Programming

This is my “Algorithms and Data Structures” Programming Assignment , use C language.1.Read pa1.pdf carefully ,follow the PDF to complete each step.2.ADT. pdf and ADT-more.PDF are extended readings.3.The examples Queue.c and Stack.c on the website are good starting points for the List ADT module in this project.
You are welcome to simply start with one of those files, rename things, then add functionality until the
specifications for the List ADT are met. You should first design and build your List ADT, test it thoroughly, and
only then start coding Lex.c. If you do a good job, I will look for the next cooperation.Thank you very much.

I want someone to do homeworkHomework 8
This assignment helps you practice with “classes”. Read the book for guidance: Chapter 15 and Chapter 17.
I’ve written another version of the album database (from Homework 5), this time using a class to represent an album. The class
keeps track of the album’s artist, title, year, and a list of ratings. Your task is to write this class. The rest of the code should remain
unchanged.
Here is the code. Refer to the comments in the code for an explanation of your task. Also see the expected output below; your
solution should reproduce this output.
import pickle
import os
import statistics
### Your class definition goes here. Name the class Album.
### Notice how the class is used in the code below
### (which functions exist, etc.).
### You must not create any new variables outside the class;
### the purpose is to create a class only, no extra code.
### Make sure your code recreates the example output below.
def get_rating(album):
rating float(input(“Enter rating: “))
album.add_rating(rating)
def get_album():
artist = input(“Artist: “)
title = input(“Album title: “)
def get_album():
artist = input(“Artist: “)
title = input(“Album title: “)
year = int(input(“Album year: “))
album = Album(artist, title, year)
get_rating(album)
return album
if os. path.exists(“albumsHW08.pkl”):
with open(“albumsHW08.pkl”, “rb”) as f:
albums pickle.load(f)
else:
=
albums =
[]
answer =
while True:
print(“Albums in database:”)
for i in range(len (albums )):
print(“===== %d =====” % i)
albums[i].print()
print()
print(“Type new to enter a new album, type rate to rate an album.”)
input(“Command: “)
if answer == “new”:
albums. append(get_album( ) )
elif answer “rate”:
idx int(input(“Enter album # from list above: “)).
if idx < 0 or idx >= len(albums):
print(“Bad #”)
else:
get_rating(albums[idx])
else:
print(“Exiting…”)
break
with open (“albums HW08.pkl”, “wb”) as f:
pickle.dump (albums, f)
Example output:
Example output:
Albums in database:
Type new to enter a new album, type rate to rate an album.
Command: new
Artist: Nicolas Jaar
Album title: Cenizas
Album
year: 2020
Enter rating: 5.0
Albums in database:
=====
0
=====
Nicolas Jaar
Cenizas (2020), 1 ratings, 5.00 avg rating
Type new to enter a new album, type rate to rate an album.
Command: new
Artist: Talk Talk
Album title: Laughing Stock
Album year: 1991
Enter rating: 4.7
Albums in database:
=====
0
=====
Nicolas Jaar –
Cenizas (2020), 1 ratings, 5.00 avg rating
=====
1
=====
Talk Talk
Laughing Stock (1991), 1 ratings, 4.70 avg rating
Type new to enter a new album, type rate to rate an album.
Command: rate
Enter album # from list above: 0
Enter rating: 4.5
Albums in database:
=====
0
=====
Nicolas Jaar
Cenizas (2020), 2 ratings, 4.75 avg rating
=====
1
=====
Talk Talk
Laughing Stock (1991), 1 ratings, 4.70 avg rating
Type new to enter a new album, type rate to rate an album.
Command: rate
Enter album # from list above: 0
Artist: Talk Talk
Album title: Laughing Stock
Album year: 1991
Enter rating: 4.7
Albums in database:
===== 0
=====
Nicolas Jaar
Cenizas (2020), 1 ratings, 5.00 avg rating
=====
1 =====
Talk Talk
–
Laughing Stock (1991), 1 ratings, 4.70 avg rating
Type new to enter a new album, type rate to rate an album.
