Question: What happens in our "Consensus-Based Total-Order Broadcast" algorithm, if the set of messages delivered in a round is not sorted deterministically after deciding in the consensus abstraction, but before it is proposed to consensus?

In the "Consensus-Based Total-Order Broadcast" algorithm, achieving a total order of messages is crucial for ensuring that all participants in a distributed system process messages in the same sequence.

When we discuss what happens if the set of messages delivered in a round is not sorted deterministically after deciding in the consensus abstraction, but before it is proposed to consensus, we are addressing a potential issue in the ordering of messages.

### Key Concepts

1. **Total-Order Broadcast**: This is a communication abstraction where all messages are delivered to all recipients in the same order. This is essential for consistency in distributed systems.

2. **Consensus**: This is a process where a group of participants must agree on a single value (or message) among multiple candidates. In this context, it is used to decide the order of messages.

3. **Deterministic Sorting**: This means that the order of messages is predictable and consistent every time the process runs, given the same input.

### Explanation of the Situation

1. **Consensus Decision**: When a round of consensus is completed, all participants agree on a set of messages that need to be delivered.

2. **Message Set Not Deterministically Sorted**: If the messages delivered in that round are not sorted in a deterministic manner, it means there is no consistent way to determine the order of these messages across different participants.

3. **Implications**:
   - If the messages are not deterministically sorted, different participants might end up delivering messages in different orders.
   - This inconsistency can lead to scenarios where some participants see a different sequence of events compared to others, which violates the principle of total-order broadcast.

### Conclusion

In summary, if the set of messages is not sorted deterministically before being proposed to consensus, it can lead to discrepancies in how different nodes perceive the order of messages. This undermines the goal of achieving a consistent and total order of message delivery, which is vital for maintaining synchronization and correctness in distributed systems.

To avoid this, the protocol must ensure that sorting occurs deterministically, so all participants agree on the order before messages are broadcast.


Question: For any function $f:\{0,1\}^p\rightarrow \{0,1\}^q$ and for any $a\in\{0,1\}^p$, we have\ldots

To analyze the notation and the question, we first need to clarify what \(\mathsf{DP}^f(a,b)\) means. Typically, in the context of functions \(f:\{0,1\}^p \rightarrow \{0,1\}^q\), \(\mathsf{DP}^f(a,b)\) represents the output of the function \(f\) when given the input \(a\) and is compared or related to \(b\) in some way.

The expression \(\Sigma _{b\in \{0,1\}^q}\mathsf{DP}^f(a,b)\) sums over all possible outputs \(b\) in \(\{0,1\}^q\).

Now, let's consider what \(\mathsf{DP}^f(a,b)\) could represent. One common interpretation is that it could be an indicator function that outputs \(1\) if \(f(a) = b\) (the function maps \(a\) to \(b\)), and \(0\) otherwise.

In this case, the sum over all \(b\) would count how many \(b\) correspond to the output of \(f(a)\). Since \(f\) can only produce one valid output for a specific input \(a\), the sum would yield:

- \(1\) if \(b\) is equal to \(f(a)\),
- \(0\) if \(b\) is not equal to \(f(a)\).

Thus, the sum \(\Sigma _{b\in \{0,1\}^q}\mathsf{DP}^f(a,b)\) would equal \(1\) for each specific input \(a\) since there is exactly one \(b\) such that \(f(a) = b\).

Therefore, the final answer is:

\(\Sigma _{b\in \{0,1\}^q}\mathsf{DP}^f(a,b)=1\)


Question: For an interactive proof system, the difference between perfect, statistical and computational zero-knowledge is based on \ldots

The question asks about the differences between perfect, statistical, and computational zero-knowledge proofs in interactive proof systems, specifically focusing on what these differences are based on.

The key concept here revolves around the degree of indistinguishability between the distributions of outputs (transcripts) generated during interactions between the prover and the verifier.

1. **Option 0: \ldots the distinguishability between some distributions.**
   - **Correct.** This option accurately captures the essence of zero-knowledge proofs. Perfect zero-knowledge guarantees that the distributions are identical, statistical zero-knowledge allows for negligible distinguishability, and computational zero-knowledge depends on the computational limits of the adversary, thus differentiating the three types of zero-knowledge.

2. **Option 1: \ldots the percentage of recoverable information from a transcript with an honest verifier.**
   - **Incorrect.** This option misrepresents the concept of zero-knowledge. The focus is not on how much information can be extracted but rather on how indistinguishable the output distributions are from each other.

3. **Option 2: \ldots the number of times the protocol is run between the prover and the verifier.**
   - **Incorrect.** The number of interactions does not define the nature of zero-knowledge proofs. The distinctions arise from the properties of the distributions, not the frequency of communication.

4. **Option 3: \ldots whether the inputs are taken in $\mathcal{P}$, $\mathcal{NP}$ or $\mathcal{IP}$.**
   - **Incorrect.** While these complexity classes relate to the types of problems addressed, they do not determine the distinctions among zero-knowledge protocols.

In conclusion, the correct answer highlights the importance of distinguishability between distributions in defining the characteristics of zero-knowledge proofs. This aspect is fundamental to understanding the security guarantees provided by different types of zero-knowledge, making Option 0 the most accurate choice.


Question:
You have been publishing a daily column for the Gazette over the last few years and have recently reached a milestone --- your 1000th column! Realizing you'd like to go skiing more often, you decide it might be easier to automate your job by training a story generation system on the columns you've already written. Then, whenever your editor pitches you a title for a column topic, you'll just be able to give the title to your story generation system, produce the text body of the column, and publish it to the website!
You initialize your model with a vocabulary $V$ with $|V|$ tokens. Given a vector of scores $S = [s_1, \ldots, s_i, \ldots, s_{|V|}]$ output by your model for each token in your vocabulary, write out the softmax function to convert score $s_1$ to a probability mass $P(s_1)$

To compute the probability mass \( P(s_1) \) using the softmax function, use the formula:

\[
P(s_1) = \frac{e^{s_1}}{\sum_{i=1}^{|V|} e^{s_i}}
\]

This transforms the score into a normalized probability.

---

To derive the probability mass \( P(s_1) \) using the softmax function, we start from the vector of scores \( S = [s_1, s_2, \ldots, s_{|V|}] \). The softmax function transforms these scores into probabilities.

Specifically, the formula for \( P(s_1) \) can be expressed as:

\[
P(s_1) = \frac{e^{s_1}}{\sum_{i=1}^{|V|} e^{s_i}}
\]

Here, \( e^{s_1} \) represents the exponential of the score for the first token, while the denominator is the sum of exponentials of all scores in the vector \( S \). This ensures that the resulting probabilities are non-negative and normalized.

In essence, the softmax function allows us to interpret the raw scores as probabilities, which is quite useful for various applications, including our story generation model. Just remember, it's all about exponentials and normalization, really!


Question: The following function(s) have a unique minimizer.

To determine which functions have a unique minimizer in their respective intervals, let's analyze each option.

(a) \( f(x) = x^{2}, x \in [-3, 2] \)

The function \( f(x) = x^{2} \) is a quadratic function that is concave up (its second derivative is positive). The minimum occurs at the vertex, which is at \( x = 0 \). Since \( 0 \) is within the interval \([-3, 2]\), this function has a unique minimizer at \( x = 0 \).

(b) \( f(x) = \log(x), x \in (0, 10] \)

The function \( f(x) = \log(x) \) is a monotonically increasing function in the interval \( (0, 10] \). Therefore, it does not have a minimum at any point in the interval; its minimum value approaches \(-\infty\) as \( x \) approaches \( 0 \). Hence, it does not have a unique minimizer.

(c) \( f(x) = \sin(x), x \in [-10, 10] \)

The function \( f(x) = \sin(x) \) is periodic and oscillates between \(-1\) and \(1\). In the interval \([-10, 10]\), it attains multiple local minima (e.g., at \( x = -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2} \), etc.). Therefore, it does not have a unique minimizer.

(d) \( f(x) = e^{3x} + x^{4} - 3x, x \in [-10, 10] \)

The function \( f(x) \) is the sum of an exponential term and polynomial terms. To find if it has a unique minimum, we can check the derivative. The first derivative \( f'(x) = 3 e^{3x} + 4x^{3} - 3 \). The behavior of this derivative can be complex, but due to the nature of the exponential term \( e^{3x} \) (which grows rapidly), combined with the polynomial terms, it's likely that the function has a unique minimum in the given interval. A more thorough analysis or numerical check would confirm this, but it appears to trend toward a unique minimizer.

Based on this analysis:

Final Answer: 1, 4.


Question: (Convex III) Let $f, g: \mathbb{R} \rightarrow \mathbb{R}$ be two convex functions. Then $h=f \circ g$ is always convex.

Answer: 2

Reasoning: The composition of two convex functions is not always convex unless additional conditions are met, such as g being non-decreasing. A counterexample can be constructed where f and g are both convex but their composition h = f(g(x)) is not convex.


Question: A shadow stack...

A shadow stack is primarily used to enhance the security of the call stack in programming languages, especially against control flow attacks such as return-oriented programming (ROP) and other stack-based exploits.

Analyzing each option:

1. **Offers stronger protection than stack canaries**: This statement is true; a shadow stack provides a more robust means of verifying return addresses compared to traditional stack canaries, which only guard against specific buffer overflow attacks.

2. **Needs to be protected against or hidden from arbitrary writes**: This is also true. The integrity of the shadow stack must be maintained, meaning it should be protected from arbitrary overwrites that could compromise its contents.

3. **Is an exact copy of the original stack allowing for a complete integrity verification**: This statement is misleading because while the shadow stack may contain return addresses or relevant data associated with function calls, it typically does not replicate all aspects of the original stack state.

4. **Is used by exception handling for stack trace generation**: While it might aid in preserving call information relevant to exceptions, this isn't its primary purpose or defining featureâhence it's not strictly correct.

5. **Is not usable in combination with CFI (Control Flow Integrity)**: This statement is false; shadow stacks can indeed complement CFI systems by providing additional layers of protection without conflicting with their mechanisms.

Based on this analysis:

The most accurate options regarding shadows stacks are 1 and 2, but if looking for one definitive answer that captures their essence effectively:

**The best choice would be Option 1**, as it emphasizes enhanced security over traditional methods like canaries while acknowledging that both options (1 and 2) are valid statements about shadow stacks' characteristics.


Question: Which of the following is/are true about fuzzing with sanitizers?

Let's evaluate each of the provided statements regarding fuzzing with sanitizers:

1. **Some fuzzers use fork servers to reduce sanitizer overhead.**
   **True.** Fork servers allow a fuzzer to spawn multiple test cases from a single process, which can significantly reduce the overhead associated with instrumentation, as the sanitizers only need to be initialized once.

2. **The set of sanitizers used during a fuzzing campaign must be carefully chosen (tradeoff between bug visibility/execution speed).**
   **True.** Different sanitizers have different performance impacts and visibility into issues. For example, using multiple sanitizers can increase bug detection, but might slow down the execution speed. Therefore, a careful selection is important for effective fuzzing.

