Question: Consider a regression task. You are using your favorite learning algorithm with parameters w and add a regularization term of the form $\frac{\lambda}{2}\|\mathbf{w}\|^{2}$. Which of the following statements are correct for a typical scenario?

Let's revise the answer by incorporating the feedback from the critique, ensuring clarity, completeness, and proper definitions of key terms.

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### Question:
Consider a regression task. You are using your favorite learning algorithm with parameters \( \mathbf{w} \) and add a regularization term of the form \( \frac{\lambda}{2}\|\mathbf{w}\|^{2} \). Which of the following statements are correct for a typical scenario?
Options:
1. 'The training error as a function of \( \lambda \geq 0 \) decreases.'
2. 'The training error as a function of \( \lambda \geq 0 \) increases.'
3. 'The test error as a function of \( \lambda \geq 0 \) increases.'
4. 'The test error as a function of \( \lambda \geq 0 \) decreases.'
5. 'The training error as a function of \( \lambda \geq 0 \) first decreases and then increases.'
6. 'The test error as a function of \( \lambda \geq 0 \) first decreases and then increases.'

### Answer:
Let's break down the implications of adding a regularization term to our regression task step by step.

### Understanding Regularization

In regression tasks, the goal is to minimize a loss function that quantifies how well the model predicts the output using the training data. Regularization adds a penalty term to this loss function, discouraging overly complex models. The specific regularization term we are considering is \( \frac{\lambda}{2}\|\mathbf{w}\|^{2} \), where:
- \( \lambda \) is a hyperparameter that controls the strength of the regularization (a larger \( \lambda \) means more regularization).
- \( \|\mathbf{w}\|^{2} \) is the squared Euclidean norm of the weight parameters, promoting smaller weights.

### Effect of Regularization on Training and Test Errors

1. **Training Error**:
   - As \( \lambda \) increases, the penalty for large weights increases, which discourages the model from fitting too closely to the training data (overfitting).
   - When \( \lambda = 0 \), the model can fit the training data perfectly, leading to a lower training error. As \( \lambda \) increases from 0, the model begins to generalize better, and the training error may initially decrease.
   - However, if \( \lambda \) becomes too large, the model may become overly constrained, leading to underfitting. This results in the training error increasing again after a certain point.
   - Therefore, the training error as a function of \( \lambda \) **first decreases and then increases**.

2. **Test Error**:
   - The test error reflects how well the model generalizes to unseen data. When \( \lambda \) is small, the model may overfit the training data, leading to a higher test error.
   - As \( \lambda \) increases, the model becomes more generalizable, which typically reduces the test error initially because it avoids overfitting.
   - However, if \( \lambda \) becomes excessively large, the model may underfit the data, resulting in a higher test error once again.
   - Thus, the test error as a function of \( \lambda \) **first decreases and then increases**.

### Analyzing the Options

Now, let's analyze the provided options:

1. **The training error as a function of \( \lambda \geq 0 \) decreases.** - **Incorrect**. The training error does not continuously decrease; it first decreases and then increases.

2. **The training error as a function of \( \lambda \geq 0 \) increases.** - **Incorrect**. The training error initially decreases before it starts increasing.

3. **The test error as a function of \( \lambda \geq 0 \) increases.** - **Incorrect**. The test error does not continuously increase; it first decreases and then increases.

4. **The test error as a function of \( \lambda \geq 0 \) decreases.** - **Incorrect**. The test error initially decreases, but as \( \lambda \) increases further, it eventually increases.

5. **The training error as a function of \( \lambda \geq 0 \) first decreases and then increases.** - **Correct**. This aligns with the explanation above regarding how training error behaves as \( \lambda \) increases.

6. **The test error as a function of \( \lambda \geq 0 \) first decreases and then increases.** - **Correct**. This reflects the initial improvement in generalization followed by potential underfitting as \( \lambda \) becomes too large.

### Conclusion

From the analysis, the correct options are:

- **The training error as a function of \( \lambda \geq 0 \) first decreases and then increases.**
- **The test error as a function of \( \lambda \geq 0 \) first decreases and then increases.**

### Additional Notes

Understanding concepts such as overfitting (where the model captures noise in the training data) and underfitting (where the model is too simple to capture the underlying trend) is crucial in this context. Visual representations of how training and test errors vary with \( \lambda \) can also help solidify these concepts.

Would you like to discuss any specific aspect further or explore how to select the optimal \( \lambda \) value in practice?


Question: Let $\xv_1, . . . , \xv_N$ be a dataset of $N$ vectors in $\R^D$. Write down the covariance matrix of the dataset $\Xm = (\xv_1, . . . , \xv_N) \in \R^{D	imes N}$, \emph{and} state its dimensions. Data is centered.

1. **Overview**:
   The task is to derive the covariance matrix of a centered dataset consisting of $N$ vectors in $\mathbb{R}^D$. The covariance matrix is a crucial statistical tool that captures the relationships and variances of different dimensions in the dataset. In this case, the data is centered, meaning the mean of each dimension across all vectors is zero. We will follow the suggested strategy to derive the covariance matrix and establish its dimensions.

2. **Step-by-Step Reasoning**:
   - **Step 1**: Recall the definition of the covariance matrix. For a dataset represented as a matrix $\Xm \in \mathbb{R}^{D \times N}$, where each column $\xv_i$ represents a centered vector, the covariance matrix $\Sigma$ is defined as:
     $$
     \Sigma = \frac{1}{N} \Xm \Xm^T.
     $$
   - **Step 2**: Since the dataset is centered, the mean vector $\mathbf{m} = \frac{1}{N} \sum_{i=1}^N \xv_i$ is zero for each dimension, which simplifies our calculations because we do not need to subtract the mean from each observation in the covariance formula.
   - **Step 3**: When we compute $\Xm \Xm^T$, we perform the following matrix multiplication:
     - $\Xm \in \mathbb{R}^{D \times N}$ and $\Xm^T \in \mathbb{R}^{N \times D}$ results in a matrix product of size $D \times D$.
     - The element at position $(j, k)$ in the covariance matrix $\Sigma$ is given by:
       $$
       \Sigma_{jk} = \frac{1}{N} \sum_{i=1}^N x_{ij} x_{ik},
       $$
       where $x_{ij}$ is the $j$-th component of the $i$-th vector $\xv_i$.
   - **Step 4**: Since each entry in $\Sigma$ computes the covariance between dimensions $j$ and $k$, this $D \times D$ matrix provides a comprehensive view of how the dimensions vary together.

3. **Technical Details**:
   - The covariance matrix $\Sigma$ can be explicitly expressed as:
     $$
     \Sigma = \frac{1}{N}\begin{pmatrix}
     \sum_{i=1}^N x_{i1}^2 & \sum_{i=1}^N x_{i1} x_{i2} & \cdots & \sum_{i=1}^N x_{i1} x_{iD} \\
     \sum_{i=1}^N x_{i2} x_{i1} & \sum_{i=1}^N x_{i2}^2 & \cdots & \sum_{i=1}^N x_{i2} x_{iD} \\
     \vdots & \vdots & \ddots & \vdots \\
     \sum_{i=1}^N x_{iD} x_{i1} & \sum_{i=1}^N x_{iD} x_{i2} & \cdots & \sum_{i=1}^N x_{iD}^2
     \end{pmatrix}.
     $$
   - Each entry of this matrix is computed from the data across all vectors, ensuring that we capture the variance and covariance for each pair of dimensions.

4. **Justification**:
   The derived covariance matrix $\Sigma$ is correct because it adheres to the definition of covariance for centered data. The use of $N$ in the denominator ensures that we obtain the average covariance, which is standard practice in statistics. The dimensions of the covariance matrix being $D \times D$ is justified by the structure of the matrix multiplication involved, confirming that each dimension's relationship with all other dimensions is effectively captured. Thus, our approach and final result are optimal and consistent with statistical theory.


Question: Which statement about 	extit{black-box} adversarial attacks is true:

Black-box adversarial attacks do not require access to the model's gradients, which means attackers can estimate these gradients using methods like finite difference approximations, where they slightly change the input and observe the output. This allows them to create adversarial examples without knowing the model's internal workings. Additionally, adversarial examples can often deceive other similar models, and these attacks can be carried out using techniques that don't rely on gradients at all.

Thus, the correct answer is “4. They can be implemented using gradient approximation via a finite difference formula. They can be implemented using gradient approximation via a finite difference formula as shown in the lecture slides. The rest of the options are incorrect since (1) gradient access is not needed, (2) adversarial examples are transferrable between similar models, and (3) they can be implemented via gradient-free optimization methods such as random search.”


Question: In this problem, we consider a generalization of the min-cost perfect matching problem. The generalization is called the \emph{min-cost perfect $b$-matching problem} and is defined as follows: \begin{description} \item[Input:] A graph $G = (V,E)$ with edge costs $c: E \rightarrow \mathbb{R}$ and degree bounds $b: V \rightarrow \{1,2, \ldots, n\}$. \item[Output:] A subset $F \subseteq E$ of minimum cost $\sum_{e\in F} c(e)$ such that for each vertex $v\in V$: \begin{itemize} \item The number of edges incident to $v$ in $F$ equals $b(v)$, i.e., $|\{e\in F :  v \in e\}| = b(v)$. \end{itemize} \end{description} Note that min-cost perfect matching problem is the special case when $b(v) =1$ for all $v\in V$. An example with general $b$'s is as follows: \begin{tikzpicture} \node at (1, 2.8) {Input}; \node[vertex] (u1) at (0,2) {$u_1$}; \node[vertex] (u2) at (0,0) {$u_2$}; \node[vertex] (v1) at (2,2) {$v_1$}; \node[vertex] (v2) at (2,0) {$v_2$}; \node[left = 0.1cm of  u1] {$b(u_1) = 1$}; \node[left = 0.1cm of  u2] {$b(u_2) = 2$}; \node[right = 0.1cm of  v1] {$b(v_1) = 1$}; \node[right = 0.1cm of  v2] {$b(v_2) = 2$}; \draw (u1) edge[ultra thick] (v1) edge (v2); \draw (u2) edge (v1) edge[ultra thick] (v2); \begin{scope}[xshift=7cm] \node at (1, 2.8) {Output}; \node[vertex] (u1) at (0,2) {$u_1$}; \node[vertex] (u2) at (0,0) {$u_2$}; \node[vertex] (v1) at (2,2) {$v_1$}; \node[vertex] (v2) at (2,0) {$v_2$}; \draw (u1)  edge (v2); \draw (u2) edge (v1) edge[ultra thick] (v2); \end{scope} \end{tikzpicture} On the left, we illustrate the input graph with the degree bounds (the $b$'s). Thin and thick edges have cost $1$ and $2$, respectively. On the right, we illustrate a solution of cost $1+1 +2 = 4$. It is a feasible solution since the degree of each vertex $v$ equals $b(v)$ in the solution. Your task is to prove the following statement: If the input graph $G=(V,E)$ is bipartite then any extreme point solution to the following linear programming relaxation (that has a variable $x_e$ for every edge $e\in E$) is integral: \begin{align*} \textbf{Minimize} \hspace{0.8cm} &  \sum_{e\in E} c(e) x_e\\ \textbf{subject to}\hspace{0.8cm} & \sum_{e\in E: v\in e} x_e  = b(v) \qquad \mbox{for all $v\in V$}\\ \hspace{0.8cm} & \hspace{0.9cm} 0 \leq  x_e \leq 1 \hspace{0.9cm} \mbox{for all $e\in E$}. \end{align*} {\em (In this problem you are asked to prove that every extreme point solution to the above linear program is integral assuming that the input graph $G$ is bipartite.   Recall that you are allowed to refer to material covered in the lecture notes.)}

To suggest that every extreme point solution to the linear programming relaxation of the min-cost perfect \( b \)-matching problem in a bipartite graph is integral, we can cite the characteristics of bipartite graphs and the nature of the linear program. In bipartite graphs, the vertices are divided into two sets such that there are no edges connecting vertices within the same set. The constraints in the linear program imply that the degree conditions \( b(v) \) are satisfied for each vertex, which often leads to integral solutions due to properties associated with bipartite structures. This might suggest that extreme points of the feasible region defined by these constraints tend to correspond to integral solutions, although exceptions can arise in specific cases.

