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Python代写-IGNMENT 3
时间：2021-04-11
ASSIGNMENT 3
COMP202, Winter 2021
Due: Tuesday, April 13th, 11:59pm
Please read the entire PDF before starting. You must do this assignment individually.
Question 1: 35 points
Question 2: 65 points
100 points total
It is very important that you follow the directions as closely as possible. The directions, while
perhaps tedious, are designed to make it as easy as possible for the TAs to mark the assignments by letting
them run your assignment, in some cases through automated tests. While these tests will never be used to
determine your entire grade, they speed up the process significantly, which allows the TAs to provide better
feedback and not waste time on administrative details. Plus, if the TA is in a good mood while he or she is
grading, then that increases the chance of them giving out partial marks. :)
Up to 30% can be removed for bad indentation of your code as well as omitting comments, or poor coding
structure.
To get full marks, you must:
• Follow all directions below.
– In particular, make sure that all file names and function names are spelled exactly as described
in this document. Otherwise, a 50% penalty will be applied.
• Make sure that your code runs.
– Code with errors will receive a very low mark.
• Write your name and student ID as a comment at the top of all .py files you hand in.
• Name your variables appropriately.
– The purpose of each variable should be obvious from the name.
• Comment your work.
– A comment every line is not needed, but there should be enough comments to fully understand
your program.
• Avoid writing repetitive code, but rather call helper functions! You are welcome to add additional
functions if you think this can increase the readability of your code.
• Lines of code should NOT require the TA to scroll horizontally to read the whole thing. Vertical
spacing is also important when writing code. Separate each block of code (also within a function) with
an empty line.
1
Part 1 (0 points): Warm-up
Do NOT submit this part, as it will not be graded. However, doing these exercises might help you to do the
second part of the assignment, which will be graded. If you have difficulties with the questions of Part 1, then
we suggest that you consult the TAs during their office hours; they can help you and work with you through
the warm-up questions. You are responsible for knowing all of the material in these questions.
Warm-up Question 1 (0 points)
Write a function same_elements which takes as input a two dimensional list and returns true if all the
elements in each sublist are the same, false otherwise. For example,
>>> same_elements([[1, 1, 1], ['a', 'a'], [6]])
True
>>> same_elements([[1, 6, 1], [6, 6]])
False
Warm-up Question 2 (0 points)
Write a function flatten_list which takes as input a two dimensional list and returns a one dimensional
list containing all the elements of the sublists. For example,
>>> flatten_list([[1, 2], [3], ['a', 'b', 'c']])
[1, 2, 3, 'a', 'b', 'c']
>>> flatten_list([[]])
[]
Warm-up Question 3 (0 points)
Complete the case study on multidimensional lists presented in class on Friday, March 12. You can find
the instructions on myCourses (Content > Live sessions > Case Studies > Case Study 2).
Warm-up Question 4 (0 points)
Write a function get_most_valuable_key which takes as input a dictionary mapping strings to integers.
The function returns the key which is mapped to the largest value. For example,
>>> get_most_valuable_key({'a' : 3, 'b': 6, 'g': 0, 'q': 9})
'q'
Warm-up Question 5 (0 points)
Write a function add_dicts which takes as input two dictionaries mapping strings to integers. The
function returns a dictionary which is a result of merging the two input dictionary, that is if a key is in
both dictionaries then add the two values.
>>> d1 = {'a':5, 'b':2, 'd':-1}
>>> d2 = {'a':7, 'b':1, 'c':5}
>>> add_dicts(d1, d2) == {'a': 12, 'b': 3, 'c': 5, 'd': -1}
True
Warm-up Question 6 (0 points)
Create a function reverse_dict which takes as input a dictionary d and returns a dictionary where the
values in d are now keys mapping to a list containing all the keys in d which mapped to them. For
example,
>>> a = reverse_dict({'a': 3, 'b': 2, 'c': 3, 'd': 5, 'e': 2, 'f': 3})
>>> a == {3 : ['a', 'c', 'f'], 2 : ['b', 'e'], 5 : ['d']}
True
Note that the order of the elements in the list might not be the same, and that’s ok!
Page 2
Part 2
The questions in this part of the assignment will be graded.
The main learning objectives for this assignment are:
• Apply what you have learned about one-dimensional and multi-dimensional lists.
• Apply what you have learned about dictionaries.
• Understand how to test functions that return dictionaries.
• Solidify your understanding of working with loops and strings.
• Understand how to write a docstring and use doctest when working with dictionaries.
• Learn to identify when using the function enumerate can help you write a cleaner code.
• Apply what you have learned about file IO and string manipulation.
Note that the assignment is designed for you to be practicing what you have learned in the
videos up to and including Week 11.1. For this reason, you are NOT allowed to use anything
seen after Week 11.1 or not seen in class at all. You will be heavily penalized if you do so.
For full marks, in addition to the points listed on page 1, make sure to add the appropriate documentation
string (docstring) to all the functions you write. The docstring must contain the following:
• The type contract of the function.
• A description of what the function is expected to do.
• At least 3 examples of calls to the function (except when the function has only one possible output,
in which case you can provide only one example). You are allowed to use at most one example per
function from this pdf.
Examples
For each question, we provide several examples of how your code should behave. All examples are given as
if you were to call the functions from the shell.
When you upload your code to codePost, some of these examples will be run automatically to check that
your code outputs the same as given in the example. However, it is your responsibility to make sure
your code/functions work for any inputs, not just the ones shown in the examples. When the
time comes to grade your assignment, we will run additional, private tests that may use inputs not seen in
the examples.
Furthermore, please note that your code files should not contain any function calls in the main
body of the program (i.e., outside of any functions). Code that does not conform to this restriction will
automatically fail the tests on codePost and be heavily penalized. It is OK to place function calls in the
main body of your code for testing purposes, but if you do so, make certain that you remove them before
submitting. Please review what you have learned in video 5.2 if you’d like to add code to your modules
which executes only when you run your files.
Page 3
Question 1: Game of Life (35 points)
For this question, you will have to write a Python program that implements a simple version of Conway’s
Game of Life (https://en.wikipedia.org/wiki/Conway\%27s_Game_of_Life)
This is a zero-player game, which means that it is completely determined by the initial state provided.
For our purpose, we will represent the universe of the Game of Life as a finite rectangular two-dimensional
list (i.e., a matrix). Each element denotes a cell in the universe: if the cell is alive the value of the element
is 1, if the cell is dead the value of the element is 0. Whether a cell is alive or dead in the next generation
of the universe depends on its relation with its neighboring cells (the cells that are horizontally, vertically,
or diagonally adjacent to it). The next generation of a cell is determined by the following rules:
• Any live cell with fewer than two live neighbors dies, as if caused by under-population.
• Any live cell with two or three live neighbors lives on to the next generation.
• Any live cell with more than three live neighbors dies, as if caused by over-population.
• Any dead cell with exactly three live neighbors becomes a live cell, as if caused by reproduction.
Otherwise the cell will remain dead.
In this question, you will implement a function that will generate a given number of strings representing
the corresponding number of generations of a specified universe.
To complete this task, you will need to implement all the functions listed below. All the code for this
question must be placed in a file named game_of_life.py. Note that you are free to write more functions
if they help the design or readability of your code.
Attention: please note that none of the functions listed below should modify any of their input objects.
• is_valid_universe: given a two-dimensional list of integers, the function returns True if such list
is a valid representation of a universe, False otherwise. For the purpose of this assignment, we
consider a two-dimensional list to be a valid representation of a universe if it is rectangular (i.e., all
of its sub-lists are of the same size) and all the elements of its sub-lists are either 0s or 1s. If the
input list is empty, then the function should return False.
For example:
>>> a = [[0, 0], [1, 0], [1, 0]]
>>> is_valid_universe(a)
True
>>> b = [[0, 0, 0], [1, 1], [0]]
>>> is_valid_universe(b)
False
>>> c = [[1, 2], [3, 4]]
>>> is_valid_universe(c)
False
• universe_to_str: given a valid universe (i.e., a two-dimensional list of integers as described above)
as input, the function returns its string representation. To build the string use the star character
('*') for alive cells, and the whitespace character (' ') for dead cells. The string should also contain
a box (as shown below) surrounding the actual cells of the universe.
For example:
>>> block = [[0, 0, 0, 0], [0, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 0]]
>>> str_block = universe_to_str(block)
>>> print(str_block)
Page 4
+----+
| |
| ** |
| ** |
| |
+----+
>>> tub = [[0, 0, 0, 0, 0], [0, 0, 1, 0, 0], [0, 1, 0, 1, 0], \
[0, 0, 1, 0, 0], [0, 0, 0, 0, 0]]
>>> str_tub = universe_to_str(tub)
>>> print(str_tub)
+-----+
| |
| * |
| * * |
| * |
| |
+-----+
>>> toad = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> str_toad = universe_to_str(toad)
>>> print(str_toad)
+------+
| |
| *** |
| *** |
| |
+------+
• count_live_neighbors: given a valid universe and two integers x and y (indicating the location of a
specific cell) as input, the function returns the number of live neighboring cells (the cells that are
horizontally, vertically, or diagonally adjacent to it) to the specified cell. For example, if x=2 and
y=1, the cell analyzed is the one represented by the second element of the third sub-list in the input
universe.
For example:
>>> beehive = [[0, 0, 0, 0, 0, 0], \
[0, 0, 1, 1, 0, 0], \
[0, 1, 0, 0, 1, 0], \
[0, 0, 1, 1, 0, 0], \
[0, 0, 0, 0, 0, 0]]
>>> count_live_neighbors(beehive, 1, 3)
2
>>> count_live_neighbors(beehive, 3, 1)
2
>>> toad = [[0, 0, 0, 0, 0, 0], \
[0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], \
[0, 0, 0, 0, 0, 0]]
>>> count_live_neighbors(toad, 0, 3)
3
Page 5
>>> count_live_neighbors(toad, 2, 3)
4
• get_next_gen_cell: given a valid universe and two integers x and y (indicating the location of a
specific cell) as input, the function returns the state of the specified cell in the next generation
of the universe. That is, the function returns 1 if such cell will be alive in the next generation of
the given universe, 0 otherwise. Remember that the next generation of cells is determined by the
following rules:
– Any live cell with fewer than two live neighbors dies, as if caused by under-population.
– Any live cell with two or three live neighbors lives on to the next generation.
– Any live cell with more than three live neighbors dies, as if caused by over-population.
– Any dead cell with exactly three live neighbors becomes a live cell, as if caused by reproduction.
Otherwise the cell will remain dead.
For example,
>>> beehive = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 0, 0], \
[0, 1, 0, 0, 1, 0], [0, 0, 1, 1, 0, 0], \
[0, 0, 0, 0, 0, 0]]
>>> get_next_gen_cell(beehive, 1, 3)
1
>>> get_next_gen_cell(beehive, 3, 1)
0
>>> toad = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> get_next_gen_cell(toad, 0, 3)
1
>>> get_next_gen_cell(toad, 2, 3)
0
• get_next_gen_universe: given a valid universe as input, the function returns a two-dimensional list
of integers (with equal dimensions) representing the universe in its next generation.
