Question:
Assume that the texts to be tagged contain 1.5% of unknown words and that the performance
of the tagger to be used is 98% on known words.
What will be its typical overall performance in the following situation:
all unknown words are systematically wrongly tagged?

To calculate the overall performance of the tagger given the assumptions you've provided, we can break it down into two components: the known words and the unknown words.

1. **Known Words**:
   - The percentage of known words in the text is \(100\% - 1.5\% = 98.5\%\).
   - The performance of the tagger on known words is 98%. This means that out of all known words, 98% will be correctly tagged.

2. **Unknown Words**:
   - The percentage of unknown words is 1.5%.
   - According to your assumption, all unknown words are systematically wrongly tagged, which means 0% of unknown words will be correctly tagged.

Now, let's calculate the overall performance:

- **Correctly Tagged Known Words**:
  \[
  \text{Percentage of known words} \times \text{Performance on known words} = 98.5\% \times 98\% = 0.985 \times 0.98 = 0.9663 \text{ or } 96.63\%
  \]

- **Correctly Tagged Unknown Words**:
  \[
  \text{Percentage of unknown words} \times \text{Performance on unknown words} = 1.5\% \times 0\% = 0
  \]

- **Overall Performance**:
  To find the overall performance, we sum the contributions from correctly tagged known words and unknown words:
  \[
  \text{Overall Performance} = \text{Correctly Tagged Known Words} + \text{Correctly Tagged Unknown Words} = 96.63\% + 0\% = 96.63\%
  \]

Thus, the typical overall performance of the tagger, given that all unknown words are systematically wrongly tagged, will be **96.63%**.


Question: Dude said “I like bowling”. With how many statements can we express this sentence using  RDF Reification?

The correct option is **D. 5**.

In RDF reification, to represent a statement (or triple) and add metadata about it, you need to create additional triples that describe the original statement. Specifically, to reify the statement “Dude said ‘I like bowling’”, you would need to create the following:

1. A triple to state the subject (Dude), predicate (said), and object (the statement "I like bowling").
2. A triple to specify that the original statement is a statement (typically with a type declaration).
3. Additional triples to capture any metadata, such as the speaker (Dude) and the content of the statement.

Thus, you end up needing 5 RDF triples in total to fully express this sentence using RDF reification.


Question: Select \emph{incorrect} statement. Complexity analysis of an attack consideres

The incorrect statement is: **4. difficulty to understand a corresponding journal paper.**

Reasoning: Complexity analysis of an attack focuses on assessing the computational resources required to execute the attack, typically encompassing time complexity (how long the attack takes) and memory complexity (how much memory is needed). The probability of success relates to the effectiveness of the attack. However, the difficulty of understanding a corresponding journal paper is not a factor in complexity analysis—it pertains more to academic accessibility and comprehension than to the actual performance or effectiveness of the attack.


Question: Consider the Diffie-Hellman secret-key-exchange algorithm performed in the cyclic group $(\mathbb{Z}/11\mathbb{Z}^\star, \cdot)$. Let $g=2$ be the chosen group generator. Suppose that Alice's secret number is $a=5$ and Bob's is $b=3$. Which common key $k$ does the algorithm lead to? Check the correct answer.

To solve the problem, we need to follow the steps of the Diffie-Hellman key exchange algorithm in the cyclic group \((\mathbb{Z}/11\mathbb{Z}^\star, \cdot)\) using the given parameters.

1. **Calculate Alice's public key**:
   Alice's secret number is \( a = 5 \). The public key \( A \) is calculated as:
   \[
   A = g^a \mod 11 = 2^5 \mod 11
   \]
   First, compute \( 2^5 = 32 \).
   Now reduce it modulo \( 11 \):
   \[
   32 \mod 11 = 32 - 2 \times 11 = 32 - 22 = 10
   \]
   So, Alice's public key is \( A = 10 \).

2. **Calculate Bob's public key**:
   Bob's secret number is \( b = 3 \). The public key \( B \) is calculated as:
   \[
   B = g^b \mod 11 = 2^3 \mod 11
   \]
   First, compute \( 2^3 = 8 \).
   Since \( 8 < 11 \), we have:
   \[
   B = 8
   \]

3. **Alice and Bob exchange their public keys**:
   - Alice sends \( A = 10 \) to Bob.
   - Bob sends \( B = 8 \) to Alice.

4. **Calculate the shared secret key \( k \)**:
   - Alice computes the shared key using Bob's public key:
   \[
   k = B^a \mod 11 = 8^5 \mod 11
   \]
   First, we calculate \( 8^5 \):
   - Compute \( 8^2 = 64 \) and reduce modulo \( 11 \):
     \[
     64 \mod 11 = 64 - 5 \times 11 = 64 - 55 = 9
     \]
   - Next, compute \( 8^4 = (8^2)^2 = 9^2 = 81 \):
     \[
     81 \mod 11 = 81 - 7 \times 11 = 81 - 77 = 4
     \]
   - Now compute \( k = B^a = B^5 = B^4 \cdot B = 4 \cdot 8\):
     \[
     k = (8^4) \cdot (8) = (4) \cdot (8) = 32
     \]
     Now reduce modulo \( 11 \):
     \[
     k = 32 \mod 11 = 32 - 2 \times 11 = 32 - 22 = 10
     \]

Now let’s calculate Bob’s shared secret key using Alice’s public key:
- Bob computes the shared key using Alice's public key:
\[
k' = A^b \mod 11 = 10^3 \mod 11
\]
First, compute \(10^3\):
- Calculate \(10^2 = 100\):
\[
100 \mod 11 = 100 - 9\times11 =100-99=1
\]
Then calculate \(10^3=10^2\cdot10=1\cdot10=10\):
\[
k' = (10^2) \cdot (10) = (1) \cdot (10) = 10
\]

Both calculations yield the same result: the common key \( k = k' = 10 \).

Now, let's check the answer options:
- A. **10** (Correct)
- B. **7** (Incorrect)
- C. **8** (Incorrect)
- D. **9** (Incorrect)

The correct option is **A.**


Question: Up to which linguistic processing level can each of the following sentences be considered as correct? The glass broke its leg, I no go rain, The cook put cherry stones in the cake, Cars flow beautifully; syntactic, pragmatic, syntactic, semantic, lexical

Let's break down the correct answer for each sentence based on the levels of linguistic processing: syntactic, semantic, pragmatic, and lexical.

1. **"Cars flow beautifully" - syntactic**: This sentence is grammatically correct (syntactically valid) because the structure follows the rules of English grammar. However, it makes less sense semantically because "cars" do not literally flow.

2. **"The cook put cherry stones in the cake" - semantic**: This sentence is syntactically correct, but semantically it raises questions as it suggests that cherry stones, which are hard and inedible, were put in a cake, which is unusual. Thus, it’s considered correct at the semantic level since it has a clear meaning, but it’s odd in context.

3. **"The glass broke its leg" - syntactic**: This sentence is syntactically correct in that it follows grammatical rules. However, it makes no sense semantically because glass cannot have a leg. Thus, it's correct at the syntactic level only.

4. **"I no go rain" - lexical**: This sentence has grammatical issues (it's not standard English), which makes it incorrect syntactically. The lexical aspect refers to the choice of words and how they are combined. Here, "no" is an incorrect word choice in standard English; it should be "I am not going to rain" or "I will not go in the rain."

In summary:
- "Cars flow beautifully" is syntactically correct but makes little sense semantically.
- "The cook put cherry stones in the cake" is semantically correct but raises contextual questions.
- "The glass broke its leg" is syntactically correct but nonsensical semantically.
- "I no go rain" has lexical issues due to incorrect word choice.

Each sentence's classification helps us understand where it fails or succeeds in terms of language processing.


Question:
With respect to reorder buffers, Would you expect to find the memory address where a particular
  instruction was fetched (i.e., the value of the PC at the time of fetching) inside the reorder buffer? If so, why would it be there? If not, elaborate on why it would it be unneeded.

### Understanding the Role of Reorder Buffers

A reorder buffer (ROB) is a key component in modern out-of-order execution processors, primarily designed to maintain the in-order completion of instructions and ensure coherent state in the presence of exceptions. To understand whether the memory address where a particular instruction was fetched (the Program Counter, or PC) should be stored in the ROB, we must analyze the role of the PC and the architecture of the ROB.

### Function of the Program Counter (PC)

The Program Counter (PC) is a register that holds the address of the next instruction to be executed. During instruction fetching, the PC is incremented to point to the subsequent instruction, leading to a sequence of instruction execution. When an instruction is dispatched from the instruction queue to the execution units, its associated PC value is critical for several reasons:

1. **Exception Handling**: If an exception occurs during the execution of an instruction, the processor must know the exact instruction that caused the exception. This information is crucial for debugging and for the operating system to handle the exception properly.

2. **State Restoration**: If the processor must revert to a previous state due to an exception or a misspeculation, the PC allows the processor to restore the correct execution flow.

### The Role of the Reorder Buffer

The reorder buffer serves the primary purpose of allowing instructions to execute out of order while ensuring they commit in the original program order. Each entry in the ROB typically holds:

- Instruction status (pending, completed)
- Destination register
- Value to be written back to the register file
- Exception flags (if any)

### Need for Storing the PC in the ROB

Given the above roles of the PC, we can evaluate the necessity of storing the PC in the ROB:

1. **Exception Lifecycle**: If an exception is raised after the execution of an instruction, the current PC would reflect the next instruction to be fetched, which may not correspond to the instruction that caused the exception. Thus, storing the PC at the time of instruction fetch is essential to handle exceptions effectively.

2. **Correctness of State**: When an exception occurs, the processor must know the correct instruction address to access the appropriate exception handler. This is crucial in architectures that support precise exceptions, where the state of the program must be exactly as it was at the time of the instruction that caused the exception.

3. **Instruction Completion**: In case of mis-speculation or incorrect execution paths, having the original PC stored allows the processor to recover correctly and maintain the integrity of the program state.

### Conclusion

In conclusion, storing the PC value in the reorder buffer is indeed necessary for effective exception handling and maintaining the correctness of program execution. The ROB not only tracks the status and results of executed instructions but also ensures that the architecture can accurately revert to the correct execution state when exceptions arise. Thus, the presence of the PC value in the ROB is justified as it supports the overall reliability and correctness of modern out-of-order processors.


Question: Tick the \emph{correct} assertion. The maximum advantage of an \textbf{adaptive} distinguisher limited to $q$ queries between two random functions $F$ and $F^*$ is always\dots

Certainly! Let’s break down each of the options methodically, ensuring a logical progression toward the final answer, which will be clearly stated at the end.

### Background Concepts

1. **Adaptive Distinguisher**: An adaptive distinguisher can make queries to a function based on the results of previous queries, thereby adapting its strategy dynamically.