3. **Some fuzzers dynamically tweak sanitizers to speed up fuzzing.**
   **True.** There are fuzzers that can adjust configurations dynamically to balance the trade-offs between detection capabilities and execution speed, which might include enabling or disabling certain sanitizer features on-the-fly.

4. **ASAN instrumentation has a negligible startup overhead.**
   **False.** AddressSanitizer (ASan) does have a noticeable startup overhead compared to running a program without it. While it is designed to be efficient in terms of runtime performance during the execution of the program, the initial overhead can be significant.

Based on the evaluations, the true statements are:

- **1. Some fuzzers use fork servers to reduce sanitizer overhead.**
- **2. The set of sanitizers used during a fuzzing campaign must be carefully chosen (tradeoff between bug visibility/execution speed).**
- **3. Some fuzzers dynamically tweak sanitizers to speed up fuzzing.**

So the correct options are **1**, **2**, and **3**.


Question: Which of the following statements on Latent Semantic Indexing (LSI) and Word Embeddings (WE) is false?

To answer the question regarding Latent Semantic Indexing (LSI) and Word Embeddings (WE), let's first understand the key concepts involved and then evaluate each option.

1. **Latent Semantic Indexing (LSI)**:
   - LSI is a technique in natural language processing that uses singular value decomposition (SVD) to reduce the dimensionality of the term-document matrix. It identifies patterns in the relationships between the terms and concepts in the documents.
   - The dimensions resulting from LSI can often be interpreted as latent concepts that the documents are related to.
   - LSI operates on the entire document and does not consider the order of words; it focuses on word co-occurrence.

2. **Word Embeddings (WE)**:
   - Word embeddings are a type of word representation that allows words to be represented as vectors in a continuous vector space. This representation captures semantic relationships between words based on their context in large corpora.
   - The order of words does matter in WE since they are often generated from models like Word2Vec or GloVe, which consider context windows or sequences of words.
   - Word embeddings can be stochastic, especially when using techniques like negative sampling, where the training can yield different embeddings each time due to random sampling.

Now, let's analyze each option:

- **A: The dimensions of LSI can be interpreted as concepts, whereas those of WE cannot**:
  - This statement is true. LSI's dimensions are often interpretable as concepts, while WE dimensions do not have a clear interpretability in terms of distinct concepts.

- **B: LSI does not depend on the order of words in the document, whereas WE does**:
  - This statement is also true. LSI ignores word order and focuses on word co-occurrence, while WE takes word order into account through context.

- **C: LSI is deterministic (given the dimension), whereas WE is not**:
  - This statement is true. LSI produces consistent results given the same input and dimensionality, while WE can produce different embeddings due to randomness in training.

- **D: LSI does take into account the frequency of words in the documents, whereas WE with negative sampling does not**:
  - This statement is false. LSI indeed considers the frequency of words through the term-document matrix. In contrast, while WE with negative sampling does utilize frequencies in some contexts (e.g., in the softmax function), it primarily focuses on predicting context words rather than directly using frequencies in the same way LSI does.

After evaluating all the options, it is clear that the false statement is D.

Action: Answer("D")


Question: In Bluetooth, the link key $K_{link}$ is ...

The link key \( K_{link} \) in Bluetooth is primarily **used to authenticate devices**. This key is established during the pairing process and is essential for ensuring that the devices can verify each other's identity before establishing a secure connection.

To elaborate further, the link key is a symmetric key that is used in various security protocols within Bluetooth, including the generation of session keys for encryption and authentication purposes. While it may play a role in generating ephemeral keys (like \( K_{init} \)) for specific sessions, its main function revolves around device authentication.

Thus, among the options provided:

- **Used to generate an ephemeral key \( K_{init} \)**: This can be seen as partially correct but is not the primary purpose of \( K_{link} \).
- **Not used to generate the encryption key**: This is incorrect, as \( K_{link} \) can indeed influence encryption key generation.
- **Used to authenticate devices**: This is the correct answer, as \( K_{link} \) is fundamentally about authenticating the devices involved in the Bluetooth connection.
- **The input to the pairing protocol**: While \( K_{link} \) is established during the pairing process, this option does not capture its primary purpose.

Therefore, the best answer is that the link key \( K_{link} \) is **used to authenticate devices**.


Question: Let $E$ be a finite ground set and let $\mathcal{I}$ be a family of ground sets. Which of the following definitions of $\mathcal{I}$ guarantees that $M = (E, \mathcal{I})$ is a matroid? \begin{enumerate} \item $E$ is the edges of an undirected bipartite graph and $\mathcal{I} = \{X \subseteq E : \mbox{$X$ is an acyclic edge set}\}$. \item $E$ is the edges of an undirected graph and $\mathcal{I} = \{X \subseteq E : \mbox{$X$ is an acyclic edge set}\}$. \item $E$ is the edges of an undirected bipartite graph and $\mathcal{I} = \{X \subseteq E : \mbox{$X$ is a matching}\}$. \item $E$ is the edges of an undirected  graph and $\mathcal{I} = \{X \subseteq E : \mbox{$X$ is a matching}\}$. \item $E = \{1, 2, \ldots, n\}$ is the set of indices of  vectors $v_1, \ldots, v_n \in \mathbb{R}^d$ and \\$\mathcal{I} = \{X \subseteq E : \mbox{the vectors $\{v_i : i \in X\}$ are linearly \emph{dependent}}\}$. \item $E = \{1, 2, \ldots, n\}$ is the set of indices of  vectors $v_1, \ldots, v_n \in \mathbb{R}^d$ and \\$\mathcal{I} = \{X \subseteq E : \mbox{the vectors $\{v_i : i \in X\}$ are linearly \emph{independent}}\}$. \end{enumerate} The definitions of $\mathcal{I}$ that guarantees that $M = (E, \mathcal{I})$ is a matroid are:

To determine which options guarantee that \( M = (E, \mathcal{I}) \) is a matroid, we need to evaluate each of the proposed definitions of \( \mathcal{I} \) based on the two matroid axioms: the hereditary property and the exchange property.

1. **Option 1: \( E \) is the edges of an undirected bipartite graph and \( \mathcal{I} = \{X \subseteq E : \text{X is an acyclic edge set}\} \).**
   - This defines a graphic matroid since acyclic edge sets correspond to forests, which satisfy both matroid axioms.
   - **Conclusion:** This option is valid.

2. **Option 2: \( E \) is the edges of an undirected graph and \( \mathcal{I} = \{X \subseteq E : \text{X is an acyclic edge set}\} \).**
   - This also defines a graphic matroid over general graphs. Acyclic edge sets correspond to forests in a graph, hence satisfying the matroid properties.
   - **Conclusion:** This option is valid.

3. **Option 3: \( E \) is the edges of an undirected bipartite graph and \( \mathcal{I} = \{X \subseteq E : \text{X is a matching}\} \).**
   - The collection of matchings does not satisfy the exchange property in general. If you have two matchings, one can’t necessarily exchange elements from them while keeping both sets as matchings.
   - **Conclusion:** This option is not valid.

4. **Option 4: \( E \) is the edges of an undirected graph and \( \mathcal{I} = \{X \subseteq E : \text{X is a matching}\} \).**
   - Similar to option 3, the collection of matchings in a general graph does not satisfy the exchange property.
   - **Conclusion:** This option is not valid.

5. **Option 5: \( E = \{1, 2, \ldots, n\} \) is the set of indices of vectors \( v_1, \ldots, v_n \in \mathbb{R}^d \) and \( \mathcal{I} = \{X \subseteq E : \text{the vectors } \{v_i : i \in X\} \text{ are linearly dependent}\} \).**
   - This set does not satisfy the hereditary property because if a set of vectors is dependent, a subset of those vectors may not be dependent.
   - **Conclusion:** This option is not valid.

6. **Option 6: \( E = \{1, 2, \ldots, n\} \) is the set of indices of vectors \( v_1, \ldots, v_n \in \mathbb{R}^d \) and \( \mathcal{I} = \{X \subseteq E : \text{the vectors } \{v_i : i \in X\} \text{ are linearly independent}\} \).**
   - This defines a vector matroid and satisfies both matroid axioms (hereditary and exchange properties).
   - **Conclusion:** This option is valid.

Now, consolidating the conclusions:

- Valid options: (1), (2), (6)
- Invalid options: (3), (4), (5)

Next, we check the provided multiple-choice options:

- **Option 1:** (a), (c), (f) — Invalid (c is not valid)
- **Option 2:** (a), (b), (c), (d), (f) — Invalid (c, d are not valid)
- **Option 3:** (a), (b), (c), (f) — Invalid (c is not valid)
- **Option 4:** (a), (b), (e) — Invalid (e is not valid)
- **Option 5:** (a), (c), (d), (f) — Invalid (c, d are not valid)
- **Option 6:** (a), (b), (c), (d), (e) — Invalid (c, d, e are not valid)
- **Option 7:** (a), (c), (d), (e) — Invalid (c, d, e are not valid)
- **Option 8:** (a), (f) — Valid (both are valid)
- **Option 9:** (a), (b), (c), (e) — Invalid (c, e are not valid)
- **Option 10:** (a), (b), (f) — Valid (a, b, f are valid)
- **Option 11:** (a), (c), (e) — Invalid (c, e are not valid)
- **Option 12:** (a), (e) — Invalid (e is not valid)

The correct options that guarantee \( M = (E, \mathcal{I}) \) is a matroid are therefore:

- **Option 8:** (a), (f)
- **Option 10:** (a), (b), (f)

Thus, the answer is that options that guarantee that \( M \) is a matroid are **Option 8 and Option 10**.


Question: /True or false:/ Is the following statement true or false? Justify your answer. "The node with the highest clustering coefficient in an undirected graph is the node that belongs to the largest number of triangles."

**True.**

The clustering coefficient of a node in an undirected graph measures the degree to which its neighbors are also connected to each other. Specifically, it quantifies the proportion of pairs of the node's neighbors that are connected by edges, thus forming triangles with the node itself.

A triangle is a complete subgraph involving three nodes, and for a node \( v \), the number of triangles it participates in can be calculated based on the connections between its neighbors. If a node has a high clustering coefficient, it indicates that many of its neighbors are interconnected, leading to a greater number of triangles involving that node.

Therefore, the node with the highest clustering coefficient indeed belongs to the largest number of triangles because a higher clustering coefficient directly reflects a higher density of connections among its neighbors, which results in more triangles being formed around that node.

Thus, the statement is **true**: the node with the highest clustering coefficient is the node that belongs to the largest number of triangles in an undirected graph.


Question: Consider the following sentences:  ```Aphrodite and Eros are Gods.```  ```Aphrodite is a parent of Eros.```  ```Aphrodite is beautiful.```  ```Aphrodite is happy.```  Specify which are the *classes*, the *instances* and the *properties* in the above statements.

