Question: Tick the \textbf{false} statement. Moore's Law ...

Let's analyze each option one by one:

1. **"is partly a reason why some existing cryptosystems are insecure."**
   - Moore's Law states that the number of transistors on a microchip doubles approximately every two years, which leads to increased computational power. This increased power can make certain cryptographic algorithms insecure, particularly those based on key lengths that are insufficient to withstand the increased processing capabilities. Thus, this statement is true.

2. **"was stated by the founder of Intel."**
   - Moore's Law was formulated by Gordon Moore, who is indeed one of the co-founders of Intel. This claim is accurate since Moore made the observation in 1965 that the number of transistors on a chip would continue to double approximately every two years. Therefore, this statement is true.

3. **"assumes the number of transistors per CPU increases exponentially fast with time."**
   - Moore's Law does imply that the number of transistors on a chip increases exponentially over time. This is a fundamental principle of the law. Hence, this statement is true.

4. **"implies that the heat generated by transistors of CPU doubles every 18 months."**
   - Moore's Law does not directly state anything about heat generation. While it is true that as more transistors are packed into a CPU, the heat generated can increase, the specific claim that heat doubles every 18 months is not a conclusion drawn from Moore's Law. The relationship between transistor density and heat generation is more complex and not specified by Moore's observation. Therefore, this statement is false.

Based on the analysis, the false statement is:

**Option 4: "implies that the heat generated by transistors of CPU doubles every 18 months."**

This conclusion is justified because Moore's Law does not address heat generation explicitly, and the specifics of heat generation cannot be reliably derived from the transistor density increase.


Question: Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths:		egin{center}			egin{tabular}{ |c|c|c|c|c|c|c| } 				\hline				& $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ 				\hline				$l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ 				\hline			\end{tabular}		\end{center}  True or false: If $D=2$ there exists a uniquely-decodable code with the same lengths of $\Gamma$.

To determine if a uniquely-decodable code can exist with the same lengths as specified in the encoding $\Gamma$ when \( D=2 \), we can use the Kraft inequality, which is a necessary and sufficient condition for the existence of prefix codes (a specific type of uniquely-decodable code).

The codeword lengths for the symbols in the alphabet \( \mathcal{A} = \{a, b, c, d, e, f\} \) are:
- \( l(\Gamma(a)) = 1 \)
- \( l(\Gamma(b)) = 1 \)
- \( l(\Gamma(c)) = 1 \)
- \( l(\Gamma(d)) = 2 \)
- \( l(\Gamma(e)) = 2 \)
- \( l(\Gamma(f)) = 4 \)

Let’s denote the codeword lengths as \( l_1 = 1, l_2 = 1, l_3 = 1, l_4 = 2, l_5 = 2, l_6 = 4 \).

We check the Kraft inequality:
\[
\sum_{i=1}^{n} D^{-l_i} \leq 1
\]
where \( D = 2 \).

Calculating the left-hand side:
\[
\sum_{i=1}^{6} 2^{-l_i} = 2^{-1} + 2^{-1} + 2^{-1} + 2^{-2} + 2^{-2} + 2^{-4}
\]

Calculating each term:
- \( 2^{-1} = \frac{1}{2} \)
- \( 2^{-1} = \frac{1}{2} \)
- \( 2^{-1} = \frac{1}{2} \)
- \( 2^{-2} = \frac{1}{4} \)
- \( 2^{-2} = \frac{1}{4} \)
- \( 2^{-4} = \frac{1}{16} \)

Summing these values:
\[
\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{16} = \frac{3}{2} + \frac{1}{2} + \frac{1}{16} = 2 + \frac{1}{16} = \frac{32}{16} + \frac{1}{16} = \frac{33}{16}
\]

Since \( \frac{33}{16} = 2.0625 > 1 \), the Kraft inequality is violated.

Thus, it is impossible to construct a uniquely-decodable code with the same lengths as \( \Gamma \) when \( D=2 \).

The answer is:
- **False**


Question: Under which condition is an element $x\in \mathbb{Z}_n$ invertible?

To determine if an element \( x \in \mathbb{Z}_n \) is invertible, it must satisfy the condition:

\[
\mathsf{gcd}(x, n) = 1
\]

This means \( x \) and \( n \) are coprime, which allows for an integer \( y \) such that:

\[
x \cdot y \equiv 1 \pmod{n}
\]

Now, let's evaluate the options:

1) \( \mathsf{gcd}(x, \varphi(n)) = 1 \) - This condition is not relevant to the invertibility of \( x \) in \( \mathbb{Z}_n \).
2) \( \mathsf{gcd}(x, n-1) = 1 \) - This condition also does not determine whether \( x \) is invertible in \( \mathbb{Z}_n \).
3) \( \mathsf{gcd}(x, n) = 1 \) - This is the correct condition! If \( x \) and \( n \) are coprime, then \( x \) is indeed invertible.
4) \( \mathsf{gcd}(x, n) \ne 1 \) - This implies that \( x \) is not invertible.

Thus, the correct answer is option 3:

\[
\mathsf{gcd}(x, n) = 1
\]


Question: Consider the problem of finding a maximum cardinality set packing in the semi-streaming model. An instance of this problem consists of a known universe $U$ of $n$ elements and sets $S \subseteq U$ are streamed one-by-one. The goal is to select  a family $\mathcal{T}$ of pairwise disjoint sets (i.e., $S\cap S' = \emptyset$ for any two distinct sets $S, S' \in \mathcal{T}$) of maximum cardinality while only using $O(n\cdot \textrm{poly}\log n)$ storage space. Devise an algorithm in this setting that returns a set packing of cardinality at least $1/k$ times that of a maximum cardinality set packing, assuming that each streamed set $S$ has cardinality at most $k$, i.e.,  $|S| \leq k$. \\[0mm] {\em (In this problem you are asked to (i) design the algorithm, (ii) show that it uses $O(n\cdot  \textrm{\textnormal{poly}}{\log n})$ space, and (iii) prove that it returns a solution of cardinality at least $1/k$ times the cardinality of a maximum cardinality set packing. Recall that you are allowed to refer to material covered in the course.) }

1. **Problem Overview**:
   - Goal: Find maximum cardinality set packing from streamed sets \( S \subseteq U \) with \( |S| \leq k \).
   - Constraints: \( n \) elements in universe \( U \); storage limited to \( O(n \cdot \textrm{poly}\log n) \).

2. **Algorithm Steps**:
   - **Initialize**: Start with empty set \( \mathcal{T} \).
   - **Process Sets**:
     - For each set \( S_i \):
       - If no intersection with \( \mathcal{T} \), add \( S_i \).
       - If intersecting, replace or skip based on a strategy (e.g., greedy).

3. **Space Complexity**:
   - Track intersections with \( O(n \cdot \log n) \) space, fitting within \( O(n \cdot \textrm{poly}\log n) \).

4. **Performance Guarantee**:
   - Provides at least \( 1/k \) of maximum cardinality due to disjoint selection.

5. **Conclusion**:
   - Efficiently finds a near-optimal solution while adhering to space constraints.


Question: When constructing a word embedding, negative samples are:

In constructing a word embedding, negative samples are best described as **"word-context word combinations that are not occurring in the document collection."**

Let’s use a simple analogy. Imagine teaching a robot what an "apple" is. You show it pictures of apples along with the word, which represent positive samples. To help the robot grasp what an "apple" is not, you might show it images of bananas, oranges, and cars—these become your negative samples.

In word embeddings, negative samples refer to pairs of words (like "apple" and "car") that don’t co-occur in the training data. This distinction helps the model learn to differentiate between related words (positive samples) and unrelated ones (negative samples).

By incorporating negative sampling, the model refines its understanding of word meanings based on the contexts in which they appear. The other options, such as context words not in the vocabulary or less frequent words, don’t accurately capture the purpose of negative sampling. Ultimately, this technique enhances the model's ability to understand the subtle relationships within language, leading to richer representations of meaning.


Question: Consider the following contains function defined on Iterable (in particular, it accepts both Vector and List).  def contains[A](l: Iterable[A], elem: A): Boolean =   val n = l.size   if n <= 5 then     for i <- l do       if i == elem then         return true     false   else     val (p0, p1) = parallel(       contains(l.take(n / 2), elem),       contains(l.drop(n / 2), elem)     )   p0 || p1 Let $n$$n$ be the size of l. Assume that drop and take run in $\Theta(1)$ on Vector and $\Theta(n)$ on List. What is the asymptotic depth of contains if it is called on a List?

To analyze the asymptotic depth of the `contains` function when called on a `List`, we must first understand how the function operates and how the underlying data structure affects its performance.

### Function Overview:

The `contains` function checks if a given element exists in an iterable collection. The key points of its implementation are:

1. **Base Case**: If the size of the collection \( n \) is less than or equal to 5, the function iterates through the entire collection and checks for equality with the target element. This operation runs in \( O(n) \) time since it must potentially examine every element.

2. **Recursive Case**: If \( n > 5 \), the function splits the collection into two halves using `take` and `drop`. It then makes two parallel calls to `contains` for each half and combines the results using logical OR (`||`).

### Complexity Analysis on List:

1. **Size Calculation**: The function starts by determining the size of the list, \( n \). For a `List`, calculating the size is \( O(n) \) because it requires traversing the entire list to count the elements.

2. **Base Case Complexity**: For \( n \leq 5 \), the time complexity is \( O(n) \), which is straightforward since it involves a simple iteration over the list.

3. **Recursive Case Complexity**:
   - The operations `l.take(n / 2)` and `l.drop(n / 2)` both run in \( O(n) \) for a `List` because these methods must traverse the list to create new lists representing the two halves.
   - The recursive calls are made in parallel, but since we focus on the asymptotic depth, we need to consider how many levels of recursion there are.

### Recursion Depth Calculation:

The recursive structure of the function can be described as follows:
- At the first call, the list of size \( n \) is split into two halves of size \( n/2 \).
- Each of those halves can be further split into \( n/4 \), then \( n/8 \), and so on.

This recursion continues until the size of the list is reduced to 5 or fewer elements. The number of levels of recursion can be calculated as:

\[
k = \log_2(n)
\]

At each level, the size of the lists passed to `contains` is halved. The recursion stops when \( n \) is at most 5, which occurs at:

\[
2^k \leq 5 \implies k \leq \log_2(5) \approx 2.32
\]

Thus, the recursion depth \( d \) is \( O(\log n) \).

### Asymptotic Depth Calculation:

However, the depth in terms of the actual time complexity must consider the cost of operations at each depth level:

- At each level of the recursion, the size of the list that is processed is reduced to half, but we incur \( O(n) \) overhead from the `take` and `drop` operations on a `List` for each of these recursive calls.
- This leads to a cost of \( O(n) \) for the initial call, followed by \( O(n) \) at each of the \( O(\log n) \) levels due to the split operations.

### Final Asymptotic Depth:

The overall asymptotic depth of `contains` on a `List` can therefore be expressed as:

\[
T(n) = O(n) + O(n) \cdot O(\log n) = O(n \log n)
\]

Thus, the asymptotic depth of the `contains` function when called on a `List` is:

\[
\boxed{O(n \log n)}
\]

This captures the total work done due to both the splitting and the traversal necessary to check for the element's presence in the list.


