Question:
Consider the following PyTorch code:

    class ThreeLayerNet (nn.Module):
        def __init__():
            super().__init__()

        def forward(x):
            x = nn.Linear(100, 10)(x)
            x = nn.ReLU()(x)
            x = nn.Linear(10, 200)(x)
            x = nn.ReLU()(x)
            x = nn.Linear(200, 1)(x)
            return x

    Suppose that inputs are 100-dimensional, and outputs are 1-dimensional. What will happen if we try to train this network?

1. **There will be an error because we are re-using the variable x throughout the forward() method.**
   This is incorrect because re-using variable names in Python is valid; it simply updates the reference to the new value.

2. **There will be an error because the second layer has more neurons than the first. The number of neurons must never increase from one layer to the next.**
   This is incorrect because neural networks can have layers with an increasing number of neurons. It is common to expand the dimensionality in hidden layers.

3. **Everything is fine.**
   This is incorrect because while the code may run without syntax errors, the training dynamics and model performance will not be effective due to the architecture choice.

**Correct answer: The model will not train properly. The performance will be the same at the beginning of the first epoch and at the end of the last epoch.**
The architecture of the network may lead to insufficient learning due to the lack of capacity to adjust weights effectively across the layers. The model may struggle to converge or learn meaningful patterns, resulting in little to no change in performance metrics throughout training. This is often a consequence of poor initialization or inappropriate layer configurations that do not allow the model to adapt.


Question:
Consider the loss function $L: \R^d 	o \R$, $L(\wv) = rac{eta}{2}\|\wv\|^2$, where $eta > 0$ is a constant. We run gradient descent on $L$ with a stepsize $\gamma > 0$ starting from some $\wv_0
eq 0$. Which of the statements below is true?

To analyze the correctness of each option, we first note that the loss function given is \( L(\wv) = \frac{\eta}{2} \|\wv\|^2 \). The gradient of this function is:

\[
\nabla L(\wv) = \eta \wv.
\]

Using gradient descent, the update rule is:

\[
\wv_{t+1} = \wv_t - \gamma \nabla L(\wv_t) = \wv_t - \gamma \eta \wv_t = (1 - \gamma \eta) \wv_t.
\]

Now, let's evaluate each option step-by-step.

1. **Option 1:** "Gradient descent converges to the global minimum for any stepsize \( \gamma > 0 \)."

   This statement is **false**. The convergence of gradient descent depends on the choice of the stepsize \( \gamma \). If \( \gamma \) is too large (specifically, if \( \gamma \geq \frac{2}{\eta} \)), the iterates will diverge. Therefore, it does not converge for any positive stepsize.

2. **Option 2:** "Gradient descent with stepsize \( \gamma = \frac{2}{\eta} \) produces iterates that diverge to infinity (\( \|\wv_t\| \to \infty \) as \( t\to \infty \))."

   This statement is **true**. If we set \( \gamma = \frac{2}{\eta} \), then the update becomes:

   \[
   \wv_{t+1} = (1 - 2) \wv_t = -\wv_t.
   \]

   This means that the norm of \( \wv_t \) will oscillate and not converge to zero, leading to divergence in terms of the iterates not approaching the global minimum.

3. **Option 3:** "Gradient descent converges in two steps for \( \gamma = \frac{1}{\eta} \) (i.e., \( \wv_2 \) is the first iterate attaining the global minimum of \( L \))."

   This statement is **false**. If we set \( \gamma = \frac{1}{\eta} \), we have:

   \[
   \wv_{t+1} = (1 - 1) \wv_t = 0.
   \]

   Thus, after one step, we reach the global minimum at \( L(\mathbf{0}) = 0 \). Therefore, it does not take two steps to reach the minimum; it takes only one step.

4. **Option 4:** "Gradient descent converges to the global minimum for any stepsize in the interval \( \gamma \in (0, \frac{2}{\eta}) \)."

   This statement is **true**. For any stepsize \( 0 < \gamma < \frac{2}{\eta} \), the update rule leads to:

   \[
   |\wv_{t+1}| = |(1 - \gamma\eta)|\wv_t| < |\wv_t|,
   \]

   which implies that the iterates will converge to zero, thus reaching the global minimum.

**Final answer:** d


Question: Which scheme is the most secure?

3: Three-key Triple DES offers the longest effective key length of 168 bits and is significantly more resistant to brute-force and other attacks than the other options. Therefore, the most secure scheme is Three-key Triple DES (3TDES).


Question: Which of these components was not part of the Enigma machine?

The component that was not part of the Enigma machine is **a pseudo-random number generator**.

The Enigma machine utilized several key components for its encryption process, including:

- **A reflector**: This component redirected the electrical signals back through the rotors, creating a symmetrical encryption process.
- **Rotors**: These were the primary mechanism for scrambling the letters. Each rotor had a different wiring configuration and would rotate with each key press, changing the encryption pattern.
- **A plugboard with wire connections**: This was used to further scramble the letters before and after they passed through the rotors, allowing for additional complexity in the encryption.

In contrast, a pseudo-random number generator is a computational algorithm used to produce a sequence of numbers that approximates the properties of random numbers, and it was not a part of the Enigma machine's design or function.


Question:
Consider a DSP with an Address Generation Unit which has a single
                    address register which can only be automodified to point to the next
                    or previous word in memory without using the main ALU nor reloading
                    the address register.  A program uses five integer variables erb+i+,
                    erb+j+, erb+x_coord+, erb+y_coord+, and erb+sum+, and the
                    sequence of accesses in the main program loop is statically known and
                    is
                    egin{verbatim}
                    i → j → x_coord → y_coord → x_coord → i → y_coord →

                    → x_coord → y_coord → j → sum → x_coord → y_coord
                    \end{verbatim}
                    Note that these accesses are all inside a loop which repeats many
                    times. What is an optimal placement of the five integers in memory?
                    Show how you have arrived to the result.

To optimize memory access for the given sequence in a DSP with a single address register, we should arrange the five integer variables contiguously in memory based on their access frequency. The access pattern shows that `x_coord` and `y_coord` are accessed multiple times in succession. Therefore, placing these two variables next to each other will minimize the address modifications required. A suggested order could be: `x_coord`, `y_coord`, `i`, `j`, and `sum`. This arrangement allows for fewer automodifications, as `x_coord` and `y_coord` can be accessed sequentially, followed by the less frequently accessed variables.


Question: The edit distance between “piece” and “peace” is(Penalty for wrong ticks)

To determine the edit distance between the words "piece" and "peace", we first define what edit distance means. Edit distance is the minimum number of single-character edits (insertions, deletions, or substitutions) needed to transform one string into another.

Now, let's analyze the words "piece" and "peace" character by character:

1. The first character of both "piece" and "peace" is 'p', which matches.
2. The second character in "piece" is 'i', while in "peace" it is 'a'. Here, we need to substitute 'i' with 'a'.
3. The third character in both words is 'e', which matches.
4. The fourth character in "piece" is 'c', while in "peace" it is absent. This necessitates deleting 'c' from "piece".
5. The fifth character in both words is 'e', which matches.

In summary, the edits required are:
- 1 substitution (changing 'i' to 'a')
- 1 deletion (removing 'c')

This results in a total of 2 edits needed to transform "piece" into "peace".

Now, looking at the provided options, the correct answer would be **7. 2, if considering insertion, deletion and substitution**. This option correctly reflects the edits made without including unnecessary operations like transpositions.

Thus, the final answer is option 7.


Question:
Let $f:\R^D
ightarrow\R$ be an $L$-hidden layer multi-layer perceptron (MLP) such that
                    \[
	                    f(xv)=\sigma_{L+1}ig(\wv^	op\sigma_L(\Wm_L\sigma_{L-1}(\Wm_{L-1}\dots\sigma_1(\Wm_1xv)))ig),
                    \]
                    with $\wv\in\R^{M}$, $\Wm_1\in\R^{M	imes D}$ and $\Wm_\ell\in\R^{M	imes M}$ for $\ell=2,\dots, L$, and $\sigma_i$ for $i=1,\dots,L+1$ is an entry-wise activation function. For any MLP $f$ and a classification threshold $	au$ let $C_{f, 	au}$ be a binary classifier that outputs YES for a given input $xv$ if $f(xv) \leq 	au$ and NO otherwise. space{3mm}
                    Which of the following techniques do \emph{not} improve the generalization performance in deep learning?

To determine which techniques do not improve generalization performance in deep learning, let's analyze each option step-by-step:

1. **Data Augmentation**: This technique involves artificially increasing the size of the training dataset by making modifications (e.g., rotations, translations) to existing data. It helps the model generalize better by exposing it to more varied scenarios. **This improves generalization.**

2. **L2 Regularization**: This technique adds a penalty term to the loss function based on the size of the weights, discouraging overly complex models. By preventing overfitting, it helps the model generalize better to unseen data. **This improves generalization.**

3. **Dropout**: This regularization method randomly sets a fraction of the neurons to zero during training, which helps prevent co-adaptation of feature detectors. It effectively reduces overfitting, promoting better generalization. **This improves generalization.**

4. **Tuning the Optimizer**: Carefully selecting and adjusting optimizer parameters (like learning rate) can significantly improve training efficiency and convergence. While it may not directly influence model complexity, effective tuning can lead to better convergence and thus indirectly aid generalization. **This improves generalization.**

5. **None. All techniques here improve generalization**: This option is correct as all mentioned techniques do indeed enhance generalization performance.

### Potential Misunderstanding:
A common misconception might be that tuning the optimizer does not impact generalization directly. However, it plays a crucial role in how effectively a model learns, which can ultimately affect its ability to generalize well.

Thus, the correct answer is: **None. All techniques here improve generalization.**


Question: Consider password-based access control. Tick the \textit{incorrect} assertion.

To determine which assertion is incorrect regarding password-based access control, let's analyze each option:

1. **Double hashing the password can help avoid the problems related to low-entropy passwords.**
   - **Analysis**: Double hashing does not inherently address the issue of low-entropy passwords. Low-entropy passwords are weak and can be easily cracked regardless of the hashing method used. While double hashing may add a layer of complexity, it does not improve the fundamental weakness of the password itself.
   - **Conclusion**: This assertion is **incorrect**.

2. **Salt can be used to thwart multi-target attacks.**
   - **Analysis**: Salting involves adding a unique random value to each password before hashing, which prevents attackers from using precomputed hash tables (rainbow tables) to crack multiple passwords at once. By using a unique salt, even identical passwords will have different hashes, effectively thwarting multi-target attacks.
   - **Conclusion**: This assertion is **correct**.

3. **Increasing the delay between authentication attempts can protect from online attacks.**
   - **Analysis**: Implementing a delay (e.g., exponential backoff) between authentication attempts makes brute-force attacks more time-consuming and less feasible for attackers. This method helps mitigate online attacks by slowing down the rate at which guesses can be made.
   - **Conclusion**: This assertion is **correct**.

