Online Computer Network Security Assignment Help

Online Computer Network Security Assignment Help

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Problem 1-1. Security Policy for Github Write a security policy for a distributed version control hosting site like Github. Make sure to define relevant roles, functions, and policies. Your policy should be of the sort that would be usable to an implementor of a clone of Github. For the purposes of this problem, assume that Github is the maintainer of Git. If you can’t find relevant material on Github as currently implemented, invent new material as appropriate. Try to be as complete as you can, but emphasize the Github-specific aspects—in particular, what security goals are the most relevant for Github? What roles does (or, rather, should) Github have, and what should each principal be allowed to do?

(This problem is a bit open-ended, but should give you excellent practice in writing a security policy. Also, you may actually care about such security policies for designing systems used by large numbers of people—or if you use Github yourself! We have included sample solutions from similar questions in previous years on the course website.)

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Solution 1 Objective The purpose of this document is to outline the rules of use and operation of GitHub by defifining the roles and permissions of its members and the policies that concern usage, protection, and sharing of its assets.

2 Definitions GitHub is a distributed revision control hosting service that makes use of Git revision control software. It facilitates collaborative software development by allowing users (as individuals and/or in teams) to track, share, and discuss source code hosted in public and private repositories. It is available to its users as a website and as a remote endpoint for Git software.

1. Repositories A repository can be best understood as a versioned folder for a software project. It contains all the fifilleess,, documentation, and history pertaining to that project. Repositories can either be private or public. A public repository is open to website visitors, who can be non­registered users.

They can view, clone, and pull from it, but must fifirst create a user account before performing any additional actions. computernetworkassignmenthelp.com

A user is an individual GitHub account holder and can create repositories, be invited to join an organization, and collaborate with other users via their repositories. A user can be authorized to perform Git actions on a repository, such as pull, push, commit, clone, branch, and merge.1 Users can either be free or paid; only paid users are allowed to create private repositories.

A full list of Git actions are available at http://gitref.org/.

3 User Roles and Access Levels 1

1. Primary Roles 1. O W N E R S Users that register an account with GitHub are known as the O W N E R of the repositories they create. (Only paid users may create private repositories.) They have full control of their account and agree to the following rules of usage: 1.Protect personal account credentials: Users must choose a strong account password and must not share it with other users. GitHub provides, and strongly recommends, two-factor authen­ tication so that users can protect their account with more than just a password.

2.Maintain repositories: Every O W N E R is responsible for all content posted under their account. They can create and delete repositories, with read/access privileges .

3.Appoint collaborators: The O W N E R of a repository can collaborate on a project by assigning read/write privileges to other users, who are known as C O L L A B O R AT O R S . computernetworkassignmenthelp.com

2.

C O L L A B O R AT O R S

When a user is assigned C O L L A B O R AT O R access by the O W N E R of a repository, the user has read/write privileges (§3.2.3, §3.2.4) for that repository.

2. Secondary Roles A registered GitHub user can create an organization account and appoint team-based permissions for the repositories created under that account.

Other users can also be added to the O W N E R S or A D M I N team of that account. In particular, a user can be a member of an organization at one of four levels in each team, described below. Note that each level includes the permissions of all the levels below it.

1.

OWNERS

The users appointed as O W N E R S of an account have full access to all repositories and teams within it. They control the account profile and settings, billing and payment information, and members. They can delete the account and all of its content.

2.

A D M I N Team Members computernetworkassignmenthelp.com

1.Create repositories within the team. Delete repositories assigned to the team.

2.Change settings.

3.Transfer team repositories to the organization account.

4.Add and remove users from the team. Add collaborators to the repositories assigned to the team.

3.

W R I T E Access Team Members

A user in a W R I T E access team can, for the repositories assigned to the team: 1.Push to the repository.

2.Apply labels and milestones. Assign, close, and re-open issues.

3.Edit commits, pull requests, and issue comments.

4.Merge and close pull requests.

4.

R E A D Access Team Members

A user in a R E A D access team can, for the repositories assigned to the team: 1.Pull from the repository.

2.Fork the repository. Send pull requests from forks.

3.Open issues. computernetworkassignmenthelp.com

4 Authentication • The service password-protects user accounts and prevents brute-force attacks via rate limiting. The passwords are encrypted and salted.

• GitHub allows users to enable two-factor authentication, which requires the user to enter a password as well as a security code sent to the user’s mobile phone or generated via a two-factor application.

• Git actions require either H TT PS authentication via GitHub account username and password or, for convenience, SSH authentication via keys.

