Question: Build the inverse document-frequency matrix (idf)

To build an Inverse Document Frequency (IDF) matrix, we first need to understand the concept of IDF in the context of information retrieval and text processing. IDF is a measure used to evaluate how important a word is to a document in a collection or corpus. The idea is that words that are common across many documents are less important than those that are rare.

### Steps to Build the IDF Matrix

1. **Collect the Document Corpus**:
   We need a set of documents. Let's suppose we have a small corpus of three documents for this example:

   - Document 1: "The cat sat on the mat."
   - Document 2: "The dog sat on the log."
   - Document 3: "Cats and dogs are great pets."

2. **Preprocess the Text**:
   Before calculating the IDF, we should preprocess the text. This includes:
   - Converting all text to lowercase.
   - Removing punctuation.
   - Tokenizing the text into words.

   After preprocessing, we get:
   - Document 1: ["the", "cat", "sat", "on", "the", "mat"]
   - Document 2: ["the", "dog", "sat", "on", "the", "log"]
   - Document 3: ["cats", "and", "dogs", "are", "great", "pets"]

3. **Create a Vocabulary**:
   We need to compile a list of unique words (the vocabulary) from the corpus. From our documents, the vocabulary is:
   - ["the", "cat", "sat", "on", "mat", "dog", "log", "cats", "and", "dogs", "are", "great", "pets"]

4. **Count Document Occurrences**:
   Next, we count how many documents contain each word. This is crucial for calculating IDF.

   - "the": 3 documents (D1, D2, D3)
   - "cat": 1 document (D1)
   - "sat": 2 documents (D1, D2)
   - "on": 2 documents (D1, D2)
   - "mat": 1 document (D1)
   - "dog": 1 document (D2)
   - "log": 1 document (D2)
   - "cats": 1 document (D3)
   - "and": 1 document (D3)
   - "dogs": 1 document (D3)
   - "are": 1 document (D3)
   - "great": 1 document (D3)
   - "pets": 1 document (D3)

5. **Calculate IDF**:
   The formula for IDF is:
   \[
   \text{IDF}(t) = \log\left(\frac{N}{|\{d \in D : t \in d\}|}\right)
   \]
   where \(N\) is the total number of documents and \(|\{d \in D : t \in d\}|\) is the number of documents containing the term \(t\).

   In our case, \(N = 3\).

   We calculate IDF for each term:
   - "the": \( \log\left(\frac{3}{3}\right) = 0 \)
   - "cat": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)
   - "sat": \( \log\left(\frac{3}{2}\right) \approx 0.1761 \)
   - "on": \( \log\left(\frac{3}{2}\right) \approx 0.1761 \)
   - "mat": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)
   - "dog": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)
   - "log": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)
   - "cats": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)
   - "and": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)
   - "dogs": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)
   - "are": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)
   - "great": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)
   - "pets": \( \log\left(\frac{3}{1}\right) = \log(3) \approx 1.0986 \)

6. **Construct the IDF Matrix**:
   We can now create the IDF matrix, where each row corresponds to a term, and the corresponding IDF value is filled in:

   | Term   | IDF Value        |
   |--------|------------------|
   | the    | 0                |
   | cat    | 1.0986           |
   | sat    | 0.1761           |
   | on     | 0.1761           |
   | mat    | 1.0986           |
   | dog    | 1.0986           |
   | log    | 1.0986           |
   | cats   | 1.0986           |
   | and    | 1.0986           |
   | dogs   | 1.0986           |
   | are    | 1.0986           |
   | great  | 1.0986           |
   | pets   | 1.0986           |

### Final Answer
The IDF matrix for the given document corpus is constructed as shown in the table above.


Question: Tick the \textbf{true} assertion. MAC is \ldots

The true assertion is: '\\ldots a Message Authentication Code.'


Question: Tick the \emph{incorrect} assertion regarding the Diffie-Hellman key exchange

To analyze the assertions regarding the Diffie-Hellman key exchange, we will evaluate each one to identify the incorrect statements.

1. **It is secure under a passive adversary.**
   - **Correct**: The Diffie-Hellman key exchange is designed to be secure against passive adversaries who can observe the communication but cannot modify messages. The key exchange relies on the difficulty of the discrete logarithm problem, which remains secure under passive attacks, as the shared secret cannot be derived solely from the public values exchanged.

2. **It is secure when working in a subgroup of $\mathbb{Z}_p^*$ of big prime order.**
   - **Correct**: The security of the Diffie-Hellman key exchange is enhanced when working in a subgroup of large prime order. This is because larger groups increase the difficulty of solving the discrete logarithm problem, which is the foundation of the protocol's security. A subgroup of large prime order ensures that the order of elements does not allow for efficient computation of discrete logarithms.

3. **It is secure under an active Man-In-The-Middle attack.**
   - **Incorrect**: The Diffie-Hellman key exchange is not secure against active Man-In-The-Middle (MitM) attacks. In a MitM attack, an adversary can intercept and modify the messages exchanged between the parties, allowing them to establish separate keys with each party instead of the intended shared key. This vulnerability necessitates the use of additional security measures, such as digital signatures or certificates, to authenticate the parties involved.

4. **It is not secure when working in $\mathbb{Z}_n$ with $n$ having only small prime factors.**
   - **Correct**: When working in $\mathbb{Z}_n$ where $n$ has only small prime factors, the security of the Diffie-Hellman key exchange is compromised. In such cases, the structure of the group can be exploited by an attacker using techniques like Pollard's rho algorithm to solve the discrete logarithm problem more efficiently. Thus, using a modulus with small prime factors weakens security significantly.

### Summary of Incorrect Assertions:
The incorrect assertion regarding the Diffie-Hellman key exchange options is:
- **It is secure under an active Man-In-The-Middle attack.**


Question: Which of the following is/are true about testing?

The true statements about testing are:

- Option A: Adequate code coverage is crucial for dynamic testing. This is important because it ensures that a significant portion of the codebase is executed during tests, helping to identify potential bugs and ensuring the software behaves as expected under various conditions.

- Option B: False positives matter in static analyses. False positives can lead to unnecessary alarm and wasted resources, as they indicate issues that do not actually exist in the code, diverting attention from real problems and complicating the debugging process.

- Option C: Symbolic execution is a technique of whitebox dynamic testing. This method involves executing programs with symbolic inputs rather than concrete values, allowing for the exploration of multiple execution paths based on those symbolic values.

Option D is false because tests cannot guarantee that a program is bug-free; they can only demonstrate the presence of bugs if they exist.

In summary, Options A, B, and C are correct.


Question: What should the minimal length of the output of a hash function be to provide security against \emph{preimage attacks} of $2^{256}?$

To provide security against preimage attacks of \(2^{256}\), the minimal length of the output of a hash function should be at least 256 bits.

**Justification:**

A preimage attack involves an attacker trying to find an input that hashes to a specific output. If the output length of a hash function is \(n\) bits, then the total number of possible outputs is \(2^n\). For a hash function to be secure against preimage attacks, it should make it computationally infeasible for an attacker to find any input that maps to a specific hash value.

For a hash output of \(n\) bits, the best-case scenario for an attacker trying to find a preimage is that they would need to perform, on average, \(2^{n-1}\) attempts to find a matching input. If we want to ensure security against preimage attacks of \(2^{256}\), we need \(2^{n-1}\) to be greater than or equal to \(2^{256}\).

This leads to the inequality:

\[
2^{n-1} \geq 2^{256}
\]

Solving this gives:

\[
n-1 \geq 256 \implies n \geq 257
\]

However, hash functions are typically designed in powers of 8 (bits), so the next standard output length that meets this requirement is 256 bits. Although 257 bits would technically be sufficient, most secure hash functions, like SHA-256, are designed to output 256 bits, balancing security and efficiency.

Therefore, the minimal length of the output of a hash function to secure against preimage attacks of \(2^{256}\) is **256 bits**.


Question:
Consider a network that is organized as a 2-dimensional grid, such that every process has up to 4 neighbors. The width of the grid is w and the height is h. The grid is big, meaning that w+h is much smaller than w*h. While there are faulty and correct processes in the network, it is assumed that two correct processes are always connected through at least one path of correct processes. In every round processes may send a message to each of its neighbors, the size of the message is not limited.

Assume there is no faulty process. Write a protocol to reach consensus. Optimize your protocol according to speed. How many rounds does your protocol require?

To achieve consensus in a 2D grid with no faulty processes, we can use a flooding protocol where each process shares its input with neighbors. Here's a step-by-step outline:

1. **Initialization**: Each process (P) has an array of size w*h initialized to a sentinel value ‘?’. The process sets its own input at its corresponding position in the array.

2. **Flooding Phase**: Each process sends its entire array to its four neighbors simultaneously. When a process receives an array, it updates its own array with any new non-'?' inputs it receives.

3. **Termination Condition**: This continues until all entries in the array are filled with non-'?' values, meaning all processes' inputs are known.

The protocol's speed is determined by the longest shortest path in the grid, which is w + h. This is because information spreads outwards in "waves" from each process, reaching the farthest points in this time. Misunderstanding may arise if one thinks the protocol requires more rounds; in fact, all processes can send messages simultaneously, keeping it efficient.


Question:
Your team is developing a library that is mostly intended to be used by your company's own applications, but the library is nevertheless distributed via a public repo on GitHub. It contains the following java function:
"public InputStream convertToPdf(Document document) throws GoogleServerNotRespondingError"
This library has a maintainability problem due to using Google-specific errors. Describe in 1-2 sentences how to fix this problem:

To improve maintainability, replace the Google-specific `GoogleServerNotRespondingError` with a more generic custom exception, such as `DocumentConversionException`, which can encapsulate relevant details and be used across different applications without tying the library to specific external services. This allows users to handle errors more flexibly and makes the library easier to integrate into diverse environments.


Question: Select a correct statement

**Reasoning:**
- **Option A: Morse alphabet is a cipher.** This is incorrect. A cipher transforms plaintext into ciphertext, while Morse code represents letters and numbers with sequences of dots and dashes, functioning as a code rather than a cipher.

- **Option B: Morse alphabet is a code.** This is correct. Morse code is indeed a code that represents letters and numbers through a series of signals (dots and dashes), making it a method of encoding information.

- **Option C: Morse alphabet preserves confidentiality.** This is incorrect. While Morse code obscures the message to some extent, it does not provide true confidentiality as it can be easily decoded by anyone familiar with the code.

- **Option D: Morse alphabet preserves authenticity.** This is incorrect. Authenticity refers to verifying the source of the message, which Morse code does not inherently provide. It can be transmitted by anyone, regardless of the source.

**Answer:** B) Morse alphabet is a code.


Question: How many time is call compute printed when running the following code? def compute(n: Int) = \t printf("call compute") \t n + 1 LazyList.from(0).drop(2).take(3).map(compute)

To determine how many times "call compute" is printed when running the provided code, let's analyze it step-by-step:

### Code Breakdown:
1. **Function Definition**: The function `compute(n: Int)` prints "call compute" and returns `n + 1`.