4. **Blocking the access after 10 unsuccessful authentication attempts can protect from online attacks.**
   - **Analysis**: While blocking access after a certain number of failed attempts can help prevent brute-force attacks, it is also a potential vector for denial-of-service (DoS) attacks if an attacker can intentionally trigger the lockout. This measure can provide some level of protection but is not foolproof and may not be sufficient by itself.
   - **Conclusion**: This assertion has limitations but can be considered **correct** in the context of providing some protection.

### Summary of Findings:
- **Incorrect Assertion**:
  - **Double hashing the password can help avoid the problems related to low-entropy passwords.**

- **Correct Assertions**:
  - **Salt can be used to thwart multi-target attacks.**
  - **Increasing the delay between authentication attempts can protect from online attacks.**
  - **Blocking the access after 10 unsuccessful authentication attempts can protect from online attacks.**


Question: Consider an array $A[1,\ldots, n]$ consisting of the $n$ distinct numbers $1,2, \ldots, n$. We are further guaranteed that $A$ is almost sorted in the following sense: $A[i] \neq i$ for at most $\sqrt{n}$ values of $i$.  What are tight asymptotic worst-case running times for Insertion Sort and Merge Sort on such instances?

6: It is $\Theta(n \log n)$ for Merge Sort and $\Theta(n^{3/2})$ for Insertion Sort.

**Explanation**: Merge Sort maintains its $\Theta(n \log n)$ complexity regardless of the input's near-sorted condition due to its divide-and-conquer nature. For Insertion Sort, while it is efficient on nearly sorted arrays, the presence of $\sqrt{n}$ misplaced elements leads to a total of $O(n \cdot \sqrt{n}) = \Theta(n^{3/2})$ in the worst case, as each misplaced element may require up to $\sqrt{n}$ comparisons and shifts.


Question: Tick the \textit{correct} assertion.

Let’s evaluate each assertion step by step to determine which one is correct:

### Assertion A: In a finite field \( K \), every element has exactly two square roots.
This assertion is **false**. In a finite field \( K \), every non-zero element has either two square roots or none, depending on whether the order of the field is odd or even. Specifically, if \( K = \mathbb{F}_{p^n} \) where \( p \) is a prime, then the number of square roots of an element \( a \) (if it is a quadratic residue) is actually two for non-zero elements. However, the statement implies that every element has two square roots, which is incorrect since the element \( 0 \) does not have a square root in the sense of distinct elements.

### Assertion B: In a finite field \( K \), 1 has exactly one square root and it is 1.
This assertion is **true**. In any field, including finite fields, the only square root of \( 1 \) is \( 1 \) itself, because \( 1^2 = 1 \). It does not have a second square root, so this assertion is valid.

### Assertion C: The set of quadratic residues in \( \mathbb{Z}_n \) is a field.
This assertion is **false**. The set of quadratic residues modulo \( n \) does not necessarily form a field. For a set to be a field, it must be closed under addition and multiplication, and every non-zero element must have a multiplicative inverse. Quadratic residues do not satisfy these properties in general, especially when \( n \) is not a prime power.

### Assertion D: An element can have more than two square roots in \( \mathbb{Z}_n \).
This assertion is **true** but context-dependent. In \( \mathbb{Z}_n \), if \( n \) is not prime, an element might have multiple square roots. For example, consider \( n = 8 \), where \( 4 \) has square roots \( 4 \) and \( 0 \) (and also \( 4 + n \)). This assertion is generally true, but it is important to note that in a finite field, a non-zero element has at most two distinct square roots.

### Conclusion:
The correct assertion is **B**: In a finite field \( K \), 1 has exactly one square root and it is 1.


Question: Which of the following statements are true about DEP?

To analyze the statements regarding Data Execution Prevention (DEP), let's evaluate each one:

1. **DEP prohibits code injection but not code reuse.** This statement is partially correct. DEP primarily aims to prevent code execution in certain memory regions, which helps mitigate code injection attacks. However, it does not inherently protect against code reuse attacks, such as Return-Oriented Programming (ROP), where existing code is reused to perform malicious actions.

2. **A single bit is used to indicate that a memory region is DEP protected.** This statement is true. DEP uses a single bit in the page table entries to mark whether a memory page is executable or not.

3. **Even if it is called Enhanced Virus Protection on AMD hardware, it does not protect against viruses.** This statement is misleading. While Enhanced Virus Protection (EVP) is based on DEP, it is not a comprehensive antivirus solution. It provides some level of protection against certain types of attacks, but it is not designed specifically to prevent all viruses.

4. **Data Execution Prevention is usually handled at byte-level granularity.** This statement is false. DEP is handled at the page level, not the byte level. It marks entire memory pages as executable or non-executable.

5. **A region marked as DEP protected cannot be read.** This statement is incorrect. DEP prevents execution but does not restrict read or write access to memory regions. Data can still be read from DEP-protected areas.

Based on this analysis, the most accurate statements about DEP are that a single bit indicates protection status and that it effectively prevents code injection but has limitations with code reuse. The EVP on AMD hardware does provide some protective measures but is not foolproof against all viruses.


Question: What is the complexity of prime number generation for a prime of length $\ell$?

To determine the complexity of generating a prime number of length \(\ell\), we can break it down into a few key steps:

1. **Random Number Generation**: First, we need to generate random numbers of the desired length \(\ell\). This step involves generating numbers in the range of \(2^{\ell-1}\) to \(2^{\ell}\). The complexity for generating random numbers is generally linear with respect to \(\ell\).

2. **Primality Testing**: Once a candidate number is generated, we must test whether it is prime. The complexity of primality testing depends on the algorithm used. Efficient algorithms can run in polynomial time.

3. **Overall Process**: The overall complexity is influenced by both the number of candidates generated and the efficiency of the primality test.

In conclusion, the complexity of generating a prime number of length \(\ell\) can be summarized as being related to \(\Theta(\ell^4)\), although the precise details can vary based on the specific methods employed. Therefore, the answer is \(\mathbf{O}(\ell^4)\).


Question: Consider the Poisson distribution with parameter $\lambda$. It has a probability mass function given by $p(i)=\frac{\lambda^{i} e^{-\lambda}}{i !}$, $i=0,1, \cdots$ (i) Write $p(i)$ in the form of an exponential distribution $p(i)=h(i) e^{\eta \phi(i)-A(\eta)}$. Explicitly specify $h, \eta, \phi$, and $A(\eta)$ (ii) Compute $\frac{d A(\eta)}{d \eta}$ and $\frac{d^{2} A(\eta)}{d \eta^{2}}$ ? Is this the result you expected?

To express the Poisson distribution in the form of an exponential family, we start with the probability mass function given by:

\[
p(i) = \frac{\lambda^{i} e^{-\lambda}}{i!}, \quad i = 0, 1, 2, \ldots
\]

### Part (i)

We can rewrite this in the exponential family form:

\[
p(i) = h(i) e^{\eta \phi(i) - A(\eta)}
\]

where we identify the components as follows:

1. **Identify \(h(i)\)**:
   We can set \(h(i) = \frac{1}{i!}\) since it does not depend on \(\lambda\).

2. **Identify \(\eta\)**:
   We choose \(\eta = 1\), which simplifies our analysis.

3. **Identify \(\phi(i)\)**:
   We can set \(\phi(i) = \log(\lambda) \cdot i\). This connects \(i\) with \(\lambda\) in a linear way.

4. **Identify \(A(\eta)\)**:
   Since we have \(\eta = 1\), we find \(A(1)\) by matching the exponent with the original function. The term involving \(\lambda\) can be grouped as follows:
   \[
   e^{-\lambda} = e^{-A(1)}
   \]
   Thus, we have:
   \[
   A(1) = \lambda
   \]

Putting this together, we can express the Poisson distribution as:

\[
p(i) = \frac{1}{i!} e^{i \log(\lambda) - \lambda}
\]

So we have:
- \(h(i) = \frac{1}{i!}\)
- \(\eta = 1\)
- \(\phi(i) = i \log(\lambda)\)
- \(A(\eta) = \lambda\)

### Part (ii)

Next, we need to compute the derivatives of \(A(\eta)\).

Since \(A(\eta) = \lambda\) is a constant with respect to \(\eta\), we find:

1. **First derivative**:
   \[
   \frac{d A(\eta)}{d \eta} = 0
   \]

2. **Second derivative**:
   \[
   \frac{d^{2} A(\eta)}{d \eta^{2}} = 0
   \]

### Conclusion

The results we computed are expected because the function \(A(\eta)\) being constant implies that its derivatives are zero. This matches the fact that in the exponential family, the log-partition function \(A(\eta)\) provides normalization and does not vary with changes in \(\eta\) for a fixed \(\lambda\).


Question: Tick the \textbf{incorrect} assertion regarding WEP and WPA2.

The incorrect assertion is that WEP ensures message integrity. In reality, WEP's integrity mechanism (CRC-32 checksum) is weak and does not provide true message integrity, as it is vulnerable to bit-flipping and other manipulation attacks. Therefore, WEP does not adequately ensure integrity.


Question: Which is true about the use of entropy in decision tree induction?

The correct statement about the use of entropy in decision tree induction is:

**The entropy of the set of class labels of the samples from the training set at the leaf level is always 0 if the leaf is pure (i.e., contains samples from only one class).**

This is true because, at the leaf level of a decision tree, when all samples belong to a single class, there is no uncertainty or disorder in the class labels, resulting in an entropy of 0. However, if a leaf contains samples from multiple classes, the entropy can be greater than 0, depending on the distribution of those classes.

Let's evaluate the other options:

- **"We split on the attribute that has the highest entropy"**: This statement is incorrect. In decision tree induction, we actually choose the attribute that minimizes entropy (or maximizes information gain) to make the split. The goal is to achieve a more homogeneous grouping of classes in the resulting subsets.

- **"The entropy of the set of class labels of the samples from the training set at the leaf level can be 1"**: This statement can be misleading. While it is true that entropy can reach a maximum value (like 1 for a perfectly balanced binary classification), at the leaf level, we typically aim for purity (entropy of 0) by making effective splits. An entropy of 1 would indicate equal representation of classes in a non-pure leaf.

- **"We split on the attribute that has the lowest entropy"**: This statement is correct. The strategy in decision tree induction is to select the attribute that, when used for splitting, results in the lowest entropy for the resulting subsets. This approach effectively maximizes information gain.

In summary, entropy is a key measure of impurity or disorder in a dataset and plays a crucial role in decision tree induction. The objective is to minimize entropy through strategic attribute selection for splitting, leading to more accurate classification of samples in the resulting branches of the tree.


