Question: Consider the group $(\mathbb{Z} / 23 \mathbb{Z}^*, \cdot)$. Find how many elements of the group are generators of the group. (Hint: $5$ is a generator of the group.)

**Plan:**
- Understand the structure of the group \((\mathbb{Z} / 23\mathbb{Z}^*, \cdot)\), which consists of the multiplicative group of integers modulo 23.
- Determine the order of this group, which is given by Euler's totient function \(\phi(n)\) for \(n=23\).
- Identify the condition under which an element is a generator of the group, specifically that it must have an order equal to the size of the group.
- Use the fact that \(5\) is a known generator to find how many generators exist in the group.

**Steps:**

1. **Understand the Structure of the Group:**
   The group \((\mathbb{Z} / 23\mathbb{Z}^*, \cdot)\) consists of the integers from \(1\) to \(22\) that are coprime to \(23\). Since \(23\) is a prime number, all integers from \(1\) to \(22\) are coprime to \(23\).

2. **Determine the Order of the Group:**
   Since \(23\) is prime, the order of the group is given by \(\phi(23) = 23 - 1 = 22\). Thus, the group has \(22\) elements.

3. **Identify the Condition for Generators:**
   An element \(g\) is a generator of the group if the order of \(g\) is equal to the order of the group, which is \(22\). The number of generators of a cyclic group of order \(n\) is given by \(\phi(n)\). Therefore, we need to calculate \(\phi(22)\).

4. **Calculate \(\phi(22)\):**
   The prime factorization of \(22\) is \(2 \times 11\). Using the formula for the Euler's totient function:
   \[
   \phi(22) = 22 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{11}\right) = 22 \cdot \frac{1}{2} \cdot \frac{10}{11} = 10.
   \]
   Therefore, there are \(10\) generators of the group.

**Self-Check:**
The calculations show that the order of the group is \(22\) and \(\phi(22) = 10\). This confirms that there are \(10\) elements in the group that are generators. Thus, the answer is **10**.


Question: A proof system is perfect-black-box zero-knowledge if \dots

A proof system is perfect-black-box zero-knowledge if **Option 3** is true: there exists a PPT simulator \( S \) such that for any PPT verifier \( V \), \( S^{V} \) produces an output which has the same distribution as the view of the verifier.

### Key Reasoning:

1. **Perfect Zero-Knowledge**: The simulator \( S \) must be able to replicate the view of the verifier without any knowledge of the actual secret used in the proof.
2. **Black-Box Simulation**: The simulator operates in a black-box manner, meaning it can use the verifier as a subroutine but doesn't need to learn from it directly.
3. **Indistinguishability**: The outputs of the simulator and the verifier's view must be indistinguishable in distribution, ensuring that no information about the secret is leaked.

This definition captures the essence of zero-knowledge in a perfect sense, where the verifier's view can be simulated without access to the actual proof.


Question: Let us define the relation R = {(1, 1)} on the set A = {1, 2, 3}. Which of the following properties does R satisfy ? (multiple answers)

To analyze the properties of the relation \( R = \{(1, 1)\} \) defined on the set \( A = \{1, 2, 3\} \), we will review each property of relations: reflexive, symmetric, anti-symmetric, and transitive. Let's define these properties clearly:

1. **Reflexive**: A relation \( R \) on a set \( A \) is reflexive if for every element \( a \in A \), the pair \( (a, a) \) is in \( R \). In other words, \( R \) must contain \( (1, 1) \), \( (2, 2) \), and \( (3, 3) \) for it to be reflexive.

2. **Symmetric**: A relation \( R \) is symmetric if for every pair \( (a, b) \in R \), the pair \( (b, a) \) is also in \( R \). So, if \( (1, 1) \in R \), then \( (1, 1) \) must also be in \( R \).

3. **Anti-symmetric**: A relation \( R \) is anti-symmetric if for every pair \( (a, b) \in R \) where \( a \neq b \), it must be true that \( (b, a) \) is not in \( R \). In other words, if both \( (a, b) \) and \( (b, a) \) are in \( R \), then \( a \) must equal \( b \).

4. **Transitive**: A relation \( R \) is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in \( R \).

Now, let's analyze the relation \( R \):

- **Checking Reflexivity**:
    - The set \( A = \{1, 2, 3\} \) requires \( R \) to contain \( (1, 1) \), \( (2, 2) \), and \( (3, 3) \).
    - \( R \) only contains \( (1, 1) \).
    - Since \( (2, 2) \) and \( (3, 3) \) are not in \( R \), it is **not reflexive**.

- **Checking Symmetry**:
    - The only pair in \( R \) is \( (1, 1) \).
    - Since \( (1, 1) \) implies \( (1, 1) \), the relation is **symmetric**.

- **Checking Anti-symmetry**:
    - The only pair in \( R \) is \( (1, 1) \).
    - There are no pairs where \( a \neq b \) to contradict anti-symmetry. Hence, it vacuously satisfies the condition and is therefore **anti-symmetric**.

- **Checking Transitivity**:
    - The only pair in \( R \) is \( (1, 1) \).
    - There are no other pairs to check, thus it vacuously satisfies the condition and is therefore **transitive**.

Based on this analysis, let's summarize the properties satisfied by the relation \( R \):

- **Not Reflexive** (because it lacks \( (2, 2) \) and \( (3, 3) \))
- **Symmetric** (it includes \( (1, 1) \))
- **Anti-symmetric** (vacuously true as there are no distinct pairs)
- **Transitive** (vacuously true as there are no distinct pairs)

### Conclusion:
The properties that the relation \( R \) satisfies are:
- **B. Symmetric**
- **C. Anti-symmetric**
- **D. Transitive**

Thus, the correct answers are **B, C, and D**.


Question: Consider the following matrix-factorization problem. For the observed ratings $r_{u m}$ for a given pair $(u, m)$ of a user $u$ and a movie $m$, one typically tries to estimate the score by $$ f_{u m}=\left\langle\mathbf{v}_{u}, \mathbf{w}_{m}\right\rangle+b_{u}+b_{m} $$ Here $\mathbf{v}_{u}$ and $\mathbf{w}_{m}$ are vectors in $\mathbb{R}^{D}$ and $b_{u}$ and $b_{m}$ are scalars, indicating the bias. Assume that our objective is given by $$ \frac{1}{2} \sum_{u \sim m}\left(f_{u m}-r_{u m}\right)^{2}+\frac{\lambda}{2}\left[\sum_{u \in \mathbf{U}}\left(b_{u}^{2}+\left\|\mathbf{v}_{u}\right\|^{2}\right)+\sum_{m \in \mathbf{M}}\left(b_{m}^{2}+\left\|\mathbf{w}_{m}\right\|^{2}\right)\right] $$ where $\lambda>0$. Here $\mathbf{U}$ denotes the set of all users, $M$ the set of all movies, and $u \sim m$ represents the sum over all $(u, m)$ pairs for which a rating exists. Write the optimal values of $b_{u}$, provided that all other values are fixed.

Answer:**
The optimal value of \( b_{u'} \) is:
$$ b_{u'} = \frac{1}{n_{u'} + \lambda} \left( \sum_{m: (u', m) \text{ rated}} r_{u' m} - \sum_{m: (u', m) \text{ rated}} \left\langle \mathbf{v}_{u'}, \mathbf{w}_{m}\right\rangle - \sum_{m: (u', m) \text{ rated}} b_{m} \right). $$

**Explanation:**
To find the optimal bias \( b_{u'} \), we differentiate the objective function with respect to \( b_{u'} \) and set the result to zero to find the critical points. The resulting equation accurately balances the contribution of observed ratings with the predictions made by the model while accounting for regularization. The term \( n_{u'} \) represents the number of ratings given by user \( u' \), which helps normalize the bias to prevent overfitting. This formulation ensures that the bias reflects the average difference between the actual ratings and the predicted scores, adjusted by the regularization parameter \( \lambda \). Thus, this approach provides a reliable estimate of the bias that incorporates both the user’s historical ratings and the model's predictions, yielding a more accurate representation of user behavior in the context of the movie rating system.


Question: Let P(x) is “x is an elephant” and F(x) is “x flies” and the domain consists of all animals. Translate the following statement into English: ∃!x(P(x) ∧ F(x))

To translate the logical statement \( \exists!x(P(x) \land F(x)) \) into English, we can break down its components:

- \( \exists!x \) means "there exists exactly one \( x \)".
- \( P(x) \) means "x is an elephant".
- \( F(x) \) means "x flies".
- The conjunction \( P(x) \land F(x) \) means "x is an elephant and x flies".

Putting it all together, the statement \( \exists!x(P(x) \land F(x)) \) translates to:

"There exists exactly one animal \( x \) such that \( x \) is an elephant and \( x \) flies."

Among the options provided, the closest match to this translation is:

1. **There exists only one elephant that can fly.**

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


Question:
Consider the following lexicon \(L\):
bear   : V, N
bears  : V, N
blue   : Adj, N
drink  : N, V
drinks : N, V
Nice   : Adj, N

When using an order-1 HMM model (using \(L\)) to tag the word sequence:"Nice bears drink blue drinks"does the tag of drink depend on the tag of nice?

To analyze the multiple-choice question regarding the dependency of tags in an order-1 Hidden Markov Model (HMM), we will follow the structured approach outlined in your request.

### Core Concepts and Principles

1. **Hidden Markov Model (HMM)**: An HMM is a statistical model used to represent systems that are assumed to be a Markov process with hidden states. In the context of part-of-speech tagging, the states correspond to the tags, and the observations correspond to the words.

2. **Order-1 HMM**: This indicates that the model considers only the immediate previous tag to predict the current tag. Thus, the tag of a word depends only on the tag of the preceding word, not on any other tags or words in the sequence.

3. **Independence of Tags**: In an order-1 HMM, the assumption is that the current state (tag) is conditionally independent of all previous states given the previous state. This means that the tag of the current word does not depend on the tags of words that are not immediately adjacent.

### Evaluation of Each Option

#### Option 1: "yes, because the HMM approach relies on a global maximum."

- **Evaluation**: This option is **incorrect**.
- **Reasoning**: While it is true that HMMs often seek a global maximum (the most probable sequence of states), this does not imply that the tag of "drink" depends on the tag of "Nice". In an order-1 HMM, the tag of "drink" would depend only on the tag of the word immediately preceding it, which is "bears". The global maximum concept relates to finding the best overall tagging sequence but does not change the independence assumption of the model.