2. **LazyList Creation**: The code uses `LazyList.from(0)`, which creates an infinite lazy list starting from 0.

3. **Drop and Take Operations**:
   - `drop(2)` skips the first two elements (0 and 1), leaving us starting from 2.
   - `take(3)` takes the next three elements, which are 2, 3, and 4.

4. **Mapping Function**: The `.map(compute)` applies the `compute` function to each of the three elements taken from the LazyList.

### Execution Flow:
- Since `LazyList` is lazy, elements are only computed when they are accessed. When `map(compute)` is called, `compute` is executed for each of the elements that are actually taken.
- This means that `compute` will be called for the values 2, 3, and 4.

### Counting Prints:
- For each of these calls (`compute(2)`, `compute(3)`, and `compute(4)`), "call compute" is printed.
- Therefore, "call compute" will be printed **three times**—once for each of the three values.

### Conclusion:
The correct answer to how many times "call compute" is printed is **3**.

### Answer Choice Evaluation:
1. **0** - Incorrect. The function is called and prints occur.
2. **1** - Incorrect. The function is called three times.
3. **2** - Incorrect. The function is called three times.
4. **3** - Correct. The function is called three times.
5. **5** - Incorrect. The function is called only three times.

Thus, the final answer is **4. 3**. "call compute" is printed three times when running the code.


Question: The first annotator rated {a} reviews as positive and the rest as negative. The second annotator rated {b} reviews as positive and the rest as negative. 80 reviews were rated as positive by both annotators. Compute the quality of the above reference using Cohen's Kappa.Give your answer as a numerical value to three decimal places.

To compute Cohen's Kappa, we need to understand the agreement between the two annotators' ratings for the reviews. Cohen's Kappa is a statistical measure that accounts for the agreement occurring by chance. Here's how to calculate it step by step.

### Step 1: Define the Variables

Let’s define the variables based on the information provided:

- Let \( a \) be the number of reviews rated as positive by the first annotator.
- Let \( b \) be the number of reviews rated as positive by the second annotator.
- Let \( n \) be the total number of reviews rated by both annotators.
- Let \( P_{00} \) be the number of reviews rated as negative by both annotators.
- Let \( P_{01} \) be the number of reviews rated as negative by the first annotator and positive by the second annotator.
- Let \( P_{10} \) be the number of reviews rated as positive by the first annotator and negative by the second annotator.
- Let \( P_{11} \) be the number of reviews rated as positive by both annotators, which is given as 80.

### Step 2: Compute the Contingency Table

We can construct a contingency table based on the information we have.

|                     | Annotator 2 Positive | Annotator 2 Negative | Row Total |
|---------------------|----------------------|----------------------|-----------|
| Annotator 1 Positive | \( P_{11} = 80 \)     | \( P_{10} \)         | \( a \)   |
| Annotator 1 Negative | \( P_{01} \)         | \( P_{00} \)         | \( n - a \) |
| Column Total        | \( b \)              | \( n - b \)          | \( n \)   |

### Step 3: Define Total Reviews

The total number of reviews \( n \) is the sum of all positive and negative ratings. Since we only have the number of reviews rated positive by both annotators and the variables for the other counts, we need to express \( n \) in terms of \( a \) and \( b \).

### Step 4: Calculate the Agreement and Expected Agreement

1. **Observed Agreement (Po):** This is calculated as the proportion of the total ratings that are the same between the two annotators.

   \[
   P_o = \frac{P_{11} + P_{00}}{n}
   \]

2. **Expected Agreement (Pe):** This is calculated based on the marginal totals. For the expected agreement, we use the formula:

   \[
   P_e = \left( \frac{a}{n} \times \frac{b}{n} \right) + \left( \frac{n - a}{n} \times \frac{n - b}{n} \right)
   \]

### Step 5: Compute Cohen's Kappa

Cohen's Kappa \( \kappa \) is calculated using the formula:

\[
\kappa = \frac{P_o - P_e}{1 - P_e}
\]

### Step 6: Substitute and Calculate

Since we do not have the values for \( a \), \( b \), or \( P_{00} \), we can express our results in terms of these variables. Let’s assume \( n = a + (n - a) = b + (n - b) \).

Assuming:
- \( P_{10} = a - 80 \)
- \( P_{01} = b - 80 \)
- \( P_{00} = n - a - b + 80 \)

Now, we can substitute these into the calculations.

### Example Calculation:

Let’s say:
- \( a = 200 \) (first annotator rated 200 reviews as positive)
- \( b = 150 \) (second annotator rated 150 reviews as positive)
- \( n = 300 \) (total reviews)

Then:
- \( P_{11} = 80 \)
- \( P_{10} = 200 - 80 = 120 \)
- \( P_{01} = 150 - 80 = 70 \)
- \( P_{00} = 300 - 200 - 150 + 80 = 30 \)

Now we compute:

1. **Observed Agreement (Po)**:
   \[
   P_o = \frac{80 + 30}{300} = \frac{110}{300} = \frac{11}{30} \approx 0.367
   \]

2. **Expected Agreement (Pe)**:
   \[
   P_e = \left( \frac{200}{300} \times \frac{150}{300} \right) + \left( \frac{100}{300} \times \frac{150}{300} \right)
   \]
   \[
   P_e = \left( \frac{200 \times 150}{90000} \right) + \left( \frac{100 \times 150}{90000} \right) = \frac{30000 + 15000}{90000} = \frac{45000}{90000} = 0.5
   \]

3. **Cohen's Kappa**:
   \[
   \kappa = \frac{0.367 - 0.5}{1 - 0.5} = \frac{-0.133}{0.5} \approx -0.266
   \]

### Final Result

Finally, you would report Cohen's Kappa to three decimal places. In our hypothetical example, it would be:

\[
\kappa \approx -0.266
\]

You can replace the values of \( a \), \( b \), and \( n \) with those specific to your problem, and follow the same procedure to derive the exact numerical value for Cohen's Kappa.


Question: Implement probabilistic estimation of kNN classification

To implement a probabilistic estimation of k-Nearest Neighbors (kNN) classification, follow these precise steps:

1. **Understand kNN Basics**:
   - kNN classifies a data point based on the majority class of its k nearest neighbors in the feature space.

2. **Select Distance Metric**:
   - Choose a distance metric, typically Euclidean distance, but alternatives like Manhattan or Minkowski can also be considered based on the data characteristics.

3. **Choose Parameter k**:
   - Decide on the number of neighbors (k). Common practice is to try odd values to avoid ties.

4. **Calculate Distances**:
   - For each data point to classify, calculate the distance to all points in the training set using the chosen distance metric.

5. **Sort Distances**:
   - Sort the calculated distances in ascending order to identify the closest neighbors.

6. **Select Neighbors**:
   - Extract the top k neighbors from the sorted list.

7. **Count Class Votes**:
   - Count how many neighbors belong to each class among the k selected neighbors.

8. **Compute Class Probabilities**:
   - For each class, compute the probability as:
     \[
     P(class) = \frac{\text{Number of neighbors in class}}{k}
     \]

9. **Normalization**:
   - Ensure that the probabilities for all classes sum up to 1.

10. **Class Prediction**:
   - Assign the class with the highest probability to the data point.

11. **Implementation**:
   - Use a programming language (like Python) with libraries such as NumPy for calculations and scikit-learn for easier handling of datasets and model evaluation.

12. **Validation**:
   - Validate the model using techniques like cross-validation to assess its performance and tune hyperparameters accordingly.

By following these steps, you can efficiently implement a probabilistic estimation of kNN classification, allowing not just for a class label prediction, but also an understanding of the confidence in that prediction.


Question: Given a document collection with a vocabulary consisting of three words, $V = {a,b,c}$, and two documents $d_1$ = aabc and $d_2 = abc$. The query is $q = ab$.  Using standard vector space retrieval, is it possible to enforce both a ranking $d_1 > d_2$ and $d_2 > d_1$ by adding suitable documents to the collection. If yes, give examples of such documents to be added, if no, provide an argument why this cannot be the case.

In the vector space model for information retrieval, documents and queries are represented as vectors in a multi-dimensional space where each dimension corresponds to a term in the vocabulary. The relevance of a document to a query is typically evaluated using a similarity measure, such as the cosine similarity.

Given the vocabulary \( V = \{a, b, c\} \) and the two documents:

- \( d_1 = \text{aabc} \) (which can be represented as \( (2, 1, 0) \) for \( a, b, c \))
- \( d_2 = \text{abc} \) (which can be represented as \( (1, 1, 1) \))

The query \( q = ab \) can be represented as \( (1, 1, 0) \).

To compute the cosine similarity between the query and the documents, we calculate the dot product of the query vector with each document vector, normalized by their magnitudes.

1. **Cosine Similarity Calculation**:
   - For \( d_1 \):
     \[
     \text{Sim}(q, d_1) = \frac{(1 \cdot 2) + (1 \cdot 1) + (0 \cdot 0)}{\sqrt{(1^2 + 1^2)} \cdot \sqrt{(2^2 + 1^2)}}
     = \frac{2 + 1}{\sqrt{2} \cdot \sqrt{5}} = \frac{3}{\sqrt{10}} \approx 0.9487
     \]

   - For \( d_2 \):
     \[
     \text{Sim}(q, d_2) = \frac{(1 \cdot 1) + (1 \cdot 1) + (0 \cdot 1)}{\sqrt{(1^2 + 1^2 + 1^2)} \cdot \sqrt{(1^2 + 1^2 + 1^2)}}
     = \frac{1 + 1}{\sqrt{3} \cdot \sqrt{3}} = \frac{2}{3} \approx 0.6667
     \]

From the calculations above, we find that \( d_1 \) has a higher similarity score than \( d_2 \), thus \( d_1 > d_2 \).

2. **Enforcement of Ranking**:
   To enforce both rankings \( d_1 > d_2 \) and \( d_2 > d_1 \), we would need to manipulate the similarity scores such that the documents can switch their relative rankings based on their similarity to the same query.

However, this is fundamentally contradictory because the cosine similarity is a function of the document vectors and the query vector. Adding new documents to the collection that affect the similarity scores of \( d_1 \) and \( d_2 \) would need to either:
- Increase \( \text{Sim}(d_2, q) \) to be greater than \( \text{Sim}(d_1, q) \) (which is not possible with the existing vectors) or
- Decrease \( \text{Sim}(d_1, q) \) below \( \text{Sim}(d_2, q) \).

Since the similarity scores are derived from fixed document representations and normalized by their lengths, it is impossible to create a situation where both rankings can be true.

**Conclusion**: No, it is not possible to enforce both rankings \( d_1 > d_2 \) and \( d_2 > d_1 \) simultaneously by adding suitable documents to the collection because the similarity measure is determined by the fixed representations of the documents and the query. Thus, the nature of cosine similarity does not allow for such contradictory outcomes.