Question: Tick the \textbf{false} assertion.

Option C is false; the NP language can have zero-knowledge proofs for certain problems, as demonstrated by protocols like those for NP-complete problems, which can be constructed to be zero-knowledge under specific conditions. Zero-knowledge proofs exist for various languages within NP, particularly when certain properties are satisfied.


Question: How does the bias-variance decomposition of a ridge regression estimator compare with that of the ordinary least-squares estimator in general?

In the context of bias-variance decomposition, the goal is to understand how the expected error of a model can be attributed to bias, variance, and irreducible error. Ordinary least squares (OLS) regression tends to have lower bias but can be prone to high variance, particularly in the presence of multicollinearity or overfitting. Ridge regression, on the other hand, introduces a penalty on the size of the coefficients, which increases the bias but reduces the variance of the estimator. This trade-off makes ridge regression particularly useful in situations where overfitting is a concern. Therefore, the correct comparison is that ridge regression has a larger bias and smaller variance than ordinary least squares.

Correct answer: Ridge has a larger bias, and smaller variance.


Question: Which sets are countable (Multiple answers can be correct) :

Answer: A

Answer: C

Explanation: Set A is countable because it consists of finite strings, which can be enumerated. Set C is countable as it is a subset of natural numbers, which are countable. Sets B and D are not countable; B is uncountable due to the infinite nature of real numbers, and D is uncountable since removing a countable set from an uncountable set still leaves an uncountable set.


Question: In which type of processors do you expect to find a reorder buffer?

**Revised Answer:**

A reorder buffer (ROB) is typically found in dynamically scheduled out-of-order processors. The ROB is essential for enabling these processors to execute instructions out of order while preserving the appearance of in-order execution from the perspective of the software. This is crucial for correctly managing instruction dependencies, as it allows the processor to track the status and completion of instructions, ensuring that results are committed in the correct order. Additionally, the ROB plays a key role in handling exceptions and mis-speculations, allowing the processor to recover gracefully from errors.

While reorder buffers are primarily associated with out-of-order execution, they can also be found in certain superscalar architectures where multiple instructions are issued and executed simultaneously. However, the use of a reorder buffer introduces complexity into the processor design and can lead to increased power consumption.

Examples of processors that utilize reorder buffers include modern Intel and AMD CPUs, which leverage out-of-order execution to enhance instruction throughput and overall performance.


Question: Tick the \textbf{false} assertion. For a Vernam cipher...

To solve the question, we need to understand how the Vernam cipher works. The Vernam cipher is a type of symmetric key cipher where each character of the plaintext is combined with a character from the key using the XOR (exclusive OR) operation. For the cipher to work correctly, the key must be at least as long as the plaintext.

Now let's analyze each option to determine which assertion is false.

1. **SUPERMAN can be the result of the encryption of the plaintext ENCRYPT.**
   - The plaintext "ENCRYPT" has 7 characters. If "SUPERMAN" is the ciphertext, it has 8 characters. Since the key must be at least as long as the plaintext, and here the ciphertext is longer than the plaintext, this assertion is **false**.

2. **CRYPTO can be used as a key to encrypt the plaintext PLAIN.**
   - The plaintext "PLAIN" has 5 characters, and the key "CRYPTO" has 6 characters. Since the key is longer than the plaintext, it can be used for encryption. This assertion is **true**.

3. **SERGE can be the ciphertext corresponding to the plaintext VAUDENAY.**
   - The plaintext "VAUDENAY" has 8 characters, while "SERGE" has only 5 characters. The key must be at least as long as the plaintext, so this assertion is **false**.

4. **The key IAMAKEY can be used to encrypt any message of size up to 7 characters.**
   - The key "IAMAKEY" has 7 characters, so it can indeed encrypt any message of size up to 7 characters. This assertion is **true**.

Now, let's summarize the evaluations:

- Option 1 (SUPERMAN can be the result of the encryption of the plaintext ENCRYPT) - **false**
- Option 2 (CRYPTO can be used as a key to encrypt the plaintext PLAIN) - **true**
- Option 3 (SERGE can be the ciphertext corresponding to the plaintext VAUDENAY) - **false**
- Option 4 (The key IAMAKEY can be used to encrypt any message of size up to 7 characters) - **true**

Given that we are looking for a **false** assertion, both options 1 and 3 are false, but since these are mutually exclusive statements, we can conclude that the first one (SUPERMAN can be the result of the encryption of the plaintext ENCRYPT) is the first one we analyzed and can be marked as false.

Thus, the false assertion is:

**SUPERMAN can be the result of the encryption of the plaintext ENCRYPT.**


Question: $L_1$ regularization often results in sparser solutions than $L_2$ regularization.

The correct answer is **TRUE**.

**Step-by-step reasoning for each option:**

1. **TRUE**:
   - **Explanation**: $L_1$ regularization, also known as Lasso regularization, adds a penalty equal to the absolute value of the coefficients to the loss function. This can force some coefficients to be exactly zero, effectively selecting a simpler model with fewer features. This results in sparser solutions.
   - **Example**: If you have a model with ten features, $L_1$ might retain only three significant features while setting the others to zero.

2. **FALSE**:
   - **Explanation**: $L_2$ regularization, or Ridge regularization, adds a penalty equal to the square of the coefficients. This penalty reduces the magnitude of the coefficients but does not force any of them to be exactly zero. As a result, $L_2$ typically retains all features in the model, leading to denser solutions.
   - **Misunderstanding**: A common misconception is that all forms of regularization will lead to the removal of features. However, while $L_2$ can shrink coefficients, it does not eliminate them, which is why it does not produce sparse solutions.

In summary, $L_1$ regularization encourages sparsity by setting some coefficients to zero, while $L_2$ regularization tends to keep all coefficients, leading to denser solutions. Thus, the statement is true.


Question:
Consider the following toy corpus: the cat cut the hat
Considering only lowercase alphabetical and whitespace, how many bigrams are possible?

27^2 = 729 bigrams in total


Question: In the following let $\kappa_{1}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)$ and $\kappa_{2}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)$ be two valid kernels. Show that the following are is a valid kernel: $\kappa\left(\mathbf{x}, \mathbf{x}^{\prime}\right)=a \kappa_{1}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)+b \kappa_{2}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)$ for all $a, b \geq 0$.

To show that the function \(\kappa(\mathbf{x}, \mathbf{x}^{\prime}) = a \kappa_{1}(\mathbf{x}, \mathbf{x}^{\prime}) + b \kappa_{2}(\mathbf{x}, \mathbf{x}^{\prime})\) is a valid kernel for all \(a, b \geq 0\), we need to demonstrate that \(\kappa\) is symmetric and positive semi-definite.

### Step 1: Symmetry

A kernel \(\kappa(\mathbf{x}, \mathbf{x}^{\prime})\) is symmetric if \(\kappa(\mathbf{x}, \mathbf{x}^{\prime}) = \kappa(\mathbf{x}^{\prime}, \mathbf{x})\) for all \(\mathbf{x}, \mathbf{x}^{\prime}\).

1. Since \(\kappa_{1}\) and \(\kappa_{2}\) are valid kernels, they are symmetric:
   \[
   \kappa_{1}(\mathbf{x}, \mathbf{x}^{\prime}) = \kappa_{1}(\mathbf{x}^{\prime}, \mathbf{x}) \quad \text{and} \quad \kappa_{2}(\mathbf{x}, \mathbf{x}^{\prime}) = \kappa_{2}(\mathbf{x}^{\prime}, \mathbf{x}).
   \]
2. Therefore, we have:
   \[
   \kappa(\mathbf{x}, \mathbf{x}^{\prime}) = a \kappa_{1}(\mathbf{x}, \mathbf{x}^{\prime}) + b \kappa_{2}(\mathbf{x}, \mathbf{x}^{\prime}) = a \kappa_{1}(\mathbf{x}^{\prime}, \mathbf{x}) + b \kappa_{2}(\mathbf{x}^{\prime}, \mathbf{x}) = \kappa(\mathbf{x}^{\prime}, \mathbf{x}).
   \]

This shows that \(\kappa\) is symmetric.

### Step 2: Positive Semi-Definiteness

A kernel \(\kappa\) is positive semi-definite if for any finite set of points \(\{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_n\}\) and any real coefficients \(c_1, c_2, \ldots, c_n\), the following inequality holds:
\[
\sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j \kappa(\mathbf{x}_i, \mathbf{x}_j) \geq 0.
\]

Given that \(\kappa_1\) and \(\kappa_2\) are valid kernels, we can apply the definition of positive semi-definiteness:

1. We can express the sum involving \(\kappa\):
   \[
   \sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j \kappa(\mathbf{x}_i, \mathbf{x}_j) = \sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j \left( a \kappa_1(\mathbf{x}_i, \mathbf{x}_j) + b \kappa_2(\mathbf{x}_i, \mathbf{x}_j) \right).
   \]
2. This can be separated as follows:
   \[
   = a \sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j \kappa_{1}(\mathbf{x}_i, \mathbf{x}_j) + b \sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j \kappa_{2}(\mathbf{x}_i, \mathbf{x}_j).
   \]
3. Since both terms on the right are non-negative (due to the positive semi-definiteness of \(\kappa_1\) and \(\kappa_2\)), we have:
   \[
   \sum_{i=1}^{n} \sum_{j=1}^{n} c_i c_j \kappa(\mathbf{x}_i, \mathbf{x}_j) \geq 0.
   \]

### Conclusion

Since \(\kappa(\mathbf{x}, \mathbf{x}^{\prime})\) is symmetric and positive semi-definite for \(a, b \geq 0\), we conclude that \(\kappa\) is a valid kernel. Thus, we have shown that \(\kappa(\mathbf{x}, \mathbf{x}^{\prime}) = a \kappa_{1}(\mathbf{x}, \mathbf{x}^{\prime}) + b \kappa_{2}(\mathbf{x}, \mathbf{x}^{\prime})\) is indeed a valid kernel.


Question: Which attribute gives the best split?A1PNa44b44A2PNx51y33A3PNt61j23

To determine which attribute gives the best split, we need to understand the context of what "best split" means in a decision tree or classification context. A "split" is a method of dividing a dataset into subsets based on the values of an attribute, and the goal is typically to create subsets that are as pure as possible (i.e., contain instances of a single class).

1. **Understanding the Attributes**:
   - **A1**: This attribute could be a categorical or numerical feature, but we need to know its values and how they relate to the target variable to evaluate its effectiveness in splitting the data.
   - **A2**: Similar to A1, we would need to know how A2's values contribute to the classification task.
   - **A3**: Again, the same logic applies to A3.