To analyze the given sentences, we must first understand the distinctions between classes, instances, and properties in the context of ontology and knowledge representation.

1. **Classes** are general categories or types that group together similar entities. They represent sets of objects that share common characteristics. In this context, the classes can be identified as follows:
   - **God**: This class encompasses deities like Aphrodite and Eros.
   - **Beautiful**: This class represents the quality of beauty, which can be possessed by entities like Aphrodite.
   - **Happy**: This class represents the quality of happiness, which can also be attributed to entities.

2. **Instances** are specific examples or individual entities that belong to a class. In our sentences, the instances are:
   - **Aphrodite**: An individual deity who is part of the class of Gods.
   - **Eros**: Another individual deity and also a part of the class of Gods.

3. **Properties** are relationships or attributes that describe how instances relate to each other or to classes. In the provided sentences, the properties can be identified as:
   - **isa**: This property indicates membership in a class (e.g., "Aphrodite is a God" can be interpreted as "Aphrodite isa God").
   - **isParentOf**: This property represents a familial relationship, specifically indicating that one instance is a parent of another (e.g., "Aphrodite is a parent of Eros" means "Aphrodite isParentOf Eros").

Now, summarizing the identification of classes, instances, and properties in the sentences:

- **Classes**: God, beautiful, happy
- **Instances**: Aphrodite, Eros
- **Properties**: isa, isParentOf

**Intuition**: The distinction between classes, instances, and properties allows us to structure knowledge in a way that captures relationships and characteristics of entities. By organizing this information, we can better understand the relationships between different entities in any given domain.

Therefore, the final answer is:
- **Classes**: God, beautiful, happy
- **Instances**: Aphrodite, Eros
- **Properties**: isa, isParentOf


Question: Consider the linear programming relaxation for minimum-weight vertex cover: \begin{align*} \text{Minimize}  \quad &\sum_{v\in V} x_v w(v)\\ \text{Subject to} \quad &x_u + x_v \geq 1 \quad \forall \{u,v\} \in E \\ &0 \leq x_v \leq 1 \quad \ \  \forall v \in V \end{align*} In class, we saw that any extreme point is integral when considering bipartite graphs. For general graphs, this is not true, as can be seen by considering the graph consisting of a single triangle. However, we have the following statement for general graphs: \begin{itemize} \item[] Any extreme point $x^*$ satisfies $x^*_v \in \{0, \frac12, 1\}$ for every $v\in V$\,. \end{itemize} Prove the above statement.

To prove that any extreme point \( x^* \) of the linear programming relaxation for the minimum-weight vertex cover satisfies \( x^*_v \in \{0, \frac{1}{2}, 1\} \) for every vertex \( v \in V \), we need to analyze the feasible region and the implications of the constraints more rigorously.

### Step 1: Understanding the Linear Program

The linear programming relaxation for the minimum-weight vertex cover is formulated as follows:

\[
\text{Minimize} \quad \sum_{v \in V} x_v w(v)
\]

Subject to:

\[
x_u + x_v \geq 1 \quad \forall \{u,v\} \in E
\]
\[
0 \leq x_v \leq 1 \quad \forall v \in V
\]

Here, \( w(v) \) is the weight of vertex \( v \), and \( x_v \) represents the fraction of vertex \( v \) included in the cover.

### Step 2: Properties of Extreme Points

In linear programming, extreme points are points in the feasible region that cannot be expressed as a convex combination of other feasible points. For our problem, we will show that at extreme points, the variables \( x_v \) take on specific values constrained by the edge-covering requirement.

### Step 3: Analyzing Edge Constraints

For each edge \( \{u, v\} \in E \), the constraint \( x_u + x_v \geq 1 \) must hold. This constraint implies that at least one of \( x_u \) or \( x_v \) must be at least \( \frac{1}{2} \) if both are less than 1.

### Step 4: Case Analysis

We will analyze three cases for each vertex \( v \):

1. **Case 1: \( x_v = 1 \)**
   If \( x_v = 1 \), vertex \( v \) is fully included in the cover. This satisfies the edge constraint for all edges incident to \( v \) since \( x_u + x_v \geq 1 \) will be satisfied regardless of the values of \( x_u \).

2. **Case 2: \( x_v = 0 \)**
   If \( x_v = 0 \), vertex \( v \) is not included in the cover. For every edge \( \{u, v\} \), the constraint \( x_u + x_v \geq 1 \) implies \( x_u \) must be at least 1, thus fulfilling the requirement.

3. **Case 3: \( 0 < x_v < 1 \)**
   In this case, since \( x_v \) is fractional, let's denote \( x_v = k \) where \( 0 < k < 1 \). The edge constraint \( x_u + k \geq 1 \) implies \( x_u \geq 1 - k \). Therefore, for any adjacent vertex \( u \) connected to \( v \), if \( k \) is not equal to \( \frac{1}{2} \), there will be a contradiction on whether \( x_u \) can satisfy the constraints while remaining less than 1 or more than 0. This leads to the conclusion that \( k \) must be exactly \( \frac{1}{2} \); otherwise, it would force \( x_u \) to take on a value that cannot satisfy the linear constraints.

### Step 5: Conclusion

From the analysis of the cases, we conclude:

- If \( x_v = 1 \), vertex \( v \) is included.
- If \( x_v = 0 \), vertex \( v \) is excluded.
- If \( 0 < x_v < 1 \), then for the constraints to hold without contradictions, \( x_v \) must equal \( \frac{1}{2} \).

Thus, every extreme point \( x^* \) of the linear programming relaxation for the minimum-weight vertex cover must satisfy:

\[
x^*_v \in \{0, \frac{1}{2}, 1\} \quad \forall v \in V,
\]

which completes our proof.


Question: You are asked to implement the following List functions using only the specified List API methods. You are also allowed to use the reverse method in any subquestion, should you see it fit. If you need another method of List, you need to reimplement it as part of your answer. Please refer to the appendix on the last page as a reminder for the behavior of the given List API methods. Implement scanLeft using only foldLeft, Nil and :: (cons). def scanLeft[A, B](xs: List[A])(z: B)(op: (B, A) => B): List[B] = ???

To implement the `scanLeft` function using only `foldLeft`, `Nil`, and `::` (the cons operator), we need to understand the purpose of `scanLeft`. The `scanLeft` function computes a running total (or accumulation) of a binary operation applied to the elements of a list, starting with an initial value.

The `foldLeft` function processes the list from the left, applying the binary operation and carrying the accumulated value forward. We can leverage this behavior to build our `scanLeft` function.

Here's how we can approach the implementation:

1. **Initialize the Result List**: Start with the initial value `z`, which will be the first element of the result list.
2. **Use `foldLeft`**: We will use `foldLeft` to traverse the input list `xs`. For each element, we will apply the operation `op` to the accumulated value and the current element, and then prepend the new accumulated value to the result list.
3. **Reverse the Result**: Since `foldLeft` processes the list from the left to the right, the accumulated values will be in reverse order. Therefore, we will need to reverse the final result list before returning it.

Here is the complete implementation of `scanLeft`:

```scala
def scanLeft[A, B](xs: List[A])(z: B)(op: (B, A) => B): List[B] = {
  // Start with the initial value in the result list
  val initialResult = List(z)

  // Use foldLeft to accumulate results
  val result = xs.foldLeft(initialResult) { (acc, elem) =>
    // Get the last accumulated value
    val lastAcc = acc.last
    // Compute the new accumulated value
    val newAcc = op(lastAcc, elem)
    // Prepend the new accumulated value to the result list
    newAcc :: acc
  }

  // Reverse the result to maintain the correct order
  result.reverse
}
```

### Explanation of the Code:
- We start with a list containing just the initial value `z`.
- We use `foldLeft` on the input list `xs`. For each element `elem`, we:
  - Retrieve the last accumulated value from the current result list `acc`.
  - Apply the operation `op` to this last accumulated value and the current element to get the new accumulated value.
  - Prepend this new accumulated value to the result list using `::`.
- Finally, we reverse the result list to restore the correct order of accumulated values.

This implementation adheres to the constraints of using only `foldLeft`, `Nil`, and `::`, and effectively computes the desired output of `scanLeft`.


Question: Let $n$ be a positive integer. An element $x \in \mathbb{Z}_n$ is \emph{always} invertible when \dots

To determine when an element \( x \in \mathbb{Z}_n \) is always invertible, we need to understand the conditions under which an element has a multiplicative inverse in the modular arithmetic system \( \mathbb{Z}_n \).

1. **Invertibility in \( \mathbb{Z}_n \)**: An element \( x \) is invertible in \( \mathbb{Z}_n \) if there exists an integer \( y \) such that \( xy \equiv 1 \mod n \). This is equivalent to the condition that \( \gcd(x, n) = 1 \). Thus, for \( x \) to have an inverse, \( x \) must be coprime with \( n \).

2. **Evaluating the options**:
   - **Option 1**: \( x \) and \( n \) are coprime. This is the correct condition for \( x \) to be invertible in \( \mathbb{Z}_n \) since if \( \gcd(x, n) = 1 \), then \( x \) has a multiplicative inverse.
   - **Option 2**: \( x \) and \( \varphi(n) \) are coprime. This option does not guarantee that \( x \) is invertible in \( \mathbb{Z}_n \). Instead, \( \varphi(n) \), Euler's totient function, gives the count of integers up to \( n \) that are coprime to \( n \), but it does not directly relate to the invertibility of \( x \).
   - **Option 3**: \( x \) is even. This condition does not guarantee invertibility because even numbers can share factors with \( n \), especially when \( n \) is even.
   - **Option 4**: \( n \) is prime. While every non-zero \( x < n \) is invertible when \( n \) is prime, this condition is not necessary for \( x \) to be invertible in general cases where \( n \) could be composite.

3. **Conclusion**: The only condition that universally guarantees that \( x \) is invertible in \( \mathbb{Z}_n \) for any positive integer \( n \) is the first option.

Correct answer: \( x \) and \( n \) are coprime.


Question: Tick the \emph{true} assertion related to the ElGamal signature scheme.

In the context of the ElGamal signature scheme, the true assertion is:

**The generator \(g\) generates a subgroup of prime order.**

### Explanation:
In the ElGamal signature scheme, we work within a cyclic group of prime order \(q\) that is generated by \(g\). The security of the scheme relies on the difficulty of the discrete logarithm problem in this subgroup, which ensures that the signatures are secure.

### Intuitive Example:
Think of a large prime \(p\) and a generator \(g\) such that \(g\) generates a subgroup of order \(q\) (where \(q\) is also prime). This means that repeatedly applying \(g\) will cycle through exactly \(q\) distinct elements before returning to the identity. This subgroup structure is crucial because it limits the possible values for the discrete logarithm problem, enhancing the security of the signatures.