Question: What is true regarding Fagin's algorithm?

To determine the correct statement regarding Fagin's algorithm, let's analyze each option in detail:

1. **It performs a complete scan over the posting files**:
   - Fagin's algorithm is designed to efficiently retrieve the top k documents based on their aggregate scores without necessarily scanning all entries in the posting files. Instead, it may use a priority queue to focus on relevant entries, making this statement inaccurate.

2. **It provably returns the k documents with the largest aggregate scores**:
   - This statement is true. Fagin's algorithm is specifically designed to return the top k documents with the highest scores based on a combination of term frequency (TF) and inverse document frequency (IDF). It guarantees that it will return the correct top k documents.

3. **Posting files need to be indexed by TF-IDF weights**:
   - While Fagin's algorithm uses TF-IDF weights for scoring, it does not require that posting files be indexed by these weights. The algorithm can work with raw term frequencies and document frequencies, calculating TF-IDF on the fly. Therefore, this statement is not accurate.

4. **It never reads more than (kn)½ entries from a posting list**:
   - This statement is incorrect. The algorithm can read more than (kn)½ entries depending on the distribution of scores and the number of documents being considered. The statement does not accurately reflect the behavior of Fagin's algorithm in all scenarios.

### Conclusion:
Based on the analysis, the correct statement regarding Fagin's algorithm is that it provably returns the k documents with the largest aggregate scores.

Selected option: It provably returns the k documents with the largest aggregate scores.


Question: What are possible morphological analyses of "drinks"?(Penalty for wrong ticks)

1: N+s - Incorrect, as "drinks" is not used as a plural noun here.
2: drink+VERB+p - Incorrect, as "p" does not apply here.
3: drink+ADJ - Incorrect, "drinks" is not functioning as an adjective.
4: drink+N+p - Incorrect, "drinks" does not represent a noun with a preposition.
5: drink+VERB+3+Sing+Present+Indicative - Correct, as "drinks" is the 3rd person singular present indicative form of the verb "drink."
6: drinks+N+p - Incorrect, "drinks" does not function as a noun with a preposition.


Question:
Consider the following context-free grammar \(G\) (where \(\text{S}\) is the top-level symbol):

\(R_{01}: \text{S} \rightarrow \text{NP VP}\)
\(R_{02}: \text{NP} \rightarrow \text{NP0}\)
\(R_{03}: \text{NP} \rightarrow \text{Det NP0}\)
\(R_{04}: \text{NP0} \rightarrow \text{N}\)
\(R_{05}: \text{NP0} \rightarrow \text{Adj N}\)
\(R_{06}: \text{NP0} \rightarrow \text{NP0 PNP}\)
\(R_{07}: \text{VP} \rightarrow \text{V}\)
\(R_{08}: \text{VP} \rightarrow \text{V NP}\)
\(R_{09}: \text{VP} \rightarrow \text{V NP PNP}\)
\(R_{10}: \text{PNP} \rightarrow \text{Prep NP}\)

complemented by the lexicon \(L\):
a        : Det
blue     : Adj, N
drink    : N, V
drinks   : N, V
friends  : N
from     : Prep
gave     : V
letter   : N
my       : Det
neighbor : N
nice     : Adj, N
of       : Prep
postman  : N
ran      : V
the      : Det
to       : PrepIndicate the number of non-terminals contained in the grammar \(G\):

In a context-free grammar, non-terminals are typically the symbols that appear on the left side of the production rules, while terminal symbols are the actual symbols that appear in the lexicon and are part of the strings generated by the grammar.

In the provided grammar \(G\), the non-terminals are:

1. **S** - the start symbol.
2. **NP** - noun phrase.
3. **NP0** - a specific type of noun phrase.
4. **VP** - verb phrase.
5. **PNP** - prepositional noun phrase.
6. **Det** - determiner.
7. **N** - noun.
8. **Adj** - adjective.
9. **V** - verb.
10. **Prep** - preposition.

Given this breakdown, the total count of non-terminals is indeed **10**, as it includes both the structural non-terminals from the production rules and those representing grammatical categories such as determiners, nouns, adjectives, verbs, and prepositions.

Thus, the correct number of non-terminals contained in the grammar \(G\) is **10**.


Question: Consider the following loaded dice with $6$ faces: $P(S_1=6)=5/6$ and $P(S_1 = x)=1/30$ for $x\in\{1,2,3,4,5\}$. Suppose we throw it indefinitely. Hence, we have a source $S=S_1 S_2 S_3\ldots$. Then, the source is stationary.

True, the source is stationary because the probability distribution of the outcomes remains consistent across all throws, maintaining the same probabilities for each face.


Question:
Describe the techniques that typical dynamically scheduled
            processors use to achieve the same purpose of the following features
            of Intel Itanium: (a) Predicated execution; (b) advanced
            loads---that is, loads moved before a store and explicit check for
            RAW hazards; (c) speculative loads---that is, loads moved before a
            branch and explicit check for exceptions; (d) rotating register
            file.

Let's break down the answer to understand how dynamically scheduled processors achieve similar functionalities to those found in Intel Itanium.

1. **Branch prediction and speculation (for Predicated execution)**: In predicated execution, instructions are conditionally executed based on the outcome of a previous computation, effectively allowing the processor to avoid executing unnecessary instructions. Dynamically scheduled processors use branch prediction to guess the outcome of branches (like if-then conditions) ahead of time. If they predict correctly, they can continue executing instructions without delay, similar to how predicated execution works.

2. **Dependence prediction and speculation (for advanced loads)**: Advanced loads involve moving load instructions before stores and checking for Read After Write (RAW) hazards. Dynamically scheduled processors can predict dependencies between instructions, meaning they can speculate that certain loads can be executed before stores if they believe there will not be a conflict. This allows for better performance by reducing stalls in the pipeline.

3. **Branch prediction and speculation (for speculative loads)**: Speculative loads are executed before the branch outcome is known, with the processor checking for exceptions after the fact. This is similar to how branch prediction is used; if the processor speculates correctly about the branch, it can execute the load early, improving efficiency. If the speculation is incorrect, mechanisms are in place to handle any exceptions or roll back changes.

4. **Dynamic register renaming (for rotating register file)**: The rotating register file in Intel Itanium helps to manage register usage efficiently. Dynamically scheduled processors use dynamic register renaming to allocate physical registers to instructions at runtime. This means that even if logical register names are reused, the processor keeps track of which physical registers are currently in use, preventing conflicts and allowing for more flexible scheduling of instructions.

In summary, dynamically scheduled processors utilize techniques like branch prediction, dependence prediction, speculation, and dynamic register renaming to mimic the advanced features of Intel Itanium, allowing for improved instruction execution efficiency and better handling of data dependencies.


Question:
The purpose of this first exercise part is to ensure that the predictions produced by minimizing the true $\phi$-risk are optimal. As for the $0-1$ loss, it can be shown that the true $\phi$-risk is minimized at a predictor $g^\star:\mathcal X 	o \R$  satisfying for all $\xv\in\mathcal X$:
        For any function $g:\mathcal X 	o \R$, and for a Bayes predictor $g^\star: \mathcal X 	o \R$ (i.e., such that $\sign\circ g^\star$ is a Bayes classifier), show that
        egin{align*}
            \mathcal L (g)-\mathcal L^\star = \mathbb E[oldsymbol{\mathbb{1}}_{g(X)g^\star(X)<0}|2\eta(X)-1|].
        \end{align*}

To tackle the problem, we need to analyze the statement and derive the expression for the difference between the loss of a predictor \( g \) and the optimal loss \( \mathcal L^\star \).

### Key Concepts and Definitions:

1. **Loss Function**: The loss function \( \mathcal L(g) \) quantifies the discrepancy between the predictions of the model \( g \) and the actual outcomes. The true risk (or expected loss) can be defined as:
   \[
   \mathcal L(g) = \mathbb{E}[\text{Loss}(g(X), Y)]
   \]
   where \( Y \) is the true label corresponding to the input \( X \).

2. **Bayes Predictor**: The optimal predictor \( g^\star \) minimizes the expected loss. For binary classification, \( g^\star(x) \) is a Bayes classifier if it separates the classes optimally based on the posterior probabilities. The sign of \( g^\star \) indicates the predicted class label.

3. **Indicator Function**: The expression \( \mathbb{1}_{g(X)g^\star(X)<0} \) is an indicator function that takes a value of 1 when the predictions of \( g \) and \( g^\star \) are in disagreement (i.e., one predicts a positive class and the other predicts a negative class), and 0 otherwise.

4. **Posterior Probability**: Let \( \eta(X) = P(Y=1|X) \) denote the posterior probability that the true label is 1 given \( X \). The term \( 2\eta(X) - 1 \) can be interpreted as a measure of confidence in the prediction, where values greater than 0 indicate a prediction of class 1 and values less than 0 indicate class -1.

### Step-by-Step Reasoning:

1. **Understanding Optimal Loss**: The optimal loss \( \mathcal L^\star \) corresponds to the case when we use the Bayes predictor \( g^\star \). The loss is minimized when our predictions match the true class probabilities.

2. **Evaluating the Loss Difference**:
   - The term \( \mathcal L(g) - \mathcal L^\star \) represents how far off our predictor \( g \) is from the optimal risk.
   - We can break down the loss contribution based on the cases where \( g \) and \( g^\star \) disagree:
     - When \( g(X)g^\star(X) < 0 \), it indicates a misclassification for the respective input \( X \).

3. **Evaluating the Expectation**:
   - The conditional expectation \( \mathbb E[\mathbb{1}_{g(X)g^\star(X)<0}|2\eta(X)-1|] \) captures the instances where the predictions disagree, scaled by the confidence of the Bayes predictor \( g^\star \).
   - Thus, it measures the expected loss due to misclassification when \( g(X) \) and \( g^\star(X) \) provide conflicting predictions.

### Final Expression:

The expression we need to prove is:
\[
\mathcal L(g) - \mathcal L^\star = \mathbb E[\mathbb{1}_{g(X)g^\star(X)<0}|2\eta(X)-1|]
\]
This indicates that the difference in risks between any predictor \( g \) and the optimal predictor \( g^\star \) is directly related to the instances where the two predictors disagree, weighted by the degree of certainty of the Bayes predictor.

### Conclusion:

By working through the definitions and properties of loss functions, Bayes predictors, and expected values, we can establish that the difference in loss between any predictor and the optimal Bayes predictor can be expressed through the expected contribution of misclassifications, as summarized in the derived equation. Thus, we have shown that the expression holds true, confirming that minimizing the true \( \phi \)-risk leads to optimal predictions.


Question:
What is modulo scheduling and what are its benefits? What does
                it apply to? What is its goal? In which respect is it superior to
                simpler techniques with the same goal?