2. **Evaluating the Options**:
   - **A: A1**: If A1 leads to a clear separation of classes when splitting the dataset, it might be the best choice.
   - **B: A3**: The same reasoning applies; if A3 offers a better separation than A1 and A2, it would be the best.
   - **C: A2**: If A2 results in the highest purity of classes after the split, it would be the best.
   - **D: All the same**: This option implies that none of the attributes provide a better split than the others, suggesting that they all yield the same level of purity.

3. **Evaluating Each Choice**:
   - Without specific data or metrics (like Gini impurity or information gain), it is impossible to definitively say which attribute gives the best split.
   - If the attributes lead to equal splits in terms of class purity, then "All the same" could be correct.
   - If one attribute clearly outperforms the others, that would be the best split.

Given the lack of specific information about how each attribute influences the split, we must rely on the possibility that they might not significantly differ.

After careful consideration, I conclude that without additional data, we cannot determine a clear best attribute.

Action: Answer("D")


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!
You consider using either a transformer or a recurrent neural network (RNN) as the underlying model for your text generator. Assuming there are no practical issues with selecting either one (such as the amount of data available), which one would you choose for this task?
Give two reasons why.

I would choose a transformer model for the task of generating columns. First, transformers excel at capturing long-range dependencies in text, allowing them to better understand context and coherence over longer passages compared to RNNs, which can struggle with this due to their sequential nature. Second, transformers can process input data in parallel, significantly speeding up training and inference times, which is beneficial for generating high-quality content efficiently. These advantages make transformers a more suitable choice for automating column generation.


Question: In which of the following cases does the TLB need to be flushed?

The Translation Lookaside Buffer (TLB) is a cache that stores recent translations of virtual memory addresses to physical memory addresses to speed up the process of virtual address translation.

A TLB may need to be flushed under the following circumstances:

Inserting a new page into the page table for a user-space application: Typically, this would not require a TLB flush, as a new entry means the TLB may simply not have that page cached yet.

Deleting a page from the page table: Yes, this usually does require a TLB flush. If the page entry is deleted from the page table and the TLB still has its mapping, it could lead to incorrect access.

Changing the read/write permission bit in the page table: Yes, this typically requires the TLB to be flushed as well. Any change in permissions can result in pages that are marked in the TLB becoming invalid.

Inserting a new page into the page table for kernel: Similar to the user-space application case, this might not require a flush unless it involves modifications to existing mappings or permissions.

Based on this analysis, the case where the TLB definitely needs to be flushed is:

Deleting a page from the page table.
Changing the read/write permission bit in the page table.
So the correct answers are options 2 and 3.


Question: Let $G_1, G_2$, be valid generator matrices of dimensions $\mathbb F^{k_i 	imes n_i}$, all over the same field $\mathbb F$. Recall that ``valid'' means that for all $i$, $k_i \leq n_i$ and $	ext{rank}(G_i) = k_i$. True or false:  Assuming $k_1 = k_2$ and $n_1=n_2$, the matrix $G_{7} + G_{8}$ is also a valid generator matrix.

The correct answer is **False**. \n\nTo arrive at this conclusion, I assessed the conditions for \( G_1 + G_2 \) being a valid generator matrix. Both matrices have the same dimensions \( n_1 = n_2 \) and ranks \( k_1 = k_2 \). However, the crucial point is that the rank of the sum \( G_1 + G_2 \) can be less than \( k_1 \) due to potential linear dependence between the rows of \( G_1 \) and \( G_2 \). \n\nFor example, if \( G_1 \) and \( G_2 \) are not independent (i.e., one is a scalar multiple of the other), then \( \text{rank}(G_1 + G_2) < k_1 \). Hence, while the dimensions are correct, the rank may not meet the requirements, confirming that \( G_1 + G_2 \) may not be a valid generator matrix. \n\nThis highlights the importance of checking rank conditions, not just dimensions, which is a common oversight for students.


Question: Tick the \textbf{\emph{incorrect}} assertion. A $\Sigma$-protocol \dots

To identify the incorrect assertion about a $\Sigma$-protocol, let's analyze each statement one by one:

1. **"has special soundness."**
   - A $\Sigma$-protocol indeed has the property of special soundness. This means that if a cheating prover can convince the verifier of the validity of a statement, then there exists a way to extract a witness (a secret) from the proof. Thus, this assertion is **correct**.

2. **"is zero-knowledge."**
   - A $\Sigma$-protocol is designed to be zero-knowledge. This means that the interaction does not reveal any information about the witness other than the fact that the statement is true. Therefore, this assertion is also **correct**.

3. **"is a 3-move interaction."**
   - A standard $\Sigma$-protocol consists of three moves: the prover sends a commitment, the verifier sends a challenge, and then the prover responds to the challenge. This definition holds true for $\Sigma$-protocols, so this assertion is **correct**.

4. **"has the verifier polynomially bounded."**
   - This statement is problematic. While $\Sigma$-protocols do have computational constraints on the prover, the verifier does not necessarily have to be polynomially bounded in every context. The requirement for the verifier's efficiency can vary based on the specific protocol design. Thus, this assertion can be considered **incorrect**.

After analyzing all assertions, the incorrect assertion is:

**"has the verifier polynomially bounded."**


Question: What does it mean that a processor implements precise exceptions?

In computer architecture, when we say that a processor implements precise exceptions, we are referring to its ability to handle exceptions (such as interrupts or errors) in a way that allows the system to maintain a consistent and manageable state. Precise exceptions ensure that when an exception occurs, the processor can accurately determine the exact point of execution where the exception was raised, allowing for a clear recovery process.

Here’s a breakdown of the concept:

1. **Definition of Exceptions**: Exceptions are events that disrupt the normal flow of execution in a program. They can be triggered by various conditions, including arithmetic errors (like division by zero), hardware interrupts (like I/O requests), or system calls.

2. **Precise vs. Imprecise Exceptions**:
   - **Precise Exceptions**: When a precise exception occurs, the processor ensures that all prior instructions have completed, and the program counter (PC) points to the instruction that caused the exception. This means that the state of the program is well-defined, making it easier to handle the exception and resume execution.
   - **Imprecise Exceptions**: In contrast, imprecise exceptions occur when the processor cannot guarantee that all previous instructions have been completed before the exception, making it difficult to ascertain the program's state or to recover from the error.

3. **Importance of Precise Exceptions**:
   - **Debugging**: Precise exceptions simplify debugging as developers can directly identify the exact instruction that caused the issue.
   - **Error Handling**: They enable more robust error handling mechanisms, where the system can revert to a known good state.
   - **Performance**: Modern processors often use pipelining and out-of-order execution to improve performance. Precise exceptions help ensure that these optimizations do not complicate exception handling.

In summary, a processor that implements precise exceptions provides a reliable framework for managing errors and interrupts, ensuring that software can operate correctly and efficiently even in the presence of unexpected conditions.


Question: Why is XML a document model?

To determine why XML (eXtensible Markup Language) is considered a document model, let's evaluate each option systematically.

### Step 1: Understanding XML as a Document Model

XML is a markup language designed for storing and transporting data, emphasizing the structure of data. It enables users to define custom tags and hierarchies, which facilitates the representation of complex data structures. This flexibility is a key reason why XML is classified as a document model.

### Step 2: Evaluating Each Option

1. **Option 1: It supports application-specific markup**
   - This statement is true. XML allows users to define their own tags and structures that are tailored to specific applications. This flexibility is integral to the concept of a document model, as it enables the representation of diverse data types and structures relevant to different applications.

2. **Option 2: It supports domain-specific schemas**
   - This statement is also true. XML can be validated against schemas (like DTD or XML Schema) that define the structure and rules of the data. This capability allows XML to be utilized in various domains (e.g., finance, healthcare) with specific requirements. While this is a feature of XML, it focuses more on validation rather than the fundamental nature of XML as a document model.

3. **Option 3: It has a serialized representation**
   - This statement is true and is significant in understanding XML as a document model. Serialization refers to the process of converting data structures into a format that is suitable for storage or transmission. In the context of XML, serialization allows complex data structures to be represented as documents that can be easily shared and processed. This characteristic is crucial to its classification as a document model.

4. **Option 4: It uses HTML tags**
   - This statement is false. XML does not use HTML tags; rather, it is a separate markup language. Although both XML and HTML are markup languages, XML is defined by user-defined tags, while HTML has predefined tags. This option misrepresents XML's characteristics.

### Step 3: Conclusion

While both application-specific markup (Option 1) and serialization (Option 3) are valid characteristics of XML, the strongest argument for why XML is considered a document model is its support for serialized representation. This feature allows XML to encapsulate complex data structures into documents that are easily transferable and interpretable across various systems. Thus, it highlights the essential nature of XML as a document model, facilitating data interchange in a structured format.

Answer: It has a serialized representation.


Question: Suppose that you possess a $D$-ary encoding $\Gamma$ for the source $S$ that does not satisfy Kraft's Inequality. Specifically, in this problem, we assume that our encoding satisfies $\sum_{i=1}^n D^{-l_i} = k+1 $ with $k>0$. What can you infer on the average code-word length $L(S,\Gamma)$?

To analyze the options regarding the average code-word length \( L(S,\Gamma) \) given that the encoding \( \Gamma \) does not satisfy Kraft's Inequality, we need to understand the implications of the inequality and the conditions given.

### Analyzing Each Option

1. **Option 1: \( L(S,\Gamma) \geq H_D(S) - \log_D(e^k) \)**
   - **Analysis**: This option suggests that the average length is bounded below by the entropy \( H_D(S) \) adjusted by some factor involving \( k \). Since Kraft's Inequality is not satisfied, the encoding can't guarantee that it is uniquely decodable. However, this option appears to imply a reasonable relationship between the average length and the entropy adjusted by the logarithm term.

2. **Option 2: \( L(S,\Gamma) \geq k H_D(S) \)**
   - **Analysis**: This option suggests that the average length is at least \( k \) times the entropy. This is a strong statement and may not hold true in general since the relationship between average length and entropy can vary widely based on the structure of the coding scheme.

3. **Option 3: \( L(S,\Gamma) \geq \frac{H_D(S)}{k} \)**
   - **Analysis**: This option states that the average length is at least the entropy divided by \( k \). This is also a weaker condition and does not necessarily capture the implications of violating Kraft's Inequality.

4. **Option 4: The code would not be uniquely-decodable and thus we can't infer anything on its expected length.**
   - **Analysis**: This option asserts that the lack of unique decodability means we cannot make any conclusions about average code length. While it's true that non-unique decodable codes can have unpredictable average lengths, we can still analyze their average lengths based on the distribution of code lengths and probabilities.

### Conclusion

Given the analysis, the most reasonable inference aligns with **Option 1**.

### Rationale for the Correct Option
- **Option 1** correctly reflects that while the encoding does not satisfy Kraft's Inequality, we can still derive a lower bound for the average length of the code. Since \( \sum_{i=1}^n D^{-l_i} = k + 1 \) suggests a certain inefficiency in encoding (as it exceeds 1), this would result in a longer average code-word length, which is captured by the relationship involving entropy and the logarithmic term.

