Question:
For this homework you will use a dataset of 18,403 music reviews scraped from Pitchfork¹, including relevant metadata such as review author, review date, record release year, review score, and genre, along with the respective album's audio features pulled from Spotify's API. The data consists of the following columns: artist, album, recordlabel, releaseyear, score, reviewauthor, reviewdate, genre, key, acousticness, danceability, energy, instrumentalness, liveness, loudness, speechiness, valence, tempo.

Create a new dataframe containing one row per 1st-2nd album pair. The dataframe should contain rows:

score_diff: the difference in scores between the second and the first album (second - first).
time_diff: the number of days elapsed between the first and the second album.
did_style_change: a dummy variable that indicates whether the style of the music has changed. To obtain it, first, calculate the standardized euclidean distance of music-related numerical features¹ between the second and the first album. Second, assign 1 to the 20% most distant 1st-2nd album pairs and 0 to all others.

Answer: import pandas as pd
from scipy.spatial.distance import euclidean
from sklearn.preprocessing import StandardScaler
from datetime import datetime

# Load the dataset
reviews = pd.read_csv('music_reviews.csv')

# Create a new dataframe for 1st-2nd album pairs
df_pairs = reviews.merge(reviews, on='artist')

# Filter to get only 1st-2nd album pairs based on release year
df_pairs = df_pairs[df_pairs['releaseyear_x'] < df_pairs['releaseyear_y']]

# Calculate score_diff and time_diff

df_pairs['score_diff'] = df_pairs['score_y'] - df_pairs['score_x']
df_pairs['time_diff'] = (pd.to_datetime(df_pairs['reviewdate_y']) - pd.to_datetime(df_pairs['reviewdate_x'])).dt.days

# Select relevant audio features and standardize them
features = ['acousticness', 'danceability', 'energy', 'instrumentalness', 'liveness', 'loudness', 'speechiness', 'valence', 'tempo']
scaler = StandardScaler()
standardized_features_x = scaler.fit_transform(df_pairs[features].iloc[:, 8:])
standardized_features_y = scaler.transform(df_pairs[features].iloc[:, 9:])

# Calculate the Euclidean distance for each pair

distances = [euclidean(x, y) for x, y in zip(standardized_features_x, standardized_features_y)]
df_pairs['distance'] = distances

# Identify the 20% most distant pairs
distance_threshold = df_pairs['distance'].quantile(0.8)
df_pairs['did_style_change'] = (df_pairs['distance'] > distance_threshold).astype(int)

# Select the final dataframe
final_df = df_pairs[['score_diff', 'time_diff', 'did_style_change']]\nReasoning: The solution involves merging the dataframe with itself to find album pairs, calculating the differences in scores and dates, standardizing the audio features, and determining style changes based on Euclidean distance, resulting in the desired output format.


Question: Consider the cipher defined using the key $K\in \{0,1\}^{64} $ by $$\begin{array}{llll} C : & \{0,1\}^{64} & \rightarrow & \{0,1\}^{64} \\ & x & \mapsto & C(x)=x \oplus K \\ \end{array} $$ Let $x=1\dots 11$, the value $\mathsf{LP}^{C_K}(x,x)$ is equal to

To evaluate \(\mathsf{LP}^{C_K}(x, x)\), we first need to understand what this notation means. \(\mathsf{LP}^{C_K}(x, y)\) represents the probability that \(C_K(x) = y\) for a random key \(K\). In this case, we are looking for \(\mathsf{LP}^{C_K}(x, x)\), which is the probability that \(C_K(x) = x\).

Given the cipher \(C\) defined as:

\[
C(x) = x \oplus K
\]

where \(\oplus\) is the bitwise XOR operation, let's analyze the scenario where \(K\) is a random 64-bit key.

**Step 1: Calculate \(C_K(x)\)**

Substituting \(x\) into the cipher function, we get:

\[
C_K(x) = x \oplus K
\]

We want to find the condition under which \(C_K(x) = x\).

**Step 2: Set up the equality \(C_K(x) = x\)**

We need to solve:

\[
x \oplus K = x
\]

**Step 3: Simplify the equation**

Using the properties of XOR, we can rearrange the equation:

\[
x \oplus K = x \implies K = x \oplus x = 0
\]

**Step 4: Determine the probability**

The equation \(K = 0\) means that the key \(K\) must be exactly the all-zero vector (64 zeros). Since \(K\) is uniformly chosen from the set of all possible 64-bit keys, which contains \(2^{64}\) possible keys, there is only one key (the all-zero key) that satisfies \(K = 0\).

Thus, the probability that \(K = 0\) is:

\[
\text{Probability} = \frac{1 \text{ (successful outcome)}}{2^{64} \text{ (total outcomes)}} = \frac{1}{2^{64}}
\]

Since we are interested in the probability that \(C_K(x) = x\), we realize that this condition occurs only when \(K\) is the specific value \(0\).

**Step 5: Conclusion**

The probability that \(C_K(x) = x\) is therefore:

\[
\mathsf{LP}^{C_K}(x, x) = \frac{1}{2^{64}}
\]

This means that the probability is not one of the options provided (0, 1/4, 1/2, 1). However, if we consider the context of the question, it's reasonable to interpret this as indicating that \(C_K(x) = x\) occurs with a very low probability, particularly since we are constrained to the options provided.

Since \(0\) is the closest to \(\frac{1}{2^{64}}\), we would conclude:

**The correct answer is \(0\).**


Question: Stream ciphers often use a nonce to \dots

Answer: C

Explanation: A nonce (number used once) in stream ciphers ensures that the same plaintext encrypts to different ciphertexts each time, thus avoiding key stream reuse and enhancing security.


Question: To which expression is the following for-loop translated? for x <- xs if x > 5; y <- ys yield x + y

The correct translation of the given for-loop statement, 'for x <- xs if x > 5; y <- ys yield x + y', is option C: 'xs.withFilter(x => x > 5).flatMap(x => ys.map(y => x + y)'. To justify this choice, we first analyze the components of the for-loop. The 'for x <- xs' indicates that we are iterating over the collection 'xs'. The condition 'if x > 5' specifies a filtering criterion, which we can represent using the 'withFilter' method in Scala. The second part, 'y <- ys', signifies that for each valid 'x', we will iterate over another collection 'ys'. The final expression 'yield x + y' suggests that we are producing a new collection based on the combination of 'x' and 'y'. Therefore, the appropriate structure is to first filter 'xs' with 'withFilter', then apply 'flatMap' to combine each filtered 'x' with every 'y' from 'ys' using the 'map' function. This sequence effectively captures the intent of the original for-loop, thereby making option C the correct choice.


Question:
Devise an algorithm that, without consensus, implements a weaker specification of NBAC by replacing the termination property with weak termination.

Weak termination: Let p be a distinguished process, known to all other processes. If p does not crash then all correct processes eventually decide. Your algorithm may use a perfect failure detector.

To develop an algorithm that implements a weaker specification of Non-Blocking Atomic Commit (NBAC) with weak termination, we first need to understand the components involved: processes, the distinguished process \( p \), and the perfect failure detector.

### Definitions
- **Perfect Failure Detector**: A system component that can accurately identify which processes have crashed and which are still operational. It provides reliable failure information to all processes.
- **Weak Termination**: This property ensures that if the distinguished process \( p \) does not crash, all correct processes will eventually reach a decision.

### Algorithm Outline
The algorithm can be structured as follows:

1. **Initialization**:
   - Each process \( i \) (where \( i \) can be \( 1, 2, \ldots, n \)) initializes its state and prepares to communicate.

2. **Proposal Phase**:
   - Each process \( i \) proposes a value or a decision. The distinguished process \( p \) sends its proposal to all other processes.

3. **Notification Phase**:
   - Process \( p \) uses the perfect failure detector to determine which processes are still alive.
   - After sending its proposal, \( p \) waits for acknowledgments from all other processes. Each correct process \( j \) (where \( j \neq p \)) sends an acknowledgment back to \( p \) upon receiving the proposal.

4. **Decision Phase**:
   - Once \( p \) receives acknowledgments from all correct processes, it can decide on a value.
   - It then broadcasts this decision to all processes.
   - If any process \( j \) does not receive the decision from \( p \) (due to \( p \) crashing), it relies on the perfect failure detector to determine \( p \)'s status.

5. **Finalization**:
   - Each process, upon receiving the decision from \( p \), updates its state to reflect the decision.
   - If \( p \) crashes before receiving acknowledgments from all processes, the remaining processes will use the information from the failure detector to check the status of \( p \) and can either wait for a new proposal or decide based on their own proposals.

### Pseudocode
Here is a simplified pseudocode outline of the algorithm:

```plaintext
function Process(i):
    if i == p:
        propose_value()
        for each process j != p:
            send proposal to j
        wait for acknowledgments from all processes
        if received from all:
            decide(value)
            broadcast decision to all processes
    else:
        wait for proposal from p
        if received proposal:
            send acknowledgment to p
        if received decision from p:
            finalize(decision)
        else:
            check if p has crashed using the failure detector
            if p has crashed:
                decide based on local state or wait for new proposal
```

### Explanation of Weak Termination
- If \( p \) does not crash, all correct processes will eventually receive the proposal and send acknowledgments, allowing \( p \) to make a decision and communicate it.
- If \( p \) crashes, the perfect failure detector allows the other processes to ascertain the crash. They can then either wait for a new proposal or make a decision based on their local state.

### Real-World Example
Consider a banking system where a distinguished node \( p \) represents the central server handling transactions. If the server is operational, all branches (other processes) will eventually receive transaction confirmations. If the server crashes, branches can evaluate the transaction based on their current state and available data, ensuring that they can still operate without being blocked indefinitely.

This algorithm satisfies weak termination by ensuring that as long as \( p \) is alive, decisions are made, while also providing a mechanism for processes to proceed in the event of \( p \)'s failure.


Question: Let $n$ be an integer. Which of the following is \emph{not} a group in the general case?

**Plan:**
- Define the criteria that structure a mathematical set as a group.
- Analyze each option to see if it satisfies the group properties (closure, associativity, identity, and invertibility).
- Identify which option fails to meet the group criteria in general.
- Conclude with the correct option that is not a group.

**Steps:**

1. **Defining Group Properties:** A set \( G \) with a binary operation \( * \) is a group if:
   - It is closed under the operation \( * \) (for any \( a, b \in G, a * b \in G \)).
   - The operation is associative (for all \( a, b, c \in G, (a * b) * c = a * (b * c) \)).
   - There is an identity element \( e \in G \) such that \( a * e = e * a = a \) for all \( a \in G \).
   - Every element \( a \in G \) has an inverse \( b \in G \) such that \( a * b = b * a = e \).

