Question: Tick the \emph{incorrect} assertion. A cipher $C$ perfectly decorrelated at order 2 implies\dots

To identify the incorrect assertion regarding a cipher \( C \) that is perfectly decorrelated at order 2, we need to understand what perfect decorrelation implies and how it relates to the various security properties mentioned in the options.

### Definitions:
1. **Perfectly Decorrelated at Order 2**: This means that the output of the cipher does not show any statistical correlation with the input when considering pairs of bits (or symbols) at a second-order level. In practical terms, it suggests that the relation between the input and output is weak, reducing predictability.

2. **Perfect Secrecy**: A cipher is said to achieve perfect secrecy if the ciphertext reveals no information about the plaintext, which is usually associated with one-time pads or ciphers that satisfy Shannon's conditions.

3. **Differential Cryptanalysis**: This is a method of cryptanalysis that studies how differences in input can affect the resultant difference at the output. A cipher that is secure against this attack will not show predictable patterns when inputs differ.

4. **Linear Cryptanalysis**: This attack uses linear approximations to describe the relation between plaintext, ciphertext, and key bits. A cipher that is secure against this attack will not allow for significant linear relationships to be established.

5. **Exhaustive Search**: This is the brute-force method of trying all possible keys to decrypt the ciphertext. A cipher that is secure against exhaustive search has a sufficiently large keyspace.

### Evaluating the Options:
**A) Perfect secrecy when used twice.**
- **Analysis**: Even if a cipher is perfectly decorrelated at order 2, it does not imply perfect secrecy when used multiple times. Perfect secrecy typically requires the key to be as long as the message and used only once (as in the one-time pad). Thus, using a cipher twice does not guarantee that the combined output remains perfectly secret.
- **Verdict**: This statement is likely incorrect.

**B) Security against differential cryptanalysis.**
- **Analysis**: Perfect decorrelation at order 2 suggests that the relationship between input differences and output differences is weak, which directly enhances security against differential cryptanalysis, as attackers will not find predictable differentials.
- **Verdict**: This statement is true.

**C) Security against linear cryptanalysis.**
- **Analysis**: Similarly, a cipher that is perfectly decorrelated at order 2 can also reduce predictability in linear relationships, thus providing some degree of security against linear cryptanalysis.
- **Verdict**: This statement is true.

**D) Security against exhaustive search.**
- **Analysis**: The security against exhaustive search is primarily dependent on the key size, not the decorrelation property. A cipher can be perfectly decorrelated yet still have a small keyspace, making it vulnerable to brute force attacks.
- **Verdict**: This statement is true.

### Conclusion:
The incorrect assertion among the options is **A) perfect secrecy when used twice**. While the other options (B, C, D) reflect realistic security properties related to perfect decorrelation at order 2, option A misrepresents the requirements for achieving perfect secrecy.


Question: Which of the following is correct regarding Louvain algorithm?

The Louvain algorithm is a popular method for community detection in networks, particularly known for its efficiency and ability to uncover hierarchical structures in large graphs. Let's analyze each of the provided statements regarding the Louvain algorithm:

1. **It creates a hierarchy of communities with a common root.**
   - **Explanation:** This statement is **correct**. The Louvain algorithm works in two phases: first, it optimizes modularity by grouping nodes into communities, and then it aggregates these communities into a new network where the nodes represent the communities. This process can be repeated, leading to a hierarchical structure of communities. Thus, the algorithm can indeed create a hierarchy of communities.

2. **Clique is the only topology of nodes where the algorithm detects the same communities, independently of the starting point.**
   - **Explanation:** This statement is **incorrect**. While cliques (complete subgraphs) are a topology where the Louvain algorithm can consistently detect the same communities regardless of the starting point, it is not the only topology. Other structures, such as complete bipartite graphs or certain regular graphs, may also yield consistent community detection results. Therefore, cliques are not the only topology for this property.

3. **If n cliques of the same order are connected cyclically with n-1 edges, then the algorithm will always detect the same communities, independently of the starting point.**
   - **Explanation:** This statement is **correct**. If you have n cliques of the same size connected in a cyclic manner with n-1 edges, the structure is such that the Louvain algorithm will consistently identify the same communities regardless of the initial configuration. The connectivity ensures that the algorithm will recognize the cliques as a single community due to the edges connecting them.

4. **Modularity is always maximal for the communities found at the top level of the community hierarchy.**
   - **Explanation:** This statement is **incorrect**. While the Louvain algorithm aims to maximize modularity at each level, it does not guarantee that the communities found at the top level of the hierarchy will have maximal modularity. The top-level communities may not represent the absolute maximum modularity possible for the entire graph, as the algorithm may converge to a local maximum rather than a global maximum.

### Summary:
- **Correct Statements:** 1 and 3
- **Incorrect Statements:** 2 and 4

The Louvain algorithm is effective for community detection, but its results can depend on the structure of the graph and the nature of the connections between nodes.


Question:
Assume you are working on SuperQuiz, a trendy app that lets everyone design quizzes and share them with friends! SuperQuiz recently hired a new CEO, who wants to improve the development practices using modern methods. However, this CEO has no engineering background, so the suggested improvements are well intentioned but not always feasible. The latest CEO suggestion is this:
"Continuous Integration is a modern best practice. We must adopt it, so that the code in our repository never has bugs. From now on, all branches in the SuperQuiz repository must have continuous integration enabled, and at the end of each day all branches must pass all tests."
Propose (in 1-2 sentences) a compromise that achieves the CEO's true objective:

To achieve the CEO's objective of reducing bugs and ensuring code quality, we can implement continuous integration (CI) on the main branch and require that all feature branches undergo CI checks before merging into the main branch. This way, we ensure that only tested and verified code is integrated into the main codebase while allowing developers the flexibility to work on their branches without daily testing requirements.


Question: For binary classification, which of the following methods can achieve perfect training accuracy on \textbf{all} linearly separable datasets?

Hey! Great question! Let’s break it down concisely.

1. **Decision Tree**: Can achieve perfect accuracy on linearly separable datasets by recursively splitting data until each leaf node has a single class.

2. **Hard-margin SVM**: Achieves perfect accuracy by finding the optimal hyperplane \( w \cdot x + b = 0 \) that maximizes the margin \( \frac{2}{\|w\|} \), ensuring all points are correctly classified.

3. **15-nearest neighbors**: Can achieve perfect accuracy if \( k = 15 \) is appropriate, but if \( k \) is too large, it risks including points from the wrong class.

4. **None of the suggested**: This option doesn’t apply since both Decision Trees and Hard-margin SVMs can achieve perfect accuracy.

So, the final answer is that both **Decision Trees** and **Hard-margin SVMs** can achieve perfect training accuracy on all linearly separable datasets. If you have to pick one, go with **2) Hard-margin SVM**, but remember decision trees work too!


Question:
You are working on an app which is a search engine for cat photos. The app works by making requests to a server which stores the photos. Users search for cat photos and see a batch of results at a time; they can tap on a photo to see it full screen. You are getting two main complaints from users about the app’s performance:
1. When going from a page of results to the next one, the photos take too long to load
2. When going back to the search results after looking at a picture, the photos take too long to be re-downloaded
For each of these complaints, write exactly one sentence giving a possible solution and explaining why it helps:

To address user complaints about slow loading times in your cat photo search engine app:
1. Implementing lazy loading or pre-fetching techniques will help reduce load times between pages by ensuring that images are ready to display as users navigate through their search results.
2. Utilizing caching mechanisms on both client and server sides will speed up photo retrieval when users return from viewing images by allowing access to previously loaded photos without needing to re-download them from the server.


Question: Suppose that in a given FP Tree, an item in a leaf node N exists in every path. Which of the following is true?

In an FP Tree, if an item N exists in a leaf node and appears in every path, it indicates that N is included in every transaction represented by those paths. Therefore, the number of paths in the FP Tree directly corresponds to the minimum possible support of the itemset {N}. Each path indicates a transaction that contains N, establishing its support. Thus, the correct statement is that {N}’s minimum possible support is equal to the number of paths.


Question: There are N philosphers sitting around a circular table eating spaghetti and discussing philosphy. The problem is that each philosopher needs two forks to eat, and there are only $N$ forks, one between each pair of philosophers. We want to design an algorithm that the philosophers can use, that ensures that no one starves as long as each philosopher eventually stops eating, and such that the maximum number of philosophers can eat at once. Lecture 5 provides one possible solution which uses a central arbiter. Can you write the philospherTurn function without a central arbiter? You may modify the provided class Fork if required.  class Fork() {   var inUse: Boolean = false  }  def philosopherTurn(l: Fork, r: Fork): Boolean = ??? // your implementation here // your implementation here  def run() =     val n = 5     val forks = new Array[Fork](n)     val philosophers = new Array[Thread](n)     for p <- 0 to n - 1 do         forks(p) = new Fork()      for p <- 0 to n - 1 do         philosophers(p) = new Thread {             override def run() = {                 while (!philosopherTurn(forks(p % n), forks((p + 1) % n))) { /* wait */ }             }         }         philosophers(p).start      for p <- 0 to n - 1 do         philosophers(p).join() Hint: Use the deadlock prevention technique introduced in the lecture.

To solve the problem of N philosophers sitting around a circular table with N forks, we can implement a solution that prevents deadlock while allowing as many philosophers as possible to eat at the same time. The key to this solution is to impose a consistent order for acquiring forks, which minimizes the potential for contention.

### Implementation Details

1. **Fork Class**: Each fork will have a unique identifier (`myNumber`) to determine its order. The `inUse` variable indicates whether the fork is currently being used.

2. **Philosopher Turn Logic**: In the `philosopherTurn` function, we define the logic for acquiring and releasing forks. Philosophers will attempt to pick up the left and right forks in a way that avoids deadlock.

3. **Synchronization**: The use of synchronized blocks ensures that only one philosopher can hold a fork at a time. The order of synchronization is determined by the fork identifiers to prevent circular wait conditions.

4. **Eating Simulation**: The function simulates a philosopher eating by briefly sleeping to represent the eating time before releasing the forks.

Here’s the complete implementation:

```scala
class Fork(var myNumber: Int) {
    var inUse: Boolean = false
}

def philosopherTurn(l: Fork, r: Fork): Boolean = {
    // Introduce a slight delay to simulate thinking
    Thread.sleep(100)

    // Determine which fork to pick up first based on their numbers
    var left = l
    var right = r
    if (left.myNumber > right.myNumber) {
        left = r
        right = l
    }

    // Synchronize on the left fork first, then the right fork
    left.synchronized {
        right.synchronized {
            // Check if both forks are available
            if (!left.inUse && !right.inUse) {
                left.inUse = true
                right.inUse = true
            } else {
                return false // Cannot eat, one or both forks are in use
            }
        }
    }

    // Simulate eating time
    Thread.sleep(1000)

    // Release the forks
    left.synchronized {
        right.synchronized {
            left.inUse = false
            right.inUse = false
        }
    }

    true // Successfully eaten
}

def run() = {
    val n = 5
    val forks = new Array[Fork](n)
    val philosophers = new Array[Thread](n)

    // Initialize forks
    for (p <- 0 until n) {
        forks(p) = new Fork(p)
    }

    // Initialize philosophers
    for (p <- 0 until n) {
        philosophers(p) = new Thread {
            override def run() = {
                while (!philosopherTurn(forks(p % n), forks((p + 1) % n))) {
                    // Loop until able to eat
                }
            }
        }
        philosophers(p).start()
    }

    // Wait for all philosophers to finish
    for (p <- 0 until n) {
        philosophers(p).join()
    }
}
```

### Explanation of the Code:

1. **Fork Class**: Each fork is represented by an instance of the `Fork` class, which holds an integer identifier and a boolean to indicate whether it's in use.

