ECE106 Lecture Notes - Lecture 39: Atomic Orbital, Electric Field, Electric Potential

1.Write a function encipher(s, n) that takes as inputs an arbitrary string s and a non-negative integer n between 0 and 25, and that returns a new string in which the letters in s have been "rotated" by n characters forward in the alphabet, wrapping around as needed. For example:

>>> encipher('hello', 1) result: 'ifmmp'

>>> encipher('hello', 2) result: 'jgnnq'

>>> encipher('hello', 4) result: 'lipps'

Upper-case letters should be "rotated" to upper-case letters, even if you need to wrap around. For example:

>>> encipher('XYZ', 3)

result: 'ABC'

Lower-case letters should be "rotated" to lower-case letters:

>>> encipher('xyz', 3)

result: 'abc'

Non-alphabetic characters should be left unchanged:

>>> encipher('#caesar!', 2)

result: '#ecguct!'

Hints/reminders:

You can use the built-in functions ord and chr convert from single-character strings to integers and back:

>>> ord('a')

result: 97

>>> chr(97)

result: 'a'

You can use the following test to determine if a character is between 'a' and 'z' in the alphabet:

if 'a' <= c <= 'z':

A similar test will work for upper-case letters.

We recommend writing a helper function rot(c, n) that rotates a single character c forward by n spots in the alphabet. We have given you a template for this helper function in ps3pr3.py that checks to ensure that c is a single-character string. We wrote rot13(c) in lecture; rot(c, n) will be very close to rot13(c)! You can test your rot(c, n) as follows:

>>> rot('a', 1)

result: 'b'

>>> rot('y', 2)

result: 'a'

>>> rot('A', 3)

result: 'D'

>>> rot('Y', 3)

result: 'B'

>>> rot('!', 4)

result: '!'

Once you have rot(c, n), you can write a recursive encipher function.

Once you think you have everything working, here are three more examples to try:

>>> encipher('xyza', 1)

result: 'yzab'

>>> encipher('Z A', 2)

result: 'B C'

>>> encipher('Caesar cipher? I prefer Caesar salad.', 25)

result: 'Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.'

2.Write a function decipher(s) that takes as input an arbitrary string s that has already been enciphered by having its characters "rotated" by some amount (possibly 0). decipher should return, to the best of its ability, the original English string, which will be some rotation (possibly 0) of the input string s. For example:

>>> decipher('Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.')

result: 'Caesar cipher? I prefer Caesar salad.'

Here are two more examples:

>>> decipher('Hu lkbjhapvu pz doha ylthpuz hmaly dl mvynla lclyfaopun dl ohcl slhyulk.')

result: 'An education is what remains after we forget everything we have learned.'

>>> decipher('python')

result: 'eniwdc'

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def rot(c, n):
    """ your docstring goes here """
    # check to ensure that c is a single character
    assert(type(c) == str and len(c) == 1)

def letter_prob(c):
    """ if c is the space character (' ') or an alphabetic character,
        returns c's monogram probability (for English);
        returns 1.0 for any other character.
        adapted from:
       
    """
    # check to ensure that c is a single character   
    assert(type(c) == str and len(c) == 1)

    if c == ' ': return 0.1904
    if c == 'e' or c == 'E': return 0.1017
    if c == 't' or c == 'T': return 0.0737
    if c == 'a' or c == 'A': return 0.0661
    if c == 'o' or c == 'O': return 0.0610
    if c == 'i' or c == 'I': return 0.0562
    if c == 'n' or c == 'N': return 0.0557
    if c == 'h' or c == 'H': return 0.0542
    if c == 's' or c == 'S': return 0.0508
    if c == 'r' or c == 'R': return 0.0458
    if c == 'd' or c == 'D': return 0.0369
    if c == 'l' or c == 'L': return 0.0325
    if c == 'u' or c == 'U': return 0.0228
    if c == 'm' or c == 'M': return 0.0205
    if c == 'c' or c == 'C': return 0.0192
    if c == 'w' or c == 'W': return 0.0190
    if c == 'f' or c == 'F': return 0.0175
    if c == 'y' or c == 'Y': return 0.0165
    if c == 'g' or c == 'G': return 0.0161
    if c == 'p' or c == 'P': return 0.0131
    if c == 'b' or c == 'B': return 0.0115
    if c == 'v' or c == 'V': return 0.0088
    if c == 'k' or c == 'K': return 0.0066
    if c == 'x' or c == 'X': return 0.0014
    if c == 'j' or c == 'J': return 0.0008
    if c == 'q' or c == 'Q': return 0.0008
    if c == 'z' or c == 'Z': return 0.0005
    return 1.0

Question 4. [4 MARKS] You decide to save all your passwords in files, one password per file. You then decide you want to update your 6-character UTORid password, but you don't remember which of your password files you saved it in. Luckily, this is the only 6-character password you have so you decide to write a function to find the password and change it. Your code runs, but the function's body contains four bugs which will result in the function execution not matching its docstring description. In the space provided, rewrite the buggy lines with the bugs fixed. Do not add or remove any lines. def change password(files_list, new-password): " (list of str, str) -> str Preconditions: (1) files list is not empty. (2) Each file in files_list contains one single-line password. (3) All files in files_list contain passwords of different lengths. (4) Your 6-character password is guaranteed to be in one of these files. In the appropriate file from files_list, replace your 6-character password with new-password, and return your old password. #Code If you think a given line contains a bug, rewrite it. Otherwise leave it blank. found = True for filename in files list[:-1]: passwd_file = open(filename, 'r') first line = passwd_file.read() password = first line.strip("2") if password == 6: passwd_file.close() passwd_file = open(filename, 'w') passwd_file.write(new_password) found = True passwd_file.close() if found: # This is correct. Do not modify. return password