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CS1010S — Programming Methodology
National University of Singapore
Practical Examination
15 April 2017
Time allowed: 2 hours
Instructions (please read carefully):
1. This is an open-book exam. You are allowed to bring in any course or reference
materials in printed form. No electronic media or storage devices are allowed.
2. This practical exam consists of three questions. The time allowed for solving this
test is 2 hours.
3. The maximum score of this test is 30 marks. Note that the number of marks
awarded for each question IS NOT correlated with the difficulty of the question.
4. You are advised to attempt all questions. Even if you cannot solve a question
correctly, you are likely to get some partial credit for a credible attempt.
5. While you are also provided with the template practical-template.py to work
with, your answers should be submitted on Coursemology. Note that you can
only run the test cases on Coursemology for a limited number of tries
because they are only for checking that your code is submitted correctly. You are
expected to test your own code for correctness using IDLE and not depend only
on the provided test cases. Do ensure that you submit your answers correctly by
running the test cases at least once.
6. In case there are problems with Coursemology.org and we are not able to upload
the answers to Coursemology.org, you will be required to name your file no>.py where is your matriculation number and leave the file on the
Desktop. If your file is not named correctly, we will choose any Python file found
at random.
7. Please note that it shall be your responsibility to ensure that your solution is
submitted correctly to Coursemology and correctly left in the desktop folder at
the end of the examination. Failure to do so will render you liable to receiving
a grade of ZERO for the Practical Exam, or the parts that are not uploaded
correctly.
8. Please note that while sample executions are given, it is not sufficient to write
programs that simply satisfy the given examples. Your programs will be
tested on other inputs and they should exhibit the required behaviours as specified
by the problems to get full credit. There is no need for you to submit test cases.
GOOD LUCK!
CS1010S Programming Methodology — 15 April 2017
Question 1 : 1337 speak [10 marks]
“Leet (or ‘1337’), also known as eleet or leetspeak, is an alternative alphabet for many
languages that is used primarily on the Internet. It uses some characters to replace others
in ways that play on the similarity of their glyphs via reflection or other resemblance.
For example, leet spellings of the word leet include 1337 and l33t; eleet may be spelled
31337 or 3l33t.”
- Source: Wikipedia
We can use Python to help us easily translate words to leet to impress0r our friends by
using a dictionary of characters to replace. The keys of the dictionary will be a single
character and the value will be the replacement characters.
A. The function l33tify takes as inputs a string and a dictionary, and outputs the
string translated to leet according to the dictionary.
Examples:
>>> l33t_dict = {
'a': '4',
'b': '8',
'c': '(',
'e': '3',
'g': '[',
'i': '1',
'o': '0',
's': '5',
't': '7',
'x': '%',
'z': '2',
}
>>> l33tify("pheer my leet skills", l33t_dict)
'ph33r my l337 5k1ll5'
>>> l33tify("python is cool", l33t_dict)
'py7h0n 15 (00l'
Implement the function l33tify. You may assume that all the characters of the
input string are lowercase and the keys of the dictionary are single characters.
[5 marks]
B. The l33tify function is not very leet because it always replaces with a fixed
character. Since there are usually multiple possible replacements for a character, we can
do better by having our function rotate through the possible replacements.
For example, the character ‘l’ can be replaced by the following list of strings: ['|_',
'|', '1']. In this case, the first replacement of ‘l’ will use '|_', the second will use
'|', the third will use '1'. The next replacement will go back to the first string '|_',
and the cycle continues.
2
CS1010S Programming Methodology — 15 April 2017
The values of the leet dictionary will now be a list of strings, and the order of the
replacement will follow the order of the list.
The function advance_l33tify takes as inputs a string and a dictionary, and outputs
the string translated to leet according to scheme described above.