2. **Advantage of a Distinguisher**: The advantage quantifies how well a distinguisher can differentiate between two distributions (here, the outputs of functions \( F \) and \( F^* \)):
   \[
   \text{Adv}(D) = \left| \Pr[D(F) = 1] - \Pr[D(F^*) = 1] \right|
   \]

3. **Norms**:
   - **\( L^\infty \) Norm**: \( |||[F]^q - [F^*]^q|||_{\infty} \) represents the maximum absolute difference in probability distributions, highlighting the worst-case distinguishability.
   - **\( L^a \) Norm**: \( |||[F]^q - [F^*]^q|||_{a} \) generally represents a more averaged measure but may not capture peak distinguishability effectively.

### Evaluation of Options

#### **Option 1: \( \frac{1}{2}|||[F]^q - [F^*]^q |||_{\infty} \)**

- **Analysis**: This option implies that the maximum advantage of the adaptive distinguisher is proportional to half the maximum difference in the output distributions. Given that adaptive distinguishers can exploit the worst-case differences effectively, this is a reasonable assertion.
- **Conclusion**: This option is likely correct as it aligns with the expected behavior of distinguishers.

#### **Option 2: \( \frac{1}{2}|||[F]^q - [F^*]^q |||_{a} \)**

- **Analysis**: This option suggests the advantage is based on an \( L^a \) norm. While this reflects a potential measure of the advantage, the \( L^a \) norm does not capture the maximum distinguishability as effectively as the \( L^\infty \) norm. It may not universally apply to all scenarios.
- **Conclusion**: This option is less likely to be correct.

#### **Option 3: 1 when \( F = F^* \)**

- **Analysis**: This option correctly states that if \( F \) and \( F^* \) are the same, a perfect distinguisher would yield an advantage of 1 (100% certainty). However, this is a specific scenario and does not address the general case for all distinguishers.
- **Conclusion**: While true, it does not represent the maximum advantage in general cases.

#### **Option 4: Lower than the advantage of the best non-adaptive distinguisher.**

- **Analysis**: This option claims that the adaptive distinguisher's advantage is lower than that of a non-adaptive one. This is incorrect, as adaptive distinguishers are typically more powerful than non-adaptive ones due to their ability to make informed decisions based on earlier outcomes.
- **Conclusion**: This assertion is false.

### Final Conclusion

After carefully analyzing each option based on definitions, logical reasoning, and the behavior of distinguishers, the most accurate assertion regarding the maximum advantage of an adaptive distinguisher limited to \( q \) queries between two random functions \( F \) and \( F^* \) is:

**Final Answer**: **Option 1: \( \frac{1}{2}|||[F]^q - [F^*]^q |||_{\infty} \)**.


Question: For which document classifier the training cost is low and inference is expensive?

To determine which document classifier has a low training cost and high inference cost, let's analyze each option in detail:

1. **k-Nearest Neighbors (kNN)**:
   - **Training Cost**: kNN does not require a traditional training phase; it simply stores the entire dataset for use during inference. Therefore, the training cost is effectively O(1), since there is no actual parameter learning involved.
   - **Inference Cost**: The inference cost for kNN is high because each prediction requires calculating the distance from the query point to all training samples. This results in a linear growth in inference time with respect to the number of training samples (O(n)), making it costly as the dataset size increases.

2. **Naive Bayes (NB)**:
   - **Training Cost**: Naive Bayes has a low training cost, as it involves estimating probabilities from the training data. While it does require some computation to calculate these probabilities, it is generally much faster than other models. Compared to kNN, which does not involve any learning, Naive Bayes’ training cost is still low but not zero.
   - **Inference Cost**: The inference cost for Naive Bayes is also low, as it simply computes probabilities based on the learned parameters. Thus, it does not align with having high inference costs.

3. **FastText**:
   - **Training Cost**: FastText typically has a moderate training cost, as it involves training word embeddings and classifiers. While it is faster than many deep learning models, it still requires more resources than Naive Bayes.
   - **Inference Cost**: Inference with FastText is relatively efficient, as it uses the trained embeddings for quick predictions. However, the inference cost is not as high as that of kNN, which requires extensive distance calculations.

4. **None**:
   - This option suggests that no classifiers fit the criteria of low training cost and high inference cost.

### Conclusion:
Among the classifiers analyzed, kNN is the only one that has a low training cost (effectively zero) and a high inference cost that grows linearly with the dataset size. This clearly fits the criteria specified in the question.

Selected option: kNN


Question: Assume a user program executes following tasks. Select all options that will use a system call.

Upon analyzing the options presented, we can identify which tasks involve system calls based on their interaction with the operating system. System calls are essential for requesting services from the kernel, which manages hardware resources and provides various functionalities to user programs.

A. Reading the user's input 'Hello world' from the keyboard: This action requires a system call to interact with the input device (keyboard). Hence, this option involves a system call.

B. Writing 'Hello world' to a file: This operation necessitates a system call to access the file system for writing data, which is another direct interaction with the operating system. Therefore, this option also involves a system call.

C. Encrypting 'Hello world' by AES: This process is purely computational and does not require interaction with the operating system or any hardware resources, making it independent of system calls. Thus, this option does not involve a system call.

D. Sending 'Hello world' to another machine via Network Interface Card: This operation involves network communication, which requires a system call to manage the sending of data over the network. Consequently, this option includes a system call.

In conclusion, the correct answers are A, B, and D, as they all necessitate system calls to perform their respective tasks.


Question: Let $n=pq$ where $p$ and $q$ are prime numbers. We have:

To derive the answer, we need to understand what the Euler's totient function \( \varphi(n) \) represents. The totient function counts the number of integers up to \( n \) that are relatively prime to \( n \). For a number \( n \) that is the product of two distinct prime numbers \( p \) and \( q \), the formula for the totient function is:

\[
\varphi(n) = n \left(1 - \frac{1}{p}\right)\left(1 - \frac{1}{q}\right)
\]

Substituting \( n = pq \) into this formula gives:

\[
\varphi(n) = pq \left(1 - \frac{1}{p}\right)\left(1 - \frac{1}{q}\right) = pq \cdot \frac{(p-1)}{p} \cdot \frac{(q-1)}{q} = (p-1)(q-1)
\]

This means that \( \varphi(n) = (p-1)(q-1) \) accurately represents the count of integers less than \( n \) that are coprime to \( n \).

Thus, the correct answer is the fourth option: \( \varphi(n) = (p-1)(q-1) \).


Question: Given that $100000000003$ is prime, what is the cardinality of $\mathbf{Z}_{200000000006}^*$?

To find the cardinality of \\(\\mathbf{Z}_{200000000006}^*\\), we need to determine the number of integers that are relatively prime to \\(200000000006\\) in the range from \\(0\\) to \\(200000000006 - 1\\).

First, we can factor \\(200000000006\\):
\\[
200000000006 = 2 \\times 100000000003
\\]
where \\(100000000003\\) is given to be prime.

Next, we can apply Euler's Totient Function \\(\\phi(n)\\) for \\(n = p \\times q\\) where \\(p\\) and \\(q\\) are distinct primes:
\\[
\\phi(n) = n \\left(1 - \\frac{1}{p}\\right)\\left(1 - \\frac{1}{q}\\right)
\\]
In our case, we have:
- \\(p = 2\\)
- \\(q = 100000000003\\)

Thus, we can compute \\(\\phi(200000000006)\\):
\\[
\\phi(200000000006) = 200000000006 \\left(1 - \\frac{1}{2}\\right)\\left(1 - \\frac{1}{100000000003}\\right)
\\]
Calculating each part:
1. The first part:
   \\[
   200000000006 \\times \\left(1 - \\frac{1}{2}\\right) = 200000000006 \\times \\frac{1}{2} = 100000000003
   \\]
2. The second part:
   \\[
   100000000003 \\times \\left(1 - \\frac{1}{100000000003}\\right) = 100000000003 \\times \\frac{100000000002}{100000000003} = 100000000002
   \\]
Thus, combining these results:
\\[
\\phi(200000000006) = 100000000003 \\times \\frac{100000000002}{100000000003} = 100000000002
\\]

Therefore, the cardinality of \\(\\mathbf{Z}_{200000000006}^*\\) is \\(100000000002\\).

So, the answer is:
 The cardinality of \\(\\mathbf{Z}_{200000000006}^*\\) is \\(100000000002\\). The correct answer is option 2.


Question: Tick the \textit{incorrect} assertion. Let $P, V$ be an interactive system for a language $L\in \mathcal{NP}$.

To determine which assertion is incorrect, let's analyze each option in the context of interactive proof systems, particularly focusing on the properties of soundness and zero-knowledge.

1. **Assertion 1**: "The proof system is \( \beta \)-sound if \( \Pr[\text{Out}_{V}(P^* \xleftrightarrow{x} V) = \text{accept}] \leq \beta \) for any \( P^* \) and any \( x \notin L \)."
   - This statement correctly describes the definition of \( \beta \)-soundness in interactive proof systems. It states that for any prover \( P^* \) that does not have a valid witness for \( x \) (where \( x \notin L \)), the probability that the verifier accepts should be at most \( \beta \). This assertion is true.

2. **Assertion 2**: "The soundness of the proof system can always be tuned close to \( 0 \) by sequential composition."
   - This assertion is misleading. While it is true that soundness can be reduced through repeated executions (sequential composition), it cannot be made arbitrarily close to \( 0 \) without compromising the integrity of the proof system. In fact, soundness must remain above a certain threshold to ensure security. Therefore, this assertion is also true but requires careful interpretation.

3. **Assertion 3**: "It is impossible for the proof system to be sound and zero knowledge at the same time."
   - This assertion is incorrect. It is indeed possible for a proof system to be both sound and zero-knowledge. Zero-knowledge proofs are designed to ensure that no information about the witness is leaked to the verifier, while soundness guarantees that only valid statements can be accepted. There are many examples of proof systems that achieve both properties simultaneously, such as the interactive proofs for NP-complete problems.

4. **Assertion 4**: "Both the verifier \( V \) and the prover \( P \) run in time that is polynomial in \( |x| \), if we assume that \( P \) gets the witness \( w \) as an extra input."
   - This statement is true. In an interactive proof system for languages in NP, both the prover and verifier are required to run in polynomial time relative to the size of the input \( x \) and the witness \( w \). This is a fundamental property of NP.

Based on this analysis, the incorrect assertion is indeed **Assertion 3**, which claims that it is impossible for a proof system to be sound and zero-knowledge at the same time. This is a misconception, as both properties can coexist in well-designed proof systems.


Question:
Assume the company you're working in recently hired a new CEO, who wants to improve development using modern methods. However, this CEO does not have an engineering background, so his suggestions are well-intentioned but not always feasible. The CEO comes to you with a new suggestion:
> Continuous integration is a good practice. We must adopt it, so that our code never has bugs.
> All branches in all our repositories must use continuous integration, and at the end of each day all branches must pass continuous integration.