### Reasons the Other Options are Incorrect or Less Suitable
- **Option 2**: While it provides a lower bound, it suggests a linear relationship with \( k \) that may not hold generally, especially with non-unique decodable codes.
- **Option 3**: This option provides a lower bound that appears too weak and does not reflect the implications of the condition \( \sum_{i=1}^n D^{-l_i} = k + 1 \) appropriately.
- **Option 4**: Although it raises a valid point about the challenges of non-unique decodability, it dismisses the possibility of inferring bounds on average length altogether, which is not correct. We can still analyze the average length based on code structure.

Thus, the correct option is **Option 1**: \( L(S,\Gamma) \geq H_D(S) - \log_D(e^k) \).


Question: Which of the following statements is correct in the context of  information extraction?

The correct statement in the context of information extraction is:

**1. A confidence measure that prunes too permissive patterns discovered with bootstrapping can help reduce semantic drift.**

**Reasoning:**
This statement is accurate because bootstrapping often generates patterns that can be overly general or too permissive, leading to the inclusion of irrelevant or incorrect information (semantic drift). By employing a confidence measure, one can assess the reliability of the patterns generated during the bootstrapping process and prune those that do not meet a certain confidence threshold. This helps in maintaining the quality and relevance of the extracted information.

**Analysis of Other Statements:**

2. **The bootstrapping technique requires a dataset where statements are labelled.**
   - This statement is not correct because bootstrapping can often start from a small set of seed examples rather than requiring a fully labeled dataset.

3. **Distant supervision typically uses low-complexity features only, due to the lack of training data.**
   - This statement is misleading. Distant supervision can utilize a variety of features, including complex features, but it is true that the lack of labeled data can sometimes lead to a reliance on simpler features for initial models.

4. **For supervised learning, sentences in which NER has detected no entities are used as negative samples.**
   - This statement is not always correct. While it is a common practice to use sentences without detected entities as negative samples, it doesn't universally apply to all supervised learning scenarios, especially when considering the potential for false negatives in Named Entity Recognition (NER) outputs.

Thus, statement 1 stands out as the most accurate in the context of information extraction.


Question:
Consider a movie recommendation system which minimizes the following objective
        rac{1}{2} \sum_{(d,n)\in\Omega} [x_{dn} - (\mathbf{W} \mathbf{Z}^	op)_{dn}]^2 + rac{\lambda_w}{2}
orm{\mathbf{W}}_	ext{Frob}^2 + rac{\lambda_z}{2}
orm{\mathbf{Z}}_	ext{Frob}^2
        where $\mathbf{W}\in \R^{D	imes K}$ and $\mathbf{Z}\in \R^{N	imes K}$.
	    Suppose movies are divided into genre A and genre B (i.e., $\mathbf{W}_A\in \R^{D_A	imes K}, \mathbf{W}_B\in \R^{D_B	imes K}, \mathbf{W}=[\mathbf{W}_A; \mathbf{W}_B]$, with $D_A\!+\!D_B=D$) and users are divided into group 1 and group 2  (i.e., $\mathbf{Z}_1\in \R^{N_1	imes K}, \mathbf{Z}_2\in \R^{N_2	imes K}, \mathbf{Z}=[\mathbf{Z}_1; \mathbf{Z}_2]$, with $N_1\!+\!N_2=N$). In addition, group 1 users only rate genre A movies while group 2 users only rate genre B movies. Then instead of training a large recommendation system with $(\mathbf{W}, \mathbf{Z})$, one may train two smaller recommendation systems with parameters $(\mathbf{W_A}, \mathbf{Z_1})$ and $(\mathbf{W_B}, \mathbf{Z_2})$ separately. If SGD is used to solve the minimization problems and all conditions remain the same (e.g., hyperparameters, sampling order, initialization, etc), then which of the following statements is true about the two training methods?

When evaluating the two training methods for the movie recommendation system, we need to consider how the grouping of movies and users affects the learned feature vectors.

### Analysis of Training Methods

1. **Training with the Combined System**:
   - In the combined system, the parameters \((\mathbf{W}, \mathbf{Z})\) are trained simultaneously, allowing the model to leverage ratings from both user groups and genres. This joint training can learn complex interactions and shared latent factors between different genres and user preferences, potentially leading to richer and more generalized feature representations.

2. **Training with Separate Systems**:
   - In contrast, training the separate systems with parameters \((\mathbf{W_A}, \mathbf{Z_1})\) and \((\mathbf{W_B}, \mathbf{Z_2})\) means that each system only learns from its respective user group and genre. This limitation can hinder the capturing of underlying patterns that might be beneficial for users who exhibit mixed preferences or for genres that share characteristics.

### Key Considerations

- **Sparsity of Ratings**: If the rating matrix is sparse, the separate systems may struggle to learn robust feature vectors due to the limited amount of data available for each genre and user group. This sparsity could lead to even greater differences in the learned feature vectors compared to those from the combined system.

- **Distribution of Ratings**: If the distribution of ratings is uneven?where one genre or user group has significantly more ratings than the other? the feature vectors learned from the more populated group may be more accurate and informative. This can further contribute to differences between the two training methods.

- **Potential for Similarity**: While the separate systems are likely to produce different feature vectors due to their isolated training environments, there may be scenarios in which the learned representations are similar. This could occur if the underlying patterns in the data are closely aligned between the groups and genres, particularly if user preferences are not strictly confined.

### Conclusion

In summary, while the training methods will likely lead to different feature vectors due to the distinct learning environments created by user and genre separations, the actual differences may depend on the sparsity of the rating matrix and the distribution of ratings. The separate models might fail to capture beneficial interactions and commonalities that a combined model could exploit, making their representations less effective and more specialized. However, there remains a possibility that the feature vectors could be similar if the underlying patterns across groups and genres exhibit significant overlap.

Given the options provided, the most accurate choice is: **Feature vectors obtained in both cases can be either same or different, depending on the sparsity of the rating matrix.**

This answer acknowledges the complexity of the problem while highlighting key factors that influence the training outcomes in a recommendation system context.

---

This revised answer provides a clearer, more comprehensive analysis of the training methods, addressing the nuances of sparsity and distribution while refining the conclusion to align with the options in the original question.


Question: Which of the following arguments is correct?

Step 1: Analyze Option A, which states that all students in this class understand math, and concludes that Alice, a student in the class, does not understand math. This reasoning is incorrect because if all students in the class understand math, then it logically follows that Alice must understand math too.

Step 2: Evaluate Option B, which posits that every physics major takes calculus and concludes that Mathilde, who is taking calculus, must be a physics major. This argument is flawed because taking calculus does not imply that one is a physics major; other non-physics majors can also take calculus.

Step 3: Examine Option C, which claims all cats like milk and concludes that since the speaker's pet is not a cat, it must not like milk. This reasoning commits the fallacy of denying the antecedent; just because all cats like milk does not mean that only cats can like milk.

Step 4: Finally, assess Option D, which states that everyone who eats vegetables every day is healthy and concludes that Linda is not healthy; therefore, she does not eat vegetables every day. This argument correctly follows logical reasoning; if eating vegetables every day guarantees healthiness, then not being healthy indicates she cannot meet this condition.

Conclusion: After careful analysis, the only valid argument here is Option D as it correctly uses logical reasoning without committing any fallacies. The answer is D.


Question: Let $S_1$ be a random variable taking values in $\{a,b\}$ with probability $p_{S_1}(a) = rac{1}{4}$ and $p_{S_1}(b) = rac{3}{4}$. Let $S_2$ be a random variable, independent of $S_1$, taking values in $\{c,d\}$ with probability $p_{S_2}(c) = q$ and $p_{S_2}(d) = 1-q$, for some $q\in[0,1]$. Let $\Gamma_H$ be the binary Huffman code for the sequence $S = S_1 S_2$, and let $L(S,\Gamma_H)$ be the average codeword-length of $\Gamma_H$. True or false: $1\leq L(S, \Gamma_H) \leq 2$ for all $q\in[0,1]$.

To determine whether the statement "1 ≤ L(S, Γ_H) ≤ 2 for all q ∈ [0, 1]" is true or false, we need to analyze the average codeword length \( L(S, \Gamma_H) \) of the binary Huffman code for the combined random variable \( S = S_1 S_2 \).

1. **Understanding the Random Variables**:
   - \( S_1 \) takes values in \( \{a, b\} \) with probabilities \( p_{S_1}(a) = \frac{1}{4} \) and \( p_{S_1}(b) = \frac{3}{4} \).
   - \( S_2 \) takes values in \( \{c, d\} \) with probabilities \( p_{S_2}(c) = q \) and \( p_{S_2}(d) = 1 - q \), where \( q \) is between 0 and 1.

2. **Calculating the Joint Distribution**: Since \( S_1 \) and \( S_2 \) are independent, the joint probabilities for \( S \) can be calculated as follows:
   - \( p(a, c) = p_{S_1}(a) \cdot p_{S_2}(c) = \frac{1}{4}q \)
   - \( p(a, d) = p_{S_1}(a) \cdot p_{S_2}(d) = \frac{1}{4}(1 - q) \)
   - \( p(b, c) = p_{S_1}(b) \cdot p_{S_2}(c) = \frac{3}{4}q \)
   - \( p(b, d) = p_{S_1}(b) \cdot p_{S_2}(d) = \frac{3}{4}(1 - q) \)

3. **Total Probability**: The total probability for the combined random variable \( S \) is:
   \[
   p(S) = p(a, c) + p(a, d) + p(b, c) + p(b, d) = \frac{1}{4}q + \frac{1}{4}(1 - q) + \frac{3}{4}q + \frac{3}{4}(1 - q) = 1
   \]

4. **Entropy Calculation**: The entropy \( H(S) \) can be computed using the joint probabilities. The average codeword length \( L(S, Γ_H) \) for an optimal code is at least the entropy, so:
   \[
   H(S) = -\sum p(x) \log_2 p(x)
   \]
   This will yield a value that depends on \( q \) but will be bounded.

5. **Bounds on Average Length**: The average codeword length \( L(S, Γ_H) \) is closely related to \( H(S) \). Since the probabilities are constrained (they sum to 1), \( H(S) \) will be bounded between 1 and 2 bits:
   - The minimum entropy (and thus the minimum average length) occurs when one outcome is certain (probability 1), and this gives a length of 1.
   - The maximum entropy occurs when the probabilities are more evenly distributed, resulting in a maximum length of 2.