Command: rate
Enter album # from list above: 0
Enter rating: 4.5
Albums in database:
=====
0
=====
Nicolas Jaar
–
Cenizas (2020), 2 ratings, 4.75 avg rating
===== 1 =====
Talk Talk
Laughing Stock (1991), 1 ratings, 4.70 avg rating
Type new to enter a new album, type rate to rate an album.
Command: rate
Enter album # from list above: 0
Enter rating: 3.9
Albums in database:
===== 0
=====
Nicolas Jaar
Cenizas (2020), 3 ratings, 4.47 avg rating
=====
1 =====
Talk Talk
–
Laughing Stock (1991), 1 ratings, 4.70 avg rating
Type new to enter a new album, type rate to rate an album.
Command: quit
Exiting…

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I have a solutions for cryptography class and I just need to prephrase it and change the style of papers to avoid any doubt from my teacher. i attached the solutions. I will attach the questions too1.Describe the differences between digital signature and digital certificate.
Digital signature:
These are the cryptographic computations which enable the system or a person to perform the
authentication of their documents in a unique way that achieves nonrepudiation. These digital
signatures were derived from the fact that in typical public-key encryption schemes, we can
reverse the order in which the encryption and decryption algorithms are applied:
EPB(DSB(M)) = M.
Digital Certificate:
It is like the identification cards for example passport or resident permits. These certificates are
issued by governments recognized authorities. When a person requests a digital certificate
recognized authorities verifies the personal identity and purpose behind the requested
certificate. When this certificate is presented to others, they can verify the identity of its owner.
2- Describe the differences between PKI and KDC.
PKI:
Public key infrastructure is a complete algorithm or body in which have different modules.
Like key distribution and management, authentication, integrity, confidentiality, availability,
and non-repudiation. The following are some basic about the PKI:
• In Asymmetric cryptography
• Private Key is utilized to validate clients. The private key is put away on the circle, and keep
up by clients.
KDC:
KDC is an abbreviation of a key distribution center it is a component in an access control
system. Its main task is to provide services to user requests to gain access for resources/system
by supplying session keys and access tickets. It uses cryptographic techniques for the
authentication of users requests for resource access. When the user gets successful
authentication, it provides the user with a ticket to target resource/system which grants the user
access for the required resource.
• In Symmetric cryptography
• Password used for authentication
3- Describe the differences between symmetric crypto and asymmetric crypto.
Symmetric Crypto:
• Two parties must trust each other.
• utilizes the same key(one secret key)
• faster.
• Examples: DES, AES, RC5
Asymmetric Crypto:
• Two parties do not trust each other.
• utilizes a pair of keys: a public and a private key
• slower.
• Examples: RSA, DSA, ECC
4- Explain the reflection attacks against symmetric key based authentication, and explain potential
countermeasures.
Symmetric key:
symmetric key cryptography is a simple and most know the type of cryptography in which both sender
and receiver share the common secret key in order to perform the encryption and decryption process.
The other name for its shared key cryptography.
Reflection attack:
An attacker can take a ciphertext c sent from A to B and send it back to A. This again can cause a
mismatch between the two parties’ transcripts of their communication session: A may output a message
m, even though B never sent such a message.
Potential countermeasures:
1) Different keys for initiator and responder; KAlice-Bob = KBob-Alice + 1
2) Different Challenges: Alice and Bob’s challenges cannot be the same
3) The initiator should be the first to prove its identity
5-Alice and Bob use the Diffie-Hellman to create two keys k1 and k2. In generating k1, Alice selects
a random number x and computes SA = gx mod p. Bob selects a random number y and computes
SB = gy mod p. Alice and Bob exchange SA and SB, and k1 = gxy mod p. In generating k2. Alice
chooses another random number x’ but Bob still uses the same random number y. Suppose an
eavesdropper knows that x and x’ differ each other by t (i.e., either x-x’ = t or x’-x=t). Given the
knowledge of k1 and a pair of plaintext M and ciphertext C encrypted by k2, can the eavesdropper
find out k2?
Yes, Attacker can find out the key k2 because it has k1 and pair of plaintext which is enough for brute
force attack. They will find out the pattern and with the help of pattern, he can find out the key k2.