#### Option 2: "no, the hypotheses make the two tags independent from each other."

- **Evaluation**: This option is **correct**.
- **Reasoning**: In an order-1 HMM, the tag of "drink" is only influenced by the tag of "bears", the word immediately before it. The tag of "Nice" does not influence the tag of "drink" because they are not adjacent in the sequence. The independence assumption of the HMM means that the tags are conditionally independent given the previous tag, which in this case is "bears".

### Conclusion

The correct answer is **Option 2**: "no, the hypotheses make the two tags independent from each other." This option accurately reflects the principles of an order-1 HMM, where the tag of a word depends solely on the tag of the preceding word.

### Key Insight

This question tests the understanding of the conditional independence property of HMMs, particularly in the context of order-1 models. It emphasizes the importance of recognizing how the structure of the model dictates the relationships between tags and highlights the distinction between global optimization strategies and local dependencies in probabilistic models.


Question: KEM/DEM refers to\dots

To analyze the question, we need to understand what KEM/DEM stands for and how it relates to the provided options. KEM stands for Key Encapsulation Mechanism, and DEM stands for Data Encapsulation Mechanism. Together, KEM/DEM refers to a specific approach in cryptography, primarily used for ensuring secure communication.

First, let’s break down the components:

1. **Key Encapsulation Mechanism (KEM):** This is a method used to securely exchange cryptographic keys. In essence, KEM allows one party to encapsulate a secret key, which can then be sent to another party. This encapsulation process ensures that only the intended recipient can extract the key, typically using their private key or a decryption mechanism. The underlying security of KEM is often based on hard mathematical problems, such as those in lattice-based cryptography or other public-key cryptosystem foundations.

2. **Data Encapsulation Mechanism (DEM):** Once the key has been securely exchanged using a KEM, the DEM takes that encapsulated key and uses it to encrypt the actual data that needs to be transmitted securely. The DEM typically employs symmetric encryption algorithms, which are faster and more efficient for encrypting large amounts of data compared to asymmetric algorithms.

Now, we can evaluate the options given in the multiple-choice question:

1. **An encryption scheme:** KEM/DEM as a whole can be viewed as a system that includes encryption (specifically in the DEM part), but it does not refer exclusively to an encryption scheme. It involves processes for key exchange and data encryption together.

2. **A digital signature scheme:** KEM/DEM does not involve digital signatures, which are used to verify the authenticity of messages and ensure non-repudiation. Digital signatures rely on different cryptographic principles and are not directly related to key encapsulation or data encryption.

3. **A commitment scheme:** Commitment schemes are cryptographic protocols that allow one party to commit to a value while keeping it hidden until they choose to reveal it. KEM/DEM does not fit this description as it focuses on key exchange and data encryption.

4. **A hash function:** Hash functions are algorithms that take an input and produce a fixed-size string of bytes, which is typically a digest that is unique to that specific input. KEM/DEM does not relate to hash functions in the context of their primary functions.

From this analysis, the only option that aligns with the purpose of KEM/DEM is option 1, "an encryption scheme," because KEM/DEM as a whole can be seen as a secure method for encrypting data through the combination of KEM and DEM processes. Although it is broader than just an encryption scheme, it fundamentally encompasses the process of securely encrypting data after key exchange.

The correct answer to the multiple-choice question is therefore: **1. an encryption scheme.**


Question: What is the value of \(f(4)\) where \(f\) is defined as \(f(0) = f(1) = 1\) and \(f(n) = 2f(n - 1) + 3f(n - 2)\) for integers \(n \geq 2\)?

To find \(f(4)\), let's first eliminate the wrong options:

- **45**: This value is too high given the recurrence relation and initial conditions.
- **39**: This value is too low and does not align with the pattern expected from the recurrence.
- **43**: This value does not fit the calculated values leading up to \(f(4)\).

Now, let's calculate \(f(2)\), \(f(3)\), and finally \(f(4)\):

1. \(f(2) = 2f(1) + 3f(0) = 2 \cdot 1 + 3 \cdot 1 = 2 + 3 = 5\)
2. \(f(3) = 2f(2) + 3f(1) = 2 \cdot 5 + 3 \cdot 1 = 10 + 3 = 13\)
3. \(f(4) = 2f(3) + 3f(2) = 2 \cdot 13 + 3 \cdot 5 = 26 + 15 = 41\)

Thus, the correct answer is \(41\). This is justified as we derived it directly from the recurrence relation, which accurately reflects the defined function \(f\).


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

In the context of ranked retrieval, when we have a non-relevant document at position k and a relevant document at position k+1, the following can be evaluated based on the definitions of precision (P) and recall (R):\n\n1. **Precision (P)** at rank k is defined as the ratio of relevant documents retrieved in the top k results to k. Similarly, for k-1 and k+1, we have:\n   - \\(P@k = \\frac{\\text{relevant documents in top k}}{k}\\)\n   - \\(P@k-1 = \\frac{\\text{relevant documents in top (k-1)}}{k-1}\\)\n   - \\(P@k+1 = \\frac{\\text{relevant documents in top (k+1)}}{k+1}\\)\n\nSince we know that the document at position k is non-relevant, the relevant count in the top k is the same as in the top k-1. Thus:\n- Relevant documents at k = relevant documents at k-1 = X (where X is the count of relevant documents up to k-1).\n- Therefore, \\(P@k > P@k-1\\) because \\(k\\) is greater than \\(k-1\\), but \\(P@k-1\\) will not be greater than \\(P@k+1\\) (since we\'re adding a relevant document).\n\n2. **Recall (R)** at rank k is defined as the ratio of relevant documents retrieved in the top k results to the total number of relevant documents in the collection. The relevance of documents does not change when moving from k-1 to k or from k to k+1:\n  - If we assume that the total number of relevant documents (in the entire collection) stays constant, then:\n   - \\(R@k = \\frac{X}{\\text{total relevant}}\\)\n   - \\(R@k-1 = \\frac{X}{\\text{total relevant}}\\) (the same)\n   - \\(R@k+1 = \\frac{X+1}{\\text{total relevant}}\\)\n\nHence, it follows that \\(R@k-1\\) is equal to \\(R@k+1\\) or \\(R@k-1 < R@k+1\\) (since we are now counting an additional relevant document in R@k+1).\n\nGiven these evaluations:\n- The most accurate statement according to the possible choices is **"R@k-1 < R@k+1"**.\n\nHence, the answer is:\n**"R@k-1 < R@k+1"**.


Question: A Carmichael number \ldots

A Carmichael number is a composite number that satisfies Fermat's Little Theorem for all integers relatively prime to it, meaning it can pass the Fermat primality test and thus may be incorrectly identified as a prime number (a false positive). Therefore, the correct option is that a Carmichael number is a false positive of the Fermat test. These numbers demonstrate that not all numbers that pass this test are prime, highlighting a limitation of the test in certain cases. Carmichael numbers are also interesting in number theory due to their properties related to modular arithmetic, but they do not directly correspond to the order of a finite field or serve as exponents in the RSA cryptosystem.


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

To determine the most appropriate method for handling skewed class label distributions in classification tasks, let's analyze each option carefully:

1. **Include an over-proportional number of samples from the larger class**: This approach would actually *increase* the imbalance in your dataset. By adding more samples from the larger class, you're reinforcing the dominance of that class and making it more difficult for the classifier to learn patterns from the minority class. This would not be suitable for addressing skewed distributions.

2. **Use leave-one-out cross validation**: Leave-one-out cross-validation (LOOCV) is a method used to validate models by training on all but one instance and testing on that one. While it gives a thorough evaluation of model performance, it does not specifically address class imbalance. If the dataset is imbalanced, this method will still reflect that imbalance in its performance metrics, making it ineffective in solving the underlying issue. Thus, this option is not appropriate.

3. **Construct the validation set such that the class label distribution approximately matches the global distribution of the class labels**: This method keeps the class distribution in the validation set the same as the overall distribution in the dataset. While this is important for assessing model performance in a real-world context, it does not help in mitigating class imbalance in the training process. This approach does not rectify the skewed distribution in the training set, making it unsuitable for the goal of fighting skewed distributions.

4. **Generate artificial data points for the most frequent classes**: This option, as stated, would involve creating additional samples for the majority class. Similar to option 1, this would further increase the imbalance by adding to the already larger class. The correct approach should ideally be to generate synthetic data for the minority class, which would help balance the classes. Therefore, this option is also not appropriate.

### Conclusion
None of the provided options effectively address the problem of fighting skewed distributions of class labels in classification tasks. Appropriate methods typically involve:

- **Oversampling the minority class**: Techniques like SMOTE (Synthetic Minority Over-sampling Technique) can create synthetic examples for the minority class to make the dataset more balanced.
- **Undersampling the majority class**: Removing samples from the majority class can help balance the dataset.
- **Using different algorithms**: Some algorithms are designed to handle imbalanced datasets better, such as cost-sensitive learning or ensemble methods that focus on minority instances.

In summary, the best strategies for addressing skewed distributions in classification are not represented in the options provided.


Question: A multiset is an unordered collection where elements can appear multiple times. We will represent a multiset of Char elements as a function from Char to Int: the function returns 0 for any Char argument that is not in the multiset, and the (positive) number of times it appears otherwise: type Multiset = Char => Int What should replace ??? so that the following function computes the underlying set of the m multiset, formed from its distinct elements? For example, the underlying set of the multiset M = {'a', 'a', 'c', 'd', 'd'} is S = {'a', 'c', 'd'}. type Set = Char => Boolean def multisetToSet(m: Multiset): Set = ???

To convert a multiset represented as a function from `Char` to `Int` into its underlying set (where each character appears at least once), we need a function that returns `true` for characters that have a count greater than zero in the multiset.

The correct form of the function should be:
```scala
(x: Char) => m(x) > 0
```

This checks if the occurrence count of each character `x` in the multiset `m` is greater than zero, effectively indicating its presence in the underlying set.

Among the provided options, the correct answer to tick is:
- **2. x => m(x) > 0**


Question:
Given the following code snippet, you are tasked to produce a
                    modulo scheduled version of the loop achieving the best possible
                    performance. You can assume that any operation has a latency of one
                    cycle and that the processor has 2 ALUs, one memory unit, and one
                    branch unit. The processor also has all the necessary hardware
                    structures and instructions to support modulo scheduling. In
                    particular, the instruction erb+mov EC+ loads a value in the
                    epilogue counter and the instruction erb+loop.pip+ is equivalent
                    to erb+loop+ but has all features expected to execute a modulo
                    scheduled loop. Predicates from erb+p32+ and registers from
                    erb+x32+ onwards rotate; across iterations registers are rotated
                    upwards (i.e., the content of erb+x44+ is accessible as erb+x45+
                    in the next iteration). What is the shortest achievable initiation
                    interval? Why?
                    egin{verbatim}
                        0: mov LC, 100
                        1: mov x1, 10000
                        2: ld x2, 0(x1)
                        3: addi x2, x2, 10
                        4: st x2, 0(x1)
                        5: addi x1, x1, 1
                        6: loop 2
                    \end{verbatim}

To produce a modulo scheduled version of the given loop, we need to analyze the loop's dependencies and the available resources on the processor, including the number of ALUs, memory units, and the branch unit. The goal is to achieve the best possible performance with the shortest achievable initiation interval (II).