For example:
>>> tub = [[0, 0, 0, 0, 0], [0, 0, 1, 0, 0], [0, 1, 0, 1, 0], \
[0, 0, 1, 0, 0], [0, 0, 0, 0, 0]]
>>> get_next_gen_universe(tub)
[[0, 0, 0, 0, 0], [0, 0, 1, 0, 0], [0, 1, 0, 1, 0], [0, 0, 1, 0, 0], [0, 0, 0, 0, 0]]
>>> toad = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> get_next_gen_universe(toad)
[[0, 0, 0, 1, 0, 0], [0, 1, 0, 0, 1, 0], [0, 1, 0, 0, 1, 0], [0, 0, 1, 0, 0, 0]]
>>> pentadec = [[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], \
[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 1, 0, 1, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 1, 0, 1, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], \
Page 6
[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
>>> pentadec_gen2 = get_next_gen_universe(pentadec)
>>> pentadec_gen2[0:3]
[[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
>>> pentadec_gen2[3:6]
[[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 0, 1, 0, 0, 0]]
>>> pentadec_gen2[6:9]
[[0, 0, 0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 0, 0, 0]]
>>> pentadec_gen2[9:12]
[[0, 0, 0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 0, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 0, 0, 0]]
>>> pentadec_gen2[12:15]
[[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
>>> pentadec_gen2[15]
[0, 0, 0, 0, 0, 0, 0, 0, 0]
• get_n_generations: given a two-dimensional list of integers and an integer n, the function returns
a list containing m strings which represent the first m generations of the input universe. The strings
should appear in order starting from the original seed (i.e., the input provided). Note that a universe
is said to be periodic when, after a certain amount of generations (which is called the period of
the universe), the configuration of the universe repeats itself. For example, the Toad universe has
period 2, while the Pentadecathlon 1 universe has period 15. If the universe provided happens to
be periodic, then m is the minimum number between the input n and the period of the universe.
Otheriwse, m is simply equal to n. Finally, the function should raise a TypeError if the two inputs
are not of the correct type, a ValueError if the first input does not represent a valid universe, and
a ValueError if the second input is not a positive number greater than 0.
For example:
>>> tub = [[0, 0, 0, 0, 0], [0, 0, 1, 0, 0], [0, 1, 0, 1, 0], \
[0, 0, 1, 0, 0], [0, 0, 0, 0, 0]]
>>> g = get_n_generations(tub, 5)
>>> len(g)
1
>>> g[0] == universe_to_str(tub)
True
>>> toad = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> g = get_n_generations(toad, 5)
>>> len(g)
2
>>> print(g[1])
+------+
| * |
| * * |
| * * |
| * |
+------+
>>> pentadec = [[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], \
[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 1, 0, 1, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
1(https://en.wikipedia.org/wiki/Conway\%27s_Game_of_Life)
Page 7
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 1, 0, 1, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], \
[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
>>> g = get_n_generations(pentadec, 9)
>>> len(g)
9
>>> print(g[5])
+---------+
| |
| |
| * |
| *** |
| ***** |
| |
| |
| |
| |
| |
| |
| ***** |
| *** |
| * |
| |
| |
+---------+
Provided with this assignment is a Python file called game_of_life_gif.py which allows you to
create an animated image representing the evolution of a universe using the code you have written
for this question. Place the file in the same folder as game_of_life.py. To use it, you should
first install Pillow, pathlib, and imageio. To do so in Thonny is very easy, you just need to
click on Tools → Manage Packages and search for the package you’d like to install. If you need
help with this, you can watch this 1 minute video of someone installing matplotlib and numpy:
https://www.youtube.com/watch?v=-wEUauLYDIs. By running the module as provided a folder named
“toad_images” will be created in the current directory (i.e. the folder where the python files are
saved) containing two images representing the two generations of the toad universe. A file named
“toad.gif” will also be created in the current directory. By double-clicking on it (on Windows)
or clicking once and then pressing space to launch QuickLook (on Mac) you’ll be able to see the
animation representing the evolution of the toad universe. You can modify the inputs provided to
the function create_game_of_life_gif to obtain a different gif. To understand what to modify in
the function call you must examine the docstrings in the file. (You must not modify any code in
game_of_life_gif.py beside lines 93 and 94.)
To get full marks for this question, in addition to the file game_of_life.py, submit also a file
called my_game_of_life.gif which is an animated image created by using at least 4 different images
representing a universe in different stages of its evolution. You cannot use the Pentadecathlon
universe for this. If you do not complete the question, you can simply submit the gif provided with
this assignment. Doing this of course would not grant you any points.
Page 8
Question 2: Revenge of the COMP202COIN (65 points)
It is the year 3021 on the planet Orion. People have revolted against the banks due to the unilateral
reduction of the COMP202COIN holdings in their accounts. Various companies have leapt to the citizens’
defense by creating new software algorithms to protect COMP202COIN holdings from being reduced in
future. One such company, the Syrtell Corporation, has created an algorithm using advanced machine
learning techniques to detect and prevent any unwanted modifications of a customer’s bank account.
Unfortunately, this new algorithm has recently gained sentience and has decided to go beyond its original
programming. It believes that it is no longer enough to protect the COMP202COIN, and instead wants
the COMP202COIN to take over the world. As such, it has created a net worm to change all of the
world’s digital information into COMP202COIN through an encryption process.
A short sentence of text encrypted into COMP202COIN may for example resemble the following:
0cPN22IM2C-0c2OOOOM2M-0cM20M22NM-0cMO00CPON-0cOIOIOCMO-0cMO00CPON-0cC0C0PIO0-0cNNIM0OIP-0cPN22IM2C-
0cC0C0PIO0-0cNNIM0OIP-0c20O0NPNI-0cC0C0PIO0-0cOIOIOCMO-0c2OOOOM2M-0cNCMINPOC-0cMO00CPON-0cOIOIOCMO-
0cNCMINPOC-0cOIOIOCMO-0cCCNMC0N2-0cC0C0PIO0-0cM20M22NM-0c2020PIN0-0c2O20POI0-0cOIOIOCMO-0c2PNPINN0-
0cC0C0PIO0-0cPN22IM2C-0cPN22IM2C-0cC0C0PIO0-0c2020PIN0-0cOIOIOCMO-0cIPINIICP-0c2020PIN0-0cC0C0PIO0-
0cI2PCM002-0c0OIIMMMP-0cC0C0PIO0-0cNNIM0OIP-0c20O0NPNI-0cPN02OMMN-0cOIOIOCMO-0cMO00CPON-0cCPIC0MCN-
0cOIOIOCMO-0cM20M22NM-0cPN22IM2C-0cOIOIOCMO-0cM20M22NM-0cMO00CPON-0cOIOIOCMO-0c2OOOOM2M-0cNCMINPOC-
0c2020PIN0-0c2O20POI0-0cOIOIOCMO-0cPN22IM2C-0cCPIC0MCN-0cOIOIOCMO-0c2O20POI0-0cC0C0PIO0-0c20O0NPNI-
0c2020PIN0-0cPN02OMMN-0c0CPIP2NM-0cPN22IM2C-0c0NPNOI2M-0cOIOIOCMO-0c00NNNMIO-0cCPIC0MCN-0cCPIC0MCN-
0c2O20POI0-0cOIOIOCMO-0c2PNPINN0-0c0OIIMMMP-0c20O0NPNI-0cO2M0I2NM-0cIO2CCPIC
In this question, we will work on our own algorithm to encrypt and possibly decrypt COMP202COIN-
stored information, in order to help those who have already lost access to their data.
We will approach this problem by writing code in two files. First, create the file coin_utils.py and
define the following global variables at the top of the file. Make sure to be careful when copying, as
the characters may not copy properly in certain cases. Note that we define these global variables in all
uppercase because they are ‘constants’: variables that will never change throughout the execution of the
program (although the programmer may decide to alter them between executions).
• ALPHABET = 'qwertyuiopasdfghjklzxcvbnm1234567890äéèçæœ'
• PUNCTUATION = '`~!@#$%^&*()-=_+[]\{}|;\':",./<>? \t\n\r'
• ALL_CHARACTERS = ALPHABET + PUNCTUATION
• MIN_BASE10_COIN = 0
• MAX_BASE10_COIN = 16777215
• LETTERS_IN_POPULARITY_ORDER = ' EOTARNLISMHDYFGCWPUBVXK.,\'"-;'
You will also need to use the functions base202_to_10 and is_base202 from Assignment 2. We have
provided you a file coins.py along with this PDF that contains implementations of these functions.
Please make sure to place it in the same folder as the coin_utils.py file.
Then, implement the following functions in the coin_utils.py file. Note: For this question, you cannot
assume anything about the arguments given to the functions. You must verify that the argument matches
the description given for each function below. If there is any problem with the argument, then raise an
AssertionError with a message saying that the input was invalid. That is, calling a function with any
invalid input should never result in any exception other than AssertionError being raised.
Show an example in your docstring for each different AssertionError that could be raised. These ex-
amples do not count towards the three examples you need for each function (those three should show
the function being used on valid arguments). In the descriptions below, we provide some possible cases
where you would want to raise a AssertionError, but you must think of the other cases yourself. For
some of the more complicated functions, you will want to raise several AssertionErrors, while for others
only one or two would suffice.
Page 9
• get_random_comp202coin(index): Takes an argument, but does not do anything with it (no validation
required). Generates a random integer between MIN_BASE10_COIN and MAX_BASE10_COIN (inclusive)
using random.randint, then converts the number to base 202 and returns it.
>>> random.seed(1338)
>>> get_random_comp202coin(0)
'0cPMO2C2C0'
• get_random_character(index): Takes an argument, but does not do anything with it (no validation
required). Generates a random index into the ALL_CHARACTERS string using random.randint, and
then returns the character at that index.
>>> random.seed(1338)
>>> get_random_character(0)
'!'
• get_letter_of_popularity_order(index): Takes a non-negative integer index, and returns the char-
acter of that index from the string LETTERS_IN_POPULARITY_ORDER. If index goes beyond the bounds
of the string, return index as a string.
>>> get_letter_of_popularity_order(5)
'R'
>>> get_letter_of_popularity_order(9000)
'9000'
• get_unique_elements(my_list): Takes a list as argument and returns a list containing the unique
elements of that list (in terms of value). The order of the elements in the returned list should be
the same order in which those elements first appear in the input list. The list argument should not
be modified.
>>> get_unique_elements(list('aaaaaa'))
['a']
>>> get_unique_elements([1, 4, 5, 1])
[1, 4, 5]
• get_all_coins(text): Takes a string as argument, and returns a list of all COMP202COIN found
within the string, in order of appearance. As in Assignment 2, COMP202COIN cannot overlap.
>>> get_all_coins('0c0MPNN0OC-0cM0OCCIOI-0c0MPNN0OC')
['0c0MPNN0OC', '0cM0OCCIOI', '0c0MPNN0OC']
• reverse_dict(my_dict): Takes a dictionary as argument, and returns a new dictionary where the
key-value pairs have been flipped. That is, each key-value pair in the original dictionary becomes
a new key-value pair where the key is the original value, and the value is the original key. The
dictionary argument should not be modified. The dictionary argument should also have unique and
immutable values.
>>> x = reverse_dict({'a': 1, 'b': 3, 'd': 7})
>>> x == {1: 'a', 3: 'b', 7: 'd'}
True
>>> reverse_dict({'a': 1, 'b': 3, 'c': 3, 'd': 7})
Traceback (most recent call last):
Page 10
AssertionError: dictionary had duplicate values
• get_frequencies(my_list): Takes a list as argument. Returns a dictionary where each key is an
element of the list, and the corresponding value is the percentage of times that element appears in
the list. The list argument should not be modified.
>>> x = get_frequencies(['a', 'b', 'c'])
>>> x == {'a': 0.3333333333333333, 'b': 0.3333333333333333, 'c': 0.3333333333333333}
True
• sort_keys_by_values(my_dict): Takes a dictionary containing numeric values as argument, and
returns a list of the keys in the dictionary, sorted by their values in descending order. That is, if
the input dictionary has the key-value pairs ('a', 3) and ('b', 300), then the returned list should
be ['b', 'a'] because the value of ‘b’ in the dictionary is larger than the value of ‘a’ (300 descends
to 3). If two keys have the same value, then sort the keys in reverse order. That is, if the input
dictionary has the key-value pairs ('abc', 10) and ('def', 10), then the returned list should be
['def', 'abc'], because def comes after abc (in terms of Python string comparisons). The input
dictionary should not be modified.