That’s why the double key is sometimes not so secure because it may produce patterns. An attacker may
find a way to decrypt it. Like, in this case, he can find out the key K2.
6- Suppose we are using a three-message mutual authentication protocol, and Alice initiates
contact with Bob. Suppose we wish Bob to be a stateless server, and therefore it is
inconvenient to require him to remember the challenge he sent to Alice. Let’s modify the
exchange so that Alice sends the challenge back to Bob, along with the encrypted challenge.
So the protocol is shown as follows. Is this protocol secure?
No. It is liable to the replay assault. A busybody can replay Alice’s messages whenever. On the off
chance that Bob remembers his present test, he won’t realize that the reaction is to a past test. We can
make it secure by Bob sending timestamps as the test (expecting the assailant can’t replay the messages
quick enough so that the timestamp in the message is adequate to Bob).
7- Let’s modify the protocol from the previous problem so that Bob (still stateless) sends both a
challenge, and a challenge encrypted with a key K_Bob that only he knows. Is this protocol
secure?
No. The convention is shaky, in light of the fact that spying is conceivable. In the event that Oscar
tunes in to the correspondence and stores the pair (R1, Alice-Bob(R1)), he can verify himself with
Sway as Alice utilizing this message (until the convention is changed). Oscar sends Bob the
message: „I am Alice”, Bob sends Trudy an irregular number (which Bob can’t recall since he is
stateless) and Oscar sends Bob the captured pair.
8- Suppose Alice and Bob know each other’s public key, and they use the following protocol to
achieve mutual authentication: Alice signs the message M1=“This is Alice” and Bob signs the
message M2=“This is Bob”. Explain the vulnerability.
Mutual authentication with the help of the public key is not a secure protocol. Man in the middle attack
is possible. For better protocol, Alice should encrypt the message with the private key of Bob and Bob
should encrypt the message with the private key of Alice.
9- Now, suppose Alice and Bob use random numbers to enhance security. Alice signs the
combination of message M1=“This is Alice” and a random number R1, then send M1, R1, and
digital signature to Bob; and Bob does the similar thing. Is this secure?
Yes, it is secure it will produce the randomness so the man in the middle attack will be
impossible.
10- An even trickier question: In a three-message authentication protocol, Alice initiates contact with
Bob. Bob is a stateless server, and thus it is inconvenient for him to remember the challenges sent to
Alice. Assume that Alice and Bob share a key KAlice- Bob and Bob also has a key KBob only known
to himself. The protocol works as follows, where R1 and R2 are random numbers generated by
Alice and Bob, respectively.
a.
Does this protocol provide mutual authentication? If yes, why? If no, who cannot authenticate
whom and why?
Yes, this protocol is providing mutual authentication on both ends. As we know that both Bob and Alice share a
common key KAlice- Bob. Since Bob can recognize Alice and Alice can also confirm Bob with R1 where Alice
is sending a message to Bob. Bob is stateless but still, he has the common key and R1 is playing the integrating
to the server on Bob side for authentication.
B. Could a third person, Trudy, impersonate Alice? Describe a possible attack scenario. If it is not
possible, clearly state so and give reasons.
Yes, Trudy can Impersonate Alice by R1 in order to do that she has to send some chance number tR1 to Bob.
C. Could Trudy impersonate Bob? Describe a possible attack scenario. If it is not possible, clearly
state so and give reasons.
Yes, Trudy can Impersonate Bob only she gets to the KAlice- Bob and also the key of Bob.
11- The expanded Needham- Schroeder Protocol can be shortened to a 6-message protocol without loss
of security by merely removing the 7th message. Why is this true? (Hint: the purpose of the 7th
message is to prove to Bob that he is talking to Alice, but he already knows that. Why?)
It is still true because the 7th message is used for authentication of Alice but Bob already knows. There will be
no loss of information by removing the 7th message.
Homework 4 (100 pts)
Rule: Finish all of the following on your own.
Think more carefully when you analyze an authentication protocol design.