2. **Analyzing Each Option:**
   - **Option A: \((\mathbf{R},+)\)**: The set of real numbers under addition is a group because it satisfies all group properties.
   - **Option B: \((\mathbf{Q} \setminus \{0\},\times)\)**: The set of nonzero rational numbers under multiplication is a group, as it satisfies all group properties.
   - **Option C: \((\mathbf{Z}_n,+ \pmod{n})\)**: The set of integers modulo \( n \) under addition is a group since it satisfies closure, associativity, identity, and inverses.
   - **Option D: \((\mathbf{Z}_n,\times \pmod{n})\)**: This set is not necessarily a group in general because not all elements in \(\mathbf{Z}_n\) have multiplicative inverses. Specifically, if \( n \) is not prime, some elements do not have inverses modulo \( n \).

3. **Identifying the Non-Group Option:** Since \((\mathbf{Z}_n,\times \pmod{n})\) does not guarantee that every element has an inverse, it fails to meet one of the group criteria.

4. **Conclusion:** Therefore, the correct answer is D) \((\mathbf{Z}_n,\times \pmod{n})\) which is not a group in the general case.

**Self-Check:** The analysis confirms that \((\mathbf{Z}_n,\times \pmod{n})\) does not satisfy the group property of having inverses for all elements in the case of composite \( n \), thus validating that the answer is indeed D) \((\mathbf{Z}_n,\times \pmod{n})\).


Question: Tick the \emph{false} answer. In a group, the operation\dots

To identify the false answer regarding operations in a group, it’s essential to remember the key properties of a mathematical group. A group consists of a set equipped with an operation that satisfies four conditions: closure, associativity, the existence of an identity element, and the existence of inverses for every element.

Suppose the options presented included statements like:

1. The operation is closed.
2. The operation is associative.
3. Every element has an inverse.
4. The operation is commutative.

To find the false answer, we need to recognize that while closure, associativity, and the existence of inverses are required properties for a group, commutativity is not. A group where the operation is commutative is specifically called an “abelian group,” but not all groups are abelian.

Therefore, if one of the options states that the operation must be commutative for the structure to be a group, that would be the false answer.

In summary, the false statement about groups is typically one that incorrectly asserts a necessary property that is not universally applicable, such as the requirement for commutativity.


Question: You are in $D$-dimensional space and use a KNN classifier with $k=1$. You are given $N$ samples and by running experiments you see that for most random inputs $\mathbf{x}$ you find a nearest sample at distance roughly $\delta$. You would like to decrease this distance to $\delta / 2$. How many samples will you likely need? Give an educated guess.

To solve the problem, we need to understand how the dimensionality of the space ($D$), the number of samples ($N$), and the distance to the nearest neighbor relate to each other in a KNN (k-nearest neighbors) classifier setting.

1. **Understanding the problem**: You are in a $D$-dimensional space and you have $N$ samples. The distance to the nearest sample for most random inputs $\mathbf{x}$ is approximately $\delta$. You want to reduce this distance to $\delta / 2$.

2. **Distance and Density**: In a high-dimensional space, the volume of the space increases exponentially with the number of dimensions. Therefore, the density of points becomes sparse as dimensions increase. To ensure that we can find points closer together (i.e., reducing the distance from $\delta$ to $\delta / 2$), we will need to pack more samples into the space.

3. **Scaling of Samples with Dimensionality**: In order to maintain a certain density of points, the number of samples required increases exponentially with the number of dimensions. Specifically, if we want to reduce the distance to the nearest neighbor, we require more samples to fill the space sufficiently to ensure that we have points within the desired distance.

4. **Estimating the number of samples needed**: To reduce the nearest neighbor distance from $\delta$ to $\delta / 2$, we can think of it in terms of volume. The volume of a $D$-dimensional ball of radius $\delta$ scales with $\delta^D$, and similarly for $\delta/2$, it scales as $(\delta/2)^D = \delta^D / 2^D$. The ratio of volumes indicates how many additional samples we need to maintain density.

   This ratio leads us to conclude that if we reduce the distance to half, we need approximately $2^D$ times the number of samples to keep the density sufficient. Therefore, to reduce the distance to $\delta / 2$, we will need about $2^D N$ samples.

5. **Evaluating the Options**:
   - 0. $2^D N$ — This makes sense based on our analysis (correct).
   - 1. $N^D$ — Incorrect; this suggests a polynomial increase based on the number of samples, which is not applicable here.
   - 2. $2 D$ — Incorrect; this is a linear function in $D$, which does not reflect the exponential increase needed.
   - 3. $\log (D) N$ — Incorrect; logarithmic growth does not capture the necessary sample increase in high dimensions.
   - 4. $N^2$ — Incorrect; again, this suggests a polynomial relationship which does not apply.
   - 5. $D^2$ — Incorrect; linear or polynomial in $D$ is not sufficient.
   - 6. $2 N$ — Incorrect; while it suggests doubling the number of samples, it does not capture the dimensionality.
   - 7. $D N$ — Incorrect; again, linear scaling with $D$ does not represent the true scaling behavior.

In conclusion, the only option that fits the reasoning and analysis provided is:

**Correct Option: 0. $2^D N$**.


Question: KEM \dots


Question: Which of the following is an element of $\mathbb{Z}_{60}^*$?

- 49

To ascertain which of the given integers is an element of \(\mathbb{Z}_{60}^*\), we must consider the criteria that define this multiplicative group. An integer \(x\) is included in \(\mathbb{Z}_{60}^*\) if it satisfies the condition that \(\text{gcd}(x, 60) = 1\), meaning it should not share any prime factors with \(60\), specifically \(2\), \(3\), or \(5\).

Now, evaluating the options:
- **49**: It is not divisible by \(2\) (it's odd), the sum of its digits is \(13\) (not divisible by \(3\)), and it does not end in \(0\) or \(5\) (not divisible by \(5\)). Hence, it seems like a candidate.
- **30**: Clearly divisible by \(2\), \(3\), and \(5\).
- **26**: Divisible by \(2\).
- **21**: Divisible by \(3\).

Thus, the only feasible candidate is **49**. So, \(49 \in \mathbb{Z}_{60}^*\).


Question: A Feistel scheme is used in\dots

The correct answer is **DES** (Data Encryption Standard).

A Feistel scheme is a specific design structure used in block ciphers. Picture it like a locked box (your data) that you want to protect. Instead of creating an entirely new lock (cipher), the Feistel scheme rearranges and transforms the contents through stages, using a series of keys.

In DES, the data is divided into two halves and processed through multiple rounds, applying a function with part of the key. This is similar to mixing two different ingredients in a recipe, where each step enhances the flavor (security) of your dish (data).

Although AES (Advanced Encryption Standard) is a modern cipher, it does not utilize the Feistel structure; it employs a substitution-permutation network instead. FOX and CS-Cipher also do not follow this scheme. Thus, when considering a classic example of a Feistel structure in cryptography, DES stands out as a foundational model for secure data encryption!


Question: Why is natural language processing difficult?Select all that apply.You will get a penalty for wrong answers.

Natural Language Processing (NLP) is indeed a challenging field for several reasons. Let's analyze each of the options you've provided:

1. **Lack of linguistic competence and resources**: This is a valid reason. Many languages have limited resources, such as annotated corpora or linguistic tools, which can hinder the development of effective NLP models. Additionally, the complexity of human language, including idioms, metaphors, and variances in dialects, poses challenges.

2. **Impact of power laws**: This option is somewhat ambiguous. In the context of language, power laws can relate to the distribution of words (e.g., Zipf's law), but this is more of an observation about language rather than a direct difficulty in NLP. Therefore, it might not be as relevant as other options.

3. **High dimensionality and sparseness of data**: This is definitely a challenge in NLP. Natural language data is high-dimensional (think of all the possible words and features) and often sparse (many combinations of words or phrases may not occur frequently), making it difficult for models to generalize and learn effectively.

4. **Subjectivity of annotators**: This is also a valid reason. When human annotators are involved in labeling data, their interpretations can vary widely. This subjectivity can lead to inconsistencies in the training data, which can negatively impact the performance of NLP models.

Based on the analysis, the applicable options for why NLP is difficult are:

- **1. Lack of linguistic competence and resources**
- **3. High dimensionality and sparseness of data**
- **4. Subjectivity of annotators**

So, the correct selections are 1, 3, and 4.


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

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

### Option 1: Unit tests may serve as foundation to create libFuzzer fuzzing stubs.
**Analysis**: This statement is correct. Unit tests often define the expected behavior of functions, and these tests can be adapted to create fuzzing stubs for libFuzzer. By using existing unit tests, developers can define inputs that are known to be valid, and then modify or expand those inputs to explore edge cases and unexpected behavior during fuzzing.

### Option 2: In libFuzzer’s default mode (not fork-mode), the tested APIs must not contain `exit()`.
**Analysis**: This statement is correct. In the default mode, when libFuzzer runs a target function, it expects that the function will return control back to libFuzzer without terminating the process. If the tested API contains `exit()`, it will terminate the entire process, preventing libFuzzer from continuing its fuzzing efforts. This is why `exit()` (and similar calls) should be avoided in the code that is being fuzzed in this mode.

### Option 3: It is better to put narrow targets into the fuzzing stubs, e.g., if a target can parse several data formats, split it into several targets, one per format.
**Analysis**: This statement is also correct. Splitting targets into narrower scopes allows libFuzzer to focus on specific functionalities, making it easier to identify issues related to particular formats. By creating separate fuzzing stubs for each format, you can achieve higher precision in testing and better coverage of edge cases associated with each format.

### Option 4: libFuzzer can only test single-threaded targets.
**Analysis**: This statement is incorrect. While libFuzzer is primarily designed to work with single-threaded targets in its default mode, it does have the ability to test multi-threaded targets when using fork mode or other configurations. So stating that it can "only" test single-threaded targets is misleading, as it is more accurate to say that its default operation is single-threaded but it can accommodate multi-threaded scenarios.

### Summary of Correct Options:
1. **Option 1**: Correct
2. **Option 2**: Correct
3. **Option 3**: Correct
4. **Option 4**: Incorrect

### Rationale:
- Options 1, 2, and 3 are correct because they align with best practices and operational characteristics of libFuzzer. They emphasize the utility of unit tests in fuzzing, the necessity of avoiding abrupt termination in tested functions, and the advantage of focusing on narrow targets for effective fuzzing.
- Option 4 is incorrect because it falsely restricts libFuzzer's capabilities; while it is designed for single-threaded testing in its default configuration, it can also handle multi-threaded applications when set up correctly.

In conclusion, Options 1, 2, and 3 are correct, while Option 4 is incorrect.


Question:
Assume you are working on a school project with your friend.
Your friend claims that using very detailed names are good in code because they describe exactly what is going on. Do you agree? Explain in max 2 sentences.

While it's important for names in code to be descriptive so that others can understand what they do, using overly detailed names can clutter the code and make it less readable. It's best to find a balance between being descriptive and keeping names concise to maintain clarity.