• Finally, the service uses SSL for the transmission of all private data.

5 Employee Access • GitHub employees cannot access private repositories unless it is required for user support.

• Employees that do access those repositories must temporarily attach their SSH key to the owner’s account.

6 Information Gathering and Usage 1.User Account Information: GitHub stores the names and email addresses of all registered users. Additionally, paid users are required to enter billing and payment information, which is securely stored on PCI compliant servers.

Cookies: GitHub uses browser cookies to record current session information, but does not use permanent cookies. To protect against unauthorized access to user accounts, users are logged out after a certain period of inactivity. computernetworkassignmenthelp.com

1.Communication with GitHub: GitHub stores all communication with the service via website visits, website registrations, surveys, and email.

The information collected by the service is used to improve its quality; it is not shared with or sold to third parties except in special circumstances, such as to comply with subpoenas from law enforcement or to respond to user violations of the Terms of Service (ToS). GitHub users retain all rights to their data, including software projects.

7 Amendments to the Security Policy So far we have defined a security policy for the main GitHub hosting service, but it is open to amendments depending on the needs of its users. To give an example, here are a couple amendments: • GitHub Enterprise: GitHub provides an external service known as GitHub Enterprise that allows users to host their repositories on their own infrastructure.

• Government Users: A user that is a U.S. government agency member and account holder of the service on behalf of an agency must agree to a modified ToS.4 This amendment primarily waives the requirements within the ToS that GitHub be used for private, personal, or non­ commercial purposes. GitHub will look solely upon the agency, and not the individual using the service on behalf of the agency, to enforce the modified ToS. computernetworkassignmenthelp.com

Problem 1-2. One-time pad with ciphertext feedback It is well known that re-using a “one-time pad” can be insecure. This problem explores this issue, with some variations.

In this problem all characters are represented as 8-bit bytes with the usual US-ASCII encoding (e.g.“A” is encoded as 0x41). The bitwise exclusive-or of two bytes x and y is denoted x ⊕ y.

Let M = (m 1 , m2 , . . . , m n ) be a message, consisting of a sequence of n message bytes, to be encrypted. Let P = (p1, p2, . . . , p n ) denote a pad, consisting of a coresponding sequence of (randomly chosen) “pad bytes” (key bytes).

In the usual one-time pad, the sequence C = (c1 , c2, . . . , c n ) of ciphertext bytes is obtained by xor-ing each message byte with the corresponding pad byte: When we talk about more than one message, we will denote the messages ci = m i ⊕ pi , for i = 1 . . . n . as M 1 , M 2 , . . . , M k and the bytes of message M j as m j i , namely M j = (m j 1 , . . . , m j n ); we’ll also use similar notation for the corresponding ciphertexts. (a)Here are two 8-character English words encrypted with the same “onetime pad”. What are the words? e9 3a e9 c5 fc 73 55 d5 f4 3a fe c7 e1 68 4a df Describe how you figured out the words. computernetworkassignmenthelp.com

(b)Ben Bitdiddle decided to fix this problem by making sure that you can’t just “cancel” pad bytes by xor-ing the ciphertext bytes.

In his scheme the key is still as long as the ciphertext. If we define c0 = 0 for notational convenience, then the ciphertext bytes c1, c2, . . . , c n are obtained as follows: ci = m i ⊕ ((pi + ci − 1 ) mod 256) .

That is, each ciphertext byte is added to the next key byte and the addition result (modulo 256) is used to encrypt to the next plaintext byte.

Ben is now confident he can reuse his pad, since (k i + c i − 1 ) mod 256 will be different for different messages, so nobody would be able to cancel the k i ’s out. You are provided with otp-feedback.py, which contains an implementation of Ben’s algorithm.

You are also given the file tenciphs.txt, containing ten ciphertexts C 1 , C 2 , . . . , C 10 produced by Ben, using the same pad P . You know that these messages contain valid English text.

Submit the messages and the pad, along with a careful explanation of how you found them, and any code you used to help find the messages.

The most important part is the explanation.

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Solution Part (a) To decode the ciphertexts, we XORed them together, which gives C1 ⊕ C2 = (M 1 ⊕ P )⊕(M 2 ⊕ P ) = M 1 ⊕ M 2 . Thus the pad P is irrelevant. Then we found a set of 8-character words from an English dictionary. For each M 1 in this set, we constructed M 2 by XORing M 1 with C1 ⊕ C2, since M 1 ⊕ (C1 ⊕ C2 ) = M 1 ⊕ (M 1 ⊕ M 2 ) = M 2 . Lastly, we checked if M 2 was also in the set. If both M 1 and M 2 were in the set (as valid English words), then we outputted them.