Examples:
>>> adv_l33t_dict = {
'a': ['4', '/-\\', '/_\\', '@', '/\\'],
'b': ['8', '|3', '13', '|}', '|:', '|8', '18', '6', '|B'],
'c': ['<', '{', '[', '('],
'd': ['|)', '|}', '|]'],
'e': ['3'],
'f': ['|=', 'ph', '|#', '|"'],
'g': ['[', '-', '[+', '6'],
'h': ['|-|', '[-]', '{-}', '|=|', '[=]', '{=}'],
'i': ['1', '|'],
'j': ['_|', '_/', '_7', '_)'],
'k': ['|<', '1<'],
'l': ['|_', '|', '1'],
'm': ['|\\/|', '^^', '/\\/\\'],
'n': ['|\\|', '/\\/', '/V', '][\\\\]['],
'o': ['0', '()', '[]', '{}'],
'p': ['|o', '|O', '|>', '|*', '|Ârˇ', '|D', '/o'],
'q': ['O_', '9', '(', ')', ''],
'r': ['|2', '12', '.-', '|^'],
's': ['5', '$', '§'],
't': ['7', '+', '7`', "'|'"],
'u': ['|_|', '\\_\\', '/_/', '\\_/', '(_)'],
'v': ['\\/'],
'w': ['\\/\\/', '(/\\)', '\\^/', '|/\\|'],
'x': ['%', '*', '><', '}{', ')('],
'y': ['`/', '¥'],
'z': ['2', '7_', '>_']
}
>>> print(advance_l33tify("Bow b4 me 4 I am root!!!", adv_l33t_dict))
80\/\/ |34 |\/|3 4 1 4^^ |2()[]7!!!
>>> print(advance_l33tify("Mississippi", adv_l33t_dict))
|\/|15$|§51|o|O|
Provide an implementation for the function advance_l33tify. Note that for this part,
the input string can be a mix of uppercase and lowercase characters, while the
dictionary keys will always be a single, lowercase character. You can also assume that
the lists in the dictionary values will not be empty.
Tip: Be mindful when you are modifying a mutable input parameter. [5 marks]
3
CS1010S Programming Methodology — 15 April 2017
Question 2 : 3G Coverage [10 marks]
Important note: You are provided with a data file 3g-coverage.csv for this question
for testing, but your code should work correctly for any data file with the same format
and it will be tested with other data files.
It is time to renew your mobile phone contract and you wonder which telco provides the
best coverage. You decided to review the historical data that IMDA has collected over
the past few years.
The first line of the data file is a header which describes each column of data. You
may assume that all data files follow this same fixed order of columns and there are no
duplicate data for a telco in a month.
A. You wish to find out the average coverage of all available telcos in any given year.
Implement the function yearly_average that takes as input a filename and a year (as
strings). The function returns a dictionary which keys are the telcos obtained from the
data file and the values are the average coverage (rounded to 4 decimal places) for the
stated year. You can use the function round(n, d) to round n to d decimal places.
Note, the data for some telcos might not be available for certain years and there might
not always be data for every month in the year. [5 marks]
Sample execution:
>>> yearly_average("3g-coverage.csv", "2015")
{'Singtel': 99.5308, 'M1': 99.5358, 'Starhub': 99.7608}
>>> yearly_average("3g-coverage.csv", "2013")
{'Singtel': 99.4078, 'M1': 98.6622, 'Starhub': 99.4544}
B. Sometimes the average does not give an accurate picture. You think it might be
better to compare the telcos on a monthly basis. Essentially, you want to know which
telco has the best coverage for each month, and how many times each telco win the “best
telco of the month” in a given year.
Implement the function best_telco that takes as input a filename and a year (as strings).
The function returns a dictionary which keys are the telcos obtained from the data file
and the values are number of times the telco was the “best telco of the month” in the
given year.
