Question: Implement weigthing estimation of kNN classification

To implement weighted estimation for kNN classification, the provided function `knn_weighting_estimate` calculates the contribution of each nearest neighbor to the classification score based on their similarity to the query vector. Here's a breakdown of how it works:

1. **Parameters**: The function takes in document vectors, their corresponding labels, a query vector, and an optional parameter `k` specifying the number of nearest neighbors.

2. **Finding Neighbors**: It utilizes a helper function `knn` (not defined here) to retrieve the indices of the top `k` nearest documents based on distance metrics.

3. **Initializing Scores**: A dictionary named `scores` is created to store cumulative scores for each unique label found in `doc_labels`.

4. **Calculating Weights**: For each of the top `k` neighbors, it computes the cosine similarity between the query vector and each document vector, adding this value to the score associated with that document's label.

5. **Return Value**: Finally, it returns the scores dictionary, which contains weighted estimates for each label/topic sorted by their contributions.

This approach allows for more nuanced predictions as it accounts for both proximity and similarity rather than just counting votes among neighbors.

Here’s the implementation again for clarity:

```python
def knn_weighting_estimate(doc_vectors, doc_labels, query_vector, k=10):
    """ Weighting estimation for kNN classification
    :param doc_vectors: Document vectors (np.array(np.array))
    :param doc_labels: Document labels/topics (list)
    :param query_vector: Query vector (np.array)
    :param k: Number of nearest neighbors to retrieve
    :return: A dictionary containing the estimation (sorted) score for each label/topic (dict)
    """
    top_k_doc_indices = knn(doc_vectors, query_vector, k)
    top_k_labels = [doc_labels[i] for i in top_k_doc_indices]

    scores = {t: 0 for t in list(set(doc_labels))}
    for i in top_k_doc_indices:
        scores[doc_labels[i]] += cosine_similarity(query_vector, doc_vectors[i])

    return scores
```

In summary, this method enhances kNN classification by incorporating similarity weights into the decision-making process.


Question: Tick the \textbf{true} assertion. In a zero-knowledge interactive proof for $L$, \ldots

1. For any ppt verifier, there is a simulator which for any \( x \in L \) produces a conversation indistinguishable from the original conversation.

This assertion is valid because, in the context of zero-knowledge proofs, one of the key properties is that the interaction between the prover and the verifier can be simulated by a simulator. The simulator must be able to generate a conversation that is indistinguishable from the real interaction between the prover and the verifier for any input \( x \) in the language \( L \). This ensures that no information about \( x \) beyond its membership in \( L \) is revealed to the verifier. The other options either misrepresent the properties of zero-knowledge proofs or are incorrect in their assertions about the roles of the simulator and the verifier.


Question:
To avoid the effect of the prefetcher, one could use a hash
            table to randomize the access order during probing so that the
            prefetcher cannot anticipate the access pattern. What is the
            potential problem with this idea?

Using a hash table to randomize access order during probing in order to avoid the effects of the prefetcher can indeed mitigate some predictability of memory access patterns. However, there are several potential problems with this approach:

1. **Increased Cache Misses**: Hash tables typically involve non-sequential access patterns due to the hash function and probing. This can lead to increased cache misses because accessing data randomly may not take advantage of spatial locality, which is a key principle in cache design. Caches are designed to optimize for sequential access patterns, and random accesses can lead to worse cache performance.

2. **Hash Function Collisions**: In a hash table, there is a possibility of collisions where multiple keys hash to the same index. This requires additional handling (probing, chaining, etc.), which can introduce overhead and complicate the access pattern further. If collisions are frequent, it can lead to longer probing times and increase the latency of accessing elements in the hash table.

3. **Overhead of Hashing**: The process of hashing itself introduces computational overhead. Depending on the complexity of the hash function and the number of collisions, this can degrade performance, especially compared to simpler data structures that provide more predictable access patterns.

4. **Increased Complexity**: Implementing a hash table with randomized probing adds complexity to the code and may lead to bugs or inefficiencies if not designed carefully. The overhead of managing the hash table, including resizing it when it becomes too full, can also introduce performance penalties.

5. **Impact on Data Structure Performance**: Depending on the load factor and the method of handling collisions, the overall efficiency of the hash table can be negatively impacted. This can lead to degraded performance not only in terms of access times but also in terms of memory usage.

6. **Prefetcher Adaptation**: Although the intention is to prevent the prefetcher from adapting to the access pattern, modern prefetchers can sometimes adapt to various patterns, including hash-based access patterns. If this occurs, the benefits of randomizing access may not be as effective as intended.

In conclusion, while randomizing access patterns through a hash table may seem like a viable solution to avoid prefetcher predictability, it introduces a host of potential problems, including increased cache misses, overhead from hashing and collisions, added complexity, and potential performance degradation. Thus, careful consideration must be given to these factors when designing data structures intended to mitigate prefetcher effects.


Question: In which group is the discrete logarithm problem believed to be hard?

The discrete logarithm problem (DLP) is believed to be hard in certain mathematical groups. Let's analyze the options provided:

1. **In a subgroup of \(\mathbb{Z}_p^*\) with large prime order**:
   - This is a well-known setting where the discrete logarithm problem is hard. The security of many cryptographic systems, such as Diffie-Hellman and ElGamal, is based on the hardness of the DLP in this context. The problem remains difficult due to the lack of efficient algorithms for solving it in groups of large prime order.

2. **In \(\mathbb{Z}_n\), where \(n = pq\) for two large primes \(p\) and \(q\)**:
   - This setting does not have a hard discrete logarithm problem. Instead, the security in this context typically relies on the hardness of integer factorization (the RSA problem). Therefore, this is not the correct answer.

3. **In a group \(G\) of smooth order**:
   - A group of smooth order (where the order of the group has only small prime factors) does not provide a hard DLP. In fact, algorithms like the Pohlig-Hellman algorithm can efficiently solve the DLP in such groups. Thus, this is not a suitable option.

4. **In \(\mathbb{Z}_2^p\), for a large prime \(p\)**:
   - The group \(\mathbb{Z}_2^p\) does not pose a hard discrete logarithm problem. The structure of this group allows for straightforward computation, and therefore, the DLP is not hard here.

### Conclusion:
The only option where the discrete logarithm problem is believed to be hard is:

**1. In a subgroup of \(\mathbb{Z}_p^*\) with large prime order.**


Question: Can we implement TRB with an eventually perfect failure detector ◇P, under the assumption that at least one process can crash?

Yes, we can implement Total Order Broadcast (TRB) using an eventually perfect failure detector ◇P, even with the possibility of process crashes. The failure detector will eventually identify correct processes, enabling reliable message delivery in total order as outlined in the reasoning.


Question: Which cipher is AES?

The correct answer is **3: RIJNDAEL**. I arrived at this by recognizing that AES (Advanced Encryption Standard) is based on the Rijndael cipher, which was selected by NIST after a competition.

Students may confuse this with other ciphers like Blowfish or RC5, which are also well-known but unrelated to AES. To avoid similar mistakes, focus on the specifics of the question, particularly the relationship between AES and Rijndael. Always verify your knowledge of definitions and relationships among cryptographic algorithms.


Question: Tick the \emph{false} assertion. Given a ring $R$, $R^\star$ is\ldots

To determine which assertion about \( R^\star \) is false, let's analyze each statement step by step.

1. **"A group."**
   The set of invertible elements in a ring \( R \), which we denote by \( R^\star \), is indeed a group under the operation of multiplication. This is because:
   - **Closure:** If \( a, b \in R^\star \), then both \( a \) and \( b \) are invertible, meaning there exist \( a^{-1} \) and \( b^{-1} \) in \( R \) such that \( aa^{-1} = 1 \) and \( bb^{-1} = 1 \). The product \( ab \) is also invertible, with inverse \( b^{-1}a^{-1} \) because \( ab \cdot b^{-1}a^{-1} = (aa^{-1})(bb^{-1}) = 1 \).
   - **Identity:** The multiplicative identity \( 1 \) is in \( R^\star \) since it is invertible (its own inverse).
   - **Inverses:** For every \( a \in R^\star \), the inverse \( a^{-1} \) is also in \( R^\star \).
   Hence, \( R^\star \) is indeed a group.

2. **"The set of invertible elements in \( R \)."**
   This statement is true by definition. The set \( R^\star \) is specifically defined as the set of all units (invertible elements) in the ring \( R \).

3. **"$R - \{0\}$."**
   This assertion is false in general. The set \( R - \{0\} \) includes all non-zero elements of the ring \( R \), but not all non-zero elements are necessarily invertible. For example, consider the ring of integers \( \mathbb{Z} \). The only invertible elements are \( 1 \) and \( -1 \); thus, \( \mathbb{Z} - \{0\} = \{ \ldots, -2, -1, 1, 2, \ldots \} \) contains many elements that are not invertible. Therefore, \( R^\star \) is a proper subset of \( R - \{0\} \) in most rings.

4. **"The set of units."**
   This statement is also true since "units" is a term used to specifically denote the invertible elements in a ring.

Now, collating our findings:
- The first assertion is true (it is a group).
- The second assertion is true (it is the set of invertible elements).
- The third assertion is false (it is not equal to \( R - \{0\} \)).
- The fourth assertion is true (it is the set of units).

Thus, the false assertion is **"$R - \{0\}$."**


Question: What is the length in bits of the input and output of a DES S-Box respectively?

The length of the input to a DES S-Box is 6 bits, and the output is 4 bits, making option 3 the correct choice. This is because the S-Box takes a 6-bit input and maps it to a 4-bit output based on a fixed substitution table.


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

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

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

One of your colleagues suggests that "add" should be changed to return a boolean indicating whether the passed value was added or not. Explain whether this breaks backward compatibility and why or why not (without worrying about whether this is a good or a bad thing).

Changing the `add` method to return a boolean indicating whether the value was added would break backward compatibility.

Backward compatibility means that existing code relying on the original method should still function correctly after changes. If the method signature changes from `public void add(int val)` to `public boolean add(int val)`, any existing code that calls `add` expecting no return value would result in a compilation error. This is because the expected return type differs from the previous implementation, which would disrupt users’ code that interacts with this method.

In summary, modifying the return type from `void` to `boolean` changes the method's interface, thus breaking backward compatibility for any existing client code expecting a void method.


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

The model might reproduce exact names, home addresses or contact details of healthcare workers or patients that appeared in the COVID-19 news corpus, thereby leaking personally identifiable information.


