CS 351: Introduction to Cyber SecurityXXXXXXXXXXAssignment 1 1) A ciphertext has been generated with an affine cipher. The most frequent letter of the...

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I have provided the textbook for the reference. It is the best to use for answers (just Ctrl+F). Please don't use any AI, as trackers are available. Provide the schema diagram as well that one of the questions requires. The textbook isComputer Security: Principles and Practice by William Stallings available in libgen.is
Here is the link:http://library.lol/main/2D22877FA0BBE6A86D497AE2B891CE4E



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CS 351: Introduction to Cyber Security Assignment 1 1) A ciphertext has been generated with an affine cipher. The most frequent letter of the ciphertext is 'B', and the second most frequent letter of the ciphertext is 'U'. Break this code. (10 pts) 2) Suppose someone suggests the following way to confirm that the two of you are both in possession of the same secret key. You create a random bit string the length of the key, XOR it with the key, and send the result over the channel. Your partner XORs the incoming block with the key (which should be the same as your key) and sends it back. You check, and if what you receive is your original random string, you have verified that your partner has the same secret key, yet neither of you has ever transmitted the key. Is there a flaw in this scheme? (15 pts) 3) Classify each of the following as a violation of confidentiality, of integrity, of availability, or of some combination thereof, and explain a bit why you think so. (20 pts) a) Carol changes the amount of Angelo's check from $100 to $1,000. b) Gina forges Roger's signature on a letter and sends the letter to Rachel. c) Rhonda registers the domain name “AddisonWesley.com" and refuses to let the publishing house buy or use that domain name. d) Jonah and Peter were original acquaintance, Jonah even told Peter his credit card number and information. However, Peter made a mischief, and inform the credit card company to cancel the card. e) Henry sends a doc containing a Trojan horse to Marry, and Marry opened the file, and sends her PIN to Henry. Henry now can have access to Marry's computer. Vashu Patel DO NOT USE CHATGPT. WE HAVE TRACKERS CS 351: Introduction to Cyber Security 4) A well designed hash function normally has multiple properties, including collision resistance, which means it is very hard for one to find a pair of inputs x≠y, such that h(x) = h(y). Could we just use a collision resistant function to do encryption? i.e., Enc(k,m)=f(k,m), where f is a collision resistant function. (15 pts) 5) Why Diffie-Helman key exchange algorithm alone is not secure against man in the middle attacks, please show step by step. How can we improve security and make sure that the algorithm is secure against MITM attacks? (Draw a schema for both problems and explain the steps) (20 pts) 6) We know that a digital signature is for the purpose of ensuring data integrity and authenticity. (20 points) a) Checksum adds all the bits of the message (or blocks), is checksum a good way to construct a digital signature scheme? How about a hash function, i.e, Sign(M)= h(M)? b) If we use a hash to generate a signature in a more complicated way as follows Sign(k,m) = σ = h(k) XOR m XOR h(m), and m, σ will be sent along. Would this be a secure signature? Briefly explain.
Answered 2 days AfterFeb 21, 2023

Answer To: CS 351: Introduction to Cyber SecurityXXXXXXXXXXAssignment 1 ...

Aditi answered on Feb 23 2023
45 Votes
Solution
1.
To break an affine cypher, we must first identify the encryption function's two unknown values: the multiplicative key (a) and indeed the additive key (b) (b). This may be accomplished by comparing the frequency distribution of letters in the ciphertext to the frequency distribution of letters in the English language.
To b
egin, we determine the frequency of the each word in the ciphertext:
'B' is the most common letter.
'U' is the second most common letter.
Additional letters: less common
The frequency of each letter in the English language is then calculated:
'E' is the most often used letter.
'T' is the second most common letter.
Additional letters: less common
We may guess that 'B' and 'U' correlate to 'E' and 'T,' respectively. This is because 'E' and 'T' are the most common letters in English, whereas 'B' and 'U' are the most common letters in the ciphertext. As a result, we may construct the following equation system:
'B' = a * 'E' + b (mod 26) (mod 26)
'U' = a * 'T' + b (mod 26) (mod 26)
This set of equations may be resolved using modular arithmetic. Then, utilising the assumption that the letters A, B,..., Z are equal to 0, 1, and 25, respectively, we convert the letters B, U, and T to their corresponding numerical values of 0 and 19, 4 and 19, and 4 and 19, respectively:
0 = 4a + b (mod 26) (mod 26)
19 = 19a + b (mod 26) (mod 26)
To get the values of "a" and "b," we can solve this system of equations. Eliminating "b" by deducting the first equation from the second is one approach to accomplish this:
19 - 0 = (19 - 4) (19 - 4)
a (mod 26) (mod 26)
15 = 15a (mod 26) (mod 26)
The modular inverse of 15 mod 26 is 7, which we can calculate since 15 and 26 are coprime.
15a = 15 * 7 * a (mod 26) (mod 26)
15a = 105a (mod 26) (mod 26)
a = 105a (mod 26) (mod 26)
a = 5 (mod 26) (mod 26)
We can use either of the original equations to calculate "b" now that we know what "a" is worth. Using the first equation as an illustration:
0 = 4 * 5 + b (mod 26) (mod 26)
b = 22 (mod 26) (mod 26)
b = -4 (mod 26) (mod 26)
b = 22 (because we want a positive number) (since we want a positive value)
As a result, the additive key (b) is 22 and the multiplicative key (a) is 5. We employ the affine cipher's inverse function, which is: to decipher the ciphertext.
x = (a^-1 * (y - b)) (mod 26)
Hence, a = 3 and b =15
2.
Indeed, an attacker may listen in on the conversation and get the key by doing the following: If A and B are partners, and K is a key of size N.
1. A generates random string R
2. Calculates S : = K XOR R
3. send S to B
4. B calculates S1:= S XOR K
5. sends S1 to A
6. A validates whether R is equal to S1 or not?
As a result, an attacker who is listening in on the chat will have both S and S1. As a result, he will compute the key as K:= S1 XOR S.
3.
a.
John copies Mary's schoolwork in a confidential and genuine manner.
Explanation:
Because "confidential information" is described as information that the public does not generally have access, there has been a violation of confidentiality. Since students are not permitted to duplicate one another's schoolwork, John is in this instance duplicating Mary's homework, which is confidential to Mary at the school level.
Because "Authenticity" is the guarantee that a message, trade, or other exchange of information originates from the source...
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