tuple created is stored over the previous one, and finally a tuple of k
length is created.
Question 3A
[4 marks]
*optional
<PRE>:
<T1>: lambda a, b: b + (ceil(b[-1] * (1+r0)),)
<T2>: lambda a: (1,)
<T3>: k
Question 4A
[3 marks]
def make_booking(room_size, start_day, end_day):
return (room_size, tuple(range(start_day, end_day)))
def size_of(booking):
return booking[0]
def booked_days(booking):
return booking[1]
6
Solutions
CS1010S Midterm Test — 30 Sep 2020
Question 4B
[2 marks]
def make_empty_reservations():
return ()
def add_booking(booking, reservations):
return reservations + (booking,)
Question 4C
[4 marks]
def current_bookings(reservations, room_size, day):
bookings = ()
for booking in reservations:
if size_of(booking) == room_size and day in booked_days(booking):
bookings += (booking,)
return bookings
Question 4D Provide an implementation.
[4 marks]
def can_accept(booking, reservations):
size = size_of(booking)
for day in booked_days(booking):
num_booked = len(current_bookings(reservations, size, day))
if hotel_capacity[size] <= num_booked:
return False
return True
7
Solutions
CS1010S Midterm Test — 30 Sep 2020
Question 4E
[4 marks]
def accept_upgrade(booking, reservations):
for size in range(size_of(booking), len(hotel_capacity)):
if can_accept(booking, reservations):
return booking
else:
booking = make_booking(size+1, booked_days(booking)[0],
booked_days(booking)[-1] + 1)
return False
8