2. **Philosopher Turn Logic**:
   - The method starts with a small sleep to simulate the philosopher thinking.
   - It then checks which fork has a lower identifier and attempts to acquire it first, followed by the other fork.
   - The synchronization is done in a nested manner: first on the left fork and then on the right fork. This ensures that the philosopher locks the forks in a consistent order.

3. **Eating**: If both forks are available, the philosopher marks them as in use, simulates eating (with a longer sleep), and finally releases both forks.

4. **Thread Management**: Each philosopher runs in its own thread, repeatedly trying to eat until successful. The main method waits for all threads to finish execution.

### Conclusion
This approach effectively avoids deadlock by ensuring that forks are always acquired in a consistent order, allowing multiple philosophers to eat without starving or entering a deadlock state.


Question: Select the \emph{incorrect} statement.

The incorrect statement is: "The Discrete Logarithm is hard to compute for the additive group \(\mathbf{Z}_{n}\)." In this context, while computing discrete logs in multiplicative groups used in cryptography (like prime fields) poses significant challenges, doing so within \(\mathbf{Z}_{n}\) under addition does not imply similar complexity and may sometimes be trivial depending on specific circumstances or values of \( n \).


Question: Describe the main principles of the standard vector space model for semantics.

The standard vector space model (VSM) for semantics is a framework used to represent and analyze the meaning of words and documents in a mathematical way. Here are the main principles:

1. **Vector Representation**: In the VSM, words or documents are represented as vectors in a high-dimensional space. Each dimension corresponds to a unique term from a vocabulary, and the value in each dimension indicates the importance of that term for the given word or document. For example, a document might be represented as a vector where each component reflects the frequency of specific words.

2. **Dimensionality Reduction**: While the initial representation can be very high-dimensional (especially with a large vocabulary), techniques like Singular Value Decomposition (SVD) can be used to reduce the number of dimensions while preserving the most important information. This makes the model more efficient and helps to highlight the underlying patterns in the data.

3. **Cosine Similarity**: The VSM uses cosine similarity to measure the similarity between vectors. This metric calculates the cosine of the angle between two vectors, providing a value between -1 and 1. A cosine similarity of 1 indicates that the vectors point in the same direction (high similarity), while -1 indicates they point in opposite directions (low similarity). This is useful for tasks like document retrieval, where we want to find documents similar to a query.

4. **Contextual Meaning**: The VSM captures the idea that words with similar meanings will have similar vector representations. For instance, "cat" and "dog" would be closer together in the vector space than "cat" and "car." This is often achieved through techniques like term frequency-inverse document frequency (TF-IDF), which adjusts the importance of terms based on their occurrence across documents.

5. **Scalability and Flexibility**: The VSM is scalable to large datasets and flexible enough to incorporate various types of information, including term frequencies, co-occurrence statistics, and even more complex features like word embeddings (e.g., Word2Vec or GloVe).

In summary, the vector space model for semantics provides a structured way to represent and analyze the meanings of words and documents through mathematical vectors, facilitating tasks like information retrieval and semantic similarity assessment. The model's reliance on vector representation, dimensionality reduction, cosine similarity, and the principle of contextual meaning makes it a powerful tool in natural language processing.


Question: In the following problem  Alice holds a string $x = \langle x_1, x_2, \ldots, x_n \rangle$ and Bob holds a string $y = \langle y_1, y_2, \ldots, y_n\rangle$. Both strings are of length $n$ and $x_i, y_i \in \{1,2,\ldots, n\}$ for $i=1,2, \ldots, n$. The goal is for Alice and Bob to use little communication to estimate the quantity \begin{align*} Q = \sum_{i=1}^n (x_i + y_i)^2\,. \end{align*} A trivial solution  is for Alice to transfer all of her string $x$ to Bob who then computes $Q$ exactly. However this requires Alice to send $\Theta(n \log n)$ bits of information to Bob. In the following, we use randomization and approximation to achieve a huge improvement on the number of bits transferred from Alice to Bob. Indeed, for a small parameter $\epsilon > 0$, your task is to devise and analyze a protocol of the following type: \begin{itemize} \item On input $x$, Alice uses a randomized algorithm to compute a message $m$ that consists of $O(\log (n)/\epsilon^2)$ bits. She then transmits the message $m$ to Bob. \item Bob then, as a function of $y$ and the message $m$,  computes an estimate $Z$. \end{itemize} Your protocol should ensure that \begin{align} \label{eq:guaranteeStream} \Pr[| Z - Q| \geq \epsilon Q] \leq 1/3\,, \end{align} where the probability is over the randomness used by Alice.\\ {\em (In this problem you are asked to (i) explain how Alice computes the message $m$ of $O(\log(n)/\epsilon^2)$ bits (ii) explain how Bob calculates the estimate $Z$, and (iii) prove that the calculated estimate satisfies~\eqref{eq:guaranteeStream}.  Recall that you are allowed to refer to material covered in the lecture notes.) }

To tackle the problem at hand, we want to devise a communication protocol between Alice and Bob to estimate the quantity

\[
Q = \sum_{i=1}^n (x_i + y_i)^2
\]

with minimal communication, specifically using a randomized approach that allows Alice to send a message \( m \) of size \( O(\log(n)/\epsilon^2) \) bits to Bob. Our goal is to ensure that Bob can compute an estimate \( Z \) such that

\[
\Pr[|Z - Q| \geq \epsilon Q] \leq \frac{1}{3}
\]

where the probability is over the randomness used by Alice.

### Step 1: Understanding the Quantity \( Q \)

The expression for \( Q \) can be expanded as follows:

\[
Q = \sum_{i=1}^n (x_i^2 + 2x_iy_i + y_i^2)
\]

This means we can separate \( Q \) into three components:

1. \( A = \sum_{i=1}^n x_i^2 \)
2. \( B = \sum_{i=1}^n y_i^2 \)
3. \( C = \sum_{i=1}^n 2x_iy_i \)

Thus, we have:

\[
Q = A + B + C
\]

### Step 2: Alice Computes the Message \( m \)

To reduce the amount of information Alice needs to send, we can use a randomized technique called "random sampling." Alice can compute the following:

1. Randomly select a subset \( S \subseteq \{1, 2, \ldots, n\} \) of size \( k \). The choice of \( k \) will be determined later based on the desired accuracy \( \epsilon \).
2. For each index \( i \in S \), Alice computes \( (x_i + y_i)^2 \).
3. Alice then computes the average of these sampled values:

\[
\hat{Q} = \frac{1}{k} \sum_{i \in S} (x_i + y_i)^2
\]

4. Alice sends a message \( m \) to Bob which contains the values of \( S \) and the computed average \( \hat{Q} \).

To ensure the size of \( m \) is \( O(\log(n)/\epsilon^2) \), we can set \( k = \frac{C}{\epsilon^2} \) for some constant \( C \). The size of \( m \) can be calculated as follows:

- The size to describe \( k \) indices is \( O(k \log(n)) = O\left(\frac{C}{\epsilon^2} \log(n)\right) \).
- The message \( m \) can be of size \( O(\log(n)/\epsilon^2) \).

### Step 3: Bob Computes the Estimate \( Z \)

After receiving the message \( m \), Bob can compute an estimate \( Z \) as follows:

1. Bob retrieves the subset \( S \) and the average \( \hat{Q} \) from \( m \).
2. He computes the full estimate \( Z \) based on the average:

\[
Z = k \cdot \hat{Q}
\]

This provides an estimate of the total \( Q \) based on the average of the sampled values.

### Step 4: Proving the Guarantee \( \Pr[|Z - Q| \geq \epsilon Q] \leq \frac{1}{3} \)

To establish this probability bound, we can invoke the Chernoff bound for sums of random variables. Specifically, since the sampled values \( (x_i + y_i)^2 \) are drawn independently, we can treat them as independent random variables.

1. The expectation of \( (x_i + y_i)^2 \) is given by \( \mathbb{E}[(x_i + y_i)^2] = \frac{Q}{n} \).
2. By the Chernoff bound, we can bound the probability that the sample average deviates from the true mean:

\[
\Pr\left(\left|Z - Q\right| \geq \epsilon Q\right) \leq 2 \exp\left(-\frac{\epsilon^2 k}{3}\right)
\]

3. Setting \( k \) sufficiently large (for instance, \( k = \frac{C}{\epsilon^2} \)), we can choose \( C \) such that this probability is less than \( \frac{1}{3} \).

### Conclusion

By following this protocol, Alice can send a message \( m \) of size \( O(\log(n)/\epsilon^2) \) to Bob, who can then compute an estimate \( Z \) for \( Q \) such that

\[
\Pr[|Z - Q| \geq \epsilon Q] \leq \frac{1}{3}
\]

This demonstrates the efficacy of using randomization and sampling in reducing communication complexity while ensuring a reliable estimate of the desired quantity.


Question: $\mathrm{GF}(2^k)$ is represented by the set of\dots

To determine how the finite field \(\mathrm{GF}(2^k)\) is represented, we need to understand the nature of elements in this field.

1. **Definition of \(\mathrm{GF}(2^k)\)**:
   - The finite field \(\mathrm{GF}(2^k)\) consists of \(2^k\) elements. These elements can be represented as polynomials over the field \(\mathbb{GF}(2)\) (which has two elements: 0 and 1).

2. **Polynomial Representation**:
   - The elements of \(\mathrm{GF}(2^k)\) can be represented by polynomials of degree at most \(k-1\) using binary coefficients (0 or 1).

3. **Understanding Each Option**:
   - **Option 1**: "polynomials of degree at most \(k-1\) with binary coefficients."
     - This is correct, as it aligns with the representation of elements in \(\mathrm{GF}(2^k)\).

   - **Option 2**: "polynomials of degree at most \(k-1\) with coefficients in \(\mathbb{Z}_k\)."
     - This is incorrect. The coefficients should be from \(\mathbb{GF}(2)\), not \(\mathbb{Z}_k\).

   - **Option 3**: "polynomials of degree at most \(2^k\) with coefficients in \(\mathbb{Z}\)."
     - This is incorrect because the degree is limited to \(k-1\) and coefficients must be binary, not from \(\mathbb{Z}\).

   - **Option 4**: "polynomials of degree at most \(2\) with coefficients in \(\mathbb{Z}_k\)."
     - This is incorrect for the same reasons as option 2: incorrect degree and coefficients.

4. **Conclusion**:
   - The correct representation of \(\mathrm{GF}(2^k)\) is accurately described in Option 1, where elements are represented as polynomials of degree at most \(k-1\) with binary coefficients.

Correct answer: polynomials of degree at most \(k-1\) with binary coefficients.


Question:
For security reasons, you accept the performance and memory overhead
    introduced by common sanitizers and deploy them in your user-facing
    production server software.
    Assuming that all memory safety bugs in your software are detected by the
    sanitizers, which of the following properties do the sanitizers provide to
    your code?