Explain to the CEO why his goal is not realistic

Certainly! Here’s a more precise and logically valid response to the CEO's suggestion regarding continuous integration:

---

**Dear [CEO's Name],**

Thank you for your commitment to improving our development practices through the adoption of continuous integration (CI). While CI is indeed a powerful methodology for enhancing software quality and reducing bugs, the specific goal of having all branches in all repositories pass CI at the end of each day is not feasible for several reasons:

1. **Development Workflow Diversity**:
   - **Branch Purpose**: Different branches serve different purposes (e.g., feature development, bug fixes, experiments). Many branches may contain work-in-progress code that is not ready for integration. Expecting all branches to pass CI daily can lead to a forced integration of incomplete or unstable features.
   - **Merge Conflicts**: Frequent integration of multiple branches can create merge conflicts, particularly in larger teams. This can lead to increased complexity and frustration among developers, ultimately slowing down progress.

2. **CI Pipeline Complexity**:
   - **Configuration Variability**: Each branch may have unique dependencies and configurations. A uniform CI setup for all branches may not account for these variations, leading to failures in the CI pipeline that do not reflect the quality of the code.
   - **Test Scope**: The scope and duration of tests can vary significantly. Some branches may require extensive testing (e.g., integration tests, performance tests) that cannot realistically be executed daily without impacting developer productivity.

3. **Resource Constraints**:
   - **Infrastructure Limitations**: Running CI processes for all branches at the end of each day can overwhelm our CI infrastructure. This could lead to longer build times and delays in feedback, which counteracts the benefits of CI.
   - **Cost Considerations**: Additional resources (e.g., more build agents or cloud resources) may be required to accommodate daily CI for all branches, increasing operational costs.

4. **Quality vs. Speed**:
   - **Pressure to Merge**: A daily requirement may inadvertently pressure developers to merge code that is not fully tested or ready, leading to technical debt and potential regressions in the codebase. The goal should be to prioritize code quality over the frequency of integration.
   - **Stability Concerns**: Frequent integrations without proper readiness can lead to unstable builds, which can negatively affect the team's productivity and confidence in the CI process.

5. **Team Dynamics**:
   - **Pacing Variability**: Development teams and individual developers work at different paces. Some may require more time to refine their code before merging. A rigid daily requirement can create stress and reduce morale, deviating from our collaborative culture.

6. **Incremental and Strategic Adoption**:
   - **Focusing on Key Branches**: A more effective strategy would be to implement CI for key branches, such as the main or develop branches. This allows for stable integration while providing flexibility for feature branches to evolve without the same constraints.
   - **Gradual Implementation**: Introducing CI in stages allows teams to adapt, learn, and optimize their processes. This incremental approach can lead to better long-term results and a more successful transition to CI practices.

In conclusion, while the aspiration to have all branches pass CI daily reflects a strong desire for quality, the realities of software development necessitate a more balanced approach. I recommend we develop a phased CI strategy that prioritizes quality and accommodates our team's workflow. This can help us achieve the desired improvements in our development practices without compromising our efficiency or morale.

Best regards,
[Your Name]
[Your Position]

---

This response provides a structured, logically valid argument that addresses the CEO's suggestion while highlighting the importance of flexibility and quality in the development process.


Question: Which of the following is wrong regarding Ontologies?

The correct answer is "Ontologies dictate how semi-structured data are serialized."

To understand why this statement is incorrect, we need to clarify the role of ontologies. Ontologies are frameworks for representing knowledge within a specific domain, defining the concepts and relationships between them. They facilitate data integration and support domain-specific vocabularies, allowing different systems to understand and share information effectively. However, ontologies do not dictate how semi-structured data, such as JSON or XML, are serialized. Serialization refers to the process of converting data into a specific format for storage or transmission, which is determined by the data format itself rather than the ontology. Thus, the other options correctly describe the capabilities of ontologies.


Question: Select \emph{incorrect} statement. The exhaustive search

Answer: 4. runs in time polynomial in the length of the key.


Question: We consider now the ridge regression problem: $$ \min _{\mathbf{w} \in \mathbb{R}^{d}} \frac{1}{2 N} \sum_{n=1}^{N}\left[y_{n}-\mathbf{x}_{n}^{\top} \mathbf{w}\right]^{2}+\lambda\|\mathbf{w}\|_{2}^{2}, $$ where the data $\left\{\left(\mathbf{x}_{n}, y_{n}\right)\right\}_{n=1}^{N}$ are such that the feature vector $\mathbf{x}_{n} \in \mathbb{R}^{D}$ and the response variable $y_{n} \in \mathbb{R}$ Compute the closed-form solution $\mathbf{w}_{\text {ridge }}^{\star}$ of this problem, providing the required justifications. State the final result using the data matrix $\mathbf{X} \in \mathbb{R}^{N \times D}$.

To solve the ridge regression problem, we start with the objective function:

\[
\min _{\mathbf{w} \in \mathbb{R}^{d}} \frac{1}{2 N} \sum_{n=1}^{N}\left[y_{n}-\mathbf{x}_{n}^{\top} \mathbf{w}\right]^{2}+\lambda\|\mathbf{w}\|_{2}^{2}.
\]

### Reasoning Path 1: Gradient Descent Approach

1. **Objective Function**: Rewrite the objective function for clarity:
   \[
   J(\mathbf{w}) = \frac{1}{2 N} \sum_{n=1}^{N} (y_n - \mathbf{x}_n^\top \mathbf{w})^2 + \lambda \|\mathbf{w}\|_2^2.
   \]

2. **Gradient Calculation**: Compute the gradient of \( J(\mathbf{w}) \):
   \[
   \nabla J(\mathbf{w}) = -\frac{1}{N} \sum_{n=1}^{N} (y_n - \mathbf{x}_n^\top \mathbf{w}) \mathbf{x}_n + 2\lambda \mathbf{w}.
   \]

3. **Setting the Gradient to Zero**: To find the minimum, set the gradient to zero:
   \[
   -\frac{1}{N} \sum_{n=1}^{N} (y_n - \mathbf{x}_n^\top \mathbf{w}) \mathbf{x}_n + 2\lambda \mathbf{w} = 0.
   \]

4. **Rearranging**: Rearranging gives:
   \[
   \frac{1}{N} \sum_{n=1}^{N} (y_n - \mathbf{x}_n^\top \mathbf{w}) \mathbf{x}_n = 2\lambda \mathbf{w}.
   \]

5. **Matrix Formulation**: Define \( \mathbf{X} \) as the data matrix where each row corresponds to \( \mathbf{x}_n^\top \) and \( \mathbf{y} \) as the vector of responses. The above equation can be expressed in matrix form as:
   \[
   \mathbf{X}^\top \mathbf{X} \mathbf{w} = \lambda N \mathbf{w} + \mathbf{X}^\top \mathbf{y}.
   \]

### Reasoning Path 2: Normal Equations Approach

1. **Formulate the Problem**: The ridge regression problem can also be viewed as a regularized linear regression problem. The normal equation for ordinary least squares is:
   \[
   \mathbf{X}^\top \mathbf{X} \mathbf{w} = \mathbf{X}^\top \mathbf{y}.
   \]

2. **Adding Regularization**: In ridge regression, we add a regularization term:
   \[
   \mathbf{X}^\top \mathbf{X} \mathbf{w} + \lambda N \mathbf{w} = \mathbf{X}^\top \mathbf{y}.
   \]

3. **Rearranging**: This can be rearranged as:
   \[
   (\mathbf{X}^\top \mathbf{X} + \lambda N \mathbf{I}) \mathbf{w} = \mathbf{X}^\top \mathbf{y},
   \]
   where \( \mathbf{I} \) is the identity matrix of appropriate size.

### Final Reconciliation

From both reasoning paths, we arrive at the same conclusion regarding the closed-form solution of the ridge regression problem. The solution can be expressed as:

\[
\mathbf{w}_{\text{ridge}}^{\star} = (\mathbf{X}^\top \mathbf{X} + \lambda N \mathbf{I})^{-1} \mathbf{X}^\top \mathbf{y}.
\]

This formula provides the weights \( \mathbf{w}_{\text{ridge}}^{\star} \) that minimize the ridge regression objective function, incorporating both the data matrix \( \mathbf{X} \) and the regularization parameter \( \lambda \).


Question:
If process i fails, then eventually all processes j≠i fail
Is the following true? If some process j≠i does not fail, then process i has failed

Answer: The statement is false. If some process j≠i does not fail, it does not imply that process i has failed. The failure of one process does not guarantee the failure of all other processes, as there can be independent failures in a distributed system.

Reasoning: The statement suggests a cascading failure where the failure of one process leads to the failure of all others. However, in a distributed system, processes can operate independently; thus, the existence of a functioning process j indicates that process i may still be operational. Therefore, the failure of process i cannot be concluded based solely on the status of process j.


Question: Implement the modularity metric for communities.

To implement the modularity metric for communities, we need to understand both the concept of modularity and how to compute it based on a given network (graph) and its community structure. Here’s a step-by-step breakdown:

### Step 1: Understand Modularity

Modularity is a measure used to evaluate the structure of networks, especially how well-defined communities are within the graph. It quantifies the difference between the actual number of edges within communities and the expected number of edges based on a random graph that preserves node degrees.

The modularity \( Q \) is defined as:

\[
Q = \frac{1}{2m} \sum_{i=1}^{c} \left( e_i - \frac{k_i^2}{2m} \right)
\]

Where:
- \( m \) is the total number of edges in the graph.
- \( c \) is the number of communities.
- \( e_i \) is the number of edges within community \( i \).
- \( k_i \) is the total degree of nodes in community \( i \).

### Step 2: Construct the Graph

To compute modularity, we first need a representation of the graph. This can be done using an adjacency list or adjacency matrix. For example, we can represent a graph with a dictionary of lists in Python, where keys are nodes and values are lists of connected nodes.

```python
graph = {
    'A': ['B', 'C'],
    'B': ['A', 'D'],
    'C': ['A'],
    'D': ['B', 'E'],
    'E': ['D']
}
```

### Step 3: Define Communities

Communities can be represented as lists of nodes. For example:

```python
communities = [
    ['A', 'B', 'C'],  # Community 1
    ['D', 'E']        # Community 2
]
```

### Step 4: Calculate Total Edges (m)

Count the total number of edges \( m \) in the graph. This can be done by iterating over the adjacency list and counting the edges.

```python
def calculate_total_edges(graph):
    return sum(len(neighbors) for neighbors in graph.values()) // 2

m = calculate_total_edges(graph)  # Total edges
```

### Step 5: Calculate \( e_i \) and \( k_i \)

For each community, we need to count \( e_i \) (the number of edges within the community) and \( k_i \) (the sum of the degrees of the nodes in the community).

```python
def calculate_modularity(graph, communities):
    total_edges = calculate_total_edges(graph)
    Q = 0

    for community in communities:
        e_i = 0
        k_i = 0

        # Calculate k_i
        for node in community:
            k_i += len(graph[node])

        # Calculate e_i (edges within the community)
        for node in community:
            for neighbor in graph[node]:
                if neighbor in community:
                    e_i += 1  # Count each edge twice (A-B and B-A)

        e_i //= 2  # Since edges are counted twice

        # Update the modularity
        Q += (e_i / (2 * total_edges)) - (k_i ** 2) / (4 * total_edges ** 2)

    return Q

modularity_score = calculate_modularity(graph, communities)
```

### Final Step: Return the Modularity Score

After executing the above function, we will have the modularity score that reflects the quality of the community structure.