Question: Assume Your project depends on the latest available minor release version of a package instead of a hardcoded exact version, what are the pros and cons of this?

Depending on the latest available minor release version of a package has both advantages and disadvantages:

**Pros:**
1. **Up-to-Date Fixes:** You benefit from the latest bug fixes, security patches, and performance improvements, which can enhance the stability and security of your project.
2. **Less Maintenance:** You reduce the need for frequent updates, as you automatically incorporate minor changes without manual intervention.

**Cons:**
1. **Unexpected Behavior:** Minor version updates may introduce subtle changes that could lead to unexpected behavior or bugs in your project, even if the versions are technically compatible.
2. **Testing Overhead:** As new minor versions are released, you may need to conduct additional testing to ensure that the new changes do not break existing functionality, which can increase development time.

Overall, while using the latest minor version can streamline updates and improve security, it also requires careful consideration of potential risks and thorough testing.


Question: Consider the exhaustive search of a uniformly distributed key in a set of size $N$. Think of the possible strategies and their complexities. Which of the following is \textbf{not} possible (We assume that memory access is constant.)

The option that is **not** possible is **C**.

Justification:

- Option A: This option suggests that we can find the key with precomputation time of $0$, using $O(1)$ memory and $O(N)$ time. This is feasible if we have a direct access to the key (for example, if we simply check each possibility one by one without prior information). Therefore, this option is possible.

- Option B: This option claims that we can find the key with precomputation of $O(N)$, memory of $O(N)$, and constant time $O(1)$. This suggests that we can store all keys in memory and instantly retrieve the correct one. This is possible if we use a hash table or similar structure, thus this option is also possible.

- Option C: For this option, we have precomputation of $O(N)$, memory of $O(N^{2/3})$, and time of $O(N^{2/3})$. The issue here is that if we are searching through keys in a uniformly distributed manner, we cannot guarantee that we can find the key within $O(N^{2/3})$ time unless we have more than $O(N^{2/3})$ memory to store precomputed results or a direct mapping strategy. The time complexity of searching through $N$ keys should be at least $O(N)$ unless we can utilize some form of efficient indexing or hashing, which is not indicated here. Thus, this option is not possible.

- Option D: This suggests that with no precomputation, we can find the key using $O(1)$ time while having $O(N)$ memory. This is feasible if we use the $O(N)$ memory to store all possible keys and can instantly access any of them (in a similar manner to option B). Hence, this option is possible.

In conclusion, the correct answer is **C**, as it presents a scenario that is not feasible given the constraints of memory and time for exhaustive search.


Question: Which of the following terms represents a mode of operation which transforms a block cipher into a stream cipher?

When comparing the options, 3DES, CBC, and ECB are modes that primarily apply to block ciphers without transforming them into stream ciphers. 3DES is a block cipher that applies the DES algorithm three times to enhance security, but it does not offer the functionality of a stream cipher. CBC (Cipher Block Chaining) and ECB (Electronic Codebook) are both modes of operation for block ciphers, where CBC provides better security through chaining block dependencies, while ECB can be vulnerable to pattern attacks due to its independent block handling. In contrast, CTR (Counter mode) effectively transforms a block cipher into a stream cipher by generating a keystream from a counter value, allowing encryption of data in a manner similar to stream ciphers. Thus, while all options involve block ciphers, CTR stands out as the only mode that achieves the transformation into a stream cipher, making it the correct answer.


Question: In class, we saw Karger's beautiful randomized algorithm for finding a minimum cut in an undirected graph $G=(V,E)$. Recall that his algorithm works by repeatedly contracting a randomly selected edge until the graph only consists of two vertices which define the returned cut. For general graphs, we showed that the returned cut is a minimum cut with probability at least $1/\binom{n}{2}$. In this problem, we are going to analyze the algorithm in the special case when the input graph is a tree.  Specifically, you should show that if the input graph $G=(V,E)$ is a spanning tree, then Karger's algorithm returns a minimum cut with probability $1$. \\ {\em (In this problem you are asked to show that Karger's min-cut algorithm returns a minimum cut with probability $1$ if the input graph is a spanning tree.   Recall that you are allowed to refer to material covered in the lecture notes.)}

To explore why Karger's algorithm returns a minimum cut with probability 1 when applied to a spanning tree, we should first consider the properties of spanning trees. A spanning tree is a connected graph that contains no cycles, and its edges are critical for maintaining connectivity. Interestingly, this characteristic of trees makes them unique compared to more complex graphs where multiple cuts might exist.

When Karger's algorithm randomly selects an edge to contract, in the case of a spanning tree, each edge removal results in exactly two components. This means that any edge we choose to contract will lead to a situation where we're always left with a cut that separates the tree into two parts. However, it’s worth noting that while Karger's algorithm is quite effective for trees, its performance could vary in graphs with more complex structures due to the presence of multiple minimum cuts.

Thus, as we contract edges in the spanning tree, we inevitably end up with a configuration that represents a minimum cut. Therefore, in this specific scenario, Karger's algorithm guarantees that the returned cut is indeed a minimum cut with high probability. This contrasts with more intricate graph structures where the randomness of edge selection could introduce variability in the minimum cut result.

1


Question: An element of the finite field $\mathrm{GF}(2^8)$ is usually represented by\dots

An element of the finite field \(\mathrm{GF}(2^8)\) is typically represented by two hexadecimal digits. This is because \(\mathrm{GF}(2^8)\) consists of \(256\) elements, ranging from \(0\) to \(255\). Each of these elements can be represented by a byte, which is \(8\) bits long. Since one byte can be expressed using two hexadecimal digits, this representation is both compact and efficient.

Here's a breakdown of the options:

A. **One hexadecimal digit.**
This is incorrect because one hexadecimal digit can only represent values from \(0\) to \(15\), which is not sufficient for the \(256\) elements in \(\mathrm{GF}(2^8)\).

B. **Eight bytes.**
This option is also incorrect. Eight bytes would correspond to \(64\) bits, far exceeding the size of elements in \(\mathrm{GF}(2^8)\), which only need one byte.

C. **Two hexadecimal digits.**
This is the correct answer. Each element in \(\mathrm{GF}(2^8)\) can indeed be represented as a byte, which is \(8\) bits, and can be expressed with two hexadecimal digits.

D. **An irreducible polynomial of degree 8.**
While irreducible polynomials are fundamental in defining the structure and operations of the field, the elements themselves are not directly represented by these polynomials.

In conclusion, the most accurate representation of an element in \(\mathrm{GF}(2^8)\) is C, which is two hexadecimal digits.

C.


Question: Why do block ciphers use modes of operation?

Block ciphers operate on fixed-size blocks of data. However, real-world applications often require encrypting data that exceeds this block size, such as messages or files of arbitrary length. Modes of operation provide a method for handling these scenarios by defining how to encrypt data larger than the block size and ensuring that the encryption remains secure and efficient. The correct answer is that block ciphers use modes of operation primarily to encrypt messages of any size.

4


Question: Design an algorithm that implements consensus using multiple TRB instances.

Run one TRB instance per process. Each process p invokes TRB[p].broadcast(vâ) with its proposal vâ. Then every correct process waits until it has either delivered a value or â¥ from each TRB instance. Let S be the multiset of delivered values (treating missing ones as â¥). Finally, apply a deterministic decision function f(S) (e.g., pick the non-â¥ value with smallest process ID) to agree on the consensus value.


Question: Pick the \emph{false} statement.

The false statement is A. A ring is not necessarily commutative, meaning that in a general ring, it is possible for \( ab \neq ba \). While all rings satisfy associativity, distributivity, and the existence of additive identities, commutativity is an additional property that only some rings (called commutative rings) possess.


Question: When computing PageRank iteratively, the computation ends when...

To determine when the PageRank computation ends in an iterative process, we need to analyze the options provided. PageRank uses a specific algorithm to rank nodes in a graph based on their importance. The process involves iterating until the rank vector converges.

### Key Definitions and Formulas:
1. **PageRank Vector**: This is a vector that represents the rank of each node in the graph. We denote it as \( \mathbf{R} \).
2. **Iteration**: In each iteration, the PageRank vector is updated based on the ranks of the nodes pointing to it and a damping factor \( d \).
3. **Convergence**: We say that the computation has converged when the change in the rank vector between successive iterations is sufficiently small.

### Convergence Criteria:
The convergence can be mathematically defined using norms, specifically:

- The **difference between two successive rank vectors** can be computed as:
  \[
  \|\mathbf{R}^{(n)} - \mathbf{R}^{(n-1)}\|
  \]
  where \( \mathbf{R}^{(n)} \) is the rank vector at the \( n \)-th iteration, and \( \mathbf{R}^{(n-1)} \) is the rank vector at the previous iteration.

- The **norm** often used is the **L1 norm** (sum of absolute differences) or the **L2 norm** (Euclidean distance).

### Evaluating Options:
1. **The difference among the eigenvalues of two subsequent iterations falls below a predefined threshold**:
   - This is not a standard criterion for PageRank convergence. PageRank does not typically focus on eigenvalues for convergence in iterations.

2. **The norm of the difference of rank vectors of two subsequent iterations falls below a predefined threshold**:
   - This is the correct criterion for convergence in the PageRank algorithm. When the change in the rank vector is small enough, we consider the computation to have converged.

3. **All nodes of the graph have been visited at least once**:
   - This is not necessary for convergence in PageRank. PageRank can converge without all nodes being visited.

4. **The probability of visiting an unseen node falls below a predefined threshold**:
   - While this could be related to certain probabilistic interpretations of PageRank, it is not a standard criterion for determining convergence.

### Conclusion:
The correct answer is:
- **The norm of the difference of rank vectors of two subsequent iterations falls below a predefined threshold**.

This option directly relates to the convergence criteria used in the PageRank computation process.


Question: Explain the difference between inflectional and derivational morphology. Illustrate your explanation with concrete examples in English or French.

Inflectional morphology refers to modifications of a word that do not change its grammatical category but rather provide grammatical information, such as tense, number, or case. For example, in English, the verb "give" can change to "gives" (present tense) or "given" (past participle), but it remains a verb throughout these changes.

On the other hand, derivational morphology involves alterations that change the grammatical category of a word. For instance, the noun "process" can be transformed into "processing" (verb), "processor" (noun for a person or thing that processes), or "processable" (adjective). Each of these derivatives belongs to a different grammatical category than the original word.

In French, an example of inflectional morphology would be "parle" (speak) changing to "parlÃ©" (spoken), while an example of derivational morphology could be "acte" (act) becoming "acteur" (actor) or "acte" to "acte de naissance" (birth certificate) to change its meaning and category.