Question: Let us assume that a kernel $K: \mathcal{X} \times \mathcal{X} \rightarrow \mathbb{R}$ is said to be valid if there exists $k \in \mathbb{N}$ and $\Phi: \mathcal{X} \rightarrow \mathbb{R}^{k}$ such that for all $\left(x, x^{\prime}\right) \in \mathcal{X} \times \mathcal{X}, K\left(x, x^{\prime}\right)=\Phi(x)^{\top} \Phi\left(x^{\prime}\right)$ Which one of the following kernels is not valid ?

To determine which kernels are not valid, we need to analyze each option to see if there exists a mapping \(\Phi: \mathcal{X} \rightarrow \mathbb{R}^k\) such that \(K(x, x') = \Phi(x)^{\top} \Phi(x')\).

1. **Option 1: \(\mathcal{X}=\mathbb{N}, K(x, x')=2\)**
   This kernel is constant and does not depend on \(x\) or \(x'\). A valid representation for this kernel can be given by \(\Phi(x) = \sqrt{2} \cdot e_i\) for any \(i\), where \(e_i\) is a basis vector in \(\mathbb{R}^k\). Thus, this kernel is valid.

2. **Option 2: \(\mathcal{X}=\mathbb{R}^d, K(x, x')=(x^{\top} x')^2\)**
   This is a polynomial kernel of degree 2. It can be represented using the map \(\Phi(x) = (x_1, x_2, \ldots, x_d, \sqrt{2} x_1 x_2, \sqrt{2} x_1 x_3, \ldots, \sqrt{2} x_{d-1} x_d)\), which allows us to express \(K(x, x')\) as \(\Phi(x)^{\top} \Phi(x')\). This kernel is valid.

3. **Option 3: \(\mathcal{X}=\mathbb{R}, K(x, x')=\cos(x-x')\)**
   The cosine kernel can be expressed as \(K(x, x') = \Phi(x)^{\top} \Phi(x')\) with a suitable mapping, such as using the Fourier transform. Thus, this kernel is valid.

4. **Option 4: All of the proposed kernels are in fact valid.**
   This option will be assessed based on the previous options.

5. **Option 5: \(\mathcal{X}=\mathbb{Q}, K(x, x')=2^{x+x'}\)**
   This kernel is an exponential kernel. However, for \(\mathcal{X} = \mathbb{Q}\), there might not always be a finite-dimensional mapping \(\Phi\) such that \(K(x, x') = \Phi(x)^{\top} \Phi(x')\). As such, this kernel is not valid.

6. **Option 6: \(\mathcal{X}=\mathbb{R}^d, K(x, x')=x^{\top} A x'\) where \(A\) is a \(d \times d\) symmetric positive semi-definite matrix.**
   This kernel can be expressed using a feature map that corresponds to the inner product in the transformed space defined by \(A\). Therefore, this kernel is valid.

In conclusion, the kernels that are not valid are:

- **Option 5: \(\mathcal{X}=\mathbb{Q}, K(x, x')=2^{x+x'}\)**

Thus, the final answer is:
- Option 5.


Question: Tick all correct answers:

The correct answers are indeed:

1. **Fuzz testing scales at least to 1'000s of lines of code.**
   This is correct because fuzz testing can efficiently identify vulnerabilities in large applications, especially when the code is modular.

2. **Compiler warnings scale to millions of lines of code.**
   This is also correct; modern compilers are designed to handle extensive codebases, providing warnings that help developers identify potential issues.

Mistake Reflection:
- I initially falsely assessed that formal verification scales effectively to high line counts. This is a common pitfall for students who might not consider the computational complexity involved in formal methods.
- Students should carefully evaluate the scalability limitations of different testing and verification techniques, focusing on their practical application contexts.

Revised Answer: The correct responses are the first two options, while the last two are incorrect due to scalability constraints in formal verification methods.


Question: The company finally decides to implement a hybrid model consisting of a 4-gram character model combined (independently) with a 3-gram word model.How many parameters would such a hybrid model have in total?Provide the answer in the form 10^A + 10^B (for instance, write "10^7 + 10^9").

To ensure security guarantees after software has been shipped, several important practices must be followed:

**1. Conduct regular security assessments**
**2. Implement a robust incident response plan**

### Explanation:

1. **Conduct regular security assessments**:
   - After shipping software, it’s vital to routinely assess its security posture. This involves performing penetration testing, vulnerability scanning, and code reviews to identify weaknesses. Regular assessments can help uncover security flaws that may have gone unnoticed during the initial development phase. By systematically evaluating the software, teams can prioritize security improvements and ensure that any vulnerabilities are addressed proactively.

2. **Implement a robust incident response plan**:
   - Even with the best preventive measures in place, security incidents can still occur. Having a well-defined incident response plan is crucial for minimizing damage. This plan should outline procedures for detecting, responding to, and recovering from security breaches. Key elements include establishing a response team, communication protocols, and post-incident analysis to learn from any security events, thereby improving future resilience.

### Why the other options are less relevant:

3. **Provide new features to attract new users**:
   - While enhancing the software with new features may attract new users, it does not contribute to maintaining existing security guarantees. In fact, adding features without adequate security considerations can introduce new vulnerabilities, undermining the software’s security posture.

4. **Ensure the software works on newer machines**:
   - While it is important for software to function on the latest hardware, this focus does not inherently address security. Compatibility updates can be important for usability, but they should not take precedence over addressing security vulnerabilities that could potentially compromise user data and trust.

### Conclusion:
In conclusion, maintaining security guarantees after software is deployed requires conducting regular security assessments and implementing a robust incident response plan. These practices help in identifying and addressing vulnerabilities while preparing the organization to respond effectively to any security incidents that may arise.


Question: Consider the min-cost perfect matching problem on a bipartite graph $G=(A \cup B, E)$ with costs $c: E \rightarrow \mathbb{R}$. Recall from the lecture that the dual linear program is \begin{align*} \text{Maximize}  \quad  & \sum_{a\in A} u_a + \sum_{b\in B} v_b\\ \text{Subject to} \quad &u_a + v_b \leq c(\{a,b\}) \qquad \mbox{for every edge $\{a,b\} \in E$.} \\ \end{align*} Show that the dual linear program is unbounded if there is a set $S \subseteq A$ such that $|S| > |N(S)|$, where $N(S) = \{ v\in B: \{u,v\} \in E \mbox{ for some $u\in S$}\}$ denotes the neighborhood of $S$. This proves (as expected) that the primal is infeasible in this case.

To show that the dual linear program is unbounded under the condition that there exists a set \( S \subseteq A \) such that \( |S| > |N(S)| \), we can consider the following.

Let us define \( v_b = 0 \) for all \( b \in B \) and set \( u_a = 0 \) for all \( a \in A \). This gives us a feasible solution to the dual, as it satisfies the constraints \( u_a + v_b \leq c(\{a,b\}) \) for every edge \( \{a,b\} \in E \).

Now, we can construct a new vector \( (u^*, v^*) \) where:

\[
u^*_a =
\begin{cases}
1 & \text{if } a \in S \\
0 & \text{otherwise}
\end{cases}
\quad
\text{and}
\quad
v^*_b =
\begin{cases}
-1 & \text{if } b \in N(S) \\
0 & \text{otherwise}
\end{cases}
\]

By choosing a scalar \( \alpha > 0 \), we find that the combination of these vectors yields:

\[
(u,v) + \alpha (u^*, v^*)
\]

is also feasible. The value of this dual solution can be expressed as:

\[
\sum_{a\in A} u_a + \sum_{b\in B} v_b + \alpha (|S| - |N(S)|)
\]

Since we have established that \( |S| > |N(S)| \), this means that as we let \( \alpha \to +\infty \), the dual value becomes unbounded. Therefore, this illustrates that the optimal solution to the dual problem is indeed unbounded.

(Note: The justification presented contains an error in stating initial values for feasibility; they should comply with actual costs, hence leading to correct conditions for showing unboundedness.)


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

Hey! So, you're asking if the addition operation for these Float8 numbers is associative, right? Basically, associative means that it doesn't matter how you group the numbers when you add them. Like, (a + b) + c should give you the same result as a + (b + c). To check this out, let's take some examples. Let's say we have a = Float8(3, 2), b = Float8(2, 3), and c = Float8(1, 2). If we do (a + b) + c first, we need to add a and b: 1. a + b: - b's exponent (3) is bigger than a's (2), so we shift a's mantissa: - Shift = 3 - 2 = 1, so a becomes (3 >> 1) = 1.5 (which is 1 since we're using integers), and we add that to b's mantissa: - 1 + 2 = 3 with exponent 3. So, (a + b) becomes Float8(3, 3). 2. Now add this result to c: - c has exponent 2, so we shift 3's mantissa: - Shift = 3 - 2 = 1, so 3 becomes (3 >> 1) = 1.5 (which is 1 again), and now we add: - 1 + 1 = 2, keeping the exponent 3. We get Float8(2, 3). Now, letâs do a + (b + c): 1. First, calculate b + c: - c's exponent (2) is less than b's (3), so we shift c's mantissa: - Shift = 3 - 2 = 1, so c becomes (1 >> 1) = 0, and we add: - 2 + 0 = 2 with exponent 3. So, b + c is Float8(2, 3). 2. Now add this to a: - a's exponent (2) is less than the result's (3), so we shift a's mantissa: - Shift = 3 - 2 = 1, so a becomes (3 >> 1) = 1.5 (which is 1), and we add: - 1 + 2 = 3 with exponent 3. We get Float8(3, 3). So, both ways we end up with Float8(2, 3) and Float8(3, 3), which are not equal. This shows that the operation is not associative because we can find different results by changing the grouping. Hope that clears things up!


Question: Tick the \emph{incorrect} assertion. In RSA with public key $(e,N)$ and private key $(d,N)$ \ldots

The correct answer is **Option 1: we can recover \(d\) if we can compute square root modulo \(N\) efficiently.**

### Explanation of the Correct Assertion (Option 1):
In RSA, the public key consists of an exponent \(e\) and a modulus \(N\), while the private key consists of an exponent \(d\) and the same modulus \(N\). The private key \(d\) is derived from \(e\) and the factorization of \(N\) (which is \(N = pq\), where \(p\) and \(q\) are large prime numbers). The relationship is given by \(d \equiv e^{-1} \mod \phi(N)\), where \(\phi(N) = (p-1)(q-1)\).

Being able to compute square roots modulo \(N\) does not provide any direct method to compute \(d\). The task of recovering \(d\) from \(e\) and \(N\) requires knowledge of the prime factors \(p\) and \(q\), and simply finding square roots does not help in this regard. Thus, this assertion is incorrect.

### Explanation of the Plausible but Incorrect Alternative (Option 4):
Option 4 states: "we must have that \(\gcd(e,d) = 1\) to be able to decrypt unambiguously."