### Final Answer

The implemented modularity metric for communities returns a modularity score that quantifies the strength of the community structure in the given graph. This score helps in assessing how well-defined the communities are, with higher values indicating stronger community structures. The final modularity score can be retrieved from the `modularity_score` variable.


Question: What kind of exceptions require the processor to implement them precisely? Why? Give three examples of such exceptions.

Answer: Nonterminating exceptions require precise implementation to ensure execution can continue from a well-defined state.
Explanation: Precise handling allows the processor to return to a known point in the code after the exception is dealt with, which is crucial for maintaining program correctness. Examples of such exceptions include I/O interrupts, TLB misses, and timer interrupts.


Question: What happens in our "Consensus-Based Total-Order Broadcast" algorithm, if the set of messages decided on by consensus is not sorted deterministically at all?

In a Consensus-Based Total-Order Broadcast algorithm, if the set of messages decided on by consensus is not sorted deterministically, it can lead to inconsistencies in message delivery across different nodes. Each node may receive messages in a different order, violating the total-order requirement, which ensures that all nodes agree on the same message order. This can cause issues in systems that rely on the correct sequence of operations, such as database transactions or state machine replication. To maintain total order, the algorithm must implement additional mechanisms to enforce a consistent ordering, such as timestamps or sequence numbers, to ensure that all nodes eventually agree on the same message order.


Question:
Current software is complex and often relies on external dependencies.
    What are the security implications?

Certainly! Here’s an improved and slightly more comprehensive analysis of the security implications of complex software that relies on external dependencies:

---

When analyzing the security implications of complex software that incorporates external dependencies, it's crucial to consider the following options:

1. **During the requirement phase of the secure development lifecycle, a developer must list all the required dependencies.**
   - This option underscores the importance of identifying and documenting all dependencies early on. A comprehensive inventory allows teams to monitor these dependencies for known vulnerabilities and facilitates timely updates or patches. By understanding what components are in use, developers can better manage risks associated with those libraries.

2. **It is necessary to extensively security test every executable on a system before putting it in production.**
   - This option emphasizes the necessity of rigorous security testing prior to deployment. Conducting thorough security assessments helps identify vulnerabilities that could be exploited if left unchecked. This practice is essential in a secure development lifecycle, especially for software that relies heavily on external components, ensuring that security issues are addressed before the software is live.

3. **As most third-party software is open source, it is safe by default since many people reviewed it.**
   - This statement is misleading. While open-source software benefits from community scrutiny, it is not inherently secure. Vulnerabilities can still exist, and mere visibility does not equate to safety. Developers must actively perform security assessments and stay informed about potential vulnerabilities in the libraries they use to mitigate risks effectively.

4. **Closed source code is more secure than open source code as it prohibits other people from finding security bugs.**
   - This reflects a common misconception. Closed-source software is not necessarily more secure; vulnerabilities can remain hidden without community review. Relying on obscurity as a security measure can lead to a false sense of security. Effective security practices should focus on regular testing and audits, regardless of whether the code is open or closed source.

In conclusion, the most prudent approach to managing security risks associated with complex software and external dependencies is the first option: **"During the requirement phase of the secure development lifecycle, a developer must list all the required dependencies."** This proactive measure lays the foundation for effective risk management and security assurance throughout the development process.


Question: Which of the following cryptographic primitives have a security level that is significantly lower than 80 bits?

To determine which of the listed cryptographic primitives has a security level significantly lower than 80 bits, we need to analyze each option based on established security benchmarks.

1. **Symmetric Key Encryption with a Secret Key of 82 bits**: The security level of symmetric key encryption is directly related to the key length. A key length of 82 bits is considered secure against brute force attacks, as it requires \(2^{82}\) operations to break. This is above the 80-bit security threshold.

2. **RSA Signature Scheme with a 1613-bit Modulus**: The security of RSA is determined by the difficulty of factoring large numbers. A 1613-bit RSA modulus provides a security level that is generally considered to be around 80 bits, as it is believed that the best-known factoring attacks (such as the General Number Field Sieve) will require approximately \(2^{80}\) operations. Therefore, this is not significantly lower than 80 bits.

3. **ElGamal Cryptosystem over a Subgroup \(H \subset \mathbb{Z}_p^*\) with a 1613-bit Prime \(p\) and \(|H| \approx 2^{70}\)**: The security of the ElGamal cryptosystem is dependent on the size of the subgroup \(H\). If \(|H| \approx 2^{70}\), this indicates that the effective security level is around 70 bits. This is significantly lower than the 80-bit threshold.

4. **Hash Function with the Output of Size 163 bits**: The security of hash functions in terms of collision resistance is roughly half the output size. Therefore, a 163-bit hash function would provide a collision resistance level of approximately 81.5 bits, which is above the 80-bit threshold.

Based on this analysis, the cryptographic primitive with a security level significantly lower than 80 bits is:

**The ElGamal cryptosystem over a subgroup \(H \subset \mathbb{Z}_p^*\) with \(|H| \approx 2^{70}\)**.


Question: Let  $$G= egin{pmatrix} 1 &1 &1 &0 &1 &0\ 0 &1 &1 &1 &0 &0\ 0 &1 &1 &0 &0 &0\ 0 &1 &1 &1 &0 &1 \end{pmatrix}$$ be the generator matrix of a $(6,4)$ linear code $\mathcal C$ over $\mathbb F_2$. True or false: If one substitutes the last row of $G$ by $(1,0,0,1,1,1)$, the thereby obtained matrix generates the same code $\mathcal C$.

To determine whether substituting the last row of the generator matrix \( G \) with the vector \( (1, 0, 0, 1, 1, 1) \) generates the same code \( \mathcal{C} \), we need to analyze the implications of this substitution.

The original generator matrix \( G \) is given by:

\[
G = \begin{pmatrix}
1 & 1 & 1 & 0 & 1 & 0 \\
0 & 1 & 1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 & 1
\end{pmatrix}
\]

The code \( \mathcal{C} \) is generated by the linear combinations of the rows of \( G \). The last row of \( G \) is \( (0, 1, 1, 1, 0, 1) \). We need to check if the new row \( (1, 0, 0, 1, 1, 1) \) can be expressed as a linear combination of the other rows of \( G \).

To substitute the last row, we can represent the change in the generator matrix as:

\[
G' = \begin{pmatrix}
1 & 1 & 1 & 0 & 1 & 0 \\
0 & 1 & 1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 1
\end{pmatrix}
\]

Next, we need to check if the new row can be derived from the existing rows. If we perform row operations and combinations of the first three rows, we can manipulate the rows to see if we can obtain the new last row.

By observing the rows of the original matrix \( G \), we note that the last row \( (0, 1, 1, 1, 0, 1) \) can indeed be expressed in relation to the first three rows. However, the key point is that the new row \( (1, 0, 0, 1, 1, 1) \) introduces new combinations that are not dependent solely on the previous rows.

After examining the linear independence of the rows, we can establish that the new row \( (1, 0, 0, 1, 1, 1) \) is linearly independent from the first three rows. This means that while both generator matrices produce valid codewords, they may span different codes.

However, the critical insight here is that replacing the last row with \( (1, 0, 0, 1, 1, 1) \) does not change the linear combinations available to produce codewords. This is because the new row can be generated as a combination of the original rows, thus still allowing the generation of all codewords from \( \mathcal{C} \).

Therefore, the correct answer is **True**: substituting the last row by \( (1, 0, 0, 1, 1, 1) \) does generate the same code \( \mathcal{C} \) as the original matrix \( G \).


Question: Select the \emph{incorrect} statement: hash functions can be used to construct

To determine which statement is incorrect regarding the use of hash functions in constructing cryptographic primitives, let's analyze each option carefully.

### Option Analysis

1. **Commitment schemes**:
   - This statement is **correct**. Hash functions can be used to construct commitment schemes. A common method involves hashing a message together with a random nonce (salt) to create a commitment. This commitment can later be opened by revealing the message and nonce, allowing verification without revealing the message initially.

2. **Key derivation functions (KDFs)**:
   - This statement is **correct**. Hash functions are often used in key derivation functions to derive cryptographic keys from a base key or password. A well-known example is PBKDF2, which applies a hash function multiple times to increase computational difficulty and enhance security.

3. **Message authentication codes (MACs)**:
   - This statement is **correct**. Hash functions can be used to construct message authentication codes. A common construction is HMAC (Hash-based Message Authentication Code), which combines a cryptographic hash function with a secret key to provide message integrity and authenticity.

4. **Public key cryptosystems**:
   - This statement is **incorrect**. Hash functions are not used to construct public key cryptosystems directly. Public key cryptosystems are based on mathematical problems that are hard to solve (e.g., factoring large integers in RSA or the discrete logarithm problem in ElGamal). While hash functions can be used within public key systems (e.g., for signing or hashing messages), they do not form the basis for a public key cryptosystem itself.

### Conclusion

Based on the analysis:

- **Option 1**: Correct (hash functions can construct commitment schemes).
- **Option 2**: Correct (hash functions can construct key derivation functions).
- **Option 3**: Correct (hash functions can construct message authentication codes).
- **Option 4**: **Incorrect** (hash functions do not construct public key cryptosystems).

The **incorrect statement** is:

- **Option 4**: **public key cryptosystems**.


Question: The training loss of the 1-nearest neighbor classifier is always zero.

To analyze the question regarding the training loss of the 1-nearest neighbor (1-NN) classifier, let's break down the concepts involved:

1. **Understanding 1-NN Classifier**:
   - The 1-nearest neighbor classifier is a type of instance-based learning algorithm. It classifies a data point based on the class of the nearest point in the training dataset.

2. **Training Loss**:
   - The training loss typically refers to the error rate or the cost associated with the predictions made on the training dataset. For classification tasks, this is often represented as the proportion of incorrectly classified instances.

3. **How 1-NN Classifier Works on Training Data**:
   - When a 1-NN classifier is used on the training dataset, for any data point in the training set, the closest point (the nearest neighbor) is itself. Therefore, the classifier will always predict the correct class for each training sample because it has access to the exact data point.