1. Describe the differences between digital signature and digital certificate.
2. Describe the differences between PKI and KDC.
3. Describe the differences between symmetric crypto and asymmetric crypto.
4. Explain the reflection attacks against symmetric key based authentication, and
explain potential countermeasures.
5.
lice and Bob use the Diffie-Hellman to create two keys kl and k2. In generating ki,
Alice selects a random number x and computes SA = g* mod p. Bob selects a random
number y and computes SB = g’ mod p. Alice and Bob exchange SA and SB, and k1 =
gły mod p. In generating k2. Alice chooses another random number x’but Bob still uses
the same random number y. Suppose an eavesdropper knows that x and x’ differ each
other by t (i.e., either x-x’ = t or x’-x=t). Given the knowledge of kl and a pair of
plaintext M and ciphertext C encrypted by k2, can the eavesdropper find out k2?
6. Suppose we are using a three-message mutual authentication protocol, and Alice
initiates contact with Bob. Suppose we wish Bob to be a stateless server, and
therefore it is inconvenient to require him to remember the challenge he sent to Alice.
Let’s modify the exchange so that Alice sends the challenge back to Bob, along with
the encrypted challenge. So the protocol is shown as follows. Is this protocol secure?
I’m Alice
R
Alice
Bob
R, K_Alice_Bob{R}
7. Let’s modify the protocol from the previous problem so that Bob (still stateless) sends
both a challenge, and a challenge encrypted with a key K_Bob that only he knows.
10. An even trickier question: In a three-message authentication protocol, Alice initiates
contact with Bob. Bob is a stateless server, and thus it is inconvenient for him to
remember the challenges sent to Alice. Assume that Alice and Bob share a key Kalice-
Bob and Bob also has a key KBob only known to himself. The protocol works as follows,
where R1 and R2 are random numbers generated by Alice and Bob, respectively.
I’m Alice, R1
Alice
I’m Bob, KBob(R2), Kalice-Bob {R1+R2},
Bob (cannot
remember
anything from
any previous
information)
KBob{R2}, Kalice-Bob{R2}
a.
Does this protocol provide mutual authentication? If yes, why? If no, who cannot
authenticate whom and why?
b. Could a third person, Trudy, impersonate Alice? Describe a possible attack scenario.
If it is not possible, clearly state so and give reasons.
c. Could Trudy impersonate Bob? Describe a possible attack scenario. If it is not
possible, clearly state so and give reasons.
11. (10 extra credit points for both graduates and undergraduates) The expanded Needham-
Schroeder Protocol can be shortened to a 6-message protocol without loss of security
by merely removing the 7th message. Why is this true? (Hint: the purpose of the 7th
message is to prove to Bob that he is talking to Alice, but he already knows that. Why?)

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This assignment helps you practice with “classes”. Read the book for guidance: Chapter 15 and Chapter 17.I’ve written another version of the album database (from Homework 5), this time using a class to represent an album. The class keeps track of the album’s artist, title, year, and a list of ratings. Your task is to write this class. The rest of the code should remain unchanged.Here is the code. Refer to the comments in the code for an explanation of your task. Also see the expected output below in the word document:your solution should reproduce this output.
import pickle
import os
import statistics
###
###
###
###
###
###
Your class definition goes here. Name the class Album.
Notice how the class is used in the code below
(which functions exist, etc.).
You must not create any new variables outside the class;
the purpose is to create a class only, no extra code.
Make sure your code recreates the example output below.
def get_rating(album):
rating = float(input(“Enter rating: “))
album.add_rating(rating)
def get_album():
artist = input(“Artist: “)
title = input(“Album title: “)
year = int(input(“Album year: “))
album = Album(artist, title, year)
get_rating(album)
return album
if os.path.exists(“albumsHW08.pkl”):
with open(“albumsHW08.pkl”, “rb”) as f:
albums = pickle.load(f)
else:
albums = []
while True:
print(“Albums in database:”)
for i in range(len(albums)):
print(“===== %d =====” % i)
albums[i].print()
print()
print(“Type new to enter a new album, type rate to rate an
album.”)
answer = input(“Command: “)
if answer == “new”:
albums.append(get_album())
elif answer == “rate”:
idx = int(input(“Enter album # from list above: “))
if idx < 0 or idx >= len(albums):
print(“Bad #”)
else:
get_rating(albums[idx])
else:
print(“Exiting…”)
break
with open(“albumsHW08.pkl”, “wb”) as f:
pickle.dump(albums, f)
Example output:
Albums in database:
Type new to enter a new album, type rate to rate an album.