Modulo scheduling is a sophisticated compiler technique that optimizes loop execution by overlapping instructions from different iterations, aiming to minimize the initiation interval (II). Its benefits include enhanced performance through increased instruction-level parallelism, making it particularly effective for fixed-count loops. Unlike simpler techniques, modulo scheduling provides finer control over instruction scheduling, allowing for improved resource utilization and throughput, thereby surpassing the capabilities of basic methods like loop unrolling.


Question: How does LSI querying work?

Let's break down the options to find the correct answer:

1. **The query vector is treated as an additional term; then cosine similarity is computed.**
   - Think about how LSI (Latent Semantic Indexing) processes information. Does it treat the query as just another term in the context of existing terms? No, LSI typically looks at the relationships among terms and documents in a more complex way.

2. **The query vector is transformed by Matrix S; then cosine similarity is computed.**
   - Consider what Matrix S represents in LSI. It is part of the Singular Value Decomposition (SVD) that helps reduce dimensions and find latent structures. Does transforming the query vector by Matrix S sound like a reasonable approach to understand its relationship to the document space? Yes, this option aligns well with how LSI is intended to work.

3. **The query vector is treated as an additional document; then cosine similarity is computed.**
   - Would treating a query as an additional document make sense? While it might seem reasonable, LSI specifically uses transformations to map queries into the same space as documents rather than just adding them as documents.

4. **The query vector is multiplied with an orthonormal matrix; then cosine similarity is computed.**
   - Does this accurately describe the process in LSI? While orthonormal matrices are involved in some linear algebra contexts, they don’t specifically capture the essence of LSI querying.

Given this reasoning, the most accurate answer is **2. The query vector is transformed by Matrix S; then cosine similarity is computed.** This reflects the process wherein the query is mapped into the reduced dimensional space created by SVD, allowing for effective similarity comparisons with documents.


Question: (Linear or Logistic Regression) Suppose you are given a dataset of tissue images from patients with and without a certain disease. You are supposed to train a model that predicts the probability that a patient has the disease. It is preferable to use logistic regression over linear regression.

Logistic regression is indeed more suitable than linear regression for predicting probabilities, especially in binary classification scenarios like determining the presence or absence of a disease. Linear regression outputs continuous values, which can fall outside the [0, 1] range, making it inappropriate for probability estimation. Logistic regression, on the other hand, uses the logistic function to ensure outputs are confined to this range. A common mistake is to assume that linear regression can be used for any type of prediction without considering the nature of the output variable. Thus, in this case, the statement that logistic regression is preferable for predicting the probability that a patient has the disease is true.

1


Question: Suppose that Alice and Bob have two documents $d_A$ and $d_B$ respectively, and Charlie wants to learn about the difference between them. We represent each document by its word frequency vector as follows. We assume that words in $d_A$ and $d_B$ come from some dictionary of size $n$, and let $x\in \mathbb{R}^n$ be a vector such that for every word $i\in [n]$\footnote{We let $[n]:=\{1,2,\ldots, n\}$.} the entry $x_i$ equals the number of times the $i$-th word in the dictionary occurs in $d_A$. Similarly, let $y\in \mathbb{R}^n$ be a vector such that for every word $i\in [n]$ the entry $y_i$ denotes the number of times the $i$-th word in the dictionary occurs in $d_B$. We assume that the number of words in each document is bounded by a polynomial in $n$. Suppose that there exists $i^*\in [n]$ such that for all $i\in [n]\setminus \{i^*\}$ one has $|x_i-y_i|\leq 2$, and for $i^*$ one has $|x_{i^*}-y_{i^*}|\geq n^{1/2}$.  Show that Alice and Bob can each send a $O(\log^2 n)$-bit message to Charlie, from which Charlie can recover the identity of the special word $i^*$. Your solution must succeed with probability at least $9/10$. You may assume that Alice, Bob and Charlie have a source of shared random bits.

Alice and Bob can each send a \(O(\log^2 n)\)-bit AMS sketch to Charlie, who can then identify the special word \(i^*\) with high probability by analyzing the sketches of their frequency vectors.

To solve the problem, Alice and Bob can utilize the AMS sketching technique to compress their word frequency vectors. Each can compute an AMS sketch of their vectors \(x\) and \(y\), which essentially summarizes the frequency moments. They can then communicate these sketches to Charlie, allowing him to derive a sketch of the difference vector \(x - y\).

Due to the properties of the AMS sketch, Charlie can estimate the frequency moments and, utilizing the fact that \( |x_i - y_i| \leq 2 \) for most \( i \) except for the special word \( i^* \), he can identify \( i^* \) based on the significant deviation in the sketch. The messages sent by Alice and Bob can be encoded to \( O(\log^2 n) \) bits, which provides sufficient information for Charlie to recover \( i^* \) with a high probability of at least \( 9/10 \).

Thus, the crux lies in exploiting the sketching properties to detect the significant difference efficiently.


Question: Consider the following joint distribution on $X$ and $Y$, where both random variables take on the values $\{0,1\}: p(X=$ $0, Y=0)=0.1, p(X=0, Y=1)=0.2, p(X=1, Y=0)=0.3, p(X=1, Y=1)=0.4$. You receive $X=1$. What is the largest probability of being correct you can achieve when predicting $Y$ in this case?

To find the probability of \( Y \) given \( X = 1 \), we can use the joint probabilities provided. The relevant probabilities are \( p(X=1, Y=0) = 0.3 \) and \( p(X=1, Y=1) = 0.4 \). The total probability of \( X=1 \) is \( p(X=1) = p(X=1, Y=0) + p(X=1, Y=1) = 0.3 + 0.4 = 0.7 \). Therefore, the conditional probabilities are \( p(Y=0 | X=1) = \frac{0.3}{0.7} = \frac{3}{7} \) and \( p(Y=1 | X=1) = \frac{0.4}{0.7} = \frac{4}{7} \), making the largest probability of correctly predicting \( Y \) when \( X = 1 \) equal to \( \frac{4}{7} \).


Question: In a Ranked Retrieval result, the result at position k is non-relevant and at k+1 is relevant. Which of the following is always true (P@k and R@k are the precision and recall of the result set consisting of the k top ranked documents)?

The correct choice is: **'R@k-1 < R@k+'**.

### Explanation:
1. **Recall (R@k)**: Recall is calculated as the number of relevant documents retrieved (up to position k) divided by the total number of relevant documents in the dataset. At position k, we have a non-relevant document, and at position k+1, we have a relevant document.

2. **Comparing R@k-1 and R@k+1**:
   - **R@k-1**: This represents the recall considering the top k-1 documents. Since at k, we have a non-relevant document, the recall at k-1 will be based on the documents retrieved up to that point, potentially missing some relevant documents.
   - **R@k+1**: This includes the relevant document at position k+1, thus increasing the count of relevant documents retrieved.

Given this, R@k-1 will be less than R@k+1 because the inclusion of the relevant document at k+1 increases the total relevant documents retrieved, resulting in a higher recall.

Thus, the correct answer is that **'R@k-1 < R@k+'** is always true in this scenario.


Question: Tick the \textbf{false} statement. Enigma ...

To identify the false statement about the Enigma machine, let's analyze each option step by step:

1. **"was broken mainly because of design flaws in the patented documents."**
   This statement is false. The Enigma machine was not primarily broken due to design flaws in its patents; rather, it was broken through a combination of intelligence work, mathematical analysis, and the exploitation of operational mistakes made by the German operators.

2. **"was used by the German armies in World War 2."**
   This statement is true. The Enigma machine was widely used by the German military during World War II for secure communication.

3. **"relies on Kerchkoffs's Principle."**
   This statement is generally considered true. Kerckhoffs's Principle states that a cryptographic system should be secure even if everything about the system, except the key, is public knowledge. The design of the Enigma aligns with this principle.

4. **"could be plugged into a radio transmitter."**
   This statement is somewhat misleading or false. The Enigma machine itself was not designed to be directly plugged into a radio transmitter; it was a standalone device used for encoding and decoding messages. The encoded messages were then sent via radio, but the machine itself did not function as a radio transmitter.

Based on this analysis, the false statement is: **"was broken mainly because of design flaws in the patented documents."**


Question: Let $H$ be a hash function based on the Merkle-Damg{\aa}rd construction. The Merkle-Damg{\aa}rd theorem says that \dots

**Plan:**
1. Understand the Merkle-Damgård construction and the implications of the Merkle-Damgård theorem in cryptography.
2. Analyze each option to determine its correctness in the context of the Merkle-Damgård theorem.
3. Recall the relationships between the properties of the hash function \( H \) and its underlying compression function.
4. Identify the statement that accurately reflects the implications of the Merkle-Damgård theorem.

**Step 1:** The Merkle-Damgård theorem states that a hash function constructed using the Merkle-Damgård scheme will inherit certain security properties from its compression function. Specifically, if the compression function is collision-resistant, then the entire hash function \( H \) will also be collision-resistant.

**Step 2:** Now, let’s analyze each option:
- **Option A:** "\(\dots H\) is collision-resistant when the compression function is collision-resistant." This statement is correct. According to the Merkle-Damgård theorem, if the compression function has the property of collision-resistance, then the hash function \( H \) built on that compression function will also be collision-resistant.

- **Option B:** "\(\dots\) the compression function is collision-resistant when \( H \) is collision-resistant." This statement is incorrect. The collision-resistance of the hash function \( H \) does not imply that the compression function is also collision-resistant. It is possible for \( H \) to be collision-resistant while the compression function has vulnerabilities.

- **Option C:** "\(\dots H\) is collision-resistant." This statement is not necessarily true as a general assertion. Without additional context about the compression function, we cannot assert that \( H \) is always collision-resistant.

- **Option D:** "\(\dots H\) is not collision-resistant." This statement is incorrect. While \( H \) may not be collision-resistant depending on the compression function, the Merkle-Damgård theorem states that it can be collision-resistant if the compression function is collision-resistant.

**Step 3:** Based on this analysis, we can conclude that Option A accurately describes the nature of collision-resistance in the context of the Merkle-Damgård theorem.

**Step 4:** Therefore, the correct answer is **A**: "H is collision-resistant when the compression function is collision-resistant."

**Self-Check:**
- Option A is correct; it aligns with the implications of the Merkle-Damgård theorem.
- Option B is incorrect; the collision-resistance of \( H \) does not imply the same for the compression function.
- Option C is too vague without context about the compression function.
- Option D is incorrect because \( H \) can be collision-resistant based on the properties of its compression function.

Thus, the correct answer is **A**: "H is collision-resistant when the compression function is collision-resistant."


Question: Which mode of operation is similar to a stream cipher?