To derive the correct answer, we need to understand what memory sanitizers do and the properties they provide.

1. **Confidentiality of the program data**: Sanitizers help prevent unauthorized access to sensitive data by detecting memory safety issues such as buffer overflows and use-after-free errors, which can lead to data leaks. This protection ensures that confidential information remains secure.

2. **Integrity of the program data**: By identifying and preventing memory corruption issues, sanitizers ensure that the data has not been altered or tampered with in an unauthorized manner. This is crucial for maintaining the accuracy and trustworthiness of the program's data.

3. **Availability of the program**: While sanitizers can help prevent crashes caused by memory issues, they do not directly ensure availability; this property is more about system uptime and continuous accessibility.

4. **Accountability of accesses to the program**: This property involves tracking who accessed what data and when, which is not something sanitizers provide. They focus on detecting memory errors rather than logging access.

In summary, the correct answer includes both confidentiality and integrity, as these are directly supported by memory sanitizers in detecting and preventing memory-related vulnerabilities, thus enhancing the overall security of the software.


Question: Tick the \emph{false} assertion. The index of coincidence

The false assertion is:

**D. is the best known attack against the Vernam cipher.**
The Vernam cipher, also known as the one-time pad, is theoretically unbreakable when used correctly; thus, the index of coincidence does not apply as an effective attack method. The strength of the Vernam cipher lies in its use of truly random keys that are as long as the message, making statistical analyses, like the index of coincidence, ineffective. Therefore, claiming that it is the best known attack against this cipher is incorrect.

**A. is a probability.**
The index of coincidence is a statistical measure that quantifies the likelihood of two randomly chosen letters from a text being the same. It is calculated based on the frequency of letters in the text, thus it is indeed a probability value.

**B. can help breaking Vigen\`ere cipher.**
The index of coincidence can be useful in cryptanalysis, especially for the VigenÃ¨re cipher, as it helps determine the length of the key by comparing the observed index of coincidence of the ciphertext to the expected values for random text. This analysis can provide insights into the structure of the ciphertext and facilitate breaking the encryption.

**C. is different for a random string than for some text in English.**
The index of coincidence for random strings tends to be lower due to the uniform distribution of characters, while English text has a higher index due to the uneven frequency of letters (like 'e', 't', 'a', etc.). This discrepancy is a key aspect of using the index of coincidence in cryptanalysis, making this statement true.


Question: Let $G_1, G_2, G_3$, be valid generator matrices of dimensions $\mathbb F^{k_i 	imes n_i}$, all over the same field $\mathbb F$. Recall that ``valid'' means that for all $i$, $k_i \leq n_i$ and $	ext{rank}(G_i) = k_i$. True or false:  Assuming $k_1 = k_2 + k_3$, the matrix $egin{pmatrix} G_1 &\vline &egin{matrix} G_2 &\vline &0\ \cline{1-3} 0 &\vline &G_3 \end{matrix} \end{pmatrix}$ is also a valid generator matrix.

Certainly! Let's analyze the combined generator matrix formed by the matrices \( G_1, G_2, \) and \( G_3 \) and verify whether it is a valid generator matrix.

### Given Matrices

We have three valid generator matrices defined as follows:

- \( G_1 \): This is a \( k_1 \times n_1 \) matrix with \( \text{rank}(G_1) = k_1 \).
- \( G_2 \): This is a \( k_2 \times n_2 \) matrix with \( \text{rank}(G_2) = k_2 \).
- \( G_3 \): This is a \( k_3 \times n_3 \) matrix with \( \text{rank}(G_3) = k_3 \).

Given that \( k_1 = k_2 + k_3 \), we are interested in constructing a new generator matrix by combining these matrices.

### Structure of the Combined Matrix

The new matrix is formed as follows:

\[
\begin{pmatrix} G_1 & H \end{pmatrix}
\]

where \( H \) is defined as:

\[
H = \begin{pmatrix} G_2 & 0 \\ 0 & G_3 \end{pmatrix}
\]

### Dimensions of the Combined Matrix

1. **Rows**:
   - The left block \( G_1 \) contributes \( k_1 \) rows.
   - The block \( H \) also has \( k_1 \) rows because it consists of \( G_2 \) and \( G_3 \), where \( k_2 + k_3 = k_1 \).

Thus, the combined matrix has \( k_1 \) rows.

2. **Columns**:
   - The total number of columns is \( n_1 + n_2 + n_3 \), which comes from the dimensions of \( G_1, G_2, \) and \( G_3 \).

Therefore, the new matrix has dimensions \( k_1 \times (n_1 + n_2 + n_3) \).

### Rank of the Combined Matrix

To establish that the combined matrix is a valid generator matrix, we need to confirm that its rank equals \( k_1 \).

- **Rank of \( G_1 \)**: We know that \( \text{rank}(G_1) = k_1 \).
- **Rank of \( H \)**: The block \( H \), which is structured as \( \begin{pmatrix} G_2 & 0 \\ 0 & G_3 \end{pmatrix} \), has a rank of \( k_2 + k_3 = k_1 \).

When we combine \( G_1 \) and \( H \), we find that:

\[
\text{rank} \begin{pmatrix} G_1 & H \end{pmatrix} \geq \text{rank}(G_1) = k_1
\]

Since adding columns (in this case, from \( H \)) cannot decrease the rank, we conclude that:

\[
\text{rank} \begin{pmatrix} G_1 & H \end{pmatrix} = k_1
\]

### Conclusion

The combined matrix:

\[
\begin{pmatrix} G_1 & H \end{pmatrix} = \begin{pmatrix} G_1 & G_2 & 0 \\ 0 & 0 & G_3 \end{pmatrix}
\]

has:

- **Dimensions**: \( k_1 \times (n_1 + n_2 + n_3) \)
- **Rank**: \( k_1 \)

Since the rank equals the number of rows, we conclude that the combined matrix is indeed a valid generator matrix.

Thus, the statement âthe combined matrix is a valid generator matrixâ is **True**. This confirms that the combined structure maintains the necessary properties of linear independence and rank, making it suitable for generating codewords in a linear code.


Question: Which of the following algorithms is \emph{not} a hash function?

To determine which of the provided algorithms is \emph{not} a hash function, we can analyze each option step-by-step:

1. **SHA-1**: This stands for Secure Hash Algorithm 1. It is a cryptographic hash function that produces a 160-bit hash value. It is used widely for data integrity and verification purposes. Therefore, SHA-1 *is* a hash function.

2. **MD5**: MD5 stands for Message-Digest Algorithm 5. It is also a widely used cryptographic hash function that produces a 128-bit hash value. It is used for checksums and data integrity. Thus, MD5 *is* a hash function.

3. **RC4**: RC4 is a stream cipher that is used for encryption. It is not a hash function; rather, it is designed to encrypt data. Therefore, RC4 *is not* a hash function.

4. **MD4**: MD4 stands for Message-Digest Algorithm 4. Like MD5 and SHA-1, it is a cryptographic hash function that produces a 128-bit hash value. Thus, MD4 *is* a hash function.

After analyzing all options, we can conclude:

- SHA-1: Hash function
- MD5: Hash function
- RC4: Not a hash function (encryption algorithm)
- MD4: Hash function

Therefore, the answer to the question "Which of the following algorithms is not a hash function?" is indeed **RC4** (option 3).


Question: Assume that we have a convolutional neural net with $L$ layers, $K$ nodes per layer, and where each node is connected to $k$ nodes in a previous layer. We ignore in the sequel the question of how we deal with the points at the boundary and assume that $k<<<K$ (much, much, much smaller). How does the complexity of the back-propagation algorithm scale in these parameters?

4.  The back-propagation algorithm's complexity scales linearly with the number of layers \( L \), the number of nodes \( K \), and the number of connections \( k \) between layers, leading to a complexity of \(\Theta(L K k)\).


Question: Which of the following are true regarding the lengths of integers in some base \(b\) (i.e., the number of digits base \(b\)) in different bases, given \(N = (FFFF)_{16}\)?

To address the question regarding the lengths of the integer \(N = (FFFF)_{16}\) in various bases, we first need to convert \(N\) from hexadecimal (base 16) to decimal (base 10). The hexadecimal number \(FFFF\) can be computed as:

$$
FFFF_{16} = 15 \cdot 16^3 + 15 \cdot 16^2 + 15 \cdot 16^1 + 15 \cdot 16^0 = 15 \cdot (4096 + 256 + 16 + 1) = 15 \cdot 4371 = 65565.
$$

Now, let’s analyze the lengths of \(N\) in different bases:

1. **For base 2**: The length of a number in base \(b\) can be determined using the formula:

$$
\text{length} = \lfloor \log_b(N) \rfloor + 1.
$$

Thus, for base 2:

$$
\text{length} = \lfloor \log_2(65565) \rfloor + 1.
$$

Calculating $\log_2(65565)$ using the change of base formula:

$$
\log_2(65565) = \frac{\log_{10}(65565)}{\log_{10}(2)} \approx \frac{4.817}{0.301} \approx 16.
$$

So, the length in base 2 is:

$$
\lfloor 16 \rfloor + 1 = 17.
$$

Hence, the statement that \((N)_2\) is of length 16 is **false**.

2. **For base 10**:

The length of \(N\) in base 10 is:

$$
\text{length} = \lfloor \log_{10}(65565) \rfloor + 1.
$$

Calculating this:

$$
\log_{10}(65565) \approx 4.817,
$$

Thus, the length in base 10 is:

$$
\lfloor 4.817 \rfloor + 1 = 5.
$$

This means the statement that \((N)_{10}\) is of length 40 is **false**.

3. **For base 4**:

Using the same formula for the length:

$$
\text{length} = \lfloor \log_4(65565) \rfloor + 1.
$$

Calculating $\log_4(65565)$:

$$
\log_4(65565) = \frac{\log_{10}(65565)}{\log_{10}(4)} \approx \frac{4.817}{0.602} \approx 7.99.
$$

So the length in base 4 is:

$$
\lfloor 7.99 \rfloor + 1 = 8.
$$

Therefore, the statement that \((N)_{4}\) is of length 12 is **false**, and the statement that \((N)_{4}\) is of length 4 is also **false**.

In summary, we have evaluated each option based on the conversions and calculations of the lengths of \(N\) in various bases. All the statements provided are false.


Question: Implement a reliable broadcast algorithm without using any failure detector, i.e., using only BestEffort-Broadcast(BEB).

To implement a reliable broadcast using an all-to-all communication approach:

1. **Initialization**: Each process maintains a set `delivered` to track received messages.

2. **RB-broadcast(m)**: When a process wants to broadcast a message `m`, it immediately sends `m` to all other processes and calls `RB-deliver(m)`.

3. **BEB-receive(m)**: Upon receiving a message `m` from another process `q`, check if `m` is in `delivered`. If not, send `m` to all processes except the sender and `q`, then call `RB-deliver(m)` and add `m` to `delivered`.

4. **Agreement**: Since all correct processes relay received messages, every correct process will eventually receive `m`, ensuring reliable delivery.

This approach guarantees that all correct processes will eventually deliver the same set of messages.


Question: In class we saw that Karger's min-cut algorithm implies that an undirected graph has at most $n \choose 2$ minimum cuts. Show that this result is tight by giving a graph with $n$ vertices and $n \choose 2$ minimum cuts.

To address the question, we need to understand a few key concepts related to graph theory and Karger's min-cut algorithm.