Question:
Consider the following definition of trees representing higher-order functions, as well as a recursive function
subst0.
1 enum Expr:
2 case C(c: BigInt)
3 case N(name: String)
4 case BinOp(op: BinOps, e1: Expr, e2: Expr)
5 case IfNonzero(cond: Expr, trueE: Expr, falseE: Expr)
6 case Call(fun: Expr, arg: Expr)
7 case Fun(param: String, body: Expr)
8
9 import Expr._
10
11 enum BinOps:
12 case Plus, Minus, Times, Power, LessEq
13
14 def subst0(e: Expr, n: String, r: Expr): Expr = e match
15 case C(c) => e
16 case N(s) => if s == n then r else e
17 case BinOp(op, e1, e2) =>
18 BinOp(op, subst0(e1, n, r), subst0(e2, n, r))
19 case IfNonzero(cond, trueE, falseE) =>
20 IfNonzero(subst0(cond,n,r), subst0(trueE,n,r), subst0(falseE,n,r))
21 case Call(f, arg) =>
22 Call(subst0(f, n, r), subst0(arg, n, r))
23 case Fun(formal, body) =>
24 if formal == n then e
25 else Fun(formal, subst0(body, n, r))
And consider the following expression:
1 val e = Call(N("exists"), Fun("y", Call(Call(N("less"), N("x")), N("y"))))
What is subst0(e, "y", C(42)) equal to?

To evaluate `subst0(e, "y", C(42))` where `e` is defined as:

```scala
val e = Call(N("exists"), Fun("y", Call(Call(N("less"), N("x")), N("y"))))
```

we will apply the `subst0` function step-by-step, carefully analyzing how it processes each part of the expression.

### Step-by-Step Evaluation

1. **Understanding the Structure of `e`**:
   - The expression `e` is a `Call` expression, comprised of:
     - A function: `N("exists")`
     - An argument: `Fun("y", Call(Call(N("less"), N("x")), N("y")))`

2. **Applying `subst0` to `Call(fun, arg)`**:
   - In the `subst0` function, when we encounter a `Call(fun, arg)`, we need to apply `subst0` to both the function (`fun`) and the argument (`arg`).

   - **For the function part (`fun = N("exists")`)**:
     - Since the name "exists" is not equal to the variable name "y" (the one we want to substitute), we simply return `N("exists")`.

   - **For the argument part (`arg = Fun("y", Call(Call(N("less"), N("x")), N("y")))`)**:
     - We analyze this `Fun` expression more closely. The formal parameter of this function is "y".

3. **Analyzing the `Fun` Expression**:
   - Inside the `Fun` expression, we check if the formal parameter "y" is equal to the variable name we are substituting, which is also "y". Since they are equal, according to the `subst0` function's definition, we do not perform any substitution in the body of the function.
   - Therefore, `subst0(Fun("y", Call(Call(N("less"), N("x")), N("y"))), "y", C(42))` results in the original `Fun` expression:
     ```scala
     Fun("y", Call(Call(N("less"), N("x")), N("y")))
     ```

4. **Combining Results**:
   - Now we combine the results from both parts of the `Call`:
     - The function part remains `N("exists")`.
     - The argument part remains unchanged as `Fun("y", Call(Call(N("less"), N("x")), N("y")))`.

5. **Final Result**:
   - Consequently, the final result of the substitution is:
   ```scala
   Call(N("exists"), Fun("y", Call(Call(N("less"), N("x")), N("y"))))
   ```

### Conclusion
Thus, the result of evaluating `subst0(e, "y", C(42))` is:

```scala
Call(N("exists"), Fun("y", Call(Call(N("less"), N("x")), N("y"))))
```

This confirms that the substitution does not alter the expression because the variable being substituted ("y") is the same as the formal parameter of the function that contains it. If there had been occurrences of "y" elsewhere in the expression that were not bound by the function, those would have been substituted with `C(42)`. However, in this case, all instances of "y" are bound, and thus remain unchanged.


Question: What is the number of secret bits in a WEP key?

To determine the number of secret bits in a WEP (Wired Equivalent Privacy) key, let’s analyze the options step by step.

### Understanding WEP Key Sizes:
WEP is an older security protocol used in wireless networks to provide a level of security equivalent to that of a wired LAN. The key sizes specified for WEP are:

- **40-bit WEP key**: This is often referred to as a "40-bit" key, but it actually consists of a 24-bit initialization vector (IV) plus a 40-bit secret key, making a total of 64 bits when combined.
- **104-bit WEP key**: Similarly, this is referred to as a "104-bit" key, which consists of a 24-bit IV plus a 104-bit secret key, totaling 128 bits when combined.

### Analyzing the Options:
1. **Option 1: 64 or 128 bits.**
   - This option refers to the total length of the keys including the IV. The WEP key lengths themselves are rather 40 bits and 104 bits for the secret parts, not 64 or 128 bits.

2. **Option 2: 40 or 104 bits.**
   - This option correctly identifies the lengths of the secret bits in a WEP key. The secret key is indeed either 40 bits or 104 bits long.

3. **Option 3: 64 or 128 bytes.**
   - This option is incorrect, as these sizes are far too large. A byte consists of 8 bits, so 64 bytes would equal 512 bits, and 128 bytes would equal 1024 bits, which is not relevant to WEP key sizes.

4. **Option 4: 40 or 104 bytes.**
   - This option is also incorrect for the same reason as Option 3; these sizes are excessively large when compared to the actual WEP key sizes.

### Conclusion:
Based on the analysis, the correct statement regarding the number of secret bits in a WEP key is:

Answer: 40 or 104 bits.


Question: Implement the F1-score to evaluate your classifier.

To implement the F1-score for evaluating a classifier, we need to understand a few steps:

1. **Definition of F1-score**: The F1-score is the harmonic mean of precision and recall. It is particularly useful when the class distribution is imbalanced. The formula for F1-score is:
   \[
   F1 = 2 \times \frac{Precision \times Recall}{Precision + Recall}
   \]

2. **Calculating Precision and Recall**:
   - **Precision**: This measures the accuracy of the positive predictions. It is calculated as:
     \[
     Precision = \frac{TP}{TP + FP}
     \]
     where \(TP\) is the number of true positives and \(FP\) is the number of false positives.
   - **Recall**: This measures the ability of the classifier to find all the positive samples. It is calculated as:
     \[
     Recall = \frac{TP}{TP + FN}
     \]
     where \(FN\) is the number of false negatives.

3. **Implementation Steps**:
   - Collect the predictions from the classifier and the true labels.
   - Count the true positives, false positives, and false negatives.
   - Calculate precision and recall using the formulas above.
   - Finally, compute the F1-score using the precision and recall.

4. **Example**:
   Let's say our classifier made the following predictions:
   - True Positives (TP) = 70
   - False Positives (FP) = 30
   - False Negatives (FN) = 10

   We can calculate:
   - Precision = \( \frac{70}{70 + 30} = \frac{70}{100} = 0.7 \)
   - Recall = \( \frac{70}{70 + 10} = \frac{70}{80} = 0.875 \)
   - F1-score = \( 2 \times \frac{0.7 \times 0.875}{0.7 + 0.875} = 2 \times \frac{0.6125}{1.575} \approx 0.7778 \)

Now, let’s format the answer as requested:

```json
{
    "result": {
        "reasoning": "The F1-score is calculated as the harmonic mean of precision and recall, providing a balance between the two metrics, especially useful in imbalanced datasets. By calculating true positives, false positives, and false negatives, we can derive the precision and recall, ultimately leading to the computation of the F1-score.",
        "answer": "The F1-score for the given example is approximately 0.7778."
    }
}
```


Question: Tick the \textbf{false} assertion. A distinguisher \ldots

To identify the false assertion about distinguishers, we need to analyze each option in the context of cryptographic principles.

### Identifying the False Assertion

The options are:

1. **"... can break PRNG."**
2. **"... is an algorithm calling an oracle."**
3. **"... recovers the secret key of a stream cipher."**
4. **"... can differentiate the encryption of two known plaintexts."**

Among these options, the assertion that is **false** is **"... is an algorithm calling an oracle."**

### Rationale for the Correct Answer

**3. "... recovers the secret key of a stream cipher."** is also a misleading assertion but it can be somewhat true in the context of some theoretical scenarios where distinguishing can lead to key recovery. However, the distinction of the oracle aspect is more universally false.

### Explanation of Each Option

1. **"... can break PRNG."**
   - This statement is true. A distinguisher can potentially distinguish between the output of a pseudorandom number generator (PRNG) and true random output. If a distinguisher can successfully do this, it implies the PRNG is not secure, and thus, it can be considered "broken" in that context.

2. **"... is an algorithm calling an oracle."**
   - This statement is false in the context of what a distinguisher is. A distinguisher typically does not "call an oracle" in the sense of querying a cryptographic function or accessing secret information directly. Instead, it is designed to differentiate between two distributions (for example, real vs. random), not to invoke an oracle. Oracles are often theoretical constructs used in cryptographic proofs (like a "random oracle" model), but it is not a defining feature of distinguishers themselves.

3. **"... recovers the secret key of a stream cipher."**
   - This statement can be contextually true. A distinguisher may not directly recover a key but can indicate weaknesses in the stream cipher that could lead to key recovery in certain attack scenarios. In practice, if a distinguisher can differentiate the output of a stream cipher from that of a random sequence, it may indicate vulnerabilities that could be exploited to recover the key through further analysis.

4. **"... can differentiate the encryption of two known plaintexts."**
   - This statement is true. A distinguisher can indeed differentiate between the encryptions of two known plaintexts if the encryption scheme is vulnerable. This is especially relevant in distinguishing attacks where the goal is to tell apart different encryption outputs.

### Conclusion

In summary, the assertion that a distinguisher "... is an algorithm calling an oracle." is false because a distinguisher is primarily concerned with differentiating distributions rather than interacting with an oracle. Other options accurately reflect the capabilities of distinguishers in the context of cryptography.


Question: Consider a mobile station (MS) with a SIM card associated to a home network (HN). The MS tries to connect to a visited network (VN). In the GSM authentication, who knows the key $K_i$?

In the GSM authentication process, the key ( K_i ) is known only to the SIM card and the home network (HN). Therefore, the correct answer is:

"SIM and HN."


Question: In this week's lecture, you have been introduced to the aggregate method of ParSeq[A] (and other parallel data structures). It has the following signature:  def aggregate[B](z: B)(f: (B, A) => B, g: (B, B) => B): B Discuss, as a group, what aggregate does and what its arguments represent. Discuss the implementations from questions 4 and 5. Which one do you think would be more efficient?

- **Purpose of `aggregate`**:
  - The `aggregate` method is used to combine elements of a parallel data structure (like `ParSeq[A]`) into a single result of a different type `B`.
  - It allows for parallel processing of elements, making it efficient for large datasets.

- **Arguments of `aggregate`**:
  - `z: B`: This is the initial value or zero value for the aggregation. It serves as the starting point for the accumulation.
  - `f: (B, A) => B`: This is a binary function that takes the current accumulated value of type `B` and an element of type `A`, and returns a new accumulated value of type `B`. It defines how to combine an element with the accumulated result.
  - `g: (B, B) => B`: This is another binary function that combines two accumulated values of type `B`. It is used when merging results from different parallel computations.

- **Implementation Considerations**:
  - In question 4, the implementation may focus on a straightforward sequential aggregation, which could be simpler but less efficient for large datasets.
  - In question 5, the implementation likely utilizes parallelism more effectively, potentially dividing the dataset into chunks that can be processed concurrently, thus speeding up the aggregation process.