### Clarification on Other Options:
- **A signature of a given message is obtained in a deterministic way**: This is false because the signature generation involves random elements, making it probabilistic.
- **The signature verification procedure is probabilistic**: This is also false; verification is deterministic.
- **The main computational cost of the signature generation is due to one modular exponentiation**: This is misleading; the signature generation typically involves multiple modular exponentiations and is not limited to just one.

Thus, the correct choice emphasizing the theory behind the ElGamal signature scheme is that **the generator \(g\) generates a subgroup of prime order**.


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: $G$ admits a systematic form (i.e., it can be put into systematic form via elementary row operations).

To determine whether the given generator matrix \( G \) of a \((6,4)\) linear code over \( \mathbb{F}_2 \) can be transformed into systematic form through elementary row operations, we will follow the outlined steps.

### 1. Define Systematic Form
A matrix is said to be in systematic form if it has the structure:

\[
G = \begin{pmatrix}
I_k & P
\end{pmatrix}
\]

where \( I_k \) is the \( k \times k \) identity matrix (in this case \( 4 \times 4 \)), and \( P \) is a \( k \times (n-k) \) matrix. For our \((6,4)\) code, this means the first four columns of \( G \) must form the \( 4 \times 4 \) identity matrix, and the last two columns represent additional information.

### 2. Elementary Row Operations
Elementary row operations include:
- Swapping two rows.
- Multiplying a row by a non-zero scalar (in \( \mathbb{F}_2\), this means multiplying by 1, as the only other element, 0, does not lead to a valid operation).
- Adding a multiple of one row to another row.

These operations can be used to manipulate the rows of a matrix to achieve the desired form.

### 3. Row Reduction Process
We will start with the given matrix \( G \):

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

#### Step 1: Make the first column a leading 1
The first row already has a leading 1 in the first column.

#### Step 2: Eliminate non-zero entries below the leading 1 in column 1
No changes are needed since the first column below the first row is already zero.

#### Step 3: Move to the second column
The second row has a leading 1 in the second column. We need to eliminate the 1s in the third and fourth rows.

- Row 3: \( R_3 \leftarrow R_3 - R_2 = (0, 1, 1, 0, 0, 0) - (0, 1, 1, 1, 0, 0) = (0, 0, 0, 1, 0, 0) \)
- Row 4: \( R_4 \leftarrow R_4 - R_2 = (0, 1, 1, 1, 0, 1) - (0, 1, 1, 1, 0, 0) = (0, 0, 0, 0, 0, 1) \)

Now the matrix looks like this:

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

#### Step 4: Move to the third column
We have a leading 1 in the third row at the fourth column, but we need to swap rows to get a leading 1 in the third column. We can swap row 3 and row 4 to move the leading 1 up.

After the swap, the matrix becomes:

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

#### Step 5: Eliminate non-zero entries in columns above the leading 1s
- Row 1: \( R_1 \leftarrow R_1 - R_4 = (1, 1, 1, 0, 1, 0) - (0, 0, 0, 1, 0, 0) = (1, 1, 1, 1, 1, 0) \)

Now, we need to focus on the last two rows to make sure we have identity structure. However, we notice that we cannot achieve the identity structure completely across the first four columns due to the presence of 1s in the first row that cannot be eliminated while maintaining the structure.

### 4. Identify Key Properties
We have leading ones in the second and fourth columns. However, we cannot get the first four columns to match the identity matrix due to the dependencies in the rows.

### 5. Conclusion
Based on our analysis, while we have manipulated the matrix through various row operations, we cannot achieve a systematic form. The dependencies among the rows prevent us from forming the required identity matrix in the first four columns. Thus, the generator matrix \( G \) cannot be transformed into systematic form using elementary row operations.


Question: In this problem we are going  to formally analyze the important median trick. Suppose that we have a streaming algorithm for distinct elements that outputs an estimate $\hat d$ of the number $d$ of distinct elements such that \begin{align*} \Pr[\hat d > 3d] \leq 47 \% \qquad \mbox{and} \qquad  \Pr[\hat d < d/3] \leq 47\%\,, \end{align*} where the probabilities are over the randomness of the streaming algorithm (the selection of hash functions). In other words, our algorithm overestimates the true value by a factor of 3 with a quite large  probability $47\%$ (and also underestimates with large probability). We want to do better! An important and useful technique for doing better is the median trick: run $t$ independent copies in parallel and output the median of the $t$ estimates (it is important that it is the median and \emph{not} the mean as a single horrible estimate can badly affect the mean). Prove that if we select $t = C \ln(1/\delta)$ for some large (but reasonable) constant $C$, then the estimate $\hat d$ given by the median trick satisfies \begin{align*} d/3 \leq \hat d \leq 3d \qquad \mbox{with probability at least $1-\delta$.} \end{align*} \emph{Hint: an important tool  in this exercise are the Chernoff Bounds, which basically say that sums of independent variables are highly concentrated.} Two such bounds can be stated as follows. Suppose  $ X_1, X_2, \dots, X_n$  are independent random variables  taking values in  $\{0,1\}$. Let $X$ denote their sum and let $\mu = \mathbb{E}[X]$ denote the sum's expected value. Then for any $\delta \in (0,1)$, \begin{align*} \Pr[ X \leq (1- \delta) \mu] \leq e^{-\frac{\delta^2 \mu }{2}} \qquad \mbox{and} \qquad \Pr[ X \geq  (1+ \delta) \mu] \leq e^{-\frac{\delta^2 \mu }{3}}\,. \end{align*}

To prove that the median of \(t\) independent estimates from the algorithm yields an estimate \(\hat d\) such that \(d/3 \leq \hat d \leq 3d\) with probability at least \(1 - \delta\), we can utilize the properties of the median and Chernoff bounds.

### Step 1: Define the Problem

We have \(t\) independent copies of the algorithm, each producing an estimate \(\hat d_i\). The probabilities of overestimating and underestimating are given as follows:
\[
\Pr[\hat d_i > 3d] \leq 0.47 \quad \text{and} \quad \Pr[\hat d_i < d/3] \leq 0.47.
\]

### Step 2: Analyzing the Median

The median of \(t\) estimates, \(\hat d\), will be within \([d/3, 3d]\) unless more than half of the estimates are bad. Specifically, for \(\hat d\) to be outside \([d/3, 3d]\), at least \(\lceil t/2 \rceil\) of the estimates must either be greater than \(3d\) or less than \(d/3\).

### Step 3: Counting Bad Estimates

Let \(X_1\) be the count of estimates where \(\hat d_i > 3d\) and \(X_2\) be the count of estimates where \(\hat d_i < d/3\). We have:
\[
\Pr[X_1 \geq \lceil t/2 \rceil] \leq 0.47 \quad \text{and} \quad \Pr[X_2 \geq \lceil t/2 \rceil] \leq 0.47.
\]

The total number of estimates that are either greater than \(3d\) or less than \(d/3\) is given by \(X = X_1 + X_2\).

### Step 4: Expected Value of Bad Estimates

Since each \(X_i\) follows a binomial distribution,
\[
\mathbb{E}[X_1] = t \cdot 0.47 \quad \text{and} \quad \mathbb{E}[X_2] = t \cdot 0.47.
\]
Thus,
\[
\mathbb{E}[X] = \mathbb{E}[X_1] + \mathbb{E}[X_2] = t \cdot 0.94.
\]

### Step 5: Applying Chernoff Bounds

We want to ensure that the total number of bad estimates is at most \(t/2\):
\[
\Pr\left[X \geq \frac{t}{2}\right].
\]

Setting \(\mu = \mathbb{E}[X] = 0.94t\), we can use the Chernoff bound:
\[
\Pr\left[X \geq \frac{t}{2}\right] = \Pr\left[X \geq (1 + \delta) \mu\right],
\]
where \((1 + \delta) \mu = \frac{t}{2}\). Solving for \(\delta\):
\[
(1 + \delta)0.94t = \frac{t}{2} \implies 1 + \delta = \frac{1/2}{0.94} \implies \delta = \frac{1 - 0.94/2}{0.94} = \frac{0.06}{0.94} \approx 0.06383.
\]

Using the Chernoff bound:
\[
\Pr[X \geq \frac{t}{2}] \leq e^{-\frac{(0.06383)^2 \cdot 0.94t}{3}}.
\]

### Step 6: Choosing \(t = C \ln(1/\delta)\)

To ensure that the probability of having more than half the estimates bad is less than \(\delta\), we need to set \(C\) large enough. For large enough \(C\), we have:
\[
e^{-\frac{(0.06383)^2 \cdot 0.94t}{3}} \leq \delta,
\]
which can be satisfied by choosing \(t = C \ln(1/\delta)\) with \(C\) sufficiently large.

### Conclusion

Thus, with \(t = C \ln(1/\delta)\) for a suitable constant \(C\), we ensure:
\[
\Pr\left[\hat d < \frac{d}{3} \text{ or } \hat d > 3d\right] \leq \delta,
\]
which completes the proof that:
\[
\frac{d}{3} \leq \hat d \leq 3d \quad \text{with probability at least } 1 - \delta.
\]


Question: Once software has been shipped, what does the Software Development Lifecycle require you to do to maintain security guarantees?

To maintain security guarantees after software has been shipped, the Software Development Lifecycle (SDLC) requires the following actions:

1. **Track the Evolution of Third Party Dependencies**:
   - Continuously monitor and assess third-party libraries and frameworks used in the software.
   - Stay informed about vulnerabilities, updates, and patches related to these dependencies to mitigate potential security risks.

2. **Deploy Updates Timely and Safely**:
   - Implement a robust update mechanism that allows for timely deployment of security patches and updates.
   - Ensure that updates are tested for stability and security before being released to users, minimizing the risk of introducing new vulnerabilities.

While providing new features to attract users and ensuring compatibility with newer machines can be important for overall software health, they do not directly relate to maintaining security guarantees post-deployment.


Question:
You have been publishing a daily column for the Gazette over the last few years and have recently reached a milestone --- your 1000th column! Realizing you'd like to go skiing more often, you decide it might be easier to automate your job by training a story generation system on the columns you've already written. Then, whenever your editor pitches you a title for a column topic, you'll just be able to give the title to your story generation system, produce the text body of the column, and publish it to the website!
Your column generation system has become quite successful and you've managed to automate most of your job simply by typing your editor's title pitches into your model to produce your column every day. Two years later, during the COVID--25 pandemic, your editor proposes to use your system to generate an information sheet about the pandemic for anyone looking for information about symptoms, treatments, testing sites, medical professionals, etc. Given the similarity to a previous pandemic many years before, COVID--19, you train your model on all news articles published about COVID--19 between the years of 2019--2022. Then, you generate the information page from your trained model.
Give an example of a potential harm that your model could produce from the perspective of human interaction harms.

A potential harm could be the generation of misleading or inaccurate information about COVID-25 symptoms or treatments, which could lead individuals to make unsafe health decisions or avoid seeking necessary medical care, thereby endangering their health and safety.