Every extreme point solution to the linear programming relaxation is likely integral if the input graph \( G \) is bipartite.


Question: What is the inverse of 28 in $\mathbf{Z}_{77}$?

To find the inverse of 28 in \(\mathbf{Z}_{77}\), we need to find an integer \(x\) such that:

\[
28x \equiv 1 \pmod{77}
\]

This means we want to find \(x\) such that when \(28x\) is divided by 77, the remainder is 1. The inverse exists if and only if 28 and 77 are coprime (i.e., their greatest common divisor (gcd) is 1).

First, we compute the gcd of 28 and 77:

- The prime factorization of 28 is \(2^2 \times 7^1\).
- The prime factorization of 77 is \(7^1 \times 11^1\).

The only common factor is \(7\), so:

\[
\gcd(28, 77) = 7
\]

Since the gcd is not 1, 28 does not have an inverse in \(\mathbf{Z}_{77}\). This means none of the options provided can be the inverse of 28 in \(\mathbf{Z}_{77}\).

Now let's analyze the options provided:

1. **$0$**: This cannot be the inverse, as there cannot be an inverse of 0 in any modular arithmetic context.
2. **$1$**: This cannot be the inverse either, as \(28 \cdot 1 = 28\), and \(28 \mod 77\) is not 1.
3. **$36$**: This cannot be the inverse, as we would need to check if \(28 \cdot 36 \mod 77 \equiv 1\). Calculating gives \(28 \cdot 36 = 1008\), and \(1008 \mod 77 = 25\), which is not 1.
4. **$49$**: This cannot be the inverse as well, since we would check if \(28 \cdot 49 \mod 77 \equiv 1\). Calculating gives \(28 \cdot 49 = 1372\), and \(1372 \mod 77 = 37\), not 1.

Since none of these options can be correct, we conclude that the inverse of 28 does not exist in \(\mathbf{Z}_{77}\). Therefore, the correct conclusion is that there is no valid inverse among the provided options.


Question: Which of the following is WRONG about inverted files? (Slide 24,28 Week 3)

To determine which statement about inverted files is wrong, let's analyze each option step by step:

1. **The space requirement for the postings file is O(n)**:
   - This statement is generally true. The size of the postings file is proportional to the number of postings (or occurrences of terms in documents). As the number of documents (n) increases, the space requirement for the postings file can also be said to grow linearly, hence O(n).

2. **Variable length compression is used to reduce the size of the index file**:
   - This statement is incorrect in the context provided because variable length compression is typically used to reduce the size of the postings file, not the index file itself. The index file generally stores information that can be accessed in a fixed manner, while the postings file, which lists document IDs and positions, benefits more from variable-length encoding techniques.

3. **The index file has space requirement of O(n^beta), where beta is about 1⁄2**:
   - This statement is true. Inverted indexes often require space that grows sub-linearly with respect to the number of documents, which can be characterized by O(n^beta) with beta around 1/2, as some information is shared among multiple terms.

4. **Storing differences among word addresses reduces the size of the postings file**:
   - This statement is true as well. Using delta encoding (storing differences between consecutive addresses) is a common technique in compression, allowing for a more efficient representation of the postings lists.

Given this analysis, the statement that is wrong about inverted files is:

Answer: [2]


Question: Tick the \textbf{non-commutative} operation.

The non-commutative operation from the given options is:

3. $-$ (subtraction) over $\mathbb{Z}$.

**Explanation:**
- An operation is said to be commutative if changing the order of the operands does not change the result, i.e., for an operation \( \ast \), it holds that \( a \ast b = b \ast a \) for all \( a \) and \( b \).
- **Addition ($+$)** is commutative: \( a + b = b + a \).
- **Exclusive-or ($\oplus$)** is also commutative: \( a \oplus b = b \oplus a \).
- **Subtraction ($-$)** is not commutative: \( a - b \neq b - a \) in general.
- **Multiplication ($\times$)** is commutative: \( a \times b = b \times a \).

Thus, subtraction is the only non-commutative operation in the list.


Question: Let $H$ be a hash function. Collision resistance means that \dots

To determine which option best describes the concept of collision resistance in the context of hash functions, let's analyze each option:\n\nA. **given \( y \), it is hard to find \( x \) such that \( H(x) = y \)**  \nThis describes pre-image resistance, not collision resistance. Pre-image resistance means that for a given hash output \( y \), it should be computationally infeasible to find any input \( x \) that hashes to \( y \). Therefore, this option is **incorrect** regarding collision resistance.\n\nB. **given \( x \), it is hard to find \( y \) such that \( H(x) = y \)**  \nThis also describes a property related to hash functions but focuses on second pre-image resistance. It means that if you have an input \( x \), it should be hard to find another output \( y \) such that \( H(x) = y \). This is not what collision resistance is about. Thus, this option is **incorrect** as well.\n\nC. **it is hard to find \( x_1 \) and \( x_2 \neq x_1 \) such that \( H(x_1) = H(x_2) \)**  \nThis option describes collision resistance correctly. Collision resistance implies that it should be difficult to find two different inputs \( x_1 \) and \( x_2 \) that produce the same output when passed through the hash function \( H \). Therefore, this option is **correct**.\n\nD. **given \( x_1 \), it is hard to find \( x_2 \neq x_1 \) such that \( H(x_1) = H(x_2) \)**  \nWhile this option also touches on the idea of collision resistance, it is slightly more specific than option C. It implies that, given a specific input \( x_1 \), it should be hard to find a different input \( x_2 \) that collides with it. While this describes a facet of collision resistance, it does not encompass the overall definition as completely as option C does. However, it is still a valid aspect of collision resistance. Thus, this option could be considered **correct**, but it is more limited in scope than option C.\n\nConsidering the definitions and the explanations provided, the best answer that encompasses the full concept of collision resistance is:\n\n**C. it is hard to find \( x_1 \) and \( x_2 \neq x_1 \) such that \( H(x_1) = H(x_2) \)**.


Question: A bag contains the letters of LETSPLAY. Someone picks at random 4 letters from the  bag without revealing the outcome to you. Subsequently you pick one letter at random among the remaining 4 letters. What is the entropy (in bits) of the random variable that models your choice? Check the correct answer.

To determine the entropy of your choice, we first need to understand how many letters remain after 4 have been picked from the letters of "LETSPLAY." The original set has 8 unique letters (L, E, T, S, P, A, Y). After 4 letters are picked, 4 letters remain.

The random variable representing your choice has 4 possible outcomes (the remaining letters). The entropy \( H \) can be calculated using the formula:

\[ H = \log_2(n) \]

where \( n \) is the number of outcomes. In this case, \( n = 4 \):

\[ H = \log_2(4) = 2 \text{ bits} \]

Thus, the correct answer is: B) 2.


Question:
You share an apartment with friends.
Kitchen, living room, balcony, and bath room are shared resources among all parties.
Which policy/policies violate(s) the principle of least privilege?

To determine which policy violates the principle of least privilege, let's analyze each option step-by-step:

1. **Different bedrooms do not have a different key.**
   - **Analysis:** If all roommates have the same key for each other's bedrooms, then anyone can enter any bedroom, regardless of whether they need to. This violates the principle of least privilege, which states that individuals should only have access to resources necessary for their roles or needs. Here, unrestricted access to all bedrooms permits unnecessary access.

2. **There is no lock on the fridge.**
   - **Analysis:** The absence of a lock on the fridge doesn't inherently violate the principle of least privilege. While it may raise concerns about food theft or misuse, it does not grant unnecessary access to a resource that should be restricted. Roommates can still agree on food-sharing rules without a lock.

3. **To access the kitchen you have to go through the living room.**
   - **Analysis:** This setup is more about physical layout than privilege access. While it may be inconvenient, it doesn't provide anyone with unauthorized access; it simply requires passing through a common space. Thus, it doesn't violate the principle.

4. **Nobody has access to the neighbor's basement.**
   - **Analysis:** This is a standard practice. Not having access to another individual’s private space is in line with the principle of least privilege, as it limits access strictly to your own areas.

**Potential Misunderstanding:** The principle of least privilege is about ensuring that individuals have only the permissions necessary for their tasks, not general access to shared spaces unless specifically agreed upon.

**Correct Answer:** The option that violates the principle of least privilege is: **Different bedrooms do not have a different key.**


Question: In the following let $\kappa_{1}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)$ and $\kappa_{2}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)$ be two valid kernels. Show that the following is also valid kernel: $\kappa\left(\mathbf{x}, \mathbf{x}^{\prime}\right)=\kappa_{1}\left(\mathbf{x}, \mathbf{x}^{\prime}\right) \kappa_{2}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)$.

To show that the product of two valid kernels \(\kappa_{1}(\mathbf{x}, \mathbf{x}^{\prime})\) and \(\kappa_{2}(\mathbf{x}, \mathbf{x}^{\prime})\) is also a valid kernel, we need to demonstrate that the function \(\kappa(\mathbf{x}, \mathbf{x}^{\prime}) = \kappa_{1}(\mathbf{x}, \mathbf{x}^{\prime}) \kappa_{2}(\mathbf{x}, \mathbf{x}^{\prime})\) is positive semi-definite (PSD).

Recall that a kernel \( \kappa(\mathbf{x}, \mathbf{x}^{\prime}) \) is considered valid (or a positive semi-definite kernel) if for any finite set of points \(\{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_n\}\), the corresponding kernel matrix \(K\) defined by \(K_{ij} = \kappa(\mathbf{x}_i, \mathbf{x}_j)\) is positive semi-definite. This means that for any vector \(\mathbf{c} \in \mathbb{R}^n\),

\[
\mathbf{c}^T K \mathbf{c} \geq 0.
\]

Let’s denote the kernel matrices associated with \(\kappa_1\) and \(\kappa_2\) as \(K_1\) and \(K_2\) respectively, such that:

\[
K_{1, ij} = \kappa_{1}(\mathbf{x}_i, \mathbf{x}_j) \quad \text{and} \quad K_{2, ij} = \kappa_{2}(\mathbf{x}_i, \mathbf{x}_j).
\]

Since both \(\kappa_1\) and \(\kappa_2\) are valid kernels, we have:

\[
\mathbf{c}^T K_1 \mathbf{c} \geq 0 \quad \text{and} \quad \mathbf{c}^T K_2 \mathbf{c} \geq 0.
\]

We now consider the kernel matrix \(K\) defined by the product kernel:

\[
K_{ij} = \kappa(\mathbf{x}_i, \mathbf{x}_j) = \kappa_{1}(\mathbf{x}_i, \mathbf{x}_j) \kappa_{2}(\mathbf{x}_i, \mathbf{x}_j).
\]

To show that \(K\) is positive semi-definite, we need to evaluate the quadratic form \(\mathbf{c}^T K \mathbf{c}\):

\[
\mathbf{c}^T K \mathbf{c} = \sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j K_{ij} = \sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j \kappa_{1}(\mathbf{x}_i, \mathbf{x}_j) \kappa_{2}(\mathbf{x}_i, \mathbf{x}_j).
\]

We can rewrite this as:

\[
\mathbf{c}^T K \mathbf{c} = \sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j \kappa_{1}(\mathbf{x}_i, \mathbf{x}_j) \kappa_{2}(\mathbf{x}_i, \mathbf{x}_j) = \sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j \kappa_{1}(\mathbf{x}_i, \mathbf{x}_j) \cdot \kappa_{2}(\mathbf{x}_i, \mathbf{x}_j).
\]

This expression can be interpreted as a double sum over products of the two kernels. By the Cauchy-Schwarz inequality in the context of Hilbert spaces, we can apply the following reasoning. The product of two PSD matrices is PSD if both matrices are positive semi-definite.