Question: Is $(\mathbb{Z} / 8\mathbb{Z}^*, \cdot)$ isomorphic to $(\mathbb{Z} / k\mathbb{Z}, +)$ for some $k$?

The correct answer is **False**. Let's analyze the reasoning step-by-step:\n\n1. **Structure of \((\mathbb{Z} / 8\mathbb{Z}^*, \cdot)\)**:\n   - The group consists of the units modulo 8: \{1, 3, 5, 7\}, which indeed has 4 elements. However, this group is non-cyclic because it cannot be generated by a single element.\n\n2. **Structure of \((\mathbb{Z} / k\mathbb{Z}, +)\)**:\n   - \((\mathbb{Z} / k\mathbb{Z}, +)\) is always cyclic for any integer \(k\), meaning it can be generated by one element.\n\n3. **Isomorphism Requirements**:\n   - For two groups to be isomorphic, they must have the same order and structure (cyclic vs. non-cyclic). Since \((\mathbb{Z} / 8\mathbb{Z}^*, \cdot)\) is non-cyclic and \((\mathbb{Z} / k\mathbb{Z}, +)\) is cyclic, they cannot be isomorphic.\n\nStudents may mistakenly focus only on the order of the groups without considering their structural properties. Therefore, the answer is **False**; no \(k\) can make these groups isomorphic.


Question: In an x86 multiprocessor with JOS, how many Bootstrap Processors (BSP) is it possible to have at most? And how many Application Processors (AP) at most?

Answer: B

Explanation: In an x86 multiprocessor system, there can be at most one Bootstrap Processor (BSP) responsible for booting the system, while the number of Application Processors (AP) can be infinite, as they can be added dynamically.


Question:
Is the processors in recoder buffers ordered (e.g., FIFO-like) or an unordered structure?
  Why?

The processors in reorder buffers are ordered (FIFO-like) because they ensure that instructions are completed in the same order they were issued to maintain program correctness and consistency.


Question:
Given the following method:
1 def mystery6(nIter: Int) (ss: List[String] ): List[String] =
2 if nIter <= 0 then ss
3 else mystery6 (nIter - 1) (
4 for
5 s <- ss
6 c <- List (’c’ , ’b’ , ’a’)
7 yield
8 s + c
9 ) ::: ss
What is the output if we call mystery6 this way:
mystery6(5)(List("")).filter(_.exists(_ == ’b’))(0)

To determine the output of the call `mystery6(5)(List(""))`, we can analyze how the `mystery6` function processes the input.

### Step-by-step Analysis:
1. **Initial Call**: `mystery6(5)(List(""))`
   - `nIter` is `5`, and `ss` is `List("")`.
   - Since `nIter` is greater than `0`, it proceeds to the `else` clause.

2. **Iteration 1**: `mystery6(4)(...)`
   - The `for` comprehension generates new strings by concatenating each string in `ss` with each character in `List('c', 'b', 'a')`.
   - For `ss = List("")`, the results will be `List("c", "b", "a")`.
   - The result of this iteration is `List("c", "b", "a") ::: List("")`, which results in `List("c", "b", "a", "")`.

3. **Iteration 2**: `mystery6(3)(List("c", "b", "a", ""))`
   - The function again generates combinations:
   - From `List("c", "b", "a", "")`, we get:
     - "c" + 'c' = "cc"
     - "c" + 'b' = "cb"
     - "c" + 'a' = "ca"
     - "" + 'c' = "c"
     - "b" + 'c' = "bc"
     - "b" + 'b' = "bb"
     - "b" + 'a' = "ba"
     - "a" + 'c' = "ac"
     - "a" + 'b' = "ab"
     - "a" + 'a' = "aa"
     - "" + 'b' = "b"
     - "" + 'a' = "a"
   - The results will be `List("cc", "cb", "ca", "c", "bc", "bb", "ba", "ac", "ab", "aa", "", "b", "a")`.
   - The result of this iteration is `List("cc", "cb", "ca", "c", "bc", "bb", "ba", "ac", "ab", "aa", "", "b", "a") ::: List("c", "b", "a", "")`.

4. **Iterations 3 to 5**: The process continues similarly, growing the list exponentially as more characters are added.

### Filtering for 'b':
After the fifth iteration, we have a list of strings generated. The filter operation `filter(_.exists(_ == 'b'))` will keep only those strings that contain the character 'b'.

### Final Output:
The output of `mystery6(5)(List(""))` will contain many strings, including "b", "bb", "ab", "ba", "bc", etc. The `filter` operation will produce a filtered list of strings containing 'b'.

Finally, calling `.apply(0)` on this filtered list will give the first element that contains 'b', which in this case will be the empty string when there are no strings present.

### Conclusion:
Since the first string containing 'b' will be `b`, the final output will be:

```scala
"b"
```

Justification: The function generates combinations of strings through recursive concatenation, and after filtering for strings containing 'b', we access the first occurrence, which is "b".


Question: Which of the following acronyms does not designate a mode of operation?

The correct answer is **3. CRC**.

### Justification:
The acronyms listed in the question refer to different concepts in cryptography:

1. **CBC (Cipher Block Chaining)** - This is a mode of operation for block ciphers that uses chaining to provide confidentiality.
2. **CTR (Counter Mode)** - This is another mode of operation for block ciphers that turns a block cipher into a stream cipher by generating a keystream.
3. **CRC (Cyclic Redundancy Check)** - This is not a mode of operation for encryption but rather an error-detecting code used to detect accidental changes to raw data.
4. **ECB (Electronic Codebook)** - This is a straightforward mode of operation for block ciphers that encrypts each block independently.

Thus, **CRC** does not designate a mode of operation for encryption, making it the correct answer.


Question: You have been hired to evaluate an email monitoring system aimed at detecting potential security issues. The targeted goal of the application is to decide whether a given email should be further reviewed or not. You have been given the results of three different systems that have been evaluated on the same panel of 157 different emails. Here are the classification errors and their standard deviations: system 1 (error=0.079, stddev=0.026) system 2 (error=0.081, stddev=0.005) system 3 (error=0.118, stddev=0.004) Which system would you recommend? Why?

**Answer:** I would recommend system 2 for the email monitoring system. While the classification error of system 2 (0.081) is slightly higher than that of system 1 (0.079), the difference in error rates is statistically non-significant given the standard deviations (stddev) of both systems. System 2 has a much lower standard deviation (0.005) compared to system 1 (0.026), indicating that its performance is more consistent and reliable across different emails. This consistency is crucial in security applications, where false negatives or positives could have serious implications. Therefore, considering both the error rates and the standard deviations, system 2 is the better choice for further review.


Question: The $n^2$ problem ...

The "$n^2$ problem" in the context of communication and symmetric ciphers refers to the challenge of managing unique keys for secure interactions among users. Here's a breakdown:

1. **Context**: The problem arises when $n$ users need to communicate securely, necessitating unique keys for each pair of users.

2. **Symmetric Ciphers**: In symmetric encryption, each pair requires a shared key. The number of unique pairs is calculated using the combination formula \( \frac{n(n-1)}{2} \), which approximates to \( n^2/2 \) for large $n$. This leads to the "$n^2$ problem."

3. **Evaluating Options**:
   - (A) Moore's Law pertains to computer processing growth.
   - (B) Murphy's Law deals with things going wrong, not key management.
   - (C) Kerckhoffs's Principle highlights cipher security, not key quantity.
   - (D) Correctly describes the need for numerous keys among $n$ users communicating with symmetric ciphers.

4. **Conclusion**: Therefore, the correct answer is (D), as it accurately represents the scenario of $n$ users needing to securely communicate via symmetric ciphers, resulting in the "$n^2$ problem."


Question: Consider a classification problem using either SVMs or logistic regression and separable data. For logistic regression we use a small regularization term (penality on weights) in order to make the optimum welldefined. Consider a point that is correctly classified and distant from the decision boundary. Assume that we move this point slightly. What will happen to the decision boundary?

To analyze the effect of moving a correctly classified point that is distant from the decision boundary on the decision boundary itself, we need to consider the fundamental differences between Support Vector Machines (SVMs) and logistic regression.

1. **Understanding SVMs**:
   - SVMs focus on maximizing the margin between classes. The decision boundary is determined by the support vectors, which are the data points that lie closest to the boundary. If a point is far from the boundary and correctly classified, it is not a support vector. Therefore, moving this point won't affect the position of the decision boundary since it doesn't influence the support vectors.

2. **Understanding Logistic Regression**:
   - Logistic regression, on the other hand, attempts to find the best-fitting line (or hyperplane) that predicts the probability of class membership. The weights in logistic regression are affected by all points in the dataset, including those that are far from the decision boundary. If we move a point slightly, even if it is far from the decision boundary, it could slightly change the weights due to the regularization term. However, since the point is distant and correctly classified, the change is likely to be small.

3. **Combining Insights**:
   - For SVMs, the decision boundary will remain unchanged because the point moved is not a support vector.
   - For logistic regression, the decision boundary may experience a minor change due to the movement of the point, but this change will not be large since the point is distant from the boundary.

Given this analysis, we can conclude:
- SVMs: No change in the decision boundary.
- Logistic Regression: Small change in the decision boundary.

Thus, the correct answer is **D. No change for SVMs and a small change for logistic regression.**


Question: What is the algorithm to perform optimization with gradient descent? Actions between Start loop and End loop are performed multiple times. (One answer)

To determine the correct order of actions to perform optimization using gradient descent, we need to understand the typical process involved in this algorithm. The process generally involves the following steps:

1. **Initialize weights**: Before starting the optimization, the weights of the model need to be initialized.
2. **Start loop**: This indicates the beginning of the iterative process where we will repeatedly adjust the weights based on computed gradients.
3. **Compute gradients**: In each iteration, we calculate the gradients of the loss function with respect to the weights. This tells us how to adjust the weights to minimize the loss.
4. **Update weights**: Using the computed gradients, we then update the weights in the direction that reduces the loss.
5. **End loop**: This signifies the end of the iterative process, which typically continues until a stopping criterion is met (like a certain number of iterations or convergence).

Now, let’s evaluate each of the provided options:

1. **Option 1**: '1 Start loop, 2 Initialize weights, 3 Compute gradients, 4 Update weights, 5 End loop'
   - This option incorrectly places "Start loop" before "Initialize weights." The loop should start after weights are initialized.