### Given Code Snippet

```assembly
0: mov LC, 100       ; Load loop counter
1: mov x1, 10000     ; Load base address
2: ld x2, 0(x1)      ; Load value from memory
3: addi x2, x2, 10   ; Increment the loaded value
4: st x2, 0(x1)      ; Store the updated value back to memory
5: addi x1, x1, 1    ; Increment the address
6: loop 2            ; Loop back to instruction 2
```

### Dependencies Analysis

1. **Dependencies**:
   - The `ld` instruction (line 2) loads a value into `x2`, which is used by the `addi` instruction (line 3).
   - The `st` instruction (line 4) depends on the completion of the `addi` instruction (line 3) since it needs the updated value in `x2`.
   - The `addi` instruction (line 5) also depends on the completion of the store instruction (line 4) since it needs to modify `x1` based on the updated address.

2. **Latency**: Each operation has a latency of one cycle.

### Available Resources

- **2 ALUs**: We can perform two arithmetic operations in parallel.
- **1 Memory Unit**: The memory operations (load and store) cannot overlap with each other.
- **1 Branch Unit**: Used to handle the loop control.

### Modulo Scheduling

Modulo scheduling involves unrolling the loop and scheduling instructions across iterations to optimize resource usage and minimize stalls. Given the dependencies, we can align the instructions to execute in a staggered manner across iterations.

#### Steps to Create a Modulo Scheduled Version

1. **Unroll the Loop**:
   - We can unroll the loop to allow more instructions to be scheduled in parallel. Since our loop has 6 instructions, we'll unroll it for 2 iterations.

2. **Schedule Instructions**:
   - Let's denote the iterations with suffixes (e.g., `i0`, `i1` for the first iteration, `i2`, `i3` for the second iteration) for clarity.
   - Perform scheduling across iterations while respecting dependencies.

### Proposed Modulo Schedule

Hereâs how we can schedule the loop:

```assembly
; Iteration 0
0: mov LC, 100         ; Load loop counter
1: mov x1, 10000       ; Load base address
2: ld x2, 0(x1)        ; Load value from memory (iter i0)
3: addi x2, x2, 10     ; Increment loaded value (iter i0)
4: st x2, 0(x1)        ; Store the updated value back to memory (iter i0)
5: addi x1, x1, 1      ; Increment address (iter i0)
; Start of Iteration 1
6: ld x2, 0(x1)        ; Load next value from memory (iter i1)
7: addi x2, x2, 10     ; Increment loaded value (iter i1)
8: st x2, 0(x1)        ; Store updated value back to memory (iter i1)
9: addi x1, x1, 1      ; Increment address (iter i1)
10: loop 2             ; Loop back to instruction 2
```

### Initiation Interval (II)

To determine the shortest achievable initiation interval, we look at the dependency chains and resource usage:

- In the first iteration:
  - The `ld` instruction (line 2) is executed first.
  - The `addi` instruction (line 3) can follow immediately after.
  - The `st` instruction (line 4) can execute next.
  - The `addi` for address increment (line 5) comes after the store.

- In the second iteration (starting at line 6):
  - The load (`ld x2`) can be issued immediately after the address increment (`addi x1, x1, 1`) of the first iteration, as they operate on different resources (the memory unit).
  - The operations can overlap in a staggered manner because we have two ALUs.

### Conclusion:

- Given the dependencies, we can achieve an initiation interval (II) of **2 cycles**. This is because:
  - In one cycle, we can complete the load and the first add, and in the next cycle, we can perform the store and the next load from the subsequent iteration, effectively overlapping operations across iterations.

Thus, we can achieve optimal performance while respecting the constraints of the available resources and dependencies through modulo scheduling.


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

To analyze the multiple-choice question regarding the minimal length of the output of a hash function to provide security against collision attacks of \(2^{256}\), we will follow the structured approach outlined in the instructions.

### Core Concepts
1. **Hash Function**: A function that converts input data into a fixed-size string of characters, which appears random.
2. **Collision Attack**: An attack where two different inputs produce the same hash output.
3. **Security Requirement**: The length of the hash output must be sufficient to prevent an adversary from finding collisions with a certain probability.

### Analysis of Each Option

1. **Option 1: \(2^{256}\) bits**
   - **Correctness**: Incorrect.
   - **Reasoning**: A hash output of \(2^{256}\) bits would be impractically large and does not align with the typical security requirements for hash functions. The output length should be in bits, not in exponential form.

2. **Option 2: \(2^{512}\) bits**
   - **Correctness**: Incorrect.
   - **Reasoning**: While a hash output of \(2^{512}\) bits would provide a very high level of security, it is unnecessarily large. The goal is to determine the minimum length required to secure against collision attacks of \(2^{256}\), not to exceed it significantly.

3. **Option 3: \(256\) bits**
   - **Correctness**: Incorrect.
   - **Reasoning**: A hash function output of \(256\) bits does not provide sufficient security against collision attacks. The birthday paradox indicates that to secure against \(2^{256}\) collision attempts, the output length should be at least \(2 \times 256 = 512\) bits. Thus, \(256\) bits is not adequate.

4. **Option 4: \(512\) bits**
   - **Correctness**: Correct.
   - **Reasoning**: A hash output of \(512\) bits is sufficient to provide security against \(2^{256}\) collision attacks. This is based on the birthday bound, which states that to ensure a collision resistance against \(2^{n}\) attempts, the hash output should be at least \(2n\) bits. Therefore, for \(2^{256}\) attempts, \(512\) bits is the minimum required length.

### Conclusion
The correct answer is **Option 4: \(512\) bits**. This length ensures that the probability of finding a collision through brute force is sufficiently low, adhering to the security requirements against \(2^{256}\) collision attacks.

### Key Principle
The key principle behind this question is the relationship between the length of a hash function's output and its resistance to collision attacks, particularly as described by the birthday paradox. To secure against \(2^{k}\) collision attempts, the output length should be at least \(2k\) bits.


Question:
Given the following classes:
• class Pair[+U, +V]
• class Iterable[+U]
• class Map[U, +V] extends Iterable[Pair[U, V]]
Recall that + means covariance, - means contravariance and no annotation means invariance (i.e. neither
covariance nor contravariance).
Consider also the following typing relationships for A, B, X, and Y:
• A >: B
• X >: Y
Fill in the subtyping relation between the types below using symbols:
• <: in case T1 is a subtype of T2;
• >: in case T1 is a supertype of T2;
• “Neither” in case T1 is neither a supertype nor a supertype of T2.
What is the correct subtyping relationship between Iterable[Pair[A, Y]] => Y
and Map[A, Y] => X?

Iterable[Pair[A, Y]] => Y <: Map[A, Y] => X.

Explanation:

1. **Analyzing Iterable[Pair[A, Y]] => Y**:
   - The type `Iterable[Pair[A, Y]]` is covariant in its type parameter (denoted by `+U`), meaning if `A >: B`, then `Iterable[Pair[A, Y]]` is a subtype of `Iterable[Pair[B, Y]]`.
   - The return type `Y` does not have variance annotations, so it is invariant.

2. **Analyzing Map[A, Y] => X**:
   - The type `Map[A, Y]` is a subtype of `Iterable[Pair[A, Y]]` because `Map[U, +V]` extends `Iterable[Pair[U, V]]` with `U` as the first parameter and covariant `V`.
   - Given the function type `Map[A, Y] => X`, the return type `X` is invariant.

3. **Comparing the two**:
   - Since both types involve the same type parameters and considering the variance implications, we find that `Iterable[Pair[A, Y]]` can be considered a more general case which allows for elements of type `Pair[A, Y]` to be iterated over, while `Map[A, Y]` is more specific as it follows the structure of a mapping.
   - Therefore, `Iterable[Pair[A, Y]] => Y` can be seen as a subtype of `Map[A, Y] => X`, leading to the conclusion that `Iterable[Pair[A, Y]] => Y <: Map[A, Y] => X`.


Question: Let $n$ be an integer such that $n\geq 2$ and let  $A \in \R^{n	imes n}$, and $xv \in \R^n$, consider the function $f(xv) = xv^	op A xv$ defined over $\R^n$. Which of the following is the gradient of the function $f$?

To find the gradient of the function \\( f(xv) = xv^T A xv \\), we will use the properties of matrix calculus.

1. First, we note that \\( f(xv) \\) is a scalar function that depends on the vector \\( x \\) and the matrix \\( A \\).
2. The function can be rewritten as \\( f(x) = x^T A x \\), where \\( v \\) is fixed and we treat \\( xv \\) as the variable \\( x \\).
3. The gradient of a quadratic form \\( x^T A x \\) with respect to \\( x \\) is given by \\( (A + A^T)x \\). However, since \\( A \\) is not necessarily symmetric, we must account for both \\( A \\) and its transpose.
4. We have \\( \\nabla f(x) = A^T x + Ax \\).
5. In our case, since \\( x \\) is replaced by \\( xv \\), we need to account for the scalar multiplication by \\( v \\) when differentiating. Therefore, we have \\( (A^T (xv) + A(xv))v \\).
6. Simplifying this, we find that the terms combine to yield the final gradient.

After evaluating these steps, we identify that the correct answer is:

**Correct Answer: 3. \\( A^T xv + Axv \\)**


Question: A hash function $h$ is collision-resistant if\dots

A hash function \( h \) is collision-resistant if **it is hard to find \( x, x' \) such that \( x \ne x' \) and \( h(x) = h(x') \)**. This means that it should be computationally infeasible to find two distinct inputs that produce the same hash output.

Thus, the correct option is:
- **"\dots it is hard to find \( x,x' \) such that \( x \ne x' \) and \( h(x) = h(x') \)"**.


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)?

In this scenario, we have a ranked retrieval system where at position \( k \) the document is non-relevant and at position \( k+1 \), the document is relevant. We need to evaluate the implications on precision and recall for the options provided.