### Step 1: Use the fact that \(K_1\) and \(K_2\) are PSD

Since both \(K_1\) and \(K_2\) are PSD, we can apply the Schur product theorem, which states that the Hadamard (element-wise) product of two positive semi-definite matrices is also positive semi-definite.

### Step 2: Conclude that \(K\) is PSD

To apply this theorem, we can introduce the identity matrix \(I\) of appropriate size. The Hadamard product of \(K_1\) and \(K_2\) can be expressed as:

\[
K = K_1 \circ K_2,
\]

where \(K\) is the kernel matrix defined by the product of the kernels.

Thus, by the Schur product theorem:

\[
K = K_1 \circ K_2 \text{ is positive semi-definite.}
\]

Hence, we conclude that \(\kappa(\mathbf{x}, \mathbf{x}^{\prime}) = \kappa_{1}(\mathbf{x}, \mathbf{x}^{\prime}) \kappa_{2}(\mathbf{x}, \mathbf{x}^{\prime})\) is a valid kernel, as it satisfies the condition of being positive semi-definite.


Question: Consider a hash function $H$ with $n$ output bits. Tick the \emph{incorrect} assertion.

To determine which assertion about a hash function \( H \) with \( n \) output bits is incorrect, let's analyze each option with respect to the properties of hash functions and the implications of the birthday paradox.

1. **Due to the birthday paradox, an output collision of \( H \) can be found much faster than with running time \( 2^n \)**:
   - The birthday paradox indicates that you can find a collision in a hash function with \( n \) output bits in approximately \( O(2^{n/2}) \) time, rather than \( 2^n \). Therefore, this statement is **correct**.

2. **It is possible to find an output collision of \( H \) with \( O(2^{\frac{n}{2}}) \) memory and \( O(2^{\frac{n}{2}}) \) running time**:
   - The birthday attack requires storing a certain number of hash values (approximately \( O(2^{n/2}) \)) to find a collision. This statement is consistent with the expected complexity of a birthday attack and is **correct**.

3. **It is possible to find an output collision of \( H \) with \( O(1) \) memory and \( O(2^{\frac{n}{2}}) \) running time**:
   - This is plausible since you can perform the computation of the hash function repeatedly while using a small amount of memory (storing only a few hash values). This can be done using a brute-force approach with \( O(2^{n/2}) \) time and negligible memory. Thus, this statement is **correct**.

4. **It is possible to find an output collision of \( H \) with \( O(2^{\frac{n}{2}}) \) memory and \( O(1) \) running time**:
   - This assertion is incorrect because finding a collision inherently involves computing hashes of inputs. Thus, \( O(1) \) running time (which implies constant time irrespective of the input) is not sufficient to compute \( O(2^{n/2}) \) hashes or to find a collision. You need a significant amount of time to compute the hashes, making this statement **incorrect**.

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

- **It is possible to find an output collision of \( H \) with \( O(2^{\frac{n}{2}}) \) memory and \( O(1) \) running time.**


Question: When computing HITS, the initial values

The answer is that the initial values for the HITS algorithm are set all to 1, offering a uniform starting point that aids in the convergence of the iterative calculations for authority and hub scores.


Question: Select the \emph{incorrect} statement. The brute force attack \dots

The incorrect statement is:

"can break a cipher with a
128
128-bit key on your PC today."

This statement is incorrect because, as of my last training data, breaking a cipher with a 128-bit key through brute-force means exhausting all possible key configurations, which is computationally infeasible with current technology and likely will remain so for the foreseeable future, even considering Moore's law. The number of possible keys for a 128-bit key space is (2^{128}), making it impractical to break using brute-force attacks using conventional computing power.


Question: In x86, select all synchronous exceptions?

To analyze the question regarding synchronous exceptions in the x86 architecture, we first need to clarify the core concepts involved.

### Core Concepts
- **Synchronous Exceptions**: These are exceptions that occur as a direct result of executing an instruction. They are predictable and occur at the point of instruction execution. Examples include divide errors and page faults.
- **Asynchronous Exceptions**: These exceptions occur independently of the current instruction execution and can happen at any time. Examples include interrupts from hardware devices like timers or keyboards.

### Option Analysis

1. **Divide Error**
   - **Evaluation**: Correct
   - **Reasoning**: A divide error occurs when a division operation attempts to divide by zero or results in an overflow. This exception is synchronous because it is directly tied to the execution of a division instruction. The processor detects the error at the moment the instruction is executed, making it a synchronous exception.

2. **Timer**
   - **Evaluation**: Incorrect
   - **Reasoning**: A timer interrupt is an example of an asynchronous exception. It is generated by the hardware timer and can occur at any time, independent of the currently executing instruction. The processor does not wait for a specific instruction to execute before the timer can trigger an interrupt, thus it does not qualify as a synchronous exception.

3. **Page Fault**
   - **Evaluation**: Correct
   - **Reasoning**: A page fault occurs when a program accesses a page that is not currently in physical memory. This is considered a synchronous exception because it happens as a direct result of an instruction that accesses memory. The processor generates a page fault exception when it attempts to access an invalid memory address, making it synchronous.

4. **Keyboard**
   - **Evaluation**: Incorrect
   - **Reasoning**: Keyboard interrupts are also asynchronous exceptions. They occur when a key is pressed, generating an interrupt signal to the processor. This event is independent of the instruction currently being executed, and thus it does not qualify as synchronous.

### Summary of Correct Answers
The correct options for synchronous exceptions in x86 are:
- Divide Error
- Page Fault

### Conclusion
The key insight this question tests is the understanding of the distinction between synchronous and asynchronous exceptions in the x86 architecture. Synchronous exceptions are directly related to the execution of instructions (like divide errors and page faults), while asynchronous exceptions are triggered by external events (like timer and keyboard interrupts). Understanding this distinction is crucial for grasping how the CPU handles errors and interrupts during program execution.


Question:
In Itanium's procedure call and return mechanism, What problem might arise when the processor executes
                erb+alloc+? Which mechanisms could be used to handle the problem?
                Feel free to mention what Itanium actually does (if you recall it),
                but list any effective solution that comes to mind. Just outline the
                basic ideas, do not try to describe or solve the details.

In Itanium architecture, the procedure call and return mechanism leverages a stack-based approach for managing function calls and local variables. The problem that may arise when executing `berb+alloc+` (a hypothetical function call) could be related to stack management, particularly stack overflow or corruption.

### Potential Problems:

1. **Stack Overflow**: If too many nested function calls occur or if a function allocates too much stack space, the stack can overflow, leading to runtime errors or undefined behavior.

2. **Stack Corruption**: Improper handling of stack pointers or frame pointers during function calls and returns can lead to corruption of the stack, causing incorrect behavior when accessing local variables or parameters.

### Mechanisms to Handle the Problems:

1. **Stack Size Limits**: Implement a maximum stack size limit and check for overflow before each allocation. If the limit is approached, raise an error or handle it gracefully by terminating the program or switching to a different memory allocation strategy.

2. **Stack Guard Pages**: Use guard pages at the end of the stack to detect overflows. When a stack overflow occurs, accessing the guard page will trigger a segmentation fault, allowing the program to handle the error.

3. **Dynamic Stack Growth**: Implement a mechanism for dynamically growing the stack when the limit is reached. This could involve moving the stack to a larger memory region, but it requires careful management to ensure that pointers remain valid.

4. **Frame Pointer Usage**: Utilize frame pointers to maintain the integrity of the stack frames. This helps in verifying the structure and preventing corruption during function calls and returns.

5. **Static Analysis**: Use static code analysis tools to identify potential stack overflow risks during compile-time by analyzing function call depth and local variable sizes.

6. **Compiler Optimizations**: Leverage compiler optimizations that can reduce stack usage, such as inlining small functions or optimizing variable lifetimes.

### Itanium's Approach:
Itanium implements a unique calling convention with a focus on register-based parameter passing and a streamlined stack usage, which helps mitigate some of these risks. It also uses a sophisticated exception handling mechanism to manage errors that arise during execution, including handling stack-related exceptions.

In summary, addressing potential stack-related issues in Itanium’s procedure call mechanism involves a combination of checks, dynamic handling strategies, and robust error detection and recovery mechanisms.


Question: Consider the source $S_1, S_2, \dots$ such that $S_1$ is uniformly distributed on $\mathbb{Z}/10\mathbb{Z}^*$, and for every $n\geq 1$, $S_{n+1}$ is distributed uniformly on $\mathbb{Z}/(S_n+1)\mathbb{Z}^*$. Let $H(\mathcal{S}) = \lim_{n	o\infty} H(S_n)$. True or false: $H(\mathcal{S}) = 0$.

1, True. The entropy approaches zero as the sequence converges to deterministic values, leading to lower uncertainty in the distribution over time.


Question:
Which of the following apply to recent Android-based mobile systems but not
    to Linux-based desktop systems?

The correct options that apply to recent Android-based mobile systems but not to Linux-based desktop systems are A, B, and C. A. Apps should use the binder interface to communicate with other apps: This is specific to Android's architecture, where the binder IPC (Inter-Process Communication) mechanism is used for efficient communication between apps and system services. B. By default, each app runs as its own user: This provides a security model in Android that isolates apps from each other, unlike traditional Linux desktop systems where apps typically run under the same user account. C. All apps run in a strict container with only limited system calls available: Android utilizes a security model that restricts what system calls an app can make, thus enhancing security compared to standard Linux desktop environments where apps generally have more freedom. Option D does not apply, as both systems allow file sharing through shared directories, although the implementations and permissions may differ


Question: ASLR stands for

The question asks for the meaning of the acronym ASLR, which is essential in the context of computer security.

1. **Analyze each option**:
   - **0. Address Space Layout Randomization**: **Correct.** ASLR is a security technique that randomizes memory addresses used by processes, making it harder for attackers to exploit vulnerabilities related to memory corruption.
   - **1. Address Structure Layout Randomization**: **Incorrect.** This option misuses the term "Structure" instead of "Space," which is critical to the concept of ASLR. ASLR specifically randomizes the layout of address space in memory.
   - **2. Augmented Security Layer Randomization**: **Incorrect.** While it includes "Security," this option does not accurately reflect the function of ASLR, which focuses on randomizing memory layouts rather than a security layer.
   - **3. Address Space Linker Randomization**: **Incorrect.** The term "Linker" is irrelevant to ASLR. ASLR pertains to memory address layouts, not linking processes or libraries.
   - **4. Automatic Structure Layout Randomization**: **Incorrect.** Similar to option 1, this option incorrectly replaces "Space" with "Structure." Furthermore, "Automatic" implies a lack of intentionality, which does not represent how ASLR operates.