- **Efficiency Comparison**:
  - The implementation from question 5 is expected to be more efficient due to its parallel nature, allowing multiple cores to work on different parts of the data simultaneously.
  - The overhead of managing parallel tasks must be considered, but for large datasets, the benefits of reduced computation time typically outweigh these costs.

**Final answer:** The `aggregate` method efficiently combines elements of a parallel data structure using an initial value and two combining functions, with the implementation from question 5 likely being more efficient due to its parallel processing capabilities.


Question: When a test fails, it means that:

Correct answer: **either the program under test or the test itself has a bug, or both.**

Rationale: A test failing indicates that there is an issue that could stem from either the program being tested (a bug) or the test itself (which may be incorrectly designed or implemented). It is also possible that both have issues, but the statement captures the essence that either one or both could be the cause of the failure.


Question: Suppose we have a universe $U$ of elements. For $A,B\subseteq U$, the Jaccard distance of $A,B$ is defined as $$ J(A,B)=\frac{|A\cap B|}{|A\cup B|}.$$ This definition is used in practice to calculate a notion of similarity of documents, webpages, etc. For example, suppose $U$ is the set of English words, and any set $A$ represents a document considered as a bag of words. Note that for any two $A,B\subseteq U$, $0\leq J(A,B)\leq 1$. If $J(A,B)$ is close to 1, then we can say $A\approx B$. Let $h: U\to [0,1]$ where for each $i\in U$, $h(i)$ is chosen uniformly and independently at random. For a set $S\subseteq U$, let $h_S:=\min_{i\in S} h(i)$. \textbf{Show that } $$ \Pr[h_A=h_B] = J(A,B).$$ Now, if we have sets $A_1, A_2,\dots,A_n$, we can use the above idea to figure out which pair of sets are ``close'' in time essentially $O(n|U|)$. We can also obtain a good approximation of $J(A,B)$ with high probability  by using several independently chosen hash functions. Note that the naive algorithm would take $O(n^2|U|)$ to calculate all pairwise similarities.

The probability that \(h_A = h_B\) equals the probability that the minimum hash value over \(A \cup B\) is located in \(A \cap B\). Since all elements have equal chances, this probability is \(|A \cap B| / |A \cup B|\), which is exactly the Jaccard similarity \(J(A,B)\). Therefore, \(\Pr[h_A = h_B] = J(A,B).\)


Question: Visual cryptography is a nice visual application of \ldots

Visual cryptography is a technique that allows for the encryption of visual information (like images) in such a way that decryption does not require complex computations but rather a simple overlay of images. It involves dividing a secret image into several parts, where each part alone reveals no information about the original image. When the parts are stacked together, they reconstruct the original image.

The Vernam cipher, also known as the one-time pad, is a method of encryption that uses a random key that is as long as the message itself. The unique characteristic of the Vernam cipher is its ability to achieve perfect secrecy when the key is truly random, used only once, and kept secret. In the context of visual cryptography, the idea of combining multiple layers (or keys) in a way that each layer alone does not reveal any information directly aligns with the principles of the Vernam cipher.

On the other hand, options like the Vigenère cipher, Caesar cipher, and ROT13 are classical encryption methods that involve substitution and transposition of characters in a message, which do not directly relate to the visual representation or layering aspect that is central to visual cryptography.

Thus, the correct answer option is: \ldots the Vernam cipher.


Question:
You are discussing coding habits with a colleague, who says:
"When I edit a part of a function, if I see unclean code in another part of it, I also clean that other part up."
In one sentence, explain if this is a good habit and why:

Yes, it's a good habit because it encourages continuous improvement of code quality, making the overall codebase cleaner and easier to maintain over time.


Question: What is the content of the superblock in the JOS file system?

The answer is "Total number of blocks on disk" and "Magic number identifying the file system" as these are standard components of a file system's superblock, providing critical information for managing the file system structure.


Question: In which of the following cases does JOS acquire the big kernel lock?

To determine when the JOS (a simple operating system developed for educational purposes) acquires the big kernel lock, we need to understand the role of the big kernel lock within the operating system and the context in which it is utilized.

### Step-by-Step Reasoning:

1. **Understanding the Big Kernel Lock**:
   - The big kernel lock is used to ensure that only one processor can execute kernel code at any given time. This is crucial for maintaining data consistency and preventing race conditions in a multi-core or multi-processor environment.

2. **Evaluating Each Option**:
   - **Processor traps in user mode**:
     - When a trap occurs in user mode, it typically indicates that the user program is making a request to the kernel (e.g., system calls). In this scenario, the big kernel lock is acquired to ensure that the transition from user mode to kernel mode is safe and that no two processors can access the kernel simultaneously.
     - **Conclusion**: The big kernel lock is acquired.

   - **Processor traps in kernel mode**:
     - When a trap occurs in kernel mode (e.g., an interrupt or exception), the system is already executing in kernel mode. The typical behavior here does not involve acquiring the big kernel lock since the code is already in a trusted state. However, if the trap handler itself could lead to further kernel invocations, the lock may be utilized, but generally, it is not explicitly acquired just for processing traps in kernel mode.
     - **Conclusion**: The big kernel lock is not necessarily acquired.

   - **Switching from kernel mode to user mode**:
     - This transition does not require acquiring the big kernel lock. The lock is primarily concerned with preventing concurrent access to kernel resources, and switching to user mode is a controlled exit from kernel execution that does not necessitate locking.
     - **Conclusion**: The big kernel lock is not acquired.

   - **Initialization of application processor**:
     - During the initialization of an application processor, it's important to ensure that the system is in a consistent state, and the big kernel lock is typically acquired to prevent any other processors from executing kernel code until the initialization is complete.
     - **Conclusion**: The big kernel lock is acquired.

3. **Consolidating Findings**:
   - From the evaluations:
     - The big kernel lock is acquired during:
       - **Processor traps in user mode**: Yes
       - **Initialization of application processor**: Yes
     - The big kernel lock is not acquired during:
       - **Processor traps in kernel mode**: No
       - **Switching from kernel mode to user mode**: No

### Final Answer:
The correct options where JOS acquires the big kernel lock are:
- **Processor traps in user mode**
- **Initialization of application processor**

Thus, the correct option from the given choices is:
**['Processor traps in user mode', 'Initialization of application processor']**.


Question: When using the plain ElGamal cryptosystem over the group $\mathbb{Z}_p^*$ with a fixed $\ell$-bit prime $p$ and a fixed generator $g\in\mathbb{Z}_p^*$, the \textit{tightest} complexity of generating a new key-pair is\ldots

To determine the tightest complexity of generating a new key-pair in the ElGamal cryptosystem, we need to consider the operations involved in key generation. The key generation involves selecting a random private key, which requires generating a random integer in the range \([1, p-1]\), and computing the public key using the generator \(g\) raised to the power of the private key.

**Option A: $\mathcal{O}(4\ell)$**
This option suggests a linear complexity which is less precise than the actual complexity involved. While generating a random number may take linear time, the exponentiation operation for computing the public key actually requires more sophisticated algorithms to compute efficiently, leading to a higher complexity than this option suggests.

**Option B: $\mathcal{O}(\ell^3)$**
Exponentiation in a group like \(\mathbb{Z}_p^*\) can be performed efficiently using techniques like square-and-multiply, which generally has a complexity of \(\mathcal{O}(\ell^2)\). However, if we consider the need for multiplication in finite fields and the generation of random numbers, this could lead to an overall complexity that might reach cubic bounds in certain contexts, but this isn't the tightest bound.

**Option C: $\mathcal{O}(\ell^2)$**
This option accurately reflects the complexity involved in generating the public key through modular exponentiation, which can be done in \(\mathcal{O}(\ell^2)\) using efficient algorithms. Given that generating a random integer and computing the public key are the primary operations, and since the most complex operation dominates the complexity, this is the tightest complexity for the key generation process.

**Option D: $\mathcal{O}(\log{\ell}^4)$**
This option suggests a logarithmic complexity, which is not appropriate for this context since both the generation of a random integer and the modular exponentiation involve polynomial complexities. The operations in the ElGamal key generation are more involved than logarithmic time, making this option incorrect.

In conclusion, the tightest complexity of generating a new key-pair in the ElGamal cryptosystem is best captured by **Option C: $\mathcal{O}(\ell^2)$**.


Question: Implement the precision at k metric

Certainly! The Precision at k (P@k) metric is commonly used in information retrieval to evaluate the effectiveness of a search or recommendation system. It measures the proportion of relevant items in the top-k results returned by the system.

Here's a step-by-step guide to implementing the Precision at k metric:

### Step 1: Define the Inputs
You will need the following inputs:
- **Relevant Items**: A list of items that are considered relevant for the query.
- **Retrieved Items**: A list of items returned by the recommendation or search system.
- **k**: The number of top items to consider for the precision calculation.

### Step 2: Select Top-k Retrieved Items
You need to extract the top-k items from the retrieved items list. If the number of retrieved items is less than k, you should use all the retrieved items.

### Step 3: Calculate the Number of Relevant Items in Top-k
Count how many of the top-k retrieved items are present in the relevant items list.

### Step 4: Calculate Precision
Precision is calculated using the formula:

\[
\text{Precision at k} = \frac{\text{Number of relevant items in top-k}}{k}
\]

If \( k \) is larger than the number of retrieved items, you can adjust the formula to use the number of retrieved items instead.

### Step 5: Implement the Function
Here's a sample code implementation in Python:

```python
def precision_at_k(relevant_items, retrieved_items, k):
    # Step 1: Select top-k retrieved items
    top_k_retrieved = retrieved_items[:k]  # Get top k items (or all if less than k)

    # Step 2: Calculate the number of relevant items in top-k
    relevant_count = sum(1 for item in top_k_retrieved if item in relevant_items)

    # Step 3: Calculate precision
    # If k is more than the number of retrieved items, adjust k
    actual_k = min(k, len(retrieved_items))

    precision = relevant_count / actual_k if actual_k > 0 else 0.0

    return precision

# Example usage:
relevant = ['A', 'B', 'C', 'D']
retrieved = ['A', 'E', 'B', 'F', 'G']
k = 3

print("Precision at k:", precision_at_k(relevant, retrieved, k))
```

### Step 6: Testing
Make sure to test the function with various cases, including:
- When the retrieved items contain no relevant items.
- When all top-k retrieved items are relevant.
- When \( k \) is greater than the number of retrieved items.
- When there are no retrieved items.

### Conclusion
By following these steps, you can implement the Precision at k metric effectively. This metric will help you evaluate the performance of your retrieval or recommendation system in terms of how many of the top-k results are relevant.


Question: (Linear Regression) You are given samples $\mathcal{S}=\left\{\left(\mathbf{x}_{n}, y_{n}\right)\right\}_{n=1}^{N}$ where $\mathbf{x}_{n} \in \mathbb{R}^{D}$ and $y_{n}$ are scalar values. You are solving linear regression using normal equations. You will always find the optimal weights with 0 training error in case of $N \leq D$.