While it is true that \(e\) and \(d\) must be coprime (i.e., \(\gcd(e, \phi(N)) = 1\) for RSA to work properly), it is not necessary for \(e\) and \(d\) themselves to be coprime. The key requirement for RSA is that \(e\) and \(\phi(N)\) are coprime, which ensures that \(d\) can be calculated as the modular inverse of \(e\) modulo \(\phi(N)\). Therefore, the assertion that \(\gcd(e, d) = 1\) is not a requirement for decryption to be successful, making this assertion misleading and incorrect in the context of RSA.

In summary, the incorrect assertion is Option 1, while Option 4 is a plausible but incorrect alternative because it misstates the relationship required between \(e\) and \(d\).


Question: Which of the following probability distributions are members of the exponential family:

To determine which of the given distributions belong to the exponential family, we can analyze their forms:

1. **Cauchy distribution**: The Cauchy distribution does not fit into the exponential family. Its probability density function does not have the structure of the form \( p(y|\theta) = h(y) \exp(\eta(\theta) T(y) - A(\theta)) \) that defines members of the exponential family.

2. **Poisson distribution**: The Poisson distribution can be expressed in the exponential family form. It can be rewritten as \( p(y|\mu) = \frac{e^{-\mu} \mu^y}{y!} = h(y) \exp(y \log(\mu) - \mu - \log(y!)) \), where \( h(y) = \frac{1}{y!} \), \( \eta(\theta) = \log(\mu) \), and \( A(\theta) = \mu \). Thus, it belongs to the exponential family.

3. **Uniform distribution**: The uniform distribution over \([0, \eta]\) does not belong to the exponential family. Its density function does not allow for a transformation into the required exponential family form, as it does not involve an exponential function dependent on a sufficient statistic.

Therefore, the correct option is:
- **Poisson distribution**: It is a member of the exponential family.


Question:
Consider the following toy learning corpus of 59 tokens (using a tokenizer that splits on whitespaces and punctuation), out of a possible vocabulary of $N=100$ different tokens:
Pulsed operation of lasers refers to any laser not classified as continuous wave, so that the optical power appears in pulses of some duration at some repetition rate. This\linebreak encompasses a wide range of technologies addressing a number of different motivations. Some lasers are pulsed simply because they cannot be run in continuous wave mode.
Using a 2-gram language model, what are the values of the parameters corresponding to "continuous wave" and to "pulsed laser" using Maximum-Likelihood estimates?

Let’s break this down step-by-step to find the Maximum-Likelihood Estimates (MLE) for the bigrams "continuous wave" and "pulsed laser" based on the provided corpus.

### Step 1: Count the Bigrams
First, we need to identify how many times each bigram appears in the text.

1. **Identify occurrences:**
   - **"continuous wave":** This bigram appears **twice** in the text.
   - **"pulsed laser":** This bigram does **not** appear at all in the text, so the count is **zero**.

### Step 2: Count Total Bigrams
In a corpus of 59 tokens, the number of bigrams (pairs of consecutive words) can be calculated as:
\[
\text{Number of Bigrams} = \text{Total Tokens} - 1 = 59 - 1 = 58
\]

### Step 3: Calculate MLE for "continuous wave"
Using the formula for MLE:
\[
P(word_2 | word_1) = \frac{C(word_1, word_2)}{C(word_1)}
\]
- For "continuous wave":
  - \( C(continuous, wave) = 2 \) (it appears twice)
  - \( C(continuous) = 58 \) (the total number of bigrams)

Thus, the probability is:
\[
P(continuous \, wave) = \frac{C(continuous, wave)}{\text{Total Bigrams}} = \frac{2}{58} \approx 0.0345
\]

### Step 4: Calculate MLE for "pulsed laser"
- For "pulsed laser":
  - \( C(pulsed, laser) = 0 \) (it never appears)
  - \( C(pulsed) = 58 \) (the total number of bigrams)

Thus, the probability is:
\[
P(pulsed \, laser) = \frac{C(pulsed, laser)}{\text{Total Bigrams}} = \frac{0}{58} = 0
\]

### Final Results
- The probability of "continuous wave" is \( P(continuous \, wave) = \frac{2}{58} \).
- The probability of "pulsed laser" is \( P(pulsed \, laser) = 0 \).

And there you have it! We’ve successfully navigated the jungle of bigrams to arrive at these probabilities!


Question: In this week's lecture, you have been introduced to the aggregate method of ParSeq[A] (and other parallel data structures). It has the following signature:  def aggregate[B](z: B)(f: (B, A) => B, g: (B, B) => B): B Discuss, as a group, what aggregate does and what its arguments represent. Under which condition(s) on z, f, and g does aggregate always lead to the same result? Come up with a formula on z, f, and g that implies the correctness of aggregate.  Hint: You may find it useful to use calls to foldLeft(z)(f) in your formula(s).

The `aggregate` method in ParSeq[A] is designed to combine elements of a parallel collection using a specified binary operation. It takes three arguments: `z`, which is the initial value or the "zero" element that serves as the starting point for the aggregation; `f`, a function that specifies how to combine the initial value or the accumulated value with each element of type A; and `g`, a function that combines results from different subcollections.

To ensure that `aggregate` yields the same result regardless of the order of operations (which can vary in a parallel execution), `z`, `f`, and `g` must satisfy certain conditions. Specifically, `f` must be associative, meaning that for any values $x, y, z$ of type B, the equation $f(f(x, y), z) = f(x, f(y, z))$ must hold. Additionally, `g` must also be associative, and `z` must act as the identity element for both functions.

A formula that captures the correctness of `aggregate` could be expressed as:
$
aggregate(z)(f, g) = foldLeft(z)(f)
$
This equality holds true if, for every partition of the collection, combining results using `g` after applying `f` across the entire collection leads back to the same result as a single sequential fold operation from the initial value `z`.


Question: The inverse document frequency of a term can increase

**Chosen Option:** 4. by adding a document to the document collection that does not contain the term.

**Rationale:**

- **Understanding Inverse Document Frequency (IDF):**
  - IDF is a measure used in information retrieval to evaluate how important a term is within a collection of documents. It is calculated using the formula:
    \[ \text{IDF}(t) = \log\left(\frac{N}{df(t)}\right) \]
    where \( N \) is the total number of documents in the collection, and \( df(t) \) is the number of documents that contain the term \( t \). Hence, the IDF increases as the number of documents containing the term decreases.

- **Effect of Adding a Document that Does Not Contain the Term:**
  - When you add a document that does not contain the term, the total document count \( N \) increases while the document frequency \( df(t) \) remains unchanged (since the new document does not contribute to \( df(t) \)).
  - This increase in \( N \) results in a higher IDF value for the term, because the term now appears to be less common across the entire set of documents, thus increasing its significance.

- **Analysis of Other Options:**
  - **Option 1 (by adding the term to a document that contains the term):** This does not affect \( df(t) \) because the term is already present in that document. Therefore, IDF remains the same.
  - **Option 2 (by removing a document from the document collection that does not contain the term):** This also does not increase IDF because removing a document that does not contribute to \( df(t) \) does not change the term's distribution across the remaining documents.
  - **Option 3 (by adding a document to the document collection that contains the term):** This would increase \( df(t) \), thereby reducing the IDF value, as the term would appear to become more common.

- **Conclusion:**
  - The correct approach to increase the inverse document frequency of a term is to add a document that does not contain that term, as this increases the total number of documents while keeping the count of documents containing the term the same, thereby enhancing the term's rarity and importance within the document collection. This aligns with the principles of how IDF is calculated and understood in the context of information retrieval.


Question:
The \textbf{parameters} (weights \textbf{W}) are learned with ...
(One answer)

To answer the question regarding how the parameters (weights W) are learned in the context of machine learning, let's evaluate each of the provided options:

1. **Training**: This is the correct answer. In machine learning, the parameters of a model are learned from the training dataset. The training dataset is used to fit the model by minimizing a loss function, thus adjusting the weights based on the input data and the corresponding output labels. The learning process involves optimization techniques such as gradient descent, where the model iteratively updates the weights to improve performance on this dataset.

2. **Validation**: The validation dataset is used to tune hyperparameters and assess the model's performance during the training process. It helps in selecting the best model configuration but does not contribute to learning the parameters themselves. The weights are not adjusted based on the validation data, which is kept separate to ensure that the model does not overfit.

3. **Test**: The test dataset is utilized to evaluate the final model's performance after training and validation. It provides an unbiased estimate of how well the model generalizes to unseen data. The parameters are not learned from the test set; rather, the test set helps to measure the effectiveness of the parameters learned during the training phase.

4. **All the data together**: While using all available data for training can be beneficial in certain contexts (e.g., in cross-validation or when refitting a model after validating), the standard practice is to divide the data into training, validation, and test sets. Learning from all data at once would compromise the ability to evaluate the model's performance on unseen data, which is crucial for assessing generalization.

**Conclusion**: The correct answer is **'training'**. The parameters (weights W) are specifically learned from the training data, which is the dataset used to fit the model. Understanding this distinction is vital in machine learning to ensure proper model evaluation and to avoid overfitting.


Question: Which of the following is correct regarding the use of Hidden Markov Models (HMMs) for entity recognition in text documents?

Let's analyze each of the answer choices regarding Hidden Markov Models (HMMs) for entity recognition in text documents:

1. **HMMs cannot predict the label of a word that appears only in the test set**: This statement is not entirely correct. While HMMs rely on the training data to estimate probabilities, they can still assign labels to unseen words based on the context (previous labels and the model's structure). However, the accuracy of those predictions may be lower.

2. **If the smoothing parameter λ is equal to 1, the emission probabilities for all the words in the test set will be equal**: This statement is incorrect. The smoothing parameter λ is used in the context of smoothing techniques, such as Laplace smoothing. If λ = 1, it does not mean that all probabilities become equal; rather, it adjusts the probabilities of unseen events but does not equalize them.

3. **When computing the emission probabilities, a word can be replaced by a morphological feature (e.g., the number of uppercase first characters)**: This statement is correct. In HMMs for entity recognition, it is common to use features derived from the words (such as morphological features) to improve the model's ability to recognize entities. Using features like the number of uppercase characters can help in identifying proper nouns and entities.

4. **The label of one word is predicted based on all the previous labels**: This statement aligns with how HMMs operate. In HMMs, the prediction of the current state (label) depends on the previous state (label), following the Markov property, which states that the future state depends only on the current state.

Based on the analysis above, the correct answer is:

**3. When computing the emission probabilities, a word can be replaced by a morphological feature (e.g., the number of uppercase first characters)**.


Question: You are given the following accident and weather data. Each line corresponds to one event:  1. car_accident rain lightning wind clouds fire 2. fire clouds rain lightning wind 3. car_accident fire wind 4. clouds rain wind 5. lightning fire rain clouds 6. clouds wind car_accident 7. rain lightning clouds fire 8. lightning fire car_accident  (b) Find all the association rules for minimal support 0.6 and minimal confidence of 1.0 (certainty). Follow the apriori algorithm.

{car_accident} -> {rain} (support: 0.625, confidence: 1.0)
{fire} -> {lightning} (support: 0.625, confidence: 1.0)
{clouds} -> {rain} (support: 0.625, confidence: 1.0)
{lightning} -> {fire} (support: 0.625, confidence: 1.0)
{rain} -> {clouds} (support: 0.625, confidence: 1.0)


Question: Give some concrete examples of NLP applications that might benefit from the semantic vectorial representations.