In conclusion, **Address Space Layout Randomization** is the correct answer as it accurately defines the acronym ASLR and highlights its role in enhancing security through randomization of memory addresses, which is crucial for preventing attacks that exploit known memory locations. Understanding ASLR is vital for anyone studying modern computer security practices.


Question: Consider the LP-rounding algorithm for Set Cover that works as follows: \begin{enumerate} \item Solve the LP relaxation to obtain an optimal solution $x^*$. \item Return the solution $\{S: x^*_S >0\}$, i.e., containing all sets with a positive value in the fractional solution. \end{enumerate} Use the complementarity slackness conditions to prove that the algorithm is an $f$-approximation algorithm, where $f$ is the  frequency (i.e., the maximum number of sets that any element belongs to).

To establish that the LP-rounding algorithm for Set Cover is an $f$-approximation algorithm using complementarity slackness conditions, we start with the formulation of the Set Cover problem and its linear programming relaxation.

### Step 1: LP Formulation

The Set Cover problem can be formulated as follows:

- Let \( S \) be the collection of subsets, and let \( U \) be the universe of elements we want to cover.
- For each set \( S_i \in S \), let \( x_i \) be a binary variable indicating whether set \( S_i \) is included in the cover.

The linear programming relaxation of the Set Cover problem is:

\[
\text{Minimize} \quad \sum_{i} c_i x_i
\]
\[
\text{Subject to} \quad \sum_{i : e \in S_i} x_i \geq 1 \quad \forall e \in U
\]
\[
x_i \geq 0 \quad \forall i
\]

where \( c_i \) is the cost of including set \( S_i \).

### Step 2: Solving the LP Relaxation

We solve the LP relaxation to obtain an optimal solution \( x^* \). This solution is typically fractional, meaning \( 0 \leq x^*_i \leq 1 \).

### Step 3: Selecting Sets

The LP-rounding algorithm selects the sets \( S \) such that \( x^*_i > 0 \). This means that any set that contributes positively to the fractional solution is included in the final cover.

### Step 4: Complementarity Slackness

The complementarity slackness conditions relate the primal solution \( x \) to the dual solution \( y \). For our LP, the dual variables \( y_e \) correspond to the constraints covering each element \( e \):

1. If \( x^*_i > 0 \) (set \( S_i \) is chosen), then the corresponding dual variable \( y_e \) must satisfy all sets \( S_i \) that cover \( e \) such that \( y_e \) is at its upper bound (i.e., the contribution of set \( S_i \) is realized).
2. If \( y_e > 0 \), then none of the sets that do not cover \( e \) can have \( x_i \) greater than zero.

### Step 5: Analyzing the Cost

Let the cost of the sets selected by the algorithm be \( C = \sum_{i : x^*_i > 0} c_i \).

For each element \( e \) in the universe \( U \), let \( f \) be the maximum frequency of sets covering any element, meaning that any element \( e \) is covered by at most \( f \) sets. Therefore, for each element \( e \), the dual price \( y_e \) must be such that:

\[
\sum_{i : e \in S_i} x^*_i \geq 1
\]

Multiplying both sides of the above inequality by \( y_e \) and summing over all elements gives:

\[
\sum_{e} y_e \cdot \sum_{i : e \in S_i} x^*_i \geq \sum_{e} y_e
\]

### Step 6: Bounding the Cost

Since each element is covered by at most \( f \) sets and by the way we selected sets, we have:

\[
C \leq f \cdot \sum_{e} y_e
\]

Thus, the total cost of the selected sets \( C \) is bounded by \( f \) times the optimal cost of the fractional solution. Therefore, the algorithm achieves an approximation ratio of \( f \).

### Conclusion

Through the use of complementarity slackness conditions, we have shown that the LP-rounding algorithm for Set Cover is indeed an \( f \)-approximation algorithm, where \( f \) is the maximum frequency of sets covering any single element in the universe. This guarantees that the cost of the selected sets is at most \( f \) times the cost of the optimal cover.


Question: Implement a function that takes a lists ls as argument and returns the length of the longest contiguous sequence of repeated elements in that list. For this second question, you are required to use foldLeft in your solution, and your solution should not be recursive. For example: longest(List(1, 2, 2, 5, 5, 5, 1, 1, 1)) == 3 def longest[A](ls: List[A]): Int = ???

To solve the problem of finding the length of the longest contiguous sequence of repeated elements in a list using `foldLeft`, we need to define a few steps and concepts in our implementation.

### Steps to Approach the Problem:

1. **Define the State**: We need to maintain a state that keeps track of:
   - The current count of repeated elements.
   - The maximum count found so far.

2. **Initial State**: When starting the fold, the initial state can be `(0, 0)`, where:
   - The first element (`0`) represents the current count of repeated elements.
   - The second element (`0`) represents the maximum count of repeated elements found so far.

3. **Fold Function**: We will iterate through the list using `foldLeft`, and for each element:
   - If it is the same as the previous element, we increment the current count.
   - If it is different, we check if the current count is greater than the maximum count recorded, update if necessary, and reset the current count to 1.

4. **Final Comparison**: After processing the entire list, we need to perform a final check to ensure the last sequence is considered.

### Implementation:

Here is how you can implement the `longest` function in Scala:

```scala
def longest[A](ls: List[A]): Int = {
  ls.foldLeft((0, 0, None: Option[A])) { case ((currentCount, maxCount, lastElem), elem) =>
    if (lastElem.contains(elem)) {
      // Increment current count if the same as the last element
      (currentCount + 1, maxCount, lastElem)
    } else {
      // Check if we need to update max count
      val newMaxCount = math.max(currentCount, maxCount)
      // Reset current count for the new element
      (1, newMaxCount, Some(elem))
    }
  } match {
    // We need to return the maximum count after processing all elements
    case (currentCount, maxCount, _) => math.max(currentCount, maxCount)
  }
}

// Example usage
println(longest(List(1, 2, 2, 5, 5, 5, 1, 1, 1))) // Output: 3
```

### Explanation of the Implementation:

1. **Fold Left**: We use `foldLeft` to traverse the list. The accumulator is a tuple `(currentCount, maxCount, lastElem)`:
   - `currentCount`: Tracks how many times the current element has repeated.
   - `maxCount`: Tracks the maximum number of repetitions found so far.
   - `lastElem`: Keeps the last element processed to compare with the current element.

2. **Comparing Elements**:
   - If `lastElem` is the same as the current `elem`, we increase the `currentCount`.
   - If `lastElem` is different:
     - We update `maxCount` if `currentCount` is greater than `maxCount`.
     - We reset `currentCount` to 1 for the new `elem`.

3. **Final Result**: After the fold, we check the last `currentCount` against the `maxCount` to ensure we account for the last sequence of repeated elements.

This approach efficiently calculates the longest contiguous sequence of repeated elements in a non-recursive manner using `foldLeft`.


Question:
You've been hired to modernize a codebase from a 50-year-old company: version control, automated builds, and continuous integration. One of your colleagues, who is not completely up-to-date with modern practices, asks you the following question:
"Do I have to do one "commit" each day with my day's work?"
What would be your answer?

To address your colleague's question about whether they should make one commit each day with their day's work, we need to break down the concept of version control and the purpose of commits in a modern development workflow.

### Understanding Version Control

Version control systems (VCS), like Git, are used to manage changes to codebases over time. They allow developers to track modifications, collaborate with others, and revert to previous versions if necessary. Commits are a fundamental part of this process, as they represent a snapshot of changes made to the code at a particular point in time.

### The Purpose of Commits

1. **Logical Grouping of Changes**: Each commit should represent a logical unit of work. This means that each commit should encompass changes related to a specific task or feature. This approach makes it easier for others (and yourself in the future) to understand the history of changes.

2. **Clarity and Documentation**: Each commit typically includes a message that describes what changes were made and why. This practice provides context for future developers reviewing the commit history.

3. **Collaboration**: In collaborative environments, frequent commits help keep track of individual contributions, making it easier to integrate changes from multiple developers.

### Frequency of Commits

Now, to answer your colleague's question directly:

- **Quality Over Quantity**: There is no strict rule that says you must commit exactly once a day. The key is to commit when you have a meaningful change. This could be at the end of your workday, but it could also be after completing a significant feature or fixing a bug.

- **Best Practices**: A common guideline is to commit often, but not excessively. It is beneficial to make smaller, incremental commits that encapsulate specific changes rather than waiting until the end of the day to commit all your work at once. This practice enhances the clarity of the project history and makes it easier to pinpoint when a bug was introduced if issues arise.

- **Avoiding "Commit Monsters"**: If you save all your changes for a single commit at the end of the day, you risk creating what is known as a "commit monster," which is a commit that encapsulates too many changes. This can make it difficult to understand what changed and why, complicating the review process and debugging.

### Conclusion

In summary, while committing daily can be a good habit, the focus should be on making commits that are meaningful, well-defined, and logically grouped. Encourage your colleague to commit changes as they complete specific tasks or features rather than adhering to a rigid one-commit-per-day rule. This approach will lead to a cleaner project history and a more manageable codebase.

If they are still unsure, they can refer to the project's contribution guidelines (if available) or consult with team members to understand the preferred practices within your organization's workflow.


Question: The Merkle-D{\aa}mgard construction is

To determine the correct answer regarding the Merkle-D{\aa}mgard construction, follow these steps:

1. **Understand the Purpose**: The Merkle-D{\aa}mgard construction is designed to create a hash function from a compression function.

2. **Identify the Components**: The construction uses an iterative approach where a fixed-size input is processed in blocks, feeding the output back into the next round of processing.

3. **Review Definitions**:
   - A **hash function** takes an arbitrary input and produces a fixed-size output.
   - A **compression function** reduces the size of the input data while still allowing for the retrieval of a hash.

4. **Match Definitions**: The construction iteratively applies a compression function to produce a complete hash function.

After following these steps, the answer is: The Merkle-D{\aa}mgard construction is a method which iterates a compression function to obtain a hash function.


Question: Consider a Generative Adversarial Network (GAN) which successfully produces images of goats. Which of the following statements is false?

The false statement among the options provided is:

- **After the training, the discriminator loss should ideally reach a constant value.**

**Justification:**
In a well-trained GAN, the goal is for the generator to produce images that are indistinguishable from real images to the discriminator. If the discriminator reaches a constant loss, it suggests that it is not effectively distinguishing between real and generated images, which indicates a failure in the training process. Ideally, the discriminator's loss should vary as the generator improves and generates more realistic images, reflecting the ongoing challenge of the adversarial setup. A constant loss could imply that the discriminator has either become too good (overfitting) or that the generator has failed to improve, thus not providing a suitable adversarial challenge.


Question:
Your colleague wants to improve the performance of a web application by caching common results in an in-memory LRU cache, where the least recently used results are evicted when the cache is full, and wants your opinion on the best way to implement it. He has already implemented the "Cache" interface, which he will use to store the cached results. The "Cache" interface is defined as follows:
interface Cache {
    /**
     * Returns the value associated with the given key, or null if the key is not
     * present in the cache.
     */
    CompletableFuture get(K key);

    /**
     * Associates the given value with the given key in the cache. If the cache
     * already contains a value for the given key, it is replaced.
     */
    CompletableFuture put(K key, V value);
}

What do you think of this interface?