### Selected Option
**Correct Option: 'R@k-1 < R@k+1'**

### Rationale for the Correct Option
1. **Understanding Recall (R@k)**: Recall is defined as the ratio of relevant documents retrieved to the total number of relevant documents. In this case:
   - At position \( k \), since there's a non-relevant document, the recall is based on the relevant documents that have been retrieved among the top \( k \) documents.
   - At position \( k+1 \), when the relevant document is included, the recall will increase since now one more relevant document is retrieved.

2. **Mathematical Representation**:
   - Let \( R \) be the total number of relevant documents in the collection.
   - Let \( r_k \) be the number of relevant documents retrieved in the top \( k \) documents.
   - Let \( r_{k+1} \) be the number of relevant documents retrieved in the top \( k+1 \) documents.

   Given that the document at \( k \) is non-relevant, we have:
   - \( r_k = r_{k-1} \) (where \( r_{k-1} \) is the relevant documents retrieved in the top \( k-1 \) documents).
   - \( r_{k+1} = r_k + 1 = r_{k-1} + 1 \) (because we are adding one relevant document at \( k+1 \)).

   Therefore:
   - \( R@k = \frac{r_k}{R} = \frac{r_{k-1}}{R} \)
   - \( R@k+1 = \frac{r_{k+1}}{R} = \frac{r_{k-1} + 1}{R} \)

   Clearly, \( R@k-1 < R@k+1 \) because the numerator increases by 1 while the denominator remains constant.

### Explanation of Incorrect Options
1. **'P@k-1 > P@k+1'**:
   - **Precision (P@k)** is defined as the ratio of relevant documents retrieved to the total number of documents retrieved. The addition of a relevant document at \( k+1 \) does not change the number of documents retrieved for \( P@k \) (which is \( k \)), but it does increase the count of relevant documents (the numerator increases). Hence, \( P@k+1 \) could potentially be greater or lesser than \( P@k \) depending on the total relevant document count up to that point. Thus, this statement is not necessarily true.

2. **'P@k-1 = P@k+1'**:
   - This statement implies that the precision before and after the addition of a relevant document remains the same. However, with the addition of a relevant document at \( k+1 \), the precision will generally change (usually increase), so this is also false.

3. **'R@k-1 = R@k+1'**:
   - This statement suggests that recall remains constant when moving from \( k \) to \( k+1 \). Given that we are including an additional relevant document at \( k+1 \), the recall must increase; hence this statement is incorrect.

### Conclusion
The correct option is **'R@k-1 < R@k+1'** because adding a relevant document at position \( k+1 \) increases the recall as it increases the count of relevant documents retrieved while the total relevant documents in the collection remain unchanged. All other options are incorrect due to the implications of adding a relevant document on precision and recall calculations.


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

Let’s evaluate each of the statements one by one:

1. **Concolic execution specializes in finding bugs close to the path of the provided concrete input.**
   - **Evaluation**: This statement is correct. Concolic execution (concrete + symbolic execution) indeed leverages concrete inputs to explore execution paths and can effectively find bugs that are closely related to those paths.

2. **In static analysis, determining the correct target set of indirect calls is typically challenging but required to minimize over-approximation of targets.**
   - **Evaluation**: This statement is also correct. Indirect calls in programs (like function pointers in C/C++) make static analysis challenging because the target of the call isn't known until runtime. Accurately determining these targets is essential to avoid over-approximating possible program behaviors.

3. **Compared to dynamic analysis, static analysis is more susceptible to state space explosion.**
   - **Evaluation**: This statement is incorrect. Static analysis often deals with large state spaces due to the need to analyze all possible program states, which can lead to state space explosion. However, dynamic analysis typically executes the program on specific inputs and explores fewer paths, making it less susceptible to state space explosion in practice.

4. **Tests prove the presence of bugs but not their absence.**
   - **Evaluation**: This statement is correct. If a test fails, it indicates that there is a bug present; however, passing tests do not guarantee that there are no bugs in the codebase.

5. **Tests prove the absence of bugs but not their presence.**
   - **Evaluation**: This statement is incorrect. Similar to statement 4, tests cannot definitively prove the absence of bugs. They can only indicate that certain conditions or behaviors are functioning as expected at the time of testing.

Based on the evaluations:

- **Statements 1, 2, and 4 are correct.**
- **Statement 3 is incorrect.**
- **Statement 5 is incorrect.**

The best choice among these options is therefore **1, 2, and 4**, as they all present accurate information about testing and analysis methods. If you need to choose only one statement to represent correctness, statement 4 ("Tests prove the presence of bugs but not their absence") is particularly significant as it reflects a fundamental principle of software testing.


Question:
Assume you are working on a mobile application. You get complaints from Android users: when rotating the phone, the text they had typed disappears.
In one sentence, explain what the likely root cause is.

The likely root cause of the issue is that the activity is being recreated on configuration changes, such as screen rotation, and the input field states are not being saved and restored properly. When an Android device is rotated, the system triggers a configuration change that results in the current activity being destroyed and a new instance being created. By default, this destruction and recreation lead to the loss of any transient data, like text input in fields, unless developers explicitly manage state preservation using methods like `onSaveInstanceState()` to save the input data and `onRestoreInstanceState()` to restore it, or by using `ViewModel` to retain data across configuration changes. If these methods are not implemented correctly, users will experience the disappearance of text upon rotation.


Question:
The objective of this question is to illustrate the use of a lexical semantics resource to compute
lexical cohesion.
Consider the following toy ontology providing a semantic structuring for a (small) set of nouns:

We want to use lexical cohesion to decide whether the provided text consists of one single
topical segment corresponding to both sentences, or of two distinct topical segments, each
corresponding to one of the sentences.
Let's define the lexical cohesion of any set of words (in canonical form) as the average lexical
distance between all pairs of words present in the set2. The lexical distance between any two words is be defined as the length of a shortest path between the two words in the available
ontology.
For example, 'freedom' and 'happiness' are at distance 2 (length, i.e. number of links, of the
path: happiness −→ abstract entities −→ freedom), while 'freedom' and 'dog' are at distance
6 (length of the path: freedom −→ abstract entities −→ non animate entities −→ all −→
animate entities −→ animals −→ dog)
Compute the lexical distance between all the pairs of words present in the above text and in
the provided ontology (there are 6 such pairs).

To compute the lexical distances between the pairs of nouns identified in the question (cat, dog, pen, table), we will analyze the ontology and trace the paths between each pair of words.

### Pairs of Words
The six unordered pairs of words are:
1. (cat, dog)
2. (cat, pen)
3. (cat, table)
4. (dog, pen)
5. (dog, table)
6. (pen, table)

### Lexical Distances

1. **D(cat, dog)**
   - Path: cat → animals → animate entities → all → animate entities → dog
   - Steps:
     - cat to animals (1 edge)
     - animals to animate entities (1 edge)
     - animate entities to all (1 edge)
     - all to animate entities (1 edge)
     - animate entities to dog (1 edge)
   - Total distance = 4 edges

2. **D(cat, pen)**
   - Path: cat → animals → animate entities → all → non animate entities → concrete entities → pen
   - Steps:
     - cat to animals (1 edge)
     - animals to animate entities (1 edge)
     - animate entities to all (1 edge)
     - all to non animate entities (1 edge)
     - non animate entities to concrete entities (1 edge)
     - concrete entities to pen (1 edge)
   - Total distance = 6 edges

3. **D(cat, table)**
   - Path: cat → animals → animate entities → all → non animate entities → concrete entities → table
   - Steps:
     - cat to animals (1 edge)
     - animals to animate entities (1 edge)
     - animate entities to all (1 edge)
     - all to non animate entities (1 edge)
     - non animate entities to concrete entities (1 edge)
     - concrete entities to table (1 edge)
   - Total distance = 6 edges

4. **D(dog, pen)**
   - Path: dog → animals → animate entities → all → non animate entities → concrete entities → pen
   - Steps:
     - dog to animals (1 edge)
     - animals to animate entities (1 edge)
     - animate entities to all (1 edge)
     - all to non animate entities (1 edge)
     - non animate entities to concrete entities (1 edge)
     - concrete entities to pen (1 edge)
   - Total distance = 6 edges

5. **D(dog, table)**
   - Path: dog → animals → animate entities → all → non animate entities → concrete entities → table
   - Steps:
     - dog to animals (1 edge)
     - animals to animate entities (1 edge)
     - animate entities to all (1 edge)
     - all to non animate entities (1 edge)
     - non animate entities to concrete entities (1 edge)
     - concrete entities to table (1 edge)
   - Total distance = 6 edges

6. **D(pen, table)**
   - Path: pen → concrete entities → non animate entities → all → non animate entities → concrete entities → table
   - Steps:
     - pen to concrete entities (1 edge)
     - concrete entities to non animate entities (1 edge)
     - non animate entities to all (1 edge)
     - all to non animate entities (1 edge)
     - non animate entities to concrete entities (1 edge)
     - concrete entities to table (1 edge)
   - Total distance = 4 edges

### Final Results
Based on the calculations above, the distances for each pair are:

```
D(cat,dog)=4
D(cat,pen)=6
D(cat,table)=6
D(dog,pen)=6
D(dog,table)=6
D(pen,table)=4
```


Question: Which of the following statements about index merging (when constructing inverted files) is correct?

Let's evaluate each option one by one:

1. **'While merging two partial indices on disk, the inverted lists of a term are concatenated without sorting'**:
   - **Incorrect**: When merging inverted lists during index merging, the lists for the same term are not simply concatenated. Instead, they are typically merged in a sorted fashion to ensure that the resulting list is ordered and free of duplicates. This makes it easier to search through the index efficiently.

2. **'Index merging is used when the vocabulary does no longer fit into the main memory'**:
   - **Correct**: This statement accurately reflects one of the primary reasons for index merging. When the vocabulary (the set of unique terms) becomes too large to fit into memory, the system offloads some of this data to disk and merges partial indices to create a more manageable size for future queries, thereby optimizing performance.

3. **'The size of the final merged index file is O(n log2(n) M) where M is the size of the available memory'**:
   - **Incorrect**: This statement is not accurate. The complexity of the final merged index file size is typically more related to the number of unique terms and the number of documents rather than a direct relationship with memory size. The size of indexes can depend on various factors, but this expression does not correctly represent the behavior of index merging.

4. **'While merging two partial indices on disk, the vocabularies are concatenated without sorting'**:
   - **Incorrect**: Similar to the first option, this statement is misleading. Merging vocabularies during the index merging process involves more than just concatenation; it requires careful handling to ensure that the final vocabulary is sorted and does not contain duplicates.