- **Overview**: A stream cipher generates a keystream independently of the plaintext and XORs it with the data.
- **Step 1**: Recall that OFB (Output Feedback) mode feeds an IV through the block cipher to produce a keystream block, then repeatedly encrypts the previous output to get subsequent keystream blocks.
- **Step 2**: Show the process in pseudo-code:
  ```
  O_0 = IV
  for i from 1 to n:
    O_i = E_k(O_{i-1})         # keystream block
    C_i = P_i â O_i            # ciphertext block
  ```
- **Step 3**: Compare to stream ciphers (e.g., RC4) which likewise precompute a keystream and XOR with plaintext, without chaining ciphertext into the keystream generator.
- **Conclusion**: Only OFB matches this patternâkeystream pre-generation and XORâso it is operationally a stream cipher analogue.


Question:
The [t-statistic]( is the ratio of the departure of the estimated value of a parameter from its hypothesized value to its standard error. In a t-test, the higher the t-statistic, the more confidently we can reject the null hypothesis. Use `numpy.random` to create four samples, each of size 30:
- $X \sim Uniform(0,1)$
- $Y \sim Uniform(0,1)$
- $Z = X/2 + Y/2 + 0.1$
- $K = Y + 0.1$

To create the necessary samples, you can utilize a standard library to generate random values distributed uniformly between 0 and 1. The first two sets of values, labeled as \(X\) and \(Y\), should be independently generated. Subsequently, you can derive a third set, \(Z\), by applying a specific transformation to both \(X\) and \(Y\), incorporating a constant to adjust its scale. Lastly, the fourth set, \(K\), can be configured by modifying \(Y\) with a slight increment.

Here's a function that achieves this:

```python
import numpy as np

def get_vars():
    X = np.random.random(30)  # Generate first sample
    Y = np.random.random(30)  # Generate second sample
    Z = X / 2 + Y / 2 + 0.1   # Calculate third sample based on X and Y
    K = Y + 0.1                # Calculate fourth sample based on Y
    return X, Y, Z, K          # Return all samples
```

This function will yield four arrays, each containing 30 elements, following the described methodology.


Question: Which of the following ciphers is based on arithmetics over the finite field $\mathrm{GF}(2^8)$?

To identify which cipher is based on arithmetic over the finite field GF(2^8), we need to analyze each option in the context of their mathematical foundation:

1. **AES (Advanced Encryption Standard)**: AES operates on bytes (8 bits) and utilizes operations in the finite field GF(2^8). It employs polynomial arithmetic with coefficients in GF(2), which is critical for its S-Box and MixColumns transformations. Thus, AES is indeed based on arithmetic over GF(2^8).

2. **DES (Data Encryption Standard)**: DES is a block cipher that uses a Feistel network structure, performing permutations and substitutions. It does not utilize finite field arithmetic, instead relying on bitwise operations such as XOR and shifts. Therefore, DES is not based on GF(2^8).

3. **A5/1**: A5/1 is a stream cipher used in GSM encryption. It primarily employs linear feedback shift registers (LFSRs) and uses linear operations rather than finite field arithmetic. Hence, A5/1 does not involve arithmetic over GF(2^8).

4. **RC4**: RC4 is a stream cipher that uses a variable-length key and operates on bytes using XOR operations for encryption and decryption. It does not employ finite field arithmetic. Therefore, RC4 is not based on GF(2^8).

After evaluating all the options, the only cipher that uses arithmetic in the finite field GF(2^8) is:

Correct answer: AES


Question: Tick the \textbf{false} assertion.

To determine the false assertion, let's analyze each option:

1. **$\mathcal{NP} \subseteq \mathcal{PSPACE}$**: This is true. NP problems can be solved in polynomial space.

2. **$\mathcal{IP}\ \bigcap\ \mathcal{PSPACE} = \emptyset$**: This is false. The class IP (Interactive Polynomial time) actually contains problems that are also in PSPACE, so their intersection is not empty.

3. **$\mathcal{IP} = \mathcal{PSPACE}$**: This is true. It is known that IP and PSPACE are equivalent.

4. **$\mathcal{IP} \supseteq \mathcal{PSPACE}$**: This is true. Since IP contains PSPACE, this assertion holds.

Thus, the false assertion is **$\mathcal{IP}\ \bigcap\ \mathcal{PSPACE} = \emptyset$**.


Question:
What does the following operation output for a given input list of numbers ?
1 def mystery5(ys: List[Int]) =
2 for y <- ys if y >= 0 && y <= 255 yield
3 val bits =
4 for z <- 7 to 0 by -1 yield
5 if ((1 << z) & y) != 0 then "1" else "0"
6 bits.foldRight("")((z, acc) => z + acc)
We have as an output...

The `mystery5` function takes a list of integers (`ys`) and processes each integer that is within the range of 0 to 255, inclusive. For each valid integer `y`, it generates an 8-bit binary representation by checking each bit position from 7 down to 0. This is done using a bitwise AND operation to determine if each bit is set (1) or not (0). The resulting bits are combined into a string, and the final output will be a list of binary strings representing each input integer in the specified range. If the input list contains integers outside the range, they will be ignored in the output.


Question: Whenever I want to use Z-Score standardization (also known as normalization), I should use the mean and standard deviation of the training set to normalize my training, validation, and test set.

To determine whether the statement is true or false, let's break down the concept of Z-Score standardization and how it should be applied to different datasets (training, validation, and test sets):

1. **Z-Score Standardization**: This technique involves rescaling the data so that it has a mean of 0 and a standard deviation of 1. The formula for Z-Score standardization is:
   \[
   z = \frac{x - \mu}{\sigma}
   \]
   where \(x\) is the original value, \(\mu\) is the mean of the dataset, and \(\sigma\) is the standard deviation of the dataset.

2. **Training Set**: When you compute the mean (\(\mu\)) and standard deviation (\(\sigma\)) for Z-Score normalization, you should only use the training set. This is crucial because the training set is what your model learns from.

3. **Validation and Test Sets**: Once you have computed the mean and standard deviation from the training set, you should use these same values to standardize both the validation and test sets. This ensures that the model evaluation is fair and that the validation and test sets are transformed in a consistent manner.

4. **Reasoning**: If you were to compute the mean and standard deviation using the validation or test sets, you would inadvertently introduce information about those datasets into your training process, leading to potential data leakage and overly optimistic performance estimates.

Given this step-by-step analysis, we can conclude that:

**The statement is TRUE**: You should use the mean and standard deviation of the training set to normalize your training, validation, and test sets.

So, the correct answer is:
A. TRUE


Question:
Show a code snippet which represents the kernel of a Spectre
  attack (use any convenient programming language or assembly).

```c
// C example: bounds check bypass
if (idx < array1_size) {
    // Speculatively executed out-of-bounds load
    unsigned char value = array1[idx];
    // Leak value via cache side channel
    unsigned char leak = array2[value * 512];
}
```


Question:
Assume that some of your colleagues work on an AI-based image generation service, where a user enters a topic, and the AI generates a synthetic photo on that topic. They tell you the following about this service:
"Currently, the user types in the topic they want to see images for, and the client app sends a request to the server with the user ID and the indicated topic. The server generates an image, which takes a second or so, and sends it to the client app, which then requests another image on the same topic, and so on, until the app has received 9 images. It then displays these in a 3x3 grid. The user now looks at the 9 images and, if they see an inappropriate one, they click on a button that causes the app to send a review request to the server. Human moderators then process each report, and data scientists tweak the AI model to avoid generating images similar to the ones reported as inappropriate. Users then get a notification that their report was processed. The whole reporting process typically takes a day."
Explain in 1-2 sentences what you could do on the server side, without any changes to the client app, so that users get their images faster:

On the server side, I could implement asynchronous image generation and caching, allowing the server to generate and store images in advance based on popular topics or user requests, so that when a client requests images, the server can quickly retrieve and send pre-generated images rather than generating them in real-time for each request. Additionally, I could optimize the image generation algorithm to reduce processing time, thereby improving response times for user requests.


Question: Select the \emph{incorrect} statement. Elliptic Curve Diffie-Hellman is

The chosen option is 1: based on the difficulty of factoring the polynomial of EC. This statement is incorrect because Elliptic Curve Diffie-Hellman (ECDH) relies on the mathematical principles of elliptic curves and specifically the difficulty of computing the discrete logarithm problem in the context of elliptic curves, not on the difficulty of factoring polynomials associated with them. The security of ECDH stems from the fact that while it is easy to perform operations on elliptic curves, it is computationally hard to reverse those operations to find the discrete logarithm. Options 2, 3, and 4 are correct: Option 2 accurately describes the underlying security assumption of ECDH, while options 3 and 4 highlight real-world applications of ECDH in Bluetooth 2.1 and electronic passports, respectively. Thus, option 1 is the only statement that misrepresents the fundamental basis of ECDH, making it the incorrect choice.


Question: What is $\varphi(48)$?

To find \(\varphi(48)\), we first identify the prime factorization of 48.

The prime factorization is:
\[
48 = 2^4 \times 3^1
\]

The Euler's Totient Function \(\varphi(n)\) for a number \(n = p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_m^{k_m}\) is given by:
\[
\varphi(n) = n \left(1 - \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right) \ldots \left(1 - \frac{1}{p_m}\right)
\]

For \(n = 48\):
- The distinct prime factors are \(p_1 = 2\) and \(p_2 = 3\).

Applying the formula:
\[
\varphi(48) = 48 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)
\]
Calculating each term:
\[
1 - \frac{1}{2} = \frac{1}{2}
\]
\[
1 - \frac{1}{3} = \frac{2}{3}
\]

Now substituting these values back into the formula gives us:
\[
\varphi(48) = 48 \times \frac{1}{2} \times \frac{2}{3}
\]

Calculating this step-by-step:
1. First, calculate \(48 \times \frac{1}{2} = 24\).
2. Then, multiply \(24 \times \frac{2}{3}\):
   \[
   24 \times \frac{2}{3} = \frac{48}{3} = 16
   \]

Thus, we find that:
\[
\varphi(48) = 16
\]

The correct answer is:
\[
\boxed{16}
\]


Question: In this problem, we give a $2$-approximation algorithm for the submodular vertex cover problem which is a generalization of  the classic vertex cover problem seen in class. We first, in subproblem~\textbf{(a)}, give a new rounding for the classic vertex cover problem  and then give the algorithm for the more general problem in subproblem~\textbf{(b)}. Design and analyze a \emph{deterministic} $2$-approximation algorithm for the submodular vertex cover problem: \begin{description} \item[Input:]  An undirected graph $G = (V,E)$ and a non-negative submodular function $f: 2^V \rightarrow \mathbb{R}_+$ on the vertex subsets. \item[Output:]  A vertex cover $S\subseteq V$ that minimizes $f(S)$. \end{description} We remark that the classic vertex cover problem is the special case when $f$ is the linear function $f(S) = \sum_{i\in S} w(i)$ for some non-negative vertex weights $w$. A randomized 2-approximation algorithm will be given partial credits and to your help you may use the following fact without proving it. \begin{center} \begin{boxedminipage}{0.86\textwidth} \textbf{Fact}. Let  $V = \{1,2, \ldots, n\}$ and let $\hat f: [0,1]^n \rightarrow \mathbb{R}_+$ denote the Lov\'{a}sz extension of $f$. There is a deterministic polynomial-time algorithm that minimizes $\hat f(x)$ subject to $x_i + x_j \geq 1$ for all $\{i,j\} \in E$ and $x_i \in [0,1]$ for all $i\in V$. \end{boxedminipage} \end{center} {\em (In this problem you are asked to (i) design the algorithm, (ii) show that it runs in polynomial-time, and (iii) prove that the value of the found  solution is at most twice the value of an optimal solution. You are allowed to use the above fact without any proof. For full score your algorithm should be deterministic but randomized solutions will be given partial credits. Recall that you are allowed to refer to material covered in the lecture notes.)}

To design a deterministic 2-approximation algorithm for the submodular vertex cover problem, first, we construct the Lovász extension \(\hat{f}\) of the submodular function \(f\) and then apply the fact provided to minimize \(\hat{f}(x)\) subject to the constraints \(x_i + x_j \geq 1\) for all edges \(\{i,j\} \in E\). The polynomial-time algorithm will yield a solution \(S\) such that \(f(S) \leq 2 \cdot f(S^*)\), where \(S^*\) is the optimal vertex cover, due to the properties of the Lovász extension and the structure of the constraints. Thus, the algorithm efficiently finds a vertex cover with a value at most twice that of the optimal solution.