C1 = [0 xe9 , 0 x3a , 0 xe9 , 0 xc5 , 0 xfc , 0 x73 , 0 x55 , 0 xd5 ] C2 = [0 xf4 , 0 x3a , 0 xfe , 0 xc7 , 0 xe1 , 0 x68 , 0 x4a , 0 xdf ] M1_XOR_M2 = [c ˆ d for c, d in zip ( C1 , C2 )] with open ('/ usr / share / dict / words ', 'r') as words_file : words = words_file . read (). split () words = set ([ word for word in words i f len ( word ) == len ( M1_XOR_M2 )]) for word1 in words : M1 = [ ord ( c) for c in word1 ] M2 = [c ˆ d for c, d in zip ( M1 , M1_XOR_M2 )] word2 = ''. join ([ chr ( c) for c in M2 ]) i f word2 in words : pad = [c ˆ d for c, d in zip ( M1 , C1 )] print ' word1 = %s, word2 = %s, pad = %s' % ( word1 , computernetworkassignmenthelp.com

Output of running the code: word1 = networks, word2 = security, pad = [135, 95, 157, 178, 147, 1, 62, 166] word1 = security, word2 = networks, pad = [154, 95, 138, 176, 142, 26, 33, 172] Though their ordering is indiscernible, the two words are security and networks.

Part (b) Messages and Pad We stand today on the brink of a revolution in cryptography.

Probabilistic encryption is the use of randomness in an encr Secure Sockets Layer (SSL), are cryptographic protocols that This document will detail a vulnerability in the ssh cryptog MIT developed Kerberos to protect network services provided NIST announced a competition to develop a new cryptographic Diffie-Hellman establishes a shared secret that can be used Public-key cryptography refers to a pad = [119, 75, 116, 51, 113, 72,used 105, to 76,sign 78, the 114,certificates 79, 84, 49,had 71, 101, cryptographic system req85,The keys 88, 116, 78, 113, 102, 113, 84, 65,others 51, 55,to99, 56, 107, 69, been71,stolen from We hope this 87, inspires work in this 116, 105, fascinating fie 110, 109, 97, 113, 79, 106, 122, 68, 66, 98, 77, 72, 112, 72, 55, 53, 104, 54, 99, 71, 87, 97, 68, 98, 112, 49] computernetworkassignmenthelp.com

Process This part’s code was more interactive because, due to Ben’s addition of feedback, we couldn’t simply XOR the 10 ciphertexts together and look up possible messages in the dictionary. Our plan of attack was to first calculate all possible pad bytes (pi’s) that would result in valid and likely English characters (letters and common punctuation) from all 10 ciphertexts. valid_chars = set ( range (65 , 65 + 26) + range (97 , 97 + 26) + # A-Z, a- z [32 , 44 , 46 , 63 , 33 , 45 , 40 , 41]) # space , ,.?! -() Let’s consider a particular index i in the 60-character messages (0 ≤ i < 60). For our purposes, pi is independent from the pad bytes surrounding it, because it only depends on mi, ci, and ci−1 (as the calculate pad function below shows). We could have tried all 28 possible pi’s, but |valid chars| is only 60. Therefore, we took each valid character, calculated which pi would result in that character in the first ciphertext, and checked if it resulted in valid characters for the other 9 ciphertexts. def calculate_pad ( ctext , msg , prev_c =0): assert len ( ctext ) == len ( msg ) pad = [] f or i i n xrange ( len ( ctext )): p = (( ctext [ i] ˆ msg [ i]) - prev_c ) % 256 pad . append ( p) prev_c = ctext [ i] return pad computernetworkassignmenthelp.com

def prev_c_at ( ciph , index ): ret urn 0 i f index == 0 el se ciph [ index - 1] def ctext_at ( ciph , index ): ret urn ciph [ index : index + 1] msglen = 60 possible_pad_bytes = [[] f or _ i n range ( msglen )] f or index i n range ( msglen ): f or c i n valid_chars : possible_pad_byte = calculate_pad ( ctext_at ( tenciphs [0] , index ), [ c], prev_c = prev_c_at ( tenciphs [0] , index )) is_valid = True f or ciph i n tenciphs : msg = ben_decrypt ( ctext_at ( ciph , index ), possible_pad_byte , prev_c = prev_c_at ( ciph , index )) i f not set ( msg ). issubset ( valid_chars ): is_valid = False break if is_valid : possible_pad_bytes [ index ]. append ( possible_pad_byte [0]) This calculated all possible pi’s at each i, which is a good start since some choice of P = p1, . . . , p60 within these pi’s would result in Ben’s messages. However, here’s the issue: computernetworkassignmenthelp.com

