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Given the following Schedule of Three processes P1,P2 and P3 : 

P1----> P2 -----> P3

Knowing that the processes arrived the same time (0ms) and the execution time for them is X,Y,Z respectively .

What is the response time for each of the three processes: 

a. P1=0, P2 = X , P3 = X+Y

b. P1 = X , P2 = Y , P3 = X+Y+Z

c. P1 = 0 , P2 = X+Y , P3 = X+Y+Z

d. None of the above

 

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21. Suppose a process, P1, is running and the operating system uses the Paged Memory Management technique. Assume P1 has the following page-map table (PMT), and that the page size and the frame size are the same, each being 1024 bytes long. Assuming all addresses start from 0, the logical address 2564 is mapped to the physical address _, where (Frame X, offset Y) means byte number Y in Frame number X. Page Frame | 22 56 98 | 12 60 a) Frame 2, offset 516 b) Frame 98, offset 516 x c) Frame 2, offset 500 d) Frame 98, offset 0 e) None of the above

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'

------------------------------------

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

33. Base10 is the set of numbers created from the digits {0,1,2,3,4,5,6,7,8,9}. Whereas base2, which is the set of numbers created with {0,1}. We could similarly define base; as the ordered set of characters created from {X,Y,Z}. What is the following basez number, Yxz, in base ? Random graph For this section you will need to look at the classes Graph and GraphNode present in the appendix. A Graph consists of a number of nodes, each of which has connections to other nodes in the graph. You can think of nodes as a representation of a person in a social network (e.g. Facebook). The ArrayList connectedTo variable in GraphNodes is the list of friends that particular person has. For example, in the following graph, the node O is friends with node 1 and 2. On the other hand, node 2 and node 1 are not friends, hence the lack of link. 0-0 In the constructor of the Graph, the process of connecting nodes is randomized. Moreover, we will not allow a node to be connected to itself and we are going to assume that the arguments passed to the Graph constructor respect the following conditions: n > 2, and n > nbConnections. Pay close attention to the character order

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