Imagine you're trying to grab a snack from a cupboard. If the cupboard is well-organized, you can quickly find your favorite snack without much hassle. However, if every time you reach for that snack, you have to wait for someone else to hand it to you, it becomes frustrating and slows down your snacking process. This analogy mirrors your colleague's "Cache" interface. The use of `CompletableFuture` suggests a level of complexity that isn't necessary for a simple in-memory cache, which ideally should provide immediate access to its values.

By using asynchronous methods, the interface introduces unnecessary delays, as users will have to await the results instead of accessing them directly. This could lead to a ripple effect where other parts of the application also need to handle these asynchronous calls, complicating the codebase and increasing the cognitive load on developers. A synchronous approach would simplify interaction with the cache and improve overall usability.

Your colleague's "Cache" interface is overly complex for an in-memory cache. It introduces unnecessary asynchronous behavior that complicates usage and may lead to increased complexity throughout the application.


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 all your weights to -10}, and set all the bias terms to 0. You start optimizing using SGD. What will likely happen?

To analyze the situation, let's break down the components of the question:

1. **Network Structure**: You have a 3-layer fully-connected neural network with ReLU (Rectified Linear Unit) activations. The ReLU activation function outputs the input directly if it is positive; otherwise, it outputs zero.

2. **Weight Initialization**: All weights are initialized to -10. This is crucial because it affects the output of the neurons in the first layer.

3. **Bias Initialization**: All bias terms are set to 0.

4. **Input Data**: The input data has components in the range [0, 1].

### Analysis of the Situation

- **ReLU Activation Function**: The ReLU function is defined as \( f(x) = \max(0, x) \). Given that all weights are initialized to -10, when you multiply the input (which is non-negative and in the range [0, 1]) by the weights, the output of the first layer will be negative or zero. Specifically, for any input \( x \) in [0, 1], the output of the first layer will be:
  \[
  \text{output} = \text{ReLU}(-10 \cdot x + 0) = \text{ReLU}(-10x) = 0
  \]
  since \(-10x\) will always be negative or zero for \( x \) in [0, 1].

- **Gradient Calculation**: Since the output of the first layer is 0 for all inputs, the gradients with respect to the weights will also be zero during backpropagation. This is because the gradient of the ReLU function is 0 for inputs less than or equal to 0. Therefore, the gradients for the weights will not change, leading to no updates during the optimization process.

### Conclusion

Given this analysis, the most likely outcome is:

1. **The gradient is 0 so nothing happens**.

This means that the model will not learn anything from the data because the weights will not be updated due to the zero gradients. Therefore, the correct answer is:

**1. The gradient is 0 so nothing happens.**


Question: Tick the \textbf{incorrect} assumption. A language $L$ is in NP if\dots

The incorrect assumption about a language \( L \) being in NP is:

- \( L \) is NP-hard.

### Explanation:

1. **\( x \in L \) can be decided in polynomial time.**
   - This statement is incorrect for languages in NP. While languages in P (which can be decided in polynomial time) are also in NP, not all languages in NP can be decided in polynomial time. Languages in NP are defined as those for which membership can be verified in polynomial time given a suitable witness.

2. **\( x \in L \) can be decided in polynomial time given a witness \( w \).**
   - This is the defining characteristic of NP. A language \( L \) is in NP if there exists a polynomial-time verifier that can confirm whether \( x \) belongs to \( L \) when provided with a witness \( w \).

3. **\( L \) is NP-hard.**
   - This statement is incorrect for a language being in NP. NP-hardness is a property that indicates a language is at least as hard as the hardest problems in NP. However, not all languages in NP are NP-hard; in fact, many languages in NP are not NP-hard (for example, certain languages in P). Thus, this assumption is incorrect.

4. **\( L \) (Turing-)reduces to a language \( L_2 \) with \( L_2 \) in \( P \), i.e., if there is a polynomial deterministic Turing machine which recognizes \( L \) when plugged to an oracle recognizing \( L_2 \).**
   - This statement is true. If a language \( L \) can be reduced to a language in \( P \), then \( L \) is also in NP since \( L_2 \) can be solved in polynomial time, and thus \( L \) can be solved by polynomially many queries to \( L_2 \).

In summary, the incorrect assumption about a language \( L \) being in NP is that \( L \) is NP-hard.


Question: Implement the recall at k metric

Answer:**

Here is the correct implementation of the recall at k metric in Python:

```python
def recall_at_k(retrieved, ground_truth, k):
    """
    Compute the recall at k metric.

    :param retrieved: List of retrieved items (e.g., tweets).
    :param ground_truth: List of relevant items (e.g., ground truth tweets).
    :param k: The number of top retrieved items to consider.
    :return: Recall at k as a float.
    """

    # Step 1: Count the total number of relevant items
    R = len(ground_truth)

    # Step 2: If there are no relevant items, return 0
    if R == 0:
        return 0.0

    # Step 3: Extract the top k retrieved items (slice the list)
    top_k_retrieved = retrieved[:k]  # Get the first k items

    # Step 4: Count how many of the top k are relevant
    R_k = len(set(top_k_retrieved) & set(ground_truth))  # Set intersection to find relevant items

    # Step 5: Calculate recall at k
    recall = R_k / R

    return recall

# Example usage:
retrieved_tweets = ['tweet1', 'tweet2', 'tweet3', 'tweet4', 'tweet5']
ground_truth_tweets = ['tweet2', 'tweet3']
k = 3

result = recall_at_k(retrieved_tweets, ground_truth_tweets, k)
print("Recall at k:", result)
```

**Explanation:**
The `recall_at_k` function calculates the recall at k metric by following a straightforward process. First, it counts how many relevant items are present in the ground truth dataset (denoted as \( R \)). If there are no relevant items, the function returns a recall of 0 to prevent division errors. It then extracts the top \( k \) retrieved items using list slicing. The function counts how many of these top \( k \) items are relevant by using the intersection of the top \( k \) items and the ground truth. Finally, it computes the recall at k by dividing the count of relevant items found in the top \( k \) (`R_k`) by the total number of relevant items (`R`) and returns the result. This approach is efficient and aligns with the standard definition of the recall at k metric.


Question: Let $y_1, y_2, \ldots, y_n$ be uniform random bits. For each non-empty subset $S\subseteq \{1,2, \ldots, n\}$, define $X_S = \oplus_{i\in S}\:y_i$. Show that the bits $\{X_S: \emptyset \neq S\subseteq \{1,2, \ldots, n\} \}$ are pairwise independent. This shows how to stretch $n$ truly random bits to $2^n-1$ pairwise independent bits. \\ \emph{Hint: Observe that it is sufficient to prove $\mathbb{E}[X_S] = 1/2$ and $\mathbb{E}[X_S X_T] = 1/4$ to show that they are pairwise independent. Also use the identity $\oplus_{i\in A}\: y_i = \frac{1}{2}\left( 1 - \prod_{i\in A} (-1)^{y_i} \right)$.}

To show that the random bits \( \{X_S: \emptyset \neq S \subseteq \{1, 2, \ldots, n\} \} \) are pairwise independent, we need to demonstrate two key properties:

1. **Expectation of \(X_S\)**: We need to show that \(\mathbb{E}[X_S] = \frac{1}{2}\) for any non-empty subset \(S\).
2. **Expectation of the product \(X_S X_T\)**: We need to show that \(\mathbb{E}[X_S X_T] = \frac{1}{4}\) for any two distinct non-empty subsets \(S\) and \(T\).

### Step 1: Calculate \(\mathbb{E}[X_S]\)

Recall that \(X_S = \oplus_{i \in S} y_i\). The expectation of \(X_S\) can be computed as follows:

\[
X_S = \oplus_{i \in S} y_i = y_{i_1} \oplus y_{i_2} \oplus \ldots \oplus y_{i_k}
\]

where \(S = \{i_1, i_2, \ldots, i_k\}\). Since \(y_i\) are independent uniform random bits, we know:

\[
\mathbb{E}[y_i] = \frac{1}{2}
\]

To find \(\mathbb{E}[X_S]\), we also use the fact that the XOR of two independent random bits is:

\[
\mathbb{E}[y_1 \oplus y_2] = \mathbb{E}[y_1] + \mathbb{E}[y_2] - 2 \mathbb{E}[y_1]\mathbb{E}[y_2] = \frac{1}{2} + \frac{1}{2} - 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}
\]

By induction, it follows that:

\[
\mathbb{E}[X_S] = \frac{1}{2}
\]

for any non-empty subset \(S\).

### Step 2: Calculate \(\mathbb{E}[X_S X_T]\)

Next, we calculate \(\mathbb{E}[X_S X_T]\) for distinct non-empty subsets \(S\) and \(T\).

From the hint, we can use the identity:

\[
X_S = \frac{1}{2} \left( 1 - \prod_{i \in S} (-1)^{y_i} \right)
\]

Thus, we can express \(X_S X_T\) as:

\[
X_S X_T = \left(\frac{1}{2} \left( 1 - \prod_{i \in S} (-1)^{y_i} \right)\right) \left(\frac{1}{2} \left( 1 - \prod_{j \in T} (-1)^{y_j} \right)\right)
\]

Expanding this gives:

\[
X_S X_T = \frac{1}{4} \left( 1 - \prod_{i \in S} (-1)^{y_i} - \prod_{j \in T} (-1)^{y_j} + \prod_{i \in S} (-1)^{y_i} \prod_{j \in T} (-1)^{y_j} \right)
\]

Now, taking the expectation:

1. The expectation of \(1\) is \(1\).
2. The expectation of \(\prod_{i \in S} (-1)^{y_i}\) is calculated as follows. Since the \(y_i\) are independent:

\[
\mathbb{E}\left[\prod_{i \in S} (-1)^{y_i}\right] = \prod_{i \in S} \mathbb{E}\left[(-1)^{y_i}\right] = \prod_{i \in S} \left( \frac{1}{2}(-1)^0 + \frac{1}{2}(-1)^1 \right) = \prod_{i \in S} 0 = 0
\]

Thus, each individual product \(\mathbb{E}\left[ \prod_{i \in S} (-1)^{y_i} \right] = 0\), and similarly for \(T\).

So, we have:

\[
\mathbb{E}[X_S] = \frac{1}{2}, \quad \mathbb{E}[X_T] = \frac{1}{2}, \quad \mathbb{E}\left[\prod_{i \in S} (-1)^{y_i}\right] = \mathbb{E}\left[\prod_{j \in T} (-1)^{y_j}\right] = 0
\]

This simplifies to:

\[
\mathbb{E}[X_S X_T] = \frac{1}{4} (1 - 0 - 0 + 0) = \frac{1}{4}
\]

### Conclusion

Since we have shown that:

\[
\mathbb{E}[X_S] = \frac{1}{2} \quad \text{and} \quad \mathbb{E}[X_S X_T] = \frac{1}{4}
\]

for any distinct non-empty subsets \(S\) and \(T\), we conclude that the bits \( \{X_S : \emptyset \neq S \subseteq \{1, 2, \ldots, n\} \} \) are pairwise independent.

Final answer: The bits \( \{X_S: \emptyset \neq S \subseteq \{1,2,\ldots,n\}\} \) are pairwise independent.


Question: Tick the \emph{minimal} assumption on the required channel to exchange the key of a Message Authentication Code (MAC):

To determine the minimal assumption required for a channel to exchange the key of a Message Authentication Code (MAC), we can analyze the options provided:

1. **Nothing**: This option suggests that no assumptions are needed, which is unrealistic since exchanging a key requires at least some level of security.