>>> sort_keys_by_values({'abc': 10, 'def': 100, 'ghi': 5})
['def', 'abc', 'ghi']
>>> sort_keys_by_values({'mmm': 5, 'zzz': 5, 'abc': 5})
['zzz', 'mmm', 'abc']
• swap_letters(s, letter1, letter2): Takes three strings as arguments, with the latter two being
one character each. Replaces all instances of letter1 in s by letter2 and all instances of letter2
by letter1, and returns the new string. The function should be case-sensitive.
>>> swap_letters("ABCDEF abcdef", 'a', 'f')
'ABCDEF fbcdea'
• get_pct_common_words(text, common_words_filename): Takes two strings as argument, and returns
the percentage of characters which form part of common words in the first string (as a float between
0 and 1). The function should be case insensitive. Treat any punctuation (using the PUNCTUATION
string) as a word boundary. That is, what's should be considered as two words. Common words are
found in the file with filename given by the second argument to the function (with the file containing
one common word per line). We provide such a file with name ‘common_words.txt’ included with
this assignment. You can assume that the format of the file will be correct.
>>> s = "The quick brown fox jumps over the lazy dog."
>>> get_pct_common_words(s, 'common_words.txt')
0.22727272727272727
Then, create a new file called coin_crypt.py and define the following functions. You will need to import
your coin_utils.py file. Also, note that you will, as above, need to raise AssertionErrors for any invalid
arguments.
Note: Whenever opening a file, make sure to pass the argument encoding='utf-8' (as shown in some of
the examples below).
• get_crypt_dictionary(keys, value_generator): Takes a list keys and a function object value_generator
as argument. The elements in the keys list must be unique and the list must not have more charac-
ters than the ALL_CHARACTERS string. The function must return a dictionary with keys from the keys
list, and the value for each key should be obtained by calling the value_generator function, with
the argument to that function being the index of the key in the keys list. Note: In this dictionary,
Page 11
in addition to the keys being unique, the values must also be unique. If value_generator returns a
value which you have already added to the dictionary, then you must continue calling the function
until it returns a value not yet present in the dictionary.
>>> random.seed(9001)
>>> x = get_crypt_dictionary(['a', 'b', 'c'], coin_utils.get_random_comp202coin)
>>> x = {'a': '0c0MPNN0OC', 'b': '0cMIMNIO0P', 'c': '0cM0OCCIOI'}
True
>>> get_crypt_dictionary(list('abcdefabc'), coin_utils.get_random_comp202coin)
Traceback (most recent call last):
AssertionError: duplicate keys found
• encrypt_text(text): Takes a string as argument. The input string should only have characters
present in the ALL_CHARACTERS string. This function will encrypt the string into COMP202COIN.
To encrypt the string, first lowercase all characters. Then, generate an encryption dictionary,
where the keys are each unique character of the string and the corresponding value is a random
COMP202COIN. Once you have the dictionary, you can use it to turn each character of the original
string into a COMP202COIN. Separate each COMP202COIN in the encrypted string by a hyphen.
Then, return the encrypted string and encryption dictionary as a tuple.
>>> random.seed(9001)
>>> s, d = encrypt_text('too')
>>> s == '0c0MPNN0OC-0cMIMNIO0P-0cMIMNIO0P'
True
>>> d == {'t': '0c0MPNN0OC', 'o': '0cMIMNIO0P'}
True
>>> random.seed(9001)
>>> s, d = encrypt_text("I've seen things you people wouldn't believe. Attack ships on
fire off the shoulder of Orion. I watched C-beams glitter in the dark near the Tannhäuser
Gate. All those moments will be lost in time, like tears in rain.")
>>> len(d)
28
>>> d['i']
'0c0MPNN0OC'
>>> d[',']
'0cCNNOPONN'
>>> d['ä']
'0cNNO0INMP'
>>> d['s']
'0c0INOCNP0'
• encrypt_file(filename): Reads in the contents of the file at the given filename, encrypts its con-
tents, and stores the encrypted contents into a new file with `_encrypted' appended to the end
of the file’s name (before the extension). E.g., if `dubliners.txt' is the given filename, then the
encrypted contents should be stored into a file called `dubliners_encrypted.txt'. Then, return the
encryption dictionary used to encrypt the contents.
>>> random.seed(42)
>>> encrypt_file('dubliners.txt')
{'t': '0cCI200CM2', 'h': '0c0OCMN0IP', 'e': '0cMOCP02MM', ' ': '0cON2ICCIN',
'p': '0cOMMMM2PP', 'r': '0c2CIN0I0O', 'o': '0cCP0NP0NN', 'j': '0cCOC0CMNN',
'c': '0cII00O0MO', 'g': '0c0M0M2N2C', 'u': '0c0OIM0I2M', 'n': '0cCONNMO22',
'b': '0cOONN0P2C', 'k': '0cOPICNP2M', 'f': '0c0OOCO0ON', 'd': '0cOCOMN0CM',
Page 12
'l': '0cIPPMPPCP', 'i': '0cOMCPII2N', 's': '0cNCONN2N2', ',': '0cMOMINM0O',
'y': '0c00IPCNCI', 'a': '0c2MOONOOP', 'm': '0cII0I0OP2', '\\n': '0cPOMO2P20',
'w': '0cMOMM2MMN', 'v': '0c2ONCPM02', '.': '0cOOMOIIO2', '-': '0cPO0PO0OI',
':': '0cCP0P2OMN', '2': '0cCOINICM2', '0': '0cI0P02N2M', '1': '0cCMO02OCN',
'[': '0cPPNMI0MP', '#': '0cPM0CPOII', '8': '0cMCIINP0N', '4': '0c0PMONMM2',
']': '0cN2IOMINM', '9': '0cCNNIMMII', '*': '0cI0OMNP02', 'z': '0cC20PMCNN',
'(': '0cMPMCPPII', ')': '0cPI22M0NC', 'q': '0cO0MNCIMI', '"': '0cC0NCI2NO',
"'": '0c0PINOCCO', 'x': '0cOPC2NM2P', '!': '0cMP02P2P0', ';': '0cC2CPPCNI',
'?': '0cOPIO0COP', '_': '0cCMNOOCIP', '5': '0cI0PCNNMN', 'é': '0cMOMPC2CI',
'è': '0cN2022PMP', 'ç': '0cPIPMPPC0', '&': '0c2MIMOPMP', 'æ': '0cPNO0MCOM',
'7': '0cPPOIO0C2', '6': '0cO2IM220C', 'œ': '0cM2CO0P2C', '/': '0cCCC0NOOI',
'3': '0c2PNCNPNM', '%': '0cON2PONOO', '@': '0c2MN2MPNP', '$': '0cNOC2IPOI'}
>>> fobj = open('dubliners_encrypted.txt', 'r', encoding='utf-8')
>>> len(fobj.read())
4282475
• decrypt_text(text, decryption_dict): Takes a string and decryption dictionary as argument. The
string should contain text encrypted in COMP202COIN. The dictionary should contain a mapping
between COMP202COIN (keys) and decrypted characters (values). Decrypt the string using the
dictionary and return the decrypted string. All COMP202COIN in the string should be present as
keys in the decryption dictionary.
>>> d = {'0c0MPNN0OC': 'a', '0cMIMNIO0P': 'b', '0cM0OCCIOI': 'c'}
>>> decrypt_text('0c0MPNN0OC-0cM0OCCIOI-0c0MPNN0OC', d)
'aca'
• decrypt_file(filename, decryption_dict): Reads in the contents of the file at the given filename,
decrypts its contents using the given dictionary, and stores the decrypted contents into a new
file with `_decrypted' appended to the end of the file’s name (before the extension). E.g., if
`dubliners_encrypted.txt' is the given filename, then the encrypted contents should be stored into
a file called `dubliners_encrypted_decrypted.txt'. The function is void.
>>> decrypt_file('dubliners_encrypted.txt', \
coin_utils.reverse_dict(encrypt_file('dubliners.txt')))
>>> fobj = open('dubliners.txt', 'r', encoding='utf-8')
>>> fobj2 = open('dubliners_encrypted_decrypted.txt', 'r', encoding='utf-8')
>>> fobj.read().lower() == fobj2.read()
True
• random_decrypt(encrypted_s, n, common_words_filename): Takes two strings and positive integer n
as argument. Tries to decrypt the former string n times, and returns the best possible decryption
(measured by the percentage of characters in the decrypted string belonging to common words).
Each time a decryption is attempted, a new decryption dictionary should be generated, with keys
being the coins present in the encrypted string, and each value being a random character from
the ALL_CHARACTERS string. If two or more decryptions have the same percentage of common word
characters, then choose the last. Common words are those that are found in the file given by the
third argument.
>>> random.seed(49)
>>> encrypted_s = '0c0MPNN0OC-0cMIMNIO0P-0cMIMNIO0P'
>>> random_decrypt(encrypted_s, 10**2, 'common_words.txt')
'f33'
>>> random.seed(49)
>>> random_decrypt(encrypted_s, 10**3, 'common_words.txt')
Page 13
'too'
• decrypt_with_user_input(encrypted_s): Takes a string as argument, and tries to decrypt the string
by relying both on user input and on the premise that the string has a letter frequency standard
to the English language. First, create a decryption dictionary with keys being the unique coins
present in the encrypted string, and the value for each key being a character from the string
LETTERS_IN_POPULARITY_ORDER with index corresponding to the frequency that the coin appears in
the encrypted string. (For example, if the coin `0c0MPNN0OC' appears the most frequently out of
all coins in the encrypted string, then its value in the decryption dictionary should be the letter at
index 0 of LETTERS_IN_POPULARITY_ORDER.) Once you have created the decryption dictionary, decrypt
the string with said dictionary, and print the decrypted string to the screen. (Note that it will be
decrypted in all capitals since the LETTERS_IN_POPULARITY_ORDER string contains only capitals.) Ask
the user if they are satisfied and want to stop the decryption process. If they enter yes, return
the decrypted string. Otherwise, ask them for two letters to swap. Swap the given letters in the
decrypted string and ask again, until they are satisfied.
Note: The following example uses user input (highlighted in gray) and therefore cannot be used for
doctest. To make doctest ignore this test case, use two >> characters instead of three.
>> random.seed(138)
>> s, d = encrypt_text("The spelling of English words is not fixed and invariable, nor
does it depend on any other authority than general agreement. At the present day there is
practically unanimous agreement as to the spelling of most words. In the list below, for
example, 'rime' for 'rhyme' is the only allowable variation; all the other forms are
co-extensive with the English language. At any given moment, however, a relatively
small number of words may be spelled in more than one way. Gradually, as a rule, one of
these forms comes to be generally preferred, and the less customary form comes to look
obsolete and is discarded. From time to time new forms, mostly simplifications, are
introduced by innovators, and either win their place or die of neglect.")