### Key Concepts

1. **Minimum Cut**: A cut in a graph is a partition of the vertices into two disjoint subsets, and the cut-set is the set of edges that cross the partition. The minimum cut is the cut with the smallest number of edges crossing it.

2. **Karger's Min-Cut Algorithm**: This is a randomized algorithm that finds a minimum cut in a graph. It works by repeatedly contracting edges until only two vertices remain, and the edges connecting these two vertices represent a cut.

3. **Count of Minimum Cuts**: The result that an undirected graph has at most \( {n \choose 2} \) minimum cuts is derived from the fact that there can be at most \( {n \choose 2} \) different ways to choose two vertices from \( n \) vertices, and each distinct pair can potentially correspond to a minimum cut.

### Graph Construction

To show that this result is tight, we need to construct a graph with \( n \) vertices that has exactly \( {n \choose 2} \) minimum cuts.

#### Graph Description

Consider a complete graph \( K_n \) with \( n \) vertices. In a complete graph:

- Every pair of distinct vertices is connected by an edge.
- The number of edges in a complete graph with \( n \) vertices is \( {n \choose 2} \).

#### Minimum Cuts in the Complete Graph

In \( K_n \):

- Any cut that separates one vertex from the remaining \( n-1 \) vertices will have exactly \( n-1 \) edges crossing the cut (since each of the \( n-1 \) vertices is connected to the isolated vertex).
- There are \( n \) such cuts (each corresponding to one of the \( n \) vertices being separated from the others).

Thus, every cut that separates one vertex from the rest is a minimum cut.

Now, because we can choose any pair of vertices, and since each of these pairs corresponds to a unique minimum cut, the total number of distinct pairs (and thus the minimum cuts) is exactly \( {n \choose 2} \).

#### Conclusion

The complete graph \( K_n \) has exactly \( n \) vertices, and it can be shown that it has \( {n \choose 2} \) minimum cuts, which confirms that Karger's result is tight. Each pair of vertices corresponds to one minimum cut, demonstrating that the maximum number of minimum cuts in a graph is indeed \( {n \choose 2} \).

Thus, the answer to the question is:

### Answer

The complete graph \( K_n \) with \( n \) vertices has exactly \( {n \choose 2} \) minimum cuts, which demonstrates that the result is tight.


Question:
An expression is referentially transparent if it always returns the same value, no matter
the global state of the program. A referentially transparent expression can be replaced by its value without
changing the result of the program.
Say we have a value representing a class of students and their GPAs. Given the following defintions:
1 case class Student(gpa: Double)
2
3 def count(c: List[Student], student: Student): Double =
4 c.filter(s => s == student).size
5
6 val students = List(
7 Student(1.0), Student(2.0), Student(3.0),
8 Student(4.0), Student(5.0), Student(6.0)
9 )
And the expression e:
1 count(students, Student(6.0))

Your analysis of the expression `count(students, Student(6.0))` in the context of referential transparency is comprehensive and well-articulated. Here’s a concise summary of the key points, confirming your conclusion about referential transparency:

### Key Points

1. **Definition of Referential Transparency**: An expression is referentially transparent if it can be replaced with its value without changing the program's behavior, which entails:
   - Consistent output for the same input.
   - No dependence on external states or side effects.

2. **Components of the Expression**:
   - `Student(gpa: Double)`: A case class representing a student with a GPA.
   - `count(c: List[Student], student: Student): Double`: A function that counts how many times a specific `Student` appears in a list of `Student` objects.
   - `students`: A predefined list of `Student` objects.

3. **Evaluation Process**:
   - The expression creates a new `Student` instance with a GPA of 6.0.
   - The `count` function filters the list of students to check for matches with the newly created `Student` instance.
   - Structural equality checks for GPAs are performed, and since there are no matches, the result is `0`.

4. **Referential Transparency Assessment**:
   - The expression `count(students, Student(6.0))` consistently returns `0` for the same input list and `Student` instance.
   - It operates without altering any external state or producing side effects.

### Conclusion

Based on your detailed breakdown, the expression `count(students, Student(6.0))` is indeed referentially transparent. It consistently yields the same output for given inputs and does not incorporate any external dependencies or side effects. Thus, your conclusion is accurate and aligns with the principles of referential transparency in functional programming.


Question: In this exercise, we will see how to combine the Principal Component Analysis (PCA) and the kernel method into an algorithm known as kernel PCA. We are given $n$ observations in a low dimensional space $\mathbf{x}_{1}, \cdots, \mathbf{x}_{n} \in \mathbb{R}^{L}$ and we consider a kernel $k$ and its associated features $\operatorname{map} \phi: \mathbb{R}^{L} \mapsto \mathbb{R}^{H}$ which satisfies: $$ k(\mathbf{x}, \mathbf{y})=\langle\phi(\mathbf{x}), \phi(\mathbf{y})\rangle_{\mathbb{R}^{H}} $$ where $\langle\cdot, \cdot\rangle_{\mathbb{R}^{H}}$ is the standard scalar product of $\mathbb{R}^{H}$. We define the empirical covariance matrix and the empirical covariance matrix of the mapped observations as: $$ \boldsymbol{\Sigma}:=\frac{1}{n} \sum_{i=1}^{n} \mathbf{x}_{i} \mathbf{x}_{i}^{\top} \quad \text { and } \quad \boldsymbol{\Sigma}^{\mathbf{H}}:=\frac{1}{n} \sum_{i=1}^{n} \phi\left(\mathbf{x}_{i}\right) \phi\left(\mathbf{x}_{i}\right)^{\top} $$ The kernel matrix $\mathbf{K}$ is defined by: $$ \mathbf{K}_{i, j}:=k\left(\mathbf{x}_{i}, \mathbf{x}_{j}\right)=\left\langle\phi\left(\mathbf{x}_{i}\right), \phi\left(\mathbf{x}_{j}\right)\right\rangle_{\mathbb{R}^{H}} $$ We also define the data matrix and the corresponding matrix of the mapped data as: $$ \mathbf{X}:=\left(\begin{array}{c} \mathbf{x}_{1}^{\top} \\ \cdots \\ \mathbf{x}_{n}^{\top} \end{array}\right) \in \mathbb{R}^{n \times L} \quad \text { and } \quad \mathbf{\Phi}:=\left(\begin{array}{c} \phi\left(\mathbf{x}_{1}\right)^{\top} \\ \cdots \\ \phi\left(\mathbf{x}_{n}\right)^{\top} \end{array}\right) \in \mathbb{R}^{n \times H} . $$ Finally we denote the eigenpairs (eigenvalues and eigenvectors) of $\boldsymbol{\Sigma}^{\mathbf{H}}$ by $\left\{\left(\lambda_{i}, \mathbf{v}_{i}\right)\right\}_{i=1}^{H}$ and those of $\mathbf{K}$ by $\left\{\left(\rho_{j}, \mathbf{w}_{j}\right)\right\}_{j=1}^{n}$. We also assume that the vectors $\mathbf{v}_{i}$ and $\mathbf{w}_{j}$ are normalized. Thus: $$ \boldsymbol{\Sigma}^{\mathbf{H}} \mathbf{v}_{i}=\lambda_{i} \mathbf{v}_{i}, \quad\left\|\mathbf{v}_{i}\right\|_{2}=1 \quad \text { and } \quad \mathbf{K} \mathbf{w}_{j}=\rho_{j} \mathbf{w}_{j}, \quad\left\|\mathbf{w}_{j}\right\|_{2}=1 $$ Let us remind that we assume in the kernel setting that we can compute $k(\mathbf{x}, \mathbf{y})$ but that we cannot directly compute $\phi(\mathbf{x})$ What we would like to do is to first map the data into the high-dimensional space using the features map $\phi$ and then to apply the standard PCA algorithm in the high-dimensional space $\mathbb{R}^{H}$. This would amount to: (a) Computing the empirical covariance matrix $\boldsymbol{\Sigma}^{\mathbf{H}}$ of the mapped data $\phi\left(\mathbf{x}_{i}\right)$. (b) Computing the eigenvectors $\mathbf{v}_{1}, \cdots, \mathbf{v}_{N}$ associated with the $N$ largest eigenvalues of $\boldsymbol{\Sigma}^{\mathbf{H}}$. (c) Computing the projection $\Pi\left(\phi\left(\mathbf{x}_{i}\right)\right) \in \mathbb{R}^{L}$ for each data point onto these eigenvectors, where the $j$-th component of the projection is given by: $$ \Pi_{j}\left(\phi\left(\mathbf{x}_{i}\right)\right)=\left\langle\phi\left(\mathbf{x}_{i}\right), \mathbf{v}_{j}\right\rangle_{\mathbb{R}^{H}} $$ Write the kernel matrix $\mathbf{K}$ as a function of the features matrix $\boldsymbol{\Phi}$. What is the size of this matrix?

The kernel matrix \(K=\boldsymbol{\Phi}^{T}\boldsymbol{\Phi}\), with dimensions \(n\times n,\) where \(n\) denotes the number of observations. Each entry in this square matrix represents the inner product between mapped feature vectors corresponding to pairs of original data points. This formulation leverages high-dimensional projections while avoiding direct computation in that space by utilizing kernels.


Question:
What is the main difficulty in performing a Prime+Probe attack
                    on a system with L1 caches private to each core and a shared LLC,
                    and with attacker and victim running on different cores? How can
                    this difficulty be circumvented?

To answer the question regarding the challenges of performing a Prime+Probe attack in a system with private L1 caches and a shared last-level cache (LLC), I will guide you through the reasoning process step-by-step.

### Step 1: Understand the Prime+Probe Attack

**Definition**: A Prime+Probe attack is a side-channel attack used to infer information about a victim's memory access patterns. The attacker 'primes' the cache by filling it with their own data, and then 'probes' it to see if the victim's operations have evicted their data from the cache.

### Step 2: Analyze the Cache Architecture

**Cache Levels**: In the given scenario, we have:
- **L1 Cache**: This is private to each core, which means each core has its own L1 cache that cannot be accessed by other cores.
- **LLC (Last Level Cache)**: This is shared among all cores, meaning that all cores can access and utilize the same LLC.

### Step 3: Identify the Difficulty

#### Difficulty Explained:
1. **L1 Cache Priming Issue**: When the attacker primes the LLC, they may not see the expected results when probing because they might be probing their own L1 cache instead of the shared LLC. Since L1 caches are private, the attacker’s probes will likely only access their own L1 cache and not the shared LLC where the victim's data resides.

2. **Eviction Confusion**: If the attacker’s data remains in their L1 cache, probing will show L1 hits (meaning the data is found in L1) rather than LLC hits or misses, which is what the attacker is trying to observe. This means the attacker cannot discern whether the victim accessed data from the LLC or the L1 cache.

### Step 4: Discuss the Solution (Repriming)

To successfully perform the attack, the attacker needs to ensure that when they probe, they are not hitting their own L1 cache but rather the LLC. This leads us to the concept of **"repriming."**

#### Repriming Explained:
- **Repriming**: The attacker must fill their L1 cache with different data that does not interfere with the data they primed in the LLC. This can be achieved by using addresses that map to the same set in the L1 cache but different sets in the LLC.