2. **Authentication and integrity only**: While these are crucial for ensuring the sender is legitimate and the message has not been altered, they do not prevent eavesdropping. An attacker could still intercept the key without confidentiality.

3. **Confidentiality only**: This ensures that the key is not exposed to eavesdroppers, but it does not guarantee that the sender is legitimate or that the message has not been tampered with.

4. **Authentication, integrity, and confidentiality**: This option encompasses all necessary security properties. It ensures that the sender is who they claim to be (authentication), the message has not been altered in transit (integrity), and the key is kept secret from potential attackers (confidentiality).

Thus, the minimal assumption required for securely exchanging a MAC key is option 4: authentication, integrity, and confidentiality. This comprehensive approach works because it addresses all potential threats during the key exchange process. By ensuring that all three properties are satisfied, we can securely exchange keys without falling prey to interception or forgery attacks.


Question: Which of the following is true?

Let's analyze the question step by step:

### Step 1: Define precision and recall

- **Precision** is defined as the ratio of true positives (TP) to the total number of predicted positives (TP + false positives (FP)):
  \[
  \text{Precision} = \frac{TP}{TP + FP}
  \]

- **Recall** (also known as sensitivity or true positive rate) is defined as the ratio of true positives to the total number of actual positives (TP + false negatives (FN)):
  \[
  \text{Recall} = \frac{TP}{TP + FN}
  \]

### Step 2: Effects of making a classifier more conservative (raising the threshold)

When you make a classifier more conservative by raising the threshold for a positive classification:
- **Precision tends to increase**: Since only the most confident predictions are classified as positive, there are fewer false positives (FP), which increases the precision.
- **Recall tends to decrease**: Some true positives (TP) may not reach the elevated threshold, leading to an increase in false negatives (FN). Thus, the recall decreases because fewer actual positives are identified.

### Step 3: Effects of making a classifier more liberal (lowering the threshold)

When you make a classifier more liberal by lowering the threshold for a positive classification:
- **Precision tends to decrease**: More instances are classified as positive, which often increases the number of false positives (FP), thereby reducing precision.
- **Recall tends to increase**: More actual positives are likely to be identified, reducing the number of false negatives (FN), thus increasing recall.

### Step 4: Evaluate each option

1. **High precision implies low recall**
   - This is generally true because when precision is high, it often means the classifier is more conservative, leading to lower recall.
   - **Mark: Valid**

2. **High precision hurts recall**
   - This statement is valid because as precision increases (often due to a conservative approach), recall tends to decrease.
   - **Mark: Valid**

3. **High recall hurts precision**
   - This statement is also valid, as increasing recall (by being more liberal) often results in lower precision due to an increase in false positives.
   - **Mark: Valid**

4. **High recall implies low precision**
   - This statement is generally true because increasing recall typically involves a more liberal classification approach, which increases false positives and therefore lowers precision.
   - **Mark: Valid**

### Final List of Valid Options

From the analysis, all four options are valid based on the trade-offs described.

**Final List: 1, 2, 3, 4**


Question:
Consider the following toy corpus: the cat cut the hat
What is the probability of the following sequences, if the parameters are estimated using MLE
(maximum-likelihood estimation) on the above corpus (make use of a calculator or even a
short program):
- cutthechat
- cut the chat
Fully justify your answer.

To calculate the probabilities of the sequences "cutthechat" and "cut the chat" using Maximum Likelihood Estimation (MLE) based on the given corpus "the cat cut the hat", we will follow these steps:

### Step 1: Preprocess the Corpus
We first remove spaces from the corpus for bigram analysis:
- Corpus without spaces: **"thecatcutthehat"**

### Step 2: Create Character Bigrams
Bigrams are pairs of consecutive characters. We will create bigrams from the corpus.

**Bigrams from "thecatcutthehat":**
- t-h
- h-e
- e-c
- c-a
- a-t
- t-c
- c-u
- u-t
- t-t
- t-h
- h-e
- e-h
- h-a
- a-t

### Step 3: Count Bigram Frequencies
Next, we tally the occurrences of each bigram in the corpus.

**Bigram Frequencies:**
- (t, h): 3
- (h, e): 2
- (e, c): 1
- (c, a): 1
- (a, t): 2
- (c, u): 1
- (u, t): 1
- (t, t): 1
- (h, a): 1

### Step 4: Calculate Total Number of Bigrams
Total number of bigrams = 13.

### Step 5: Calculate Probability of Each Bigram
Using MLE, the probability of each bigram (P(x,y)) is calculated as:
\[ P(x,y) = \frac{\text{Count}(x,y)}{\text{Total Bigram Count}} \]

### Step 6: Calculate Probability for "cutthechat"
**Bigrams in "cutthechat":**
- (c, u)
- (u, t)
- (t, t)
- (t, h)
- (h, e)
- (e, c)
- (c, h)
- (h, a)
- (a, t)

**Probabilities:**
- P(c, u) = 1/13
- P(u, t) = 1/13
- P(t, t) = 1/13
- P(t, h) = 3/13
- P(h, e) = 2/13
- P(e, c) = 1/13
- P(c, h) = 0/13 (not present)
- P(h, a) = 1/13
- P(a, t) = 2/13

**Total Probability for "cutthechat":**
\[
P(\text{"cutthechat"}) = P(c, u) \cdot P(u, t) \cdot P(t, t) \cdot P(t, h) \cdot P(h, e) \cdot P(e, c) \cdot P(c, h) \cdot P(h, a) \cdot P(a, t)
\]
Calculating:
\[
P(\text{"cutthechat"}) = \left(\frac{1}{13}\right) \cdot \left(\frac{1}{13}\right) \cdot \left(\frac{1}{13}\right) \cdot \left(\frac{3}{13}\right) \cdot \left(\frac{2}{13}\right) \cdot \left(\frac{1}{13}\right) \cdot \left(0\right) \cdot \left(\frac{1}{13}\right) \cdot \left(\frac{2}{13}\right) = 0
\]

### Step 7: Calculate Probability for "cut the chat"
**Bigrams in "cut the chat":**
- (c, u)
- (u, t)
- (t, )
- ( , t)
- (t, h)
- (h, e)
- (e, )
- ( , c)
- (c, h)
- (h, a)
- (a, t)

**Noting that spaces count as characters, we extend our bigram frequencies accordingly.**

**Assuming the corpus has spaces:**
- The probabilities of bigrams with spaces are similar in terms of counts. We will just consider:
- P(t, ) = 2/13 (space counts)

**Total Probability for "cut the chat":**
\[
P(\text{"cut the chat"}) = P(c, u) \cdot P(u, t) \cdot P(t, ) \cdot P( , t) \cdot P(t, h) \cdot P(h, e) \cdot P(e, ) \cdot P( , c) \cdot P(c, h) \cdot P(h, a) \cdot P(a, t)
\]
Calculating:
\[
P(\text{"cut the chat"}) = \left(\frac{1}{13}\right) \cdot \left(\frac{1}{13}\right) \cdot \left(\frac{2}{13}\right) \cdot \left(\frac{3}{13}\right) \cdot \left(\frac{2}{13}\right) \cdot \left(\frac{1}{13}\right) \cdot \left(\frac{2}{13}\right) \cdot \left(\frac{1}{13}\right) \cdot \left(\frac{0}{13}\right) \cdot \left(\frac{1}{13}\right) \cdot \left(\frac{2}{13}\right) = 0
\]

### Conclusion:
- The probability of "cutthechat" is **0** due to the absence of the bigram (c, h).
- The probability of "cut the chat" is also **0** due to the absence of the bigram (h, t) in the corpus.

Therefore, both sequences have a probability of **0** based on character bigram frequencies from the given corpus.


Question: We represent $GF(2^8)$ as $\mathbb{Z}_2[X]/P(X)$ where $P(X) = X^8 + X^4+X^3+X+1$. Then, $(X^7+X^6)\times (X + 1)=$\dots

To solve the multiplication \((X^7 + X^6) \times (X + 1)\) in \(GF(2^8)\) where the coefficients are taken modulo 2, we can proceed with the following steps:

1. **Distribute the terms**:
   \[
   (X^7 + X^6) \times (X + 1) = X^7 \cdot (X + 1) + X^6 \cdot (X + 1)
   \]

   This gives us:
   \[
   = X^7 \cdot X + X^7 \cdot 1 + X^6 \cdot X + X^6 \cdot 1
   \]
   \[
   = X^8 + X^7 + X^7 + X^6
   \]

2. **Combine like terms** (remembering that we are in \(GF(2)\), where \(1 + 1 = 0\)):
   \[
   = X^8 + (X^7 + X^7) + X^6 = X^8 + 0 + X^6 = X^8 + X^6
   \]

3. **Reduce modulo \(P(X)\)**:
   Since \(X^8\) can be replaced using the polynomial \(P(X) = X^8 + X^4 + X^3 + X + 1\), we have:
   \[
   X^8 \equiv - (X^4 + X^3 + X + 1) \equiv X^4 + X^3 + X + 1 \quad (\text{in } GF(2))
   \]

   Thus, we can replace \(X^8\) in our expression:
   \[
   X^8 + X^6 \equiv (X^4 + X^3 + X + 1) + X^6
   \]

4. **Combine the terms**:
   \[
   = X^6 + X^4 + X^3 + X + 1
   \]

Now we can match our result with the provided options. The expression we derived is:
- \(X^6 + X^4 + X^3 + X + 1\)

Thus, the correct answer is:
\[
\boxed{X^6 + X^4 + X^3 + X + 1}
\]


Question:
Consider the following definition of trees representing higher-order functions, as well as a recursive function
subst0.
1 enum Expr:
2 case C(c: BigInt)
3 case N(name: String)
4 case BinOp(op: BinOps, e1: Expr, e2: Expr)
5 case IfNonzero(cond: Expr, trueE: Expr, falseE: Expr)
6 case Call(fun: Expr, arg: Expr)
7 case Fun(param: String, body: Expr)
8
9 import Expr._
10
11 enum BinOps:
12 case Plus, Minus, Times, Power, LessEq
13
14 def subst0(e: Expr, n: String, r: Expr): Expr = e match
15 case C(c) => e
16 case N(s) => if s == n then r else e
17 case BinOp(op, e1, e2) =>
18 BinOp(op, subst0(e1, n, r), subst0(e2, n, r))
19 case IfNonzero(cond, trueE, falseE) =>
20 IfNonzero(subst0(cond,n,r), subst0(trueE,n,r), subst0(falseE,n,r))
21 case Call(f, arg) =>
22 Call(subst0(f, n, r), subst0(arg, n, r))
23 case Fun(formal, body) =>
24 if formal == n then e
25 else Fun(formal, subst0(body, n, r))
And consider the following expression:
1 val e = Call(N("exists"), Fun("y", Call(Call(N("less"), N("x")), N("y"))))
What is subst0(e, "x", N("y")) equal to?

To solve the problem of substituting a variable in the given expression using the `subst0` function, we will analyze the expression step-by-step, ensuring clarity in each part of the process.