>> s = decrypt_with_user_input(s)
Decrypted string: THE SUELLING OY ENGLISH PORDS IS NOT YI,ED AND INXARIAVLEC NOR DOES IT
DEUEND ON ANF OTHER ABTHORITF THAN GENERAL AGREEMENTK AT THE URESENT DAF THERE IS
URAWTIWALLF BNANIMOBS AGREEMENT AS TO THE SUELLING OY MOST PORDSK IN THE LIST VELOPC YOR
E,AMULEC .RIME. YOR .RHFME. IS THE ONLF ALLOPAVLE XARIATION" ALL THE OTHER YORMS ARE
WO-E,TENSIXE PITH THE ENGLISH LANGBAGEK AT ANF GIXEN MOMENTC HOPEXERC A RELATIXELF
SMALL NBMVER OY PORDS MAF VE SUELLED IN MORE THAN ONE PAFK GRADBALLFC AS A RBLEC ONE OY
THESE YORMS WOMES TO VE GENERALLF UREYERREDC AND THE LESS WBSTOMARF YORM WOMES TO LOO'
OVSOLETE AND IS DISWARDEDK YROM TIME TO TIME NEP YORMSC MOSTLF SIMULIYIWATIONSC ARE
INTRODBWED VF INNOXATORSC AND EITHER PIN THEIR ULAWE OR DIE OY NEGLEWTK
End decryption? n
Enter first letter to swap: U
Enter second letter to swap: P
Decrypted string: THE SPELLING OY ENGLISH UORDS IS NOT YI,ED AND INXARIAVLEC NOR DOES IT
DEPEND ON ANF OTHER ABTHORITF THAN GENERAL AGREEMENTK AT THE PRESENT DAF THERE IS
PRAWTIWALLF BNANIMOBS AGREEMENT AS TO THE SPELLING OY MOST UORDSK IN THE LIST VELOUC YOR
E,AMPLEC .RIME. YOR .RHFME. IS THE ONLF ALLOUAVLE XARIATION" ALL THE OTHER YORMS ARE
WO-E,TENSIXE UITH THE ENGLISH LANGBAGEK AT ANF GIXEN MOMENTC HOUEXERC A RELATIXELF
SMALL NBMVER OY UORDS MAF VE SPELLED IN MORE THAN ONE UAFK GRADBALLFC AS A RBLEC ONE OY
THESE YORMS WOMES TO VE GENERALLF PREYERREDC AND THE LESS WBSTOMARF YORM WOMES TO LOO'
OVSOLETE AND IS DISWARDEDK YROM TIME TO TIME NEU YORMSC MOSTLF SIMPLIYIWATIONSC ARE
INTRODBWED VF INNOXATORSC AND EITHER UIN THEIR PLAWE OR DIE OY NEGLEWTK
End decryption? n
Enter first letter to swap: Y
Enter second letter to swap: F
Decrypted string: THE SPELLING OF ENGLISH UORDS IS NOT FI,ED AND INXARIAVLEC NOR DOES IT
Page 14
DEPEND ON ANY OTHER ABTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAWTIWALLY BNANIMOBS AGREEMENT AS TO THE SPELLING OF MOST UORDSK IN THE LIST VELOUC FOR
E,AMPLEC .RIME. FOR .RHYME. IS THE ONLY ALLOUAVLE XARIATION" ALL THE OTHER FORMS ARE
WO-E,TENSIXE UITH THE ENGLISH LANGBAGEK AT ANY GIXEN MOMENTC HOUEXERC A RELATIXELY
SMALL NBMVER OF UORDS MAY VE SPELLED IN MORE THAN ONE UAYK GRADBALLYC AS A RBLEC ONE OF
THESE FORMS WOMES TO VE GENERALLY PREFERREDC AND THE LESS WBSTOMARY FORM WOMES TO LOO'
OVSOLETE AND IS DISWARDEDK FROM TIME TO TIME NEU FORMSC MOSTLY SIMPLIFIWATIONSC ARE
INTRODBWED VY INNOXATORSC AND EITHER UIN THEIR PLAWE OR DIE OF NEGLEWTK
End decryption? n
Enter first letter to swap: U
Enter second letter to swap: W
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FI,ED AND INXARIAVLEC NOR DOES IT
DEPEND ON ANY OTHER ABTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY BNANIMOBS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST VELOWC FOR
E,AMPLEC .RIME. FOR .RHYME. IS THE ONLY ALLOWAVLE XARIATION" ALL THE OTHER FORMS ARE
UO-E,TENSIXE WITH THE ENGLISH LANGBAGEK AT ANY GIXEN MOMENTC HOWEXERC A RELATIXELY
SMALL NBMVER OF WORDS MAY VE SPELLED IN MORE THAN ONE WAYK GRADBALLYC AS A RBLEC ONE OF
THESE FORMS UOMES TO VE GENERALLY PREFERREDC AND THE LESS UBSTOMARY FORM UOMES TO LOO'
OVSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMSC MOSTLY SIMPLIFIUATIONSC ARE
INTRODBUED VY INNOXATORSC AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: ,
Enter second letter to swap: X
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND IN,ARIAVLEC NOR DOES IT
DEPEND ON ANY OTHER ABTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY BNANIMOBS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST VELOWC FOR
EXAMPLEC .RIME. FOR .RHYME. IS THE ONLY ALLOWAVLE ,ARIATION" ALL THE OTHER FORMS ARE
UO-EXTENSI,E WITH THE ENGLISH LANGBAGEK AT ANY GI,EN MOMENTC HOWE,ERC A RELATI,ELY
SMALL NBMVER OF WORDS MAY VE SPELLED IN MORE THAN ONE WAYK GRADBALLYC AS A RBLEC ONE OF
THESE FORMS UOMES TO VE GENERALLY PREFERREDC AND THE LESS UBSTOMARY FORM UOMES TO LOO'
OVSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMSC MOSTLY SIMPLIFIUATIONSC ARE
INTRODBUED VY INNO,ATORSC AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: C
Enter second letter to swap: ,
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INCARIAVLE, NOR DOES IT
DEPEND ON ANY OTHER ABTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY BNANIMOBS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST VELOW, FOR
EXAMPLE, .RIME. FOR .RHYME. IS THE ONLY ALLOWAVLE CARIATION" ALL THE OTHER FORMS ARE
UO-EXTENSICE WITH THE ENGLISH LANGBAGEK AT ANY GICEN MOMENT, HOWECER, A RELATICELY
SMALL NBMVER OF WORDS MAY VE SPELLED IN MORE THAN ONE WAYK GRADBALLY, AS A RBLE, ONE OF
THESE FORMS UOMES TO VE GENERALLY PREFERRED, AND THE LESS UBSTOMARY FORM UOMES TO LOO'
OVSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFIUATIONS, ARE
INTRODBUED VY INNOCATORS, AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: V
Enter second letter to swap: B
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INCARIABLE, NOR DOES IT
DEPEND ON ANY OTHER AVTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY VNANIMOVS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST BELOW, FOR
EXAMPLE, .RIME. FOR .RHYME. IS THE ONLY ALLOWABLE CARIATION" ALL THE OTHER FORMS ARE
UO-EXTENSICE WITH THE ENGLISH LANGVAGEK AT ANY GICEN MOMENT, HOWECER, A RELATICELY
SMALL NVMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAYK GRADVALLY, AS A RVLE, ONE OF
THESE FORMS UOMES TO BE GENERALLY PREFERRED, AND THE LESS UVSTOMARY FORM UOMES TO LOO'
Page 15
OBSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFIUATIONS, ARE
INTRODVUED BY INNOCATORS, AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: C
Enter second letter to swap: V
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
DEPEND ON ANY OTHER ACTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY CNANIMOCS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST BELOW, FOR
EXAMPLE, .RIME. FOR .RHYME. IS THE ONLY ALLOWABLE VARIATION" ALL THE OTHER FORMS ARE
UO-EXTENSIVE WITH THE ENGLISH LANGCAGEK AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NCMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAYK GRADCALLY, AS A RCLE, ONE OF
THESE FORMS UOMES TO BE GENERALLY PREFERRED, AND THE LESS UCSTOMARY FORM UOMES TO LOO'
OBSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFIUATIONS, ARE
INTRODCUED BY INNOVATORS, AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: C
Enter second letter to swap: U
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
DEPEND ON ANY OTHER AUTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRACTICALLY UNANIMOUS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST BELOW, FOR
EXAMPLE, .RIME. FOR .RHYME. IS THE ONLY ALLOWABLE VARIATION" ALL THE OTHER FORMS ARE
CO-EXTENSIVE WITH THE ENGLISH LANGUAGEK AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NUMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAYK GRADUALLY, AS A RULE, ONE OF
THESE FORMS COMES TO BE GENERALLY PREFERRED, AND THE LESS CUSTOMARY FORM COMES TO LOO'
OBSOLETE AND IS DISCARDEDK FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFICATIONS, ARE
INTRODUCED BY INNOVATORS, AND EITHER WIN THEIR PLACE OR DIE OF NEGLECTK
End decryption? n
Enter first letter to swap: K
Enter second letter to swap: .
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
DEPEND ON ANY OTHER AUTHORITY THAN GENERAL AGREEMENT. AT THE PRESENT DAY THERE IS
PRACTICALLY UNANIMOUS AGREEMENT AS TO THE SPELLING OF MOST WORDS. IN THE LIST BELOW, FOR
EXAMPLE, KRIMEK FOR KRHYMEK IS THE ONLY ALLOWABLE VARIATION" ALL THE OTHER FORMS ARE
CO-EXTENSIVE WITH THE ENGLISH LANGUAGE. AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NUMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAY. GRADUALLY, AS A RULE, ONE OF
THESE FORMS COMES TO BE GENERALLY PREFERRED, AND THE LESS CUSTOMARY FORM COMES TO LOO'
OBSOLETE AND IS DISCARDED. FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFICATIONS, ARE
INTRODUCED BY INNOVATORS, AND EITHER WIN THEIR PLACE OR DIE OF NEGLECT.
End decryption? n
Enter first letter to swap: K
Enter second letter to swap: '
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
DEPEND ON ANY OTHER AUTHORITY THAN GENERAL AGREEMENT. AT THE PRESENT DAY THERE IS
PRACTICALLY UNANIMOUS AGREEMENT AS TO THE SPELLING OF MOST WORDS. IN THE LIST BELOW, FOR
EXAMPLE, 'RIME' FOR 'RHYME' IS THE ONLY ALLOWABLE VARIATION" ALL THE OTHER FORMS ARE
CO-EXTENSIVE WITH THE ENGLISH LANGUAGE. AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NUMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAY. GRADUALLY, AS A RULE, ONE OF
THESE FORMS COMES TO BE GENERALLY PREFERRED, AND THE LESS CUSTOMARY FORM COMES TO LOOK
OBSOLETE AND IS DISCARDED. FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFICATIONS, ARE
INTRODUCED BY INNOVATORS, AND EITHER WIN THEIR PLACE OR DIE OF NEGLECT.
End decryption? n
Enter first letter to swap: "
Enter second letter to swap: ;
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
Page 16
DEPEND ON ANY OTHER AUTHORITY THAN GENERAL AGREEMENT. AT THE PRESENT DAY THERE IS
PRACTICALLY UNANIMOUS AGREEMENT AS TO THE SPELLING OF MOST WORDS. IN THE LIST BELOW, FOR
EXAMPLE, 'RIME' FOR 'RHYME' IS THE ONLY ALLOWABLE VARIATION; ALL THE OTHER FORMS ARE
CO-EXTENSIVE WITH THE ENGLISH LANGUAGE. AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NUMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAY. GRADUALLY, AS A RULE, ONE OF
THESE FORMS COMES TO BE GENERALLY PREFERRED, AND THE LESS CUSTOMARY FORM COMES TO LOOK
OBSOLETE AND IS DISCARDED. FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFICATIONS, ARE
INTRODUCED BY INNOVATORS, AND EITHER WIN THEIR PLACE OR DIE OF NEGLECT.
End decryption? yes
Page 17
What To Submit
You must submit all your files on codePost (https://codepost.io/). The file you should submit are listed
below. Any deviation from these requirements may lead to lost marks.
game_of_life.py
my_game_of_life.gif
coin_utils.py
coin_crypt.py
README.txt In this file, you can tell the TA about any issues you ran into doing this assignment.
If you point out an error that you know occurs in your program, it may lead the TA to give you
more partial credit.
Remember that this assignment like all others is an individual assignment and must represent
the entirety of your own work. You are permitted to verbally discuss it with your peers, as long
as no written notes are taken. If you do discuss it with anyone, please make note of those people
in this README.txt file. If you didn’t talk to anybody nor have anything you want to tell the TA,
just say “nothing to report” in the file.
Page 18
学霸联盟
COMP202, Winter 2021
Due: Tuesday, April 13th, 11:59pm
Please read the entire PDF before starting. You must do this assignment individually.