Note, for this part, you may assume that every month will have all the same telcos
represented and that there is always one best telco. However, it may still be the case
that data for some months are not available. You should also still assume that names of
the telcos are not fixed. [5 marks]
Sample execution:
>>> best_telco("3g-coverage.csv", "2015")
{'Singtel': 0, 'M1': 1, 'Starhub': 11}
>>> best_telco("3g-coverage.csv", "2015")
{'Singtel': 5, 'M1': 0, 'Starhub': 4} # only 9 months in 2013
4
CS1010S Programming Methodology — 15 April 2017
Question 3 : Villains [10 marks]
Warning: Please read the entire question carefully and plan well before start-
ing to write your code.
Villains are evil people out to do bad things in the world. They accomplish their evil
doings using Gadgets which increases their reputation and evilness.
Figure 1: Gru, a villain with his Freeze Ray.
The class Villain models a villain and the class Gadget models a gadget.
Gadget is created with two inputs, its name (which is a string) and an awesomeness value
(which is an integer). A Gadget can only be owned by at most one Villain at any time.
Gadget supports the following methods:
• get_description() returns the string " has level awesomeness> awesomeness"
• owned_by() returns the string " belongs to " if
it is currently owned by a villain. Otherwise, the string " is
unowned" is returned.
Villain is created with one input, its name, which is a string. Every villain starts out
with an evilness of 0, has no proficiency in any gadgets and owns no gadgets. Each time
a villain does an evil deed with a gadget, the gadget’s awesomeness will decrease by 1
and the villain’s proficiency in that gadget will increase by 1.
It is possible for different gadgets to have the same name. In which case they are con-
sidered separate gadgets and a villain’s proficiency in them are independent. It is also
possible for a villain to own more than one gadget of the same name.
Villain supports the following methods:
• get_evilness() returns the current evilness of the villain, which is an integer.
5
CS1010S Programming Methodology — 15 April 2017
• gadgets_owned() returns a tuple of the names of the gadgets the villain currently
possesses.
• get_proficiency(gadget) takes as input a Gadget. If the villain’s proficiency
for the gadget is more than 0, the string "'s proficiency
with is " is returned, where
is the villain’s proficiency value for that gadget. If the villain is not proficient
with the gadget at all, the string " is not proficient with
" is returned.
• do_evil(action, gadget) takes as inputs an action (which is a string) and a
Gadget. It returns a string and performs some actions based on the following
conditions:
– If the gadget is not in the villain’s possession, the string "
does not have " is returned.
– Otherwise, the following actions will occur: 1) The villain’s evilness increases
by the sum of the gadget’s awesomeness and the villain’s proficiency of the
gadget; 2) The gadget’s awesomeness will decrease by 1, but to not less than
0; and 3) the villain’s proficiency of the gadget will increase by 1.
The string " with " is returned.
• steals(gadget) takes as input a Gadget and returns a string while performing
some actions based on the following conditions:
– If the villain already owns the gadget, the string " already
has " is returned.
– Otherwise, the villain steals the gadget and is now owns it. The gadget’s
awesomeness is also restored to its original value.
If the gadget was previously owned by another villain, the other villain’s evil-
ness will be halved (rounded down to the nearest integer) and the string
" steals from "
is returned. Else the gadget had no previous owner and the string "name> steals " is returned.
• envy(other) takes as input either a Gadget or Villain and returns a string based
on the following conditions:
– If other is a Gadget without an owner, the string " envies
" is returned.
– If other is a Gadget currently owned by another villain, the string "name> envies 's " is returned.
– If other is a Gadget currently owned the villain, the string "name> already has " is returned.
– If other is the villain himself, the string " cannot envy
himself" is returned.
– If other is another Villain who has the same evilness as the villain, the
string "Nobody is envious" is returned.
6
CS1010S Programming Methodology — 15 April 2017
– If other is another Villain with a different evilness, the string name> envies " is returned, where Villain A has a lower
evilness than Villain B.
Provide an implementation for the classes Villain and Gadget.