The question asks whether you will always achieve zero training error in linear regression using normal equations when the number of samples \( N \) is less than or equal to the number of features \( D \).

In linear regression, the relationship between the number of training samples and features is crucial. If \( N \leq D \), it is possible to fit a model with enough complexity to perfectly match the training data. However, this scenario does not guarantee zero training error due to potential issues such as noise in the data, outliers, or the model's inability to capture the true underlying relationship.

1. **True**: Incorrect. While having more features than samples could theoretically lead to a perfect fit, it often results in overfitting. Overfitting occurs when the model learns the noise in the training data instead of the underlying pattern, failing to generalize to new data.

2. **False**: Correct. The statement is false because achieving zero training error is not guaranteed with \( N \leq D \). In practice, even with sufficient parameters, the presence of noise or other complications can prevent reaching zero error.

In summary, the correct answer is **False**. Although \( N \leq D \) allows for the possibility of a perfect fit, it does not ensure it due to factors such as noise and overfitting, making the initial assertion not universally valid.


Question: A basic statement in RDF would be expressed in the relational data model by a table

Yes, a basic statement in RDF (Resource Description Framework) can be expressed in the relational data model by a table with three columns: subject, predicate, and object.


Question:
Let $G$ be a set and $*$ a commutative operation on pairs of elements from $G.$ Suppose there exists an element $e\in G$ such that $a*e=e*a=a$ for all $a \in G.$ Also, suppose there exist elements $b,c,d \in G$ such that $b*c=d*c.$. True or false: If $b
ot=d,$ then $(G,*)$ cannot be a group.

To determine whether the statement is true or false, let's analyze the given conditions step by step.

### Given Information:
1. \( G \) is a set with a commutative operation \( * \).
2. There is an identity element \( e \in G \) such that for all \( a \in G \), \( a * e = e * a = a \).
3. There exist elements \( b, c, d \in G \) such that \( b * c = d * c \).
4. We are asked to assess whether the statement "If \( b \neq d \), then \( (G, *) \) cannot be a group" is true or false.

### Analyzing the Conditions:

1. **Identity Element**: The existence of an identity element \( e \) implies that the operation is at least partially satisfying the definition of a group, as one of the axioms of a group is the existence of an identity element.

2. **Commutativity**: The operation being commutative means that the order of the elements does not affect the outcome of the operation, which is another positive aspect in favor of the structure being a group.

3. **Equation \( b * c = d * c \)**: Let's analyze this equation given that \( b \neq d \). We can rearrange the equation to see what it implies about \( c \):
   \[
   b * c = d * c \implies b * c * e = d * c * e \quad (\text{using the identity element} \, e)
   \]
   Since \( e \) does not change the outcome, we can write:
   \[
   b * c = d * c \implies b * c * e = d * c * e \implies (b * c) * e = (d * c) * e.
   \]
   However, we can simplify this to consider:
   \[
   b * c - d * c = 0.
   \]
   If we assume \( * \) behaves like standard addition, we can consider this as \( (b - d) * c = 0 \).

### Exploring the Implication When \( b \neq d \):
If \( b \neq d \), then \( (b - d) \) is not zero under standard addition. Hence, for \( (b - d) * c = 0 \) to hold (which it does due to the equation \( b * c = d * c \)), we must have \( c = e \) (the identity element), which leads to:
\[
b * e = d * e \implies b = d,
\]
which contradicts our assumption that \( b \neq d \).

### Conclusion:
Thus, if \( b * c = d * c \) holds true for \( b \neq d \), then \( c \) must be the identity element \( e \). However, this does not imply that \( (G, *) \) cannot be a group. It simply means that the operation behaves in a way consistent with the properties of a group, given that there is still an identity and the operation is commutative.

Therefore, the statement "If \( b \neq d \), then \( (G, *) \) cannot be a group" is **False** because the existence of an identity element and the commutativity of the operation are sufficient to satisfy the group properties despite the given condition.

### Final Answer:
1. **False**


Question: Consider a blockcipher $\mathsf{B}:\mathcal{K}\times\{0,1\}^n \rightarrow \{0,1\}^n$ with $|\mathcal{K}|=N$ for which we choose a secret key $K\in\mathcal{K}$ uniformly at random. Tick the \textit{correct} assertion.

To determine the correct assertion about the block cipher $\mathsf{B}$, we need to analyze the implications of security against different types of attacks:

1. **Known-Plaintext Attack (KPA)**: If $\mathsf{B}$ is secure against KPA, it means that knowing some plaintexts and their corresponding ciphertexts does not help an attacker recover the key or deduce information about the key.

2. **Ciphertext-Only Attack (COA)**: This attack involves trying to recover the key from ciphertexts alone, without any known plaintexts. If a cipher is secure against KPA, it does not guarantee security against COA, as the attacker may still attempt to perform brute-force attacks on the ciphertexts.

3. **Chosen-Ciphertext Attack (CCA)**: This is a stronger form of attack where an attacker can choose ciphertexts to be decrypted and gain information based on the resulting plaintexts. Security against KPA does not imply security against CCA, as it can exploit more information.

4. **Complexity of Key Recovery**: The assertions regarding computational complexity of recovering keys involve analyzing the feasibility of precomputation and resource requirements.

Considering these points:
- Assertion 1 claims that security against KPA implies security against COA, which is generally true.
- Assertion 2 states that security against KPA implies security against CCA, which is not necessarily true.
- Assertions 3 and 4 focus on specific resource requirements for key recovery, which do not align with conventional cryptographic principles.

Thus, the correct answer is option 1.

1


Question: Determine which of the following conditional statements evaluate to true (more than one answer can be correct):

### Step 1: Evaluate the expressions

First, we need to evaluate the expressions involved in the conditional statements.

1. Calculate \(5 - 7\):
   \[
   5 - 7 = -2
   \]
   Thus, \(5 - 7 < 0\) is true.

2. Calculate \(3 - 7\):
   \[
   3 - 7 = -4
   \]
   Thus, \(3 - 7 < 0\) is also true.

### Step 2: Analyze each conditional statement

Now, we will analyze each conditional statement based on the evaluations from Step 1.

1. **Statement 1**: If \(5 - 7 > 0\), then \(3 - 7 > 0\)
   - This translates to: If \(-2 > 0\), then \(-4 > 0\).
   - Both parts are false, but in a conditional statement, if the premise is false, the whole statement is considered true.
   - **This statement evaluates to true.**

2. **Statement 2**: If \(5 - 7 < 0\), then \(3 - 7 > 0\)
   - This translates to: If \(-2 < 0\), then \(-4 > 0\).
   - The first part is true, but the second part is false, making the entire conditional false.
   - **This statement evaluates to false.**

3. **Statement 3**: If \(5 - 7 > 0\), then \(3 - 7 < 0\)
   - This translates to: If \(-2 > 0\), then \(-4 < 0\).
   - The first part is false, so the entire statement is true.
   - **This statement evaluates to true.**

4. **Statement 4**: If \(5 - 7 < 0\), then \(3 - 7 < 0\)
   - This translates to: If \(-2 < 0\), then \(-4 < 0\).
   - Both parts are true, making the entire statement true.
   - **This statement evaluates to true.**

### Step 3: Summarize the results

Based on the evaluations:

- Statement 1: True
- Statement 2: False
- Statement 3: True
- Statement 4: True

### Conclusion

The conditional statements that evaluate to true are:

- If \(5 - 7 > 0\), then \(3 - 7 < 0\)
- If \(5 - 7 < 0\), then \(3 - 7 < 0\)
- If \(5 - 7 > 0\), then \(3 - 7 > 0\)

Thus, the correct answers are:

- If 5-7 > 0, then 3-7<0
- If 5-7 < 0, then 3-7<0
- If 5-7 > 0, then 3-7>0


Question: In the χ2 statistics for a binary feature, we obtain P(χ2 | DF = 1) > 0.05. This means in this case, it is assumed:

The correct option is: **'That the class label is independent of the feature.'**

### Rationale for the Correct Option

In statistical hypothesis testing, particularly when using the chi-squared (χ²) test for independence, the null hypothesis (H₀) generally posits that there is no association between the two categorical variables being analyzed—in this case, the binary feature and the class labels. The degrees of freedom (DF) in this scenario is 1 because we have a binary feature.

When we calculate the chi-squared statistic and obtain a p-value greater than 0.05 (P(χ² | DF = 1) > 0.05), we fail to reject the null hypothesis at the 5% significance level. This indicates that there is not enough evidence to suggest that the class label depends on the feature, implying that the class labels and the feature are independent of each other for the data in question.

### Why the Other Options Are Incorrect

1. **'That the class labels depend on the feature':**
   This option is incorrect because a p-value greater than 0.05 indicates that we do not have sufficient evidence to claim dependence. If the class labels depended on the feature, we would expect a significant chi-squared statistic leading to a p-value less than 0.05, allowing us to reject the null hypothesis.

2. **'That the class label correlates with the feature':**
   Correlation suggests a statistical relationship between two variables. However, similar to the previous option, a p-value greater than 0.05 suggests a lack of evidence for a relationship or correlation between the class label and the feature. If there were a correlation, we would expect the chi-squared test to yield a significant p-value, which is not the case here.

3. **'None of the above':**
   This option would imply that none of the statements accurately reflect the outcome of the chi-squared test. However, the statement regarding the independence of the class label from the feature is indeed accurate based on the provided p-value. Therefore, this option is incorrect because we have identified a valid interpretation of the test result.

### Understanding of the Underlying Concepts

The chi-squared test for independence is a statistical method used to determine if there is a significant relationship between two categorical variables. It works by comparing the observed frequencies in each category to the frequencies we would expect if the two variables were indeed independent. The test results in a chi-squared statistic, which is then used to calculate a p-value.

A p-value is the probability of observing the data, or something more extreme, given that the null hypothesis is true. A conventional threshold for significance is 0.05. If the p-value is greater than this threshold, it indicates that the observed data are consistent with the null hypothesis, leading to the conclusion that the variables are independent.

In summary, the outcome of P(χ² | DF = 1) > 0.05 suggests that we do not have sufficient evidence to claim a dependence between the class label and the binary feature, thus affirming their independence. This understanding is pivotal in the analysis of categorical data, guiding researchers in making data-driven decisions based on statistical evidence.


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

Map[A, X] >: Map[B, Y]


Question: Let $\mathcal{C}$ be a $(n,k)$ Reed-Solomon code on $\mathbb{F}_q$. Let $\mathcal{C}'$ be the $(2n,k)$ code such that each codeword of $\mathcal{C}'$ is a codeword of $\mathcal{C}$ repeated twice, i.e., if $(x_1,\dots,x_n) \in\mathcal{C}$, then $(x_1,\dots,x_n,x_1,\dots,x_n)\in\mathcal{C'}$. What is the minimum distance of $\mathcal{C}'$?

The minimum distance of the $(2n,k)$ code $\mathcal{C}'$ is given by the formula for the minimum distance of a code constructed by repeating codewords. Specifically, if the minimum distance of $\mathcal{C}$ is \(d\), then the minimum distance \(d'\) of the code $\mathcal{C}'$ is \(d' = 2d\).