1. **Search Engines**: Semantic vectorial representations can improve search results by understanding the meaning behind queries rather than just matching keywords. For example, a search for "best places to eat" can return results about restaurants without explicitly mentioning "restaurants."

2. **Chatbots and Conversational Agents**: These systems can leverage semantic vectors to better understand user intents and provide more relevant responses. For instance, if a user asks about "weather," the system can recognize related terms like "forecast" or "temperature."

3. **Sentiment Analysis**: By using semantic vectors, models can gauge the sentiment behind complex phrases that may not contain explicit sentiment words, such as "not bad" being interpreted as positive.

4. **Recommendation Systems**: Semantic representations can enhance product suggestions by understanding user preferences in a more nuanced way, such as suggesting books similar in theme or style rather than just genre.

5. **Machine Translation**: By capturing semantic meaning, these representations can improve translation quality by ensuring that idiomatic expressions or phrases with similar meanings in different languages are accurately translated.

6. **Information Extraction**: Extracting entities and relationships from text becomes more effective with semantic vectors, as they can identify synonyms and contextually relevant terms, enhancing tasks like named entity recognition.

7. **Text Summarization**: Semantic vectors can facilitate better summarization by identifying key concepts and themes in the text, ensuring that the summary captures the essence of the original content.

8. **Document Clustering**: Grouping similar documents based on their semantic content becomes more efficient, allowing for better organization and retrieval of information relevant to user queries.

Each of these applications can leverage the ability of semantic vectorial representations to capture the nuanced meanings of words and phrases, leading to improved performance and user satisfaction.


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

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

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

Your colleagues were changing the parameter type of "add" to an "Integer". Explain whether this breaks backward compatibility and why or why not (also without worrying about whether this is a good or a bad thing).

Changing the parameter type of the `add` method from `int` to `Integer` affects compatibility in two ways: binary compatibility and source compatibility.

1. **Binary Compatibility**: This change breaks backward binary compatibility. Existing compiled code that calls the original `add(int val)` method will fail at runtime if the method signature is changed to `add(Integer val)`. The JVM will not be able to find a matching method for the original bytecode, leading to a `NoSuchMethodError`.

2. **Source Compatibility**: The change maintains source compatibility due to Java's autoboxing feature. Existing code that calls `add(int val)` will still compile correctly because the Java compiler can automatically convert `int` values to `Integer` objects. This means that while the method signature has changed, the existing source code does not need to be modified to compile.

In summary, changing the parameter type from `int` to `Integer` breaks backward binary compatibility (existing compiled code will not work) but retains source compatibility (existing source code will still compile).


Question:
Assume you are working in a company on the back-end of a mobile application. One colleague explains that the team had originally configured the repo so that only code with >80% path coverage could be merged, but have now dropped this requirement.
In one sentence, give one possible reason behind this removal.

One possible reason for the removal of the >80% path coverage requirement could be to accelerate the development process and allow for more rapid feature delivery, especially if the team felt that the strict coverage standard was hindering productivity or causing delays in merging essential code.


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

"It is an electro-mechanical encryption device used by German armies in World War 2."

True. The Enigma machine was indeed used for military communication during World War 2.
"Its specifications are secret."

False. While the Enigma's code was initially a secret, its specifications, including the machine's design, became known after the war.
"Its secret parameters are: ordered permutations, an involution, and a number."

True. The Enigma used rotor arrangements (ordered permutations), the reflection mechanism (involution), and a number to define settings.
"It was patented in 1918."

True. The first patent for the Enigma machine was indeed filed in 1918.
False Statement: "Its specifications are secret."


Question: Tick the \textbf{true} assertion. Assume an arbitrary $f:\{0,1\}^p \rightarrow \{0,1\}^q$, where $p$ and $q$ are integers.

To determine the true assertion among the given options, let's evaluate each one step by step:

1. **First Assertion:**
   \[
   \mathsf{DP}^f(a,b)=\Pr_{X\in_U\{0,1\}^p}[f(X\oplus a)\oplus f(X)\oplus b=1]
   \]
   - This defines the differential privacy metric. It states that the probability of a certain output, conditioned on the input being altered by \(a\) and compared to \(b\), is equal to \(\mathsf{DP}^f(a,b)\). This is a common definition in differential privacy contexts and appears valid.

2. **Second Assertion:**
   \[
   \Pr[f(x\oplus a)\oplus f(x)\oplus b=0]=E(\mathsf{DP}^f(a,b))
   \]
   - This states a relationship between the probability of a specific output being 0 and the expected value of the differential privacy measure. However, it is not generally true that the expectation of the differential privacy metric equates to a probability. The relationship between these two would need to be established through specific conditions, which is not universally applicable.

3. **Third Assertion:**
   \[
   2^p\mathsf{DP}^f(a,b) \text{ is odd, for all } a \in \{0,1\}^p, b \in \{0,1\}^q.
   \]
   - This is a claim about the oddness of a product involving the differential privacy measure. There is no inherent reason for this assertion to hold true universally without specific properties of \(f\). It seems unlikely to be true in a general sense.

4. **Fourth Assertion:**
   \[
   \sum_{b\in\{0,1\}^q} \mathsf{DP}^f(a,b)=1, \text{ for all } a \in \{0,1\}^p.
   \]
   - This states that the sum of probabilities over all possible outputs \(b\) should equal 1 for any input \(a\). This is a property of probability distributions, and if \(\mathsf{DP}^f(a,b)\) is interpreted as a probability distribution over \(b\) for fixed \(a\), this statement should be true.

After evaluating all assertions, the **fourth assertion** emerges as the one that is universally true, as it aligns with the fundamental properties of probabilities.

Thus, the true assertion is:

\[
\sum_{b\in\{0,1\}^q} \mathsf{DP}^f(a,b)=1, \text{ for all } a \in \{0,1\}^p.
\]


Question: Tick the \textbf{true} assertion. Let $X,Y$ be two random variables over the same probability space. Then,

The correct answer is option 2: \( E(XY) = E(X) \times E(Y) \), if \( X \) and \( Y \) are independent.

Here's why this is the true assertion:

1. **Independence of Random Variables**: When we say that two random variables \( X \) and \( Y \) are independent, it means that the occurrence of one does not affect the occurrence of the other. This independence has specific mathematical implications.

2. **Expectation of the Product**: For independent random variables, the expected value of the product \( XY \) is the same as the product of their expected values. This is a fundamental property of independent random variables. Therefore, if \( X \) and \( Y \) are independent, we can conclude that \( E(XY) = E(X) \times E(Y) \).

3. **Other Options**:
   - The first option states that \( X \) is always independent from \( Y \), which is false; they can be dependent.
   - The third option about probabilities also deals with independence, but it's not a true assertion in the context of expectation.
   - The fourth option, stating that \( X + Y \) does not make sense, is incorrect as adding random variables is a valid operation.

In summary, option 2 correctly expresses a property of independent random variables regarding their expected values.


Question:
A sequential object is a tuple T = (Q, q0, O, R, ∆), where:
● Q is a set of states.
● q0 ∈ Q is an initial state.
● O is a set of operations.
● R is a set of responses.
● ∆ ⊆ (Q × Π × O) × (Q × R) is a relation that associates a state, a process,
and an operation to a set of possible new states and responses.
Processes invoke operations on the object. As a result, they get responses back, and the state of the object is updated to a new value, following from ∆.

Define a sequential object representing Asset Transfer, i.e., an object that allows processes to exchange units of currency.

The sequential object for Asset Transfer can be formally defined as follows:

Let \( A \) be an arbitrary set of accounts. Define a state function \( q: A \rightarrow \mathbb{N} \), where \( q(a) \) represents the balance of account \( a \in A \).

The initial state is denoted as \( q_0 \), such that \( q_0(a) = b_a \) for each account \( a \in A \), where \( b_a \) is the initial balance of account \( a \).

The set of operations \( O \) comprises:
1. **Transfer Operation**: \( \text{transfer}(u, a_1, a_2) \), where \( u \in \mathbb{N}^+ \) represents the amount to transfer from account \( a_1 \) to account \( a_2 \).
2. **Read Operation**: \( \text{read}(a) \), which retrieves the balance of account \( a \).

The set of responses \( R \) includes:
- \( \text{success} \)
- \( \text{insufficient\_funds} \)
- \( \text{invalid\_account} \)
- \( \text{balance}(b) \), where \( b \in \mathbb{N} \).

The transition relation \( \Delta \) is defined as:
\[
\Delta \subseteq (Q \times O) \times (Q \times R)
\]
For any state \( q \) and operation \( \text{transfer}(u, a_1, a_2) \):
- If \( a_1 \in A \) and \( a_2 \in A \) and \( q(a_1) \geq u \):
  \[
  (q, \text{transfer}(u, a_1, a_2)) \rightarrow (q', \text{success}) \text{, where } q'(a) =
  \begin{cases}
  q(a) - u & \text{if } a = a_1 \\
  q(a) + u & \text{if } a = a_2 \\
  q(a) & \text{otherwise}
  \end{cases}
  \]

- If \( a_1 \) or \( a_2 \) is invalid or \( q(a_1) < u \):
  \[
  (q, \text{transfer}(u, a_1, a_2)) \rightarrow (q, \text{insufficient\_funds} \text{ or } \text{invalid\_account})
  \]

For the read operation:
\[
(q, \text{read}(a)) \rightarrow (q, \text{balance}(q(a)))
\]

This formalization encapsulates the behavior of an asset transfer system and balance inquiries for accounts in a rigorous manner.


Question:
Your team is developing a library that is mostly intended to be used by your company's own applications, but the library is nevertheless distributed via a public repo on GitHub. It contains the following java function:
"public InputStream convertToPdf(Document document) throws GoogleServerNotRespondingError"
A security consultant points out that passing an arbitrary document can lead to security issues since the document could link to malicious content from the Internet, and your own application does not actually use this flexibility as it only prints plain text.
The consultant suggests replacing the "Document" parameter by a "String". You agree, but identify some risk associated with this. Explain in one sentence the main problem that such a parameter change could lead to:

Changing the parameter from "Document" to "String" could lead to risks related to input validation, as it may allow for unfiltered or improperly sanitized user input to be processed, potentially resulting in security vulnerabilities like injection attacks or malformed content.


Question: Recall the Jaccard index that we saw in Exercise Set 10: Suppose we have a universe $U$. For non-empty sets $A,B \subseteq U$, the Jaccard  index is defined as \begin{align*} J(A,B) = \frac{|A \cap B|}{|A \cup B|}\,. \end{align*} Design a locality sensitive hash (LSH) family $\mathcal{H}$ of functions $h: 2^U \rightarrow [0,1]$ such that  for any non-empty sets $A, B\subseteq U$, \begin{align*} \Pr_{h \sim \mathcal{H}}[h(A) \neq h(B)] \begin{cases} \leq 0.01 & \mbox{if $J(A,B) \geq 0.99$,}\\ \geq 0.1 & \mbox{if $J(A,B) \leq 0.9$.} \end{cases} \end{align*} {\em (In this problem you are asked to explain the hash family and argue that it satisfies the above properties. Recall that you are allowed to refer to material covered in the course.)}

To design a locality sensitive hash (LSH) family \(\mathcal{H}\) that satisfies the given properties with respect to the Jaccard index, we need to leverage the characteristics of the Jaccard similarity in our hashing functions.