2. **Option 2**: '1 Initialize weights, 2 Compute gradients, 3 Start loop, 4 Update weights, 5 End loop'
   - This option places "Compute gradients" before "Start loop," which is incorrect because gradients should be computed within the loop after it has started.

3. **Option 3**: '1 Initialize weights, 2 Start loop, 3 Update weights, 4 End loop, 5 Compute gradients'
   - This option incorrectly places "Compute gradients" after "End loop," which would not allow for any gradient computation to update weights.

4. **Option 4**: '1 Initialize weights, 2 Start loop, 3 Compute gradients, 4 Update weights, 5 End Loop'
   - This option correctly initializes weights first, starts the loop, computes gradients, updates weights, and then ends the loop. This follows the logical order of operations for gradient descent.

After analyzing the options, Option 4 is the only one that correctly represents the algorithm for performing optimization with gradient descent.

Final Answer: 4.


Question: Tick the \emph{incorrect} assertion regarding the security of the Diffie-Hellman key exchange over a subgroup $\langle g \rangle \subset \mathbb{Z}_p^*$.

The incorrect assertion regarding the security of the Diffie-Hellman key exchange is:

**3. The binary representation of the output of the key exchange is a uniformly distributed bitstring.**


Question: Tick the \textbf{false} assertion. In Linear Cryptanalysis, the corresponding mask circuit of \ldots

To identify the false assertion among the given statements related to Linear Cryptanalysis, let's evaluate each one:

1. **Assertion 1**: For an XOR gate \(X \oplus Y = Z\), the corresponding mask circuit is \(a \cdot Z = (a \cdot X) \oplus (a \cdot Y)\).

   This assertion is **true**. The masking of the output \(Z\) correctly reflects the masked inputs \(X\) and \(Y\).

2. **Assertion 2**: For an XOR to constant gate \(Y = X \oplus K\), the corresponding mask circuit is \(a \cdot Y = (a \cdot X) \oplus (a \cdot K)\).

   This assertion is **true**. The output \(Y\) correctly reflects the masked input \(X\) and the constant \(K\).

3. **Assertion 3**: For a linear circuit \(Y = M \times X\), the corresponding mask circuit is \(a \cdot Y = (M \times a) \cdot X\).

   This assertion is **false**. The correct expression for masking a linear transformation is \(a \cdot Y = M \cdot (a \cdot X)\), not \((M \times a) \cdot X\). The masking should be applied to the input \(X\) before the linear transformation \(M\).

4. **Assertion 4**: For a duplicate gate \(X = Y = Z\), the corresponding mask circuit is \((a \oplus b) \cdot X = (a \cdot Y) \oplus (b \cdot Z)\).

   This assertion is **true**. The masking of \(X\) (which equals \(Y\) and \(Z\)) correctly represents the masked outputs.

Based on the analysis, the **false assertion** is **Assertion 3**.


Question: Select the \emph{correct} answer.

A dictionary attack is a method used to guess passwords by trying every word in a predefined list, called a dictionary. The larger the dictionary, the higher the chances of successfully guessing the correct password because it includes more possible options that a user might have chosen. Therefore, the success rate of this attack directly depends on how big the dictionary is.

Thus, the correct answer is “4. The success probability of the dictionary attack depends on the size of the dictionary.”


Question: Let the first four retrieved documents be N N R R, where N denotes a non-relevant and R a relevant document. Then the MAP (Mean Average Precision) is:

To solve the problem of calculating the Mean Average Precision (MAP) for the retrieved documents represented as N N R R, we will follow the suggested strategy step by step.

1. **Identify Relevant Documents**:
   In the sequence N N R R, we have two relevant documents (R) and two non-relevant documents (N). The relevant documents appear at the 3rd and 4th positions.

2. **Calculate Precision at Each Relevant Document**:
   Precision is calculated as the number of relevant documents retrieved up to that point divided by the total number of documents retrieved up to that point.

   - For the **first relevant document (R)** at position 3:
     - Total documents retrieved = 3 (N N R)
     - Relevant documents retrieved = 1 (the first R)
     - Precision = $\frac{1}{3}$

   - For the **second relevant document (R)** at position 4:
     - Total documents retrieved = 4 (N N R R)
     - Relevant documents retrieved = 2 (the two Rs)
     - Precision = $\frac{2}{4} = \frac{1}{2}$

3. **Average the Precision Values**:
   To calculate the Mean Average Precision (MAP), we take the average of the precision values calculated at the points where relevant documents were found.

   \[
   \text{MAP} = \frac{1}{2} \left( \frac{1}{3} + \frac{1}{2} \right)
   \]

   To find this average, we need a common denominator. The least common multiple of 3 and 2 is 6.

   - Convert $\frac{1}{3}$ to have a denominator of 6:
     \[
     \frac{1}{3} = \frac{2}{6}
     \]

   - Convert $\frac{1}{2}$ to have a denominator of 6:
     \[
     \frac{1}{2} = \frac{3}{6}
     \]

   Now, we can add these:
   \[
   \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}
   \]

   Now, to find the average:
   \[
   \text{MAP} = \frac{1}{2} \cdot \frac{5}{6} = \frac{5}{12}
   \]

4. **Compare with Options**:
   The calculated MAP is $\frac{5}{12}$. Now, we check this against the given options:
   1. 1/2
   2. 5/12
   3. 3/4
   4. 7/24

   The correct answer is clearly option **2. 5/12**.

Thus, the final answer is:

**5/12**


Question: Use the integrality of the bipartite perfect matching polytope (as proved in class) to show the following classical result: \begin{itemize} \item[] The edge set of a $k$-regular bipartite graph $G=(A\cup B, E)$ can in polynomial time be partitioned into $k$ disjoint perfect matchings. \end{itemize} \noindent A graph is $k$-regular if the degree of each vertex equals $k$. Two matchings are disjoint if they do not share any edges.

To show that the edge set of a \( k \)-regular bipartite graph \( G = (A \cup B, E) \) can be partitioned into \( k \) disjoint perfect matchings, we will leverage the properties of perfect matchings in bipartite graphs and the concept of integrality of the bipartite perfect matching polytope.

### Step 1: Understanding the Structure of the Graph
A \( k \)-regular bipartite graph has equal numbers of vertices in both sets, \( |A| = |B| = n \), and each vertex has degree \( k \). This means every vertex in both \( A \) and \( B \) is connected to exactly \( k \) vertices in the opposite set.

### Step 2: Perfect Matching Definition
A perfect matching in a bipartite graph is a subset of edges that matches every vertex in \( A \) with a unique vertex in \( B \) and vice versa. Our goal is to find \( k \) such matchings that are disjoint, meaning no edge can be used in more than one matching.

### Step 3: Utilizing the Bipartite Perfect Matching Polytope
The perfect matching polytope for a bipartite graph is defined as the convex hull of the incidence vectors of all perfect matchings. This polytope is integral, which means that any linear programming solution to it will yield integer solutions corresponding to the number of matchings each edge is used in.

### Step 4: Linear Programming Formulation
We can formulate a linear programming problem to maximize the total usage of edges across \( k \) matchings.

- Let \( x_e \) be a variable representing the number of matchings that edge \( e \) is included in.
- The objective is to maximize \( \sum_{e \in E} x_e \).
- The constraints ensure that each vertex in both sets \( A \) and \( B \) is included in exactly \( k \) matchings:
  \[
  \sum_{e \text{ incident on } v} x_e = k \quad \text{for all } v \in A \cup B
  \]
- Additionally, \( x_e \geq 0 \) and should be integer-valued.

### Step 5: Finding Disjoint Perfect Matchings Iteratively
1. **First Matching**: Solve the linear program to find a perfect matching. Since the solution to the linear program is integral, we can find a perfect matching \( M_1 \) using edges from \( E \).

2. **Graph Reduction**: After obtaining the first perfect matching \( M_1 \), we remove the edges of \( M_1 \) from \( G \). This results in a new graph \( G' \) that is \( (k-1) \)-regular (since each vertex was originally \( k \)-regular and we have used one edge from each vertex).

3. **Repeat**: We repeat the process on the new graph \( G' \) to find a second perfect matching \( M_2 \). We continue this process, finding matchings and reducing the graph, until we have found \( k \) disjoint perfect matchings \( M_1, M_2, ..., M_k \).

### Conclusion
By iteratively finding perfect matchings and reducing the degree of the graph, we can guarantee that each edge in the original graph is assigned to exactly one of the \( k \) matchings. Thus, we conclude that the edge set of a \( k \)-regular bipartite graph \( G \) can indeed be partitioned into \( k \) disjoint perfect matchings in polynomial time. This construction is efficient and relies on the integral nature of the bipartite perfect matching polytope, ensuring that all matchings are valid and disjoint.


Question:
For each of the sub-questions of this question (next page), tick/check the corresponding box if the presented sentence is correct
at the corresponding level (for a human). There will be a penalty for wrong boxes ticked/checked.The mouse lost a feather as it took off.

At the lexical level all words are valid English tokens, and syntactically the sentence is well-formed; semantically itâs false (mice have no feathers), and pragmatically itâs odd, so only lexical and syntactic are correct.


Question:
Let \( P(n) \) be a proposition for a positive integer \( n \) (positive integers do not include 0).

You have managed to prove that \( orall k > 2, \left[ P(k-2) \wedge P(k-1) \wedge P(k)
ight]
ightarrow P(k+1) \).

You would like to prove that \( P(n) \) is true for all positive integers. What is left for you to do ?

Show that \( P(1)\), \(P(2)\), and potentially use other values like \(P(3)\) are true; then apply strong induction starting from there to conclude that valid implications lead us through all positive integers effectively asserting statement holds universally across them due being able to satisfy conditions repeatedly with each leap made by employing previous truths derived through valid extension of established inductions.

Thus: "Show that \(P(1)\)," \(P(2)\)", etc., are true; conclude via strong induction."


Question: Suppose that q is density reachable from p. The chain of points that ensure this relationship are {t,u,g,r} Which one is FALSE?

In the context of density reachability in clustering algorithms, particularly with DBSCAN, if point \( q \) is density reachable from point \( p \), it means that there is a sequence of points connecting \( p \) to \( q \) such that each point in the sequence is within a specified distance from the previous point and has a sufficient number of nearby points (based on a defined minimum density).

Given the chain of points {t, u, g, r} that ensures the relationship between \( p \) and \( q \), the statement that is likely false would be something that contradicts the definition of density reachability. For instance, if one of the points in the chain is not within the required distance from the previous point or does not have enough nearby points to satisfy the density condition, then that point would break the chain.