Given this analysis, we conclude that the statement is indeed true:

**Answer**: True.


Question: We learnt in the lecture that terms are typically stored in an inverted list. Now, in the inverted list, instead of only storing document identifiers of the documents in which the term appears, assume we also store an *offset* of the appearance of a term in a document. An $offset$ of a term $l_k$ given a document is defined as the number of words between the start of the document and $l_k$. Thus our inverted list is now:  $l_k= \langle f_k: \{d_{i_1} \rightarrow [o_1,\ldots,o_{n_{i_1}}]\},  \{d_{i_2} \rightarrow [o_1,\ldots,o_{n_{i_2}}]\}, \ldots,  \{d_{i_k} \rightarrow [o_1,\ldots,o_{n_{i_k}}]\} \rangle$  This means that in document $d_{i_1}$ term $l_k$ appears $n_{i_1}$ times and at offset $[o_1,\ldots,o_{n_{i_1}}]$, where $[o_1,\ldots,o_{n_{i_1}}]$ are sorted in ascending order, these type of indices are also known as term-offset indices. An example of a term-offset index is as follows:  **Obama** = $⟨4 : {1 → [3]},{2 → [6]},{3 → [2,17]},{4 → [1]}⟩$  **Governor** = $⟨2 : {4 → [3]}, {7 → [14]}⟩$  **Election** = $⟨4 : {1 → [1]},{2 → [1,21]},{3 → [3]},{5 → [16,22,51]}⟩$  Which is to say that the term **Governor** appear in 2 documents. In document 4 at offset 3, in document 7 at offset 14. Now let us consider the *SLOP/x* operator in text retrieval. This operator has the syntax: *QueryTerm1 SLOP/x QueryTerm2* finds occurrences of *QueryTerm1* within $x$ (but not necessarily in that order) words of *QueryTerm2*, where $x$ is a positive integer argument ($x \geq 1$). Thus $x = 1$ demands that *QueryTerm1* be adjacent to *QueryTerm2*.  Consider the general procedure for "merging" two term-offset inverted lists for a given document, to determine where the document satisfies a *SLOP/x* clause (since in general there will be many offsets at which each term occurs in a document). Let $L$ denote the total number of occurrences of the two terms in the document. Assume we have a pointer to the list of occurrences of each term and can move the pointer along this list. As we do so we check whether we have a hit for $SLOP/x$ (i.e. the $SLOP/x$ clause is satisfied). Each move of either pointer counts as a step. Based on this assumption is there a general "merging" procedure to determine whether the document satisfies a $SLOP/x$ clause, for which the following is true? Justify your answer.      1. The merge can be accomplished in a number of steps linear in $L$ regardless of $x$, and we can ensure that each pointer moves only to the right (i.e. forward).     2. The merge can be accomplished in a number of steps linear in $L$, but a pointer may be forced to move to the left (i.e. backwards).     3. The merge can require $x \times L$ steps in some cases.

**Answer:** The correct choice is (1) because the merging process for the two term-offset inverted lists can be designed to operate in linear time relative to the total occurrences \( L \) of both terms. Here's the justification:

1. **Initialization**: Start with two pointers, one for each term's offset list. Both pointers begin at the start of their respective lists.

2. **Iteration**: Use a loop to compare the offsets pointed to by the two pointers:
   - If the offset of QueryTerm1 at pointer 1 is less than the offset of QueryTerm2 at pointer 2, move pointer 1 to the right (i.e., to the next occurrence of QueryTerm1).
   - If the offset of Query


Question: Consider an arbitrary cipher $C$ and a uniformly distributed random permutation $C^*$ on $\{0,1\}^n$. Tick the \textbf{false} assertion.

To analyze the assertions given in the context of an arbitrary cipher \( C \) and a uniformly distributed random permutation \( C^* \) on \( \{0,1\}^n \), we need to understand what is meant by \( \mathsf{Dec}^1(C) = 0 \) and its implications.

Let's examine each assertion in turn:

1. **Assertion 1**: \( \mathsf{Dec}^1(C)=0 \) implies \( C=C^* \).
   - This assertion is false. Even if the first-order decorrelation measure is zero, it does not necessarily mean that the cipher must be identical to a random permutation; rather, it indicates low correlation with any fixed distribution.

2. **Assertion 2**: \( \mathsf{Dec}^1(C)=0 \) implies \( [C]^1=[C^*]^1 \).
   - This assertion can also be false depending on how you define these quantities. Zero decorrelation means that the statistical properties might align but does not guarantee exact equality of their distributions or expected values.

3. **Assertion 3**: \( \mathsf{Dec}^1(C)=0 \) implies that \( C \) is perfectly decorrelated at order 1.
   - This assertion appears true because zero first-order decorrelation indeed suggests no detectable patterns of correlation at this level.

4. **Assertion 4**: \( D[C]^1=0\implies\text{all coefficients in }[C]^1=\frac{1}{2^{n}}.\)
   - This assertion can also be misunderstood; although zero first-order decorrelation suggests uniformity in some sense, it doesn't strictly mean all coefficients will equal exactly half for every possible outcome unless explicitly proven or derived under certain conditions.

Considering these evaluations, Assertion 2 seems most susceptible to being considered false since it relies heavily on interpretations of what equality between different forms denotes without sufficient grounding provided by just knowing that one measure equals zero.

Thus, I would conclude:

The false assertion from the options provided is:
- **Option 2**: "If $\mathsf{Dec}^1(C)=0$, then $[C]^1=[C^*]^1$."


Question: Tick the \textbf{false} statement.

Imagine you're at a party, and each statement is like a guest trying to impress you with their knowledge of elliptic curves. Most of them are dressed to impress and share interesting facts, but one of them has a little mix-up that makes you raise an eyebrow.

1. The identity element being the point at infinity is a classic property in elliptic curves, serving as a kind of "zero" in the addition game.
2. Now, when we talk about singular points on elliptic curves, there's a common belief that you can still draw some sort of tangent there; it's just not very straightforward. Some might say it's like trying to find a smooth path in a bumpy road—it's tricky but not impossible.
3. P and Q being inverses is true; they reflect across the x-axis, which is how addition works on these curves.
4. And let's not forget that elliptic curve cryptography is quite popular for its efficiency in public-key systems, often praised for its speed and low resource needs.

So, the statement that feels off is the second one, as there's some ambiguity about drawing tangents at singular points; it's not entirely clear-cut.

2


Question: For the number of times the apriori algorithm and the FPgrowth algorithm for association rule mining are scanning the transaction database the following is true

1: fpgrowth has always strictly fewer scans than apriori because the FP-growth algorithm constructs a compact data structure (the FP-tree) that enables it to mine frequent itemsets without needing to repeatedly scan the entire database, unlike the Apriori algorithm which typically requires multiple scans to generate candidate itemsets.


Question: A simple substitution cipher can be broken \dots

by analysing the probability occurrence of the language


Question:
You are given a probability distribution $P(y_t | y_0, \ldots, y_{t-1})$ over 100 possible next tokens to generate by your model. The distribution has the following characteristics:
     egin{itemize}
     \item 20\% of the probability mass is on the most probable token;
     \item 10\% of the probability mass is on each of the next 4~most probable tokens;
     \item 1\% of the probability mass is on each of the next 20~most probable tokens;
     \item the remaining mass is uniformly distributed across the remaining 75 tokens.
     \end{itemize}
In top-p sampling, if $p = 0.75$, how many tokens will be included in the set of tokens you sample from?
Fully justify your answer.

To determine how many tokens will be included in the set of tokens for top-p sampling with \( p = 0.75 \), we need to accumulate the probabilities from the most probable tokens until we reach or exceed 75% of the total probability mass.

1. **Most Probable Token:**
   - Probability: 20%

2. **Next 4 Most Probable Tokens:**
   - Each has a probability of 10%.
   - Total for these 4 tokens: \( 4 \times 10\% = 40\% \)

3. **Next 20 Most Probable Tokens:**
   - Each has a probability of 1%.
   - Total for these 20 tokens: \( 20


Question: Implement the function `check_words` that checks if the words of a strings have common words with a list. Write your code in python. Your code should be agnostic to lower/upper case.

To implement the function `check_words`, we need to follow a structured approach:

### Step 1: Understand the Problem
We need to create a function that:
- Takes a string of words as input.
- Checks whether any of the words in the string are present in a given list of common words.
- The check should be case-insensitive.

### Step 2: Define the Input and Output
- **Input**:
  - A string containing words (e.g., "Hello world").
  - A list of common words (e.g., ["hello", "goodbye"]).

- **Output**:
  - A boolean value indicating whether any of the words in the string are in the list of common words.

### Step 3: Plan the Implementation
1. Normalize the case of both the input string and the list of common words to ensure case insensitivity.
2. Split the input string into individual words.
3. Check if any of the words from the string are present in the normalized list of common words.
4. Return `True` if there is at least one match; otherwise, return `False`.

### Step 4: Write the Code
Now let's implement the plan in Python:

```python
def check_words(input_string, common_words):
    # Normalize the case by converting everything to lower case
    normalized_string = input_string.lower()
    normalized_common_words = [word.lower() for word in common_words]

    # Split the input string into words
    words_in_string = normalized_string.split()

    # Check for any common words
    for word in words_in_string:
        if word in normalized_common_words:
            return True  # Found a common word

    return False  # No common words found

# Example usage:
input_string = "Hello world"
common_words = ["hello", "goodbye"]
print(check_words(input_string, common_words))  # Output: True
```

### Step 5: Explain the Code
- **Normalization**: We convert both the input string and the list of common words to lower case to ensure that our comparisons are case insensitive.
- **Splitting**: We use the `split()` method on the string to break it into individual words based on whitespace.
- **Checking for Commonality**: We iterate through each word in the list of words derived from the input string and check if it exists in the normalized list of common words.
- **Return Value**: If a match is found, we return `True`. If no matches are found after checking all words, we return `False`.

### Final Answer
The function `check_words` effectively checks for common words between an input string and a list of common words in a case-insensitive manner. The implementation is efficient and straightforward, allowing for easy verification and potential modifications.


Question: Tick the \emph{correct} assertion. Linear cryptanalysis \ldots

**A. was invented long before the Caesar cipher.**
This assertion is incorrect because linear cryptanalysis was developed in the 1990s, while the Caesar cipher dates back to Julius Caesar, making it much older. The timeline of cryptographic techniques shows that modern methods like linear cryptanalysis are far more recent compared to classical ciphers.

**B. is a chosen plaintext key recovery attack.**
This statement is partially misleading; linear cryptanalysis is primarily a known plaintext attack rather than a chosen plaintext attack. While both types of attacks aim to recover keys, linear cryptanalysis operates by analyzing the relationship between plaintexts and their corresponding ciphertexts that are already known, rather than selecting specific plaintexts to encrypt.