### Initial Interpretation of the Problem

The Jaccard index \(J(A, B)\) measures the similarity between two sets \(A\) and \(B\). It is defined as the size of the intersection of the sets divided by the size of the union of the sets. We want to construct a family of hash functions that will help us determine whether two sets are similar or not based on this index:

1. If \(J(A, B) \geq 0.99\), we want to ensure that the probability that the hash values are different \(h(A) \neq h(B)\) is at most 0.01.
2. If \(J(A, B) \leq 0.9\), we want the probability that the hash values are different to be at least 0.1.

### Key Information

1. **Jaccard Similarity**: \(J(A, B) = \frac{|A \cap B|}{|A \cup B|}\)
2. **Hash Values**: Our hash function \(h\) will output values in the range \([0,1]\).
3. **Requirements**: We need to ensure certain probabilities based on the Jaccard index.

### Possible Options for Hash Function Design

A common approach for locality sensitive hashing for sets is to use random projections or random sampling. For our case, we can use the following ideas:

1. **Random Sampling of Elements**: We can randomly choose a subset of elements from the universe \(U\). The size of the subset can influence the hash value.
2. **Thresholding**: We can define a threshold to determine the hash value based on the fraction of common elements versus the total number of elements.

### Proposed Hash Function Family

Let’s define our hash function \(h\) as follows:

1. **Choose a random subset** \(S\) of \(U\) with a fixed size \(k\) (for example, \(k = 10\)).
2. **Hash Function Definition**: Define \(h(A)\) as the fraction of elements in \(S\) that are contained in \(A\):
   \[
   h(A) = \frac{|A \cap S|}{k}
   \]
   Similarly, define \(h(B)\) for set \(B\).

### Justification of Probabilities

1. **When \(J(A, B) \geq 0.99\)**:
   - If \(J(A, B) \geq 0.99\), this means that almost all elements in \(B\) are also in \(A\) and vice versa. Hence, the intersection \(A \cap B\) is very large relative to the union.
   - The probability that a random sample \(S\) will contain a representative portion of elements from both sets is very high. Thus, \(h(A)\) and \(h(B)\) will likely yield the same values, leading to \(\Pr[h(A) \neq h(B)] \leq 0.01\).

2. **When \(J(A, B) \leq 0.9\)**:
   - In this case, the overlap between sets \(A\) and \(B\) is significantly smaller, meaning \(A\) and \(B\) share fewer elements.
   - The likelihood of a random sample \(S\) containing similar fractions of elements from both sets decreases significantly. Hence, the probability that \(h(A) \neq h(B)\) increases, leading to \(\Pr[h(A) \neq h(B)] \geq 0.1\).

### Conclusion

Thus, the proposed locality sensitive hash family \(\mathcal{H}\) defined by the random sampling of elements in the universe \(U\) satisfies the required conditions on the probabilities related to the Jaccard index. This LSH scheme effectively captures the similarity properties we want to exploit while adhering to the specified thresholds.


Question:
Let us remind that we define the max-margin $M_\star$ as
            egin{align*}
                M_\star = \max_{\wv\in\mathbb R^D, \| \wv\|_2=1} M 	ext{ such that } y_n \xv_n^	op \wv \geq M 	ext{ for } n=1,\cdots, N
            \end{align*}
            and a max-margin separating hyperplane $ar \wv$ as a solution of this problem:
            egin{align*}
                ar \wv \in rg \max_{\wv\in\mathbb R^D, \| \wv\|_2=1} M 	ext{ such that } y_n \xv_n^	op \wv \geq M 	ext{ for } i=1,\cdots, N
            \end{align*}
            Bound the number of perceptron updates $t$ using the quantities $R$ and $M_\star$. Prove your result.

To bound the number of perceptron updates \( t \), we can use the following result from the theory of the perceptron algorithm:

Given a dataset with \( N \) points \( (\xv_n, y_n) \) where \( y_n \in \{-1, +1\} \) and assuming \( \|\xv_n\| \leq R \) for all \( n \), the number of updates \( t \) of the perceptron can be bounded as:

\[
t \leq \frac{R^2}{M_\star^2}
\]

**Proof Outline:**

1. **Geometric Interpretation**: The max-margin \( M_\star \) represents the largest margin by which the hyperplane can separate the classes. The radius \( R \) is the maximum distance from the origin to the data points.

2. **Perceptron Update Rule**: The perceptron updates its weights \( \wv \) whenever a misclassification occurs, moving in the direction of the misclassified point. Each update can be viewed as a movement in the direction of the feature vector \( \xv_n \).

3. **Distance Moved by Updates**: Each time the perceptron makes an update, it moves the hyperplane closer to correctly classifying the point. The distance that the hyperplane can be adjusted is related to the margin \( M_\star \).

4. **Bounding Updates**: The perceptron will continue to update until the margin \( M_\star \) is achieved. The maximum possible distance between the hyperplane and the closest point (the margin) gives us a way to relate the radius \( R \) and the margin \( M_\star \).

5. **Final Bound**: The maximum number of updates before convergence can be derived from the fact that each update decreases the distance to the optimal separating hyperplane by at least \( M_\star \). Thus, the total squared distance \( R^2 \) can be divided by the squared margin \( M_\star^2 \) leading to:

\[
t \leq \frac{R^2}{M_\star^2}
\]

This completes the proof of the upper bound on the number of perceptron updates in terms of \( R \) and \( M_\star \).


Question: Implement RSME score based on the following formula.  \begin{equation} \mathit{RMSE} =\sqrt{\frac{1}{N} \sum_i (r_i -\hat{r_i})^2} \end{equation}  You can use the mean_squared_error function from sklearn.metrics.

To understand the implementation of the RMSE (Root Mean Square Error) score using the provided formula and Python code, we should break down the components involved in both the mathematical formula and the coding process.

### Understanding RMSE

RMSE is a widely used metric for evaluating the accuracy of a predictive model. The formula for RMSE is given by:

\[
\mathit{RMSE} = \sqrt{\frac{1}{N} \sum_{i} (r_i - \hat{r_i})^2}
\]

Where:
- \( r_i \) is the actual observed value (ground truth).
- \( \hat{r_i} \) is the predicted value from the model.
- \( N \) is the number of observations.

#### Interpretation:
1. **Error Calculation**: For each observation \( i \), the difference between the actual value and the predicted value is computed, \( (r_i - \hat{r_i}) \). This difference represents the error for that observation.
2. **Squaring the Error**: Squaring the error \( (r_i - \hat{r_i})^2 \) ensures that all errors are non-negative and emphasizes larger errors more than smaller ones.
3. **Mean of Squared Errors**: The average of these squared errors is computed by dividing the sum of squared errors by \( N \).
4. **Square Root**: Finally, taking the square root of the average squared error provides the RMSE, bringing the units back to the original measurement scale.

### Implementation in Python

Now let's explain the following Python code that solves this question step by step.

```python
from sklearn.metrics import mean_squared_error
from math import sqrt

def rmse(prediction, ground_truth):
    prediction = prediction[ground_truth.nonzero()].flatten()
    ground_truth = ground_truth[ground_truth.nonzero()].flatten()

    return sqrt(mean_squared_error(prediction, ground_truth))
```

#### Importing Libraries
- `mean_squared_error`: This function from the `sklearn.metrics` library computes the mean squared error directly, which simplifies our implementation.
- `sqrt`: This function from the `math` library is used to compute the square root, which is necessary for obtaining the RMSE.

#### Function Definition
- The function `rmse` takes two arguments: `prediction` (the predicted values) and `ground_truth` (the actual observed values).

#### Handling Non-Zero Entries
- The line `prediction = prediction[ground_truth.nonzero()].flatten()` selects only those predicted values where the corresponding entries in `ground_truth` are non-zero. This is important because RMSE is often computed only on valid observations, ignoring any zeros that may represent missing or irrelevant data points.
- `flatten()` converts the selected values into a 1-dimensional array, making subsequent calculations easier.

#### Calculating RMSE
1. **Mean Squared Error Calculation**: `mean_squared_error(prediction, ground_truth)` computes the mean squared error between the filtered predictions and ground truth.
2. **Square Root**: The `sqrt` function is then applied to the mean squared error to yield the RMSE.

### Summary

The provided code succinctly implements the RMSE calculation using established libraries in Python. It effectively filters out invalid (zero) entries before calculating the mean squared error, ensuring that the RMSE reflects only meaningful data. The use of `mean_squared_error` simplifies the implementation, allowing us to focus on the mathematical operations without manually computing the sum and mean. Thus, the code is both efficient and adheres to the theoretical definition of RMSE, providing a robust measure of predictive accuracy.


Question: Which statement is true for the Mean Squared Error (MSE) loss MSE( $\mathbf{x}, y):=\left(f_{\mathbf{w}}(\mathbf{x})-y\right)^{2}$, with $f_{\mathrm{w}}$ a model parametrized by the weights $\mathbf{w}$ ?

Let's evaluate each statement regarding the Mean Squared Error (MSE) loss function:

1. **MSE is not necessarily convex with respect to the weights of the model $\mathbf{w}$.**
   - **Justification:** This statement is **true**. The MSE loss is convex with respect to the model parameters (weights) $\mathbf{w}$ if the model $f_{\mathbf{w}}(\mathbf{x})$ is a linear function of $\mathbf{w}$. However, if $f_{\mathbf{w}}(\mathbf{x})$ is a non-linear function (like a neural network with non-linear activation functions), the loss landscape may contain non-convex regions. Therefore, MSE can be non-convex depending on the model.

2. **MSE is more robust to outliers than Mean Absolute Error (MAE).**
   - **Justification:** This statement is **false**. MSE is actually less robust to outliers compared to Mean Absolute Error (MAE). This is because MSE squares the errors, which means that larger errors (which occur with outliers) have a disproportionately larger effect on the MSE value. In contrast, MAE treats all errors linearly and therefore is less influenced by outliers.

3. **For any ML task you are trying to solve, minimizing MSE will provably yield the best model.**
   - **Justification:** This statement is **false**. Minimizing MSE will yield the best model only under specific circumstances, particularly when the underlying distribution of the target variable $y$ given the input $\mathbf{x}$ is Gaussian, and when the model is appropriate for the task. In other cases, especially when the target variable distribution is not Gaussian or when outliers are present, other loss functions (like MAE or Huber loss) may provide better performance.

### Summary of Correct Answers:
- The only correct answer is: **MSE is not necessarily convex with respect to the weights of the model $\mathbf{w}$.**


Question:
Now let $\xv$ be a random vector distributed according to the uniform distribution over the finite centered dataset $\xv_1, . . . , \xv_N$ from above. %
Consider the problem of finding a unit vector, $\wv \in \R^D$, such that the random variable $\wv^	op \xx$ has \emph{maximal} variance. What is the variance of the random variable $\wv^	op \xx$ over the randomness of $\xx$?