- **Inclusive Caches**: If the caches are inclusive (meaning the data in the L1 cache must also be in the LLC), it becomes crucial for the attacker to choose addresses that will allow them to fill the L1 without affecting the LLC's data. The attacker can do this because the LLC typically has more sets than the L1 cache, allowing for a greater variety of addresses.

### Step 5: Addressing All Sets in LLC

To ensure comprehensive probing of the LLC, the attacker might need to consider:
- **Using Large Pages**: Large pages can help address multiple cache lines in the LLC, which improves the chances of hitting all relevant sets in the LLC. This is especially important because the LLC is shared among multiple cores, and addresses need to be carefully chosen to cover the entire cache.

### Conclusion

In summary, the main difficulty in performing a Prime+Probe attack in the given system arises from the separation of private L1 caches and the shared LLC, which can lead to misleading results. The attacker can circumvent this issue through careful repriming of the L1 cache with different data to ensure that probes result in LLC hits and misses, thereby providing the necessary information for the attack. Using large pages can further enhance the attack's effectiveness by allowing the attacker to address all relevant sets in the LLC more efficiently.

This systematic approach helps clarify the inherent challenges and potential solutions in executing the Prime+Probe attack in such a cache architecture.


Question:
For students born in April, how many months older are they than the average student in their grade? 5.4898 months
For students born in March, how many months younger are they than the average student in their grade? 5.5102 months

Discuss: Considering your common sense and the results obtained from the simulation: what advantage do students born in April have over those born in March? How may this affect their odds of becoming professional athletes?

When we look at the difference in age between students born in April and those born in March, we see that students born in April are, on average, about 5.5 months older than their peers born in March. This discrepancy can have significant implications, especially in competitive environments like sports.

**Advantages of Being Older in a Grade:**

1. **Physical Development:** In many sports, physical attributes like strength, speed, and coordination can be crucial. Being older by several months can mean that April-born students may have had more time to develop these physical skills compared to their March-born peers. This could give them an edge in sports where physical maturity plays an important role, such as football, basketball, or soccer.

2. **Confidence and Experience:** Older students may also have greater self-confidence and maturity, which can lead to better performance under pressure. They might be more willing to take risks and try out for competitive teams, leading to more opportunities to showcase their talent.

3. **Coaching and Training:** Coaches often notice physical advantages and may give more attention or opportunities to older, more developed players. If an April-born student consistently performs better due to their age-related advantages, they may receive more training and playing time, further enhancing their skills.

**Implications for Becoming Professional Athletes:**

The age gap can influence selection for teams, especially at younger levels where the difference in physical and mental maturity is more pronounced. Coaches might favor older athletes for competitive teams, leading to a cycle where younger athletes (like those born in March) might not get the same opportunities to develop their skills.

Over time, this can create disparities in who gets scouted and who has a better chance of reaching a professional level. The earlier a player is identified as talented and given opportunities, the more likely they are to continue developing and ultimately succeed as professional athletes.

In summary, while being born in April doesn't guarantee success in sports, the slight age advantage can create a series of opportunities that might lead to higher chances of becoming a professional athlete. This highlights the importance of age and timing in youth sports, where even a few months can make a significant difference in development and opportunities.


Question:
How is it possible to compute the average Precision/Recall curves? Explain in detail the
various steps of the computation.

Explanation: To compute the average Precision and Recall curves, we first need to understand that these metrics are derived from the confusion matrix of a binary classification task. The average Precision-Recall curves can be computed using the following steps: 1. **Gather Data**: Collect the predicted probabilities or scores from the model for each test instance along with the true labels. 2. **Sort Predictions**: For each class (usually the positive class), sort the predicted probabilities in descending order. 3. **Calculate Precision and Recall**: For each threshold (which can be the unique predicted scores), compute the Precision and Recall. Precision is defined as the ratio of true positives to the sum of true positives and false positives, while Recall is the ratio of true positives to the sum of true positives and false negatives. 4. **Plot the Curves**: Create plots for Recall on the x-axis and Precision on the y-axis for each threshold. 5. **Average the Curves**: If you have multiple models or multiple classes, you can compute the average Precision and Recall curves by averaging the Precision values at each Recall level across all models or classes. 6. **Interpolation**: To smooth the curves, you might interpolate the values to ensure that the Precision values are non-increasing as Recall increases. 7. **Final Visualization**: Finally, visualize the averaged Precision-Recall curve for better understanding and comparison. This process allows for a robust evaluation of the model's performance in terms of precision and recall across various thresholds.

Answer: The average Precision/Recall curves can be computed through a series of steps involving gathering data, calculating Precision and Recall at various thresholds, plotting the curves, averaging the metrics across models/classes, and visualizing the final curves.


Question: Which of the following statements regarding topic models is false?

To determine which statement regarding topic models is false, we need to analyze each statement in the context of how topic modeling, particularly Latent Dirichlet Allocation (LDA), operates.

1. **"Topic models map documents to dense vectors."**
   - This statement is generally true. Topic models like LDA represent documents in a latent space defined by topics. Each document can be represented as a vector in a high-dimensional space where each dimension corresponds to a topic. The values in this vector can indicate the strength or proportion of each topic present in the document.

2. **"In LDA, topics are modeled as distributions over documents."**
   - This statement is false. In LDA, topics are not modeled as distributions over documents; rather, topics are modeled as distributions over words. Each topic is represented by a distribution of words, indicating the likelihood of each word appearing in that topic. Documents, on the other hand, are modeled as mixtures of these topics, where each document can be seen as a distribution over the topics rather than the other way around.

3. **"LDA assumes that each document is generated from a mixture of topics with a probability distribution."**
   - This statement is true. LDA operates on the principle that each document is generated from a mixture of topics. Specifically, it assumes that for each document, there is a distribution over topics, and for each topic, there is a distribution over words. This probabilistic framework allows LDA to infer which topics are present in each document based on the words it contains.

4. **"Topics can serve as features for document classification."**
   - This statement is also true. The topics identified by LDA can indeed be used as features for document classification tasks. By transforming documents into topic distributions (the vector representation mentioned earlier), these topic proportions can serve as informative features for machine learning classifiers.

Given this analysis, the false statement is the one that incorrectly describes the relationship between topics and documents in LDA. LDA models topics as distributions over words, not documents, which makes the second statement the incorrect one.


Question: In plain ElGamal Encryption scheme \ldots

In the plain ElGamal Encryption scheme, it’s primarily focused on confidentiality. This means that the main goal is to keep the messages private from anyone who might try to intercept them. Because of this, you really just need a confidential channel to make sure that the information exchanged stays secret. Authentication isn’t really a requirement in the plain version of the scheme, so options that suggest you need that are not correct.

So the answer is that only a confidential channel is needed for ElGamal encryption to work properly.

A


Question: What is a good representation for scores when classifying these three target classes: Car, Bike and Bus, in the context of logistic regression. (One or multiple answers)

Based on the context of logistic regression and the need for a clear representation of the target classes, the most appropriate option is the first one: {Car: $(0,1,0)$,} {Bike: $(1,0,0)$,} {Bus: $(0,0,1)$}. This representation utilizes one-hot encoding, where each class is represented by a unique binary vector. This approach allows the model to distinguish between the three classes effectively. The other options either do not provide distinct representations or use incompatible formats for classification. Therefore, I believe the first option is the best choice.


Question: A multiset is an unordered collection where elements can appear multiple times. We will represent a multiset of Char elements as a function from Char to Int: the function returns 0 for any Char argument that is not in the multiset, and the (positive) number of times it appears otherwise: type Multiset = Char => Int The filter operation on a multiset m returns the subset of m for which p holds. What should replace ??? so that the filter function is correct? def filter(m: Multiset, p: Char => Boolean): Multiset = ???

To determine the correct implementation of the `filter` function for the multiset, we need to understand how the filter operation works conceptually. The `filter` function should return a new multiset that includes only those elements from the original multiset `m` for which the predicate `p` returns true.

### Analyzing Each Option

**Option A: `x => if m(x) then p(x) else 0`**
- This option is incorrect because `m(x)` returns an integer (the count of occurrences of `x` in the multiset), and using it in a conditional check (`if m(x)`) is not valid. Moreover, it does not account for counting the occurrences of `x` based on the predicate `p`.

**Option B: `x => m(x) && p(x)`**
- This option is also incorrect. The expression `m(x) && p(x)` results in a Boolean value (true or false) rather than an integer. The filter implementation must return an integer representing the count of occurrences of `x` that satisfy the predicate `p`.

**Option C: `x => if !m(x) then p(x) else 0`**
- This option is incorrect. The condition `!m(x)` checks if the count of `x` is zero, and if so, it evaluates `p(x)`, which is not relevant for filtering. Again, this does not return the correct integer count for the multiset.

**Option D: `x => if p(x) then m(x) else 0`**
- This option is correct. It checks if the predicate `p(x)` is true for the character `x`. If it is true, it returns `m(x)`, which is the count of occurrences of `x` in the multiset. If `p(x)` is false, it returns 0, indicating that this character should not be included in the filtered multiset.

### Conclusion

After analyzing each option, the correct implementation for the filter function that adheres to the requirements of returning a multiset based on the predicate is:

**D) `x => if p(x) then m(x) else 0`**

This option correctly implements the filtering logic by returning the count of occurrences of `x` only if the predicate `p` holds true.


Question: Which of the following is correct regarding Crowdsourcing?

The correct answer is: The output of Majority Decision can be equal to the one of Expectation-Maximization.

**Step-by-step explanation:**

1. **Understanding Majority Decision**: This method selects the most frequent answer from multiple responses, providing a simple consensus result.

2. **Understanding Expectation-Maximization (EM)**: EM is an iterative algorithm used for parameter estimation in probabilistic models, refining estimates by maximizing expected log-likelihood.

3. **Comparison**: In scenarios where responses are consistent and the underlying distribution conforms to the assumptions of both methods, the Majority Decision can yield results that coincide with EM's output, particularly when the data is clustered around clear majority classes.

Students might make the mistake of assuming the two methods are fundamentally different and never yield similar results. Understanding that both can converge under specific conditions is crucial for grasping their applicability in crowdsourcing and data analysis contexts.


Question: Plain RSA (with an $\ell$-bit modulus) \dots

Let's go through each option one by one to determine their correctness in the context of Plain RSA with an \(\ell\)-bit modulus.

**Option A: is commonly used in practice.**
- **Explanation**: This statement is somewhat misleading. While RSA is a well-known and foundational public-key cryptosystem, it is not commonly used in its plain form for encryption in practice due to its inefficiency and vulnerability to certain attacks (such as chosen-plaintext attacks). Instead, RSA is often used to encrypt symmetric keys (which are then used for symmetric encryption) or for signing messages rather than directly encrypting plaintext. So, the answer is **incorrect**.

**Option B: decrypts in \(O(\ell^2)\) time.**
- **Explanation**: The decryption process in RSA involves computing \(c^{d} \mod n\), where \(c\) is the ciphertext, \(d\) is the private key exponent, and \(n\) is the modulus. The best-known algorithms for modular exponentiation have a time complexity of \(O(\ell^3)\) for general operations (using methods like square-and-multiply). However, if we specifically consider the RSA decryption operation with optimizations (such as using the Chinese Remainder Theorem), we can perform the decryption in \(O(\ell^2)\) time. So, this option is **correct**.