For example, if we say "point g is not within the required distance from point u," this would be false in the context of density reachability because it violates the necessary condition for \( q \) to be density reachable from \( p \) through the sequence of points.

Thus, to identify which statement about the points in the chain is false, look for one that indicates a failure in maintaining the required conditions of proximity or density between consecutive points in the chain leading from \( p \) to \( q \).


Question:
Consider the following grammar:
S -> NP VP
NP -> Det N
VP -> VBe Adj
NP -> NP PP
VP -> V
N -> Adj N
VP -> VP PP
Adj -> Adj PP
V -> VBe
Adj -> Ving
PP -> Prep NP
and the following lexicon:
at:Prep is:VBe old:Adj
black:Adj looking:Ving the:Det
cat:N mouse:N under:Prep
former:Adj nice:Adj with:Prep

The above grammar over-generates. One reason is that some adjectives, e.g. former, can only
occur before a noun. For instance
the cat is former
is incorrect in English (but accepted by the above grammar).
Another reason for over-generation is that PPs do not combine with adjectives occurring before
a noun. For instance:
the looking at the mouse cat is black
is incorrect in English (but accepted by the above grammar).
Explain how the above grammar might be modified to prevent these two types of over-generation.

To address the issue of over-generation in the given grammar, we need to establish a clear distinction between different types of adjectives and their placement in sentence structures. The first step involves recognizing that some adjectives, such as 'former', are inherently restricted to occur only before nouns and cannot function as predicates in a verb phrase. Therefore, we can modify the grammar to differentiate adjectives based on their syntactic roles. This can be achieved by introducing a new category for adjectives that can appear predicatively (Adj+) and those that can only appear attributively (Adj-).

For example, we can define a new production rule that specifically categorizes adjectives:
- Adj- -> Adj
- Adj+ -> Adj | Ving

This distinction allows us to specify that 'former' is an Adj- and can only occur in the construction NP -> Det Adj- N, thereby preventing sentences like "the cat is former" from being generated.

Additionally, we need to address the second issue regarding the syntactic combination of prepositional phrases (PPs) with adjectives that appear before nouns. The current grammar does not restrict the combination of PPs with adjectives, leading to incorrect constructions such as "the looking at the mouse cat is black." To remedy this, we can modify the grammar rules to ensure that PPs cannot attach to Adj-.

Thus, we can introduce constraints like:
- VP -> VBe Adj+
- NP -> Det Adj- N

By doing so, we ensure that when adjectives appear before nouns, they cannot be followed by a PP, thus maintaining correct grammatical structure.

In summary, the modifications involve clearly separating adjectives into two categories based on their syntactic roles and enforcing constraints on how PPs interact with adjectives. By implementing these changes, we can significantly reduce over-generation and improve the accuracy of the grammar.


Question:
Consider two Information Retrieval systems S1 and S2 that produced the following outputs for
the 4 reference queries q1, q2, q3, q4:
S1:                                     | referential:
q1: d01 d02 d03 d04 dXX dXX dXX dXX     | q1: d01 d02 d03 d04
q2: d06 dXX dXX dXX dXX                 | q2: d05 d06
q3: dXX d07 d09 d11 dXX dXX dXX dXX dXX | q3: d07 d08 d09 d10 d11
q4: d12 dXX dXX d14 d15 dXX dXX dXX dXX | q4: d12 d13 d14 d15
S2::                                    | referential:
q1: dXX dXX dXX dXX d04                 | q1: d01 d02 d03 d04
q2: dXX dXX d05 d06                     | q2: d05 d06
q3: dXX dXX d07 d08 d09                 | q3: d07 d08 d09 d10 d11
q4: dXX d13 dXX d15                     | q4: d12 d13 d14 d15
where dXX refer to document references that do not appear in the referential. To make the
answer easier, we copied the referential on the right.
For each of the two systems, compute the mean Precision and Recall measures (provide the
results as fractions). Explain all the steps of your computation.

To calculate the mean Precision and Recall for the two information retrieval systems (S1 and S2), we will follow a systematic approach for each query:

### Definitions:
- **Precision**: The proportion of relevant documents retrieved out of all documents retrieved.
  \[
  \text{Precision} = \frac{\text{True Positives}}{\text{True Positives} + \text{False Positives}}
  \]
- **Recall**: The proportion of relevant documents retrieved out of all relevant documents in the reference set.
  \[
  \text{Recall} = \frac{\text{True Positives}}{\text{True Positives} + \text{False Negatives}}
  \]

### Step-by-Step Calculation

#### System S1:

1. **For query q1**:
   - Retrieved: \(d01, d02, d03, d04\)
   - Relevant: \(d01, d02, d03, d04\)
   - True Positives (TP): 4, False Positives (FP): 0, False Negatives (FN): 0
   \[
   \text{Precision} = \frac{4}{4} = 1, \quad \text{Recall} = \frac{4}{4} = 1
   \]

2. **For query q2**:
   - Retrieved: \(d06\)
   - Relevant: \(d05, d06\)
   - TP: 1 (only \(d06\)), FP: 0, FN: 1 (missing \(d05\))
   \[
   \text{Precision} = \frac{1}{1} = 1, \quad \text{Recall} = \frac{1}{2} = \frac{1}{2}
   \]

3. **For query q3**:
   - Retrieved: \(d07, d09, d11\)
   - Relevant: \(d07, d08, d09, d10, d11\)
   - TP: 3 (\(d07, d09, d11\)), FP: 0, FN: 2 (\(d08, d10\))
   \[
   \text{Precision} = \frac{3}{3} = 1, \quad \text{Recall} = \frac{3}{5} = \frac{3}{5}
   \]

4. **For query q4**:
   - Retrieved: \(d12, d14, d15\)
   - Relevant: \(d12, d13, d14, d15\)
   - TP: 3 (\(d12, d14, d15\)), FP: 0, FN: 1 (\(d13\))
   \[
   \text{Precision} = \frac{3}{3} = 1, \quad \text{Recall} = \frac{3}{4} = \frac{3}{4}
   \]

**Mean Precision for S1**:
\[
\text{Mean Precision} = \frac{1 + 1 + 1 + 1}{4} = 1
\]

**Mean Recall for S1**:
\[
\text{Mean Recall} = \frac{1 + \frac{1}{2} + \frac{3}{5} + \frac{3}{4}}{4} = \frac{1 + 0.5 + 0.6 + 0.75}{4} = \frac{2.85}{4} = \frac{57}{80}
\]

---

#### System S2:

1. **For query q1**:
   - Retrieved: \(d04\)
   - Relevant: \(d01, d02, d03, d04\)
   - TP: 1 (\(d04\)), FP: 0, FN: 3 (\(d01, d02, d03\))
   \[
   \text{Precision} = \frac{1}{1} = 1, \quad \text{Recall} = \frac{1}{4} = \frac{1}{4}
   \]

2. **For query q2**:
   - Retrieved: \(d05, d06\)
   - Relevant: \(d05, d06\)
   - TP: 2 (\(d05, d06\)), FP: 0, FN: 0
   \[
   \text{Precision} = \frac{2}{2} = 1, \quad \text{Recall} = \frac{2}{2} = 1
   \]

3. **For query q3**:
   - Retrieved: \(d07, d08, d09\)
   - Relevant: \(d07, d08, d09, d10, d11\)
   - TP: 3 (\(d07, d08, d09\)), FP: 0, FN: 2 (\(d10, d11\))
   \[
   \text{Precision} = \frac{3}{3} = 1, \quad \text{Recall} = \frac{3}{5} = \frac{3}{5}
   \]

4. **For query q4**:
   - Retrieved: \(d13, d15\)
   - Relevant: \(d12, d13, d14, d15\)
   - TP: 2 (\(d13, d15\)), FP: 0, FN: 2 (\(d12, d14\))
   \[
   \text{Precision} = \frac{2}{2} = 1, \quad \text{Recall} = \frac{2}{4} = \frac{1}{2}
   \]

**Mean Precision for S2**:
\[
\text{Mean Precision} = \frac{1 + 1 + 1 + 1}{4} = 1
\]

**Mean Recall for S2**:
\[
\text{Mean Recall} = \frac{\frac{1}{4} + 1 + \frac{3}{5} + \frac{1}{2}}{4} = \frac{\frac{1}{4} + \frac{4}{4} + \frac{12}{20} + \frac{10}{20}}{4} = \frac{\frac{1 + 4 + 3 + 2}{4}}{4} = \frac{10}{20} = \frac{1}{2}
\]

### Summary of Results:
- **S1 Mean Precision:** \(1\), **Mean Recall:** \(\frac{57}{80}\)
- **S2 Mean Precision:** \(1\), **Mean Recall:** \(\frac{1}{2}\)

These results indicate that both systems have perfect precision, but S1 retrieves a higher proportion of relevant documents compared to S2.


Question: Tick the \emph{incorrect} assertion. In \emph{all} finite Abelian groups $(G,+)$ \ldots

Answer:**
2) all the elements are generators.

**Explanation:**
The incorrect assertion is that "all the elements are generators." While certain elements may serve as generators for the group, this is not universally true for all elements in a finite Abelian group. For instance, in the group \( \mathbb{Z}/6\mathbb{Z} \), the element \( 2 \) is not a generator of the entire group, as it only generates the subgroup \( \{0, 2, 4\} \). Hence, not every element can generate the entire group, which makes assertion 2 the incorrect one. Assertions 1, 3, and 4 are true based on established properties of finite Abelian groups.


Question: Select the \emph{incorrect} statement

To identify the incorrect statement among the options provided, let's analyze each statement step-by-step:

**A) RC4, A5/1, E0 are stream ciphers.**
- **True.** RC4 is a widely known stream cipher used in various encryption protocols. A5/1 is a stream cipher used in GSM (Global System for Mobile Communications) for encrypting voice and data. E0 is the stream cipher used in Bluetooth for encrypting data. Therefore, this statement is correct.

**B) MD4, MD5, SHA0, SHA1 are hash functions.**
- **True.** MD4, MD5, SHA0, and SHA1 are all cryptographic hash functions. MD4 was designed by Ronald Rivest, MD5 is a widely used hash function that produces a 128-bit hash value, SHA0 was an early version of SHA (Secure Hash Algorithm), and SHA1 produces a 160-bit hash value. Thus, this statement is also correct.

**C) DES, AES are block ciphers.**
- **True.** DES (Data Encryption Standard) and AES (Advanced Encryption Standard) are indeed block ciphers. DES processes data in 64-bit blocks, while AES works with block sizes of 128 bits, 192 bits, or 256 bits. This statement is correct.