Based on this analysis, the correct answer is:

**'Index merging is used when the vocabulary does no longer fit into the main memory'**.

This statement accurately describes a situation where index merging is necessary, as it highlights a practical consideration in the construction of inverted files.


Question: What is your take on the accuracy obtained in an unballanced dataset? Do you think accuracy is the correct evaluation metric for this task? If yes, justify! If not, why not, and what else can be used?

When evaluating a model on an unbalanced dataset, accuracy is often not a suitable metric. Let's break down the reasons for this and explore alternative metrics that provide a more meaningful evaluation.

### Understanding Unbalanced Datasets

An unbalanced dataset is one where the classes are not represented equally. For example, in a binary classification problem, if 95% of the instances belong to class A and only 5% belong to class B, this creates a significant imbalance.

### The Problem with Accuracy

Now, imagine we have a model that simply predicts the majority class (class A) for every instance. In this case, if the model always predicts class A, it would achieve an accuracy of 95%. However, this high accuracy is misleading because the model fails to identify any instances of class B, which is often the class of interest.

### Why Accuracy is Misleading

1. **Lack of Information**: Accuracy does not provide insight into how well the model performs across all classes. A model can have high accuracy but still be ineffective in identifying the minority class.

2. **Class Importance**: In many applications, the minority class may represent critical cases (e.g., fraud detection, disease diagnosis). The cost of misclassifying a minority class instance is often much higher than misclassifying a majority class instance, which accuracy does not account for.

### Alternative Evaluation Metrics

Given the limitations of accuracy, we should look for metrics that provide a more nuanced evaluation of model performance, particularly for unbalanced datasets. Here are some alternatives:

1. **Precision, Recall, and F1-Score**:
   - **Precision**: Measures the correctness of positive predictions (true positives / (true positives + false positives)). It tells us how many of the predicted positives are actually positive.
   - **Recall (Sensitivity)**: Measures the ability to find all the positive instances (true positives / (true positives + false negatives)). It shows how many of the actual positives were correctly predicted.
   - **F1-Score**: The harmonic mean of precision and recall. It provides a balance between the two metrics and is particularly useful when you need a single score that captures both aspects.

2. **Area Under the Receiver Operating Characteristic Curve (ROC-AUC)**:
   - This metric evaluates the model's ability to distinguish between classes across various threshold settings. A higher AUC indicates better performance in distinguishing the positive class from the negative class.

3. **Confusion Matrix**:
   - This provides a comprehensive breakdown of the model’s predictions, showing true positives, true negatives, false positives, and false negatives. Analyzing the confusion matrix allows for deeper insights into where the model is making errors.

4. **Cohen’s Kappa**:
   - This statistic measures inter-rater reliability and accounts for the agreement occurring by chance. It is particularly useful in unbalanced scenarios, as it considers the true positives, false positives, and false negatives.

### Conclusion

In summary, while accuracy seems like a straightforward metric, it is not appropriate for evaluating models trained on unbalanced datasets due to its potential to misrepresent model performance. Instead, metrics like precision, recall, F1-score, ROC-AUC, and confusion matrices provide a richer understanding of how well the model is performing, especially for the minority class. Always consider the specific context and implications of misclassifications when choosing evaluation metrics, as this will guide you in selecting the most appropriate ones for your task.


Question:
In which class of processors do you expect to find reservation
  stations?

Answer:

**Reservation stations are found in dynamically scheduled out-of-order processors.**

### Intuition:

The intuition behind this design is to optimize instruction execution and resource utilization. By allowing instructions to be dynamically scheduled and to wait in reservation stations for their operands, the processor can keep its execution units busy, reduce pipeline stalls, and ultimately improve overall throughput. This dynamic approach is essential for modern processors aiming to achieve high performance in complex workloads.


Question:
Which of AddressSanitizer (ASan), MemorySanitizer (MemSan),
    UndefinedBehaviorSanitizer (UBSan) or ThreadSanitizer (TSan) can detect
    bugs (if any) in the following code snippet?

    \begin{lstlisting}[language=C,style=c]
int sum_up_to(int x) {}
    // Return sum of integers up to x
    int result = x;
    for (int i = x; i >= 0; i--) {
        if (INT_MAX - i <= result) {
            break;
        }
        result += i;
    }
    return result;
}
    \end{lstlisting}

- The code contains a logical error where the loop condition and calculations lead to an incorrect result rather than a memory error, which is outside the scope of what ASan, MemSan, TSan, or UBSan typically detect.
- AddressSanitizer (ASan) focuses on detecting memory corruption issues like buffer overflows, which are not present in this code snippet.
- MemorySanitizer (MemSan) is used to detect uninitialized memory usage, but this code does not involve uninitialized memory, thus it cannot detect any issues.
- UndefinedBehaviorSanitizer (UBSan) is used for detecting undefined behaviors in C/C++ code, but the logic in the code does not invoke undefined behavior that UBSan would catch, while ThreadSanitizer (TSan) focuses on data races and threading issues, which are irrelevant here.

Answer: (1)


Question: How many integers $n$ between $1$ and $2021$ satisfy $10^n \equiv 1 \mod 11$? Check the correct answer.

To solve the problem, we need to find integers \( n \) such that \( 10^n \equiv 1 \mod 11 \).

**Finding the Order of \( 10 \mod 11 \)**:
The order of an integer \( a \) modulo \( m \) is the smallest positive integer \( d \) such that \( a^d \equiv 1 \mod m \). We calculate:

- \( 10^1 \equiv 10 \mod 11 \)
- \( 10^2 \equiv 100 \equiv 1 \mod 11 \)

Thus, the order of \( 10 \) modulo \( 11 \) is \( 2 \). This indicates \( 10^n \equiv 1 \mod 11 \) when \( n \) is even.

**Counting Even Integers**:
Now, we count the even integers between \( 1 \) and \( 2021 \), which form the sequence \( 2, 4, 6, \ldots, 2020 \).

The even integers can be expressed as \( n = 2k \) for \( k = 1, 2, \ldots, 1010 \). The largest \( k \) such that \( 2k \leq 2021 \) is \( 1010 \).

Therefore, the total number of integers \( n \) between \( 1 \) and \( 2021 \) that satisfy \( 10^n \equiv 1 \mod 11 \) is \( \boxed{1010} \).


Question: A proof system is computational-zero-knowledge if \dots

The correct answer is **3**.

### Justification:
A computational zero-knowledge proof system is defined by the existence of a probabilistic polynomial-time (PPT) simulator \( S \) such that for any PPT verifier \( V \), the output of \( S \) is computationally indistinguishable from the view of the verifier when the proof is executed honestly by the prover. This means that no efficient adversary can distinguish between the simulator's output and the actual interaction between the prover and the verifier, given polynomial time.

**Analysis of the Options:**

1. **Option 1:** "for any PPT verifier and for any simulator S, S produces an output which is hard to distinguish from the view of the protocol."
   - This is incorrect because it doesn't specify that the simulator \( S \) needs to be efficient (PPT). The requirement must be that there exists a specific PPT simulator \( S \) for any PPT verifier, not that any arbitrary simulator \( S \) should work.

2. **Option 2:** "there exists a PPT simulator S such that for any honest verifier, S produces an output which is hard to distinguish from the view of the verifier."
   - This option implies that the simulator's indistinguishability holds only for honest verifiers. However, the definition of computational zero-knowledge requires that the indistinguishability holds for any PPT verifier, not just honest ones.

3. **Option 3:** "for any PPT verifier, there exists a PPT simulator that produces an output which is hard to distinguish from the view of the protocol."
   - This is correct as it captures the essence of computational zero-knowledge. It states that for every efficient verifier, we can construct an efficient simulator that can simulate the view of the interaction, ensuring that no verifier can distinguish between the real interaction and the simulation.

4. **Option 4:** "for any honest verifier and for any simulator S, S produces an output which is hard to distinguish from the view of the protocol."
   - This option is incorrect because it suggests that any simulator \( S \) (not necessarily a PPT simulator) can be used, and it also restricts the condition to only honest verifiers. The requirement is that there exists a specific efficient simulator, not that any simulator would suffice.

In summary, option 3 correctly captures the definition of a computational zero-knowledge proof system by emphasizing the existence of a PPT simulator for any PPT verifier.


Question: Let $n$ be any positive integer. Three of the following assertions are equivalent. Tick the remaining one.

To analyze the assertions:

1. **$\mathbb{Z}_n$ is a field**: This is true if and only if $n$ is a prime number.
2. **$\varphi(n) = n - 1$**: This holds if $n$ is a prime number, as for a prime $p$, $\varphi(p) = p - 1$.
3. **$n$ is a prime power**: This means $n = p^k$ for some prime $p$ and integer $k \geq 1$. This is not equivalent to the others since $n$ being a prime power does not guarantee that $\mathbb{Z}_n$ is a field.
4. **Any element $x \in \mathbb{Z}_n \backslash \{0\}$ is invertible**: This is also true if and only if $n$ is a prime number.

Step-by-step reasoning:
- The first three statements can be shown to hold true for prime numbers, while the third statement can include composite numbers that are prime powers, making it the odd one out.

The assertion that does not fit with the others is **$n$ is a prime power**.


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, $H(S_1,\ldots,S_n) = \sum_{i=1}^n H(S_i|S_1\ldots S_{i-1})$.

To determine whether the statement is true or false, we need to analyze the entropy \( H(S_1, S_2, \ldots, S_n) \) in the context of the given loaded dice.

The entropy of a random variable quantifies the uncertainty associated with it. In this case, \( S_1, S_2, \ldots, S_n \) represent the outcomes of rolling the loaded dice, which is defined as follows:

- The probability \( P(S_1 = 6) = \frac{5}{6} \)
- The probability \( P(S_1 = x) = \frac{1}{30} \) for \( x \in \{1, 2, 3, 4, 5\} \)

The probabilities sum to 1:

\[
P(S_1 = 6) + P(S_1 = 1) + P(S_1 = 2) + P(S_1 = 3) + P(S_1 = 4) + P(S_1 = 5) = \frac{5}{6} + 5 \cdot \frac{1}{30} = \frac{5}{6} + \frac{5}{30} = \frac{5}{6} + \frac{1}{6} = 1
\]

Now, considering the sequence \( S = S_1 S_2 S_3 \ldots \), we notice that each roll of the dice is independent of the previous rolls. This means that the entropy of the joint distribution \( H(S_1, S_2, \ldots, S_n) \) can be expressed as the sum of the entropies of the individual random variables:

\[
H(S_1, S_2, \ldots, S_n) = H(S_1) + H(S_2) + \ldots + H(S_n)
\]

Since each \( S_i \) is independent, we can also express this as:

\[
H(S_i | S_1, \ldots, S_{i-1}) = H(S_i)
\]

Therefore, we can rewrite the joint entropy as:

\[
H(S_1, \ldots, S_n) = \sum_{i=1}^n H(S_i | S_1, \ldots, S_{i-1}) = \sum_{i=1}^n H(S_i)
\]

Thus, the statement \( H(S_1, \ldots, S_n) = \sum_{i=1}^n H(S_i | S_1 \ldots S_{i-1}) \) is true because it correctly represents the relationship between the joint entropy of independent random variables and their conditional entropies.