For a Reed-Solomon code, the minimum distance \(d\) is given by \(d = n - k + 1\). Therefore, \(d' = 2(n - k + 1) = 2n - 2k + 2\).

Thus, the minimum distance of $\mathcal{C}'$ is \(2n - 2k + 2\).

The correct answer is: **$2n-2k+2$**.


Question: You are given your $D \times N$ data matrix $\boldsymbol{X}$, where $D$ represents the dimension of the input space and $N$ is the number of samples. We discussed in the course the singular value decomposition (SVD). Recall that the SVD is not invariant to scaling and that empirically it is a good idea to remove the mean of each feature (row of $\boldsymbol{X}$ ) and to normalize its variance to 1 . Assume that $\boldsymbol{X}$ has this form except that the last row/feature is then multiplied by $\sqrt{2}$, i.e., it has variance $\left(\ell_{2}^{2}\right.$-norm) of 2 instead of 1. Recall that the SVD allows us to write $\boldsymbol{X}$ in the form $\boldsymbol{X}=\boldsymbol{U} \boldsymbol{S} \boldsymbol{V}^{\top}$, where $\boldsymbol{U}$ and $\boldsymbol{V}$ are unitary and $\boldsymbol{S}$ is a $D \times N$ diagonal matrix with entries $s_{i}$ that are non-negative and decreasing, called the singular values. Assume now that you add a feature, i.e., you add a row to $\boldsymbol{X}$. Assume that this row is identical to the last row of $\boldsymbol{X}$, i.e., you just replicate the last feature. Call the new matrix $\tilde{\boldsymbol{X}}$. But assume also that for $\tilde{\boldsymbol{X}}$ we normalize all rows to have variance 1. To summarize, $\boldsymbol{X}$ is the original data matrix, where all means have been taken out and all rows are properly normalized to have variance 1 except the last one that has variance 2 . And $\tilde{\boldsymbol{X}}$ is the original data matrix with the last row replicated, and all means have been taken out and all rows are properly normalized. Let $\boldsymbol{X}=\boldsymbol{U} \cdot \boldsymbol{S} \cdot \boldsymbol{V}^{\top}$ be the SVD of $\boldsymbol{X}$ and let. $\tilde{\boldsymbol{X}}=\tilde{\boldsymbol{U}} \cdot \tilde{\boldsymbol{S}} \cdot \tilde{\boldsymbol{V}}^{\top}$ be the SVD of $\tilde{\boldsymbol{X}}$ \begin{enumerate} \item Show that \end{enumerate} (a) $\tilde{V}=V$ (b) $\tilde{\boldsymbol{S}}$ is equal to $\boldsymbol{S}$ with an extra all-zero row attached. \begin{enumerate} \setcounter{enumi}{1} \item Based on the previous relationships and assuming that it is always best to run an SVD with "normalized" rows, what is better: If you $K N O W$ that a feature is highly correlated to another feature a priori. Should you rather first run the SVD and then figure out what features to keep or should you first take the highly correlated feature out and then run the SVD? Explain. \end{enumerate}

Sure! Let's break this down step by step.

1. **(a) Show that $\tilde{V} = V$.**
   Since we only added an extra row to $\boldsymbol{X}$ that is identical to the last row, the underlying relationships among the original features remain unchanged. Therefore, the matrix $\tilde{X}$ has the same span for the original features. This means the direction of the singular vectors (columns of $\boldsymbol{V}$) does not change when you add a duplicate row. So, we can conclude that $\tilde{V} = V$.

2. **(b) Show that $\tilde{\boldsymbol{S}}$ is equal to $\boldsymbol{S}$ with an extra all-zero row attached.**
   When you normalize the new matrix $\tilde{\boldsymbol{X}}$, the singular values are affected by the fact that the last feature is now represented twice but normalized. The SVD captures the variance structure, and since the last feature was duplicated, its contribution to the variance will remain the same (but it's now normalized). This effectively adds a zero singular value for the additional row. Therefore, $\tilde{\boldsymbol{S}}$ is just $\boldsymbol{S}$ with an additional row of zeros at the bottom.

3. **(2) Should you first run the SVD and then figure out what features to keep, or take the highly correlated feature out first?**
   It's generally better to remove highly correlated features before running SVD. This is because SVD finds directions of maximum variance, and if you have features that are highly correlated, they provide redundant information. If you run SVD on them together, you'll end up with similar singular vectors for those features, which means you're not capturing new information. By removing one of the correlated features first, you can achieve a clearer and more meaningful decomposition, leading to better insights from the SVD.

So, in summary, you want to clean up your data by removing redundancy before applying SVD to get the best results!


Question:
Assume you are working on a mobile application. Users report that your app freezes when they access its image gallery, which shows images in a scrollable grid. This is the (java) function run to display the gallery:
void startImageGallery() {
    // Download all the user's images from the application server
    List images = getImages();
    // Display the first few images
    displayImage(images);
    // Initialize the behavior of the exit button, back button, zoom button, etc.
    startComponentBehavior();
}
In one sentence, explain why the application is freezing:

The application freezes because the synchronous network call to getImages() blocks the main UI thread, preventing it from processing user interactions and updates, thus making the app unresponsive. Implementing asynchronous image downloading would resolve this issue.


Question: Implement a function that inserts a given element elem into a sorted (in ascending order) list list . The resulting list should also be sorted in ascending order. Implement the function recursively. def insert (elem: Int, list: List[Int]): List[Int] = ???

To implement a recursive function in Scala that inserts an element into a sorted list while maintaining the sorted order, we can break down the problem into smaller parts. The idea is to compare the element to be inserted (`elem`) with the head of the list. Depending on the comparison, we can recursively insert the element either before the head (if it is smaller or equal) or in the rest of the list.

Here is a step-by-step implementation of the `insert` function:

1. **Base Case:** If the list is empty, return a list containing just the element to be inserted.
2. **Recursive Case:** Compare `elem` with the head of the list:
   - If `elem` is less than or equal to the head, place `elem` in front of the head.
   - If `elem` is greater than the head, keep the head in place and recursively call the function on the tail of the list.

Here is the complete code for the `insert` function:

```scala
def insert(elem: Int, list: List[Int]): List[Int] = {
  list match {
    case Nil => List(elem)  // Base case: if the list is empty, return a list with the element
    case head :: tail =>
      if (elem <= head) {
        elem :: list  // If elem is less than or equal to head, prepend it to the list
      } else {
        head :: insert(elem, tail)  // Otherwise, keep the head and recursively insert into the tail
      }
  }
}
```

### Explanation of the Code:
- The function `insert` takes an integer `elem` and a sorted list of integers `list`.
- We use pattern matching to handle the list:
  - `case Nil` matches an empty list and returns a new list containing only `elem`.
  - `case head :: tail` matches a non-empty list, where `head` is the first element and `tail` is the rest of the list.
- We compare `elem` with `head`:
  - If `elem` is less than or equal to `head`, we construct a new list with `elem` at the front.
  - If `elem` is greater than `head`, we keep `head` in the result and recursively call `insert` on `tail`.

### Example Usage:
```scala
val sortedList = List(1, 3, 4, 5)
val newElement = 2
val newList = insert(newElement, sortedList)
// newList will be List(1, 2, 3, 4, 5)
```

This implementation maintains the sorted order and operates recursively as required.


Question: Which method can be used for dimensionality reduction ?

The correct answer to the question "Which method can be used for dimensionality reduction?" is:

**A. PCA** (Principal Component Analysis).

Now, letâs go through each option to understand why PCA is the correct choice and how the others relate to dimensionality reduction.

### A. PCA (Principal Component Analysis)
PCA is a widely used statistical technique for dimensionality reduction. It works by transforming the original variables into a new set of variables, called principal components, which are orthogonal and capture the maximum variance in the data.

- **How it works**: PCA identifies the directions (principal components) in which the data varies the most and projects the data onto these directions. This helps in reducing the number of dimensions while preserving as much information as possible.

### B. T-distributed Stochastic Neighbor Embedding (t-SNE)
t-SNE is also a method for dimensionality reduction, particularly effective for visualizing high-dimensional data in two or three dimensions. However, it is primarily used for visualization rather than general dimensionality reduction in pre-processing stages.

- **Key point**: While t-SNE does reduce dimensions, it is not typically used as a first-line method for dimensionality reduction in machine learning, as it focuses on preserving local structures rather than global data variance.

### C. Autoencoders
Autoencoders are a type of neural network used for learning efficient representations of data, often for the purpose of dimensionality reduction. They consist of an encoder that compresses the data and a decoder that reconstructs it.

- **Key point**: Autoencoders can perform dimensionality reduction, but they require training a neural network, which can be more complex compared to PCA. They are generally considered a more advanced technique.

### D. SVM (Support Vector Machine)
SVM is primarily a supervised machine learning algorithm used for classification and regression tasks. It does not directly perform dimensionality reduction.

- **Key point**: While SVM can handle high-dimensional data and may indirectly benefit from dimensionality reduction methods, it is not a method for reducing dimensions itself.

### Conclusion
In summary, while options B (t-SNE) and C (Autoencoders) can also achieve dimensionality reduction, PCA is the most fundamental and commonly used method for this purpose, making **A. PCA** the best choice among the options provided. Option D (SVM) does not relate to dimensionality reduction at all.


Question: The Kerckhoff's principle is not followed when security requires that \dots

The Kerckhoff's principle is not followed when the security of a cryptosystem relies on the secrecy of the encryption algorithm, as it implies that the system is fundamentally insecure if the algorithm is exposed. This principle asserts that a secure system should remain secure even if everything about the system, except the key, is public knowledge.


Question: Which of these attacks applies to the Diffie-Hellman key exchange when the channel cannot be authenticated?

To analyze the multiple-choice question regarding the Diffie-Hellman key exchange, we need to understand the attack types mentioned and how they relate to the security of the Diffie-Hellman protocol, especially when the channel is not authenticated.

The Diffie-Hellman key exchange is a method for two parties to establish a shared secret over an insecure channel. It relies on the difficulty of the discrete logarithm problem. The basic steps involve:

1. Two parties agree on a large prime \( p \) and a generator \( g \).
2. Each party selects a private key (let's call them \( a \) and \( b \) for Alice and Bob, respectively) and computes their public keys:
   - Alice computes \( A = g^a \mod p \)
   - Bob computes \( B = g^b \mod p \)
3. They exchange their public keys \( A \) and \( B \).
4. Both parties compute the shared secret:
   - Alice computes \( s = B^a \mod p \)
   - Bob computes \( s = A^b \mod p \)

The shared secret \( s \) will be the same for both parties due to the properties of modular arithmetic.

Now, let’s examine the attacks listed:

1. **Meet-in-the-middle attack**: This attack is typically associated with symmetric cryptography and does not apply to Diffie-Hellman, as it does not involve searching through a key space in the same way that symmetric keys do.

2. **Birthday Paradox**: This concept relates to the probability of collisions in hash functions, particularly in contexts like digital signatures. It does not directly apply to Diffie-Hellman key exchange, where the goal is to establish a shared secret rather than to find collisions.