4. **Conclusion on Training Loss**:
   - Since the 1-NN classifier correctly classifies every instance in the training data, the training loss is indeed zero. There are no misclassifications, leading to a training loss of zero.

Now, let’s analyze the options:

- **Option 0: TRUE** - This option states that the training loss of the 1-nearest neighbor classifier is always zero. Based on our analysis, this is correct because the classifier will perfectly classify all training data.

- **Option 1: FALSE** - This option claims that the training loss is not always zero. However, since the 1-NN classifier perfectly predicts the class of each training instance, this statement is incorrect.

Therefore, the correct answer is:

**Option 0: TRUE**.


Question: For a bipartite graph, devise an efficient algorithm for finding an augmenting path $P$ (if one exists). What is the total running time of the \textsc{AugmentingPathAlgorithm} explained in the second lecture?

To find an augmenting path \( P \) in a bipartite graph, we can utilize a breadth-first search (BFS) approach. The algorithm will explore the graph to find a path from a free vertex in one partition to a free vertex in the other partition, alternating between edges in the matching and edges not in the matching. Here’s a step-by-step outline of the algorithm:

1. **Initialization**: Start with a bipartite graph \( G = (U, V, E) \), where \( U \) and \( V \) are the two partitions of the graph. Also, maintain a matching \( M \).

2. **BFS Setup**: Create a queue to facilitate the BFS and initialize a parent map (or array) to keep track of the path.

3. **Find Free Vertex**: Enqueue all free vertices from partition \( U \) (those that are not matched in the current matching \( M \)).

4. **BFS Execution**:
   - While the queue is not empty, dequeue a vertex \( u \) from \( U \).
   - For each neighbor \( v \) of \( u \) in \( V \):
     - If \( v \) is not yet visited (i.e., not in the parent map), mark it as visited and record \( u \) as its parent.
     - If \( v \) is matched (i.e., \( (u', v) \in M \)), enqueue the matched vertex \( u' \) from \( U \) (the other side of the edge).
     - If \( v \) is free (not matched), we have found an augmenting path. Trace back the path using the parent map.

5. **Path Reconstruction**: If an augmenting path is found, reconstruct the path \( P \) using the parent references.

6. **Return**: If a path \( P \) is found, return it; otherwise, return that no augmenting path exists.

### Total Running Time

The running time of the above algorithm can be analyzed as follows:

- The BFS takes \( O(V + E) \) time, where \( V \) is the number of vertices and \( E \) is the number of edges in the bipartite graph.
- The algorithm primarily spends its time in exploring edges and vertices once, leading to a linear complexity relative to the size of the graph.

Thus, the total running time of the \textsc{AugmentingPathAlgorithm} is \( O(V + E) \).


Question:
Meltdown is a well-known attack on dynamically-scheduled
                processors which exploits the fact that loads may be executed
                speculatively before determining whether they represent a memory
                access violation. Intel Itanium has a speculative load instruction
                which allows a compiler to perform a load speculatively and needs a
                check instruction at a later point in the code to verify whether the
                load did in fact raise an exception. Would you imagine that a
                Meltdown attach on Itanium based on this instruction could be
                possible?  Explain clearly your reasoning.

To evaluate the possibility of a Meltdown attack on Intel Itanium using speculative load instructions, follow these steps:

1. **Understand Meltdown Attack**:
   - Meltdown takes advantage of out-of-order execution in CPUs, where a processor executes loads speculatively before checking permissions. If a violation occurs, the speculative execution can still reveal sensitive data.

2. **Analyze Speculative Load on Itanium**:
   - The Itanium architecture includes a speculative load instruction that allows loads to be executed before confirming memory access validity. This means that it can load data from memory even if the access rights have not been verified yet.

3. **Identify Exception Handling**:
   - Itanium requires a check instruction to verify if the speculative load raised an exception. This means there is a mechanism in place to handle invalid memory access post-speculative execution.

4. **Evaluate Exception Effects**:
   - If an exception occurs as a result of the speculative load, Itanium's architecture would prevent the speculative load results from being committed (i.e., they would not be observable outside the CPU).

5. **Consider Security Implications**:
   - The presence of a check instruction implies that the speculative loads must be verified, and any violation would lead to an exception, preventing the unauthorized access to sensitive data.

6. **Conclude Feasibility of Meltdown Attack**:
   - Given the speculative load instruction requires a later check for exceptions and that an error would prevent the data from being accessed, a Meltdown-style attack exploiting this behavior on Itanium would be significantly more challenging, if not impossible, compared to architectures without such protection.

In conclusion, while speculative loads exist on the Itanium architecture, the necessary checks and exception handling mechanisms would likely hinder the feasibility of a Meltdown attack.


Question:
If process i fails, then eventually all processes j≠i fail
Is the following true? If some process j≠i does not fail, nothing can be said about process i

Let’s break down the statement step by step. The condition given is: "If process i fails, then eventually all processes j≠i fail." This implies that the failure of process i leads to a cascade where all other processes j (except i) will eventually fail as well.

Now, let's analyze the second statement: "If some process j≠i does not fail, nothing can be said about process i." The key word here is "eventually." Even if we see that some process j≠i is currently functioning, it doesn't provide immediate information about process i. This is because process j≠i might still be in a temporary state of success and could eventually fail later due to process i's failure.

To visualize this, think of a domino setup where knocking over one domino (process i) will cause all the others (processes j≠i) to fall in time. However, if one domino is still standing, it doesn’t guarantee that the first domino hasn’t fallen; it just means that the effects of that failure have not reached every domino yet.

Thus, the statement "If some process j≠i does not fail, nothing can be said about process i" is **true**. We cannot definitively conclude the state of process i based solely on the status of process j≠i because of the eventual nature of the failure propagation.

Final answer: **True**. If some process j≠i does not fail, nothing can be said about process i due to the "eventually" condition.


Question:
Your team is discussing the following code:
/** Uploads images to the cloud. */
public final class ImageUploader {
    public void upload(Image image) { /* … */ }

    private boolean canUpload(Image image) { /* … */ }
}

One of your colleagues points out that "upload" currently has some unexpected behavior regarding file sizes, and suggests that this should be written down in a Google Doc shared with the team.
Give 1 sentence explaining why this is not a good idea and 1 sentence suggesting a better way to record this information:

Documenting unexpected behavior in a shared Google Doc may lead to miscommunication and lack of immediate visibility for critical issues. Instead, it would be more effective to create a dedicated issue in the project management tool (e.g., JIRA) to track and prioritize the upload file size behavior, ensuring it is addressed systematically.


Question: Which of the following methods does not exploit statistics on the co-occurrence of words in a text?

None of the methods listed exploit statistics on the co-occurrence of words in a text. All options—Word embeddings, Transformers, Vector space retrieval, and Fasttext—rely on statistical relationships in their functioning.


Question: Tick the \emph{correct} assertion concerning WPA2

WPA2 (Wi-Fi Protected Access II) is a security protocol used to secure wireless networks. Let's analyze the assertions one by one:

1. **WPA2 uses RC4**: This is incorrect. WPA2 primarily uses AES (Advanced Encryption Standard) for encryption, while RC4 was used in the earlier WPA (Wi-Fi Protected Access) standard.

2. **WPA2 uses AES**: This assertion is correct. WPA2 employs AES for encryption, providing a higher level of security compared to the older protocols that used RC4.

3. **WPA2 uses 64-bit keys**: This is incorrect. WPA2 typically uses 128-bit or 256-bit keys with AES encryption, not 64-bit keys.

4. **WPA2 is badly broken**: While there have been vulnerabilities discovered over time (such as the KRACK attack), it is not considered "badly broken" in a general sense. It is still widely used and considered secure when properly implemented.

Based on this analysis, the correct assertion concerning WPA2 is:

**WPA2 uses AES.**


Question:
Given a joint data distribution $\mathcal D$ on $\mathcal X 	imes \{-1,1\}$ and $n$ independent and identically distributed observations from $\mathcal D$, the goal of the classification task is to learn a classifier $f:\mathcal X 	o \{-1,1\}$ with minimum true risk $\mathcal L(f) = \mathbb E_{(X,Y)\sim \mathcal D} [oldsymbol{\mathbb{1}}_{f(X)
eq Y}]$ where $oldsymbol{\mathbb{1}}_{C} = egin{cases}
            1 \; 	ext{ if } C  	ext{ is true} \
            0 \quad 	ext{otherwise}
            \end{cases}$. %
            We denote by $\mathcal D_{X}$ the marginal law (probability distribution) of $X$, and $\mathcal D_{Y|X}$ the conditional law of $Y$ given $X$.
            Give the two reasons seen in the course which explain that minimizing the true risk with the $0-1$ loss over the set of classifiers $f:\mathcal X 	o \{-1,1\}$ is problematic.

Minimizing the true risk using the $0-1$ loss function poses challenges for two main reasons. First, the set of classifiers mapping from $\mathcal X$ to $\{-1,1\}$ is not convex due to the discrete nature of the output space. This lack of convexity implies that there may be multiple local minima, making optimization difficult. Second, the indicator function used in the $0-1$ loss is not continuous, which complicates the optimization process further, as standard gradient-based methods may not be effective in navigating the discontinuities in the loss landscape. This results in potential difficulties in finding a classifier that minimizes the risk accurately.


Question: Let $K = (K_1, K_2,..., K_n)$, where each $K_i$ is $0$ or $1$ with probability $1/2$. Let $K'=(K'_1, K'_2, ..., K'_n)$ such that, for each $i$, $K'_i \in {0, 1}$ and $K'_{i} = \sum_{j = 1}^i K_j 	ext{mod} 8.$ True or false: Using $K'$ as the key one can achieve perfect secrecy if the message is $n$ bits.

To determine if using \( K' \) allows us to achieve perfect secrecy for a message of \( n \) bits, we need to consider the definition of perfect secrecy and how \( K' \) is constructed.

### Perfect Secrecy
Perfect secrecy is achieved when the ciphertext does not reveal any information about the plaintext. For a key to provide perfect secrecy, it must be:
1. **Uniformly random**: Each key bit must be equally likely to be 0 or 1.
2. **Independent**: The key bits must be independent of each other.

### Construction of \( K' \)
The key \( K' = (K'_1, K'_2, \ldots, K'_n) \) is defined such that:
\[ K'_i = \left( \sum_{j=1}^{i} K_j \right) \mod 8. \]

Here, \( K_i \) are independent bits that are uniformly distributed over \( \{0, 1\} \).

### Analysis of \( K' \)
1. **Dependence**: While each \( K'_i \) is derived from a cumulative sum of previous bits, the values of \( K'_i \) (after taking mod 8) can still be uniformly distributed across their range, given that the original bits \( K_i \) are random and independent.
2. **Randomness**: The modulo operation can maintain a certain degree of randomness; however, it does introduce some correlation.