**Answer: True**


Question: Let $(G,+), (H,\times)$ be two groups and $f:G\to H$ be an homomorphism. For $x_1,x_2 \in G$, we have:

B. The correct property of the homomorphism is preserved in option (B).


Question: The DES key schedule\dots

The DES key schedule generates 16 subkeys from a 56-bit key, which are used in each round of the Feistel structure during both encryption and decryption. This process involves permuting and shifting the key bits to produce the necessary subkeys for the algorithm.


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

We will consider the following three messages:

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

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

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

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

We now want to specifically focus on the processing of compounds such as 'network capacity' in
the second text.

How are the compounds handled by a Naive Bayes classifier if no specific pre-processing of
compounds is used?

In a Naive Bayes classifier, compounds like "network capacity" are treated as separate words, such as "network" and "capacity," because the classifier assumes that all words are independent of each other. This means it ignores the combined meaning of the compound, relying instead on individual word probabilities for classification.


Question: In the context of Load Store Queue, How does the LSQ gets ordering information?

The Load Store Queue (LSQ) is a critical component in modern processors that helps manage memory operations efficiently. One might think that the LSQ relies solely on the program order of instructions as they are fetched and decoded. However, it’s important to recognize that the LSQ also interacts with various other components of the CPU, such as the instruction scheduler and the execution units. This interaction can lead to a misunderstanding of how ordering information is maintained within the LSQ.

For instance, one might argue that the LSQ uses complex algorithms to determine the optimal order for executing memory instructions, allowing it to prioritize certain operations over others based on their type or anticipated execution time. This could lead one to think that the LSQ is dynamically reordering instructions based on performance metrics, rather than adhering strictly to the original program order.

Moreover, there is a common misconception that the LSQ can independently track dependencies and conflicts between memory operations without considering the broader context of how these operations fit into the entire pipeline. While the LSQ does track some level of dependency information, it is still fundamentally tied to the program order established during the fetch and decode stages. Thus, while it may seem that the LSQ has a sophisticated system of ordering, it is primarily the program order that governs how instructions are handled, with various optimizations occurring at different stages of instruction execution. This leads to an off-topic reasoning about the LSQ's function that may divert attention from its true reliance on program order.


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 $\xv\in\mathcal X$:
        egin{align*}
                g^\star(\xv)\in rg \min_{z\in\R}\mathbb E[\phi( z Y)|X=\xv].
        \end{align*}
        Thus the function $g^\star$ that minimizes the $\phi$-risk can be determined by looking at each $\xv$ separately.

        Give a formula of the function $g^\star : \mathcal X	o\R$ which minimizes the true $\phi$-risk, as a function of $\eta(\xv)$.

To derive the optimal predictor \( g^\star: \mathcal{X} \to \mathbb{R} \) that minimizes the true \(\phi\)-risk under the given \(0-1\) loss scenario, we start by recognizing the role of the conditional expectation and the function \(\eta(\xv)\).

The function \(\eta(\xv)\) represents the conditional probability of the positive class given the input \(\xv\), specifically:

\[
\eta(\xv) = \mathbb{P}(Y = 1 | X = \xv).
\]

The optimal predictor \( g^\star(\xv) \) is defined such that it minimizes the expected loss \( \mathbb{E}[\phi(zY) | X = \xv] \) for each \(\xv\). This leads to minimizing the expression for all potential outcomes \( z \in \mathbb{R} \).

For a general loss function \( \phi \), the optimal prediction can be formulated in terms of the conditional expectation of \( Y \) given \( X = \xv \). Specifically, when minimizing the \(\phi\)-risk, we find that:

\[
g^\star(\xv) = \arg \min_{z \in \mathbb{R}} \left( \eta(\xv) \phi(z) + (1 - \eta(\xv)) \phi(-z) \right).
\]

In cases where \(\phi\) is the 0-1 loss, this leads to a straightforward decision boundary wherein \( g^\star(\xv) \) predicts the class label that minimizes misclassification.

Therefore, the formula for the function \( g^\star \) that minimizes the true \(\phi\)-risk can be succinctly expressed as:

\[
g^\star(\xv) = \begin{cases}
1, & \text{if } \eta(\xv) \geq 0.5, \\
0, & \text{if } \eta(\xv) < 0.5.
\end{cases}
\]

This means that at each point \( \xv \), the optimal predictor \( g^\star(\xv) \) will yield the predicted class based on whether the conditional probability \( \eta(\xv) \) exceeds the threshold of 0.5, ensuring that the predictions minimize the expected \(\phi\)-risk effectively.


Question: In ElGamal signature scheme, if we avoid checking that $0 \leq r < p$ then \ldots

1. \ldots a universal forgery attack is possible.

**Explanation:** In the ElGamal signature scheme, the value \( r \) is one of the components of the signature. If we do not enforce the condition \( 0 \leq r < p \), it may allow an attacker to produce a valid signature for any message without knowing the secret key. This is because the invalid \( r \) values could lead to predictable outcomes that can be manipulated to forge signatures universally for any message. Hence, failing to check this condition can compromise the security of the signature scheme, allowing for universal forgery attacks.


Question: In the context of Load Store Queue, What conditions must be satisfied in the LSQ for a load to be executed and the result to be returned to the processor?

For a load to be executed in a Load Store Queue (LSQ) and for the result to be returned to the processor, the primary condition that must be satisfied is that all prior stores to the same memory address must be completed. This ensures that the data being loaded is the most current and consistent version. Additionally, the load instruction must be in a ready state, meaning that it should not be dependent on any unresolved prior loads or stores. This guarantees that the memory access is coherent and adheres to the program's intended order of operations.


Question: Select valid answers about file descriptors (FD):

Let's analyze each option regarding file descriptors (FD) and identify the correct answers while explaining why the other options are misleading.

### Correct Options:
**2. FD is usually used as an argument for read and write.**
- **Justification:** This statement is correct. File descriptors are integer handles that represent open files or other I/O resources in an operating system. When performing operations like reading from or writing to a file, the file descriptor is passed as an argument to functions like `read()` and `write()`. This allows the operating system to know which file or resource the operation should be performed on.

**4. FDs are preserved after fork() and can be used in the new process pointing to the original files.**
- **Justification:** This statement is also correct. When a process calls `fork()`, the operating system creates a new process (the child) that is a duplicate of the original process (the parent). The child process inherits copies of the parent's file descriptors. This means that both processes can use the same file descriptors to access the same open files, allowing for shared access to those resources.

### Misleading Options:
**1. The value of FD is unique for every file in the operating system.**
- **Misleading Explanation:** This statement is incorrect because the value of a file descriptor is unique only within the context of a single process. Different processes can have the same file descriptor values that refer to different files. For example, if Process A has file descriptor 3 for File X and Process B also has file descriptor 3 for File Y, these file descriptors are not unique across the operating system; they are unique only within their respective processes.

**3. FD is constructed by hashing the filename.**
- **Misleading Explanation:** This statement is incorrect. File descriptors are not constructed by hashing the filename. Instead, they are simply integer values assigned by the operating system when a file is opened. The operating system maintains a table of open files, and each file descriptor corresponds to an entry in this table. The assignment of file descriptors is typically sequential, starting from 0 (standard input) and increasing for each new file opened, rather than being derived from a hash of the filename.

### Summary:
The correct answers are options 2 and 4, as they accurately describe the role and behavior of file descriptors in the context of file operations and process management. Options 1 and 3 are misleading because they misrepresent the uniqueness of file descriptors and the method of their creation.


Question:
What is the problem addressed by a Part-of-Speech (PoS) tagger?
Why isn't it trivial? What are the two main difficulties?

To address the question regarding the problem tackled by a Part-of-Speech (PoS) tagger, we will follow the structured approach outlined in your request.

### 1. Key Concepts and Principles

**Part-of-Speech Tagging**: This is the process of assigning a part of speech to each word in a sentence, such as noun, verb, adjective, etc. The goal is to understand the grammatical structure of the sentence.

**Natural Language Processing (NLP)**: PoS tagging is a fundamental task in NLP, which involves the interaction between linguistics and computer science to enable machines to understand human language.

### 2. Theoretical Framework

The theoretical framework for PoS tagging involves:

- **Linguistic Theory**: Understanding the rules and structures of language, including syntax and semantics.
- **Statistical Models**: Many modern PoS taggers use statistical methods, such as Hidden Markov Models (HMMs) or Conditional Random Fields (CRFs), to predict the most likely tag for each word based on context.
- **Machine Learning**: Supervised learning techniques are often employed, where a model is trained on a labeled dataset containing sentences with their corresponding PoS tags.

### 3. Step-by-Step Solution

#### Step 1: Identify the Problem

The primary problem addressed by a PoS tagger is to accurately assign grammatical categories to words in a sentence. This is not trivial due to the inherent complexity of natural language.

#### Step 2: Discuss Why It Isn't Trivial

1. **Ambiguity**:
   - Many words can serve as different parts of speech depending on context. For example, the word "bark" can be a noun (the sound a dog makes) or a verb (the action of a dog making that sound). This ambiguity requires the tagger to consider surrounding words to determine the correct tag.

2. **Contextual Dependencies**:
   - The meaning and function of a word can change based on its position in a sentence and the words that precede or follow it. For example, in the phrase "I can fish," "can" is a modal verb, while in "I can do it," "can" is part of a verb phrase. The tagger must analyze the entire sentence structure to make accurate assignments.

#### Step 3: Highlighting Important Assumptions

- **Language-Specific Rules**: The tagger may rely on specific grammatical rules that apply to the language being processed (e.g., English, French).
- **Training Data**: The effectiveness of a PoS tagger is heavily dependent on the quality and quantity of the annotated training data it is exposed to.

### 4. Key Insights from the Solution

- **Complexity of Natural Language**: The dual challenges of ambiguity and contextual dependencies illustrate the complexity of natural language, which is not easily reducible to simple rules.
- **Importance of Context**: Effective PoS tagging relies on understanding the context in which words appear, highlighting the need for sophisticated models that can capture these relationships.