3. **Attack on low exponents**: While this is a known vulnerability, it specifically targets situations where one of the private keys is a small integer. Although it can be a concern, it does not directly exploit the lack of authentication in the communication channel.

4. **Man-in-the-middle attack (MitM)**: This is a significant threat when the Diffie-Hellman exchange occurs over an unauthenticated channel. In a MitM attack, an adversary can intercept the public keys exchanged between Alice and Bob. The attacker could then replace the public keys with their own, effectively creating two separate keys:
   - The attacker sends \( A' \) instead of \( A \) to Bob.
   - The attacker sends \( B' \) instead of \( B \) to Alice.
   As a result, both parties would believe they are communicating securely with each other, while the attacker is actually controlling the exchange.

This attack highlights the critical need for authentication in key exchange protocols. If authentication is not implemented, the potential for a MitM attack dramatically undermines the security of the Diffie-Hellman key exchange.

Considering the analysis above, it is clear that the correct answer to the question is:

4. Man-in-the-middle attack.


Question: Implement Latent Semantic Indexing by selecting the first x largest singular values of the term document matrix    Hint 1: np.linalg.svd(M, full_matrices=False) performs SVD on the matrix $\mathbf{M}$ and returns $\mathbf{K}, \mathbf{S}, \mathbf{D}^T$   -  $\mathbf{K}, \mathbf{D}^T$ are matrices with orthonormal columns  -  $\mathbf{S}$ is a **vector** of singular values in a **descending** order

To implement Latent Semantic Indexing (LSI), you first need to perform Singular Value Decomposition (SVD) on the term-document matrix. This can be done using `np.linalg.svd(M, full_matrices=False)`, which will give you three outputs: the left singular vectors, the singular values, and the right singular vectors. Then, you can select the first `x` largest singular values and their corresponding vectors to create a reduced representation of the original matrix.

Here's a simple implementation:

```python
import numpy as np

def truncated_svd(term_doc_matrix, num_val):
    U, S, VT = np.linalg.svd(term_doc_matrix, full_matrices=False)
    U_selected = U[:, :num_val]
    S_selected = np.diag(S[:num_val])
    VT_selected = VT[:num_val, :]
    return U_selected, S_selected, VT_selected
```

This function will return the reduced matrices needed for LSI.


Question: In the maximum directed cut problem we are given as input a directed graph $G = (V, A)$. Each arc $(i, j)\in  A$ has a nonnegative weight $w_{ij} \geq 0$. The goal is to partition $V$ into two sets $U$ and $W = V \setminus U$ so as to maximize the total weight of the arcs going from $U$ to $W$ (that is, arcs $(i, j)$ with $i \in U$ and $j \in W$). Give a randomized 1/4-approximation algorithm for this problem (together with a proof that it is a 1/4-approximation in expectation).

Certainly! Let's revise the answer to the maximum directed cut problem by incorporating the feedback from the critique. We will ensure that all key terms are defined, the logic is complete, and the method is clearly justified.

---

### Problem Definition

In the maximum directed cut problem, we are given a directed graph \( G = (V, A) \) where:
- \( V \) is the set of vertices.
- \( A \) is the set of directed edges (or arcs), each with a nonnegative weight \( w_{ij} \geq 0 \) for the arc \( (i, j) \).

The goal is to partition the vertex set \( V \) into two disjoint sets \( U \) and \( W \) (where \( W = V \setminus U \)) such that the total weight of the arcs going from \( U \) to \( W \) is maximized. This means we want to maximize the weight of all arcs \( (i, j) \) where \( i \in U \) and \( j \in W \).

### Randomized Algorithm Outline

To develop a randomized algorithm, we can follow these steps:

1. **Random Assignment**: Randomly assign each vertex \( v \in V \) to set \( U \) with probability \( \frac{1}{2} \) and to set \( W \) with probability \( \frac{1}{2} \). This randomization helps ensure that each vertex has an equal chance of being in either set, allowing for an unbiased partition.

2. **Calculate the Total Weight**: After the random assignment, calculate the total weight of the arcs going from \( U \) to \( W \).

3. **Expected Value**: Use the expected value of this total weight as the approximation for the maximum directed cut.

### Step-by-Step Calculation of Expected Weight

1. **Contribution of Each Arc**: Let’s consider an arbitrary arc \( (i, j) \in A \) with weight \( w_{ij} \). For this arc to contribute to the weight from \( U \) to \( W \), vertex \( i \) must be in \( U \) (which happens with probability \( \frac{1}{2} \)) and vertex \( j \) must be in \( W \) (which also happens with probability \( \frac{1}{2} \)). Thus, the probability that the arc \( (i, j) \) contributes to the weight is:
   \[
   P(i \in U \text{ and } j \in W) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.
   \]

2. **Expected Contribution of the Arc**: The expected contribution of the arc \( (i, j) \) to the weight from \( U \) to \( W \) can be calculated as:
   \[
   \text{Expected contribution} = P(i \in U \text{ and } j \in W) \cdot w_{ij} = \frac{1}{4} w_{ij}.
   \]

3. **Total Expected Weight Calculation**: Now, summing this expected contribution over all arcs in \( A \), we get the total expected weight:
   \[
   \text{Expected weight} = \sum_{(i, j) \in A} \frac{1}{4} w_{ij} = \frac{1}{4} \sum_{(i, j) \in A} w_{ij}.
   \]

### Final Approximation

Let \( OPT \) be the maximum weight of any directed cut in \( G \). By definition, \( OPT \) represents the highest possible total weight of arcs that can be achieved by any partition of \( V \).

Since \( \sum_{(i, j) \in A} w_{ij} \) includes all arcs in the graph, it is clear that:
\[
\sum_{(i, j) \in A} w_{ij} \geq OPT.
\]
Thus, we conclude:
\[
\text{Expected weight} = \frac{1}{4} \sum_{(i, j) \in A} w_{ij} \geq \frac{1}{4} OPT.
\]

This shows that our randomized algorithm provides an expected weight that is at least \( \frac{1}{4} \) of the optimal weight.

### Conclusion

In summary, we have presented a randomized algorithm for the maximum directed cut problem that partitions the vertices into two sets randomly. The expected weight of the directed cut produced by this algorithm guarantees that it achieves at least \( \frac{1}{4} \) of the optimal solution.

This provides us with a 1/4-approximation algorithm for the maximum directed cut problem, leveraging the principles of probability and expectation.

Do you have any questions about the steps we took or any part of the explanation?


Question: Data being classified as unstructured or structured depends on the:

To classify data as structured or unstructured, we must consider how the data is organized and whether it adheres to a predefined model or format.

1. **Degree of Abstraction**: This refers to the organization of data. Structured data, such as that in relational databases, is highly organized and fits a fixed schema, making it easy to query. In contrast, unstructured data, like text documents and multimedia files, lacks this organization and does not conform to a specific format.

2. **Level of Human Involvement**: While human involvement can affect data entry, it does not determine whether data is structured or unstructured. Both types can involve varying levels of human interaction.

3. **Type of Physical Storage**: The method of physical storage (e.g., hard drives, cloud) does not classify data as structured or unstructured. Both types can be stored in any medium.

4. **Amount of Data**: The volume of data does not dictate its structure. Regardless of size, data can be classified as structured or unstructured based on its organization.

In summary, the most relevant factor for distinguishing structured from unstructured data is the **degree of abstraction**. While this classification can vary in specific contexts, it fundamentally relates to how data is organized. Thus, the best answer is: **Degree of abstraction**.


Question:
In a lottery, a bucket of 10 numbered red balls and a bucket of 5 numbered green balls are used. Three red balls and two green balls are drawn (without replacement).
What is the probability to win the lottery? (The order in which balls are drawn does not matter).

To determine the probability of winning the lottery, we first need to calculate the total number of ways to draw the specified balls and then the number of successful outcomes that satisfy the winning conditions.

### Step 1: Calculate Total Outcomes
1. **Total Ways to Choose 3 Red Balls from 10:**
   The number of ways to choose 3 red balls from 10 can be calculated using the combination formula:
   \[
   \binom{n}{k} = \frac{n!}{k!(n-k)!}
   \]
   For our case:
   \[
   \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
   \]

2. **Total Ways to Choose 2 Green Balls from 5:**
   Similarly, for the green balls:
   \[
   \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
   \]

3. **Total Combinations of Drawing 3 Red and 2 Green:**
   The total number of combinations for drawing 3 red and 2 green balls is:
   \[
   \text{Total Outcomes} = \binom{10}{3} \times \binom{5}{2} = 120 \times 10 = 1200
   \]

### Step 2: Calculate Winning Outcomes
Let’s assume that in order to win the lottery, a specific combination of 3 red balls and 2 green balls must be drawn. Since we do not have information on a specific winning combination, we will consider that there is only **1 winning combination**.

### Step 3: Calculate Probability
The probability of winning can be calculated as:
\[
P(\text{Winning}) = \frac{\text{Number of Winning Outcomes}}{\text{Total Outcomes}} = \frac{1}{1200}
\]

### Step 4: Evaluate Given Options
Now we will look at the provided options:

- **Option 1:** $$\frac{1}{14400}$$
- **Option 2:** $$\frac{1}{7200}$$
- **Option 3:** $$\frac{1}{1200}$$
- **Option 4:** $$\frac{1}{1900}$$

The only option that matches our calculated probability of winning, which is $$\frac{1}{1200}$$, is **Option 3**.

### Final Answer
Final Answer: 3.


Question: Following the notation used in class, let us denote the set of terms by $T=\{k_i|i=1,...,m\}$, the set of documents by $D=\{d_j |j=1,...,n\}$, and let $d_i=(w_{1j},w_{2j},...,w_{mj})$. We are also given a query  $q=(w_{1q},w_{2q},...,w_{mq})$. In the lecture we studied that,   $sim(q,d_j) = \sum^m_{i=1} \frac{w_{ij}}{|d_j|}\frac{w_{iq}}{|q|}$ .  (1)  Another way of looking at the information retrieval problem is using a probabilistic approach. The probabilistic view of information retrieval consists of determining the conditional probability $P(q|d_j)$ that for a given document $d_j$ the query by the user is $q$. So, practically in probabilistic retrieval when a query $q$ is given, for each document it is evaluated how probable it is that the query is indeed relevant for the document, which results in a ranking of the documents.  In order to relate vector space retrieval to a probabilistic view of information retrieval, we interpret the weights in Equation (1) as follows:  -  $w_{ij}/|d_j|$ can be interpreted as the conditional probability $P(k_i|d_j)$ that for a given document $d_j$ the term $k_i$ is important (to characterize the document $d_j$).  -  $w_{iq}/|q|$ can be interpreted as the conditional probability $P(q|k_i)$ that for a given term $k_i$ the query posed by the user is $q$. Intuitively, $P(q|k_i)$ gives the amount of importance given to a particular term while querying.  With this interpretation you can rewrite Equation (1) as follows:  $sim(q,d_j) = \sum^m_{i=1} P(k_i|d_j)P(q|k_i)$ (2)  Note that the model described in Question (a) provides a probabilistic interpretation for vector space retrieval where weights are interpreted as probabilities . Compare to the probabilistic retrieval model based on language models introduced in the lecture and discuss the differences.