>>> [ len ( p) f or p i n possible_pad_bytes ] [20 , 6 , 2 , 1 , 2 , 1 , 1 , 1 , 1 , 5 , 5 , 4 , 1 , 1 , 1 , 8 , 3 , 2 , 2 , 2 , 1 , 1 , 1, 3, 3, 1, 1, 1, 1, 5, 4, 1, 1, 2, 1, 2, 2, 9, 1, 4, 2, 1, 1, 2, 1, 1, 1, 1 ,There 1 , 2 ,is a combinatorial explosion of 20 × 6 × · · · × 1, over 247, choices 2 , 2 , 1 , 1there , 1 , 3are , 2 ,a1] for, 1P ,. 2However, tractable 480 choices for the first 9 bytes.

We figured that we could restrict the problem to first choosing those: listing all 480 p1, . . . , p9 pads, decrypting the first 9 bytes of the 10 ciphertexts with each pad, and checking which pads gave intelligible English plaintext.

While we could scan the plaintext manually, we preferred to have the computer do it and score its intelligibility. So, we loaded an English dictionary (the Ubuntu dictionary is excellent; it even contains Hellman). For each set of 10 plaintext messages, we checked how many valid English words from the dictionary appeared in it and gave it |word| 2 points for each word that did. This scoring function strongly favors longer words like cryptography. Lastly, we outputted the best-scoring messages and pad. def recursively_expand_pad ( cur_pad , cur_index , words ): i f cur_index == msglen : # To start , msglen is 9 instead of 60. # Reached the leaves of our search , so decrypt the 10 ciphertexts and score the # resulting text . texts = [ bytes_to_text ( ben_decrypt ( ciph [: msglen ], cur_pad )) for ciph in tenciphs ] text_to_score = '\ n '. join ( texts ). lower () score = sum ( len ( word )**2 f or word i n words i f word i n computernetworkassignmenthelp.com

el se : best_score = 0; best_texts = None ; best_pad = None for p in possible_pad_bytes [ cur_index ]: score , texts , pad = recursively_expand_pad ( cur_index + 1 , cur_pad + [ p], words ) i f best_score < score : best_score = score ; best_texts = texts ; best_pad = pad re u ( best_score , t_te ts be _p d ) t r bes x , st a n with open ('/ usr / share / dict / 'r ') as words_file : words ', words = set ([ word . for word in words_file . read (). split lower () pad = % s ' ()]) pri nt ' messages = %s, % ( texts , pad ) _, tex ts , pad = ly_e pand pa (0 , [], words ) recursive x _ d Output of running the code: messages = ['We stand ', 'Probabili', 'Secure So', 'This docu', 'MIT devel', 'NIST anno', 'Diffie-He', 'Public-ke', 'The keys ', 'We hope t'], pad = [119, 75, 116, 51, 85, 113, 72, 105, 76] We could have searched these plaintext prefixes on Google, but we decided to continue running our code to choose p9, . . . , p18 (often the messages we want to decrypt won’t be available online). computernetworkassignmenthelp.com

First we wrote the pad above to possible pad bytes[0:9] (so there is only 1 choice for pi, 0 ≤ i < 9), and then we increased msglen to 18.

Output of rerunning the code: messages = ['We stand today on ', 'Probabilistic encr', 'Secure Sockets Lay', 'This document will', 'MIT developed Kerb', 'NIST announced a c', 'Diffie-Hellman est', 'Public-key cryptog', 'The keys used to s', 'We hope this inspi'], pad = [119, 75, 116, 51, 85, 113, 72, 105, 76, 78, 114, 79, 84, 49, 71, 101, 71, 88] Repeating this process 4×, we eventually got the desired output (pasted at the beginning) for all 60 characters. At this point, we manually checked it with online articles to be confident in our decryption. For instance, We stand today. . . appears in a well-cited 1976 paper by Whitfield Diffie and Martin Hellman.

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Problem 1-3. Detecting Pad Reuse In the previous problem, we saw how to attack a scheme in which a one-time pad, or a scheme like it, is reused. This problem will walk you through how rare pad reuse for the standard one-time pad (not the previous part’s scheme) can be efficiently detected. We will look at the extreme case in which a pad is reused just once.