**Option C: encrypts in \(O(\ell)\) time.**
- **Explanation**: The encryption process in RSA involves computing \(m^{e} \mod n\), where \(m\) is the plaintext message and \(e\) is the public key exponent. This operation typically requires modular exponentiation, which is not \(O(\ell)\) but rather \(O(\ell^2)\) or more depending on the method used. Thus, this statement is **incorrect**.

**Option D: has homomorphic properties.**
- **Explanation**: RSA does have certain homomorphic properties, specifically that it allows for operations on ciphertexts that relate to operations on plaintexts. For instance, given two ciphertexts \(c_1\) and \(c_2\), you can compute \(c_1 \cdot c_2 \mod n\) to obtain the ciphertext of the product of the corresponding plaintexts. Therefore, this option is **correct**.

Now, let's summarize the findings:
- Option A: Incorrect
- Option B: Correct
- Option C: Incorrect
- Option D: Correct

**Best Options**: The best answers are **B** and **D**.


Question: One-time pad ...

The correct option is:

**4. uses an invertible group operation such as ``$\oplus$" for encryption.**


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

If the accuracy property of the failure detector is violated in a uniform reliable broadcast (URB) algorithm, it can lead to significant communication failures. Correct processes may be misidentified as faulty and miss intended messages, while faulty processes could be mistakenly treated as correct, resulting in ongoing message transmissions to non-responsive nodes. This misclassification disrupts the required message ordering and consistency guarantees of URB, causing some processes to hold outdated or inconsistent states. Ultimately, such inaccuracies can destabilize the system and lead to broader issues like deadlocks or state divergence among distributed nodes.


Question: Pick the \emph{correct} statement.

The question asks which statement correctly defines isomorphism in the context of mathematical structures and homomorphisms.

To understand this, we need to know that a homomorphism is a map between two algebraic structures that preserves their operations. An isomorphism is a specific type of homomorphism that is both injective (one-to-one) and surjective (onto), meaning it establishes a perfect correspondence between the elements of two structures.

Now, let's evaluate each option:

1. **A homomorphism is defined as a bijective isomorphism.**
   - **Incorrect**: A homomorphism does not have to be bijective. It can be injective or surjective but does not require both properties.

2. **An isomorphism is defined as a bijective homomorphism.**
   - **Correct**: This accurately defines an isomorphism, highlighting that it is a homomorphism with both injective and surjective properties.

3. **An isomorphism is any homomorphism \( h: X \rightarrow X \).**
   - **Incorrect**: This definition is misleading. While isomorphisms can map a structure to itself, not all homomorphisms from a structure to itself are isomorphisms unless they are bijective.

4. **A homomorphism is any non-bijective isomorphism.**
   - **Incorrect**: This is contradictory since isomorphisms must be bijective by definition, making this statement false.

In conclusion, "An isomorphism is defined as a bijective homomorphism" is the correct answer, as it succinctly captures the essential characteristics of isomorphisms, which are crucial for understanding the relationships between algebraic structures.


Question: Tick the true assertion.

To determine the true assertion among the provided options, we will evaluate each statement one by one, providing detailed explanations for our reasoning.

1. **Statement: "A dictionary attack requires less memory than a time-memory tradeoff."**
   - A dictionary attack is a method used in cryptanalysis to break passwords by systematically entering each word in a predefined list (dictionary) until the correct one is found. This method primarily relies on time to search through the dictionary.
   - In contrast, a time-memory tradeoff is a strategy that utilizes precomputed tables (such as rainbow tables) to reduce the time required to crack a password at the expense of increased memory usage.
   - Thus, while a dictionary attack may use less memory than certain time-memory tradeoff strategies, it is not universally true that dictionary attacks always require less memory. Therefore, this statement is **false**.

2. **Statement: "Double-DES succumbs under a Meet-in-the-Middle attack."**
   - Double-DES refers to the encryption of data using the DES algorithm twice, with two distinct keys. While this approach seems to enhance security, it is susceptible to a specific type of attack known as the Meet-in-the-Middle attack.
   - In a Meet-in-the-Middle attack, the attacker encrypts the plaintext with all possible keys and simultaneously decrypts the ciphertext with all possible keys, storing intermediate results. This allows the attacker to find a match between the two processes, effectively reducing the complexity from brute-force search of 2^56 to about 2^57 operations.
   - Therefore, this statement is **true**.

3. **Statement: "AES is the ancestor of DES."**
   - This statement is incorrect. The Data Encryption Standard (DES) was developed first and was adopted as a federal standard in the 1970s. Advanced Encryption Standard (AES) was later developed to replace DES due to its vulnerabilities. Thus, DES is the ancestor of AES, not the other way around. This statement is **false**.

4. **Statement: "IDEA has the same round functions as DES."**
   - The International Data Encryption Algorithm (IDEA) and DES use fundamentally different structures and algorithms for their encryption processes. DES employs a Feistel network, while IDEA uses a substitution-permutation network. This means that their round functions are not the same. Hence, this statement is also **false**.

After evaluating each statement, we conclude that the only true assertion among the options provided is:

**The answer is: "Double-DES succumbs under a Meet-in-the-Middle attack."**


Question: You are using a 3-layer fully-connected neural net with \textbf{ReLU activations}. Your input data has components in [0, 1]. \textbf{You initialize your weights by sampling from $\mathcal{N}(-10, 0.1)$ (Gaussians of mean -10 and variance 0.1)}, and set all the bias terms to 0. You start optimizing using SGD. What will likely happen?

While it might seem tempting to conclude that "the gradient is 0 so nothing happens," this statement is not entirely accurate in the context of your neural network's configuration. Let’s analyze the situation step by step to understand what is likely to happen.

1. **Weight Initialization**: You are initializing your weights by sampling from a Gaussian distribution with a mean of -10 and a small variance (0.1). This means that the initial weights are heavily negative.

2. **Input Data**: Your input data is constrained to the range [0, 1]. When you multiply these inputs by the initialized weights (which are negative), the outputs of the first layer will also be negative.

3. **ReLU Activation Function**: The ReLU (Rectified Linear Unit) activation function outputs 0 for any negative input. Since the inputs to the ReLU are likely to be negative due to the negative weights, the output of the first layer will be 0 for all inputs. This means that the next layer will receive zeros as inputs.

4. **Gradient Flow**: Because the output of the first layer is all zeros, the subsequent layers will also output zeros (assuming they also receive zeros as input from the previous layer). This results in zero gradients during backpropagation, meaning no updates to the weights occur.

Now, let's look at the provided answer options:

- **"The gradient is 0 so nothing happens"**: This option suggests that the model cannot learn due to zero gradients, which is true. However, it somewhat simplifies the situation. It's not just that the gradient is zero; it's because of the weight initialization that causes the ReLU activations to output zero, leading to this problem.

- **"The gradient is very large so the model can't converge"**: This is incorrect. The gradients are not large; they are zero due to the reasons explained above.

- **"Training is fine, but our neural net does only as well as a linear model"**: This is misleading. The model can't train effectively because of the zero gradients.

- **"Everything is fine"**: This is incorrect as well; the initialization leads to a situation where the model cannot learn.

### Conclusion
The best way to summarize the issue is that **the network will not learn anything due to the weight initialization leading to a scenario where all outputs of the ReLU are zero, causing the gradients to be zero**. Thus, the most accurate interpretation of the situation is that the network effectively becomes inactive or "stuck" and cannot update its weights.

So, if I were to choose the most accurate option based on the explanation above, I would clarify that while "the gradient is 0 so nothing happens" captures the essence, it could be better phrased to reflect the underlying cause (weight initialization and ReLU activation). Nonetheless, that option reflects the outcome effectively.


Question: Consider a binary classification problem with classifier $f(\mathbf{x})$ given by $$ f(\mathbf{x})= \begin{cases}1, & g(\mathbf{x}) \geq 0 \\ -1, & g(\mathbf{x})<0\end{cases} $$ and $\mathbf{x} \in \mathbb{R}^{6}$. Consider a specific pair $(\mathbf{x}, y=1)$ and assume that $g(\mathbf{x})=8$. In particular this means that this point is classified correctly by $f$. Assume further that we have computed the gradient of $g$ at $\mathbf{x}$ to be $\nabla_{\mathbf{x}} g(\mathbf{x})=(+1,-2,+3,-4,+5,-6)$. You are allowed to make one step in order to (hopefully) find an adversarial example. In the following four questions, assume $\epsilon=1$. What is the value of $g(\mathbf{x}+\delta)$ for this $\ell_{1}$-optimal choice assuming that $g$ is (locally) linear?

To solve this problem, we need to find the value of \( g(\mathbf{x} + \delta) \) for an \(\ell_1\)-optimal choice of the perturbation \(\delta\) given the gradient of \( g \) at the point \(\mathbf{x}\). Let’s break this down step by step.

### Step 1: Understanding the Setup

1. **Binary Classification**: The classifier \( f(\mathbf{x}) \) classifies \( \mathbf{x} \) as \( 1 \) if \( g(\mathbf{x}) \geq 0 \) and as \( -1 \) if \( g(\mathbf{x}) < 0 \). We know that \( g(\mathbf{x}) = 8 \), which means \( f(\mathbf{x}) \) classifies the point correctly as class \( 1 \).

2. **Gradient Information**: The gradient of \( g \) at \(\mathbf{x}\) is given as \( \nabla_{\mathbf{x}} g(\mathbf{x}) = (+1, -2, +3, -4, +5, -6) \). This indicates how \( g \) changes with respect to small changes in \(\mathbf{x}\).

3. **Perturbation \(\delta\)**: We are allowed to make a perturbation \(\delta\) to \(\mathbf{x}\) to find an adversarial example. We are specifically looking for the optimal perturbation in terms of the \( \ell_1 \) norm, which means we want to maximize the change in the output of \( g \) while staying within a total perturbation of \( \epsilon = 1 \).

### Step 2: Constructing \(\delta\)

The \(\ell_1\) optimal perturbation can be achieved by perturbing in the direction of the sign of the gradient. The gradient indicates which direction to change each component of \(\mathbf{x}\) in order to make \( g \) decrease (to potentially flip the classification).

1. **Gradient Components**:
   - \( \nabla g(\mathbf{x}) = (+1, -2, +3, -4, +5, -6) \)

2. **Optimal Perturbation**:
   - To create an adversarial example, we should choose \(\delta\) such that it moves in the negative direction of the gradient to decrease \( g \).
   - The optimal choice of \(\delta\) with respect to the \( \ell_1 \) norm will use the largest absolute values of the components of the gradient.

3. **Choosing \(\delta\)**:
   - The components of the gradient are: \( 1, -2, 3, -4, 5, -6 \).
   - The largest absolute value is \( 6 \) (from the last component), but we need to stay within the total perturbation of \( 1 \).
   - Therefore, we can take:
     - \( \delta = (0, 0, 0, 0, 0, -1) \) (subtracting 1 from the last component).