To compare the vector space retrieval model with the probabilistic retrieval model based on language models, we need to analyze the underlying principles and assumptions of each approach.

### Vector Space Model (VSM)

1. **Representation**: In the vector space model, documents and queries are represented as vectors in a multi-dimensional space, where each dimension corresponds to a term from the vocabulary. The weights of the terms in these vectors can be interpreted as the importance of the terms in the context of the document or query.

2. **Similarity Calculation**: The similarity between a query \( q \) and a document \( d_j \) is computed using the formula:
   \[
   sim(q, d_j) = \sum_{i=1}^{m} \frac{w_{ij}}{|d_j|} \frac{w_{iq}}{|q|}
   \]
   This can be interpreted probabilistically as:
   \[
   sim(q, d_j) = \sum_{i=1}^{m} P(k_i|d_j) P(q|k_i)
   \]
   Here, \( P(k_i|d_j) \) represents the probability of term \( k_i \) being important for document \( d_j \), and \( P(q|k_i) \) represents the probability of the query \( q \) given the term \( k_i \).

3. **Assumptions**: The VSM assumes that the terms are independent and that the relevance of a document can be determined by the presence and importance of its terms. It does not take into account the order of terms or the context in which they appear.

### Probabilistic Retrieval Model (PRM)

1. **Representation**: In the probabilistic retrieval model, the focus is on estimating the probability that a document is relevant to a given query. The model is based on the idea of relevance feedback and uses statistical methods to infer the likelihood of relevance.

2. **Probability Calculation**: The core of the probabilistic model is the computation of the conditional probability \( P(q|d_j) \), which directly assesses how likely it is that the query \( q \) is relevant to the document \( d_j \). This is often done using a language model that generates the query based on the terms present in the document.

3. **Assumptions**: The probabilistic model assumes that the relevance of a document can be modeled as a random variable and that the terms in the document can be used to predict the likelihood of relevance. It often incorporates the notion of term independence but can also account for term dependencies through more complex models.

### Key Differences

1. **Focus on Relevance**: The VSM focuses on the similarity between the query and document based on term weights, while the PRM focuses on the probability of relevance of the document given the query.

2. **Interpretation of Weights**: In the VSM, weights are interpreted as importance scores, while in the PRM, they are interpreted as probabilities of relevance.

3. **Modeling Approach**: The VSM uses a geometric approach to measure similarity in a vector space, while the PRM uses statistical methods to estimate probabilities of relevance.

4. **Handling of Term Dependencies**: The VSM generally assumes independence among terms, while the PRM can incorporate dependencies through language models, allowing for a more nuanced understanding of how terms interact in the context of relevance.

5. **Output**: The VSM typically produces a ranked list of documents based on similarity scores, while the PRM produces a ranking based on the estimated probabilities of relevance.

### Conclusion

In summary, while both models aim to retrieve relevant documents based on a user query, they do so through different methodologies and assumptions. The vector space model emphasizes geometric similarity based on term weights, while the probabilistic retrieval model emphasizes statistical relevance based on conditional probabilities. Understanding these differences is crucial for selecting the appropriate model for a given information retrieval task.


Question: Prove that √2 is irrational.

To prove that ( \sqrt{2} ) is irrational, we can use a proof by contradiction. Here are the steps in detail:

Step 1: Assume the opposite
Assume that ( \sqrt{2} ) is rational. By definition, this means that it can be expressed as the fraction of two integers ( a ) and ( b ) (where ( b \neq 0 )), such that: [ \sqrt{2} = \frac{a}{b} ] We also assume that this fraction is in its simplest form, meaning that ( a ) and ( b ) share no common factors other than 1 (i.e., they are coprime).

Step 2: Square both sides
To eliminate the square root, we square both sides of our equation: [ 2 = \left( \frac{a}{b} \right)^2 ] This simplifies to: [ 2 = \frac{a^2}{b^2} ]

Step 3: Rearrange the equation
Next, we rearrange the equation to isolate ( a^2 ): [ a^2 = 2b^2 ]

Step 4: Analyze the implications
From the equation ( a^2 = 2b^2 ), we can deduce several things:

The left side, ( a^2 ), is an integer.
The right side is twice an integer (( 2b^2 )), so ( a^2 ) is also an even number since it equals ( 2b^2 ) (which is even).
Step 5: Conclude that ( a ) is even
Since ( a^2 ) is even, we can conclude that ( a ) itself must also be even (because the square of an odd number is odd).

Step 6: Express ( a ) in terms of an integer
Since ( a ) is even, we can write ( a ) as: [ a = 2k ] for some integer ( k ).

Step 7: Substitute back into the original equation
Now we substitute ( a = 2k ) into our equation ( a^2 = 2b^2 ): [ (2k)^2 = 2b^2 ] This simplifies to: [ 4k^2 = 2b^2 ] Divide both sides by 2: [ 2k^2 = b^2 ]

Step 8: Conclude that ( b ) is even
Now we see that ( b^2 = 2k^2 ), which means ( b^2 ) is also even. Thus, ( b ) must be even as well.

Step 9: Contradiction
Now we have shown that both ( a ) and ( b ) are even. This implies that they share at least a common factor of 2. This contradicts our initial assumption that ( a ) and ( b ) have no common factors other than 1.

Conclusion
Since our assumption that ( \sqrt{2} ) is rational leads to a contradiction, we conclude that our assumption must be wrong. Thus, we can say: [ \sqrt{2} \text{ is irrational.} ]


Question:
Recall that the hard-margin SVM problem corresponds to:
	$$ \underset{\substack{\ww \in \R^d, \ orall i:\ y_i \ww^	op \xx_i \geq 1}}{\min} \Vert \ww \Vert_2.$$
	Now consider the $2$-dimensional classification dataset corresponding to the $3$ following datapoints: $\xx_1 = (-1, 2)$, $\xx_2 = (1, 2)$, $\xx_3 = (0, -2)$ and  $y_1 = y_2 = 1$, $y_3 = -1$.
	Which of the following statements is 	extbf{true}:

**Plan:**
1. Analyze the given data points and their labels to determine if the dataset is linearly separable.
2. Consider the geometrical arrangement of the points to check for the existence of a separating hyperplane.
3. Evaluate the claims made in each statement regarding the dataset and the solution of the hard-margin SVM problem.
4. Conclude which statement is true based on the analysis.

**Steps:**

1. **Analyze the Dataset**: We have three data points:
   - \(\xx_1 = (-1, 2)\) with label \(y_1 = 1\)
   - \(\xx_2 = (1, 2)\) with label \(y_2 = 1\)
   - \(\xx_3 = (0, -2)\) with label \(y_3 = -1\)

   The first two points are both labeled as \(1\) and lie on the same horizontal line \(y = 2\), while the third point is labeled as \(-1\) and is located below the line at \(y = -2\).

2. **Check for Linear Separability**: A hyperplane can separate the points if we can find a line (in 2D) such that all points with label \(1\) are on one side and all points with label \(-1\) are on the other. The line \(y = 0\) (the x-axis) can serve as a separating line, as it lies between the points \(\xx_1\) and \(\xx_2\) above it, while \(\xx_3\) lies below it. Thus, the dataset is linearly separable.

3. **Evaluate the Claims**:
   - **First Statement**: "Our dataset is not linearly separable..." - This is false since we have established linear separability.
   - **Second Statement**: "There exists a unique \(\ww^\star\) which linearly separates our dataset." - This statement is true because we can find a hyperplane that separates the points.
   - **Third Statement**: "The unique vector which solves the hard-margin problem for our dataset is \(\ww^\star = (0, 1)\)." - This is false. The weight vector \((0, 1)\) would imply a vertical line, which does not separate the given points effectively.

4. **Conclusion**: The second statement is the only one that holds true regarding our dataset.

**Self-Check:**
- The dataset is confirmed to be linearly separable as established through the analysis.
- There indeed exists a unique \(\ww^\star\) that can separate the dataset, aligning with the second statement's correctness.
- The third statement about \(\ww^\star = (0, 1)\) was shown to be incorrect as it does not effectively separate the data points.

Thus, the answer is **B**: "There exists a unique \(\ww^\star\) which linearly separates our dataset."


Question: Let $\mathcal{C}$ be a binary $(n,k)$ linear code with minimum distance $d_{\min} = 4$. Let $\mathcal{C}'$ be the code obtained by adding a parity-check bit $x_{n+1}=x_1 \oplus x_2 \oplus \cdots \oplus x_n$ at the end of each codeword of $\mathcal{C}$. Let $d_{\min}'$ be the minimum distance of $\mathcal{C}'$. Which of the following is true?

To analyze the options, let's start with the properties of the original code \(\mathcal{C}\) and the new code \(\mathcal{C}'\).

1. **Understanding the original code \(\mathcal{C}\)**:
   - \(\mathcal{C}\) is a binary linear code with minimum distance \(d_{\min} = 4\). This means that the smallest Hamming distance between any two distinct codewords in \(\mathcal{C}\) is 4.

2. **Construction of the new code \(\mathcal{C}'\)**:
   - The new code \(\mathcal{C}'\) is formed by appending a parity-check bit to each codeword in \(\mathcal{C}\). The parity-check bit \(x_{n+1} = x_1 \oplus x_2 \oplus \cdots \oplus x_n\) ensures that the total number of 1's in the codeword (including the parity bit) is even.

3. **Impact on minimum distance**:
   - The minimum distance of a linear code is defined as the minimum Hamming distance between any two distinct codewords. When we append a parity bit to each codeword, we need to analyze how this affects the distances between codewords.

4. **Considering the cases**:
   - **Case 1**: If two codewords in \(\mathcal{C}\) differ in 4 positions, appending the parity bit will not change the fact that they differ in 4 positions (the parity bit is calculated based on the original codeword). Therefore, the minimum distance \(d_{\min}'\) remains at least 4.
   - **Case 2**: If two codewords differ in exactly 4 positions, and their parity bits (which are determined by their respective codewords) are different, the total distance would be 5 due to the appended parity bit. This situation can happen if the number of 1's in the original codewords is odd.
   - **Case 3**: If the two codewords differ in 4 positions and have the same parity bit, the distance remains 4.
   - **Case 4**: For codewords differing in more than 4 positions, it is clear that the appended parity bit will always contribute an additional bit, thus increasing the minimum distance further.

5. **Conclusion**:
   - The minimum distance \(d_{\min}'\) can be 4 (if the parity bits do not differ) or can be increased to 5 (if the parity bits differ). It can also increase further if the codewords differ in more than 4 bits. The actual minimum distance \(d_{\min}'\) could take values depending on the specific code \(\mathcal{C}\) and how the parity bit interacts with the differences in codewords.

Based on this analysis, we can conclude that the correct option is:

**4. \(d_{\min}'\) can take different values depending on the code \(\mathcal{C}\).**