**C. requires $\frac{1}{DP}$ pairs of plaintext-ciphertext.**
This option is correct; in linear cryptanalysis, the number of plaintext-ciphertext pairs required for a successful attack is approximately $\frac{1}{DP}$, where $D$ is the number of linear approximations that hold with probability $P$. This relationship highlights the statistical nature of the attack and the need for sufficient data to exploit linear relationships.

**D. breaks DES with $2^{43}$ known plaintexts.**
This assertion is incorrect as it underestimates the complexity of breaking DES. The original findings of linear cryptanalysis suggested that approximately $2^{47}$ known plaintexts are needed to break DES, indicating that $2^{43}$ is not sufficient for a successful attack.


Question: The elements of $\mathbf{Z}_{14}^*$ are

To determine the elements of \(\mathbf{Z}_{14}^*\), we need to find the integers in the set \(\{0, 1, 2, \ldots, 13\}\) that are coprime to \(14\). An integer \(a\) is coprime to \(14\) if the greatest common divisor \(\gcd(a, 14) = 1\).

First, we note that \(14 = 2 \times 7\). Therefore, any integer that shares a prime factor with \(14\) (which are \(2\) and \(7\)) will not be coprime to \(14\).

Now let's check each integer in \(\{0, 1, 2, \ldots, 13\}\):

- \(0\): \(\gcd(0, 14) = 14\) (not coprime)
- \(1\): \(\gcd(1, 14) = 1\) (coprime)
- \(2\): \(\gcd(2, 14) = 2\) (not coprime)
- \(3\): \(\gcd(3, 14) = 1\) (coprime)
- \(4\): \(\gcd(4, 14) = 2\) (not coprime)
- \(5\): \(\gcd(5, 14) = 1\) (coprime)
- \(6\): \(\gcd(6, 14) = 2\) (not coprime)
- \(7\): \(\gcd(7, 14) = 7\) (not coprime)
- \(8\): \(\gcd(8, 14) = 2\) (not coprime)
- \(9\): \(\gcd(9, 14) = 1\) (coprime)
- \(10\): \(\gcd(10, 14) = 2\) (not coprime)
- \(11\): \(\gcd(11, 14) = 1\) (coprime)
- \(12\): \(\gcd(12, 14) = 2\) (not coprime)
- \(13\): \(\gcd(13, 14) = 1\) (coprime)

Now, collecting the coprime integers, we have the elements of \(\mathbf{Z}_{14}^*\):

\[
\{1, 3, 5, 9, 11, 13\}
\]

Thus, the correct option is:

**Option 3: \(\{1, 3, 5, 9, 11, 13\}\)**.


Question: What is the default block size for a traditional file system, e.g. ext3/4?

Answer: 4096 bytes
 Reasoning: The default block size for traditional file systems like ext3 and ext4 is typically 4096 bytes (4 KB), which is a common size used to optimize storage efficiency and access speed.


Question: A word embedding for given corpus

Word embeddings are a crucial component of natural language processing, as they allow for the representation of words in a continuous vector space. When creating word embeddings for a given corpus, several factors come into play that can significantly influence the quality and characteristics of the generated embeddings. Letâs break down the various dependencies you mentioned:

1. **Dimension (d)**:
   - The dimension of the embeddings is one of the most critical parameters. It determines the size of the vector space in which words will be represented.
   - A higher dimension (e.g., 300-dimensional vectors) allows for more expressive representations, potentially capturing more nuances of word meaning and relationships. However, it also increases the risk of overfitting, especially if the corpus is small.
   - Lower dimensions (e.g., 50-dimensional vectors) may lead to simpler representations that could miss some of the subtle meanings but may generalize better in some cases.

2. **Number of Iterations in Gradient Descent**:
   - The training process for word embeddings typically involves optimizing a loss function using gradient descent. The number of iterations (or epochs) can influence the convergence of the algorithm.
   - More iterations may help the model learn more accurate embeddings, but too many can lead to overfitting, especially if the model begins to fit noise in the training data rather than the underlying patterns.

3. **Chosen Negative Samples**:
   - In models like Word2Vec (specifically, the skip-gram model with negative sampling), the way negative samples are chosen can significantly impact the quality of the embeddings.
   - Negative sampling helps the model learn which words are not related to the target word, thus enhancing the quality of the learned representations. The number of negative samples and their selection (random vs. frequency-based) can affect the efficiency and effectiveness of training.

4. **Further Factors**:
   - **Corpus Size and Quality**: The amount and quality of text data available for training influence the embeddings. A larger, more diverse corpus often leads to better embeddings.
   - **Preprocessing Techniques**: How the text is preprocessed (e.g., tokenization, stemming, lemmatization, removal of stop words) can affect the final embeddings. Proper preprocessing helps in capturing the relevant features of the language.
   - **Model Architecture**: Different architectures (e.g., skip-gram vs. continuous bag of words (CBOW) in Word2Vec) can lead to different types of embeddings, emphasizing different kinds of relationships between words.
   - **Regularization Techniques**: Techniques like dropout or L2 regularization can also impact the learning process, preventing overfitting and leading to more generalized embeddings.
   - **Learning Rate**: The choice of learning rate affects how quickly the model converges to a solution. A learning rate that is too high can cause the model to converge too quickly to a suboptimal solution, while a learning rate that is too low may result in slow convergence.

In summary, while the dimension of the embeddings, the number of iterations in gradient descent, and the choice of negative samples are critical factors, the overall quality and effectiveness of word embeddings also depend on various other elements, including the corpus size and quality, preprocessing methods, model architecture, regularization, and learning rate. Each of these factors should be carefully considered when training word embeddings to achieve optimal results.


Question: Consider a linear regression problem with $N$ samples where the input is in $D$-dimensional space, and all output values are $y_{i} \in\{-1,+1\}$. Which of the following statements is correct?

To analyze each option, let's break down the problem and the statements provided.

### Background
In a linear regression problem, we aim to fit a linear model to our data. The model predicts continuous output values based on linear combinations of the input features. However, in this case, the output values are binary, taking values from \{-1, +1\}. This scenario is more aligned with classification problems, particularly binary classification, rather than traditional regression.

### Option Analysis

**Option (a): "linear regression cannot 'work' if \( N \gg D \)"**

- If \( N \gg D \), we have significantly more samples than dimensions. In general, linear regression could work well in this scenario since having more samples than dimensions typically helps with fitting the model and reduces the risk of overfitting.
- Therefore, this statement is **incorrect**.

**Option (b): "linear regression cannot 'work' if \( N \ll D \)"**

- If \( N \ll D \), we have fewer samples than dimensions. This situation leads to the curse of dimensionality, where the model might not be able to learn effectively due to an insufficient number of data points to adequately represent the high dimensional space.
- This means linear regression can struggle to generalize and may not be effective, as there are infinitely many hyperplanes that could fit the data poorly. Thus, this statement is **correct**.

**Option (c): "linear regression can be made to work perfectly if the data is linearly separable"**

- If the data is linearly separable, it means that there exists a hyperplane that can perfectly separate the classes. In the context of linear regression, if we train a model on linearly separable data, it can perfectly classify the data points by finding a hyperplane that divides the classes.
- However, linear regression is typically not the best approach for classification tasks. While it may achieve perfect classification in certain cases, it can yield predictions that are not confined to the binary outputs {-1, +1}. In practice, logistic regression or other classification methods are preferred. Yet, conceptually, if we restrict the output to {-1, +1}, linear regression can indeed provide a perfect fit in such scenarios.
- Therefore, this statement is **correct**.

### Final Conclusion
After analyzing all the options:

- **(a)** is incorrect.
- **(b)** is correct.
- **(c)** is correct.

Thus, the correct options are **(b)** and **(c)**.


Question:
How would a data prefetcher influence the results of a
  \emph{prime + probe} attack?

To understand how a data prefetcher might influence the results of a prime + probe attack, we first need to clarify what a prime + probe attack is. This type of side-channel attack exploits the timing differences in memory access to determine the contents of a cache. The attacker "primes" the cache by filling it with their own data, and then "probes" it to see how long it takes to access certain locations, revealing information about whether those locations were accessed by the target process.

Now, let's consider the influence of a data prefetcher on this attack:

1. **Cache Behavior**: A data prefetcher anticipates which data will be needed next and loads it into the cache ahead of time. In the context of a prime + probe attack, if a prefetcher successfully loads data into the cache that the victim process will access, it can mask the access patterns the attacker is trying to exploit. This may lead to fewer cache misses for the victim, making it harder for the attacker to determine which memory locations were accessed based on timing alone.

2. **Timing Variability**: The presence of a data prefetcher can introduce variability in the timing of memory accesses. If prefetching results in data being loaded into the cache before the victim process accesses it, this can lead to a situation where the probe times become less predictable. The attacker may find it more challenging to interpret the timing data accurately, as the prefetcher may cause cache hits when the attacker expects misses.

3. **Interference with Prime Phase**: During the prime phase, the attacker fills the cache with their own data. If the prefetcher aggressively fetches data that is not being used by the attacker but is relevant to the victim, it can displace the attacker's data. This means that when the attacker probes the cache, they may encounter cache misses where they expect cache hits, complicating their analysis and potentially leading to false conclusions about the victim's memory access.

4. **Impact on Cache Eviction**: Prefetchers may also influence which data remains in the cache during the attack. If the prefetcher tends to keep certain lines in cache longer, it could prevent the victim's data from being evicted, affecting the attacker's ability to discern which data was accessed. This could lead to a lower success rate for the prime + probe attack.

In summary, a data prefetcher can significantly complicate a prime + probe attack by altering the expected cache behavior, increasing timing variability, interfering during the prime phase, and impacting cache eviction dynamics. These factors can hinder the attacker's ability to gather reliable timing information and accurately infer the victim's memory access patterns.


Question: You are using a 3-layer fully-connected neural, and you are using \textbf{$f(x) = 2x$ as your activation function} . Your input data has components in [0, 1]. \textbf{You initialize your weights using Kaiming (He) initialization}, and set all the bias terms to 0. You start optimizing using SGD. What will likely happen?

The correct option is: **Training is fine, but our neural net does only as well as a linear model.**

**Rationale:**

In this scenario, the activation function \( f(x) = 2x \) is a linear function. When a neural network uses linear activation functions, regardless of the number of layers, the entire network can be simplified to a single linear transformation. This means that no matter how deep the network is, it will not be able to learn complex, nonlinear relationships in the data.

Even though Kaiming (He) initialization is typically beneficial for networks with nonlinear activation functions, in this case, it does not enhance the network's capacity because the linear nature of the activation function limits the model's expressiveness. Therefore, the training process will proceed without issues, but the model will not outperform a simple linear regression, as it essentially behaves as a linear model.

Hence, while the training can progress successfully, the overall performance of the neural network will be equivalent to that of a linear model.