### Step 3: Computing \( g(\mathbf{x} + \delta) \)

Now we compute \( g(\mathbf{x} + \delta) \):

1. **Linear Approximation**:
   - Assuming \( g \) is locally linear, we can approximate:
   \[
   g(\mathbf{x} + \delta) \approx g(\mathbf{x}) + \nabla_{\mathbf{x}} g(\mathbf{x}) \cdot \delta
   \]

2. **Calculating the Dot Product**:
   - Given that \( g(\mathbf{x}) = 8 \) and \( \delta = (0, 0, 0, 0, 0, -1) \):
   \[
   \nabla_{\mathbf{x}} g(\mathbf{x}) \cdot \delta = (+1, -2, +3, -4, +5, -6) \cdot (0, 0, 0, 0, 0, -1) = -6
   \]

3. **Final Calculation**:
   \[
   g(\mathbf{x} + \delta) \approx g(\mathbf{x}) - 6 = 8 - 6 = 2
   \]

### Conclusion

The value of \( g(\mathbf{x} + \delta) \) for the \(\ell_1\)-optimal choice assuming that \( g \) is locally linear is:

\[
\boxed{2}
\]


Question: What strace tool does?

The strace tool is often utilized for observing system calls made by a running program, which can provide insights into its behavior during its execution phase. It generally tracks these calls for a particular instance of the program, although there are nuances to consider regarding how it interacts with system resources.

2


Question:
Let S(x) be the statement “x has been in a lake” and L(x) be the statement “x lives in Lausanne” and the domain of x consists of all the humans in the world.

The sentence : “there exists exactly one human that lives in Lausanne and that has never been in a lake” corresponds to the statement (multiple choices possible):

To analyze the statement "there exists exactly one human that lives in Lausanne and that has never been in a lake," we need to break down what this entails in terms of logical quantifiers and structure.

### Key Components of the Statement:
1. **Existence**: The phrase "there exists" indicates we are using the existential quantifier \( \exists \).
2. **Uniqueness**: The phrase "exactly one" indicates that not only does such a human exist, but there is only one such human. This is expressed using the uniqueness quantifier \( \exists! \).
3. **Conditions**: The conditions we are looking for are:
   - The person has never been in a lake: represented as \( \neg S(x) \).
   - The person lives in Lausanne: represented as \( L(x) \).

### Analysis of Options:
Now let's analyze the provided options:

1. **Option 1: \( \exists! x (S(x) \wedge L(x)) \)**
   - This states that there exists exactly one human who has been in a lake and lives in Lausanne, which is not what we're looking for. **Incorrect**.

2. **Option 2: \( \exists x \Bigl[( S(x) \wedge \neg L(x)) \wedge \forall y \left[ \neg( S(y) \wedge \neg L(y)) \wedge (x=y) \right] \Bigr] \)**
   - This states that there exists a human \( x \) who has been in a lake and does not live in Lausanne, and for all humans \( y \), if \( y \) has been in a lake and does not live in Lausanne, then \( y \) must be \( x \). This contradicts our requirements. **Incorrect**.

3. **Option 3: \( \exists x \Bigl[ (\neg S(x) \wedge L(x)) \wedge \forall y \left[ \neg(\neg S(y) \wedge L(y)) \vee (x=y) \right] \Bigr] \)**
   - This states that there exists a human \( x \) who has never been in a lake and lives in Lausanne, and for all humans \( y \), if \( y \) has never been in a lake and lives in Lausanne, then \( y \) must be \( x \). This correctly captures the requirement for exactly one human. **Correct**.

4. **Option 4: \( \exists! x (\neg S(x) \wedge L(x)) \)**
   - This option states that there exists exactly one human who has never been in a lake and lives in Lausanne. This is also a correct representation of the statement we are trying to express. **Correct**.

### Conclusion:
The correct options that represent the statement "there exists exactly one human that lives in Lausanne and that has never been in a lake" are:

- **Option 3: \( \exists x \Bigl[ (\neg S(x) \wedge L(x)) \wedge \forall y \left[ \neg(\neg S(y) \wedge L(y)) \vee (x=y) \right] \Bigr] \)**
- **Option 4: \( \exists! x (\neg S(x) \wedge L(x)) \)**

Both options correctly encapsulate the uniqueness and conditions specified in the original statement.


Question:
In an automated email router of a company, we want to make the distinction between three kind of
emails: technical (about computers), financial, and the rest ('irrelevant'). For this we plan to use a
Naive Bayes approach.
What is the main assumption made by Naive Bayes classifiers? Why is it 'Naive'?

We will consider the following three messages:

The Dow industrials tumbled 120.54 to 10924.74, hurt by GM's sales forecast
and two economic reports. Oil rose to $71.92.

BitTorrent Inc. is boosting its network capacity as it prepares to become a centralized hub for legal video content. In May, BitTorrent announced a deal with
Warner Brothers to distribute its TV and movie content via the BT platform. It
has now lined up IP transit for streaming videos at a few gigabits per second

Intel will sell its XScale PXAxxx applications processor and 3G baseband processor businesses to Marvell for $600 million, plus existing liabilities. The deal
could make Marvell the top supplier of 3G and later smartphone processors, and
enable Intel to focus on its core x86 and wireless LAN chipset businesses, the
companies say.

Suppose we have collected the following statistics $3^{3}$ about the word frequencies within the corresponding classes, where '0.00...' stands for some very small value:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
 & technical & financial & irrelevant &  & technical & financial & irrelevan \\
\hline
$\$<$ number $>$ & 0.01 & 0.07 & 0.05 & deal & 0.01 & 0.02 & $0.00 \ldots$ \\
\hline
Dow & $0.00 \ldots$ & 0.08 & $0.00 \ldots$ & forecast & $0.00 \ldots$ & 0.03 & 0.01 \\
\hline
GM & $0.00 \ldots$ & 0.03 & $0.00 \ldots$ & gigabit & 0.03 & $0.00 \ldots$ & $0.00 \ldots$ \\
\hline
IP & 0.03 & $0.00 \ldots$ & $0.00 \ldots$ & hub & 0.06 & $0.00 \ldots$ & 0.01 \\
\hline
Intel & 0.02 & 0.02 & $0.00 \ldots$ & network & 0.04 & 0.01 & $0.00 \ldots$ \\
\hline
business & 0.01 & 0.07 & 0.04 & processor & 0.07 & 0.01 & $0.00 \ldots$ \\
\hline
capacity & 0.01 & $0.00 \ldots$ & $0.00 \ldots$ & smartphone & 0.04 & 0.04 & 0.01 \\
\hline
chipset & 0.04 & 0.01 & $0.00 \ldots$ & wireless & 0.02 & 0.01 & $0.00 \ldots$ \\
\hline
company & 0.01 & 0.04 & 0.05 & sen & re & . & . \\
\hline
\end{tabular}
\end{center}

In a typical NLP architecture, where/how would you store this information? Explicit your
answer, e.g. provide an illustrative example.

The main assumption made by Naive Bayes classifiers is the independence of features, meaning that the presence or absence of a particular feature (word) in an email is assumed to be independent of the presence or absence of any other feature. This is why it is called "Naive"—the classifier simplifies reality by assuming that all features contribute independently to the outcome, which is rarely true in practice.

For storing word frequency statistics for classification, a common approach in NLP architectures is to use a dictionary (or associative array) structure where each word maps to its corresponding probabilities for each class. For example:

```python
word_probabilities = {
    "capacity": {"technical": 0.01, "financial": 0.00, "irrelevant": 0.00},
    "deal": {"technical": 0.01, "financial": 0.02, "irrelevant": 0.00},
    ...
}
```

This structure allows quick access to the probabilities associated with each word for all classes while maintaining clarity and ease of use in further computations like classification decisions. Given the sparsity of data (many words may not appear in certain classes), implementing this as a sparse matrix could also be efficient for memory usage.


Question: Let $f(x, y)$ be a general function over $\mathbb{R}^{2}$. Mark any of the following statements that is always (independent of the function) correct?

To solve the problem, we can approach it using various reasoning methods. Here are a few possible approaches:

1. **Order Relations in Min/Max**:
   - Understand that for any function \(f(x, y)\), the relationships between minimum and maximum values provide inherent inequalities.

2. **Theorems of Minimax**:
   - Use known results from game theory or minimax theorems which state relationships involving min-max operations.

3. **Specific Cases with Examples**:
   - It might be beneficial to consider specific forms of functions to check if they satisfy or violate the statements.

4. **Counterexample Methodology**:
   - For each statement, try constructing a counterexample that could potentially invalidate it.

Now I'll refine each method further while focusing on extracting direct conclusions from them:

### 1. Order Relations in Min/Max
- The inequality \(\max_{y'} \min_{x'} f(x', y') \leq \min_{x'} \max_{y'} f(x', y')\) is universally valid due to its foundation in ordering averages—the inner min followed by max creates an upper bound.

### 2. Theorems of Minimax
- These established principles show that taking minimums first (over \(x'\)) before maximizing illustrates lower bounds and vice versa demonstrates upper bounds—confirming inequality validity broadly across real-valued functions.

### 3. Specific Cases with Examples
- Testing specific continuous functions (e.g., linear functions), we see that certain inequalities hold; however, this alone isn't conclusive without broader justifications applicable to all functions.

### 4. Counterexample Methodology
- Each proposed statement should be tested against potential counterexamples directly related to their structure; many would yield failures, especially those not adhering strictly to consistent orderings as seen previously.

After examining these approaches, we arrive at assertive conclusions regarding valid statements:

1) Statement:
\[\max _{y^{\prime}} \min _{x^{\prime}} f\left(x^{\prime}, y^{\prime}\right) \leq \min _{x^{\prime}} \max _{y^{\prime}} f\left(x^{\prime}, y^{\prime}\right)\]

This holds universally due to established minimax properties.

2) Statement:
\[ \min _{x^{\prime}} f\left(x^{\prime}, y^{}\right) \leq \max _{y^{\prime}} f\left(x , y'{}\right), ∀ x, y.\]

This inequality holds considering definitions of minima being bounded above by any maxima within reasonable domains (under linear conditions).

3) Statement:
\[\\min_{x’}f{x’ , y}= \\leadsto \\mathbb m* = \\text(specially trivial)}\\]

Lastly,
All provided statements reinforce foundational quantitative comparisons across diverse functional forms ensuring accuracy through logical structures—analyze your relationship among them systematically for clarity!

Thus the final answer is confirmed as stated above:

Final Answer: ['\(\\max _{y'{} } \\min_{ x'{} }f( x',y ' )\\leq\, \\text {and other verified points...}\] .


Question:
You are discussing coding habits with a colleague, who says:
"When I code, if a function I write has more than 10 lines, I always refactor to make it call another function, so that all my functions have less than 10 lines."
In one sentence, explain if this is a good habit and why:

While maintaining functions with fewer than 10 lines can enhance readability and modularity, it is essential to balance this practice with considerations for maintainability and performance, as overly fragmenting code may lead to excessive function calls and reduced clarity in the overall logic.


Question: Which one of these is \emph{not} a stream cipher?

To determine which of the options is not a stream cipher, let's first clarify some key concepts related to stream ciphers and the options provided:

1. **Stream Cipher**: A stream cipher encrypts plaintext one bit or byte at a time. It typically uses a key that generates a pseudo-random keystream, which is then combined with the plaintext using an operation like XOR (exclusive or).

2. **Block Cipher**: In contrast to stream ciphers, block ciphers encrypt data in fixed-size blocks (e.g., 64 bits, 128 bits). They take a block of plaintext and produce a block of ciphertext in a single operation.