**D) ECB, KDF, PRF are commitment schemes.**
- **False.** ECB (Electronic Codebook) is a mode of operation for block ciphers, not a commitment scheme. KDF (Key Derivation Function) is used to derive cryptographic keys from a secret value, and PRF (Pseudorandom Function) is used in various cryptographic applications but is not a commitment scheme either. Commitment schemes are cryptographic protocols that allow one party to commit to a value while keeping it hidden, with the ability to reveal it later.

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

**D) ECB, KDF, PRF are commitment schemes.**


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

2. State space explosion is a common challenge for symbolic execution.
1. Symbolic execution can efficiently handle and solve constraints in programs with simple logics but large input space.

These two options are correct. State space explosion refers to the overwhelming growth of possible execution paths in complex programs, which can make analyzing them quite difficult. On the other hand, symbolic execution is indeed capable of managing constraints effectively when the logical conditions are not too complicated, allowing it to deal with a large number of inputs. However, this doesn't mean it's always the best choice for all types of programs. The other options present misunderstandings about its capabilities, particularly regarding environmental modeling and the need for program execution.


Question: Consider a matrix factorization problem of the form $\mathbf{X}=\mathbf{W Z}^{\top}$ to obtain an item-user recommender system where $x_{i j}$ denotes the rating given by $j^{\text {th }}$ user to the $i^{\text {th }}$ item . We use Root mean square error (RMSE) to gauge the quality of the factorization obtained. Select the correct option.

**Plan:**
1. Analyze the implications of adding a new item to the recommender system and the need for retraining.
2. Evaluate the regularization terms applied to the matrices \(\mathbf{W}\) and \(\mathbf{Z}\) and their effect on the convexity of the objective function.
3. Assess the relationship between the latent dimension \(\mathrm{K}\), the number of rows \(D\), and the number of elements \(N\) in the context of matrix factorization.
4. Determine which statement among the options is true based on the analysis.

**Steps:**

1. **Analyzing Retraining with New Items**: When introducing a new item into a recommender system, it is often necessary to update the model to incorporate the new information. However, complete retraining from scratch is typically not required if the existing model can effectively utilize user ratings for the new item. Techniques like incremental learning may allow the system to adapt without full retraining. Therefore, the first statement is likely incorrect.

2. **Evaluating Regularization and Convexity**: In matrix factorization problems, regularization terms are added to the objective function to prevent overfitting. The regularization terms often take the form of the Frobenius norms of \(\mathbf{W}\) and \(\mathbf{Z}\). While adding these terms can help stabilize the optimization process, it does not guarantee that the resulting objective function is convex in all cases. The statement that it becomes convex is generally misleading because the overall optimization landscape may still be non-convex due to the interactions between \(\mathbf{W}\) and \(\mathbf{Z}\). Thus, this statement is not necessarily true.

3. **Assessing the Latent Dimension \(\mathrm{K}\)**: The latent dimension \(\mathrm{K}\) represents the number of features that capture the underlying structure of the data. In general, for robust factorization, \(\mathrm{K}\) should be much smaller than \(D\) (the number of rows) but can be larger than \(N\) (the number of elements). If \(N\) is much smaller than \(D\), having \(\mathrm{K}\) between \(D\) and \(N\) does not make practical sense, as it would lead to overfitting and not capture the complexity of the data effectively. Therefore, this statement is also misleading.

4. **Conclusion**: Since all analyzed statements have shown inaccuracies or misleading claims, the correct option must be that none of the statements are correct.

**Self-Check:**
- The first statement about needing to retrain from scratch is incorrect; incremental updates are possible.
- The second statement about regularization making the objective convex is misleading since it does not guarantee convexity.
- The third statement about the latent dimension \(\mathrm{K}\) being between \(D\) and \(N\) is incorrect as it contradicts the principles of dimensionality in matrix factorization.

Therefore, the answer is **D**: "None of the other options are correct."


Question:
Consider the following snippet used to produce a
                    high-performance circuit using a statically scheduled HLS tool, such
                    as Xilinx Vivado HLS. Assume that a erb+double+ multiplication
                    takes several cycles (latency) to compute.
                    egin{verbatim}
                        double a[ARRAY_SIZE] = ...;
                        int b = 1;
                        for (int i = 0; i < ARRAY_SIZE; i++)
                            if (a[i] * (double) b >= CONST)
                                b++;
                    \end{verbatim}
                    For the same snippet, would a dynamically scheduled circuit
                    naturally achieve better performance? If so, does the answer depend
                    on some particular technique applied in the design of the
                    dynamically scheduled circuit? Which one? Explain.

Answer: 1. Yes, a dynamically scheduled circuit would likely achieve better performance due to its ability to handle varying latencies and dependencies more flexibly. The performance gain depends on techniques such as loop unrolling and pipelining, which allow for overlapping computations and better utilization of hardware resources, reducing idle cycles during the execution of the multiplication operation.

2. In this case, dynamic scheduling can better manage the latency of the multiplication operation by allowing other operations to proceed while waiting for the multiplication result, thus improving throughput.


Question: When are paired t-tests helpful? Justify.

Paired t-tests are a statistical method used to determine whether there is a significant difference between the means of two related groups. This test is particularly useful in specific scenarios, which I will outline step by step:

### 1. **Understanding the Context**:
   - A paired t-test is applicable when the samples are not independent. This means that each observation in one sample is linked to an observation in another. Common situations include before-and-after studies, matched subjects, or repeated measurements on the same subjects.

### 2. **Key Conditions**:
   - **Related Groups**: The data must consist of two sets of related observations. For example, measuring the weight of individuals before and after a diet program.
   - **Normality**: The differences between paired observations should be approximately normally distributed, especially important in small sample sizes (n < 30).
   - **Scale of Measurement**: The data should be continuous (interval or ratio scale).

### 3. **When to Use a Paired t-test**:
   - **Before-and-After Studies**: When you want to assess the impact of an intervention. For instance, measuring blood pressure before and after treatment in the same group of patients.
   - **Matched Samples**: When each subject in one sample is paired with a subject in the other sample based on specific characteristics. For example, comparing test scores of students who received different teaching methods but are similar in age and background.
   - **Repeated Measures**: When the same subjects are measured multiple times under different conditions. An example is measuring student performance at multiple points throughout a semester.

### 4. **Justification**:
   - **Reduction of Variability**: Since paired observations are related, the variability due to individual differences is minimized, allowing for a more powerful test.
   - **Focus on Changes**: The paired t-test directly assesses the changes between two related measurements, which is often of primary interest in research contexts.
   - **Simplicity in Interpretation**: The results of a paired t-test are straightforward, as they provide a clear indication of whether the mean difference is statistically significant.

### 5. **Mathematical Foundation**:
The paired t-test statistic is calculated using the formula:

\[
t = \frac{\bar{d}}{s_d/\sqrt{n}}
\]

Where:
- \( \bar{d} \) is the mean of the differences between paired observations,
- \( s_d \) is the standard deviation of the differences,
- \( n \) is the number of pairs.

The computed t-value is then compared to a critical value from the t-distribution with \( n-1 \) degrees of freedom to determine statistical significance.

### Conclusion:
In summary, paired t-tests are particularly helpful in scenarios involving related samples where the aim is to evaluate the mean differences due to interventions or conditions, thereby providing insights that independent tests may not adequately reveal. They are crucial in experimental design, clinical research, and many fields where comparisons of paired data are relevant.


Question:
Give well chosen examples of applications that can be evaluated with the single metric derived from Precision/Recall and illustrate:
• a situation where more weight should be given to Precision;
• a situation where more weight should be given to Recall.

In evaluating applications with respect to Precision and Recall metrics, we can consider two distinct situations that illustrate when to prioritize each metric:

1. **Situation Favoring Precision**: Spam Email Filtering
   - In scenarios like spam detection systems, itâs crucial to minimize false positives since misclassifying important emails can lead to severe repercussions in communication and workflow. Therefore, maximizing precision ensures that fewer legitimate emails are incorrectly classified as spam.

2. **Situation Favoring Recall**: Disease Screening
   - Conversely, in contexts such as medical diagnostics for serious diseases like cancer or infectious diseases, recall takes precedence. The primary concern here is ensuring that almost all actual cases are identified; missing a true positive could result in grave health risks for patientsâeven if this means accepting a higher rate of false positives among healthy individuals.

By considering these examples against their respective contextsâspam detection prioritizing Precision due to potential communication failures and disease screening emphasizing Recall due to life-critical outcomesâwe gain insight into how best to apply these evaluation metrics effectively in real-world scenarios.


Question: We want to generate a $\ell$-bit prime. The complexity is roughly\dots


Question: Which of the following is WRONG for Ontologies?

1. "The question asks which statement about ontologies is wrong."
2. "Option A is correct because ontologies are essential for interoperability."
3. "Option B is correct as ontologies facilitate the integration of data."
4. "Option C is correct because ontologies allow the specification of schemas."
5. "Option D is incorrect because ontologies do not dictate how data is serialized; they focus on the meaning of the data."

**Correct Answer:**
- D: "They dictate how semi-structured data are serialized."


Question: Tick the \textbf{false} assertion. The SEI of the distribution $P$ of support $G$ \ldots

To determine which assertion about the SEI (Squared Euclidean Imbalance) of the distribution \( P \) of support \( G \) is false, we will analyze each option step by step, considering potential edge cases.

### Assertion 1:
**"is equal to \( \# G \cdot \sum_{x \in G} \left( P(x) - \frac{1}{\# G} \right)^2 \)"**

**Analysis:**
- The SEI is indeed defined as the sum of squared deviations of the probabilities from the uniform distribution, scaled by the size of the support \( G \).
- The expression \( \sum_{x \in G} \left( P(x) - \frac{1}{\# G} \right)^2 \) calculates the variance of the distribution \( P \) compared to the uniform distribution.
- Since \( \# G \) is a constant (the size of the support), multiplying by \( \# G \) gives the correct scaling for the SEI.

**Conclusion:** This assertion is **true**.

### Assertion 2:
**"is the advantage of the best distinguisher between \( P \) and the uniform distribution."**

**Analysis:**
- The advantage of a distinguisher \( D \) is defined as \( | \Pr[D(P) = 1] - \Pr[D(U) = 1]| \), where \( U \) is the uniform distribution.
- The SEI captures how much \( P \) deviates from uniformity but does not directly measure the advantage of the best distinguisher.
- In general, the advantage can be related to the SEI, but they are not equivalent; the best distinguisher's advantage depends on the specific strategy used, not just the SEI.