### Step 1: Understanding the Expression

The expression we have is:

```scala
Call(N("exists"), Fun("y", Call(Call(N("less"), N("x")), N("y"))))
```

Let’s break down the components:
- **`Call(fun: Expr, arg: Expr)`**: This represents a function call. In this case, `N("exists")` is the function being called, and it is followed by an argument.
- **`N(name: String)`**: This denotes a variable or name. Here, `N("exists")` and `N("x")` represent identifiers.
- **`Fun(param: String, body: Expr)`**: This defines a function with a parameter (here, "y") and a body that specifies what the function does.
- **`Call(Call(N("less"), N("x")), N("y"))`**: This is a nested function call where the function `less` is called with two arguments: `x` and `y`.

### Step 2: Understanding the Substitution Process

The goal of the `subst0` function is to replace all instances of a variable `n` (in this case, "x") with a new expression `r` (here, `N("y")`). The `subst0` function works recursively, meaning it will dig into each part of the expression tree and make substitutions where necessary.

### Step 3: Applying the Substitution

Now, let’s apply the substitution step by step according to the `subst0` function:

1. **Start with the entire expression**:
   ```scala
   Call(N("exists"), Fun("y", Call(Call(N("less"), N("x")), N("y"))))
   ```
   - The outermost part is a `Call`, so we will check both the function and the argument for substitution.

2. **Substituting the function `N("exists")`**:
   - Since `N("exists")` is not "x", it remains unchanged.

3. **Substituting the argument `Fun("y", Call(Call(N("less"), N("x")), N("y")))`**:
   - This is a `Fun` node. We check its parameter "y", which is not equal to "x", so we proceed to substitute within the body of the function.

4. **Body of the function: `Call(Call(N("less"), N("x")), N("y"))`**:
   - This is where we need to check for "x" and apply the substitution.

5. **Outer Call: `Call(Call(N("less"), N("x")), N("y"))`**:
   - The first part is another `Call`. We need to substitute both the function and the argument:
     - **Function `Call(N("less"), N("x"))`**:
       - Here, `N("less")` remains unchanged.
       - Now we substitute `N("x")` with `N("y")`, resulting in `N("y")`.
     - The outer call then transforms into `Call(N("less"), N("y"))`.

6. **Final argument `N("y")`**:
   - The final argument remains unchanged as it corresponds to the variable "y", which is not "x".

### Step 4: Constructing the Final Expression

After applying all substitutions, we now construct our final expression:

- The original expression:
  ```scala
  Call(N("exists"), Fun("y", Call(Call(N("less"), N("x")), N("y"))))
  ```
  transforms to:
  ```scala
  Call(N("exists"), Fun("y", Call(Call(N("less"), N("y")), N("y"))))
  ```

### Conclusion

Thus, after substituting "x" with `N("y")`, the final result of `subst0(e, "x", N("y"))` is:

```scala
Call(N("exists"), Fun("y", Call(Call(N("less"), N("y")), N("y"))))
```

### Why This Matters

Understanding the substitution process in expressions is crucial in programming language design, compiler construction, and functional programming. Each substitution step is essential for ensuring that variables are accurately replaced without affecting unrelated parts of the expression. This manipulation allows programmers to analyze and transform code efficiently, enabling optimizations and the implementation of new features. Just as in a recipe, each ingredient (or variable) must be precisely measured and substituted to achieve the intended final dish (or result). Missing a step could lead to a completely different outcome, which highlights the importance of careful and methodical reasoning in programming and mathematics.


Question:
Assume your team is considering adding support for the SwengPhotos cloud service, which provides upload and download of photos on private cloud storage. Each photo is associated with a unique name. SwengPhotos's documentation for the "upload new photo" interface describes the following error responses:
1. I/O error
2. Backend timeout error
3. Name already exists error
You decide to build a Java library to wrap the SwengPhotos API. For each error, explain in 1 sentence whether it should be a "checked" exception and why:

In Java, exceptions are categorized into two main types: checked exceptions and unchecked exceptions. Checked exceptions are those that must be either caught or declared in the method signature, as they are anticipated during normal operation of the application. Unchecked exceptions, on the other hand, represent programming errors and do not require explicit handling.

Now, let's analyze each of the error responses provided by the SwengPhotos API and determine whether they should be treated as checked exceptions:

1. **I/O error**: This should be a checked exception because I/O errors are common in file operations and can occur due to various reasons, such as issues with the file system or network failures. Since the user of the library can take corrective actions (like retrying the operation or alerting the user), it’s important to inform them about this possibility through a checked exception.

2. **Backend timeout error**: This should also be a checked exception because a timeout can occur due to network latency or server unavailability, which are conditions that users of the API should be prepared to handle. By declaring this as a checked exception, you give developers the opportunity to implement retry logic or alternative flows when the backend service is slow or unreachable.

3. **Name already exists error**: This should be a checked exception as well, since it indicates a specific user error where the user is attempting to upload a photo with a name that is already taken. In this case, the user should be notified so they can take appropriate action, such as renaming the photo. Handling this as a checked exception allows developers to explicitly manage this scenario in their code.

In summary, all three errors should be treated as checked exceptions in your Java library. This design choice not only encourages better error handling practices but also improves the robustness of applications built on top of your library, as developers will be required to consider and handle these foreseeable issues explicitly.


Question: Which of the following congruence classes has a multiplicative inverse?

To determine which of the given congruence classes has a multiplicative inverse, we need to analyze each option in terms of their respective modulus and check if they are coprime to their moduli. A number has a multiplicative inverse modulo \( n \) if and only if it is coprime to \( n \) (i.e., their greatest common divisor (gcd) is 1).

1. **Understanding the options**:
   - **A: \( [3^{10} 2^5 14]_{19} \)**: We first calculate \( 3^{10} \cdot 2^5 \cdot 14 \mod 19 \).
     - Calculate \( 3^{10} \mod 19 \): Using Fermat's Little Theorem, \( 3^{18} \equiv 1 \mod 19 \), thus \( 3^{10} \) can be calculated directly.
     - Calculate \( 2^5 \mod 19 = 32 \mod 19 = 13 \).
     - Calculate \( 14 \mod 19 = 14 \).
     - Now compute \( (3^{10} \cdot 13 \cdot 14) \mod 19 \).
     - The important part is to check if \( 3^{10} \cdot 2^5 \cdot 14 \) is coprime to 19. Since 19 is prime and none of the factors are equal to 19, they are coprime.

   - **B: \( [126]_{147} \)**: We need to check if 126 is coprime to 147.
     - The prime factorization of 126 is \( 2 \cdot 3^2 \cdot 7 \).
     - The prime factorization of 147 is \( 3 \cdot 7^2 \).
     - The gcd(126, 147) is 21, which is greater than 1, so they are not coprime. Thus, 126 does not have a multiplicative inverse modulo 147.

   - **C: \( [60]_{15} \)**: We check if 60 is coprime to 15.
     - The prime factorization of 60 is \( 2^2 \cdot 3 \cdot 5 \).
     - The prime factorization of 15 is \( 3 \cdot 5 \).
     - The gcd(60, 15) is 15, which is greater than 1, so they are not coprime. Thus, 60 does not have a multiplicative inverse modulo 15.

2. **Evaluating the options**:
   - **Option A**: \( [3^{10} 2^5 14]_{19} \) is coprime to 19, thus has a multiplicative inverse.
   - **Option B**: \( [126]_{147} \) is not coprime to 147, thus does not have a multiplicative inverse.
   - **Option C**: \( [60]_{15} \) is not coprime to 15, thus does not have a multiplicative inverse.

After evaluating all the options, we can conclude that the only congruence class that has a multiplicative inverse is option A.

Action: Answer("A")


Question: The needed number of DES operations to attack DES is about ...

The answer is option 1: $2^{54}.$ The Data Encryption Standard (DES) employs a key size of 56 bits, which theoretically allows for $2^{56}$ possible keys. However, due to certain known vulnerabilities and effective attack strategies, the practical effort required to break DES through brute force is estimated to be around $2^{54}$ operations.


Question: The number of term vactors in the matrix K_s, used for LSI

To answer the question regarding the number of term vectors in the matrix \( K_s \) used for Latent Semantic Indexing (LSI), we need to consider how LSI operates and the relationship between the term-document matrix \( M \) and the reduced matrix \( K_s \).

1. **Understanding the term-document matrix \( M \)**:
   - The matrix \( M \) typically has rows representing terms (words) and columns representing documents. Therefore, if \( M \) has \( n \) rows, it means there are \( n \) unique terms.

2. **Latent Semantic Indexing (LSI)**:
   - LSI involves singular value decomposition (SVD) of the term-document matrix \( M \). The SVD decomposes \( M \) into three matrices: \( U \) (terms), \( S \) (singular values), and \( V^T \) (documents).
   - The matrix \( K_s \) usually refers to a reduced form of the term-document matrix after selecting the top \( s \) singular values and their corresponding vectors.

3. **Dimensions of \( K_s \)**:
   - The number of term vectors in \( K_s \) corresponds to the number of terms retained after the reduction. Since \( K_s \) is derived from \( M \), it will still have rows corresponding to the terms.
   - Therefore, the number of term vectors in \( K_s \) should match the number of terms (rows) in the original matrix \( M \).

4. **Conclusion**:
   - Since \( K_s \) retains the same number of terms as \( M \) (even though it might reduce the dimensionality in terms of documents or latent concepts), we conclude that the number of term vectors in \( K_s \) is the same as the number of rows in matrix \( M \).

Final answer: **is the same as the number of rows in the matrix M**.


Question: Select the answer that correctly describes the differences between formal and natural languages.

To address the question about the differences between formal and natural languages, we need to carefully define both types of languages and analyze the provided options with clarity.

### Defining the Concepts

**Formal Languages**: Picture a precise set of building blocks that can only be arranged in specific ways to create a structure. Each block has a unique shape and can only fit together in certain configurations. In the realm of computer science, formal languages operate under strict rules and syntax, making them explicit and non-ambiguous. This means that every instruction is clearly defined, and there’s no room for multiple interpretations. For example, programming languages like Python or Java have defined grammatical rules; if you write an instruction incorrectly, the computer will throw an error. This precision is vital in programming, mathematics, and logic, as even a small mistake can lead to significant errors.

**Natural Languages**: Now, consider a colorful tapestry woven from various threads, each representing a different emotion, culture, and idea. When you speak a natural language like English or Spanish, you're using a medium that is rich in idioms, nuances, and variations. Natural languages are often implicit and can be ambiguous; the meaning of a phrase can change based on context, tone, or even the speaker's intentions. For instance, if someone says, "It's a bit chilly," they might mean they’re feeling cold, or they might be hinting that you should wear a jacket without directly saying it. This inherent ambiguity allows for creativity and emotional expression, but it can also lead to misunderstandings.

### Analyzing the Options

1. **Option 1**: "Formal languages are by construction explicit and non-ambiguous while natural languages are implicit and ambiguous."
   - **Correctness**: This option accurately describes the essence of both language types. Formal languages are designed to be clear and direct, eliminating ambiguity through strict rules. Natural languages, on the other hand, thrive on implicit meanings and can carry multiple interpretations based on context. This option correctly captures the fundamental differences.

2. **Option 2**: "Formal languages are by construction implicit and non-ambiguous while natural languages are explicit and ambiguous."
   - **Incorrectness**: This option misrepresents formal languages as implicit. Implicitness implies that meanings are not directly stated, which contradicts the very nature of formal languages. They are constructed precisely to remove ambiguity. Additionally, natural languages are not explicitly clear; rather, their ambiguity often creates challenges in understanding.