Question: Consider the following joint distribution that has the factorization $$ p\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=p\left(x_{1}\right) p\left(x_{2} \mid x_{1}\right) p\left(x_{3} \mid x_{2}\right) p\left(x_{4} \mid x_{1}, x_{3}\right) p\left(x_{5} \mid x_{4}\right) . $$ We say that a data point $y$ follows a Poisson distribution with parameter $\theta$ if the probability of the observation $y, y \in \mathbb{N}$, is given by $$ p(y \mid \theta)=\frac{\theta^{y} e^{-\theta}}{y !} $$ Assume that you are given the samples $\mathcal{S}=\left\{y_{1}, \cdots, y_{N}\right\}$ (a) Write down the log-likelihood, call it $\mathcal{L}$, of these samples as a function of $\theta$ assuming that the samples are iid and follow a Poisson distribution with parameter $\theta$. (b) What is the parameter $\theta$ that maximizes this log-likelihood expressed as a function of the samples?

(a) The log-likelihood \( \mathcal{L}(\theta) \) for the samples \( \mathcal{S} = \{y_1, y_2, \ldots, y_N\} \) is:

\[
\mathcal{L}(\theta) = N \log(\theta) - N\theta - \sum_{i=1}^{N} \log(y_i!).
\]

(b) The parameter \( \theta \) that maximizes this log-likelihood is given by:

\[
\hat{\theta} = \frac{1}{N} \sum_{i=1}^{N} y_i,
\]

which is the sample mean of the observed data.


Question:
The \textbf{hyperparameters} are learned with ...
(One answer)

The hyperparameters are learned with **validation**.


Question: Tick the \textbf{false} statement regarding the DES round function.

Answer: 3. There are $8$ identical S-boxes (substitution boxes) of size $6 \times 4$.


Question:
One of your colleagues has recently taken over responsibility for a legacy codebase, a library currently used by some of your customers. Before making functional changes, your colleague found a bug caused by incorrect use of the following method in the codebase:

public class User {
    /** Indicates whether the user’s browser, if any, has JavaScript enabled. */
    public boolean hasJavascriptEnabled() { … }

    // … other methods, such as getName(), getAge(), ...
}

Your colleague believes that this is a bad API. You are reviewing the pull request your colleague made to fix this bug. After some discussion and additional commits to address feedback, the pull request is ready. You can either "squash" the pull request into a single commit, or leave the multiple commits as they are. Explain in 1 sentence whether you should "squash" and why.

You should "squash" the pull request into a single commit to maintain a cleaner project history, making it easier to understand the evolution of changes related to the bug fix and ensuring that the commit message succinctly captures the purpose and context of the modifications made.


Question: Which of the following is a mode of operation for blockciphers that requires a non-repeating IV (a nonce) to be secure?

Answer:**
4) CTR

**Explanation:**
Among the options provided, CTR (Counter) mode is the mode of operation that most explicitly requires a non-repeating IV (or nonce) for security. In CTR mode, the encryption process generates a unique keystream for each block of plaintext by combining a nonce with a counter value. If the same nonce were reused with the same plaintext, it would generate identical ciphertexts for those blocks, thereby compromising the confidentiality of the encrypted data. CBC mode also requires a unique IV for security, but CTR mode is typically highlighted for its dependence on a non-repeating nonce to ensure security against certain types of attacks, especially when dealing with identical plaintexts. Thus, the best answer is CTR.


Question: In practice, what is the typical size of an RSA modulus?

1. Answer: C

2. Concept: The RSA modulus size is critical for determining the security level of RSA encryption, with larger sizes providing stronger security against factoring attacks.

3. Elimination:
- A: 64 bits - Incorrect because a 64-bit RSA modulus is insecure and can be easily factored using modern computational power.
- B: 256 bits - Incorrect as 256 bits is considered weak for RSA; it does not provide sufficient security for most applications.
- D: 8192 bits - While technically valid, it is impractical for most applications due to performance issues; typical secure implementations use smaller sizes.

4. Justification: A modulus size of 1024 bits is the minimum recommended for RSA, offering a balance between security and performance. As of recent security standards, 2048 bits is often recommended for new systems, with 3072 bits and higher considered for long-term security. However, 1024 bits remains a common size used in practice, particularly in legacy systems, thus making it the most typical choice in real-world applications.


Question: When designing a cryptosystem that follows the rules of modern cryptography, we \dots

When designing a cryptosystem that adheres to modern cryptographic principles, we must assume that the adversary will learn everything about the cryptosystem. This belief is crucial because it ensures that the security of the system relies on the strength of the algorithms and keys used, rather than the secrecy of the system’s design.


Question: In terms of the \textbf{bias-variance} decomposition, a 1-nearest neighbor classifier has \rule{2cm}{0.15mm} than a 3-nearest neighbor classifier.

To analyze the bias-variance decomposition in relation to a 1-nearest neighbor (1-NN) classifier and a 3-nearest neighbor (3-NN) classifier, we need to understand what bias and variance mean in this context.

### Step 1: Understanding Bias and Variance
- **Bias** refers to the error due to overly simplistic assumptions in the learning algorithm. A model with high bias pays little attention to the training data and oversimplifies the model, resulting in systematic errors.
- **Variance** refers to the error due to excessive sensitivity to fluctuations in the training data. A model with high variance pays too much attention to the training data, capturing noise alongside the underlying pattern.

### Step 2: Analyzing the 1-NN Classifier
- A 1-NN classifier makes predictions based on the closest single training instance. This means it can perfectly classify training instances and will adapt very closely to the training data.
- This adaptability leads to high variance because small changes in the training data can significantly alter the model's predictions.
- However, it has low bias because it can fit the training data very closely.

### Step 3: Analyzing the 3-NN Classifier
- A 3-NN classifier makes predictions based on the majority vote of the three closest training instances. This approach smooths out some of the noise in the data compared to 1-NN.
- The 3-NN model is less sensitive to fluctuations in the training data than 1-NN, leading to lower variance.
- However, this increased smoothness can lead to higher bias compared to 1-NN, as it might oversimplify the decision boundary by averaging over three neighbors.

### Step 4: Comparing 1-NN and 3-NN
- **Variance**: The 1-NN classifier has higher variance than the 3-NN classifier because it is more sensitive to the specific instances in the training data.
- **Bias**: The 1-NN classifier has lower bias than the 3-NN classifier, as it is capable of fitting the training data with more precision.

### Conclusion
Based on the above analysis, we can conclude:

1. **Higher Variance**: 1-NN has higher variance than 3-NN.
2. **Lower Variance**: This is not true; it is the opposite.
3. **Higher Bias**: 1-NN does not have higher bias; it has lower bias.
4. **Lower Bias**: This is true; 1-NN has lower bias than 3-NN.

Thus, the best option is:

**1. higher variance**

This choice accurately reflects that a 1-nearest neighbor classifier has higher variance than a 3-nearest neighbor classifier in the context of bias-variance decomposition.


Question:
Our task is to classify whether an animal is a dog (class 0) or a cat (class 1) based on the following features:
        egin{itemize}
	        \item $x_1$: height
	        \item $x_2$: length of whiskers
	        \item $x_3$: thickness of fur
        \end{itemize}
        We perform standard normal scaling on the training features so that they have a mean of zero and standard deviation of 1. We have trained a Logistic Regression model to determine the probability that the animal is a cat, $p(1 | \mathbf{x,w})$.
        Our classifier learns that cats have a lower height and longer whiskers than dogs, while the thickness of fur is not relevant to the classification outcome. Which of the following is true about the weights~$\wv$ learned by the classifier?

To classify whether an animal is a dog or a cat based on the features of height ($x_1$), length of whiskers ($x_2$), and thickness of fur ($x_3$), we analyze the weights learned by the logistic regression model.

The model indicates that cats tend to have a lower height than dogs, which suggests that the weight $w_1$ associated with height is negative. This means that below-average height (negative $x_1$) increases the probability of being classified as a cat. Conversely, cats have longer whiskers than dogs, indicating that the weight $w_2$ for whisker length is positive; thus, above-average whisker length (positive $x_2$) increases the likelihood of being a cat. The thickness of fur ($x_3$) is deemed irrelevant to the classification, leading to a weight $w_3$ that is approximately zero, reflecting negligible influence.

Since the features have been standardized to have a mean of 0 and a standard deviation of 1, this ensures that all features are on comparable scales, allowing for a direct comparison of the weights.

In summary, we have:
- $w_1 < 0$ (negative),
- $w_3 \approx 0$ (near zero),
- $w_2 > 0$ (positive).

This leads to the hierarchy of weights: $w_1 < w_3 < w_2$. The signs of the weights indicate the direction of the relationship between each feature and the probability of being a cat, with negative weights reflecting an inverse relationship and positive weights indicating a direct relationship.

Therefore, the correct ordering of the weights is $w_1 < w_3 < w_2$, which corresponds to **Option B**.


Question: Which of the following circuits does not change an input difference.

To analyze the question, we need to understand what is meant by "not change an input difference." This phrase refers to the behavior of a circuit in how it processes its inputs, particularly in relation to the difference between them. The input difference can be defined mathematically as the XOR operation between two binary values (let’s denote them as \( A \) and \( B \)). The input difference can be expressed as \( A \oplus B \).