To address your question, we need to delve into some concepts related to random vectors, variance, and the properties of projections in linear algebra. Let’s break this down step-by-step.

### Understanding the Problem

We have a random vector \(\xv\) that is uniformly distributed over a finite dataset \(\{\xv_1, \xv_2, \ldots, \xv_N\}\). We want to find a unit vector \(\wv\) such that the variance of the random variable \(\wv^\top \xv\) is maximized.

### Key Concepts

1. **Variance**: The variance of a random variable \(Y\) is defined as:
   \[
   \text{Var}(Y) = \mathbb{E}[(Y - \mathbb{E}[Y])^2] = \mathbb{E}[Y^2] - (\mathbb{E}[Y])^2
   \]
   For our case, \(Y = \wv^\top \xv\).

2. **Projections and Linear Transformations**: The expression \(\wv^\top \xv\) represents a linear transformation of the random vector \(\xv\). The variance of this linear transformation is influenced by the direction of \(\wv\).

3. **Covariance Matrix**: The variance of the projection of \(\xv\) can be determined using the covariance matrix of \(\xv\). If we denote the covariance matrix of \(\xv\) as \(\Sigma\), then:
   \[
   \text{Var}(\wv^\top \xv) = \wv^\top \Sigma \wv
   \]

### Finding the Maximum Variance

To maximize the variance \(\text{Var}(\wv^\top \xv)\), we need to choose \(\wv\) appropriately. The key insight here is that the maximum variance will occur when \(\wv\) aligns with the principal component of the covariance matrix, which corresponds to the largest eigenvalue.

1. **Eigenvalues and Eigenvectors**: The covariance matrix can be decomposed into its eigenvalues and eigenvectors. Let’s denote the eigenvalues of \(\Sigma\) as \(\lambda_1, \lambda_2, \ldots, \lambda_D\) (assuming \(D\) dimensions), where we can order them such that:
   \[
   \lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_D
   \]
   The corresponding eigenvectors are denoted by \(\wv_1, \wv_2, \ldots, \wv_D\).

2. **Maximizing Variance**: To maximize the variance:
   - Choose \(\wv = \wv_1\), the eigenvector associated with the largest eigenvalue \(\lambda_1\).
   - Therefore, the maximum variance is given by:
   \[
   \text{Var}(\wv_1^\top \xv) = \lambda_1
   \]

### Conclusion

In summary, to maximize the variance of the random variable \(Y = \wv^\top \xv\), you should select the unit vector \(\wv\) to be the eigenvector corresponding to the largest eigenvalue of the covariance matrix of your dataset. Thus, the variance of the random variable \(Y\) is equal to that largest eigenvalue:

\[
\text{Var}(\wv^\top \xv) = \lambda_1
\]

In practice, this means performing an eigendecomposition of your dataset's covariance matrix to find this value. This fundamental concept is rooted in Principal Component Analysis (PCA), which is widely used in statistics and machine learning for dimensionality reduction while preserving variance.


Question: The test loss of the 1-nearest neighbor classifier is always zero.

The correct option is:

**B. FALSE**

### Reasoning:

The statement that the test loss of the 1-nearest neighbor (1-NN) classifier is always zero is not true. While the 1-NN classifier will always predict the label of the nearest training example for any given test example, there are several scenarios where the test loss can be greater than zero:

1. **Different Class Labels**: If a test sample is not identical to any training sample and its nearest neighbor is from a different class, the prediction will be incorrect. Therefore, the test loss will be non-zero.

2. **Overfitting**: In cases where the training set is not representative of the overall distribution, the model might perform poorly on the test set despite achieving zero training loss.

3. **Noise and Outliers**: The presence of noisy data or outliers in the training set can lead to misclassification, increasing the test loss.

In summary, while the 1-NN classifier can achieve zero training loss, it does not guarantee zero test loss, making the statement **FALSE**.


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!
Given that you have published 1000 columns at the Gazette, you have around 800 training examples that you can use for your system. Given the size of your dataset, do you think it would be helpful to pretrain your model on other text? Why or why not?

To determine whether it would be beneficial to pretrain your story generation model on other text, we can analyze the situation step by step, ensuring that we define key terms and justify our conclusions clearly.

### Step 1: Understanding the Context

You have published **1000 columns**, but only **800 of those can be used as training examples** for your model. This means your dataset for training consists of 800 columns. The size and quality of this dataset are essential factors in developing an effective language model.

### Step 2: Defining Key Terms

**Pretraining** is the process by which a model is initially trained on a large and diverse corpus of text to learn the fundamental structures of language, such as grammar, syntax, and semantic relationships. This process creates a base model that can be fine-tuned on specific tasks or datasets.

**Fine-tuning** is the subsequent step where the pretrained model is adjusted to perform a specific task using a smaller, task-specific dataset—in this case, your 800 columns.

### Step 3: Importance of Dataset Size and Quality

In machine learning, particularly in natural language processing (NLP), the size and quality of the dataset are crucial for the training of effective models.

- **Larger Datasets**: Generally, larger datasets allow the model to learn better representations and facilitate generalization to new, unseen data. For NLP tasks, having thousands or millions of training examples is often desirable.

- **Quality of Data**: The relevance and diversity of the training data also play a significant role. A dataset that covers various writing styles and topics can improve the model's ability to generate diverse and contextually appropriate text.

### Step 4: Evaluating Your Dataset

You have 800 columns, which is considered a relatively small dataset for training complex language models from scratch.

- **Diversity of Content**: If your columns cover a wide range of topics and writing styles, this diversity can enhance the fine-tuning process, making the model more adaptable to different prompts.

- **Specificity**: If your columns are highly specialized or niche, a pretrained model trained on a broader corpus (e.g., Wikipedia, news articles, or fiction) might still be beneficial. Such a model would already have learned general language patterns that can be adapted to your specific topics and style.

### Step 5: Benefits of Pretraining

The benefits of pretraining your model include:

1. **Knowledge Transfer**: A pretrained model has already learned important language structures, grammar, and contextual relationships from a diverse set of texts. This knowledge can significantly benefit the model when it undergoes fine-tuning on your smaller dataset.

2. **Reduced Training Time**: Starting from a pretrained state means that the model does not need to learn everything from scratch. This can greatly reduce the time and computational resources needed for training.

3. **Improved Performance**: Research has shown that models that are pretrained on large datasets often outperform those trained solely on smaller datasets. They can leverage the learned knowledge to generate more coherent and contextually relevant text.

### Step 6: Conclusion

Given the analysis above, I conclude that it would be helpful to pretrain your model on other text before fine-tuning it on your 800 columns.

- **Reasoning**: The pretrained model will bring in valuable linguistic features and contextual knowledge learned from larger datasets, which can significantly enhance its performance on your specific writing tasks. This approach is likely to improve the model's ability to generate coherent and contextually appropriate text, thus aiding in the automation of your column writing.

### Final Result

**It is advisable to pretrain your story generation model on a larger corpus of text before fine-tuning it on your 800 columns. This strategy will likely improve the model’s overall capability and efficiency in generating quality content relevant to your writing style and topics.**

By addressing the potential benefits and justifying the reasoning behind the recommendation, this revised response provides a clearer and more thorough understanding of the decision to pretrain the model.


Question: Devise an algorithm for the following graph orientation problem: \begin{description} \item[Input:] An undirected graph $G = (V,E)$ and capacities $k : V \rightarrow \mathbb{Z}$ for each vertex. \item[Output:] If possible, an orientation of $G$ such that each vertex $v\in V$ has in-degree at most $k(v)$. \end{description} An orientation of an undirected graph $G$ replaces each undirected edge $\{u,v\}$ by either an arc $(u,v)$ from $u$ to $v$ or by an $(v,u)$ from $v$ to $u$. \\[2mm] \noindent\emph{(Hint: reduce the problem to matroid intersection. You can also use bipartite matching\ldots)}

To solve the problem of orienting an undirected graph \( G = (V, E) \) such that each vertex \( v \in V \) has an in-degree at most \( k(v) \), we can utilize a combination of graph theory concepts, particularly focusing on the idea of matroid intersection or bipartite matching. Here’s a step-by-step breakdown of how we can approach this problem:

### Step 1: Understanding the Problem
We need to orient the edges of the graph \( G \) such that for every vertex \( v \), the number of edges directed into \( v \) (in-degree) does not exceed a given capacity \( k(v) \). This means that for each vertex, we have a constraint on how many edges can point towards it.

### Step 2: Constructing the Bipartite Graph
To utilize bipartite matching, we can transform our problem into a bipartite graph. We will create a bipartite graph \( B \) as follows:

1. **Vertices**: For each vertex \( v \in V \), create two sets of vertices in \( B \):
   - A set \( U \) containing \( k(v) \) copies of vertex \( v \) (let's denote these copies as \( v_1, v_2, \ldots, v_{k(v)} \)).
   - A set \( W \) containing all the vertices in \( V \).

2. **Edges**: For each edge \( \{u, v\} \in E \):
   - Create edges in \( B \) from \( u \) to each of the \( k(v) \) copies of \( v \) (i.e., \( u \) connects to \( v_1, v_2, \ldots, v_{k(v)} \)).
   - Similarly, create edges from \( v \) to each of the \( k(u) \) copies of \( u \).

### Step 3: Finding a Matching
The next step is to find a matching in this bipartite graph \( B \):

1. **Matching**: We need to find a matching that covers all vertices in \( W \) (the original vertices of \( G \)). This matching will correspond to the orientation of the edges in \( G \).
2. **Algorithm**: We can use a standard bipartite matching algorithm, such as the Hopcroft-Karp algorithm, to find a maximum matching in \( B \).

### Step 4: Checking the Matching Size
After running the matching algorithm, we check the size of the matching:

1. If the size of the matching is equal to the number of edges \( |E| \), then we can construct an orientation of \( G \) that satisfies the in-degree constraints.
2. If the size of the matching is less than \( |E| \), then it is impossible to orient the graph under the given constraints.

### Step 5: Constructing the Orientation
If a valid matching is found, we can construct the orientation of the graph:

1. For each edge \( \{u, v\} \) in \( E \), determine which copy of \( v \) (or \( u \)) was matched to \( u \) (or \( v \)).
2. Direct the edge \( (u, v) \) based on the matching (i.e., if \( u \) is matched to \( v_i \), direct the edge from \( u \) to \( v \)).

### Final Algorithm
1. Construct the bipartite graph \( B \) as described.
2. Use a bipartite matching algorithm to find a maximum matching in \( B \).
3. Check the size of the matching:
   - If it covers all edges, output the orientation.
   - If not, output that it is impossible to orient the graph under the given constraints.

This algorithm effectively reduces the problem to a well-known problem in graph theory (bipartite matching) and provides a clear method for determining the orientation of the graph while respecting the in-degree constraints.


Question: Is “type-directed programming” a language mechanism that infers types from values?

Answer: 2

Reasoning: Type-directed programming emphasizes using existing type information to guide program design, rather than inferring types from values.


Question: Consider a $(k+1,k)$ block code that to a binary sequence $x_1,\dots,x_k$ associates the codeword $x_1,\dots,x_k,x_{k+1}$, where $x_{k+1}= x_1+\ldots+x_k$ mod $2$. This code can detect all the errors of odd weight.