Question: Thick the \emph{incorrect} assertion.

To analyze the assertions regarding SAS-based cryptography, let's break down each option step-by-step:

1. **The goal of SAS-based cryptography is to reduce the length of the string that has to be authenticated.**
   - This statement is misleading. The primary goal of SAS (Short Authentication String) is to provide a way to authenticate a connection or a session using a short string, but it does not inherently aim to reduce the length of the string that needs to be authenticated. Instead, it focuses on creating a secure and user-friendly method for authentication. Therefore, this assertion is likely incorrect.

2. **One way to authenticate a SAS is to use your phone.**
   - This statement is correct. SAS can be authenticated through various means, including using a mobile device. For example, users can verify a SAS displayed on a device by comparing it with a SAS shown on their phone, which is a common practice in secure communications.

3. **One can obtain a secure channel from a narrowband authenticated channel using SAS-based cryptography.**
   - This statement is generally correct. SAS-based cryptography can help establish a secure channel by authenticating the parties involved, even if the initial channel is narrowband or limited in capacity.

4. **SAS-based cryptography always requires the SAS to be collision-resistant.**
   - This statement is incorrect. While collision resistance is a desirable property in cryptographic functions, SAS-based cryptography does not necessarily require the SAS itself to be collision-resistant. The focus is more on the secure authentication process rather than the collision resistance of the SAS.

Based on this analysis, the **incorrect assertion** is:

**1. The goal of SAS-based cryptography is to reduce the length of the string that has to be authenticated.**

This option misrepresents the primary goal of SAS-based cryptography.


Question:
Assume that your team is discussing the following java code:

public final class DataStructure {
    public void add(int val) { /*...*/ }

    private boolean isFull() { /*...*/ }
}

One of your colleagues thinks that "isFull" should be made public. Explain whether this breaks backward compatibility and why or why not (also without worrying about whether this is a good or a bad thing)

The answer is correct. In Java, if a method is declared as private, it means that it can only be accessed within the same class. No other class, including subclasses, can call or see this method. Therefore, if you change the visibility of the `isFull` method from private to public, it does not break backward compatibility because no existing code outside of the `DataStructure` class relies on that method being public.

Backward compatibility typically refers to the ability of existing code to function without modification when the codebase is updated. Since `isFull` was never accessible to any external code, changing its visibility does not affect any existing functionality or behavior of the class as used by other code. Thus, while it may change how the method can be accessed in the future, it doesn't break any existing uses of the class.


Question: For his awesome research, Tugrulcan is going to use the PageRank with teleportation and HITS algorithm, not on a network of webpages but on the retweet network of Twitter! The retweet network is a directed graph, where nodes are users and an edge going out from a user A and to a user B means that "User A retweeted User B". Which one is false about a Twitter bot that retweeted other users frequently but got never retweeted by other users or by itself?

To answer this question, we first need to understand a few key concepts related to the PageRank and HITS algorithms, as well as the structure of the retweet network on Twitter.

**Key Definitions:**

1. **Directed Graph:** A structure consisting of nodes (users, in this case) connected by edges (retweets), where the direction of the edges matters.

2. **HITS Algorithm:** This algorithm assigns two values to each node:
   - **Hub Value:** Measures the quality of a node as a source of information (in this case, a user who retweets).
   - **Authority Value:** Measures the quality of a node as a destination of information (a user who is retweeted).

3. **PageRank:** This algorithm assigns a score to each node based on the number and quality of incoming links (retweets). A higher score indicates a more important node.

Now, letâs analyze each choice in relation to a Twitter bot that retweeted other users frequently but was never retweeted by any users or itself.

**A. It will have a non-zero hub value.**
This statement is **true**. Since the bot frequently retweets other users, it acts as a source of information. Therefore, it will have a non-zero hub value as it is contributing to the retweet network.

**B. It will have an authority value of zero.**
This statement is **true**. The bot is not being retweeted by anyone, meaning it does not receive any incoming links. Hence, its authority value, which depends on being retweeted, will be zero.

**C. It will have a PageRank of zero.**
This statement is **false**. The PageRank is based on the structure of the graph and the bot's outgoing edges (retweets). Since the bot is actively retweeting, it will have some non-zero PageRank score, although it might be low, it cannot be zero because it contributes retweets to the network.

**D. Its authority value will be equal to the hub value of a user who never retweets other users.**
This statement is **true**. A user who never retweets other users will also have an authority value of zero (as they do not receive any retweets). Since the bot has an authority value of zero, it will indeed be equal to the hub value of this user, which is also zero.

**Conclusion:**

The **correct answer is C**. It is false to say that the bot will have a PageRank of zero because, despite not being retweeted, its outgoing retweets contribute to its PageRank. The other statements (A, B, and D) are true based on the bot's behavior in the retweet network.


Question:
Consider an IR engine, which uses an indexing mechanism implementing the following 3 consecutive filters:

a morpho-syntactic filter that restricts indexing term candidates to only nouns, and reduces them to their root forms;
a frequencial filter parameterized with \(f_\text{min}=0.06\) (resp. \(f_\text{max}=0.20\)) as lower (resp. upper) cut-off value, expressed as relative frequencies;
a stop word filter using the following stop list: {a, in, mouse, the}.

and the following document \(d\):

Cats are the worst enemies of rodents. After all, a cat is a cat: as soon as it can, it rushes into the bushes with only one target in mind: mice, mice and mice! Naturally, the cats of houses are less frightening, as for them croquette loaded dressers have replaced prey hiding bushes. Cat's life in the house is easy!...

What is the multi-set resulting from the indexing of document \(d\) by the above described IR engine?
Format your answer as an alphabetically ordered list of the form: "lemma1(tf1), lemma2(tf2), ...", where tfi is the term frequency of indexing term i.
For instance: dog(2), frog(3), zebra(1)

To find the multi-set resulting from the indexing of document \(d\) using the described IR engine, we will follow the three filters step by step.

1. **Morpho-syntactic filter**: We will extract only nouns from the document and reduce them to their root forms (lemmas).

   From the document, the nouns are:
   - Cats → cat
   - enemies → enemy
   - rodents → rodent
   - cat (repeated)
   - target → target
   - mice (repeated)
   - cats → cat (repeated)
   - dressers → dresser
   - prey → prey
   - bushes (implied but not a noun in this context)
   - life → life (not a noun in this context)

   The unique nouns extracted and reduced to their root forms are:
   - cat (3 occurrences)
   - enemy (1 occurrence)
   - rodent (1 occurrence)
   - target (1 occurrence)
   - mouse (3 occurrences)
   - dresser (1 occurrence)
   - prey (1 occurrence)

2. **Frequencial filter**: We need to calculate the relative frequencies of each term. The total number of words in \(d\) is 59. We calculate the frequency of each lemma:

   - cat: \( \frac{3}{59} \approx 0.0508 \)
   - enemy: \( \frac{1}{59} \approx 0.0169 \)
   - rodent: \( \frac{1}{59} \approx 0.0169 \)
   - target: \( \frac{1}{59} \approx 0.0169 \)
   - mouse: \( \frac{3}{59} \approx 0.0508 \)
   - dresser: \( \frac{1}{59} \approx 0.0169 \)
   - prey: \( \frac{1}{59} \approx 0.0169 \)

   Now applying the frequencial filter with \(f_\text{min}=0.06\) and \(f_\text{max}=0.20\):
   - cat: 0.0508 (discard)
   - enemy: 0.0169 (discard)
   - rodent: 0.0169 (discard)
   - target: 0.0169 (discard)
   - mouse: 0.0508 (discard)
   - dresser: 0.0169 (discard)
   - prey: 0.0169 (discard)

   All terms are below the lower cutoff \(f_\text{min}=0.06\).

3. **Stop word filter**: The stop words are {a, in, mouse, the}. The word "mouse" is part of the stop words, but since it was discarded in the frequencial filter, we do not need to check it further.

Since all terms have been discarded due to the frequencial filter, the final multi-set is empty.

### Answer ###
(no terms)


Question: Calculate the mean of individuals who remain alive in the data. The data is stored in a pandas.DataFrame and the respective column is "alive".

To calculate the mean of the values in the 'alive' column, where `1` represents individuals who are alive and `0` represents those who are not, follow these steps:

1. **Calculate the Mean**: You can directly calculate the mean of the 'alive' column in the DataFrame. This will give you the proportion of individuals who are alive.

Here’s how to do it in code:

```python
import pandas as pd

# Assuming df is your DataFrame
mean_alive = df['alive'].mean()
```

This code calculates the mean of the 'alive' column, which will yield a value between `0` and `1`. If the mean is close to `1`, it indicates a higher proportion of individuals are alive; if it’s closer to `0`, then fewer individuals are alive.


Question: Consider an undirected graph $G=(V,E)$ and let $s\neq t\in V$. In the minimum (unweighted) $s,t$-cut problem, we wish to find a set $S\subseteq V$ such that $s\in S$, $t\not \in S$ and the number of edges crossing the cut is minimized. We shall use a linear program to solve this problem. Let ${P}$ be the set of all paths between $s$ and $t$ in the graph $G$. The linear program has a variable $y_e$ for each edge $e\in E$ and is defined as follows: \begin{equation*} \begin{array}{ll@{}ll} \text{minimize}  & & \displaystyle\sum_{e \in E} y_e &\\ \text{subject to}& & \displaystyle\sum_{e \in p} y_e \ge 1  &\forall p \in P,\\ & & y_e \ge 0 & \forall e \in E. \end{array} \end{equation*} For example, consider the following graph where the numbers on the edges depict the $y_e$-values of a feasible solution to the linear program: \begin{center} \input{cutExample} \end{center} The values on the edges depict a feasible but not optimal solution to the linear program. That it is feasible follows because each $y_e$ is non-negative and $\sum_{e\in p} y_e \geq 1$ for all $p\in P$. Indeed, for the path $s, b, a, t$ we have $y_{\{s,b\}}+ y_{\{b,a\}} + y_{\{a,t\}} = 1/4 + 1/4 + 1/2 = 1$, and similar calculations for each path $p$ between $s$ and $t$ show that $\sum_{e\in p} y_e \geq 1$. That the solution is not optimal follows because its value is $2.5$ whereas an optimal solution has value $2$. Let $\opt$ denote the number of edges crossing a minimum $s,t$-cut and let $\optlp$ denote the value of an optimal solution the linear program. Prove that $\optlp \leq \opt$. \\ {\em (In this problem you are asked to prove $\optlp \leq \opt$. Recall that you are allowed to refer to material covered in the lecture notes.)}

To prove that \(\optlp \leq \opt\), we need to establish a relationship between the optimal value of the linear program \(\optlp\) and the minimum number of edges crossing an \(s,t\)-cut, denoted as \(\opt\). Hereâs a step-by-step breakdown of the reasoning:

1. **Understanding the Linear Program**:
   - The linear program is designed to minimize the sum of variables \(y_e\) corresponding to edges \(e\) in the graph \(G\).
   - The constraints require that for each path \(p\) from \(s\) to \(t\), the sum of \(y_e\) for all edges \(e\) in that path must be at least 1. This ensures that at least one unit of flow is assigned to each path from \(s\) to \(t\).