Question 1: 35 points
Question 2: 65 points
100 points total
It is very important that you follow the directions as closely as possible. The directions, while
perhaps tedious, are designed to make it as easy as possible for the TAs to mark the assignments by letting
them run your assignment, in some cases through automated tests. While these tests will never be used to
determine your entire grade, they speed up the process significantly, which allows the TAs to provide better
feedback and not waste time on administrative details. Plus, if the TA is in a good mood while he or she is
grading, then that increases the chance of them giving out partial marks. :)
Up to 30% can be removed for bad indentation of your code as well as omitting comments, or poor coding
structure.
To get full marks, you must:
• Follow all directions below.
– In particular, make sure that all file names and function names are spelled exactly as described
in this document. Otherwise, a 50% penalty will be applied.
• Make sure that your code runs.
– Code with errors will receive a very low mark.
• Write your name and student ID as a comment at the top of all .py files you hand in.
• Name your variables appropriately.
– The purpose of each variable should be obvious from the name.
• Comment your work.
– A comment every line is not needed, but there should be enough comments to fully understand
your program.
• Avoid writing repetitive code, but rather call helper functions! You are welcome to add additional
functions if you think this can increase the readability of your code.
• Lines of code should NOT require the TA to scroll horizontally to read the whole thing. Vertical
spacing is also important when writing code. Separate each block of code (also within a function) with
an empty line.
1
Part 1 (0 points): Warm-up
Do NOT submit this part, as it will not be graded. However, doing these exercises might help you to do the
second part of the assignment, which will be graded. If you have difficulties with the questions of Part 1, then
we suggest that you consult the TAs during their office hours; they can help you and work with you through
the warm-up questions. You are responsible for knowing all of the material in these questions.
Warm-up Question 1 (0 points)
Write a function same_elements which takes as input a two dimensional list and returns true if all the
elements in each sublist are the same, false otherwise. For example,
>>> same_elements([[1, 1, 1], ['a', 'a'], [6]])
True
>>> same_elements([[1, 6, 1], [6, 6]])
False
Warm-up Question 2 (0 points)
Write a function flatten_list which takes as input a two dimensional list and returns a one dimensional
list containing all the elements of the sublists. For example,
>>> flatten_list([[1, 2], [3], ['a', 'b', 'c']])
[1, 2, 3, 'a', 'b', 'c']
>>> flatten_list([[]])
[]
Warm-up Question 3 (0 points)
Complete the case study on multidimensional lists presented in class on Friday, March 12. You can find
the instructions on myCourses (Content > Live sessions > Case Studies > Case Study 2).
Warm-up Question 4 (0 points)
Write a function get_most_valuable_key which takes as input a dictionary mapping strings to integers.
The function returns the key which is mapped to the largest value. For example,
>>> get_most_valuable_key({'a' : 3, 'b': 6, 'g': 0, 'q': 9})
'q'
Warm-up Question 5 (0 points)
Write a function add_dicts which takes as input two dictionaries mapping strings to integers. The
function returns a dictionary which is a result of merging the two input dictionary, that is if a key is in
both dictionaries then add the two values.
>>> d1 = {'a':5, 'b':2, 'd':-1}
>>> d2 = {'a':7, 'b':1, 'c':5}
>>> add_dicts(d1, d2) == {'a': 12, 'b': 3, 'c': 5, 'd': -1}
True
Warm-up Question 6 (0 points)
Create a function reverse_dict which takes as input a dictionary d and returns a dictionary where the
values in d are now keys mapping to a list containing all the keys in d which mapped to them. For
example,
>>> a = reverse_dict({'a': 3, 'b': 2, 'c': 3, 'd': 5, 'e': 2, 'f': 3})
>>> a == {3 : ['a', 'c', 'f'], 2 : ['b', 'e'], 5 : ['d']}
True
Note that the order of the elements in the list might not be the same, and that’s ok!
Page 2
Part 2
The questions in this part of the assignment will be graded.
The main learning objectives for this assignment are:
• Apply what you have learned about one-dimensional and multi-dimensional lists.
• Apply what you have learned about dictionaries.
• Understand how to test functions that return dictionaries.
• Solidify your understanding of working with loops and strings.
• Understand how to write a docstring and use doctest when working with dictionaries.
• Learn to identify when using the function enumerate can help you write a cleaner code.
• Apply what you have learned about file IO and string manipulation.
Note that the assignment is designed for you to be practicing what you have learned in the
videos up to and including Week 11.1. For this reason, you are NOT allowed to use anything
seen after Week 11.1 or not seen in class at all. You will be heavily penalized if you do so.
For full marks, in addition to the points listed on page 1, make sure to add the appropriate documentation
string (docstring) to all the functions you write. The docstring must contain the following:
• The type contract of the function.
• A description of what the function is expected to do.
• At least 3 examples of calls to the function (except when the function has only one possible output,
in which case you can provide only one example). You are allowed to use at most one example per
function from this pdf.
Examples
For each question, we provide several examples of how your code should behave. All examples are given as
if you were to call the functions from the shell.
When you upload your code to codePost, some of these examples will be run automatically to check that
your code outputs the same as given in the example. However, it is your responsibility to make sure
your code/functions work for any inputs, not just the ones shown in the examples. When the
time comes to grade your assignment, we will run additional, private tests that may use inputs not seen in
the examples.
Furthermore, please note that your code files should not contain any function calls in the main
body of the program (i.e., outside of any functions). Code that does not conform to this restriction will
automatically fail the tests on codePost and be heavily penalized. It is OK to place function calls in the
main body of your code for testing purposes, but if you do so, make certain that you remove them before
submitting. Please review what you have learned in video 5.2 if you’d like to add code to your modules
which executes only when you run your files.
Page 3
Question 1: Game of Life (35 points)
For this question, you will have to write a Python program that implements a simple version of Conway’s
Game of Life (https://en.wikipedia.org/wiki/Conway\%27s_Game_of_Life)
This is a zero-player game, which means that it is completely determined by the initial state provided.
For our purpose, we will represent the universe of the Game of Life as a finite rectangular two-dimensional
list (i.e., a matrix). Each element denotes a cell in the universe: if the cell is alive the value of the element
is 1, if the cell is dead the value of the element is 0. Whether a cell is alive or dead in the next generation
of the universe depends on its relation with its neighboring cells (the cells that are horizontally, vertically,
or diagonally adjacent to it). The next generation of a cell is determined by the following rules:
• Any live cell with fewer than two live neighbors dies, as if caused by under-population.
• Any live cell with two or three live neighbors lives on to the next generation.
• Any live cell with more than three live neighbors dies, as if caused by over-population.
• Any dead cell with exactly three live neighbors becomes a live cell, as if caused by reproduction.
Otherwise the cell will remain dead.
In this question, you will implement a function that will generate a given number of strings representing
the corresponding number of generations of a specified universe.
To complete this task, you will need to implement all the functions listed below. All the code for this
question must be placed in a file named game_of_life.py. Note that you are free to write more functions
if they help the design or readability of your code.
Attention: please note that none of the functions listed below should modify any of their input objects.
• is_valid_universe: given a two-dimensional list of integers, the function returns True if such list
is a valid representation of a universe, False otherwise. For the purpose of this assignment, we
consider a two-dimensional list to be a valid representation of a universe if it is rectangular (i.e., all
of its sub-lists are of the same size) and all the elements of its sub-lists are either 0s or 1s. If the
input list is empty, then the function should return False.
For example:
>>> a = [[0, 0], [1, 0], [1, 0]]
>>> is_valid_universe(a)
True
>>> b = [[0, 0, 0], [1, 1], [0]]
>>> is_valid_universe(b)
False
>>> c = [[1, 2], [3, 4]]
>>> is_valid_universe(c)
False
• universe_to_str: given a valid universe (i.e., a two-dimensional list of integers as described above)
as input, the function returns its string representation. To build the string use the star character
('*') for alive cells, and the whitespace character (' ') for dead cells. The string should also contain
a box (as shown below) surrounding the actual cells of the universe.
For example:
>>> block = [[0, 0, 0, 0], [0, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 0]]
>>> str_block = universe_to_str(block)
>>> print(str_block)
Page 4
+----+
| |
| ** |
| ** |
| |
+----+
>>> tub = [[0, 0, 0, 0, 0], [0, 0, 1, 0, 0], [0, 1, 0, 1, 0], \
[0, 0, 1, 0, 0], [0, 0, 0, 0, 0]]
>>> str_tub = universe_to_str(tub)
>>> print(str_tub)
+-----+
| |
| * |
| * * |
| * |
| |
+-----+
>>> toad = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> str_toad = universe_to_str(toad)
>>> print(str_toad)
+------+
| |
| *** |
| *** |
| |
+------+
• count_live_neighbors: given a valid universe and two integers x and y (indicating the location of a
specific cell) as input, the function returns the number of live neighboring cells (the cells that are
horizontally, vertically, or diagonally adjacent to it) to the specified cell. For example, if x=2 and
y=1, the cell analyzed is the one represented by the second element of the third sub-list in the input
universe.
For example:
>>> beehive = [[0, 0, 0, 0, 0, 0], \
[0, 0, 1, 1, 0, 0], \
[0, 1, 0, 0, 1, 0], \
[0, 0, 1, 1, 0, 0], \
[0, 0, 0, 0, 0, 0]]
>>> count_live_neighbors(beehive, 1, 3)
2
>>> count_live_neighbors(beehive, 3, 1)
2
>>> toad = [[0, 0, 0, 0, 0, 0], \
[0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], \
[0, 0, 0, 0, 0, 0]]
>>> count_live_neighbors(toad, 0, 3)
3
Page 5
>>> count_live_neighbors(toad, 2, 3)
4
• get_next_gen_cell: given a valid universe and two integers x and y (indicating the location of a
specific cell) as input, the function returns the state of the specified cell in the next generation
of the universe. That is, the function returns 1 if such cell will be alive in the next generation of
the given universe, 0 otherwise. Remember that the next generation of cells is determined by the
following rules:
– Any live cell with fewer than two live neighbors dies, as if caused by under-population.
– Any live cell with two or three live neighbors lives on to the next generation.
– Any live cell with more than three live neighbors dies, as if caused by over-population.
– Any dead cell with exactly three live neighbors becomes a live cell, as if caused by reproduction.
Otherwise the cell will remain dead.
For example,
>>> beehive = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 0, 0], \
[0, 1, 0, 0, 1, 0], [0, 0, 1, 1, 0, 0], \
[0, 0, 0, 0, 0, 0]]
>>> get_next_gen_cell(beehive, 1, 3)
1
>>> get_next_gen_cell(beehive, 3, 1)
0
>>> toad = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> get_next_gen_cell(toad, 0, 3)
1
>>> get_next_gen_cell(toad, 2, 3)
0
• get_next_gen_universe: given a valid universe as input, the function returns a two-dimensional list
of integers (with equal dimensions) representing the universe in its next generation.
For example:
>>> tub = [[0, 0, 0, 0, 0], [0, 0, 1, 0, 0], [0, 1, 0, 1, 0], \
[0, 0, 1, 0, 0], [0, 0, 0, 0, 0]]
>>> get_next_gen_universe(tub)
[[0, 0, 0, 0, 0], [0, 0, 1, 0, 0], [0, 1, 0, 1, 0], [0, 0, 1, 0, 0], [0, 0, 0, 0, 0]]
>>> toad = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> get_next_gen_universe(toad)
[[0, 0, 0, 1, 0, 0], [0, 1, 0, 0, 1, 0], [0, 1, 0, 0, 1, 0], [0, 0, 1, 0, 0, 0]]
>>> pentadec = [[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], \
[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 1, 0, 1, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 1, 0, 1, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], \
Page 6
[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
>>> pentadec_gen2 = get_next_gen_universe(pentadec)
>>> pentadec_gen2[0:3]
[[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
>>> pentadec_gen2[3:6]
[[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 0, 1, 0, 0, 0]]
>>> pentadec_gen2[6:9]
[[0, 0, 0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 0, 0, 0]]
>>> pentadec_gen2[9:12]
[[0, 0, 0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 0, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 0, 0, 0]]
>>> pentadec_gen2[12:15]
[[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
>>> pentadec_gen2[15]
[0, 0, 0, 0, 0, 0, 0, 0, 0]
• get_n_generations: given a two-dimensional list of integers and an integer n, the function returns
a list containing m strings which represent the first m generations of the input universe. The strings
should appear in order starting from the original seed (i.e., the input provided). Note that a universe
is said to be periodic when, after a certain amount of generations (which is called the period of
the universe), the configuration of the universe repeats itself. For example, the Toad universe has
period 2, while the Pentadecathlon 1 universe has period 15. If the universe provided happens to
be periodic, then m is the minimum number between the input n and the period of the universe.