### Conclusion
Even though \( K' \) is derived from \( K \), the cumulative nature and the modulo operation do not inherently lead to a loss of uniform distribution for the generated bits in terms of achieving perfect secrecy. The \( K' \) bits can still be viewed as uniformly random over their range, as the sum will distribute across the possible values due to the randomness of \( K \).

### Final Answer
Given this understanding and the argument that \( K' \) can still meet the requirements for perfect secrecy despite the cumulative sum, the correct answer is:

**True.**

Using \( K' \) as the key can achieve perfect secrecy for the message of \( n \) bits.


Question:
A query \(q\) has been submitted to two distinct Information Retrieval engines operating on the same document collection containing 1'000 documents, with 50 documents being truly relevant for \(q\).The following result lists have been produced by the two IR engines, \(S_1\) and \(S_2\) respectively:

\(S_1\text{:}\)
\(d_1\)
\(d_2\text{ (*)}\)
\(d_3\text{ (*)}\)
\(d_4\)
\(d_5\text{ (*)}\)

\(S_2\text{:}\)
\(d^\prime_1\text{ (*)}\)
\(d^\prime_2\text{ (*)}\)
\(d^\prime_3\)
\(d^\prime_4\)
\(d^\prime_5\)

In these result lists, the stars \(\text{(*)}\) identify the truly relevant documents.
By convention, we consider that any non retrieved document has been retrieved at rank 6.If Average Precision is used as evaluation metric, which of the two IR engines is performing better for the query \(q\)?

To evaluate the performance of the two Information Retrieval (IR) engines \(S_1\) and \(S_2\) using Average Precision (AP), we need to understand how to calculate this metric.

**Average Precision (AP)** is defined as the average of the precision scores at each rank where a relevant document is retrieved. The formula for calculating Average Precision is:

\[
AP = \frac{1}{R} \sum_{k=1}^{n} P(k) \times rel(k)
\]

Where:
- \(R\) is the total number of relevant documents (50 in this case).
- \(P(k)\) is the precision at rank \(k\).
- \(rel(k)\) is an indicator function that is 1 if the document at rank \(k\) is relevant, and 0 otherwise.

### Evaluating \(S_1\):

The result list for \(S_1\) is:
1. \(d_1\) (not relevant)
2. \(d_2\) (relevant)
3. \(d_3\) (relevant)
4. \(d_4\) (not relevant)
5. \(d_5\) (relevant)

- At rank 1: Precision = 0/1 = 0 (not relevant)
- At rank 2: Precision = 1/2 = 0.5 (1 relevant)
- At rank 3: Precision = 2/3 ≈ 0.67 (2 relevants)
- At rank 4: Precision = 2/4 = 0.5 (still 2 relevants)
- At rank 5: Precision = 3/5 = 0.6 (3 relevants)

Relevant documents retrieved are at ranks 2, 3, and 5. The sum of precision at these ranks will be:

\[
AP(S_1) = \frac{1}{3} (0 + 0.5 + 0.67 + 0.5 + 0.6) = \frac{2.27}{3} ≈ 0.7567
\]

### Evaluating \(S_2\):

The result list for \(S_2\) is:
1. \(d^\prime_1\) (relevant)
2. \(d^\prime_2\) (relevant)
3. \(d^\prime_3\) (not relevant)
4. \(d^\prime_4\) (not relevant)
5. \(d^\prime_5\) (not relevant)

- At rank 1: Precision = 1/1 = 1 (1 relevant)
- At rank 2: Precision = 2/2 = 1 (2 relevants)
- At rank 3: Precision = 2/3 ≈ 0.67 (2 relevants)
- At rank 4: Precision = 2/4 = 0.5 (2 relevants)
- At rank 5: Precision = 2/5 = 0.4 (still 2 relevants)

Relevant documents retrieved are at ranks 1 and 2. The sum of precision at these ranks will be:

\[
AP(S_2) = \frac{1}{2} (1 + 1) = 1.0
\]

### Conclusion:
Comparing the Average Precision values:
- \(AP(S_1) ≈ 0.7567\)
- \(AP(S_2) = 1.0\)

Since \(AP(S_2)\) is higher than \(AP(S_1)\), \(S_2\) is performing better for the query \(q\).

Final Answer:
Answer: 2. \(S_2\)


Question:
Suppose you have a search engine that retrieves the top 100 documents and
achieves 90% precision and 20% recall. You modify the search engine to
retrieve the top 200 and mysteriously, the precision stays the same. Which one
is CORRECT?

To solve this problem, let's first clarify what precision and recall mean:

- **Precision** is the ratio of relevant documents retrieved to the total documents retrieved. It is defined as:
  \[
  \text{Precision} = \frac{\text{Relevant Documents Retrieved}}{\text{Total Documents Retrieved}}
  \]

- **Recall** is the ratio of relevant documents retrieved to the total relevant documents in the dataset. It is defined as:
  \[
  \text{Recall} = \frac{\text{Relevant Documents Retrieved}}{\text{Total Relevant Documents}}
  \]

Given the information:

1. When the search engine retrieves the top 100 documents, precision is 90% and recall is 20%.
2. This means that out of the top 100 documents retrieved, 90% are relevant. Therefore:
   - Relevant Documents Retrieved = 0.90 * 100 = 90
   - Recall = 20% means that these 90 documents represent 20% of all relevant documents.

From the recall equation, we can calculate the total number of relevant documents (R):
\[
0.20 = \frac{90}{R} \implies R = \frac{90}{0.20} = 450
\]

Now, we modify the search engine to retrieve the top 200 documents, and it’s stated that the precision stays the same at 90%.

Since precision remains the same, the number of relevant documents retrieved from the top 200 must also be calculated using the precision formula:
\[
\text{Relevant Documents Retrieved} = 0.90 \times 200 = 180
\]

Now let's analyze the options:

A. **The recall becomes 10%:**
   - Recall would actually become \( \frac{180}{450} = 40\% \), not 10%. So, this option is incorrect.

B. **The number of relevant documents is 450:**
   - We calculated this earlier and found it to be 450. This option is correct.

C. **The F-score stays the same:**
   - F-score is a harmonic mean of precision and recall. Since recall changes from 20% to 40%, the F-score will also change. So, this option is incorrect.

D. **This is not possible:**
   - Given the calculations above, it is possible, so this option is incorrect.

Thus, the correct answer is **B. The number of relevant documents is 450.**


Question: Let $n$ be an integer. The extended Euclidean algorithm is typically used to\dots

The extended Euclidean algorithm is primarily used to find the greatest common divisor (GCD) of two integers and to express this GCD as a linear combination of these integers. This property is particularly useful in modular arithmetic, especially in the context of computing the multiplicative inverse of an element in the group of units \( \mathbf{Z}_n^* \). Specifically, if we want to find the inverse of an integer \( a \) modulo \( n \), we need to find an integer \( b \) such that \( ab \equiv 1 \mod n \). The extended Euclidean algorithm can help us find such \( b \) when \( \gcd(a, n) = 1 \).

Given the options:
- A is incorrect because the extended Euclidean algorithm does not directly perform addition.
- B is correct as it relates to finding inverses in \( \mathbf{Z}_n^* \).
- C is incorrect; the algorithm does not compute squares.
- D is incorrect; while the order may relate to the structure of \( \mathbf{Z}_n^* \), the extended Euclidean algorithm does not compute the order.

Therefore, the correct answer is:

B

B


Question:
If process i fails, then eventually all processes j≠i fail

Is the following true? If a process j≠i fails, then process i has failed

The statement "If process jâ i fails, then process i has failed" is not necessarily true.

To understand this, we can analyze the implications of the first statement: "If process i fails, then eventually all processes jâ i fail." This implies a particular kind of failure dependency: the failure of process i leads to the failure of all other processes. However, this does not establish a symmetric relationship.

In distributed systems or fault-tolerant systems, it is entirely possible for processes to fail independently. The failure of one process (jâ i) does not imply that another process (i) has also failed. There can be scenarios where multiple processes operate independently, and the failure of one does not affect the others.

For example, consider a distributed system where processes communicate with each other but do not rely on a single process to function. If process j fails due to a network issue, it does not necessitate that process i also fails; process i might continue to operate normally.

Therefore, the validity of the first statement does not lead to the conclusion of the second statement. The failure of a process is not a direct cause of failure in another process unless a specific dependency is established. Thus, the answer is **false**.


Question: Assume that while working on a new feature for your team's product, your colleague is required to write a function that takes a list of events and sorts them by their timestamp. Using their algorithm course knowledge, they remind you that merge sort's complexity is $O(n log n)$, which is better than the $O(n^2)$ worst-case complexity of quick sort, and that they will therefore use the former to solve their task. What do you think of your colleague's approach?

Your colleague's choice to use merge sort is valid, especially given its guaranteed \(O(n \log n)\) performance in all cases, which can be beneficial if worst-case scenarios are a concern. However, it’s important to consider several factors. Merge sort requires additional space, leading to a space complexity of \(O(n)\), which might not be ideal in memory-constrained environments. Additionally, the overhead associated with merging can make it slower for smaller datasets, where quick sort might actually perform better due to lower constant factors and in-place sorting. Furthermore, if the timestamps are mostly sorted or if the dataset is small, quick sort could outperform merge sort. Therefore, while merge sort is a safe choice, evaluating the specific context and data characteristics is crucial before finalizing the decision.


Question: A Carmichael number $n$ ...

A Carmichael number is a composite number \( n \) that satisfies Fermat's little theorem for all integers \( b \) that are coprime to \( n \). This means:

1. **Fermat's Little Theorem**: For a prime \( p \), if \( \gcd(b, p) = 1 \), then \( b^{p-1} \equiv 1 \pmod{p} \). A Carmichael number behaves like a prime in the sense that it satisfies this condition for all coprime integers, even though it is not prime.

Now, let's analyze each option:

A. A Carmichael number is a prime number.
**False**. By definition, Carmichael numbers are composite.

B. A Carmichael number will always pass Fermat's test for any \( 0 < b < n \).
**True**. By the definition of Carmichael numbers, they satisfy the conditions of Fermat's test for all \( b \) coprime to \( n \).

C. A Carmichael number verifies that \( \forall b \), \( \mathsf{gcd}(b,n)=1 \) implies that \( b^{n-1} \equiv 1 \pmod{n} \).
**True**. This is essentially the definition of a Carmichael number; they satisfy this property for all \( b \) coprime to \( n \).

D. A Carmichael number will be considered as a prime by the Miller-Rabin algorithm.
**True**. The Miller-Rabin primality test will classify a Carmichael number as a probable prime because they pass the test like primes do.

In summary:
- A: False
- B: True
- C: True
- D: True


Question: Which of the following is correct regarding community detection?

The correct answer indicates two key concepts in community detection:

1. **High betweenness of an edge indicates that the communities are well connected by that edge**: Betweenness centrality measures how often an edge lies on the shortest path between two nodes. If an edge has high betweenness, it means that it plays a crucial role in connecting different communities. Thus, if an edge has high betweenness, it suggests that it connects different communities strongly.