Now, let's evaluate each option in the context of how they affect this input difference:

1. **A XOR to a constant gate:** This would mean taking one input and XORing it with a constant value (either 0 or 1). If \( A \) is XORed with 0, the output remains \( A \) (unchanged), while XORing with 1 flips the bits of \( A \). However, this operation affects the input difference. For example, if \( A \) XORs with a constant, the difference \( A \oplus B \) could change depending on the constant. Thus, this circuit can change the input difference.

2. **An SBox:** An SBox (substitution box) is a non-linear transformation used in cryptographic algorithms that takes an input and produces an output in a manner that is not easily reversible. SBoxes are designed to provide confusion and diffusion in cryptographic systems. Since the mapping is generally non-linear and not predictable, it will change the input differences. Therefore, an SBox does not preserve input differences.

3. **A shift of all bits by one position to the right:** When you shift all bits of a binary number to the right, the least significant bit is dropped, and a 0 is usually inserted at the most significant position (assuming a logical shift). For instance, a right shift of \( A \) changes its value, and hence the difference \( A \oplus B \) will also generally change. Therefore, this operation alters the input difference.

4. **A non-linear circuit:** Non-linear circuits are circuits that do not have a linear relationship between their input and output. This means that the output cannot be expressed as a linear combination of the inputs. Since non-linear circuits can introduce complex transformations, they do not preserve the input difference either.

After analyzing all options, we see that all of them change the input difference. However, we are tasked with finding which circuit does not change the input difference.

Upon careful consideration, the first option, "A XOR to a constant gate," can be interpreted incorrectly. If the constant is 0, it does not change the input, while if it is 1, it flips the input. However, when we analyze how XOR operations affect differences, we see that applying XOR with 0 specifically maintains the difference unchanged (except in cases where both inputs are equal, which does not apply here).

Thus, the correct answer is:
3. **A shift of all bits by one position to the right.**


Question: Tick the \textbf{true} assertion. In RSA \ldots

**Ways to Solve the Problem:**

1. Evaluate each statement's relationship to RSA principles.
2. Recall definitions and characteristics of RSA operation (encryption/decryption).
3. Analyze implications of public key transmission in cryptographic contexts.

**Refining Each Approach:**

1. **Evaluate Each Statement:**
   - **Statement 1:** Decryption being equivalent to factoring is incorrect; decryption does not require factoring but relies on modular arithmetic with private keys.
   - **Statement 2:** Key recovery is indeed not equivalent to factoring, but this doesn't directly relate to standard practices in RSA.
   - **Statement 3:** Decryption in RSA is deterministic, given the same ciphertext and private key; hence it’s not probabilistic.
   - **Statement 4:** Public keys must be transmitted securely and can benefit from authentication mechanisms, ensuring integrity during distribution.

2. **Recalling Definitions:**
   - In RSA, encryption and decryption are mathematical processes involving exponentiation mod a product of primes (not related directly to factoring).
   - Authentication ensures that the public key received corresponds correctly with its owner.

3. **Analyzing Public Key Transmission Implications:**
   - The necessity for secure transmission aligns well with cryptography fundamentals — preventing man-in-the-middle attacks requires both authentication and integrity checks when exchanging public keys.

**Final Answer:**
The correct answer is **4**, as it reflects the importance of security in transferring public keys due to threats such as interception or impersonation in cryptographic communications.


Question: Decision trees...

Answer:** 1, 2

**Explanation:** Decision trees are effective for both classification and regression problems due to their hierarchical structure, which allows for easy interpretation of decisions. Options 3 and 4 are irrelevant as they pertain to biological plants rather than machine learning concepts.


Question: The Moore law

To analyze the question regarding Moore's Law, let's clarify what it is: Moore's Law is the observation made by Gordon Moore in 1965 that the number of transistors on a microchip doubles approximately every two years, leading to an exponential increase in computing power and a decrease in relative cost.

Now, let's consider potential answer choices (though you haven't provided them, I'll illustrate how to evaluate them):

1. **If one choice states that Moore's Law predicts a linear increase in computing power**, we can eliminate this choice because Moore's Law specifically describes an exponential growth pattern, not linear.

2. **If another choice claims Moore's Law applies to non-computing technologies**, we would also eliminate this option. Moore's Law is specifically related to semiconductor technology and the performance of integrated circuits.

3. **If a choice mentions that Moore's Law is a guarantee of future performance**, we would eliminate it, as Moore's Law is an observation and a trend, not a guaranteed outcome.

4. **If a choice correctly states that Moore's Law refers to the doubling of transistors on a chip every two years**, we can keep this option. This aligns with the original definition of Moore's Law.

After walking through these potential answer choices and eliminating the incorrect ones, we would arrive at the correct answer regarding Moore's Law.

Please provide the answer choices if you want a specific evaluation.


Question:
Consider the following sentence:
High-energy pulsed laser beams are used in soft-tissue surgery.
Using a 2-gram language model and a tokenizer that splits on whitespaces and punctuation (including hyphens (-)), what is the probability of the above sentence? Provide your answer as a formula, but clearly explaining each variable.

To calculate the probability of the sentence "High-energy pulsed laser beams are used in soft-tissue surgery" using a 2-gram language model, we need to break down the sentence into its component parts and understand how a 2-gram model works.

### Step 1: Tokenization
First, we need to tokenize the sentence based on the provided tokenizer rules (splitting on whitespaces and punctuation). The tokens for the sentence would be:

1. High
2. energy
3. pulsed
4. laser
5. beams
6. are
7. used
8. in
9. soft
10. tissue
11. surgery

### Step 2: Creating 2-grams
Next, we form 2-grams from the tokens. A 2-gram is a sequence of two adjacent tokens. The 2-grams for our sentence will be:

1. (High, energy)
2. (energy, pulsed)
3. (pulsed, laser)
4. (laser, beams)
5. (beams, are)
6. (are, used)
7. (used, in)
8. (in, soft)
9. (soft, tissue)
10. (tissue, surgery)

### Step 3: Probability Calculation
In a 2-gram language model, the probability of a sentence is calculated as the product of the conditional probabilities of each token given the previous token. The formula for the probability of a sentence \( S \) with tokens \( t_1, t_2, \ldots, t_n \) is given by:

\[
P(S) = P(t_1) \times P(t_2 | t_1) \times P(t_3 | t_2) \times \ldots \times P(t_n | t_{n-1})
\]

For our specific sentence, it can be expressed as:

\[
P(S) = P(High) \times P(energy | High) \times P(pulsed | energy) \times P(laser | pulsed) \times P(beams | laser) \times P(are | beams) \times P(used | are) \times P(in | used) \times P(soft | in) \times P(tissue | soft) \times P(surgery | tissue)
\]

### Step 4: Variables Explanation
- \( P(t_i) \): The probability of the first token \( t_1 \) occurring in the language model.
- \( P(t_i | t_{i-1}) \): The conditional probability of the token \( t_i \) given that the previous token was \( t_{i-1} \).
- Each term \( P(t_i | t_{i-1}) \) is calculated using frequency counts from a training corpus, where:
  - \( P(t_i | t_{i-1}) = \frac{C(t_{i-1}, t_i)}{C(t_{i-1})} \)
    - \( C(t_{i-1}, t_i) \): The count of occurrences of the 2-gram \( (t_{i-1}, t_i) \).
    - \( C(t_{i-1}) \): The count of occurrences of the token \( t_{i-1} \).

### Final Formula
Thus, the final formula for calculating the probability of our sentence using a 2-gram language model is:

\[
P(S) = P(High) \times P(energy | High) \times P(pulsed | energy) \times P(laser | pulsed) \times P(beams | laser) \times P(are | beams) \times P(used | are) \times P(in | used) \times P(soft | in) \times P(tissue | soft) \times P(surgery | tissue)
\]

This formula captures the entire probability of the sentence based on the observed frequencies of each word and their pairings in the context of a training corpus.


Question: For each of the following pairs, what kind of morphology is involved? cat+N => cats, break+V => breakable , freeze+V => frozen , translate+V => translation, modify+V => modifies ; inflectional, inflectional, derivational, inflectional, derivational

What is morphology in linguistics? Morphology is the study of the structure and formation of words, including how morphemes—the smallest units of meaning—combine to create new words or modify existing ones.

What are inflectional morphemes? Inflectional morphemes are added to words to express grammatical relationships without changing the word's category or core meaning. For example, adding “-s” to "cat" to form "cats" indicates plurality.

What about derivational morphemes? Derivational morphemes create new words by altering the meaning or changing the part of speech of the original word. For instance, adding “-able” to "break" to form "breakable" changes the verb into an adjective.

Now, let's analyze each pair:

1. **cat + N → cats**: This involves the addition of the inflectional morpheme “-s” to indicate plural form, maintaining the noun category.

2. **break + V → breakable**: Here, the addition of the derivational morpheme “-able” changes the verb "break" into an adjective, indicating a capability or characteristic.

3. **freeze + V → frozen**: The formation of the past participle "frozen" involves a change in the verb, indicating a different grammatical function (the state resulting from the action), which is typically considered a form of derivation.

4. **translate + V → translation**: The suffix “-tion” turns the verb "translate" into a noun, indicating the process or result of the action, which is a clear example of derivational morphology.

5. **modify + V → modifies**: The addition of the inflectional morpheme “-s” here indicates third-person singular present tense, which is a grammatical change but does not alter the core meaning or category of the word.

What conclusions can we draw from this analysis?

The pairs demonstrate a mix of inflectional and derivational morphology. The first and fifth pairs are inflectional, while the second, third, and fourth pairs are derivational.

Therefore, the answer to the question of the type of morphology involved in each pair is: inflectional, derivational, derivational, derivational, inflectional.


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

To implement User-based collaborative filtering using the provided formula, we will follow a systematic approach. Let’s break down the process into clear steps:

### Step 1: Understand the Formula
The formula for predicting the rating of item \( a \) for user \( x \) is given by:

\[
r_{x}(a) = \bar{r}_{x} + \frac{\sum\limits_{y \in N_{U}(x)} sim(x, y) (r_{y}(a) - \bar{r}_{y})}{\sum\limits_{y \in N_{U}(x)} |sim(x, y)|}
\]

Where:
- \( r_{x}(a) \) = predicted rating of item \( a \) by user \( x \)
- \( \bar{r}_{x} \) = average rating of user \( x \)
- \( N_U(x) \) = set of neighbors (other users) of user \( x \) who have rated item \( a \)
- \( sim(x, y) \) = similarity between users \( x \) and \( y \)
- \( r_{y}(a) \) = rating given by user \( y \) to item \( a \)
- \( \bar{r}_{y} \) = average rating of user \( y \)

### Step 2: Define Inputs and Outputs
- **Input:**
  - A matrix of ratings, where rows represent users and columns represent items.
  - A similarity matrix, where each element represents the similarity between two users.

- **Output:**
  - A vector of predicted ratings for a specific user for all items, or for a specific item.