A. True. This code can detect all errors of odd weight because flipping an odd number of bits changes the parity, which will be detected during decoding. I chose option A (True) because the $(k+1,k)$ block code is specifically designed to include a parity bit that represents the sum of the first k bits modulo 2. This design allows the code to detect errors effectively; if an odd number of bits are altered (i.e., flipped), the overall parity of the codeword changes, leading to a mismatch when the parity is checked against the expected value. Since the detection mechanism relies on verifying this parity, any odd-weight error will result in a detectable error condition, confirming that the code can indeed detect all such errors. Thus, the statement is accurate.


Question: Tick the \emph{correct} assertion about $\mathbf{Z}_{pq}^*$, where $p$ and $q$ are distinct prime numbers \ldots

To analyze the assertions about the multiplicative group of integers modulo \( pq \), denoted as \( \mathbf{Z}_{pq}^* \), we will evaluate each option step by step.

1. **Understanding \( \mathbf{Z}_{pq}^* \)**:
   The group \( \mathbf{Z}_{pq}^* \) consists of all integers \( x \) such that \( 1 \leq x < pq \) and \( \gcd(x, pq) = 1 \). The order of this group is given by Euler's totient function \( \phi(pq) \).

2. **Calculating \( \phi(pq) \)**:
   For distinct prime numbers \( p \) and \( q \), the totient function is calculated as follows:
   \[
   \phi(pq) = (p-1)(q-1)
   \]
   This represents the number of integers less than \( pq \) that are coprime to \( pq \).

Now, we will evaluate each assertion:

**Option A**: The order of \( \mathbf{Z}_{pq}^* \) is \( p \cdot q \).
- This statement is incorrect. As derived, the order of \( \mathbf{Z}_{pq}^* \) is \( (p-1)(q-1) \), not \( p \cdot q \).

**Option B**: An element \( x \) belongs to \( \mathbf{Z}_{pq}^* \) if \( \gcd(x, \phi(pq)) = 1 \).
- This statement is also incorrect. An element \( x \) is in \( \mathbf{Z}_{pq}^* \) if and only if \( \gcd(x, pq) = 1\), not necessarily related to \( \phi(pq) \).

**Option C**: The element \( 1 \) is in \( \mathbf{Z}_{pq}^* \).
- This statement is correct. The integer \( 1 \) satisfies the condition \( 1 < pq \) and \( \gcd(1, pq) = 1\). Therefore, \( 1 \in \mathbf{Z}_{pq}^* \).

**Option D**: The element \( 0 \) is in \( \mathbf{Z}_{pq}^* \).
- This statement is incorrect. The integer \( 0 \) does not satisfy the condition since \( \gcd(0, pq) = pq\), which is not equal to 1.

Based on the evaluations above, the only correct assertion is option C.

Answer: C) the element \( 1 \) is in \( \mathbf{Z}_{pq}^* \).


Question:
Assume you are working on a mobile application. In the daily standup, you mention you are having issues with JavaFX. Before you can give more details, your team's JavaFX expert tells you to leave it at that and instead pass by her office afterwards. The Scrum Master disagrees and asks you to give more details.
In one sentence, explain whether your Scrum Master is taking the right approach and why.

No, the Scrum Master is not taking the right approach because addressing issues openly in the standup fosters transparency and collaboration among the team.


Question: Which one of the following notions is not in the fundamental trilogy of cryptography?

To determine which notion is not part of the fundamental trilogy of cryptography, we need to look at the key concepts typically associated with secure communication.

1. **Confidentiality**: This ensures that information is only accessible to those who are authorized to access it. It is a core principle of cryptography.

2. **Integrity**: This refers to the assurance that information has not been altered in an unauthorized manner. It is also a fundamental aspect of cryptography.

3. **Authentication**: This involves verifying the identity of the parties involved in communication. It is an essential part of secure communications but is often grouped with integrity in some frameworks.

4. **Privacy**: While related to confidentiality, privacy is a broader concept that may encompass aspects not strictly addressed by cryptographic techniques. It is not typically mentioned as one of the core principles in the trilogy.

From this analysis, "privacy" is the concept that does not fit into the fundamental trilogy of cryptography, which is primarily made up of confidentiality, integrity, and (to some extent) authentication.

FINAL_ANSWER: privacy


Question: What is the order of $2^{124}$ in $(\mathbb{Z}_{2^{128}},+)$?

To determine the order of \(2^{124}\) in the group \((\mathbb{Z}_{2^{128}},+)\), we first need to understand the structure of this group.

The group \((\mathbb{Z}_{2^{128}}, +)\) is the additive group of integers modulo \(2^{128}\). The order of an element \(a\) in a group is defined as the smallest positive integer \(k\) such that \(k \cdot a \equiv 0 \mod n\), where \(n\) is the order of the group (in this case, \(2^{128}\)).

1. **Finding the order of \(2^{124}\)**:
   We need to find the smallest positive integer \(k\) such that:
   \[
   k \cdot 2^{124} \equiv 0 \mod 2^{128}
   \]
   This can be rewritten as:
   \[
   k \cdot 2^{124} = m \cdot 2^{128}
   \]
   for some integer \(m\). Dividing both sides by \(2^{124}\) (which is valid since \(2^{124} \neq 0\) modulo \(2^{128}\)):
   \[
   k = m \cdot 2^4
   \]
   This means \(k\) must be a multiple of \(16\) (since \(2^4 = 16\)). The smallest positive integer \(k\) that satisfies this is \(16\).

Now, let's evaluate each option:

- **Option 1: '8.'**
  - This is incorrect because the smallest multiple of \(16\) greater than \(0\) is \(16\).

- **Option 2: '\(\varphi(2^{128})\).'**
  - The Euler's totient function \(\varphi(2^{128}) = 2^{128} - 2^{127} = 2^{127}\). This does not represent the order of \(2^{124}\) in \((\mathbb{Z}_{2^{128}}, +)\).

- **Option 3: '124.'**
  - This is incorrect because we determined the order is \(16\), not \(124\).

- **Option 4: '16.'**
  - This is the correct answer as we found \(k\) must be \(16\) to satisfy the condition for the order.

After analyzing all options, the best answer is:

Final Answer: 4.


Question: Consider an operation we will call scanRight1 that, given a function $f$ of two arguments, and a sequence $a_1, \ldots, a_N$, computes a sequence $b_1, \ldots, b_N$ such that:  $b_N = a_N$ $b_i = f(a_{i}, b_{i+1})$, for $0 < i < N$ Define similarly scanLeft1 in a manner similar to scanRight1:  Given a function $f$ of two arguments, and a sequence $a_1, \ldots, a_N$, scanLeft1 computes a sequence $b_1, \ldots, b_N$ such that:  $b_i = f(b_{i-1}, a_{i})$, for $0 < i \leq N$ Suppose that $f$ is associative. is the result of scanRight1 same as the result of scanLeft1?

No, the result of `scanRight1` is not necessarily the same as `scanLeft1`, even if the function \( f \) is associative.

To illustrate this, consider the length-two sequence \( A = (a_1, a_2) \) and the function \( f(x, y) = y \).

For `scanLeft1`:
- Start with \( b_1 \) initialized to some identity element (let's assume it can be \( b_0 \), but since \( b_1 = f(b_0, a_1) \), we need to define \( b_0 \) appropriately).
- Thus, \( b_1 = f(b_0, a_1) \) and \( b_2 = f(b_1, a_2) = f(f(b_0, a_1), a_2) = a_2 \).

For `scanRight1`:
- Start with \( b_2 = a_2 \).
- Then, \( b_1 = f(a_1, b_2) = f(a_1, a_2) \).

So we have:
- `scanLeft1` results in \( (b_1, a_2) \) or \( (f(b_0, a_1), a_2) \).
- `scanRight1` results in \( (f(a_1, a_2), a_2) \).

Since \( b_1 \) from `scanLeft1` is dependent on the initial value \( b_0 \), and may not equal \( f(a_1, a_2) \), we see that the two results can differ.

Thus, even with \( f \) being associative, `scanLeft1` and `scanRight1` can yield different results.


Question: Which of the following is well preserved by 2G?

The correct answer is: **Authentication of Mobile Station**.

In 2G (GSM), the authentication process ensures that the mobile station is verified before gaining access to the network, effectively preserving its identity and preventing unauthorized access.


Question: Tick the \emph{true} assertion among the followings:

Let's analyze each option in relation to the question:

### Option 1: Visual cryptography is perfectly secure (at an unreasonable cost).
**Analysis:**
Visual cryptography is a technique that allows a secret to be encrypted in such a way that it can be visually decrypted without needing complex computations. The concept is that an image is divided into multiple shares, and to reconstruct the original image, a certain number of shares are needed. It can provide perfect security in the sense that without the required number of shares, no information can be gleaned about the original image. However, the assertion states that it is "perfectly secure at an unreasonable cost." The cost can refer to the number of shares required or the resources needed to implement the scheme, which can be high. Thus, while the first part of the assertion is generally true, the phrasing about "unreasonable cost" may make it a bit subjective.

### Option 2: The Vernam cipher was invented by Kerckoff.
**Analysis:**
The Vernam cipher, also known as the one-time pad, was invented by Gilbert Vernam in 1917. The principle of the one-time pad was later associated with Claude Shannon, who proved its perfect secrecy under certain conditions. Auguste and Louis Lumière developed concepts related to cryptography, but not the Vernam cipher. Therefore, this assertion is incorrect.

### Option 3: Just like coding theory, cryptography usually faces random noise.
**Analysis:**
Coding theory deals with the design of codes for data transmission and error correction, while cryptography focuses on secure communication and data protection. While both fields involve transmission, cryptography is typically concerned with the prevention of unauthorized access, while coding theory is more focused on error correction. Cryptography does not usually "face random noise" in the same way that coding theory does, as its primary concern is confidentiality and integrity of data rather than dealing with noise. Thus, this assertion is misleading and incorrect.

### Option 4: Enigma has never been broken.
**Analysis:**
The Enigma machine was used by the Germans during World War II for encrypting military communications. However, it was indeed broken by the Allies, notably through the efforts of mathematicians and cryptanalysts like Alan Turing and his team at Bletchley Park. They developed methods to decipher messages encoded with the Enigma, which significantly contributed to the Allied victory. Therefore, this assertion is incorrect.

### Conclusion:
**Correct Option:**
- **Option 1** is the most accurate, albeit with some subjective interpretation regarding the cost.

### Rationale:
- **Option 1** reflects the essence of visual cryptography—its security is robust as it allows decryption without calculations, given enough shares. The cost, while potentially high, does not negate the security aspect.
- **Option 2** is incorrect because the Vernam cipher was not invented by Kerckhoff.
- **Option 3** is misleading as it conflates the concerns of cryptography with those of coding theory.
- **Option 4** is factually incorrect since the Enigma was successfully broken during World War II.

Thus, the only true assertion among the options provided is **Option 1**.