2. **Defining the Minimum \(s,t\)-cut**:
   - An \(s,t\)-cut in the graph is defined as a partition of the vertices \(V\) into two disjoint sets \(S\) and \(V \setminus S\) such that \(s \in S\) and \(t \notin S\).
   - The edges crossing this cut are those that have one endpoint in \(S\) and the other in \(V \setminus S\). The number of such edges is \(\opt\).

3. **Constructing a Cut Based on the Flow**:
   - We can use the values of \(y_e\) from an optimal solution of the linear program to construct a feasible flow that corresponds to an \(s,t\)-cut.
   - For each edge \(e\), we consider the value \(y_e\) as a measure of flow through that edge. If we think of \(y_e\) as contributing to the "capacity" or "flow" through that edge, we can analyze how many edges are effectively utilized in the flow from \(s\) to \(t\).

4. **Flow and Cuts**:
   - By the Max-Flow Min-Cut Theorem, the maximum value of flow from \(s\) to \(t\) in a flow network is equal to the capacity of the minimum cut that separates \(s\) from \(t\).
   - In our case, the linear programâs objective function \(\sum_{e \in E} y_e\) can be interpreted as the total "flow" sent across the edges, which directly relates to the edges crossing a cut.

5. **Relating LP Value to Minimum Cut**:
   - Since \(\optlp\) represents the minimum total flow that can satisfy the path constraints, and \(\opt\) represents the minimum number of edges that can be cut to separate \(s\) from \(t\), we conclude that \(\optlp\) cannot exceed \(\opt\).
   - Formally, this is because any feasible flow solution (i.e., any assignment of \(y_e\) values that meets the path constraints) must correspond to a cut that captures the minimum number of edges required to separate \(s\) and \(t\).

6. **Conclusion**:
   - By establishing that the feasible solutions to the linear program correspond to cuts in the graph and that the value of the linear program can never exceed the minimum edge cut required to separate \(s\) from \(t\), we arrive at the conclusion:
   \[
   \optlp \leq \opt.
   \]

Thus, we have proved that the value of the linear program is less than or equal to the size of the minimum \(s,t\)-cut in the graph.

Final Answer: \(\optlp \leq \opt\).


Question: Vectorize the input with the Vector Space Model

Answer: To vectorize the input using the Vector Space Model, represent each document as a vector in a multi-dimensional space where each dimension corresponds to a unique term from the vocabulary. The vector components can be term frequency (TF) or term frequency-inverse document frequency (TF-IDF). The formula for TF-IDF is: \( TF-IDF(t, d) = TF(t, d) \times IDF(t) \), where \( IDF(t) = \log(\frac{N}{DF(t)}) \) and \( N \) is the total number of documents, and \( DF(t) \) is the number of documents containing term \( t \).

Explanation: I explained the process of vectorization in the Vector Space Model and provided the relevant formulas for TF and TF-IDF.


Question: Which of the following is true regarding the random forest classification algorithm?

1) **It is not suitable for parallelization.**
False. Random forests can build each tree independently, making them suitable for parallelization.

2) **It uses only a subset of features for learning in each weak learner.**
True! Each decision tree is built using a random subset of features, typically \( k \) out of \( m \) total features, which enhances diversity and reduces overfitting.

3) **We compute a prediction by randomly selecting the decision of one weak learner.**
False. Predictions are made by aggregating the outputs of all trees through majority voting.

4) **It produces a human interpretable model.**
False. While individual trees are interpretable, a random forest is complex and less interpretable as a whole.

In conclusion, the correct answer is **2)**: It uses only a subset of features for learning in each weak learner.


Question:
Consider two fully connected networks, A and B, with a constant width for all layers, inputs and outputs.
	Network A has depth $3L$ and width $H$, network B has depth $L$ and width $2H$.
	Everything else is identical for the two networks and both $L$ and $H$ are large.
	In this case, performing a single iteration of backpropagation requires fewer scalar multiplications for network A than for network B.

To determine whether the statement is true or false, we need to analyze the computational cost of performing a single iteration of backpropagation for both networks A and B.

1. **Network A** has:
   - Depth: \(3L\)
   - Width: \(H\)

2. **Network B** has:
   - Depth: \(L\)
   - Width: \(2H\)

### Backpropagation Cost Analysis

The computational cost of backpropagation in a fully connected network primarily depends on the number of layers (depth) and the number of neurons in each layer (width). The cost can be approximated as follows:

- For each layer, the number of scalar multiplications required to compute the outputs is proportional to the product of the number of neurons in the current layer and the number of neurons in the next layer.

For a fully connected layer with \(W\) neurons and \(H\) neurons in the next layer, the cost is \(W \times H\).

#### Network A:
- The first layer has \(H\) inputs and \(H\) outputs.
- The second layer has \(H\) inputs and \(H\) outputs.
- The third layer has \(H\) inputs and \(H\) outputs.
- This pattern continues for \(3L\) layers.

The total number of scalar multiplications for backpropagation in Network A can be approximated as:
\[
\text{Cost}_A \approx (H \times H) \times (3L) = 3L H^2
\]

#### Network B:
- The first layer has \(2H\) inputs and \(2H\) outputs.
- The second layer has \(2H\) inputs and \(2H\) outputs.
- This pattern continues for \(L\) layers.

The total number of scalar multiplications for backpropagation in Network B can be approximated as:
\[
\text{Cost}_B \approx (2H \times 2H) \times L = 4L H^2
\]

### Comparison
Now we compare the costs:
- Cost for Network A: \(3L H^2\)
- Cost for Network B: \(4L H^2\)

Since \(3L H^2 < 4L H^2\), it follows that:
\[
\text{Cost}_A < \text{Cost}_B
\]

### Conclusion
The statement "performing a single iteration of backpropagation requires fewer scalar multiplications for network A than for network B" is **True**.

Thus, the correct answer is:
1. **True**


Question: Let $X$, $Y$, and $K$ be respectively the plaintext, ciphertext, and key distributions. $H$ denotes the Shannon entropy. The consequence of perfect secrecy is \dots

Perfect secrecy is a fundamental concept in cryptography that ensures that a ciphertext provides no information about the plaintext without the key. This concept is primarily defined in the context of the one-time pad (OTP), where the key is as long as the message, truly random, and used only once.

To define perfect secrecy mathematically, we use the following notation:

- Let \( P \) be the set of plaintext messages.
- Let \( C \) be the set of ciphertext messages.
- Let \( K \) be the set of keys.
- The encryption function is denoted as \( E: P \times K \rightarrow C \).
- The decryption function is denoted as \( D: C \times K \rightarrow P \).

For a cryptographic system to achieve perfect secrecy, the requirement is that for every plaintext \( p \in P \) and every ciphertext \( c \in C \):

\[
P(C = c | P = p) = P(C = c)
\]

This equation states that the conditional probability of observing a specific ciphertext given a specific plaintext must be equal to the overall probability of observing that ciphertext. In simpler terms, knowing the ciphertext should not provide any information about the plaintext.

### Entropy in Perfect Secrecy

Entropy, denoted by \( H \), measures the uncertainty or randomness in a random variable. In the context of cryptography, the entropy of a plaintext message is crucial because it reflects the amount of information contained in that message. The higher the entropy, the more unpredictable the message is, which is essential for achieving perfect secrecy.

The entropy \( H(X) \) of a discrete random variable \( X \) is defined as:

\[
H(X) = -\sum_{x \in X} P(x) \log P(x)
\]

where \( P(x) \) is the probability of each outcome \( x \). In perfect secrecy, the key must have at least as much entropy as the plaintext to ensure that every possible plaintext can map uniquely to a ciphertext.

### Evaluation of Options

Now, considering the conditions for perfect secrecy, let's evaluate the following options:

A. If the key length is shorter than the plaintext length, perfect secrecy cannot be achieved.
- This statement is true. The key must be at least as long as the plaintext to ensure that every plaintext can correspond to every ciphertext, maintaining the condition of perfect secrecy.

B. Perfect secrecy ensures that the ciphertext contains no information about the plaintext.
- This is accurate. Perfect secrecy guarantees that the ciphertext reveals no information about the plaintext, as per the definition provided.

C. Entropy of the key must equal the entropy of the plaintext for perfect secrecy.
- This statement is not strictly necessary. While the key must have sufficient entropy to cover the plaintext, it does not have to be equal. The key can have higher entropy than the plaintext, but it must not be lower.

D. Perfect secrecy can be achieved with non-uniform distributions of plaintext.
- This is misleading. While perfect secrecy can still technically be achieved, the key must still be uniformly distributed and as long as the plaintext to maintain the condition of equal probability for all ciphertexts.

In summary, for perfect secrecy to hold, the key must have sufficient entropy, and the relationship between the key and the plaintext must be maintained to ensure that no information is leaked through the ciphertext.

The correct options based on the evaluation are: A, B.


Question:
Assume that you are part of a team developing a mobile app using Scrum.
When using the app, you identified multiple bugs and features which you think should be implemented, and took some notes. You want to
share these with the Product Owner. Your backlog of tasks includes the following task:
- [ ] Login
Is this item suitable to be submitted to the Product Backlog? Why?

Yes, the item "Login" is suitable to be submitted to the Product Backlog. Here’s a clear explanation of why this is the case:

### 1. **Definition of Product Backlog:**
The Product Backlog is a prioritized list of features, enhancements, bug fixes, and technical work that needs to be completed for the product. It serves as the single source of truth for what needs to be done to improve the product.

### 2. **Nature of the Task:**
The task "Login" represents a fundamental feature of most mobile applications. It is essential for user authentication and access control, which are critical for user experience and security. As such, it is a necessary component that should be included in the Product Backlog.

### 3. **User Needs and Requirements:**
If users are expected to log in to use the app, then this feature directly addresses user needs. Including it in the Product Backlog ensures that it is considered during the development process and prioritized appropriately based on user feedback and business goals.

### 4. **Prioritization:**
Once the task is added to the Product Backlog, the Product Owner can prioritize it against other tasks and features based on factors such as user demand, business value, and technical feasibility. This prioritization is crucial for effective Scrum practices.

### 5. **Collaboration and Transparency:**
By submitting the "Login" task to the Product Backlog, you foster collaboration and transparency within the Scrum team. The Product Owner can discuss this task with the team during backlog refinement sessions, ensuring everyone is aligned on the importance of the feature.

### Conclusion:
In summary, the "Login" task is suitable for submission to the Product Backlog because it is a critical feature that addresses user needs, contributes to the overall functionality of the app, and requires prioritization and discussion within the Scrum framework.