For simplicity, you do not have to worry about data abstraction and can access the
properties of both classes directly. Take careful note of the characters in the returned
strings, especially spaces and punctuation.
Sample Execution:
>>> gru = Villain("Gru")
>>> vector = Villain("Vector")
>>> freeze_ray = Gadget("Freeze Ray", 5)
>>> lava_gun = Gadget("Lava Lamp Gun", 3)
>>> gru.get_evilness()
0
>>> gru.gadgets_owned()
()
>>> freeze_ray.get_description()
"Freeze Ray has level 5 awesomeness"
>>> freeze_ray.owned_by()
"Freeze Ray is unowned"
>>> gru.steals(freeze_ray)
"Gru steals Freeze Ray"
>>> gru.gadgets_owned()
('Freeze Ray',)
>>> freeze_ray.owned_by()
"Freeze Ray belongs to Gru"
>>> gru.get_proficiency(freeze_ray)
"Gru is not proficient with Freeze Ray"
>>> gru.do_evil("robs a bank", freeze_ray)
"Gru robs a bank with Freeze Ray"
>>> gru.get_evilness()
5 # Gru evilness increase by Freeze Ray's awesomeness
>>> gru.get_proficiency(freeze_ray)
"Gru's proficiency with Freeze Ray is 1"
7
CS1010S Programming Methodology — 15 April 2017
>>> freeze_ray.get_description()
"Freeze Ray has level 4 awesomeness" # Freeze Ray awesomeness
decreases
>>> gru.do_evil("steals candy", freeze_ray)
"Gru steals candy with Freeze Ray"
>>> gru.get_proficiency(freeze_ray)
"Gru's proficiency with Freeze Ray is 2"
>>> gru.get_evilness()
10
>>> freeze_ray.get_description()
"Freeze Ray has level 3 awesomeness"
>>> gru.envy(freeze_ray)
"Gru already has Freeze Ray"
>>> vector.envy(freeze_ray)
"Vector envies Gru's Freeze Ray"
>>> gru.envy(vector)
"Vector envies Gru"
>>> vector.steals(freeze_ray)
"Vector steals Freeze Ray from Gru"
>>> gru.get_evilness()
5 # Gru got his evilness halved
>>> freeze_ray.get_description()
"Freeze Ray has level 5 awesomeness" # awesomeness restored to
original
>>> freeze_ray.owned_by()
"Freeze Ray belongs to Vector"
>>> gru.gadgets_owned()
()
>>> vector.do_evil("freezes Miami", freeze_ray)
"Vector freezes Miami with Freeze Ray"
>>> vector.get_evilness()
5
8
CS1010S Programming Methodology — 15 April 2017
>>> gru.envy(vector)
"Nobody is envious"
>>> gru.envy(lava_gun)
"Gru envies Lava Lamp Gun"
>>> gru.do_evil("steals Freeze Ray", lava_gun)
"Gru does not have Lava Lamp Gun"
>>> gru.envy(lava_gun)
"Gru envies Lava Lamp Gun"
>>> gru.steals(lava_gun)
"Gru steals Lava Lamp Gun"
>>> gru.do_evil("steals the Queen's crown", lava_gun)
"Gru steals the Queen's crown with Lava Lamp Gun"
>>> gru.get_evilness()
8
>>> gru.envy(vector)
"Vector envies Gru"
>>> gru.steals(freeze_ray)
"
Gru steals Freeze Ray from Vector"
>>> vector.get_evilness()
2
>>> gru.get_evilness()
8
>>> gru.gadgets_owned()
('Lava Lamp Gun', 'Freeze Ray')
>>> gru.do_evil("freezes Vector", freeze_ray)
"Gru freezes Vector with Freeze Ray"
>>> gru.get_evilness()
15
You are advised to solve this problem incrementally. Even if you cannot fulfill all the
desired behaviours, you can still get partial credit for fulfilling the basic behaviours.
— E N D O F P A P E R —
9

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