The following fact may be useful in this problem: Let R p (n) be the longest run of heads in n coin flips, each of

log 1 n + log 1 ln n. Note: you are not responsible for this paper;

it is provided as a reference only. which is heads with probability p. With high probability as n grows, R1 1 (n) is between log n−log ln ln n and p p p length of the longest repeated bitstring (i.e. a substring of both c i and c j for some i /= j) is less than (a) Suppose you are given poly(n) n-bit ciphertexts c1, . . . , cl , each of which are properly encrypted with an independently uniformly randomly chosen one-time pad. Show that with high probability, the

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pp

log2 n + log2 ln n. (b) We have collected data on the average length of the longest run of identical characters in identical positions in passages of English text by randomly sampling from the collected works of American writer Winston Churchill. You can find a graph of run length (in characters) versus log of the passage length in characters at http://courses.csail.mit.edu/6.857/2014/fi les/run_lengths.png. How do these run lengths compare to the previous part’s upper bound on the (c) Assume that each pair ifofEnglish plaintexts does share a long common longest repeated bitstring plaintext is US-ASCII encoded?run of identical characters, and any pair of ciphertexts with independently chosen pads does not. Show that given poly(n) n-bit ciphertexts with total length N and one instance of pad reuse, you can find the two ciphertexts which share a key in time which is O˜(N ) in the total ciphertext length. This means that your solution should run in O(N logc N ) time for some constant c.

Hint: You may wish to read about and use suffix trees in your solution.

They are a type of tree which store all of the suffixes of an n-character string, can be constructed in O(n log n) time and can be used, among other things, to find a string’s longest repeated substring in linearithmic time by finding the deepest non-leaf node. Note that longest repeated substring isn’t exactly what you need to solve this problem. If you’re not familiar with suffix trees, you may or may not find suffix arrays simpler and easier to understand. computernetworkassignmenthelp.com

Solution Part (a) We are given f (n) = poly(n) n-bit ciphertexts c1, . . . , c f ( n ) , each from a unique plaintext/pad pair. Our goal is to show that the upper bound on the longest run of matching bits between any two ciphertexts, at any offset, is log2 n + log2 ln n = log2 (n ln n) with high probability [Piazza:18].

First, let’s consider a particular pair of ciphertexts ci and c j (i =/ j) with no offset. We can model the longest run of matching bits within the n bits of ci and c j as the longest run of heads in n coin flips, where a coin flip represents matching bits at the same position. Since each ciphertext is encrypted from a one-time pad (chosen independently and uniformly at Next, let's 0consider a particular pair of ciphertexts offset by k bits (0 < k < random), and 1 bits are equally n). There are 2n — 1 possible offsets. We can model this as above, finding the upper bound for the longest run of matching bits within the n — k bits of 4 and c1. Since n — k < n, Ro.s(n — k) < R0.5(n) = log2 (n Inn), so the upper bound from above still holds.

Finally, let's put it all together for the f(n) ciphertexts. The longest run may occur in any of (f 2) pairs of ciphertexts, each at any of 2n — 1 offsets. The number of pairs/offsets is f'(n) = (2n — 1)(f (271)) = (2n — 1) - if(n)(f (n) — 1) which, since f(n) is poly(n), is also poly(n). We next show that the log2 (n Inn) upper bound holds for all these pairs/offsets with high probability. For each possible pair c.i and 4 and offset, the longest run common to 4 and 4 is less than log2 (n Inn) with high probability p(n). That means that for any asymptotically positive polynomial q(n), p(n) > 1— 44, for all sufficiently large n [Piazza:141. Let a polynomial 4(n) = 12(r(n)). Then p(n) > 1 — . The computernetworkassignmenthelp.com probability that the upper bound does not hold for a particular pair/offset is 1

a. The probability that it does not hold for at least one pair/offset among the f'(n) is, by union bound, less than f'(n) dryi = wriou , where 4'(n) = n. 41(n) is an asymptotically positive polynomial because we chose 4(n) to asymptotically dominate f'(n) and both are positive. The probability that the upper bound holds for all pairs/offsets is p'(n) > 1 —. In fact, for any asymptotically positive polynomial q(n), p'(n) > 1 — ik because we can repeat the argument above for a sufficiently large 4(n) > Therefore, the upper bound on the longest run of matching bits between any two ciphertexts, at any offset, is log2 (n Inn) with high probability p'(n).