**Edge Case:** Consider distributions where the SEI is low, but a clever distinguisher might still find significant advantage due to specific patterns in \( P \).

**Conclusion:** This assertion is **false**.

### Assertion 3:
**"denotes the Squared Euclidean Imbalance."**

**Analysis:**
- The term "Squared Euclidean Imbalance" is often used synonymously with the SEI, encapsulating the idea of measuring the imbalance (or deviation from uniformity) in a squared form.
- Therefore, this definition is consistent with the established terminology in probability theory.

**Conclusion:** This assertion is **true**.

### Assertion 4:
**"is positive."**

**Analysis:**
- The SEI is defined as a sum of squared terms, \( \sum_{x \in G} \left( P(x) - \frac{1}{\# G} \right)^2 \), and since squares of real numbers are non-negative, the sum will also be non-negative.
- The SEI equals zero only when \( P \) is the uniform distribution, in which case there is no imbalance.

**Edge Case:** The only case where SEI would not be positive is when the distribution \( P \) is uniform, making the SEI exactly zero.

**Conclusion:** This assertion is generally **true**, but it can be zero in the uniform case.

### Final Conclusion:
The **false** assertion is the second one:
- **Assertion 2**: "is the advantage of the best distinguisher between \( P \) and the uniform distribution."


Question: Which of the following statements are correct?

Let's evaluate each of the statements carefully:

1. **One iteration of standard SGD for SVM costs roughly \(\Theta(D)\), where \(D\) is the dimension.**
   - **Correct.** In standard stochastic gradient descent (SGD) for a support vector machine (SVM), each iteration involves computing the gradient of the loss function with respect to the model parameters. This gradient calculation typically requires operations that scale linearly with the number of features (dimensions) \(D\), resulting in a cost of \(\Theta(D)\).

2. **Unions of convex sets are convex.**
   - **Incorrect.** The union of convex sets is not necessarily convex. For instance, if you take two convex sets that do not overlap, their union will not be convex because there exist points in the union that cannot be connected by a straight line without leaving the union.

3. **Hinge loss (as in SVMs) is typically preferred over L2 loss (least squares loss) in classification tasks.**
   - **Correct.** Hinge loss is specifically designed for classification problems, particularly for maximum-margin classifiers like SVMs. It emphasizes the margin between classes and is generally more suitable for classification than L2 loss, which is more appropriate for regression tasks.

4. **In PCA, the first principal direction is the eigenvector of the data matrix \(\boldsymbol{X}\) with the largest associated eigenvalue.**
   - **Incorrect.** While it is true that the first principal component is related to the largest eigenvalue, it is more accurate to say that it is the eigenvector of the covariance matrix of the data (or the centered data matrix) that corresponds to the largest eigenvalue. The statement is misleading because it refers directly to the data matrix \(\boldsymbol{X}\) instead of the covariance matrix derived from it.

5. **MSE (mean squared error) is typically more sensitive to outliers than MAE (mean absolute error).**
   - **Correct.** The mean squared error calculates the square of the differences between predicted and actual values, which means that larger errors are disproportionately weighted. This makes MSE more sensitive to outliers compared to mean absolute error, which treats all errors linearly.

6. **One iteration of standard SGD for logistic regression costs roughly \(\Theta(N D)\), where \(N\) is the number of samples and \(D\) is the dimension.**
   - **Incorrect.** In standard SGD for logistic regression, each iteration typically processes one sample (or a small batch), resulting in a cost of \(\Theta(D)\) for each sample, not \(\Theta(N D)\). The latter would imply processing all samples simultaneously, which is not how SGD operates.

To summarize, the correct statements are:

- **1.** Correct
- **3.** Correct
- **5.** Correct

Thus, the correct answers are: **Hinge loss (as in SVMs) is typically preferred over L2 loss (least squares loss) in classification tasks; MSE (mean squared error) is typically more sensitive to outliers than MAE (mean absolute error); One iteration of standard SGD for SVM costs roughly \(\Theta(D)\), where \(D\) is the dimension.**


Question:
Recall that we say that a kernel $K: \R 	imes \R
ightarrow \R $ is
		valid if there exists $k \in \mathbb{N}$ and $\Phi: \R
ightarrow \R^k$
		such that for all $(x, x') \in \R 	imes \R $, $K(x, x') =  \Phi(x)^	op \Phi(x')$. The kernel $K(x, x') = \cos(x + x')$ is a valid kernel.

To determine if the kernel \( K(x, x') = \cos(x + x') \) is a valid kernel, we need to verify whether it can be expressed in the form \( K(x, x') = \Phi(x)^\top \Phi(x') \) for some mapping \( \Phi: \mathbb{R} \to \mathbb{R}^k \).

1. **Understanding the Kernel Definition**:
   A kernel is considered valid if it can be represented as an inner product of two feature vectors \( \Phi(x) \) and \( \Phi(x') \).

2. **Analyzing the Function**:
   The function \( K(x, x') = \cos(x + x') \) can be rewritten using the cosine addition formula:
   \[
   \cos(x + x') = \cos x \cos x' - \sin x \sin x'
   \]
   This suggests a relationship between the kernel and trigonometric functions.

3. **Finding the Feature Mapping**:
   We can define the mapping \( \Phi \) as:
   \[
   \Phi(x) = \begin{pmatrix} \cos x \\ \sin x \end{pmatrix}
   \]
   This is a mapping from \( \mathbb{R} \) to \( \mathbb{R}^2 \).

4. **Computing the Inner Product**:
   Using this mapping, we compute:
   \[
   K(x, x') = \Phi(x)^\top \Phi(x') = \begin{pmatrix} \cos x \\ \sin x \end{pmatrix}^\top \begin{pmatrix} \cos x' \\ \sin x' \end{pmatrix} = \cos x \cos x' + \sin x \sin x' = \cos(x - x')
   \]
   This does not yield the original function \( K(x, x') = \cos(x + x') \).

5. **Conclusion**:
   The original kernel \( K(x, x') = \cos(x + x') \) does not appear to be representable as an inner product of the form \( \Phi(x)^\top \Phi(x') \). Therefore, we conclude that it does not satisfy the criteria to be a valid kernel.

Final answer: **False**.


Question: Compared to the plain RSA cryptosystem and for equivalent key sizes, the plain Elgamal cryptosystem has\dots

To compare the plain RSA cryptosystem with the Elgamal cryptosystem, we need to analyze their key characteristics, particularly focusing on key generation, encryption, decryption, and ciphertext size.

1. **Key Generation**: The RSA key generation involves selecting two large prime numbers, calculating their product, and deriving public and private keys from that product. This process can be complex due to the need for efficient primality testing and large number operations. In contrast, the Elgamal key generation involves selecting a large prime and a generator, which is generally seen as simpler.

2. **Encryption Algorithm**: The RSA encryption algorithm is relatively straightforward, involving modular exponentiation. However, Elgamal encryption involves more steps, including randomly choosing a number for each encryption, which makes it more complex.

3. **Decryption Algorithm**: RSA decryption is also straightforward, requiring modular exponentiation with the private key. Elgamal decryption, while not overly complex, involves an additional step of using the multiplicative inverse, which adds complexity.

4. **Ciphertext Size**: RSA typically produces shorter ciphertexts because the ciphertext is just the result of the modular exponentiation of the plaintext. In Elgamal, the ciphertext size is larger because it consists of two components, each of which is the size of the original plaintext.

Considering these points, we can conclude that the plain Elgamal cryptosystem has a simpler key generation algorithm compared to RSA for equivalent key sizes.

Correct answer: 1. a simpler key generation algorithm.


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

Thank you for the opportunity to revisit the analysis of index merging in the context of constructing inverted files. Let's go through the options again with a more focused lens on their accuracy and implications.

1. **"While merging two partial indices on disk, the inverted lists of a term are concatenated without sorting."** - This statement can be misleading. While it is true that inverted lists can be combined, they must be sorted to maintain the correct order for efficient search operations. Therefore, this statement is not accurate as it does not reflect the standard requirement for sorted lists.

2. **"Index merging is used when the vocabulary does no longer fit into the main memory."** - This statement is indeed correct. As the size of the vocabulary grows with the addition of more documents, it's common for it to exceed the capacity of the main memory. Merging indices allows for efficient management of large datasets and ensures that the system can handle larger vocabularies effectively.

3. **"The size of the final merged index file is O(nlog2(n)*M), where M is the size of the available memory."** - This statement is misleading. The size of the final merged index is generally not expressed in this manner. The size is more closely related to the number of unique terms and documents rather than being dependent on memory size in this logarithmic format.

4. **"While merging two partial indices on disk, the vocabularies are concatenated without sorting."** - Like the first statement, this suggests that vocabularies can be joined without sorting, which is not the standard procedure. Proper merging requires sorting and organizing terms to ensure the efficiency of retrieval operations.

After this careful re-evaluation, the most accurate statement about index merging remains:

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

This highlights a fundamental reason for performing index merges in information retrieval systems, ensuring that they can efficiently manage and query large sets of data without exceeding memory constraints.


Question:
A service is an application component that performs long-running operations, usually in the background. A service doesn't provide a UI.
While reviewing the pull request of a friend you notice that he periodically fetches data from the cloud in his activity? What potential problem this could lead to, and how can you fix it?

In reviewing the pull request of my friend, I noticed that he is periodically fetching data from the cloud directly within the main activity thread. This can lead to several potential problems, primarily due to the role of the main thread in a mobile application.

### Implications of Network Operations in the Main Thread

1. **UI Unresponsiveness**: The main thread is responsible for handling UI updates and user interactions. When long-running operations, such as network requests, are executed on this thread, they can block UI updates. This results in the application becoming unresponsive to user inputs, leading to a poor user experience.

2. **Application Crashes**: If the network request takes too long or if there is a delay in the response due to network issues, the application may eventually throw an `Application Not Responding (ANR)` error. This can lead to crashes, which not only frustrate users but can also affect the application's reputation.

### Alternative Methods to Handle Long-Running Tasks

To mitigate these issues, it is essential to move network operations off the main thread. Here are some alternative approaches:

1. **Using AsyncTask**: This class allows you to perform background operations and publish results on the UI thread without having to manipulate threads and handlers.

2. **Using Loaders**: Loaders can be used to manage asynchronous loading of data in an activity or fragment, automatically handling configuration changes.