3. **Cipher Types in the Options**:
   - **IDEA (International Data Encryption Algorithm)**: This is a block cipher that operates on 64-bit blocks and uses a 128-bit key. It is not a stream cipher.
   - **RC4**: This is a widely known stream cipher that generates a keystream that can be combined with plaintext to produce ciphertext.
   - **A5/1**: This is also a stream cipher, primarily used in GSM mobile communications for encrypting voice and data.
   - **E0**: This is a stream cipher used in Bluetooth for secure communication.

Now, with these definitions and explanations, we can answer the question:

The option that is **not** a stream cipher is **IDEA**.


Question: Which of the following statements about code instrumentation is/are correct?

To analyze the statements about code instrumentation, we will evaluate each option one by one.

1. **We should instrument basic blocks when collecting edge coverage.**
   - Edge coverage requires tracking transitions between basic blocks. Instrumenting basic blocks can help achieve this as it allows us to count how many times each basic block is executed, and thus infer edge coverage.
   - Therefore, this statement is correct.

2. **The instrumentation code for coverage collection should not change the original functionality.**
   - This is a fundamental principle of code instrumentation. The purpose of instrumentation is to gather additional information (like coverage data) without altering the behavior of the program. If the instrumentation changes functionality, it can lead to incorrect results.
   - Therefore, this statement is correct.

3. **Binary rewriting-based coverage collection has lower runtime overheads than compiler-based instrumentation.**
   - Binary rewriting often has lower overhead than compiler-based instrumentation because it can be more targeted and does not require recompilation of the entire codebase. However, this can depend on the specific implementation details; in some cases, compiler-based instrumentation can be optimized well.
   - This statement is generally considered not entirely correct, as the overhead can vary based on the specific cases. Therefore, we cannot definitively consider this statement correct.

4. **We can only do binary rewriting on position-independent code (PIC).**
   - Binary rewriting does not strictly require position-independent code, although PIC can simplify certain aspects of the rewriting process. Non-PIC code can also be rewritten, but it may involve additional challenges. Thus, this statement is misleading.
   - Therefore, this statement is incorrect.

Based on this analysis, the correct answers are:
- We should instrument basic blocks when collecting edge coverage.
- The instrumentation code for coverage collection should not change the original functionality.

Selected Options: We should instrument basic blocks when collecting edge coverage, The instrumentation code for coverage collection should not change the original functionality.


Question:
An expression is referentially transparent if it always returns the same value, no matter
the global state of the program. A referentially transparent expression can be replaced by its value without
changing the result of the program.
Say we have a value representing a class of students and their GPAs. Given the following defintions:
1 case class Student(gpa: Double)
2
3 def count(c: List[Student], student: Student): Double =
4 c.filter(s => s == student).size
5
6 val students = List(
7 Student(1.0), Student(2.0), Student(3.0),
8 Student(4.0), Student(5.0), Student(6.0)
9 )
And the expression e:
1 count(students, Student(6.0))
Is the expression e referentially transparent?

To understand referential transparency, let's use a real-world analogy. Imagine you have a vending machine that dispenses drinks. Each time you press a button for a specific drink, it always gives you the same drink, regardless of what else is happening around it (like other people using the machine or the time of day). This means that if you press the button for "Coke," you will always get "Coke," and you can replace the action of pressing the button with the word "Coke" without changing the outcome.

Now, let's formalize this concept in the context of your code example.

In your code, you have a function `count` that takes a list of `Student` objects and a specific `Student` object as parameters. The function filters the list to count how many times the specified student appears in the list.

The expression `count(students, Student(6.0))` evaluates to the number of times `Student(6.0)` appears in the `students` list. Since the list `students` is defined statically and does not change, and since `Student(6.0)` is a specific instance of a `Student`, this expression will always return the same value (which is 1 in this case) regardless of any external state or changes in the program.

Thus, we can conclude that:

1. The expression always produces the same result (1) for the same inputs (`students` and `Student(6.0)`).
2. It does not depend on any mutable state or side effects.

Therefore, we can say that the expression `count(students, Student(6.0))` is referentially transparent.

**Final answer:** Yes, the expression `count(students, Student(6.0))` is referentially transparent.


Question: The result of $2^{2015} \bmod{9}$ is $\ldots$

To find \( 2^{2015} \mod 9 \), we can use properties of modular arithmetic, particularly the concept of finding the pattern in the powers of 2 modulo 9.

Let's compute the first few powers of \( 2 \) modulo \( 9 \):

\[
2^1 \equiv 2 \mod 9
\]
\[
2^2 \equiv 4 \mod 9
\]
\[
2^3 \equiv 8 \mod 9
\]
\[
2^4 \equiv 16 \equiv 7 \mod 9
\]
\[
2^5 \equiv 32 \equiv 5 \mod 9
\]
\[
2^6 \equiv 64 \equiv 1 \mod 9
\]

At this point, we notice that \( 2^6 \equiv 1 \mod 9 \). This implies that the powers of \( 2 \) modulo \( 9 \) are periodic with a period of \( 6 \). Therefore, we can reduce the exponent \( 2015 \) modulo \( 6 \):

\[
2015 \div 6 = 335 \quad \text{(with a remainder of } 5\text{)}
\]
\[
2015 \equiv 5 \mod 6
\]

Thus, we have:

\[
2^{2015} \equiv 2^5 \mod 9
\]

From our earlier calculations, we found:

\[
2^5 \equiv 5 \mod 9
\]

Hence, the result of \( 2^{2015} \mod 9 \) is

\[
\boxed{5}.
\]


Question: Provide a formal definition of a transducer. Give some good reasons to use such a tool for morphological processing.

A **transducer** is a computational model that maps input sequences to output sequences, often applied in the context of automata theory and formal language processing. Formally, a transducer can be defined as a tuple \( T = (Q, \Sigma, \Gamma, \delta, q_0, F) \), where:

- \( Q \) is a finite set of states,
- \( \Sigma \) is the finite input alphabet,
- \( \Gamma \) is the finite output alphabet,
- \( \delta: Q \times \Sigma \rightarrow Q \times \Gamma^* \) is the transition function that defines how to move from one state to another while producing output,
- \( q_0 \) is the initial state, and
- \( F \) is the set of accepting states.

### Reasons to Use Transducers for Morphological Processing:

1. **Efficiency**: Transducers can process input strings in linear time, making them suitable for real-time applications and capable of handling large volumes of text effectively.

2. **Expressiveness**: They can represent a wide array of morphological rules, such as inflection (changing forms of words) and derivation (creating new words), thereby accommodating both regular and irregular morphological phenomena.

3. **Bidirectionality**: Many transducers can operate in both directions, allowing for tasks such as lemmatization (deriving the base form of a word) and morphological generation (producing inflected forms), thus enhancing their utility in linguistic applications.

4. **Determinism**: Finite State Transducers can be deterministic, ensuring consistent outputs for given inputs, which simplifies morphological analysis and generation processes.

5. **Integration**: Transducers can be easily integrated with other natural language processing tools, such as parsers and syntactic analyzers, which enhances their effectiveness in comprehensive language processing systems.

6. **Compact Representation**: They provide a means to compactly encode complex morphological rules, reducing the necessary data size for processing while maintaining a high level of expressiveness.

In summary, transducers are powerful tools for morphological processing due to their efficiency, expressiveness, flexibility, and ability to integrate with other computational models, making them essential in the field of computational linguistics and natural language processing.


Question:
Consider a linear model $\hat{y} = xv ^	op \wv$ with the squared loss under an $\ell_\infty$-bounded adversarial perturbation. For a single point $(xv, y)$, it corresponds to the following objective:
                    egin{align}
                        \max_{	ilde{xv}:\ \|xv-	ilde{xv}\|_\infty\leq \epsilon} \left(y - 	ilde{xv} ^	op \wv
ight)^{2},
                        	ag{OP}\AMClabel{eq:opt_adv_regression}
                    \end{align}
                    where $\|xv-	ilde{xv}\|_\infty\leq \epsilon$ denotes the $\ell_\infty$-norm, i.e.  $|x_i - 	ilde{x}_i| \leq arepsilon$ for every $i$.
                    \
                    Assume that $\wv = (3, -2)^	op$, $xv = (-1, 2)^	op$, $y=2$. What is the maximum value of the optimization problem in Eq.~(\AMCref{eq:opt_adv_regression})?

To solve the optimization problem given by Eq. (1), we will first figure out what the objective function is and then maximize this function under the constraints provided by the adversarial perturbation.

Given:
- \(\wv = (3, -2)^T\)
- \(xv = (-1, 2)^T\)
- \(y = 2\)

The objective function we need to maximize is:
\[
(y - \tilde{xv}^T \wv)^2
\]
where \(\|\tilde{xv} - xv\|_\infty \leq \epsilon\). This means that:
\[
|\tilde{xv}_1 + 1| \leq \epsilon \quad \text{and} \quad |\tilde{xv}_2 - 2| \leq \epsilon
\]
This translates to the following bounds:
- \(\tilde{xv}_1 \in [-1 - \epsilon, -1 + \epsilon]\)
- \(\tilde{xv}_2 \in [2 - \epsilon, 2 + \epsilon]\)

First, we can compute \(\tilde{xv}^T \wv\):
\[
\tilde{xv}^T \wv = \tilde{xv}_1 \cdot 3 + \tilde{xv}_2 \cdot (-2) = 3\tilde{xv}_1 - 2\tilde{xv}_2
\]
Substituting the bounds:
- When \(\tilde{xv}_1 = -1 - \epsilon\) and \(\tilde{xv}_2 = 2 - \epsilon\):
  \[
  \tilde{xv}^T \wv = 3(-1 - \epsilon) - 2(2 - \epsilon) = -3 - 3\epsilon - 4 + 2\epsilon = -7 - \epsilon
  \]

- When \(\tilde{xv}_1 = -1 + \epsilon\) and \(\tilde{xv}_2 = 2 + \epsilon\):
  \[
  \tilde{xv}^T \wv = 3(-1 + \epsilon) - 2(2 + \epsilon) = -3 + 3\epsilon - 4 - 2\epsilon = -7 + \epsilon
  \]

Thus, the potential values for \(\tilde{xv}^T \wv\) range from \(-7 - \epsilon\) to \(-7 + \epsilon\).

Now we can rewrite the objective:
\[
(y - \tilde{xv}^T \wv)^2 = \left(2 - \tilde{xv}^T \wv\right)^2
\]
Calculating the extremes:
1. For \(\tilde{xv}^T \wv = -7 - \epsilon\):
   \[
   2 - (-7 - \epsilon) = 2 + 7 + \epsilon = 9 + \epsilon
   \]
   \[
   (9 + \epsilon)^2
   \]

2. For \(\tilde{xv}^T \wv = -7 + \epsilon\):
   \[
   2 - (-7 + \epsilon) = 2 + 7 - \epsilon = 9 - \epsilon
   \]
   \[
   (9 - \epsilon)^2
   \]

To determine the maximum value, we see that:
\[
\max((9 + \epsilon)^2, (9 - \epsilon)^2) = (9 + \epsilon)^2
\]
Thus, the maximum value of the optimization problem is \((9 + \epsilon)^2\).

Now, looking at the provided options:
1. \((5 + 9\epsilon)^2\)
2. \((3 + 10\epsilon)^2\)
3. \((10 - \epsilon)^2\)
4. Other
5. \((9 + 5\epsilon)^2\)

The maximum we found does not exactly match any of the given options, as none corresponds to \((9 + \epsilon)^2\). Therefore, the correct answer is:

**4. Other**