Question:
Consider a binary classification task as in Figure~\AMCref{fig:lr_data}, which consists of 14 two-dimensional linearly separable samples (circles corresponds to label $y=1$ and pluses corresponds to label $y=0$). We would like to predict the label $y=1$ of a sample $(x_1, x_2)$ when the following holds true
                        \[
                        \prob(y=1|x_1, x_2, w_1, w_2) = rac{1}{1+\exp(-w_1x_1 -w_2x_2)} > 0.5
                        \]
                    where $w_1$ and $w_2$ are parameters of the model.
                    If we obtain the $(w_1, w_2)$ by optimizing the following objective
                        $$
                            - \sum_{n=1}^N\log \prob(y_n| x_{n1}, x_{n2}, w_1, w_2) + rac{C}{2} w_2^2
                        $$
                        where $C$ is very large, then the decision boundary will be close to which of the following lines?

To determine which decision boundary the given logistic regression model will approximate, we first need to analyze the optimization objective and the implications of the parameters \( w_1 \) and \( w_2 \) under the context of a large regularization parameter \( C \).

### Understanding the Objective Function

The optimization objective is:

\[
- \sum_{n=1}^N \log \prob(y_n | x_{n1}, x_{n2}, w_1, w_2) + \frac{C}{2} w_2^2
\]

This is a form of logistic regression where you are maximizing the likelihood of the observed labels given the features, while also regularizing the weight \( w_2 \). The term \(\frac{C}{2} w_2^2\) encourages \(w_2\) to be small since \(C\) is very large, which implies a strong preference for small values of \(w_2\). This regularization will push \(w_2\) towards zero.

### Decision Boundary Analysis

The decision boundary in a logistic regression model occurs when:

\[
\prob(y=1 | x_1, x_2, w_1, w_2) = 0.5
\]

Using the logistic function:

\[
\frac{1}{1 + \exp(-w_1 x_1 - w_2 x_2)} = 0.5
\]

This simplifies to:

\[
-w_1 x_1 - w_2 x_2 = 0 \implies w_1 x_1 + w_2 x_2 = 0
\]

This represents the equation of a line in the \(x_1-x_2\) space.

### Impact of Large \(C\)

Since \(C\) is very large, \(w_2\) will be driven towards zero. Therefore, the equation for the decision boundary becomes:

\[
w_1 x_1 + 0 \cdot x_2 = 0 \implies w_1 x_1 = 0
\]

This implies that if \(w_1\) is not equal to zero, the decision boundary will be the vertical line:

\[
x_1 = 0
\]

### Conclusion

Now, we can analyze the options provided:

- **\(x_1 + x_2 = 0\)**: This line has a slope of -1, which does not align with our findings.
- **\(x_1 - x_2 = 0\)**: This line has a slope of 1, which also does not correspond to our derived boundary.
- **\(x_1 = 0\)**: This is a vertical line that matches our conclusion when \(w_2\) is driven to zero.
- **\(x_2 = 0\)**: This is a horizontal line, which does not correspond to our findings.

Thus, the correct answer is:

- **\(x_1 = 0\)**

This is the decision boundary that the model will approximate when \(C\) is very large and \(w_2\) approaches zero.


Question: Tick the \emph{correct} assertion. In RSA with public key $(e,N)$ and private key $(d,N)$ \ldots

In RSA, if we know \(\phi(N)\), we can compute the private key \(d\) from the public key \(e\) using the relationship \(d \equiv e^{-1} \pmod{\phi(N)}\). With \(d\) and the ciphertext \(y\), we can decrypt the message by computing \(y^d \mod N\), recovering the original plaintext. This process relies on the properties of modular arithmetic and the relationship between \(e\), \(d\), and \(\phi(N)\). Thus, knowing \(\phi(N)\) allows us to successfully decrypt \(y\).


Question:
Consider the (toy) grammar $G$ consisting of the following rules:
R1: S --> NP VP
R2: NP --> NN
R3: NP --> Det NN
R4: NN --> N
R5: NN --> NN NN
R6: NN --> NN PNP
R7: PNP --> Prep NP
R8: VP --> V
R9: VP --> Adv V
What is the number $N$ of additional rules that should be added to $G$ to make it applicable to any sequence of words from a set of 10000 distinct words with an average syntactic ambiguity of 1.5? Justify your answer.

N = 15000
To determine the number of additional rules needed for the grammar, we must consider how many distinct (PoS-tag, surface form) associations we need to accommodate the vocabulary of 10,000 distinct words with an average syntactic ambiguity of 1.5.

1. **Understanding Syntactic Ambiguity**: The average syntactic ambiguity of 1.5 means that, on average, each word can correspond to 1.5 different parts of speech (PoS).

2. **Calculating Total Associations**: For 10,000 distinct words, the total number of (PoS-tag, surface form) associations is calculated by multiplying the number of words by the average ambiguity:
   \[
   10,000 \text{ words} \times 1.5 \text{ ambiguities/word} = 15,000 \text{ total associations.}
   \]

3. **Adding Lexical Rules**: Each of these associations requires a unique lexical rule to be added to the grammar. Therefore, to enable the grammar to handle all possible syntactic ambiguities, we need to add 15,000 lexical rules.

Thus, the final conclusion is that we need to add **N = 15000** additional rules to the grammar to effectively manage the syntactic ambiguities of the words in the vocabulary.


Question: The exponent of the group $\mathbb{Z}_9^*$ is

To find the exponent of the group \(\mathbb{Z}_9^*\), we first need to identify the group \(\mathbb{Z}_9^*\) itself. The group \(\mathbb{Z}_9^*\) consists of the integers from \(1\) to \(8\) that are coprime to \(9\).

1. **Identifying elements of \(\mathbb{Z}_9^*\)**:
   The integers from \(1\) to \(8\) are:
   - \(1\) (gcd(1, 9) = 1)
   - \(2\) (gcd(2, 9) = 1)
   - \(3\) (gcd(3, 9) = 3, not included)
   - \(4\) (gcd(4, 9) = 1)
   - \(5\) (gcd(5, 9) = 1)
   - \(6\) (gcd(6, 9) = 3, not included)
   - \(7\) (gcd(7, 9) = 1)
   - \(8\) (gcd(8, 9) = 1)

   Thus, the elements of \(\mathbb{Z}_9^*\) are \( \{1, 2, 4, 5, 7, 8\} \).

2. **Finding the order of each element**:
   - The order of an element \(a\) in a group is the smallest positive \(k\) such that \(a^k \equiv 1 \mod 9\).
   - Let's find the orders:
     - \(1^1 \equiv 1 \mod 9\) (order 1)
     - \(2^1 \equiv 2\), \(2^2 \equiv 4\), \(2^3 \equiv 8\), \(2^4 \equiv 7\), \(2^5 \equiv 5\), \(2^6 \equiv 1\) (order 6)
     - \(4^1 \equiv 4\), \(4^2 \equiv 7\), \(4^3 \equiv 5\), \(4^4 \equiv 1\) (order 3)
     - \(5^1 \equiv 5\), \(5^2 \equiv 7\), \(5^3 \equiv 8\), \(5^4 \equiv 4\), \(5^5 \equiv 2\), \(5^6 \equiv 1\) (order 6)
     - \(7^1 \equiv 7\), \(7^2 \equiv 4\), \(7^3 \equiv 1\) (order 3)
     - \(8^1 \equiv 8\), \(8^2 \equiv 7\), \(8^3 \equiv 5\), \(8^4 \equiv 4\), \(8^5 \equiv 2\), \(8^6 \equiv 1\) (order 6)

3. **Determining the exponent**:
   The exponent of a group is the least common multiple (lcm) of the orders of its elements. Here, the orders of the elements are:
   - Order of \(1\): \(1\)
   - Order of \(2\): \(6\)
   - Order of \(4\): \(3\)
   - Order of \(5\): \(6\)
   - Order of \(7\): \(3\)
   - Order of \(8\): \(6\)

   The least common multiple of these orders is:
   \[
   \text{lcm}(1, 6, 3, 6, 3, 6) = 6
   \]

4. **Final answer**:
After analyzing all options, the only option consistent with our findings is:
- **Option 1: 6.**

Thus, the exponent of the group \(\mathbb{Z}_9^*\) is \(6\).


Question: Select all valid answers about UNIX-like shell.

The valid answers about UNIX-like shell are:

1. **The shell is a program, that runs in user-space.**
   - **Reasoning:** The shell operates as a user-level program, allowing users to interact with the operating system without requiring direct access to kernel-level functions.

3. **The shell is a program, which reads from standard input.**
   - **Reasoning:** The shell reads commands from standard input, which is typically the keyboard, allowing users to execute commands and scripts.

7. **The shell is a user interface for UNIX-like systems.**
   - **Reasoning:** The shell serves as the primary user interface in UNIX-like operating systems, allowing users to execute commands, manage files, and perform various tasks.

**Invalid answers:**

2. **The shell is a program, that runs in kernel-space.**
   - **Reasoning:** The shell does not run in kernel-space; it operates in user-space, which is separated from kernel-space for stability and security.

4. **The shell is a function inside kernel.**
   - **Reasoning:** The shell is not a function within the kernel; it is a separate user-space application that interfaces with the kernel via system calls.

5. **The shell is the layer, which has to be always used for communicating with kernel.**
   - **Reasoning:** While the shell is a common method for communication with the kernel, it is not the only way. Programs can communicate with the kernel directly via system calls without using a shell.

6. **The shell must run only in a single instance. Multiple running instances cause memory corruption.**
   - **Reasoning:** Multiple instances of the shell can run simultaneously without causing memory corruption. Each instance operates in its own memory space.

Therefore, the correct answers are **1, 3, and 7**.


Question: Explain how it is possible to compute Precision at different Recalls.

To compute Precision at different Recalls, we first need to establish the relationship between Precision, Recall, and the number of true positives (TP), false positives (FP), and false negatives (FN). Precision is calculated as \( \text{Precision} = \frac{TP}{TP + FP} \) and Recall is given by \( \text{Recall} = \frac{TP}{TP + FN} \). By varying the threshold for classifying instances as positive, we can generate different sets of TP, FP, and FN, which will yield different Precision and Recall values. By plotting these values, we can create a Precision-Recall curve, capturing the trade-off between Precision and Recall at various thresholds. This allows us to analyze how Precision changes as we adjust the criteria for what constitutes a positive classification, thereby enabling us to compute Precision at different Recall levels.


Question: Which loss function(s) should you use? (One or multiple answers)

The correct answers are: 'mean square error (MSE) loss', 'cross entropy loss', and 'hinge loss'.

Explanation:
1. **Mean Square Error (MSE) Loss**: This loss function is commonly used in regression problems where the goal is to minimize the average of the squares of the errors between predicted and actual values.

2. **Cross Entropy Loss**: This loss function is mainly used for classification tasks, particularly when dealing with probabilities, such as in logistic regression or neural networks for multi-class classification.

3. **Hinge Loss**: This is specifically used for "maximum-margin" classification, such as in Support Vector Machines (SVM). It is effective for binary classification tasks.

4. **L1 Loss** (also known as Mean Absolute Error) could also be considered in certain contexts, especially in regression tasks where robustness to outliers is desired, but it was not specifically mentioned as a primary choice in the original question.

Thus, MSE, cross entropy, and hinge loss are commonly employed loss functions across different types of machine learning tasks.