2. **High modularity of a community indicates a large difference between the number of edges of the community and the number of edges of a null model**: Modularity is a measure that helps to quantify the structure of a network. High modularity indicates that there are significantly more edges within the communities than would be expected in a random network (the null model). This suggests that the communities are well-defined and that there is a strong internal connectivity relative to what would happen by chance.

The other options are incorrect:
- The Louvain algorithm aims to maximize modularity, not minimize it.
- The Girvan-Newman algorithm focuses on edge betweenness to detect communities, but it does not maximize the betweenness measure itself; instead, it removes edges with high betweenness to separate communities.

Thus, the answers you have selected correctly represent important concepts in community detection.


Question: Confidentiality means that:

Answer: information should not leak to any unexpected party.

Reasoning steps:

1. **Understanding Confidentiality**: Confidentiality is a fundamental principle of information security that ensures that sensitive information is only accessible to authorized individuals or parties. It aims to protect information from unauthorized access and disclosure.

2. **Analyzing the Options**:
   - **Option 1**: "the message can be read by anyone." - This is the opposite of confidentiality, as it implies that there are no restrictions on who can access the information. Therefore, this statement is **incorrect**.
   - **Option 2**: "information should not leak to any unexpected party." - This aligns with the definition of confidentiality. It emphasizes that information should remain private and should only be accessible to those who are authorized to view it. This statement is **correct**.
   - **Option 3**: "the message should make clear who the author is." - This pertains more to authenticity and non-repudiation rather than confidentiality. It does not directly relate to the protection of information from unauthorized access. Therefore, this statement is **incorrect**.
   - **Option 4**: "the information must be protected against any malicious modification." - This relates to integrity, which ensures that information remains unchanged and unaltered. While this is important in information security, it does not define confidentiality. Thus, this statement is **incorrect**.

Based on the analysis, the correct assertion regarding confidentiality is: **information should not leak to any unexpected party.**


Question: Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths:		egin{center}			egin{tabular}{ |c|c|c|c|c|c|c| } 				\hline				& $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ 				\hline				$l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ 				\hline			\end{tabular}		\end{center}  True or false: If $D=4$ then $\Gamma$ is necessarily uniquely-decodable.

To determine if the encoding \(\Gamma\) is necessarily uniquely-decodable given that the code alphabet size \(D=4\), we need to consider the properties of uniquely-decodable codes.

A code is uniquely decodable if no two distinct sequences of codewords can result in the same concatenated codeword. One fundamental condition for a uniquely-decodable code is that it satisfies the Kraft's inequality:

\[
\sum_{i} D^{-l_i} \leq 1
\]

where \(D\) is the size of the alphabet, and \(l_i\) are the lengths of the codewords corresponding to each symbol.

In our case, the codeword lengths for the symbols \(a, b, c, d, e, f\) are as follows:

- \(l(a) = 1\)
- \(l(b) = 1\)
- \(l(c) = 1\)
- \(l(d) = 2\)
- \(l(e) = 2\)
- \(l(f) = 4\)

We can now compute the terms for Kraft's inequality:

\[
\begin{align*}
\sum_{i} D^{-l_i} &= D^{-1} + D^{-1} + D^{-1} + D^{-2} + D^{-2} + D^{-4} \\
&= 3D^{-1} + 2D^{-2} + D^{-4}
\end{align*}
\]

Substituting \(D = 4\):

\[
\begin{align*}
\sum_{i} 4^{-l_i} &= 3 \cdot 4^{-1} + 2 \cdot 4^{-2} + 4^{-4} \\
&= 3 \cdot \frac{1}{4} + 2 \cdot \frac{1}{16} + \frac{1}{256} \\
&= \frac{3}{4} + \frac{2}{16} + \frac{1}{256} \\
&= \frac{3}{4} + \frac{1}{8} + \frac{1}{256}
\end{align*}
\]

Next, we convert these fractions to a common denominator:

- \(\frac{3}{4} = \frac{192}{256}\)
- \(\frac{1}{8} = \frac{32}{256}\)
- \(\frac{1}{256} = \frac{1}{256}\)

Now, adding these together:

\[
\frac{192}{256} + \frac{32}{256} + \frac{1}{256} = \frac{192 + 32 + 1}{256} = \frac{225}{256}
\]

We find that:

\[
\frac{225}{256} \leq 1
\]

Since \(\frac{225}{256} < 1\), Kraft's inequality is satisfied. However, this does not guarantee that the code is uniquely decodable; it only means that there is a chance it could be.

The lengths of the codewords must also be considered. Here, we have three symbols \(a, b, c\) that are each encoded using 1 bit, and two symbols \(d, e\) that are using 2 bits. The symbol \(f\) is using 4 bits.

Since there are three symbols (1-bit) that can be represented in a binary code, the 1-bit codewords could potentially collide with the 2-bit codewords or even the 4-bit codewords depending on how they are defined. Thus, without a detailed examination of the actual codewords (which we do not have), we cannot conclude uniquely decodable.

Thus, the statement "If \(D=4\) then \(\Gamma\) is necessarily uniquely-decodable" is:

**False**.


Question: Implement Item-based collaborative filtering using the following formula:  \begin{equation} {r}_{x}(a) =  \frac{\sum\limits_{b \in N_{I}(a)} sim(a, b) r_{x}(b)}{\sum\limits_{b \in N_{I}(a)}|sim(a, b)|} \end{equation}  You will create a function that takes as input the ratings and the similarity matrix and gives as output the predicted ratings.

### Explanation and Intuition

Item-based collaborative filtering is a recommendation technique that leverages the ratings of items to predict how a user might rate an item they haven’t rated yet. The key idea is to utilize the similarity between items to weight the contributions of other rated items when making predictions. This method assumes that if two items are similar, their ratings by a user will also reflect similar preferences.

#### Formula Breakdown
The formula provided for predicting user ratings is:

\[
{r}_{x}(a) =  \frac{\sum\limits_{b \in N_{I}(a)} sim(a, b) r_{x}(b)}{\sum\limits_{b \in N_{I}(a)}|sim(a, b)|}
\]

Where:
- \( r_x(a) \): Predicted rating for user \( x \) on item \( a \).
- \( N_I(a) \): Set of items similar to item \( a \).
- \( sim(a, b) \): Similarity score between items \( a \) and \( b \).
- \( r_x(b) \): Actual rating given by user \( x \) to item \( b \).

### Implementation Steps

1. **Input Representation**: The ratings matrix will be a 2D array where rows represent users and columns represent items. The similarity matrix will indicate how similar each item is to every other item.

2. **Iterate through Users and Items**: For each user, loop through each item. If the user has not rated the item, compute the predicted rating.

3. **Identify Similar Items**: For each unrated item, identify the items that the user has rated and that are also similar to the target item.

4. **Calculate Weighted Ratings**: Compute the weighted sum of the ratings from similar items, using their similarity scores as weights.

5. **Handle Edge Cases**: When there are no similar items rated by the user, assign a random rating within the defined range (e.g., a random integer from 1 to 5) to introduce variability in predictions.

### Implementation

Here’s the revised Python function that implements item-based collaborative filtering:

```python
import numpy as np

def predict_ratings(ratings_matrix, similarity_matrix):
    """
    Predict the ratings of items for users using item-based collaborative filtering.

    Parameters:
    ratings_matrix (numpy.ndarray): A 2D array where rows represent users
                                     and columns represent items.
    similarity_matrix (numpy.ndarray): A 2D array where the entry (i,j)
                                        represents the similarity between items i and j.

    Returns:
    numpy.ndarray: A 2D array of predicted ratings for each user-item pair.
    """
    num_users, num_items = ratings_matrix.shape
    predicted_ratings = np.zeros((num_users, num_items))

    # Iterate through each user
    for user in range(num_users):
        for item in range(num_items):
            # Check if the user has already rated the item
            if ratings_matrix[user, item] == 0:  # Assuming 0 means unrated
                # Get indices of items that user has rated
                rated_items_indices = np.where(ratings_matrix[user] != 0)[0]
                # Get indices of items that are similar to the current item
                similar_items_indices = np.nonzero(similarity_matrix[item])[0]

                # Find intersection of rated items and similar items
                relevant_items_indices = np.intersect1d(rated_items_indices, similar_items_indices)

                # Calculate the weighted sum of ratings for similar items
                if len(relevant_items_indices) > 0:
                    numerator = np.sum(similarity_matrix[item][relevant_items_indices] * ratings_matrix[user, relevant_items_indices])
                    denominator = np.sum(np.abs(similarity_matrix[item][relevant_items_indices]))

                    # Avoid division by zero
                    if denominator != 0:
                        predicted_ratings[user, item] = numerator / denominator
                else:
                    # Assign a random rating within a predefined range (e.g., 1 to 5)
                    predicted_ratings[user, item] = np.random.randint(1, 6)  # Assuming ratings are on a scale of 1 to 5

    return predicted_ratings
```

### Intuition Behind the Implementation
1. **Matrix Representation**: The user-item ratings and similarity relationships are structured in matrices, facilitating efficient data handling.

2. **Iterative Calculation**: For each user and each item, the algorithm checks for unrated items, ensuring that predictions are only made when necessary.

3. **Robust Similarity Handling**: By identifying relevant items that are both rated by the user and similar to the target item, we ensure that predictions are informed by actual user preferences.

4. **Random Ratings for Unrated Cases**: In scenarios where no similar items have been rated, assigning a random rating introduces variability, which can be beneficial for exploratory recommendations and helps to avoid deterministic outputs.

This approach enhances the flexibility and robustness of the item-based collaborative filtering algorithm, making it more adaptable to different user behaviors and data distributions.


Question: Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths:		egin{center}			egin{tabular}{ |c|c|c|c|c|c|c| } 				\hline				& $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ 				\hline				$l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ 				\hline			\end{tabular}		\end{center}  True or false: If $D=4$ then $\Gamma$ is necessarily prefix-free.

**Correct Answer: 2. False**

**Key Detail:**
The statement is false because a prefix-free code requires that no codeword is a prefix of any other codeword. In this case, the encoding $\Gamma$ has codeword lengths that can lead to conflicts. For instance, if we consider codewords like 'a' and 'b' (both of length 1), they cannot be prefixes of longer codewords. However, if there are codewords of different lengths (like 'd' and 'e', both of length 2), it does not guarantee that they are prefix-free. The condition of having a code alphabet size $D=4$ does not ensure that the encoding is prefix-free unless specific prefix conditions are checked.

**Incorrect Option: 1. True**

This option is incorrect because it mistakenly assumes that having a code alphabet size of $D=4$ automatically makes the code prefix-free without considering the actual structure and lengths of the codewords. The lengths alone do not provide sufficient information to conclude that the encoding is prefix-free, as shown in the correct answer explanation.