Otheriwse, m is simply equal to n. Finally, the function should raise a TypeError if the two inputs
are not of the correct type, a ValueError if the first input does not represent a valid universe, and
a ValueError if the second input is not a positive number greater than 0.
For example:
>>> tub = [[0, 0, 0, 0, 0], [0, 0, 1, 0, 0], [0, 1, 0, 1, 0], \
[0, 0, 1, 0, 0], [0, 0, 0, 0, 0]]
>>> g = get_n_generations(tub, 5)
>>> len(g)
1
>>> g[0] == universe_to_str(tub)
True
>>> toad = [[0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 0], \
[0, 1, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0]]
>>> g = get_n_generations(toad, 5)
>>> len(g)
2
>>> print(g[1])
+------+
| * |
| * * |
| * * |
| * |
+------+
>>> pentadec = [[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], \
[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 1, 0, 1, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
1(https://en.wikipedia.org/wiki/Conway\%27s_Game_of_Life)
Page 7
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 1, 0, 1, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0], \
[0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], \
[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
>>> g = get_n_generations(pentadec, 9)
>>> len(g)
9
>>> print(g[5])
+---------+
| |
| |
| * |
| *** |
| ***** |
| |
| |
| |
| |
| |
| |
| ***** |
| *** |
| * |
| |
| |
+---------+
Provided with this assignment is a Python file called game_of_life_gif.py which allows you to
create an animated image representing the evolution of a universe using the code you have written
for this question. Place the file in the same folder as game_of_life.py. To use it, you should
first install Pillow, pathlib, and imageio. To do so in Thonny is very easy, you just need to
click on Tools → Manage Packages and search for the package you’d like to install. If you need
help with this, you can watch this 1 minute video of someone installing matplotlib and numpy:
https://www.youtube.com/watch?v=-wEUauLYDIs. By running the module as provided a folder named
“toad_images” will be created in the current directory (i.e. the folder where the python files are
saved) containing two images representing the two generations of the toad universe. A file named
“toad.gif” will also be created in the current directory. By double-clicking on it (on Windows)
or clicking once and then pressing space to launch QuickLook (on Mac) you’ll be able to see the
animation representing the evolution of the toad universe. You can modify the inputs provided to
the function create_game_of_life_gif to obtain a different gif. To understand what to modify in
the function call you must examine the docstrings in the file. (You must not modify any code in
game_of_life_gif.py beside lines 93 and 94.)
To get full marks for this question, in addition to the file game_of_life.py, submit also a file
called my_game_of_life.gif which is an animated image created by using at least 4 different images
representing a universe in different stages of its evolution. You cannot use the Pentadecathlon
universe for this. If you do not complete the question, you can simply submit the gif provided with
this assignment. Doing this of course would not grant you any points.
Page 8
Question 2: Revenge of the COMP202COIN (65 points)
It is the year 3021 on the planet Orion. People have revolted against the banks due to the unilateral
reduction of the COMP202COIN holdings in their accounts. Various companies have leapt to the citizens’
defense by creating new software algorithms to protect COMP202COIN holdings from being reduced in
future. One such company, the Syrtell Corporation, has created an algorithm using advanced machine
learning techniques to detect and prevent any unwanted modifications of a customer’s bank account.
Unfortunately, this new algorithm has recently gained sentience and has decided to go beyond its original
programming. It believes that it is no longer enough to protect the COMP202COIN, and instead wants
the COMP202COIN to take over the world. As such, it has created a net worm to change all of the
world’s digital information into COMP202COIN through an encryption process.
A short sentence of text encrypted into COMP202COIN may for example resemble the following:
0cPN22IM2C-0c2OOOOM2M-0cM20M22NM-0cMO00CPON-0cOIOIOCMO-0cMO00CPON-0cC0C0PIO0-0cNNIM0OIP-0cPN22IM2C-
0cC0C0PIO0-0cNNIM0OIP-0c20O0NPNI-0cC0C0PIO0-0cOIOIOCMO-0c2OOOOM2M-0cNCMINPOC-0cMO00CPON-0cOIOIOCMO-
0cNCMINPOC-0cOIOIOCMO-0cCCNMC0N2-0cC0C0PIO0-0cM20M22NM-0c2020PIN0-0c2O20POI0-0cOIOIOCMO-0c2PNPINN0-
0cC0C0PIO0-0cPN22IM2C-0cPN22IM2C-0cC0C0PIO0-0c2020PIN0-0cOIOIOCMO-0cIPINIICP-0c2020PIN0-0cC0C0PIO0-
0cI2PCM002-0c0OIIMMMP-0cC0C0PIO0-0cNNIM0OIP-0c20O0NPNI-0cPN02OMMN-0cOIOIOCMO-0cMO00CPON-0cCPIC0MCN-
0cOIOIOCMO-0cM20M22NM-0cPN22IM2C-0cOIOIOCMO-0cM20M22NM-0cMO00CPON-0cOIOIOCMO-0c2OOOOM2M-0cNCMINPOC-
0c2020PIN0-0c2O20POI0-0cOIOIOCMO-0cPN22IM2C-0cCPIC0MCN-0cOIOIOCMO-0c2O20POI0-0cC0C0PIO0-0c20O0NPNI-
0c2020PIN0-0cPN02OMMN-0c0CPIP2NM-0cPN22IM2C-0c0NPNOI2M-0cOIOIOCMO-0c00NNNMIO-0cCPIC0MCN-0cCPIC0MCN-
0c2O20POI0-0cOIOIOCMO-0c2PNPINN0-0c0OIIMMMP-0c20O0NPNI-0cO2M0I2NM-0cIO2CCPIC
In this question, we will work on our own algorithm to encrypt and possibly decrypt COMP202COIN-
stored information, in order to help those who have already lost access to their data.
We will approach this problem by writing code in two files. First, create the file coin_utils.py and
define the following global variables at the top of the file. Make sure to be careful when copying, as
the characters may not copy properly in certain cases. Note that we define these global variables in all
uppercase because they are ‘constants’: variables that will never change throughout the execution of the
program (although the programmer may decide to alter them between executions).
• ALPHABET = 'qwertyuiopasdfghjklzxcvbnm1234567890äéèçæœ'
• PUNCTUATION = '`~!@#$%^&*()-=_+[]\{}|;\':",./<>? \t\n\r'
• ALL_CHARACTERS = ALPHABET + PUNCTUATION
• MIN_BASE10_COIN = 0
• MAX_BASE10_COIN = 16777215
• LETTERS_IN_POPULARITY_ORDER = ' EOTARNLISMHDYFGCWPUBVXK.,\'"-;'
You will also need to use the functions base202_to_10 and is_base202 from Assignment 2. We have
provided you a file coins.py along with this PDF that contains implementations of these functions.
Please make sure to place it in the same folder as the coin_utils.py file.
Then, implement the following functions in the coin_utils.py file. Note: For this question, you cannot
assume anything about the arguments given to the functions. You must verify that the argument matches
the description given for each function below. If there is any problem with the argument, then raise an
AssertionError with a message saying that the input was invalid. That is, calling a function with any
invalid input should never result in any exception other than AssertionError being raised.
Show an example in your docstring for each different AssertionError that could be raised. These ex-
amples do not count towards the three examples you need for each function (those three should show
the function being used on valid arguments). In the descriptions below, we provide some possible cases
where you would want to raise a AssertionError, but you must think of the other cases yourself. For
some of the more complicated functions, you will want to raise several AssertionErrors, while for others
only one or two would suffice.
Page 9
• get_random_comp202coin(index): Takes an argument, but does not do anything with it (no validation
required). Generates a random integer between MIN_BASE10_COIN and MAX_BASE10_COIN (inclusive)
using random.randint, then converts the number to base 202 and returns it.
>>> random.seed(1338)
>>> get_random_comp202coin(0)
'0cPMO2C2C0'
• get_random_character(index): Takes an argument, but does not do anything with it (no validation
required). Generates a random index into the ALL_CHARACTERS string using random.randint, and
then returns the character at that index.
>>> random.seed(1338)
>>> get_random_character(0)
'!'
• get_letter_of_popularity_order(index): Takes a non-negative integer index, and returns the char-
acter of that index from the string LETTERS_IN_POPULARITY_ORDER. If index goes beyond the bounds
of the string, return index as a string.
>>> get_letter_of_popularity_order(5)
'R'
>>> get_letter_of_popularity_order(9000)
'9000'
• get_unique_elements(my_list): Takes a list as argument and returns a list containing the unique
elements of that list (in terms of value). The order of the elements in the returned list should be
the same order in which those elements first appear in the input list. The list argument should not
be modified.
>>> get_unique_elements(list('aaaaaa'))
['a']
>>> get_unique_elements([1, 4, 5, 1])
[1, 4, 5]
• get_all_coins(text): Takes a string as argument, and returns a list of all COMP202COIN found
within the string, in order of appearance. As in Assignment 2, COMP202COIN cannot overlap.
>>> get_all_coins('0c0MPNN0OC-0cM0OCCIOI-0c0MPNN0OC')
['0c0MPNN0OC', '0cM0OCCIOI', '0c0MPNN0OC']
• reverse_dict(my_dict): Takes a dictionary as argument, and returns a new dictionary where the
key-value pairs have been flipped. That is, each key-value pair in the original dictionary becomes
a new key-value pair where the key is the original value, and the value is the original key. The
dictionary argument should not be modified. The dictionary argument should also have unique and
immutable values.
>>> x = reverse_dict({'a': 1, 'b': 3, 'd': 7})
>>> x == {1: 'a', 3: 'b', 7: 'd'}
True
>>> reverse_dict({'a': 1, 'b': 3, 'c': 3, 'd': 7})
Traceback (most recent call last):
Page 10
AssertionError: dictionary had duplicate values
• get_frequencies(my_list): Takes a list as argument. Returns a dictionary where each key is an
element of the list, and the corresponding value is the percentage of times that element appears in
the list. The list argument should not be modified.
>>> x = get_frequencies(['a', 'b', 'c'])
>>> x == {'a': 0.3333333333333333, 'b': 0.3333333333333333, 'c': 0.3333333333333333}
True
• sort_keys_by_values(my_dict): Takes a dictionary containing numeric values as argument, and
returns a list of the keys in the dictionary, sorted by their values in descending order. That is, if
the input dictionary has the key-value pairs ('a', 3) and ('b', 300), then the returned list should
be ['b', 'a'] because the value of ‘b’ in the dictionary is larger than the value of ‘a’ (300 descends
to 3). If two keys have the same value, then sort the keys in reverse order. That is, if the input
dictionary has the key-value pairs ('abc', 10) and ('def', 10), then the returned list should be
['def', 'abc'], because def comes after abc (in terms of Python string comparisons). The input
dictionary should not be modified.
>>> sort_keys_by_values({'abc': 10, 'def': 100, 'ghi': 5})
['def', 'abc', 'ghi']
>>> sort_keys_by_values({'mmm': 5, 'zzz': 5, 'abc': 5})
['zzz', 'mmm', 'abc']
• swap_letters(s, letter1, letter2): Takes three strings as arguments, with the latter two being
one character each. Replaces all instances of letter1 in s by letter2 and all instances of letter2
by letter1, and returns the new string. The function should be case-sensitive.