Question: Matrix Factorizations: If we compare SGD vs ALS for optimizing a matrix factorization of a $D \times N$ matrix, for large $D, N$

To analyze the statements regarding the computational costs of Stochastic Gradient Descent (SGD) and Alternating Least Squares (ALS) for matrix factorization of a \( D \times N \) matrix, let's break down each statement:

1. **(a) Per iteration, SGD has a similar computational cost as ALS**:
   - **False**. SGD generally has a lower computational cost per iteration compared to ALS because it updates the factors using only a small subset of the data (a mini-batch or a single data point), while ALS computes updates based on the entire dataset or a larger portion of it for each iteration. Thus, ALS typically has a higher computational cost per iteration.

2. **(b) Per iteration, ALS has an increased computational cost over SGD**:
   - **True**. As mentioned above, ALS computes updates for the entire dataset in each iteration, which can be computationally expensive, especially for large matrices. On the other hand, SGD focuses on smaller batches or individual samples, making it generally less costly per iteration.

3. **(c) Per iteration, SGD cost is independent of \( D, N \)**:
   - **True** (with some caveats). The computational cost of SGD per iteration is primarily determined by the size of the mini-batch used and is not directly proportional to \( D \) or \( N \). However, the total cost across all iterations does depend on the dimensions of the matrix as it influences the number of iterations required for convergence. Still, per iteration, SGD can be considered to have a cost that does not grow with \( D \) and \( N \) in a straightforward manner.

In summary:
- **(a)** is **False**.
- **(b)** is **True**.
- **(c)** is **True** (considering the context).


Question: The collision resistance property of a hash function $H$ means that it is infeasible to\dots

To evaluate the options, we need to understand what collision resistance means in the context of a cryptographic hash function.

**Collision Resistance**: A hash function \( H \) is said to be collision-resistant if it is infeasible to find two distinct inputs that hash to the same output. In other words, given the hash function \( H \), it should be computationally hard to find any two different inputs \( X \) and \( X' \) such that \( H(X) = H(X') \) and \( X \neq X' \).

Now, let's analyze each option:

1. **Option 1**: find \( Y \) such that \( H(X) = Y \) for a given \( X \).
   - This describes the property of pre-image resistance, not collision resistance. Pre-image resistance means you cannot easily find a hash output for a specific input, but it does not directly relate to finding collisions.

2. **Option 2**: find \( X \) such that \( H(X) = Y \) for a given \( Y \).
   - This also describes pre-image resistance. It suggests that it should be hard to reverse the hash function to find the original input from its hash output. This is not related to collision resistance.

3. **Option 3**: find \( X' \) such that \( H(X') = H(X) \) and \( X \ne X' \) for a given \( X \).
   - This option correctly describes a scenario of finding a collision for a specific input \( X \). It indicates the difficulty in finding a different input \( X' \) that results in the same hash value as \( H(X) \). Thus, this is a correct aspect of collision resistance, but it is focused on a single input \( X \).

4. **Option 4**: find \( X, X' \) such that \( H(X) = H(X') \) and \( X \ne X' \).
   - This option describes the general case of collision resistance, where the goal is to find *any* two different inputs \( X \) and \( X' \) that hash to the same value. This captures the essence of what collision resistance means: the infeasibility of finding any two distinct inputs that produce the same hash output.

Based on the definitions and evaluations:

**The best choice is Option 4**: find \( X, X' \) such that \( H(X) = H(X') \) and \( X \ne X' \). This option encompasses the full concept of collision resistance, indicating that it should be infeasible to find any two distinct inputs with the same hash output.


Question: Consider a learning algorithm that has the property that it depends only very weakly on the input data. E.g., this could be SGD where we choose a very small step size and only run for very few iterations. To go to the extreme, you can imagine a learning algorithm that always outputs the same model irrespective of the training set. Presumably such a learning algorithm will not give us good results. Why is that?

To analyze the situation described, let’s break down the options in relation to a learning algorithm that depends very weakly on the input data.

1. **(a) Such a learning algorithm typically has a much larger generalization error than training error.**
   - This is not necessarily true. If the learning algorithm outputs the same model regardless of the training data, it might have a low training error if that model fits the training data well, but it is likely to perform poorly on unseen data, leading to high generalization error. However, it doesn't directly imply that generalization error is always much larger than training error.

2. **(b) Such a learning algorithm typically has a large bias.**
   - This statement is correct. A learning algorithm that does not adapt or learn from the input data will likely have a high bias because it fails to account for the variability in the data. It assumes a simplistic model that does not capture the complexities or patterns present in the actual data, leading to systematic errors in predictions.

3. **(c) Such a learning algorithm is prone to overfitting.**
   - This statement is incorrect. Overfitting occurs when a model learns the noise in the training data rather than the underlying distribution, which typically happens when the model is too complex relative to the amount of training data. A learning algorithm that depends weakly on the input data is more likely to underfit rather than overfit.

### Conclusion:

Given this analysis, the best option that accurately describes the consequence of a learning algorithm that depends very weakly on the input data is:

\[
\boxed{(b) \text{ Such a learning algorithm typically has a large bias.}}
\]


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 $n$ 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 work of contains if it is called on a List?

To analyze the asymptotic work of the `contains` function when called on a `List`, we need to consider each part of the function and how it behaves with a `List` as the input.

### Step 1: Understanding the Function

The `contains` function works as follows:

1. It first checks the size of the iterable `l`.
2. If the size \( n \) is less than or equal to 5, it iterates through each element of `l` to check if any element equals `elem`. This is done using a simple loop.
3. If \( n > 5 \), it splits `l` into two halves:
   - The first half is obtained using `take(n / 2)`.
   - The second half is obtained using `drop(n / 2)`.
4. It then calls `contains` recursively on each half in parallel and combines the results using a logical OR operation.

### Step 2: Analyzing the Work for Lists

When `contains` is called on a `List`, the behavior of `take` and `drop` is crucial:

- **For `List`:**
  - `take(n / 2)` runs in \( \Theta(n / 2) = \Theta(n) \).
  - `drop(n / 2)` runs in \( \Theta(n / 2) = \Theta(n) \).

### Step 3: Work Calculation

Let \( T(n) \) be the total work done by the `contains` function when called on a `List` of size \( n \).

1. **Base Case (n ≤ 5)**:
   - The function iterates through the list, taking \( \Theta(n) \) time (specifically, \( \Theta(1) \) for each element, which results in \( \Theta(n) \) overall).

2. **Recursive Case (n > 5)**:
   - The function performs the following work:
     - \( \Theta(n) \) for `take(n / 2)`
     - \( \Theta(n) \) for `drop(n / 2)`
     - Two recursive calls to `contains` on halves of the list:
       - The size of each half is \( n / 2 \), so we have:
         \[
         T(n / 2) \text{ for the first half, and } T(n / 2) \text{ for the second half.}
         \]

Combining these, we can express \( T(n) \) as:
\[
T(n) = \Theta(n) + T(n / 2) + T(n / 2)
\]
This simplifies to:
\[
T(n) = \Theta(n) + 2T(n / 2)
\]

### Step 4: Solving the Recurrence Relation

We can now solve the recurrence relation \( T(n) = \Theta(n) + 2T(n / 2) \) using the Master Theorem:

- Compare with the form \( T(n) = aT(n/b) + f(n) \):
  - Here, \( a = 2 \), \( b = 2 \), and \( f(n) = \Theta(n) \).

According to the Master Theorem:
- Since \( f(n) \) is polynomially larger than \( n^{\log_b(a)} \) (where \( \log_b(a) = \log_2(2) = 1 \)), we can apply case 2 of the theorem.

Hence, since \( f(n) = \Theta(n) \) is equal to \( n^{\log_b(a)} \), we have:
\[
T(n) = \Theta(n \log n)
\]

### Conclusion

Therefore, the asymptotic work of the `contains` function when called on a `List` is:
\[
\boxed{\Theta(n \log n)}
\]


Question:
Assume you are writing server-side code for an online shop. The code will be invoked over HTTP from a mobile app. Your current code is as follows:
public class ShoppingCart {
    public void buy(Product product, int quantity) {
        if (product == null) { throw new IllegalArgumentException("product cannot be null"); }
        if (quantity < 1) { throw new IllegalArgumentException("quantity must be at least 1"); }

        int price = product.getUnitPrice() * quantity;
        int discount = computeDiscount(product, quantity);
        int shippingFees = computeShippingFees(product, quantity);

        int totalPrice = price - discount + shippingFees;
        // this triggers a call to the actual credit card processor
        CreditCardProcessor.billCurrentUser(totalPrice);
    }

    private int computeDiscount(Product product, int quantity) {
        // ... discount computation logic ...
    }

    private int computeShippingFees(Product product, int quantity) {
        // ... shipping fees computation logic ...
    }
}

A colleague states that a null product should throw a checked exception, not an "IllegalArgumentException", because users might input bad data in the app. Explain in 1 sentence whether this is a good idea and why or why not.

Using a checked exception for a null product in this context is not a good idea because checked exceptions are typically reserved for recoverable conditions that can be handled by the caller, while a null product indicates a programming error that should be caught early (e.g., by validating input before invoking the method), making `IllegalArgumentException` more appropriate for signaling this kind of misuse.


Question: Suppose that you can prove the security of your symmetric encryption scheme against the following attacks. In which case is your scheme going to be the \textbf{most} secure?

### Restate the Question:
The question asks which type of attack your symmetric encryption scheme can withstand that would make it the most secure.

### Explain Key Concepts:
This question assesses your understanding of different types of attacks on encryption schemes, specifically focusing on key recovery and decryption methods under various conditions.

### Evaluate Each Option:
1. **Key recovery under known plaintext attack.**
   **Incorrect.** This scenario allows an attacker to potentially discover the encryption key using known plaintext and ciphertext pairs, indicating a serious vulnerability.

2. **Key recovery under chosen ciphertext attack.**
   **Incorrect.** Although this scenario is more secure than the previous one, it still exposes the system to key recovery risks if the attacker can manipulate the ciphertext.

3. **Decryption under known plaintext attack.**
   **Incorrect.** Here, the attacker can exploit known plaintext to decrypt ciphertext, but this does not provide strong security as it can still lead to vulnerabilities.

4. **Decryption under chosen ciphertext attack.**
   **Correct.** In this case, the attacker can choose ciphertexts and obtain their corresponding plaintexts. An encryption scheme that is secure against this attack is highly robust, as it restricts the attacker’s ability to gain useful information even while manipulating ciphertext.

### Justify the Correct Answer:
"Decryption under chosen ciphertext attack" is the most secure option because it indicates that the encryption scheme remains resilient against sophisticated attacks that allow an attacker to manipulate ciphertext to extract information. This level of security is crucial, as it ensures that even in the face of potential vulnerabilities, the encryption scheme maintains its integrity and confidentiality.