3. **Using Thread and Handler**: Creating a separate thread for network operations and using a handler to post results back to the main thread can help maintain UI responsiveness.

4. **Using RxJava**: This library allows for reactive programming and can simplify the handling of asynchronous data streams.

### Recommendation for Implementation

The most effective solution in this scenario is to implement a **Service**. A Service is an application component that runs in the background to perform long-running operations. Here’s how to properly implement this:

1. **Create a Background Service**: Implement a Service that handles the network operation. This service can run independently of the main thread, ensuring that the UI remains responsive.

2. **Use Threads within the Service**: Inside the Service, utilize a separate thread or an `IntentService` to perform the network request. This allows the Service to manage its own thread and effectively handle the long-running operation without blocking the UI.

3. **Communicate Results**: Use `BroadcastReceiver` or `LocalBroadcastManager` to send the results back to the main activity once the data is successfully fetched.

### Conclusion

In conclusion, periodically fetching data from the cloud in the main activity thread can lead to UI unresponsiveness and application crashes. To fix this, it is recommended to implement a Service to handle background operations while ensuring that the main thread remains free for user interactions. By using threading within the Service, we can maintain a smooth user experience while performing long-running tasks.


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

Protocol: the sender BEB-broadcasts m; any process that receives m for the first time immediately BEB-rebroadcasts it (implicit ACK). Each node tracks ack[m], the set of distinct processes from which it has received m. A process delivers m once |ack[m]| > N/2. Because a majority of correct nodes eventually ACK, if one correct process delivers, that majorityâs echoes reach every other correct process, ensuring uniform delivery. No failure detector is neededâjust majority ACKs with BEB.


Question:
Ignoring their different evaluation characteristics in this exercise, we consider here
that filter and withFilter are equivalent. To which expression is the following for-loop translated ?
1 def mystery7(xs : List[Int], ys : List[Int]) : List[Int] =
2 for
3 y <- ys if y < 100
4 x <- xs if x < 20
5 yield
6 if y < x then 0 else y - x

Answer:
```scala
ys.filter(y => y < 100).flatMap(y => xs.filter(x => x < 20).map(x => if (y < x) 0 else y - x))
```

### Explanation:
In this translation, we first filter the `ys` list to include only those elements less than 100, preserving the outer loop structure. For each valid `y`, we then apply `flatMap` to `xs` after filtering it to include only elements less than 20, maintaining the inner loop structure. The `map` function takes each valid `x` and applies the transformation defined in the `yield`, which checks if `y` is less than `x` and returns 0 if true; otherwise, it computes `y - x`. This accurately reflects the nested iteration behavior of the original for-loop, ensuring the correct sequence and structure of operations.


Question: Consider the following sequence of random variables $S_1,\ldots,S_n,\ldots$ Assume that the limit $H^\star(\mathcal{S})=k$ exists and is finite. Suppose that there exists $\hat{n}>0$ such that for all $i\geq \hat{n}$ one has that the marginal distributions of $S_{i+1}$ and $S_i$ satisfy $p_{S_{i+1}}=p_{S_i}$. Denote with $\mathcal{Y}_{\hat{n}}$ the alphabet of the source $S_{\hat{n}}$. True or false: Can one use this information to infer that the following holds: $|\mathcal{Y}_{\hat{n}}| \geq 2^k $?

To determine whether the statement "one can use this information to infer that \( |\mathcal{Y}_{\hat{n}}| \geq 2^k \)" is true or false, we can analyze the components involved.

1. **Definition of \( H^\star(\mathcal{S}) \)**: The limit \( H^\star(\mathcal{S}) = k \) represents the asymptotic entropy rate of the sequence of random variables. This value indicates the average amount of information produced per symbol in the long run.

2. **Marginal distributions**: The problem states that for all \( i \geq \hat{n} \), the marginal distributions of \( S_{i+1} \) and \( S_i \) are equal, meaning that the sequence has reached a stationary distribution from that point onward. This implies the randomness and distribution of the variables do not change as we go further along the sequence.

3. **Alphabet size**: The alphabet of the source \( S_{\hat{n}} \), denoted \( \mathcal{Y}_{\hat{n}} \), consists of the distinct symbols that the source can produce. The size \( |\mathcal{Y}_{\hat{n}}| \) corresponds to how many unique symbols are present.

4. **Connection to entropy**: In information theory, the Shannon entropy \( H \) of a discrete random variable is maximized when the variable is uniformly distributed over its possible values. For a random variable taking values in an alphabet of size \( m \), the maximum entropy is \( \log_2 m \). Therefore, if the entropy \( H \) is \( k \), we can conclude that:
   \[
   H \leq \log_2 |\mathcal{Y}_{\hat{n}}|
   \]
   This implies:
   \[
   k \leq \log_2 |\mathcal{Y}_{\hat{n}}|
   \]
   From this, we can deduce:
   \[
   |\mathcal{Y}_{\hat{n}}| \geq 2^k
   \]

Given these points, we can conclude that the information about the stationary distribution and the limit of the entropy does indeed allow us to infer that \( |\mathcal{Y}_{\hat{n}}| \geq 2^k \).

Final answer: **A. True**


Question: What is WRONG regarding the Transformer model?

The correct answer to the question about what is WRONG regarding the Transformer model is: **"Its computation cannot be parallelized compared to LSTMs and other sequential models."**

Let's evaluate each option:

1. **"It uses a self-attention mechanism to compute representations of the input and output."**
   - **Reasoning:** This statement is true. The Transformer model employs self-attention, allowing it to weigh the importance of different words in a sentence when forming representations.

2. **"Its computation cannot be parallelized compared to LSTMs and other sequential models."**
   - **Reasoning:** This statement is false, which makes it the correct answer. In fact, one of the key advantages of Transformers is their ability to process input data in parallel, unlike LSTMs, which are sequential and must process data step-by-step. This parallelization leads to faster training times.

3. **"Its complexity is quadratic to the input size."**
   - **Reasoning:** This statement is true. The self-attention mechanism has a complexity of O(n²), where n is the number of tokens in the input, because every token attends to every other token.

4. **"It captures the semantic context of the input."**
   - **Reasoning:** This statement is also true. Transformers effectively capture semantic relationships through their attention mechanisms, allowing them to understand context better than previous models.

**Potential misunderstanding:** The misconception may arise from comparing Transformers to LSTMs in terms of processing speed. While LSTMs are inherently sequential, Transformers excel at parallel processing, a key strength that enhances their efficiency and effectiveness in handling large datasets.


Question:
To which expression is the following for-loop translated ?
1 def mystery8(xs : List[List[Int]]) =
2 for
3 x <- xs
4 y <- x
5 yield
6 y

Answer: xs.flatMap(x => x)
 Reasoning: The for-loop iterates over a list of lists, first extracting each sublist (x) from the outer list (xs) and then iterating over each element (y) in the sublist. The result is a flattened list containing all elements from the inner lists, which is exactly what the flatMap function does by applying the function (identity in this case) to each sublist and concatenating the results.


Question: Fitting a Gaussian Mixture Model with a single Gaussian ($K=1$) will converge after one step of Expectation-Maximization.

**Answer: True**

Rationale: When fitting a Gaussian Mixture Model (GMM) with a single Gaussian component ($K=1$), the model effectively reduces to a standard Gaussian distribution. The Expectation-Maximization (EM) algorithm will converge in the first step because the expectation step (E-step) will calculate the responsibilities based entirely on the single Gaussian, and the maximization step (M-step) will update the parameters (mean and covariance) directly based on the entire dataset. Since there is only one Gaussian, the parameters are optimized immediately, leading to convergence after one iteration.


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

The primary reason we cannot directly apply the standard PCA algorithm in the high-dimensional space $\mathbb{R}^{H}$ as described is that we cannot explicitly compute the feature mapping $\phi(\mathbf{x})$ for our data points $\mathbf{x}_1, \ldots, \mathbf{x}_n$. The kernel method relies on the ability to compute the kernel function $k(\mathbf{x}, \mathbf{y}) = \langle \phi(\mathbf{x}), \phi(\mathbf{y}) \rangle_{\mathbb{R}^H}$, which allows us to evaluate the inner product of the mapped features without having to compute the actual mapping $\phi$ itself.

In practice, the dimension $H$ can be extremely large (potentially infinite) when the feature space is derived from a non-linear kernel, making it computationally infeasible to perform operations in that space directly. Specifically, the empirical covariance matrix $\boldsymbol{\Sigma}^{\mathbf{H}} = \frac{1}{n} \sum_{i=1}^{n} \phi(\mathbf{x}_i) \phi(\mathbf{x}_i)^{\top}$ cannot be constructed directly since it requires access to the mapped features $\phi(\mathbf{x}_i)$.

Instead, we can use the kernel trick, which allows us to work with the kernel matrix $\mathbf{K}$ instead of the empirical covariance matrix in the high-dimensional space. The kernel matrix is defined as $\mathbf{K}_{i,j} = k(\mathbf{x}_i, \mathbf{x}_j)$, which captures the inner products of the mapped features without requiring explicit computation of those features.

### Steps to Apply Kernel PCA:

1. **Center the Kernel Matrix**: Compute the centered kernel matrix using the formula:
   \[
   \mathbf{K}_{\text{centered}} = \mathbf{K} - \mathbf{1}_n \mathbf{K} - \mathbf{K} \mathbf{1}_n + \frac{1}{n} \mathbf{1}_n \mathbf{K} \mathbf{1}_n
   \]
   where $\mathbf{1}_n$ is a column vector of ones of size $n$.

2. **Eigenvalue Decomposition**: Compute the eigenvalue decomposition of the centered kernel matrix $\mathbf{K}_{\text{centered}}$. This will give us eigenvalues $\rho_j$ and corresponding eigenvectors $\mathbf{w}_j$.

3. **Project Data**: The projection of the original data points onto the principal components can be computed using the eigenvectors of the centered kernel matrix. Specifically, if we want to project onto the $j$-th component, the projection can be calculated as:
   \[
   \Pi_j(\mathbf{x}_i) = \sum_{k=1}^{n} \mathbf{w}_{jk} k(\mathbf{x}_i, \mathbf{x}_k)
   \]
   where $\mathbf{w}_{jk}$ is the $k$-th entry of the eigenvector corresponding to the $j$-th largest eigenvalue.

By using the kernel trick, we can perform PCA in the high-dimensional feature space implicitly, avoiding the need to compute or store the high-dimensional representations directly. This makes kernel PCA a powerful tool for non-linear dimensionality reduction.