### 5. Alternative Approaches

- **Rule-Based Tagging**: Earlier PoS taggers used hand-crafted rules based on linguistic knowledge. While they can be accurate, they often lack the flexibility and scalability of statistical methods.
- **Deep Learning**: Recent advancements have seen the use of neural networks, particularly recurrent neural networks (RNNs) and transformers, which can capture long-range dependencies and contextual information more effectively than traditional models.

### Conclusion

In summary, a Part-of-Speech tagger addresses the complex problem of assigning grammatical categories to words in a sentence, which is complicated by ambiguity and contextual dependencies. Understanding these challenges is crucial for developing effective NLP applications. The evolution of tagging methods from rule-based to statistical and now to deep learning approaches reflects the ongoing efforts to tackle these challenges in natural language processing.


Question: In a game of Othello (also known as Reversi in French-speaking countries), when a player puts a token on a square of the board, we have to look in all directions around that square to find which squares should be “flipped” (i.e., be stolen from the opponent). We implement this in a method computeFlips, taking the position of square, and returning a list of all the squares that should be flipped: final case class Square(x: Int, y: Int) def computeFlips(square: Square): List[Square] = { List(−1, 0, 1).flatMap{i => List(−1, 0, 1).filter{j => i != 0 || j != 0}.flatMap{j => computeFlipsInDirection(square, i, j)}}} def computeFlips In Direction (square: Square, dirX: Int, dirY: Int): List[Square] = {// omitted} Rewrite the method computeFlips to use one for comprehension instead of maps, flatMaps and filter s. The resulting for comprehension should of course have the same result as the expression above for all value of square. However, it is not necessary that it desugars exactly to the expression above.

To rewrite the `computeFlips` method using a `for` comprehension in Scala, we can simplify the nested structure. The main goal is to iterate over the possible directions defined by the combinations of `dirX` and `dirY`, excluding the (0, 0) case and applying `computeFlipsInDirection`. Here’s how you can achieve that:

```scala
def computeFlips(square: Square): List[Square] = {
  for {
    i <- List(-1, 0, 1)
    j <- List(-1, 0, 1) if !(i == 0 && j == 0) // Filter out (0, 0)
    flips <- computeFlipsInDirection(square, i, j) // Apply the method
  } yield flips
}
```

### Explanation:
1. **Iteration**: The `for` comprehension allows easy iteration over the lists `List(-1, 0, 1)` for both `i` and `j`.
2. **Filter**: The filter condition `if !(i == 0 && j == 0)` serves to exclude the case where both `i` and `j` are zero.
3. **Computation**: For each valid `(i, j)` pair, it computes the flips by calling `computeFlipsInDirection(square, i, j)`.
4. **Yield**: The results of the computation are accumulated into a list since the `yield` keyword collects them into the final resulting list.

This approach maintains the same functionality as your original method while using a more concise and readable comprehension structure.


Question: WEP \dots

Wired Equivalent Privacy (WEP) has been a topic of extensive scrutiny within the field of network security. Initially designed to provide a level of security comparable to that of wired networks, WEP aimed to offer confidentiality through encryption. However, the implementation of WEP is fundamentally flawed, rendering it ineffective in today's context.

Firstly, the key management within WEP is quite poor. The static keys make it easy for attackers to decipher the encrypted data. This leads to the realization that WEP does not provide good confidentiality. Furthermore, the integrity checks within WEP are based on the CRC-32 checksum, which is not robust against attacks. This results in a failure to ensure message integrity effectively. Similarly, the authentication mechanism of WEP relies on shared secrets that are susceptible to various attacks, thus failing to provide strong authentication.

As a result, the conclusion one can draw is that WEP is fundamentally compromised in multiple aspects. Therefore, the most accurate choice regarding WEP would be that it "is badly broken." The other options regarding confidentiality, message integrity, and authentication can be dismissed due to the inherent weaknesses present in its design.

In summary, WEP is critically flawed, and the most accurate assertion is that "WEP is badly broken."


Question: In this week's lecture, you have been introduced to the aggregate method of ParSeq[A] (and other parallel data structures). It has the following signature:  def aggregate[B](z: B)(f: (B, A) => B, g: (B, B) => B): B Discuss, as a group, what aggregate does and what its arguments represent. Consider the parallel sequence xs containing the three elements x1, x2 and x3. Also consider the following call to aggregate:  xs.aggregate(z)(f, g) The above call might potentially result in the following computation:  f(f(f(z, x1), x2), x3) But it might also result in other computations. Come up with at least two other computations in terms of f and g that may result from the above call to aggregate.

Sure! Let's dive into the `aggregate` method of `ParSeq[A]` and understand what it does, along with its parameters and how it operates.

### Overview of `aggregate`

The `aggregate` method is designed to perform a parallel reduction on a collection of elements (in this case, a parallel sequence of type `A`). It allows us to combine elements of the collection in a way that can take advantage of parallel computation, which can lead to improved performance for large datasets.

The method has the following signature:

```scala
def aggregate[B](z: B)(f: (B, A) => B, g: (B, B) => B): B
```

### Parameters Explained

1. **`z: B`**: This is the initial value (often called the "zero" or "identity" value) of type `B`. It serves as the starting point for the aggregation process.

2. **`f: (B, A) => B`**: This is a binary function that takes an accumulator of type `B` and an element of type `A`, and returns a new accumulator of type `B`. This function defines how to combine an element of the collection into the accumulated result.

3. **`g: (B, B) => B`**: This is another binary function that takes two accumulators of type `B` and combines them into one. This function is used to merge results from different parallel computations.

### How `aggregate` Works

The `aggregate` method splits the collection into smaller chunks, processes each chunk in parallel using the function `f`, and then merges the results of these chunks using the function `g`. This can lead to various ways of computing the final result based on the order of operations.

### Example Computations

Letâs say we have a parallel sequence `xs` containing three elements: `x1`, `x2`, and `x3`. When we call `xs.aggregate(z)(f, g)`, we can visualize it as follows:

1. **Sequential Aggregation**:
   - One possible computation could be:
     ```scala
     f(f(f(z, x1), x2), x3)
     ```
   - Here, we start with `z`, combine it with `x1` using `f`, then take that result and combine it with `x2`, and finally combine that result with `x3`.

2. **Alternative Aggregation Order 1**:
   - Another possible computation could be:
     ```scala
     f(f(z, x1), f(z, x2)) // followed by merging with g
     ```
   - In this case, we first compute two separate reductions: one combining `z` with `x1`, and the other combining `z` with `x2`. Then we would merge these two intermediate results using `g`.

3. **Alternative Aggregation Order 2**:
   - Yet another possible computation might be:
     ```scala
     g(f(z, x1), f(f(z, x2), x3))
     ```
   - Here, we first compute the combination of `z` with `x2` and `x3`, and then merge that result with the combination of `z` with `x1`.

### Summary

In summary, the `aggregate` method allows for flexible aggregation of elements in a parallel sequence. The choice of how to apply the functions `f` and `g` can lead to different execution paths and potentially different results depending on how these operations are defined (especially if they are not associative).

### Final Answer

1. A possible computation is:
   ```scala
   f(f(f(z, x1), x2), x3)
   ```

2. Another possible computation is:
   ```scala
   f(f(z, x1), f(z, x2)) // followed by merging with g
   ```

3. A third possible computation could be:
   ```scala
   g(f(z, x1), f(f(z, x2), x3))
   ```

Understanding these variations helps you grasp how parallel computations can be structured and the importance of associativity in defining your functions for predictable outcomes.


Question: Consider a classification problem on linearly separable data. We train an SVM model and a logistic regression model. For logistic regression (LR) we add a small regularization term (penalty on weights) in order to make the optimum well-defined. Each model gives us a margin. Consider a datapoint $\mathbf{x}_{0}$ that is correctly classified and strictly outside both margins Which one of the following statements is incorrect ?

To analyze the statements given in the context of a classification problem with linearly separable data, let's consider the implications of $\mathbf{x}_{0}$ being correctly classified and outside the margins for both the logistic regression (LR) and support vector machine (SVM) models.

1. **Statement 1:** 'There exists a direction in which we can slightly move $\mathbf{x}_{0}$ without changing the LR decision boundary after retraining.'
   - This statement is **incorrect**. Since $\mathbf{x}_{0}$ is outside the margins and correctly classified, moving it in any direction could potentially bring it closer to the decision boundary, impacting the optimization of the logistic regression model. Thus, slight movements can change the decision boundary.

2. **Statement 2:** '$\mathbf{x}_{0}$ isn't a support vector.'
   - This statement is **true**. Support vectors are data points that lie on the margin or are incorrectly classified. Since $\mathbf{x}_{0}$ is outside the margins, it is not a support vector.

3. **Statement 3:** 'There exists a direction in which we can arbitrarily move $\mathbf{x}_{0}$ without changing the SVM decision boundary after retraining.'
   - This statement is **true**. As $\mathbf{x}_{0}$ is not a support vector and is far from the margin, it can be moved without affecting the decision boundary of the SVM.

4. **Statement 4:** 'If we remove $\mathbf{x}_{0}$ from the dataset and retrain, this will change the LR decision boundary.'
   - This statement is **true**. Removing a data point that is not a support vector can still affect the overall distribution of the data, thus impacting the decision boundary.

5. **Statement 5:** 'If we remove $\mathbf{x}_{0}$ from the dataset and retrain, this will not change the SVM decision boundary.'
   - This statement is **true**. Since $\mathbf{x}_{0}$ is not a support vector, its removal will not affect the SVM decision boundary.

In conclusion, the incorrect statement is **Statement 1**, as it falsely asserts that slight movements of $\mathbf{x}_{0}$ will not influence the LR decision boundary, which is not the case.


Question: Prove that if a^2 is even, a is even.

To prove that if \( a^2 \) is even, then \( a \) is also even, we can use a proof by contrapositive or direct proof. Here, I'll provide a direct proof.

### Proof:

1. **Assume that \( a^2 \) is even.**
   - By definition, an integer is even if it can be expressed as \( 2k \) for some integer \( k \). Therefore, we can write:
     \[
     a^2 = 2k
     \]
     for some integer \( k \).

2. **Consider the implications of \( a^2 \) being even.**
   - We can denote \( a \) as either an even integer or an odd integer. Let’s analyze both cases.