Question:
Given a matrix $\Xm$ of shape $D	imes N$ with a singular value decomposition (SVD), $X=USV^	op$, suppose $\Xm$ has rank $K$ and  $\Am=\Xm\Xm^	op$.
        Which one of the following statements is 	extbf{false}?

The statement that is **false** is: 'The eigenvalues of A are the singular values of X' because the eigenvalues of A are the square of the singular values of X.


Question:
You are responsible for a project aiming at providing on-line recommendations to the customers of
a on-line book selling company.
The general idea behind this recommendation system is to cluster books according to both customers
and content similarities, so as to propose books similar to the books already bought by a given
customer. The core of the recommendation system is a clustering algorithm aiming at regrouping
books likely to be appreciate by the same person. This clustering should not only be achieved
based on the purchase history of customers, but should also be refined by the content of the books
themselves. It's that latter aspect we want to address in this exam question.

Consider the following six 'documents' (toy example):
d1: 'Because cows are not sorted as they return from the fields to their home pen, cow flows
are improved.'
d2: 'He was convinced that if he owned the fountain pen that he'd seen in the shop window for years, he could write fantastic stories with it. That was the kind of pen you cannot forget.'
d3: 'With this book you will learn how to draw humans, animals (cows, horses, etc.) and flowers with a charcoal pen.'
d4: 'The cows were kept in pens behind the farm, hidden from the road. That was the typical kind of pen made for cows.'
d5: 'If Dracula wrote with a fountain pen, this would be the kind of pen he would write with, filled with blood red ink. It was the pen she chose for my punishment, the pen of my torment. What a mean cow!'
d6: 'What pen for what cow? A red pen for a red cow, a black pen for a black cow, a brown pen for a brown cow, ... Understand?'

and suppose (toy example) that they are indexed only by the two words: pen and cow.

What is the result of the dendrogram clustering algorithm on those six documents, using
the cosine similarity and single linkage?
Explain all the steps.
Hint: $5 / \sqrt{34}<3 / \sqrt{10}<4 / \sqrt{17}$.

To perform clustering on the six documents (d1 to d6) using cosine similarity and single linkage, we will follow a systematic approach. The steps include calculating cosine similarities between the documents, forming a distance matrix, and then applying the single linkage clustering method to create a dendrogram.

### Step 1: Understand Cosine Similarity

Cosine similarity is a measure that calculates the cosine of the angle between two non-zero vectors in a multi-dimensional space. It is defined as:

\[
\text{cosine similarity} = \frac{A \cdot B}{\|A\| \|B\|}
\]

Where \(A\) and \(B\) are the vectors representing the documents, \(A \cdot B\) is the dot product, and \(\|A\|\) and \(\|B\|\) are the magnitudes (norms) of the vectors.

### Step 2: Create the Document Vectors

Since the documents are indexed only by the two words "pen" and "cow", we can represent each document as a vector in a 2D space. The vectors can be constructed as follows (counting the occurrences of "pen" and "cow"):

- **d1**: (0, 2)
- **d2**: (1, 0)
- **d3**: (1, 1)
- **d4**: (0, 2)
- **d5**: (1, 1)
- **d6**: (1, 1)

### Step 3: Calculate Cosine Similarities

We will calculate the cosine similarity for the pairs of documents where it is necessary. We can focus on the pairs that are likely to be the most similar based on the context of the documents.

1. **d2 and d5**:
   - \(D(2,5) = \frac{1 \cdot 1 + 0 \cdot 1}{\sqrt{(1^2 + 0^2)(1^2 + 1^2)}} = \frac{1}{\sqrt{1 \cdot 2}} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\)

2. **d3 and d5**:
   - \(D(3,5) = \frac{1 \cdot 1 + 1 \cdot 1}{\sqrt{(1^2 + 1^2)(1^2 + 1^2)}} = \frac{2}{\sqrt{2 \cdot 2}} = 1\)

3. **d1 and d3**:
   - \(D(1,3) = \frac{0 \cdot 1 + 2 \cdot 1}{\sqrt{(0^2 + 2^2)(1^2 + 1^2)}} = \frac{2}{\sqrt{4 \cdot 2}} = \frac{2}{\sqrt{8}} = \frac{1}{\sqrt{2}}\)

### Step 4: Create the Distance Matrix

From the calculations, we can summarize the distances as follows:

- \(D(2, 5) = \frac{4}{\sqrt{17}}\)
- \(D(3, 5) = \frac{5}{\sqrt{34}}\)
- \(D(1, 3) = \frac{3}{\sqrt{10}}\)

### Step 5: Apply Single Linkage Clustering

1. **Initial Clusters**: Each document starts in its own cluster:
   - {d1}, {d2}, {d3}, {d4}, {d5}, {d6}


Question: What happens in the uniform reliable broadcast algorithm if the completeness property of the failure detector is violated?

To address the question regarding the implications of a violation of the completeness property in a uniform reliable broadcast algorithm, we first need to clarify some fundamental concepts related to distributed systems, particularly focusing on failure detectors and broadcast algorithms.

### Step 1: Understanding Uniform Reliable Broadcast

Uniform reliable broadcast is a communication protocol used in distributed systems to ensure that messages sent by a process are delivered reliably and uniformly to all correct processes. The key properties of this algorithm include:

1. **Validity**: If a correct process broadcasts a message, then some correct process eventually delivers that message.
2. **Uniformity**: If a correct process delivers a message, then all correct processes eventually deliver that message.
3. **Integrity**: A message is delivered at most once by each correct process.

### Step 2: The Role of Failure Detectors

Failure detectors are mechanisms used in distributed systems to provide information about the liveness of processes. They can be categorized based on their properties:

- **Completeness**: If a process is correct, then eventually, the failure detector will suspect it as correct.
- **Accuracy**: If a process is suspected, it is actually faulty.

In the context of uniform reliable broadcast, failure detectors help in determining which processes are operational and which may have failed, thereby guiding the broadcast decisions.

### Step 3: Implications of Violating Completeness

When the completeness property of the failure detector is violated, it means that there may be correct processes that are incorrectly suspected of having failed. This can lead to several issues in the context of uniform reliable broadcast:

1. **Loss of Messages**: If a correct process is suspected to have failed, other processes may stop sending messages to it. As a result, messages intended for that process may not be delivered, violating the validity property.

2. **Inconsistent State**: Different processes may reach different conclusions about which processes are alive or dead. This inconsistency can lead to scenarios where some processes believe they have successfully delivered a message while others do not, violating the uniformity property.

3. **Increased Complexity in Recovery**: If processes are incorrectly suspected of failure, recovery mechanisms may be triggered unnecessarily. This can lead to wasted resources and increased complexity in managing state among processes.

### Step 4: Conclusion

In summary, if the completeness property of the failure detector is violated in a uniform reliable broadcast algorithm, it can lead to significant disruptions in communication among processes. Specifically:

- Messages may not be delivered to all correct processes.
- The system may enter an inconsistent state regarding which processes are operational.
- Recovery efforts may be misguided, leading to inefficiencies.

Thus, maintaining the completeness property is crucial for ensuring the reliability and consistency of message broadcasting in distributed systems.


Question: Choose the \emph{correct} statement

To analyze each option regarding elliptic curves, we need to understand the mathematical structure of elliptic curves in the context of algebraic geometry and number theory.

1. **Elliptic curves form a field.**
   - A field is a set equipped with two operations (usually addition and multiplication) that satisfy certain properties (commutativity, associativity, distributivity, etc.). Elliptic curves themselves are not fields; they are sets defined by specific equations (usually in the form \(y^2 = x^3 + ax + b\)). Therefore, this option is incorrect.

2. **Elliptic curves form a ring.**
   - A ring is a set equipped with two operations and satisfies certain properties, such as closure under addition and multiplication and the existence of an additive identity and inverses. While elliptic curves can be associated with various algebraic structures, they generally do not form a ring in the traditional sense. Therefore, this option is also incorrect.

3. **Elliptic curves form an Abelian group.**
   - An Abelian group is a set equipped with a single operation that is associative, commutative, and has an identity element and inverses for every element. The points on an elliptic curve do form an Abelian group when we define a suitable addition operation for the points. This is a fundamental property of elliptic curves, making this option correct.

4. **Elliptic curves form an ideal.**
   - An ideal is a special subset of a ring that absorbs multiplication by any element of the ring. Since elliptic curves are not rings, the concept of an ideal does not directly apply here. Hence, this option is incorrect.

Considering the evaluations of all options, the only accurate statement about elliptic curves is:

**Correct answer:** 3. **Elliptic curves form an Abelian group.**

This reflects the structure that arises when we define an appropriate point addition on the curve, making the set of points on the elliptic curve (including the point at infinity) satisfy the properties of an Abelian group.


Question: Which of the following is an acceptable commitment scheme, i.e., one that verifies the hiding and binding property (for a well chosen primitive and suitable $x$ and $r$):

To determine which of the provided options forms an acceptable commitment scheme that satisfies the hiding and binding properties, we analyze each one:

Option 0: ( Commit(x;r) = Enc_r(x) ), where ( Enc_r ) is a symmetric encryption scheme with key ( r ).

Hiding: This holds as long as the symmetric encryption scheme is secure; the encryption of ( x ) under a secret key ( r ) would be computationally indistinguishable from random noise.
Binding: This holds because, given the key ( r ), a decryption of ( Enc_r(x) ) will yield a unique ( x ), thus preventing an adversary from generating another ( x' ) such that ( Enc_r(x') = Enc_r(x) ) without knowing ( r ).
Conclusion: Acceptable commitment scheme.
Option 1: ( Commit(x;r) = H(x) ), where ( H ) is a hash function.

Hiding: This scheme does not guarantee hiding, as knowing ( x ) allows computation of ( H(x) ), and without knowledge of ( x ), it might be possible for an attacker to find ( x ) such that ( H(x) ) outputs the same result.
Binding: While it is difficult to find two different ( x ) values that hash to the same output (collision resistance), there is no randomness involved that enforces a binding against multiple ( x ). An adversary can simply choose any ( x ) and compute ( H(x) ).
Conclusion: Not an acceptable commitment scheme.
Option 2: ( Commit(x;r) = x \oplus r ).

Hiding: This scheme fails to hide ( x ). Knowing ( r ) allows one to easily recover ( x = Commit(x;r) \oplus r ), making it easily detectable. There's no secrecy in the committed value.
Binding: While the XOR operation is a one-to-one function with respect to both inputs, without hiding, the lack of a commitment to one particular value makes this scheme unsuitable.
Conclusion: Not an acceptable commitment scheme.
Option 3: ( Commit(x;r) = H(r|x) ), where ( H ) is a hash function and ( | ) denotes concatenation.