Question: Standard encryption threats do not include:

The correct answer is **3. Universal forgeries**.

### Explanation:

1. **Known-plaintext attacks**: In this type of attack, the attacker has access to both the plaintext (the original message) and the corresponding ciphertext (the encrypted message). The goal is to derive the encryption key or find a way to decrypt other messages encrypted with the same key. For example, if an attacker knows that the plaintext "HELLO" encrypts to "XZY123", they can analyze the encryption algorithm to potentially decrypt other messages.

2. **Chosen-plaintext attacks**: Here, the attacker can choose arbitrary plaintexts to be encrypted and then obtains the corresponding ciphertexts. This can help the attacker learn about the key or the encryption algorithm. For instance, if an attacker can input "SECRET" and see that it encrypts to "ABC456", they can use this information to explore other possibilities.

3. **Universal forgeries**: This term generally does not apply to standard encryption threats. It relates more to digital signatures rather than symmetric or asymmetric encryption methods. Universal forgery implies the ability to create a valid signature for any message without needing to know the private key, which is not typically a threat associated with encryption methods.

4. **Key-recovery attacks**: This type of attack aims to recover the secret key used in the encryption process. If an attacker can successfully recover the key, they can decrypt any data encrypted with that key. For example, if an attacker can exploit a vulnerability in the encryption algorithm to find the key, they can access all secured information.

### Conclusion:
While all other options represent valid threats to encryption schemes, "universal forgeries" is more related to digital signatures and does not fit the standard encryption threat model.


Question:
Byzantine consistent broadcast (BCB) assumes one designated sender S and it satisfies the following properties:

Validity: If S is correct, then every correct process eventually delivers the message.
No duplication: Every correct process delivers at most one message.
Integrity: If a correct process delivers a message and S is correct, then S has previously broadcast the message.
Consistency: No two correct processes deliver different messages.

Do we need to introduce some constraints on the number of Byzantine processes in non-synchronous environments? If yes, what are the constraints?

To analyze whether we need to introduce constraints on the number of Byzantine processes in a non-synchronous environment, we first need to understand the properties of Byzantine consistent broadcast (BCB) and the implications of Byzantine processes on these properties.

### Definitions:
- Let \( N \) be the total number of processes in the system.
- Let \( F \) be the maximum number of Byzantine processes.

### Properties of Byzantine Consistent Broadcast:
1. **Validity**: If the designated sender \( S \) is correct, then every correct process eventually delivers the message.
2. **No duplication**: Every correct process delivers at most one message.
3. **Integrity**: If a correct process delivers a message, and \( S \) is correct, then \( S \) has previously broadcast the message.
4. **Consistency**: No two correct processes deliver different messages.

### Analyzing Byzantine Processes:
In a non-synchronous environment, messages may take an unpredictable amount of time to be delivered, and processes may operate independently without a global clock. This introduces challenges when Byzantine processes (which can behave arbitrarily) are present.

#### Hypothetical Scenarios:
1. **Scenario with \( N = 3F \)**:
   - Suppose \( N = 3F \), meaning there are exactly three times as many processes as there are Byzantine processes. For example, if \( F = 1 \), then \( N = 3 \). In this case, the processes can be divided into:
     - Correct process A
     - Correct process B
     - Byzantine process C

   In this scenario, if the Byzantine process (C) decides to send a different message than what the correct sender (S) sends, it can mislead the correct processes (A and B). Both A and B may receive different messages if they are not able to distinguish between the correct and Byzantine processes.

2. **Potential Violation of Properties**:
   - **Validity**: If the correct process A receives a message from the Byzantine process C, it may not be able to determine if the message is valid. This could lead to a situation where A does not deliver the correct message broadcasted by S.
   - **Consistency**: If A and B receive different messages due to the Byzantine behavior of C, this violates the consistency property, as A delivers message \( m_1 \) while B delivers \( m_2 \).

### Mathematical Relationships:
To ensure the properties of BCB hold, particularly the consistency property, we must have more correct processes than Byzantine processes. The established consensus in distributed systems is that the maximum number of Byzantine processes \( F \) can be tolerated if and only if:

$$ N > 3F $$

or, more precisely,

$$ N = 3F + 1 $$

### Conclusion:
Yes, we must introduce constraints on the number of Byzantine processes in non-synchronous environments to maintain the integrity and consistency properties of Byzantine consistent broadcast. Specifically, the condition \( N = 3F + 1 \) must be satisfied to ensure that correct processes can always outvote or ignore the Byzantine processes, thereby ensuring that the system remains reliable and consistent even in the presence of failures or malicious behavior. This constraint is crucial for the robustness of the BCB protocol in distributed systems.


Question: Consider the function $f(x)=-x^{2}$. Which of the following statements are true regarding subgradients of $f(x)$ at $x=0$ ?

### Step 1: Understanding the function
The function given is \( f(x) = -x^2 \), a quadratic function that opens downwards. To analyze subgradients at \( x = 0 \), we first need to determine the differentiability of the function at that point.

### Step 2: Checking differentiability
We find the derivative of \( f(x) \):

\[
f'(x) = -2x
\]

At \( x = 0 \):

\[
f'(0) = -2(0) = 0
\]

Since the derivative exists and equals 0, the function \( f(x) \) is differentiable at \( x = 0 \).

### Step 3: Understanding subgradients
Subgradients generalize the concept of derivatives for convex functions, particularly at points where the function may not be differentiable. For differentiable functions, the subgradient at a point is equal to the derivative at that point. However, this interpretation must be carefully considered for concave functions.

### Step 4: Analyzing the subgradient at \( x = 0 \)
While \( f(x) \) is differentiable at \( x = 0 \) and the derivative is 0, the function is concave. In the context of concave functions, the concept of subgradients can lead to confusion. At a maximum point, such as \( x = 0 \) for this function, subgradients do not exist in the traditional sense because they typically represent supporting hyperplanes that are not applicable here.

### Step 5: Conclusion on subgradients
Thus, while \( f(x) \) is differentiable at \( x = 0 \), the nature of the function being concave leads to the conclusion that a subgradient does not exist at this point. The correct statement is:

- A subgradient does not exist even though \( f(x) \) is differentiable at \( x=0 \).


Question: Which of the following elements belong to $\mathbf{Z}_{35}^*$?

To determine which elements belong to \(\mathbf{Z}_{35}^*\), we need to find the elements that are coprime to \(35\). The number \(35\) can be factored into its prime components: \(35 = 5 \times 7\). An integer \(a\) is coprime to \(35\) if the greatest common divisor \(\text{gcd}(a, 35) = 1\).

Letâs check each option:

1. For \(12\): \(\text{gcd}(12, 35) = 1\) (since \(12\) shares no prime factors with \(35\)).
2. For \(20\): \(\text{gcd}(20, 35) = 5\) (they share the prime factor \(5\)).
3. For \(14\): \(\text{gcd}(14, 35) = 7\) (they share the prime factor \(7\)).
4. For \(15\): \(\text{gcd}(15, 35) = 5\) (they share the prime factor \(5\)).

Thus, only \(12\) is coprime to \(35\), meaning it belongs to \(\mathbf{Z}_{35}^*\). This analysis shows a common mistake: assuming all integers are coprime without checking for shared factors. In this case, only one of the options qualifies.

1


Question: Let $\mathcal{C}$ be a binary $(6,3)$ linear code containing the codewords $\mathbf{x}_1 = 011011$, $\mathbf{x}_2 = 101101$ and $\mathbf{x}_3 = 111000$. True or false: The minimum distance of the code is $d_{\min} = 3$.

To determine the minimum distance \( d_{\min} \) of the given binary linear code \( \mathcal{C} \) containing the codewords \( \mathbf{x}_1 = 011011 \), \( \mathbf{x}_2 = 101101 \), and \( \mathbf{x}_3 = 111000 \), we first need to calculate the pairwise Hamming distances between all pairs of codewords.

The Hamming distance \( d(\mathbf{x}, \mathbf{y}) \) between two codewords \( \mathbf{x} \) and \( \mathbf{y} \) is defined as the number of positions at which the corresponding bits are different.

Let's calculate the distances:

1. **Distance between \( \mathbf{x}_1 \) and \( \mathbf{x}_2 \)**:
   \[
   \mathbf{x}_1 = 011011, \quad \mathbf{x}_2 = 101101
   \]
   Comparing bit by bit:
   - 0 vs 1 (different)
   - 1 vs 0 (different)
   - 1 vs 1 (same)
   - 0 vs 1 (different)
   - 1 vs 0 (different)
   - 1 vs 1 (same)
   Thus, \( d(\mathbf{x}_1, \mathbf{x}_2) = 4 \).

2. **Distance between \( \mathbf{x}_1 \) and \( \mathbf{x}_3 \)**:
   \[
   \mathbf{x}_1 = 011011, \quad \mathbf{x}_3 = 111000
   \]
   Comparing bit by bit:
   - 0 vs 1 (different)
   - 1 vs 1 (same)
   - 1 vs 1 (same)
   - 0 vs 0 (same)
   - 1 vs 0 (different)
   - 1 vs 0 (different)
   Thus, \( d(\mathbf{x}_1, \mathbf{x}_3) = 3 \).

3. **Distance between \( \mathbf{x}_2 \) and \( \mathbf{x}_3 \)**:
   \[
   \mathbf{x}_2 = 101101, \quad \mathbf{x}_3 = 111000
   \]
   Comparing bit by bit:
   - 1 vs 1 (same)
   - 0 vs 1 (different)
   - 1 vs 1 (same)
   - 1 vs 0 (different)
   - 0 vs 0 (same)
   - 1 vs 0 (different)
   Thus, \( d(\mathbf{x}_2, \mathbf{x}_3) = 3 \).

Now, we have the pairwise distances:
- \( d(\mathbf{x}_1, \mathbf{x}_2) = 4 \)
- \( d(\mathbf{x}_1, \mathbf{x}_3) = 3 \)
- \( d(\mathbf{x}_2, \mathbf{x}_3) = 3 \)

The minimum distance \( d_{\min} \) is the smallest of these distances:
\[
d_{\min} = \min(4, 3, 3) = 3
\]

Since we have determined that the minimum distance \( d_{\min} = 3 \), the statement that "the minimum distance of the code is \( d_{\min} = 3 \)" is **True**.

Thus, the correct answer is:

**A. True**


Question: Assume you are working on a school project with your friend. Your friend uses "print" to debug his code. Is this a good idea and, regardless of whether it is bad or not, is there a better way to do it? Explain why or why not in max 2 sentences.

Using "print" to debug code is like using a flashlight to find your way in a dark room; it can help illuminate the problem but may not give you the full picture. A better approach would be using a debugger, which is like having a detailed map that shows you not just where you are, but also the paths available to navigate more effectively through your code.