### Step 3: Implement the Function
We will write a function to compute the predicted ratings based on the given inputs.

#### Step-by-Step Implementation:

1. **Calculate the Average Ratings**
   - Create a function to calculate the average ratings for each user.

2. **Identify Neighbors**
   - For a given user, retrieve all users who have rated the item.

3. **Calculate the Predicted Rating**
   - Implement the formula using the neighbors, their ratings, and the similarity values.

4. **Handle Edge Cases**
   - Ensure that the function handles cases where there are no neighbors or no ratings.

Here’s a sample implementation in Python:

```python
import numpy as np

def user_based_collaborative_filtering(ratings, similarity_matrix, user_index, item_index):
    """
    Predicts the rating of a specific item for a specific user using User-based collaborative filtering.

    :param ratings: A 2D numpy array (users x items) containing the ratings.
    :param similarity_matrix: A 2D numpy array (users x users) containing the similarity scores.
    :param user_index: The index of the user for whom we are predicting the rating.
    :param item_index: The index of the item for which we are predicting the rating.
    :return: The predicted rating for the item by the user.
    """

    # Step 1: Calculate average ratings for all users
    user_avg_ratings = np.nanmean(ratings, axis=1)

    # Step 2: Identify neighbors who rated the item
    neighbors = np.where(~np.isnan(ratings[:, item_index]))[0]

    # Step 3: If no neighbors have rated the item, return None or a default value
    if len(neighbors) == 0:
        return None  # or return some default value

    # Step 4: Calculate the numerator and denominator for the prediction formula
    numerator = 0
    denominator = 0

    for neighbor in neighbors:
        if neighbor != user_index:  # Do not include self
            similarity = similarity_matrix[user_index, neighbor]
            rating_diff = ratings[neighbor, item_index] - user_avg_ratings[neighbor]
            numerator += similarity * rating_diff
            denominator += abs(similarity)

    # Step 5: Calculate the predicted rating
    predicted_rating = user_avg_ratings[user_index] + (numerator / denominator if denominator != 0 else 0)

    return predicted_rating
```

### Final Answer
The function `user_based_collaborative_filtering` takes a ratings matrix, a similarity matrix, a user index, and an item index as input and returns the predicted rating for that item by the specified user using user-based collaborative filtering.


Question: Bluetooth pairing v2.0 is based on\dots


Question:
Let $S_{0},S_{1},S_{2},\dots$ be an infinite sequence produced by a source $\mathcal{S}$. All $S_{n}$ take values in $\{0,1\}$, and $S_{n+1}$ depends only on $S_n$, that is,   $p_{S_{n+1} | S_0, \dots, S_n}(s_{n+1} | s_0, \dots, s_n) = p_{S_{n+1} | S_n}(s_{n+1} | s_n)$.   The probability $p_{S_{n+1}|S_{n}}$ is schematically represented in the graph below:     egin{center}    	ikzset{      state/.style = {        draw,        circle,        minimum size = 20pt,        font = ontsize{12}{12}\selectfont,      }    }    egin{tikzpicture}[> = latex]
ode[state] (a) {$0$};
ode[state, right of = a] (b) {$1$};      \path[->] (a) edge[bend left, above] node {$1/2$} (b);      \path[->] (a) edge[loop left] node {$rac{1}{2}$} ();      \path[->] (b) edge[bend left, below] node {$1/4$} (a);      \path[->] (b) edge[loop right] node {$rac{3}{4}$} ();    \end{tikzpicture}  \end{center}  For instance, the edge from $0$ to $1$ means that $p_{S_{n+1}|S_{n}}(1|0) = rac{1}{2}$. We also have that $p_{S_0}(0)=1$. True or false: For every $n\geq 0$, $H(S_n|S_0,\ldots,S_{n-1})
eq H(S_n|S_{n-1}) $.

To analyze the statement "For every $n\geq 0$, $H(S_n|S_0,\ldots,S_{n-1}) \neq H(S_n|S_{n-1})$," we will carefully apply the properties of conditional entropy in the context of a Markov process, as described in the problem.

### Step 1: Understanding Conditional Entropy
Conditional entropy, $H(X|Y)$, measures the uncertainty of a random variable $X$ given another random variable $Y$. In this case, we have a sequence produced by a source where $S_{n+1}$ depends only on $S_n$. This means that the future state is conditionally independent of past states given the most recent state.

### Step 2: Exploring the Case for $n=1$
For $n=1$, we find $H(S_1|S_0)$. Since $S_0 = 0$ with probability 1, we have:
$$H(S_1 | S_0) = H(S_1 | 0)$$

The transition probabilities are:
- $p_{S_1 | S_0}(0|0) = \frac{1}{2}$
- $p_{S_1 | S_0}(1|0) = \frac{1}{2}$

This results in:
$$H(S_1 | 0) = -\left(\frac{1}{2} \log \frac{1}{2} + \frac{1}{2} \log \frac{1}{2}\right) = 1$$

Now, considering $H(S_1 | S_0, S_1)$, since $S_0$ gives us complete information about $S_1$, we have:
$$H(S_1 | S_0, S_1) = 0$$

### Step 3: Analyzing for General $n \geq 1$
For $n \geq 1$, we need to consider the relationship between $H(S_n | S_{n-1})$ and $H(S_n | S_0, \ldots, S_{n-1})$. Given that the process is Markovian, the future state $S_n$ is conditionally independent of all previous states given the most recent state $S_{n-1}$. Thus, we can apply the property of conditional independence:

$$H(S_n | S_0, \ldots, S_{n-1}) = H(S_n | S_{n-1})$$

This means that conditioning on all previous states provides no additional information beyond conditioning on just the last state.

### Step 4: Conclusion
From our analysis, we find that the statement provided is **False**. Specifically, for every $n \geq 1$, we have:

$$H(S_n | S_0, \ldots, S_{n-1}) = H(S_n | S_{n-1})$$

Thus, we conclude that the condition specified in the original statement does not hold; the entropies can indeed be equal due to the properties of conditional entropy in the context of a Markov process.


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 6 (indexed as in the lectures)?

To determine the content of the cell at row 3, column 6 in the CYK chart for the input sentence "the red cat is meowing on the roof," we need to analyze the sentence structure according to the given context-free grammar (CFG) and the lexicon.

### Step 1: Identify the Input Sentence
The input sentence is:
```
the red cat is meowing on the roof
```
This sentence consists of:
- **NP**: "the red cat"
- **VP**: "is meowing on the roof"

### Step 2: Breakdown the Sentence into Constituents
We can break down the sentence into its constituents based on the grammar rules:

1. **NP** (Noun Phrase):
   - "the red cat" can be derived from the rules:
     - `NP -> Det Adj N` using `the:Det`, `red:Adj`, `cat:N`
   - Therefore, we have `NP = {the red cat}`.

2. **VP** (Verb Phrase):
   - "is meowing" can be derived from the rules:
     - `VP -> Aux Ving` using `is:Aux`, `meowing:V`
   - "on the roof" can be derived as follows:
     - `PNP -> Prep NP`, where "on" is the prep and "the roof" is another NP:
       - "the roof" can be derived using `NP -> Det N` with `the:Det` and `roof:N`.
   - So we have `VP = {is meowing on the roof}`.

### Step 3: Identify the Components and Their Positions
The breakdown of the sentence leads to the following structure:
- 1-3: NP "the red cat"
- 4: Aux "is"
- 5: Ving "meowing"
- 6: PNP "on the roof"

### Step 4: Fill in the Chart
The CYK algorithm fills the chart based on the derivations possible:

- **Row 1** corresponds to individual words:
  - Column 1: {Det} from "the"
  - Column 2: {Adj} from "red"
  - Column 3: {N} from "cat"
  - Column 4: {Aux} from "is"
  - Column 5: {Ving} from "meowing"
  - Column 6: {Prep} from "on"
  - Column 7: {N} from "roof"

- **Row 2** corresponds to combinations of two words:
  - Column 1-2: Empty (no rules apply for a two-word combination)
  - Column 2-3: {NP} from "the red cat"
  - Column 3-4: Empty
  - Column 4-5: Empty
  - Column 5-6: Empty
  - Column 6-7: {NP} from "on the roof"

- **Row 3** corresponds to combinations of three words:
  - Column 1-3: Empty
  - Column 2-4: Empty
  - Column 3-5: Empty
  - Column 4-6: {VP} from "is meowing on the roof"
  - Column 5-7: Empty

### Step 5: Identify the Content of Row 3, Column 6
Following the above reasoning, the content of row 3, column 6 is derived from the combination of "on the roof" which is a valid PNP. Hence, this cell contains:

**Answer: PNP**


Question: Which of the following is correct regarding crowdsourcing?

To evaluate the correctness of the statements regarding crowdsourcing:

1. **Uniform spammers randomly select answers.** - This statement is generally true. Uniform spammers provide answers without any genuine contribution, often randomly.

2. **The accuracy of majority voting is never equal to the one of Expectation Maximization.** - This statement is misleading. While Expectation Maximization (EM) can provide better accuracy in some scenarios, there are instances where majority voting can be just as accurate or even better, depending on the context and the nature of the data.

3. **Honey pots can detect uniform spammers, random spammers, and sloppy workers.** - This statement is accurate. Honey pots are techniques used to identify different types of spammers and unreliable participants in crowdsourcing.

4. **Majority Decision and Expectation Maximization both give less weight to spammersâ answers.** - This statement is generally true. Both methods aim to mitigate the influence of erroneous or spam answers on the final result.

Based on this analysis, statements 1, 3, and 4 are correct, while statement 2 is not universally correct. If you need to choose the most accurate statement, option 3 stands out as it encompasses detection capabilities in crowdsourcing effectively.


Question:
What is the output of the following block of Python code? (one answer) \\
\verb|my_string = `computational'| \\
\verb|print(my_string[1])|\\
\verb|print(my_string[3:5])|
\vspace{0.25cm}

The output of the code will be 'o' from the first print statement (my_string[1]) and 'pu' from the second print statement (my_string[3:5]), so the correct option is 4: o\\pu. This is because my_string is defined as 'computational', where indexing starts at 0.


Question: Would it make sense to add the total-order property to the best-effort broadcast?

Yes, it would make sense to add the total-order property to a best-effort broadcast, especially in scenarios where the order of messages is crucial for the application.

Best-effort broadcast ensures that messages sent by a source are received by all recipients, but it does not guarantee that all recipients will receive the messages in the same order. This can lead to inconsistencies, especially in distributed systems where the state of the system depends on the order of operations.

By adding the total-order property to best-effort broadcast, we ensure that all nodes in the system receive messages in the exact same order. This is important for maintaining consistency across distributed systems, such as databases or collaborative applications, where actions depend on the sequence of events.

In summary, adding total-order to best-effort broadcast enhances consistency and reliability in message delivery, making it a sensible enhancement for applications requiring strict message ordering.