Hiding: This is hiding due to the random ( r ) that is concatenated with ( x ). Given ( r ), the output ( H(r|x) ) will look random and should not reveal information about ( x ) without ( r ).
Binding: Assuming ( H ) is a collision-resistant hash function, it is hard to find any two different pairs ( (x, r) ) and ( (x', r) ) such that ( H(r|x) = H(r|x') ) for the same ( r ). Thus, it binds the value of ( x ) to ( r ).
Conclusion: Acceptable commitment scheme.
Based on this analysis, we conclude the acceptable commitment schemes are:

Option 0: ( Commit(x;r) = Enc_r(x) )
Option 3: ( Commit(x;r) = H(r|x) )
Thus, the answer is 0 and 3.


Question:
Let $\xv_1, . . . , \xv_N$ be a dataset of $N$ vectors in $\R^D$. What does it mean for the data vectors $\xv_1, . . . , \xv_N$ to be centered, as for principle component analysis (PCA) to be meaningful?
Use the notation $x_{nd}$ for individual entries.

For the data vectors \(\xv_1, \ldots, \xv_N\) in \(\mathbb{R}^D\) to be considered centered, it means that the mean of each dimension across all vectors is zero. Mathematically, this can be expressed as follows: for each dimension \(d\) (where \(d = 1, 2, \ldots, D\)), the mean of the \(d\)-th entries of all vectors must satisfy the condition \(\frac{1}{N} \sum_{n=1}^{N} x_{nd} = 0\). This implies that if we denote the mean vector as \(\bar{\xv} = \left(\frac{1}{N} \sum_{n=1}^{N} x_{n1}, \frac{1}{N} \sum_{n=1}^{N} x_{n2}, \ldots, \frac{1}{N} \sum_{n=1}^{N} x_{nD}\right)\), then we require \(\bar{\xv} = \mathbf{0}\). Centering the data is crucial for Principal Component Analysis (PCA) because PCA seeks to identify the directions (principal components) along which the data varies the most. If the data is not centered, the principal components will be influenced by the mean of the data rather than the structure of the data itself. As a result, PCA might identify directions that do not truly represent the underlying variance of the data. By centering the data, we ensure that the first principal component corresponds to the direction of maximum variance around the origin, thus allowing PCA to effectively reveal the intrinsic properties of the dataset. This transformation helps in simplifying the analysis and interpretation of the data's structure in lower-dimensional spaces, making PCA a powerful tool for data exploration and dimensionality reduction.


Question: Which statement is true for linear regression?

Let's analyze each statement regarding linear regression:

### Option A:
**A linear regression model can be expressed as an inner product between feature vectors and a weight vector.**

This statement is true. In linear regression, we model the relationship between the dependent variable \( y \) and independent variables \( X \) using the equation:

\[
y = X\beta + \epsilon
\]

where \( y \) is the vector of outputs, \( X \) is the matrix of input feature vectors, \( \beta \) is the vector of weights (coefficients), and \( \epsilon \) represents the error term. The expression \( X\beta \) is indeed the inner product between the feature vectors (contained in \( X \)) and the weight vector \( \beta \) when we consider the prediction of a single observation.

### Option B:
**Linear regression, when using 'usual' loss functions, works fine when the dataset contains many outliers.**

This statement is false. Linear regression is sensitive to outliers because it typically uses the mean squared error (MSE) as a loss function. When outliers are present, they can disproportionately influence the estimation of the coefficients, leading to a poor fit. Robust regression techniques or different loss functions (like Huber loss) are often used in cases where outliers are a concern.

### Option C:
**A good fit with linear regression implies a causal relationship between inputs and outputs.**

This statement is also false. A good fit in linear regression (indicated by metrics like \( R^2 \) or low error) suggests that the model explains the variance in the output well, but it does not imply causation. Correlation does not imply causation, and there could be confounding variables or other factors influencing the relationship. Establishing causality requires more rigorous experimental or longitudinal study designs.

### Conclusion:
The only true statement among the provided options is:

**A. A linear regression model can be expressed as an inner product between feature vectors and a weight vector.**


Question:
we'd like to do some sentence topic classification using a Naive-Bayes model. Consider the following toy learning corpus, where each sentence has been assigned a topic, either "Medical" or "Computer":
\item Medical: plastic surgery process initial consultation can be scheduled by sending an email to the administration.
\item Medical: in the process, the laser beam comes into contact with soft tissues.
\item Medical: laser eye surgery process reshapes parts of the cornea by removing tiny amount of tissues.
\item Computer: the team behind the laser based quantum computer includes scientists from the US, Australia and Japan.
\item Computer: the optical laser barcode scanner was plugged on the USB port.
\item Computer: cdrom laser lens cleaning process starts with opening the tray.
\item Computer: laser was a computer trademark.
The parameters are learned using some appropriate additive smoothing with the same value for all parameters. In the above learning corpus, there are $42$ token occurrences in "Medical" documents and $42$ token occurrences in "Computer" documents (punctuation is ignored).
How would the following short sentence:
"pulsed laser used for surgery process"
be classified by this model?

To classify the sentence "pulsed laser used for surgery process" using the Naive-Bayes model, we need to calculate the probabilities for each topic ("Medical" and "Computer") given the sentence. We will use the following steps:

1. **Calculate the prior probabilities** for each class:
   - P(Medical) = Number of Medical sentences / Total number of sentences
   - P(Computer) = Number of Computer sentences / Total number of sentences

   There are 3 Medical sentences and 4 Computer sentences, so:
   - P(Medical) = 3 / 7
   - P(Computer) = 4 / 7

2. **Calculate the likelihoods** of the words in


Question:
Given the following function sums:
1 def add(c: Int, acc: List[(Int, Int)]): List[(Int, Int)] = acc match
2 case Nil => List((c, 1))
3 case x :: xs => if x._1 == c then (c, x._2+1) :: xs else x :: add(c, xs)
4
5 def sums(digits: List[Int]): List[(Int, Int)] =
6 digits.foldRight(List[(Int, Int)]())(add)
Your task is to identify several operations on lists of digits:
What does the following operation implement, for a given input list of digits?
1 def mystery3(digits: List[Int]): Int = sums(digits) match
2 case Nil => 0
3 case t => t.reduceLeft((a, b) => (a._1 * a._2 + b._1 * b._2, 1))._1

### Overview of the Functions

To understand the operation implemented by the `mystery3` function, we first need to analyze the helper function `sums` and the function `add` that it uses.

### Understanding the `add` Function

The `add` function takes two parameters: an integer `c` and a list of tuples `acc`, where each tuple consists of an integer and its count in the list. The purpose of `add` is to either update the count of `c` if it already exists in `acc` or to add a new entry for `c` if it does not.

1. **Base Case**: If `acc` is empty (`Nil`), it returns a list containing the tuple `(c, 1)`, indicating that `c` appears once.
2. **Recursive Case**: The function checks the head of the list `x`.
   - If `x._1` (the integer part of the tuple) is equal to `c`, it creates a new tuple with the count increased by one and prepends it to the tail `xs`.
   - If `c` is not equal to `x._1`, it keeps `x` in place and recursively calls `add` with `c` and the tail `xs`.

This effectively counts occurrences of each integer in the list when `sums` calls `add` for each digit.

### Understanding the `sums` Function

The `sums` function applies `add` to a list of digits using `foldRight`.

- It initializes the accumulator as an empty list of tuples.
- For each digit in the input list, it calls `add`, which builds a list of tuples where each tuple represents a digit and its count.

### Analyzing the `mystery3` Function

Now, we consider what `mystery3` does with the output of `sums`.

1. **Pattern Matching on Output**: It matches the result of `sums(digits)`.
   - If the result is `Nil` (meaning the input list was empty), it returns `0`.
   - If there are tuples present, it proceeds to the next step.

2. **Using `reduceLeft`**: It applies `reduceLeft` to the non-empty list of tuples.
   - For each pair of tuples `(a, b)` in the list, it computes a new tuple `(a._1 * a._2 + b._1 * b._2, 1)`.
   - Here, `a._1` and `b._1` are the digits, while `a._2` and `b._2` are their respective counts.

### Mathematical Interpretation of `reduceLeft`

The operation `(a._1 * a._2 + b._1 * b._2, 1)` computes a weighted sum of the digits, where:
- `a._1 * a._2` gives the contribution of the first digit multiplied by its count.
- `b._1 * b._2` gives the contribution of the second digit multiplied by its count.

As `reduceLeft` processes the entire list, it effectively accumulates the total sum of all digits weighted by their counts.

### Final Result Extraction

Once `reduceLeft` has processed all the tuples, it results in a single tuple of the form `(total_sum, 1)`, where `total_sum` is the total contribution of all digits in the list. The function `mystery3` then takes this tuple and extracts the first element, which is the total sum of all digits.

### Conclusion

Thus, `mystery3` returns the sum of all elements in the input list of digits. The operations performed by `add` and `sums` establish the counts of each digit, while `reduceLeft` computes the weighted sum. The final extraction step retrieves the total sum, confirming that the function correctly computes the sum of all digits provided in the input list.


Question:
Which is an appropriate method for fighting skewed distributions of class labels in
classification?

To analyze the multiple-choice question regarding methods for addressing skewed distributions of class labels in classification, we will follow the structured approach outlined in the instructions.

### Core Concepts
The question tests knowledge of techniques used to handle imbalanced class distributions in classification tasks. Skewed distributions can lead to biased models that favor the majority class, thus affecting the model's performance.

### Option Analysis

1. **Include an over-proportional number of samples from the larger class**
   - **Correctness**: Incorrect.
   - **Reasoning**: Including an over-proportional number of samples from the larger class would exacerbate the imbalance rather than mitigate it. This approach would lead to a model that is biased towards the majority class, further skewing the results.

2. **Use leave-one-out cross validation**
   - **Correctness**: Incorrect.
   - **Reasoning**: Leave-one-out cross-validation (LOOCV) is a method for estimating the performance of a model by training on all but one sample. While it can be useful for small datasets, it does not specifically address class imbalance. It may still yield biased results if the class distribution is skewed, as the model may not learn adequately from the minority class.

3. **Construct the validation set such that the class label distribution approximately matches the global distribution of the class labels**
   - **Correctness**: Incorrect.
   - **Reasoning**: While this option suggests maintaining the global distribution, it does not address the issue of imbalance. If the global distribution is skewed, the validation set will also be skewed, which does not help in evaluating the model's performance on minority classes.

4. **Generate artificial data points for the most frequent classes**
   - **Correctness**: Incorrect.
   - **Reasoning**: This option suggests generating artificial data points for the most frequent classes, which would further increase the imbalance rather than alleviate it. The goal should be to generate synthetic data for the minority class to balance the dataset.

### Conclusion
After analyzing all options, none of them provide an appropriate method for fighting skewed distributions of class labels in classification. A more effective approach would be to generate synthetic samples for the minority class (e.g., using techniques like SMOTE - Synthetic Minority Over-sampling Technique) or to use cost-sensitive learning methods.

### Key Principle
The key principle behind this question is the understanding of how to effectively manage class imbalance in classification tasks to ensure that models are trained to recognize and predict minority classes adequately, thus improving overall model performance.