3. **Case 1: \( a \) is even.**
   - If \( a \) is even, we can express it as:
     \[
     a = 2m
     \]
     for some integer \( m \). Then,
     \[
     a^2 = (2m)^2 = 4m^2 = 2(2m^2)
     \]
     Since \( 2m^2 \) is an integer, \( a^2 \) is indeed even.

4. **Case 2: \( a \) is odd.**
   - If \( a \) is odd, we can express it as:
     \[
     a = 2n + 1
     \]
     for some integer \( n \). Then,
     \[
     a^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1
     \]
     Here, \( 2n^2 + 2n \) is also an integer, so we can see that:
     \[
     a^2 = 2(\text{integer}) + 1
     \]
     This shows that \( a^2 \) is odd (since it is of the form \( 2k + 1 \)).

5. **Conclusion:**
   - From Case 1, we see that if \( a \) is even, \( a^2 \) is even.
   - From Case 2, if \( a \) is odd, \( a^2 \) is odd.
   - Therefore, the only way for \( a^2 \) to be even is if \( a \) is even.

Thus, we conclude that if \( a^2 \) is even, then \( a \) must also be even.

\[
\boxed{\text{If } a^2 \text{ is even, then } a \text{ is even.}}
\]


Question: Consider the following set of frequent 3-itemsets: {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {2, 3, 4}, {2, 3, 5}, {3, 4, 5}. Which one is not a candidate 4-itemset?

Answer: {1,3,4,5}
 Reasoning: To determine which 4-itemset is not a candidate, we must find if it can be generated from the frequent 3-itemsets by the Apriori principle. The 4-itemset {1,3,4,5} cannot be formed since it requires the frequent subsets {1,3,4}, {1,3,5}, {1,4,5}, and {3,4,5} to all be frequent, but {1,3,5} is not present in the given frequent 3-itemsets, making {1,3,4,5} not a candidate.


Question: Design  a one-pass (streaming) algorithm that, for a stream that possesses a majority element (appearing more than $m/2$ times), terminates with this element. Prove the correctness of your algorithm.

'To design a one-pass streaming algorithm that identifies the majority element in a stream of data, we can utilize a strategy similar to the Boyer-Moore Voting Algorithm. The key idea is to maintain a candidate for the majority element along with a counter that keeps track of how many times we have seen that candidate compared to other elements in the stream.

### Algorithm Description

1. **Initialization**: Start with an empty candidate and a counter set to zero.

2. **Processing the Stream**: As we read each element from the stream one by one:
   - If the counter is zero, we select the current element as the new candidate and set the counter to one.
   - If the counter is greater than zero and the current element is the same as the candidate, we increment the counter.
   - If the counter is greater than zero but the current element is different from the candidate, we decrement the counter by one.

3. **Termination**: After processing all elements in the stream, the candidate will be the majority element.

### Explanation of the Algorithm

The algorithm works based on the properties of the majority element. By the definition, the majority element appears more than half the time in the stream. When processing the elements:

- When we encounter the candidate, we reinforce our belief that it may indeed be the majority element by increasing the counter.
- When we encounter a different element, we decrease our belief in the current candidate. If the candidate is truly the majority element, it would still have a net positive count compared to non-majority elements, allowing it to survive through the decrements applied.

### Proof of Correctness

To prove that this algorithm correctly identifies the majority element, we can reason through its behavior:

1. **Counter Logic**: If an element is a majority element, it will appear more than half the time in the stream. This means that for every time we encounter a non-majority element, the majority element will still have enough occurrences to maintain a positive count.

2. **Candidate Selection**: The moment the counter reaches zero, we have effectively 'reset' our candidate. This means that we've encountered an equal number of the majority and non-majority elements, allowing us to select a new candidate. If the majority element exists, it will eventually become the candidate again.

3. **Final Candidate**: At the end of the stream, the remaining candidate must be the majority element. If we were to count the occurrences of this candidate through the stream, it will indeed have appeared more than half the time, confirming its status as the majority.

4. **Single Pass**: The algorithm only requires a single traversal of the stream, maintaining a constant space complexity (just two variables: one for the candidate and one for the counter). This efficiency is crucial for processing large data streams.

### Conclusion

This one-pass algorithm efficiently identifies the majority element by leveraging the properties of counts and the majority's dominance in occurrences. By carefully managing the candidate and counter throughout the streaming process, the algorithm guarantees that it will end with the correct majority element if one exists. The simplicity and effectiveness of this approach make it a powerful tool for solving majority problems in data streams.


Question: Which of the following models for generating vector representations for text require to precompute the frequency of co-occurrence of words from the vocabulary in the document collection

To determine which model requires the precomputation of the frequency of co-occurrence of words from the vocabulary in the document collection, we’ll analyze each option in detail:

1. **LSI (Latent Semantic Indexing)**: LSI uses a term-document matrix to represent the relationships between terms and documents. It typically involves applying singular value decomposition (SVD) to reduce dimensionality. While LSI does not directly use co-occurrence counts, it relies on term frequencies extracted from the term-document matrix. Therefore, it does not specifically require precomputing co-occurrence frequencies.

2. **CBOW (Continuous Bag of Words)**: CBOW is part of the Word2Vec framework, where the model predicts a target word based on its context words. It is a predictive model that learns embeddings directly from the training data through context windows, without requiring precomputed co-occurrence counts. The learning happens through neural networks based on the input and output words during training.

3. **FastText**: Like CBOW, FastText is also part of the Word2Vec family and incorporates subword information by considering character n-grams. It predicts words based on their context as well and does not require precomputation of co-occurrence frequencies. It learns directly from the training data.

4. **GloVe (Global Vectors for Word Representation)**: GloVe specifically focuses on capturing global statistical information in the corpus. It constructs a co-occurrence matrix where each entry counts how often words co-occur within a certain context. The model then uses this matrix to learn word vectors. Therefore, GloVe explicitly requires the precomputation of the frequency of co-occurrence of words.

### Conclusion:
The model that **requires precomputing the frequency of co-occurrence of words from the vocabulary in the document collection** is **4. GloVe**.


Question: Consider the standard linear programming relaxation of Set Cover that we saw in class.  We gave a randomized rounding algorithm for the Set Cover problem. Use similar techniques to give an algorithm that, with probability at least a positive constant, returns a collection of sets that cover at least $90\%$ of the elements and has cost at most a constant factor larger than the LP solution.

To solve the Set Cover problem using randomized rounding, we start with the standard linear programming (LP) relaxation of the Set Cover formulation. The LP provides us with fractional solutions representing the coverage of each set.

**Step 1: Solve the LP**
First, we solve the LP relaxation, which gives us a fractional solution \( x_j \) for each set \( S_j \). This solution indicates the proportion of each set that we can consider in our rounding process.

**Step 2: Randomized Rounding**
Next, we randomly select each set \( S_j \) to include in our final cover based on the values of \( x_j \). Specifically, we include set \( S_j \) in our solution with probability \( p_j = \min(1, \frac{10}{x_j}) \). The purpose of this adjustment is to ensure that we are more likely to include sets that cover more elements.

**Step 3: Analyzing Coverage**
With this rounding technique, we can analyze the expected coverage. For each element \( e \), let \( C(e) \) be the number of sets covering \( e \). The expected number of sets covering \( e \) after rounding is \( \sum_{S_j \ni e} p_j \). Using linearity of expectation, we can show that the expected coverage of each element is at least \( 0.9 \) of its original LP coverage, which is at least \( 0.9 \).

**Step 4: Cost Analysis**
The expected cost of our selected sets is proportional to the LP solution. Specifically, if the LP solution has cost \( C \), our expected cost is at most \( 10C \) due to the way we selected \( p_j \).

**Conclusion**
Thus, with high probability, our algorithm yields a set cover that covers at least \( 90\% \) of the elements at a cost no more than \( 10 \) times the LP solution. This method leverages both the LP's structure and probabilistic techniques to ensure a high-quality solution.


Question:
Given the following code snippet, explain if it would be better
                    scheduled with static HLS or dynamic HLS. Assume erb+acc+ to be a
                    floating point variable; floating-point multiplications have a
                    latency of four and all other operations a latency of one. If good
                    static scheduling requires some typical trasformations, explain
                    clearly which ones.
                    egin{verbatim}
                        0:  for i = 0 to N do
                        1:      acc = 0.0
                        2:      for j = 0 to M do
                        3:          cond = (c[j] % 5) == 0
                        4:          if (cond) then
                        5:              acc = acc * acc
                        6:          else
                        7:              acc = acc + a[i * M + j] + 1.0
                        8:          end if
                        9:     end for
                        10: b[i] = acc
                        11: end for
                    \end{verbatim}

In analyzing the provided code snippet, we can evaluate whether static or dynamic High-Level Synthesis (HLS) is more appropriate based on the characteristics of the code and the implications of each scheduling approach.

### Code Overview

The code consists of nested loops where:
- The outer loop iterates over `N`.
- The inner loop iterates over `M`.
- A conditional statement within the inner loop determines how the variable `acc` is updated based on the evaluation of `cond`, which depends on the value of `c[j]`.

### Key Observations

1. **Conditional Dependence**: The condition within the inner loop (`if (cond) then`) relies on runtime information, specifically the modulo operation involving `c[j]`. This means that the control flow cannot be determined at compile time, making static scheduling more challenging.

2. **Branch Execution Latency**: The execution paths of the `if` statement lead to two different operations:
   - If true, `acc` is updated by squaring its current value (a multiplication operation with a latency of 4).
   - If false, `acc` is updated by adding values from the array (addition with a latency of 1).

Given this structure, the worst-case branch execution latency (the multiplication case) will dictate the initiation interval due to the loop-carried dependency on `acc`. Therefore, for static HLS, the best achievable initiation interval would be 4, which is determined by the latency of the multiplication operation.

### Static HLS vs. Dynamic HLS

- **Static HLS**: While it offers potential optimization opportunities through compile-time scheduling, the reliance on runtime information for branching in this code can hinder static scheduling efficiency. The initiation interval would be constrained by the highest latency operation, leading to potential underutilization of resources.

- **Dynamic HLS**: This approach schedules operations at runtime, allowing for more flexible management of conditional paths and data dependencies. In this case, dynamic scheduling would not face the same constraints as static scheduling. It can adapt based on the data being read from memory and the control path taken during execution. The average initiation interval could vary based on runtime conditions, likely falling between 1 and 4, depending on how frequently each branch is executed.

### Conclusion

In this specific scenario, dynamic HLS is preferred due to its ability to handle the runtime-dependent conditional logic effectively. It can achieve a better average initiation interval than static HLS, which would be constrained by the worst-case latency scenario. Thus, for this code, dynamic HLS would generally provide superior performance and efficiency.