Part (b) We plotted the upper bound of ciphertext from part (a) and superimposed it on the provided graph. We assume that the English plaintext is encoded as US-ASCII characters, each 7 bits long. Since n here is the text length (in characters), the upper bound of ciphertext (in characters) is 1 7 log 2 (7n ln log 2 n vs. longest run.

7n).

Blue curve: English plaintext (average). Orange curve: ciphertext (upper bound).

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The “random” ciphertext dominates at the beginning because there is more flexibility in matching bits than matching 7-bit US-ASCII characters: (1) the bits can match across 7-bit boundaries; (2) there are more characters to match in the plaintext: 26 English letters, plus punctuation; and (3) the ciphertext is an upper bound, while the plaintext is just an average. But, interestingly, the plaintext average dominates the ciphertext upper bound when log2 n > 9. We reason that matching characters occur more frequently there because the English language has a lot of redundancy: high frequency of e, the strong tendency for h to follow t or for u to follow q, etc. (formalized by Claude Shannon in “Prediction and Entropy of Printed English” in 1951).

Therefore, for sufficiently large n, the average longest runs within English plaintext are larger than even the upper bound on longest runs within ciphertext. On the graph above, it appears as exponential vs. linear.

Part (c) We are given f (n) = poly(n) n-bit ciphertexts c1, . . . , c f ( n ) , and two of them reuse the same pad. Our goal is to find those those ciphertexts in quasilinear time in N = |c1 ·· ·· ·· c f (n) | = n · f (n).

Idea We assume that each pair of ciphertexts, at any offset, with independently chosen one-time pads do not share a long common substring [Piazza:11]. From part (a), the length of a common substring is less than log2 (n ln n) bits with high probability. We assume that each pair of plaintexts share a long computernetworkassignmenthelp.com common substring [Piazza:11]. computernetworkassignmenthelp.com

Therefore, if ciphertexts corresponding to the plaintexts are encrypted with the same pad, then they also share a long common substring. The graph from part (b) shows that, for sufficiently large n, its length is greater than log2 (n ln n). Therefore, we can take log2 (n ln n) as the separator between pad reuse and no pad reuse.

Algorithm In the following algorithm, we find the longest common substring between any two ciphertexts and detect pad reuse if its length is > log2 (n ln n).

1.Let the alphabet Σ = {0, 1, $1, . . . , $ f ( n ) }, which represents the bits and f (n) special termi- nators. Append the terminators to the ciphertexts, and concatenate them into one string S = c1$1 · · · c f ( n ) $f ( n ) .

2.Construct a suffix tree for S. 3.Traverse the suffix tree to find the internal node with the highest depth d.

4.If d ≤ log2 (n ln n), then output no pad reuse. Otherwise, follow a path of edges from the node down (i.e., toward a leaf) until an edge label contains a special terminator $i . This path corresponds to a suffix of ciphertext ci . Since an internal node must have ≥ 2 children, repeat this process along a different path—until an edge label contains $ j . Output ci and c j .

Correctness Using computernetworkassignmenthelp.com the procedure stated in the problem, Step 3 finds

1.In different ciphertexts, a substring of ci and a substring of c j : This case is desired, because it corresponds to the longest common substring of a pair of ciphertexts. Correctness follows from the idea paragraph. If there is no pad reuse, then d ≤ log2 (n ln n). Otherwise there is, and the special terminators $i and $ j identify the ciphertexts ci and c j that reuse the same pad.

2.In the same ciphertext, a substring of ci and a substring of ci : This case is not desired, because it corresponds to longest common substring between a ciphertext and itself at an offset of k bits.

However, we can model this using part (a), since the successive bits of the pad are independent. The longest run of matching bits within the n − k bits of ci and the offset ci is less than R 0 . 5 (n − k) ≤ R 0 . 5 (n) = log2 (n ln n). This correctly corresponds to no pad reuse.

3.Across ciphertexts, a substring of c i $ i c j and a substring of S: This case never occurs because the special terminator $i only appears once in S. Running Time Step 1 (concatenating the ciphertexts) takes O(N ) time. Step 2 (building the suffix tree) takes O(N log |Σ|) = O(N log f (n)) = O(N log N ) time [Piazza:25]. Step 3 (traversing for longest repeated substring) takes O(N ) time, as stated in the problem. Step 4 (traversing to identify ci and c j ) takes 2 · O(n) time, since the internal node is ≤ n edges away from an edge label containing the special terminator. This dominates computing log2 (n ln n), which involves operations on log2 n-bit numbers. Therefore, the total time computernetworkassignmenthelp.com