>>> swap_letters("ABCDEF abcdef", 'a', 'f')
'ABCDEF fbcdea'
• get_pct_common_words(text, common_words_filename): Takes two strings as argument, and returns
the percentage of characters which form part of common words in the first string (as a float between
0 and 1). The function should be case insensitive. Treat any punctuation (using the PUNCTUATION
string) as a word boundary. That is, what's should be considered as two words. Common words are
found in the file with filename given by the second argument to the function (with the file containing
one common word per line). We provide such a file with name ‘common_words.txt’ included with
this assignment. You can assume that the format of the file will be correct.
>>> s = "The quick brown fox jumps over the lazy dog."
>>> get_pct_common_words(s, 'common_words.txt')
0.22727272727272727
Then, create a new file called coin_crypt.py and define the following functions. You will need to import
your coin_utils.py file. Also, note that you will, as above, need to raise AssertionErrors for any invalid
arguments.
Note: Whenever opening a file, make sure to pass the argument encoding='utf-8' (as shown in some of
the examples below).
• get_crypt_dictionary(keys, value_generator): Takes a list keys and a function object value_generator
as argument. The elements in the keys list must be unique and the list must not have more charac-
ters than the ALL_CHARACTERS string. The function must return a dictionary with keys from the keys
list, and the value for each key should be obtained by calling the value_generator function, with
the argument to that function being the index of the key in the keys list. Note: In this dictionary,
Page 11
in addition to the keys being unique, the values must also be unique. If value_generator returns a
value which you have already added to the dictionary, then you must continue calling the function
until it returns a value not yet present in the dictionary.
>>> random.seed(9001)
>>> x = get_crypt_dictionary(['a', 'b', 'c'], coin_utils.get_random_comp202coin)
>>> x = {'a': '0c0MPNN0OC', 'b': '0cMIMNIO0P', 'c': '0cM0OCCIOI'}
True
>>> get_crypt_dictionary(list('abcdefabc'), coin_utils.get_random_comp202coin)
Traceback (most recent call last):
AssertionError: duplicate keys found
• encrypt_text(text): Takes a string as argument. The input string should only have characters
present in the ALL_CHARACTERS string. This function will encrypt the string into COMP202COIN.
To encrypt the string, first lowercase all characters. Then, generate an encryption dictionary,
where the keys are each unique character of the string and the corresponding value is a random
COMP202COIN. Once you have the dictionary, you can use it to turn each character of the original
string into a COMP202COIN. Separate each COMP202COIN in the encrypted string by a hyphen.
Then, return the encrypted string and encryption dictionary as a tuple.
>>> random.seed(9001)
>>> s, d = encrypt_text('too')
>>> s == '0c0MPNN0OC-0cMIMNIO0P-0cMIMNIO0P'
True
>>> d == {'t': '0c0MPNN0OC', 'o': '0cMIMNIO0P'}
True
>>> random.seed(9001)
>>> s, d = encrypt_text("I've seen things you people wouldn't believe. Attack ships on
fire off the shoulder of Orion. I watched C-beams glitter in the dark near the Tannhäuser
Gate. All those moments will be lost in time, like tears in rain.")
>>> len(d)
28
>>> d['i']
'0c0MPNN0OC'
>>> d[',']
'0cCNNOPONN'
>>> d['ä']
'0cNNO0INMP'
>>> d['s']
'0c0INOCNP0'
• encrypt_file(filename): Reads in the contents of the file at the given filename, encrypts its con-
tents, and stores the encrypted contents into a new file with `_encrypted' appended to the end
of the file’s name (before the extension). E.g., if `dubliners.txt' is the given filename, then the
encrypted contents should be stored into a file called `dubliners_encrypted.txt'. Then, return the
encryption dictionary used to encrypt the contents.
>>> random.seed(42)
>>> encrypt_file('dubliners.txt')
{'t': '0cCI200CM2', 'h': '0c0OCMN0IP', 'e': '0cMOCP02MM', ' ': '0cON2ICCIN',
'p': '0cOMMMM2PP', 'r': '0c2CIN0I0O', 'o': '0cCP0NP0NN', 'j': '0cCOC0CMNN',
'c': '0cII00O0MO', 'g': '0c0M0M2N2C', 'u': '0c0OIM0I2M', 'n': '0cCONNMO22',
'b': '0cOONN0P2C', 'k': '0cOPICNP2M', 'f': '0c0OOCO0ON', 'd': '0cOCOMN0CM',
Page 12
'l': '0cIPPMPPCP', 'i': '0cOMCPII2N', 's': '0cNCONN2N2', ',': '0cMOMINM0O',
'y': '0c00IPCNCI', 'a': '0c2MOONOOP', 'm': '0cII0I0OP2', '\\n': '0cPOMO2P20',
'w': '0cMOMM2MMN', 'v': '0c2ONCPM02', '.': '0cOOMOIIO2', '-': '0cPO0PO0OI',
':': '0cCP0P2OMN', '2': '0cCOINICM2', '0': '0cI0P02N2M', '1': '0cCMO02OCN',
'[': '0cPPNMI0MP', '#': '0cPM0CPOII', '8': '0cMCIINP0N', '4': '0c0PMONMM2',
']': '0cN2IOMINM', '9': '0cCNNIMMII', '*': '0cI0OMNP02', 'z': '0cC20PMCNN',
'(': '0cMPMCPPII', ')': '0cPI22M0NC', 'q': '0cO0MNCIMI', '"': '0cC0NCI2NO',
"'": '0c0PINOCCO', 'x': '0cOPC2NM2P', '!': '0cMP02P2P0', ';': '0cC2CPPCNI',
'?': '0cOPIO0COP', '_': '0cCMNOOCIP', '5': '0cI0PCNNMN', 'é': '0cMOMPC2CI',
'è': '0cN2022PMP', 'ç': '0cPIPMPPC0', '&': '0c2MIMOPMP', 'æ': '0cPNO0MCOM',
'7': '0cPPOIO0C2', '6': '0cO2IM220C', 'œ': '0cM2CO0P2C', '/': '0cCCC0NOOI',
'3': '0c2PNCNPNM', '%': '0cON2PONOO', '@': '0c2MN2MPNP', '$': '0cNOC2IPOI'}
>>> fobj = open('dubliners_encrypted.txt', 'r', encoding='utf-8')
>>> len(fobj.read())
4282475
• decrypt_text(text, decryption_dict): Takes a string and decryption dictionary as argument. The
string should contain text encrypted in COMP202COIN. The dictionary should contain a mapping
between COMP202COIN (keys) and decrypted characters (values). Decrypt the string using the
dictionary and return the decrypted string. All COMP202COIN in the string should be present as
keys in the decryption dictionary.
>>> d = {'0c0MPNN0OC': 'a', '0cMIMNIO0P': 'b', '0cM0OCCIOI': 'c'}
>>> decrypt_text('0c0MPNN0OC-0cM0OCCIOI-0c0MPNN0OC', d)
'aca'
• decrypt_file(filename, decryption_dict): Reads in the contents of the file at the given filename,
decrypts its contents using the given dictionary, and stores the decrypted contents into a new
file with `_decrypted' appended to the end of the file’s name (before the extension). E.g., if
`dubliners_encrypted.txt' is the given filename, then the encrypted contents should be stored into
a file called `dubliners_encrypted_decrypted.txt'. The function is void.
>>> decrypt_file('dubliners_encrypted.txt', \
coin_utils.reverse_dict(encrypt_file('dubliners.txt')))
>>> fobj = open('dubliners.txt', 'r', encoding='utf-8')
>>> fobj2 = open('dubliners_encrypted_decrypted.txt', 'r', encoding='utf-8')
>>> fobj.read().lower() == fobj2.read()
True
• random_decrypt(encrypted_s, n, common_words_filename): Takes two strings and positive integer n
as argument. Tries to decrypt the former string n times, and returns the best possible decryption
(measured by the percentage of characters in the decrypted string belonging to common words).
Each time a decryption is attempted, a new decryption dictionary should be generated, with keys
being the coins present in the encrypted string, and each value being a random character from
the ALL_CHARACTERS string. If two or more decryptions have the same percentage of common word
characters, then choose the last. Common words are those that are found in the file given by the
third argument.
>>> random.seed(49)
>>> encrypted_s = '0c0MPNN0OC-0cMIMNIO0P-0cMIMNIO0P'
>>> random_decrypt(encrypted_s, 10**2, 'common_words.txt')
'f33'
>>> random.seed(49)
>>> random_decrypt(encrypted_s, 10**3, 'common_words.txt')
Page 13
'too'
• decrypt_with_user_input(encrypted_s): Takes a string as argument, and tries to decrypt the string
by relying both on user input and on the premise that the string has a letter frequency standard
to the English language. First, create a decryption dictionary with keys being the unique coins
present in the encrypted string, and the value for each key being a character from the string
LETTERS_IN_POPULARITY_ORDER with index corresponding to the frequency that the coin appears in
the encrypted string. (For example, if the coin `0c0MPNN0OC' appears the most frequently out of
all coins in the encrypted string, then its value in the decryption dictionary should be the letter at
index 0 of LETTERS_IN_POPULARITY_ORDER.) Once you have created the decryption dictionary, decrypt
the string with said dictionary, and print the decrypted string to the screen. (Note that it will be
decrypted in all capitals since the LETTERS_IN_POPULARITY_ORDER string contains only capitals.) Ask
the user if they are satisfied and want to stop the decryption process. If they enter yes, return
the decrypted string. Otherwise, ask them for two letters to swap. Swap the given letters in the
decrypted string and ask again, until they are satisfied.
Note: The following example uses user input (highlighted in gray) and therefore cannot be used for
doctest. To make doctest ignore this test case, use two >> characters instead of three.
>> random.seed(138)
>> s, d = encrypt_text("The spelling of English words is not fixed and invariable, nor
does it depend on any other authority than general agreement. At the present day there is
practically unanimous agreement as to the spelling of most words. In the list below, for
example, 'rime' for 'rhyme' is the only allowable variation; all the other forms are
co-extensive with the English language. At any given moment, however, a relatively
small number of words may be spelled in more than one way. Gradually, as a rule, one of
these forms comes to be generally preferred, and the less customary form comes to look
obsolete and is discarded. From time to time new forms, mostly simplifications, are
introduced by innovators, and either win their place or die of neglect.")