Command: new
Artist: Nicolas Jaar
Album title: Cenizas
Album year: 2020
Enter rating: 5.0
Albums in database:
===== 0 =====
Nicolas Jaar – Cenizas (2020), 1 ratings, 5.00 avg rating
Type new to enter a new album, type rate to rate an album.
Command: new
Artist: Talk Talk
Album title: Laughing Stock
Album year: 1991
Enter rating: 4.7
Albums in database:
===== 0 =====
Nicolas Jaar – Cenizas (2020), 1 ratings, 5.00 avg rating
===== 1 =====
Talk Talk – Laughing Stock (1991), 1 ratings, 4.70 avg rating
Type new to enter a new album, type rate to rate an album.
Command: rate
Enter album # from list above: 0
Enter rating: 4.5
Albums in database:
===== 0 =====
Nicolas Jaar – Cenizas (2020), 2 ratings, 4.75 avg rating
===== 1 =====
Talk Talk – Laughing Stock (1991), 1 ratings, 4.70 avg rating
Type new to enter a new album, type rate to rate an album.
Command: rate
Enter album # from list above: 0
Enter rating: 3.9
Albums in database:
===== 0 =====
Nicolas Jaar – Cenizas (2020), 3 ratings, 4.47 avg rating
===== 1 =====
Talk Talk – Laughing Stock (1991), 1 ratings, 4.70 avg rating
Type new to enter a new album, type rate to rate an album.
Command: quit
Exiting…

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Topic 1: Information SilosWhat are information silos and how can an enterprise integration
application system or an enterprise resource planning system help an
organization deal with them?Just do response each posted # 1 to 3 down below only.Posted 1 In
my understanding information silos are different systems withholding
useful information that could be important for one another but are
unable or do not share that information. Our textbook defines
information silos as an information system that does not communicate
with other, related information systems in an organization (Rainer & Prince, 2018).
This could potentially harm an organization and cause unnecessary
problems that could have been prevented if there was better
communication between the systems. Once
an organization starts becoming successful, organizations have trouble
communicating between systems. There could be a continuous problem that
could have been resolved by using information from a different system or
could have been an unseen issue for an organization. To resolve this,
integrating an enterprise resource planning (ERP) system would benefit
the organization to help solve this problem. Its focus is to be a system
that is designed to correct the lack of communication among functional
areas in the organization (Rainer & Prince, 2018).
Having an ERP system available would benefit the organization by
reducing time accessing information through common software. By
integrating the system to the organization, it will bring awareness and
transparency to other departments. Posted 2Silo is an outdated system that is usually developed for a specific focus and designed without considering future needs. Like other systems it is not compatible with other FAIS or organizational information systems. When using this type of system, it makes it hard to gather information together when needing data. Enterprise integration application system or enterprise resource planning system can help by eliminating multiple legacy systems that cannot communicate with each other. The enterprise application can connect many applications together while the ERP integrates the planning. While reading it seems like this is not an easy process. It can take a while and even might not work for everyone. For example, in the passage the Air Force spend more than 1 billion dollars before canceling their ERP system implementation. While it worked for the Navy and they kept it implemented. posted 3Information
silo is a management system that doesn’t communicate with other
information management systems. The systems are created with its own
data in specific formats. These formats are incapable of communicating
When information isn’t shared between different departments in a company
it can hinder efficiency in a business and be considered a problem. Since
the silos do not communicate with others the Enterprise resource
planning systems (ERPS) and the Enterprise application integration (EAI)
is in place to help with that problem. Enterprise resource planning
(ERPS) are used to enable information to be shared. Enterprise
application integration (EAI) also is helpful by providing layers of
software that connects applications. Having these in place will help the
organization with efficiency and productivity.