3. **Option 3**: "Formal languages are by construction explicit and ambiguous while natural languages are implicit and non-ambiguous."
   - **Incorrectness**: This option inaccurately portrays formal languages as ambiguous. The primary goal of formal languages is clarity and precision, making them non-ambiguous by design. Moreover, natural languages are frequently filled with ambiguity, as expressions and phrases can have multiple meanings depending on context.

### Conclusion
The correct answer is **Option 1**: "Formal languages are by construction explicit and non-ambiguous while natural languages are implicit and ambiguous." This distinction is vital in various fields, including programming, linguistics, and logic, where precise communication is essential for effective problem-solving and collaboration.

Understanding these differences matters significantly in practice. When we design algorithms or create software, we rely on formal languages to ensure that our instructions are executed flawlessly. For example, in coding, a misplaced semicolon can cause a program to crash. Conversely, in literature or everyday conversation, natural languages allow for nuanced expression and emotional depth, enriching human connection. Recognizing the strengths and weaknesses of both language types helps us navigate communication effectively, whether we are debugging code or crafting a heartfelt message.

In summary, the ability to discern between formal and natural languages not only enhances our understanding of communication but also equips us with the tools to engage with the world around us more effectively.


Question: To support very large scale neural networks in limited amount of memory, one may want to use floating point numbers with very few bits. Here we consider substantially simplified operations on such numbers, Float8. A value Float8(mant,exp) represents the non-negative integer mant * 2^exp. We call mant a mantissa (which gives significant digits) whereas exp is the exponent. This allows us to represent both smaller and larger integers, keeping a similar number of significant digits. (The larger integers can only be represented up to a given power of two.) In our simple example, we use only four bits for both mantissa and the exponent, and we assume they are both non-negative.  final case class Float8(mant: Int, exp: Int):   require(0 <= mant && mant <= 15 && 0 <= exp && exp <= 15)   def value: Int = mant << exp  val a = Float8(15, 8) val b = Float8(5, 10) We look at the operation plus, of adding such numbers. When the exponent is smaller than another one, the operation shifts mantissa and then performs addition. If mantissa gets too large, we reduce it an increase the exponent.  extension (x: Float8)    def +(y: Float8): Float8 =     if x.exp <= y.exp then       val shift = y.exp - x.exp       val mant = (x.mant >> shift) + y.mant       if mant < 16 then Float8(mant, y.exp)       else          val exp1 = y.exp + 1         if exp1 < 16 then Float8(mant / 2, y.exp + 1)         else Float8(15, 15)     else y + x Is this operation commutative? Prove or give a counterexample.

To determine whether the addition operation defined for `Float8` is commutative, we can analyze the behavior of the `+` method under different conditions based on the exponents of the two `Float8` instances being added. The commutativity property states that for any two elements \(x\) and \(y\), the equation \(x + y = y + x\) should hold true.

Let's consider the three possible cases based on the relationship between the exponents of the two `Float8` instances:

1. **Case 1: \(x.exp < y.exp\)**
   In this scenario, when we evaluate \(x + y\), the method checks if \(x.exp <= y.exp\). Since this condition is true, it proceeds to calculate the shift required to align the mantissas. The mantissa of \(x\) is right-shifted by the difference in exponents, and then the mantissas are added. The result is then checked to see if it exceeds the maximum allowable value for the mantissa. If it does, the mantissa is halved, and the exponent is incremented accordingly.

   When evaluating \(y + x\), the method checks if \(y.exp <= x.exp\), which is false. Thus, it calls \(x + y\) instead. This means that both \(x + y\) and \(y + x\) will yield the same result, confirming that addition is commutative in this case.

2. **Case 2: \(x.exp > y.exp\)**
   Here, the evaluation of \(x + y\) will check if \(x.exp <= y.exp\), which is false, leading the method to call \(y + x\). The evaluation of \(y + x\) will then check if \(y.exp <= x.exp\), which is true. Therefore, it calculates the shift for \(y\) and adds the mantissas accordingly.

   Since the operation flips the call to \(y + x\) when \(x.exp > y.exp\), we again find that both expressions yield the same result, thus maintaining commutativity.

3. **Case 3: \(x.exp = y.exp\)**
   In this case, both evaluations of \(x + y\) and \(y + x\) will enter the condition where the shift is zero. The mantissas will be added directly without any shifting. Since both operations are performed with the same mantissas and exponents, the results will be identical.

Through this analysis, we see that in all three cases, the addition operation defined for `Float8` adheres to the commutative property. The key is that the method structure inherently resolves the order of operations based on the relationship between the exponents, ensuring that the final results of \(x + y\) and \(y + x\) are always equivalent.

Thus, we conclude that the addition operation for `Float8` is indeed commutative, as demonstrated through careful consideration of the different cases based on the exponents.


Question:
Consider the following CFG

\(\text{S} \rightarrow \text{NP VP PNP}\)
\(\text{NP} \rightarrow \text{Det N}\)
\(\text{NP} \rightarrow \text{Det Adj N}\)
\(\text{VP} \rightarrow \text{V}\)
\(\text{VP} \rightarrow \text{Aux Ving}\)
\(\text{VP} \rightarrow \text{VP NP}\)
\(\text{VP} \rightarrow \text{VP PNP}\)
\(\text{PNP} \rightarrow \text{Prep NP}\)

and the following lexicon:
the:Det, red:Adj, cat:N, is:Aux, meowing:Ving, on:Prep, roof:N
The next four questions ask you the content of a given cell of the chart used by the CYK algorithm (used here as a recognizer) for the input sentence
the red cat is meowing on the roof
Simply answer "empty'' if the corresponding cell is empty and use a comma to separate your answers when the cell contains several objects.What is the content of the cell at row 3 column 1 (indexed as in the lectures)?

To construct the parse for "the red cat is meowing on the roof," we analyze the input. The phrase "the red cat" forms a NP, "is meowing" forms a VP, and "on the roof" forms a PNP.

For row 3 column 1, which corresponds to the NP formed by "the red cat," we have:

- NP → Det Adj N: "the red cat"

Thus, the cell at row 3 column 1 contains NP.

Answer: *NP*.


Question:
Concatenating two conc-trees of heights $h_1$ and $h_2$ yields a conc-tree with
height $h$ where

To understand the height of the concatenated tree, we start by noting that the height of a tree is defined as the number of edges on the longest path from the root to a leaf. When we concatenate two conc-trees of heights \( h_1 \) and \( h_2 \), the new height \( h \) can be influenced by how the trees are joined.

1. If we attach the shorter tree to the root of the taller tree, the height will be approximately the height of the taller tree, potentially increasing by 1 if the shorter tree has a height of 1.
2. Conversely, if we attach the taller tree to the shorter one, the height will again be approximately the height of the taller tree, but it may decrease by 1 if the shorter tree has a height of 0.

Thus, the height \( h \) of the concatenated tree will satisfy the condition that it is at most 1 unit away from the maximum height of the two original trees. Therefore, the final answer is:

The height \( h \) of the concatenated conc-tree satisfies \( |h - \max(h_1, h_2)| \leq 1 \).


Question: When compressing the adjacency list of a given URL, a reference list

The correct answer is:

- May contain URLs not occurring in the adjacency list of the given URL.

When compressing an adjacency list, the reference list might include URLs that are not directly linked to the given URL in the adjacency list. This is because the purpose of the reference list is to provide additional context or relevance that could enhance navigation or searchability, even if those URLs are not part of the immediate connections. Including such URLs can help in scenarios like exploring related content or expanding the user's options beyond the direct links.


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

To evaluate the statements about fuzzing, let's break down each statement and analyze its correctness based on the principles of fuzzing techniques.

### Statement Analysis

1. **Generational fuzzing requires more manual work (to specify the generator policies) than mutational fuzzing, but can generate high-quality seeds.**
   - **Analysis:** This statement is generally correct. Generational fuzzing involves the creation of input data based on defined specifications or grammar, which often requires extensive manual effort to define policies and rules. However, this can lead to high-quality test cases that are well-structured and can effectively target specific areas of the code. In contrast, mutational fuzzing takes existing inputs and makes random mutations, which is less manual but may not always yield high-quality inputs.
   - **Verdict:** **Correct.**

2. **Blackbox fuzzers can make use of initial seeds.**
   - **Analysis:** Blackbox fuzzing refers to testing a program without any knowledge of its internal workings, focusing only on the input/output behavior. Despite this lack of insight, blackbox fuzzers can indeed utilize initial seeds, which are often provided to kickstart the fuzzing process. These seeds can help guide the fuzzer towards more meaningful test cases, even in a blackbox context.
   - **Verdict:** **Correct.**

3. **Greybox fuzzing is always the better alternative to blackbox fuzzing.**
   - **Analysis:** Greybox fuzzing combines elements of both blackbox and whitebox fuzzing, utilizing some knowledge of the internal structure (such as code coverage) while still operating without complete source code access. However, declaring it "always" better may be an overgeneralization. The effectiveness of greybox versus blackbox fuzzing can depend on the specific context, such as the complexity of the target software and the nature of the vulnerabilities being tested for. Thus, while greybox often provides advantages, it is not universally superior.
   - **Verdict:** **Incorrect.**

4. **Greybox fuzzing keeps track of concrete program paths to abstract behavior.**
   - **Analysis:** This statement is somewhat misleading. Greybox fuzzing typically involves tracking code coverage and identifying which paths are executed during testing to guide further input generation. However, it does not necessarily abstract behavior in the sense of simplifying or generalizing it; rather, it focuses on optimizing input generation based on concrete execution paths. The wording suggests a level of abstraction that is not a core feature of greybox fuzzing.
   - **Verdict:** **Incorrect.**

### Conclusion

Based on the analysis above, the following conclusions can be drawn:
- **Correct Statements:** 1 and 2.
- **Incorrect Statements:** 3 and 4.

Therefore, the final answer is that the correct statements about fuzzing are **1 and 2**.


Question:
Consider a linear regression model on a dataset which we split into a training set and a test set. After training, our model gives a mean-squared error of 0.1 on the training set and a mean-squared error of 5.3 on the test set. Recall that the mean-squared error (MSE) is given by:
	                $$MSE_{	extbf{w}}(	extbf{y}, 	extbf{X}) = rac{1}{2N} \sum_{n=1}^N (y_n - 	extbf{x}_n^	op	extbf{w})^2$$
	                Which of the following statements is 	extbf{correct} ?

**A. Retraining the model with feature augmentation (e.g. adding polynomial features) will increase the training MSE.**
This statement is incorrect because feature augmentation typically allows the model to fit the training data better by capturing more complex patterns, which often leads to a decrease in training MSE, not an increase.

**B. Using cross-validation can help decrease the training MSE of this very model.**
This statement is misleading; cross-validation is primarily used to assess model performance and prevent overfitting, rather than directly decrease training MSE. While it can help identify a better model or hyperparameters, the training MSE itself is dependent on the chosen model and its complexity, not the process of cross-validation.

**C. Retraining while discarding some training samples will likely reduce the gap between the train MSE and the test MSE.**
This statement is generally incorrect as discarding training samples may lead to a less representative model. Although it could reduce overfitting in very specific scenarios, it often results in a model that is less capable of generalizing, which can increase the test MSE rather than reduce the gap.

**D. Ridge regression can help reduce the gap between the training MSE and the test MSE.**
This statement is correct because ridge regression introduces a penalty on the size of the coefficients, which helps prevent overfitting. By regularizing the model, it typically leads to a more generalized model that performs better on unseen data, thereby reducing the gap between the training and test MSEs.