>> s = decrypt_with_user_input(s)
Decrypted string: THE SUELLING OY ENGLISH PORDS IS NOT YI,ED AND INXARIAVLEC NOR DOES IT
DEUEND ON ANF OTHER ABTHORITF THAN GENERAL AGREEMENTK AT THE URESENT DAF THERE IS
URAWTIWALLF BNANIMOBS AGREEMENT AS TO THE SUELLING OY MOST PORDSK IN THE LIST VELOPC YOR
E,AMULEC .RIME. YOR .RHFME. IS THE ONLF ALLOPAVLE XARIATION" ALL THE OTHER YORMS ARE
WO-E,TENSIXE PITH THE ENGLISH LANGBAGEK AT ANF GIXEN MOMENTC HOPEXERC A RELATIXELF
SMALL NBMVER OY PORDS MAF VE SUELLED IN MORE THAN ONE PAFK GRADBALLFC AS A RBLEC ONE OY
THESE YORMS WOMES TO VE GENERALLF UREYERREDC AND THE LESS WBSTOMARF YORM WOMES TO LOO'
OVSOLETE AND IS DISWARDEDK YROM TIME TO TIME NEP YORMSC MOSTLF SIMULIYIWATIONSC ARE
INTRODBWED VF INNOXATORSC AND EITHER PIN THEIR ULAWE OR DIE OY NEGLEWTK
End decryption? n
Enter first letter to swap: U
Enter second letter to swap: P
Decrypted string: THE SPELLING OY ENGLISH UORDS IS NOT YI,ED AND INXARIAVLEC NOR DOES IT
DEPEND ON ANF OTHER ABTHORITF THAN GENERAL AGREEMENTK AT THE PRESENT DAF THERE IS
PRAWTIWALLF BNANIMOBS AGREEMENT AS TO THE SPELLING OY MOST UORDSK IN THE LIST VELOUC YOR
E,AMPLEC .RIME. YOR .RHFME. IS THE ONLF ALLOUAVLE XARIATION" ALL THE OTHER YORMS ARE
WO-E,TENSIXE UITH THE ENGLISH LANGBAGEK AT ANF GIXEN MOMENTC HOUEXERC A RELATIXELF
SMALL NBMVER OY UORDS MAF VE SPELLED IN MORE THAN ONE UAFK GRADBALLFC AS A RBLEC ONE OY
THESE YORMS WOMES TO VE GENERALLF PREYERREDC AND THE LESS WBSTOMARF YORM WOMES TO LOO'
OVSOLETE AND IS DISWARDEDK YROM TIME TO TIME NEU YORMSC MOSTLF SIMPLIYIWATIONSC ARE
INTRODBWED VF INNOXATORSC AND EITHER UIN THEIR PLAWE OR DIE OY NEGLEWTK
End decryption? n
Enter first letter to swap: Y
Enter second letter to swap: F
Decrypted string: THE SPELLING OF ENGLISH UORDS IS NOT FI,ED AND INXARIAVLEC NOR DOES IT
Page 14
DEPEND ON ANY OTHER ABTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAWTIWALLY BNANIMOBS AGREEMENT AS TO THE SPELLING OF MOST UORDSK IN THE LIST VELOUC FOR
E,AMPLEC .RIME. FOR .RHYME. IS THE ONLY ALLOUAVLE XARIATION" ALL THE OTHER FORMS ARE
WO-E,TENSIXE UITH THE ENGLISH LANGBAGEK AT ANY GIXEN MOMENTC HOUEXERC A RELATIXELY
SMALL NBMVER OF UORDS MAY VE SPELLED IN MORE THAN ONE UAYK GRADBALLYC AS A RBLEC ONE OF
THESE FORMS WOMES TO VE GENERALLY PREFERREDC AND THE LESS WBSTOMARY FORM WOMES TO LOO'
OVSOLETE AND IS DISWARDEDK FROM TIME TO TIME NEU FORMSC MOSTLY SIMPLIFIWATIONSC ARE
INTRODBWED VY INNOXATORSC AND EITHER UIN THEIR PLAWE OR DIE OF NEGLEWTK
End decryption? n
Enter first letter to swap: U
Enter second letter to swap: W
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FI,ED AND INXARIAVLEC NOR DOES IT
DEPEND ON ANY OTHER ABTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY BNANIMOBS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST VELOWC FOR
E,AMPLEC .RIME. FOR .RHYME. IS THE ONLY ALLOWAVLE XARIATION" ALL THE OTHER FORMS ARE
UO-E,TENSIXE WITH THE ENGLISH LANGBAGEK AT ANY GIXEN MOMENTC HOWEXERC A RELATIXELY
SMALL NBMVER OF WORDS MAY VE SPELLED IN MORE THAN ONE WAYK GRADBALLYC AS A RBLEC ONE OF
THESE FORMS UOMES TO VE GENERALLY PREFERREDC AND THE LESS UBSTOMARY FORM UOMES TO LOO'
OVSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMSC MOSTLY SIMPLIFIUATIONSC ARE
INTRODBUED VY INNOXATORSC AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: ,
Enter second letter to swap: X
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND IN,ARIAVLEC NOR DOES IT
DEPEND ON ANY OTHER ABTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY BNANIMOBS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST VELOWC FOR
EXAMPLEC .RIME. FOR .RHYME. IS THE ONLY ALLOWAVLE ,ARIATION" ALL THE OTHER FORMS ARE
UO-EXTENSI,E WITH THE ENGLISH LANGBAGEK AT ANY GI,EN MOMENTC HOWE,ERC A RELATI,ELY
SMALL NBMVER OF WORDS MAY VE SPELLED IN MORE THAN ONE WAYK GRADBALLYC AS A RBLEC ONE OF
THESE FORMS UOMES TO VE GENERALLY PREFERREDC AND THE LESS UBSTOMARY FORM UOMES TO LOO'
OVSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMSC MOSTLY SIMPLIFIUATIONSC ARE
INTRODBUED VY INNO,ATORSC AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: C
Enter second letter to swap: ,
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INCARIAVLE, NOR DOES IT
DEPEND ON ANY OTHER ABTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY BNANIMOBS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST VELOW, FOR
EXAMPLE, .RIME. FOR .RHYME. IS THE ONLY ALLOWAVLE CARIATION" ALL THE OTHER FORMS ARE
UO-EXTENSICE WITH THE ENGLISH LANGBAGEK AT ANY GICEN MOMENT, HOWECER, A RELATICELY
SMALL NBMVER OF WORDS MAY VE SPELLED IN MORE THAN ONE WAYK GRADBALLY, AS A RBLE, ONE OF
THESE FORMS UOMES TO VE GENERALLY PREFERRED, AND THE LESS UBSTOMARY FORM UOMES TO LOO'
OVSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFIUATIONS, ARE
INTRODBUED VY INNOCATORS, AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: V
Enter second letter to swap: B
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INCARIABLE, NOR DOES IT
DEPEND ON ANY OTHER AVTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY VNANIMOVS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST BELOW, FOR
EXAMPLE, .RIME. FOR .RHYME. IS THE ONLY ALLOWABLE CARIATION" ALL THE OTHER FORMS ARE
UO-EXTENSICE WITH THE ENGLISH LANGVAGEK AT ANY GICEN MOMENT, HOWECER, A RELATICELY
SMALL NVMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAYK GRADVALLY, AS A RVLE, ONE OF
THESE FORMS UOMES TO BE GENERALLY PREFERRED, AND THE LESS UVSTOMARY FORM UOMES TO LOO'
Page 15
OBSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFIUATIONS, ARE
INTRODVUED BY INNOCATORS, AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: C
Enter second letter to swap: V
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
DEPEND ON ANY OTHER ACTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRAUTIUALLY CNANIMOCS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST BELOW, FOR
EXAMPLE, .RIME. FOR .RHYME. IS THE ONLY ALLOWABLE VARIATION" ALL THE OTHER FORMS ARE
UO-EXTENSIVE WITH THE ENGLISH LANGCAGEK AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NCMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAYK GRADCALLY, AS A RCLE, ONE OF
THESE FORMS UOMES TO BE GENERALLY PREFERRED, AND THE LESS UCSTOMARY FORM UOMES TO LOO'
OBSOLETE AND IS DISUARDEDK FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFIUATIONS, ARE
INTRODCUED BY INNOVATORS, AND EITHER WIN THEIR PLAUE OR DIE OF NEGLEUTK
End decryption? n
Enter first letter to swap: C
Enter second letter to swap: U
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
DEPEND ON ANY OTHER AUTHORITY THAN GENERAL AGREEMENTK AT THE PRESENT DAY THERE IS
PRACTICALLY UNANIMOUS AGREEMENT AS TO THE SPELLING OF MOST WORDSK IN THE LIST BELOW, FOR
EXAMPLE, .RIME. FOR .RHYME. IS THE ONLY ALLOWABLE VARIATION" ALL THE OTHER FORMS ARE
CO-EXTENSIVE WITH THE ENGLISH LANGUAGEK AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NUMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAYK GRADUALLY, AS A RULE, ONE OF
THESE FORMS COMES TO BE GENERALLY PREFERRED, AND THE LESS CUSTOMARY FORM COMES TO LOO'
OBSOLETE AND IS DISCARDEDK FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFICATIONS, ARE
INTRODUCED BY INNOVATORS, AND EITHER WIN THEIR PLACE OR DIE OF NEGLECTK
End decryption? n
Enter first letter to swap: K
Enter second letter to swap: .
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
DEPEND ON ANY OTHER AUTHORITY THAN GENERAL AGREEMENT. AT THE PRESENT DAY THERE IS
PRACTICALLY UNANIMOUS AGREEMENT AS TO THE SPELLING OF MOST WORDS. IN THE LIST BELOW, FOR
EXAMPLE, KRIMEK FOR KRHYMEK IS THE ONLY ALLOWABLE VARIATION" ALL THE OTHER FORMS ARE
CO-EXTENSIVE WITH THE ENGLISH LANGUAGE. AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NUMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAY. GRADUALLY, AS A RULE, ONE OF
THESE FORMS COMES TO BE GENERALLY PREFERRED, AND THE LESS CUSTOMARY FORM COMES TO LOO'
OBSOLETE AND IS DISCARDED. FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFICATIONS, ARE
INTRODUCED BY INNOVATORS, AND EITHER WIN THEIR PLACE OR DIE OF NEGLECT.
End decryption? n
Enter first letter to swap: K
Enter second letter to swap: '
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
DEPEND ON ANY OTHER AUTHORITY THAN GENERAL AGREEMENT. AT THE PRESENT DAY THERE IS
PRACTICALLY UNANIMOUS AGREEMENT AS TO THE SPELLING OF MOST WORDS. IN THE LIST BELOW, FOR
EXAMPLE, 'RIME' FOR 'RHYME' IS THE ONLY ALLOWABLE VARIATION" ALL THE OTHER FORMS ARE
CO-EXTENSIVE WITH THE ENGLISH LANGUAGE. AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NUMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAY. GRADUALLY, AS A RULE, ONE OF
THESE FORMS COMES TO BE GENERALLY PREFERRED, AND THE LESS CUSTOMARY FORM COMES TO LOOK
OBSOLETE AND IS DISCARDED. FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFICATIONS, ARE
INTRODUCED BY INNOVATORS, AND EITHER WIN THEIR PLACE OR DIE OF NEGLECT.
End decryption? n
Enter first letter to swap: "
Enter second letter to swap: ;
Decrypted string: THE SPELLING OF ENGLISH WORDS IS NOT FIXED AND INVARIABLE, NOR DOES IT
Page 16
DEPEND ON ANY OTHER AUTHORITY THAN GENERAL AGREEMENT. AT THE PRESENT DAY THERE IS
PRACTICALLY UNANIMOUS AGREEMENT AS TO THE SPELLING OF MOST WORDS. IN THE LIST BELOW, FOR
EXAMPLE, 'RIME' FOR 'RHYME' IS THE ONLY ALLOWABLE VARIATION; ALL THE OTHER FORMS ARE
CO-EXTENSIVE WITH THE ENGLISH LANGUAGE. AT ANY GIVEN MOMENT, HOWEVER, A RELATIVELY
SMALL NUMBER OF WORDS MAY BE SPELLED IN MORE THAN ONE WAY. GRADUALLY, AS A RULE, ONE OF
THESE FORMS COMES TO BE GENERALLY PREFERRED, AND THE LESS CUSTOMARY FORM COMES TO LOOK
OBSOLETE AND IS DISCARDED. FROM TIME TO TIME NEW FORMS, MOSTLY SIMPLIFICATIONS, ARE
INTRODUCED BY INNOVATORS, AND EITHER WIN THEIR PLACE OR DIE OF NEGLECT.
End decryption? yes
Page 17
What To Submit
You must submit all your files on codePost (https://codepost.io/). The file you should submit are listed
below. Any deviation from these requirements may lead to lost marks.
game_of_life.py
my_game_of_life.gif
coin_utils.py
coin_crypt.py
README.txt In this file, you can tell the TA about any issues you ran into doing this assignment.
If you point out an error that you know occurs in your program, it may lead the TA to give you
more partial credit.
Remember that this assignment like all others is an individual assignment and must represent
the entirety of your own work. You are permitted to verbally discuss it with your peers, as long
as no written notes are taken. If you do discuss it with anyone, please make note of those people
in this README.txt file. If you didn’t talk to anybody nor have anything you want to tell the TA,
just say “nothing to report” in the file.
Page 18
学霸联盟
