Question:
Consider the following context-free grammar \(G\) (where \(\text{S}\) is the top-level symbol):

\(R_{01}: \text{S} \rightarrow \text{NP VP}\)
\(R_{02}: \text{NP} \rightarrow \text{NP0}\)
\(R_{03}: \text{NP} \rightarrow \text{Det NP0}\)
\(R_{04}: \text{NP0} \rightarrow \text{N}\)
\(R_{05}: \text{NP0} \rightarrow \text{Adj N}\)
\(R_{06}: \text{NP0} \rightarrow \text{NP0 PNP}\)
\(R_{07}: \text{VP} \rightarrow \text{V}\)
\(R_{08}: \text{VP} \rightarrow \text{V NP}\)
\(R_{09}: \text{VP} \rightarrow \text{V NP PNP}\)
\(R_{10}: \text{PNP} \rightarrow \text{Prep NP}\)

complemented by the lexicon \(L\):
a        : Det
blue     : Adj, N
drink    : N, V
drinks   : N, V
friends  : N
from     : Prep
gave     : V
letter   : N
my       : Det
neighbor : N
nice     : Adj, N
of       : Prep
postman  : N
ran      : V
the      : Det
to       : PrepIf the notation \(T(w)\) is used to refer to the rule \(T \rightarrow w\), which of the following correspond to valid derivations according to the grammar \(G\)?(Penalty for wrong ticks.)

To determine which of the provided derivations correspond to valid constructions according to the grammar \(G\), we will examine each derivation step by step, checking if they adhere to the rules specified in the grammar and the lexicon.

### Derivation 1:
1. **\(R_{01}\)**: \( \text{S} \rightarrow \text{NP VP} \)
2. **\(R_{08}\)**: \( \text{VP} \rightarrow \text{V NP} \)
3. **\(R_{02}\)**: \( \text{NP} \rightarrow \text{NP0} \)
4. **\(R_{04}\)**: \( \text{NP0} \rightarrow \text{N} \) (N = letter)
5. **\(N(\text{letter})\)**: This is valid since "letter" is defined as an N in the lexicon.
6. **\(V(\text{ran})\)**: This is valid since "ran" is defined as a V in the lexicon.
7. **\(R_{03}\)**: \( \text{NP} \rightarrow \text{Det NP0} \)
8. **\(Det(\text{the})\)**: This is valid since "the" is defined as a Det in the lexicon.
9. **\(R_{04}\)**: \( \text{NP0} \rightarrow \text{N} \) (N = drinks)
10. **\(N(\text{drinks})\)**: This is valid since "drinks" is defined as an N in the lexicon.

This derivation is valid according to the grammar.

### Derivation 2:
1. **\(R_{01}\)**: \( \text{S} \rightarrow \text{NP VP} \)
2. **\(R_{03}\)**: \( \text{NP} \rightarrow \text{Det NP0} \)
3. **\(Det(\text{a})\)**: This is valid since "a" is defined as a Det in the lexicon.
4. **\(R_{05}\)**: \( \text{NP0} \rightarrow \text{Adj N} \)
5. **\(Adj(\text{blue})\)**: This is valid since "blue" is defined as an Adj in the lexicon.
6. **\(N(\text{drink})\)**: This is valid since "drink" is defined as an N in the lexicon.
7. **\(R_{07}\)**: \( \text{VP} \rightarrow \text{V} \)
8. **\(V(\text{ran})\)**: This is valid since "ran" is defined as a V in the lexicon.

This derivation is also valid according to the grammar.

### Derivation 3:
1. **\(R_{01}\)**: \( \text{S} \rightarrow \text{NP VP} \)
2. **\(R_{02}\)**: \( \text{NP} \rightarrow \text{NP0} \)
3. **\(R_{04}\)**: \( \text{NP0} \rightarrow \text{N} \) (N = friends)
4. **\(N(\text{friends})\)**: This is valid since "friends" is defined as an N in the lexicon.
5. **\(R_{09}\)**: \( \text{VP} \rightarrow \text{V NP PNP} \)
6. **\(V(\text{gave})\)**: This is valid since "gave" is defined as a V in the lexicon.
7. **\(R_{02}\)**: \( \text{NP} \rightarrow \text{NP0} \)
8. **\(R_{04}\)**: \( \text{NP0} \rightarrow \text{N} \) (N = postman)
9. **\(N(\text{postman})\)**: This is valid since "postman" is defined as an N in the lexicon.

This derivation is valid according to the grammar.

### Conclusion:
All three derivations are valid according to the grammar \(G\):
- **Derivation 1**: Valid
- **Derivation 2**: Valid
- **Derivation 3**: Valid

Thus, each derivation corresponds to a valid construction according to the grammar \(G\).


Question: Tick the \emph{incorrect} assertion. The Diffie-Hellman key agreement protocol \ldots

To determine the incorrect assertion regarding the Diffie-Hellman key agreement protocol, we will analyze each of the provided options in detail.

1. **Assertion:** "allows two participants to set up a key so that they can communicate securely."
   - **Explanation:** The primary purpose of the Diffie-Hellman key agreement protocol is indeed to allow two parties to establish a shared secret key over an insecure channel. This key can then be used to encrypt their communication, thus ensuring its confidentiality. This assertion is correct.

2. **Assertion:** "requires the hardness of the Discrete Logarithm problem."
   - **Explanation:** The security of the Diffie-Hellman protocol relies on the difficulty of solving the Discrete Logarithm problem. In essence, even if an adversary knows the public keys exchanged between the two parties, calculating the shared secret key without knowledge of the private keys is computationally infeasible. Thus, this assertion is also correct.

3. **Assertion:** "uses ElGamal encryption in order to establish the key."
   - **Explanation:** The Diffie-Hellman key agreement protocol does not utilize ElGamal encryption for key establishment. Instead, it directly allows two parties to generate a shared secret without employing any encryption scheme like ElGamal. ElGamal encryption is a separate cryptographic scheme that can utilize the shared key generated by the Diffie-Hellman protocol for encrypting messages, but it is not a part of the Diffie-Hellman protocol itself. Therefore, this assertion is incorrect.

4. **Assertion:** "is easy to break when working on the group \( \mathbf{Z}_{n} \)."
   - **Explanation:** The Diffie-Hellman protocol can indeed be vulnerable if implemented over certain groups, especially if \( n \) is not chosen properly. However, when using a suitable group, particularly a group of large prime order or well-defined elliptic curves, the protocol is considered secure. Thus, this assertion can be somewhat misleading, as the security of Diffie-Hellman does depend on the choice of group, but it is not universally "easy to break."

Based on the analysis of the assertions presented, the assertion that is clearly incorrect is:

**"uses ElGamal encryption in order to establish the key."**

This statement misrepresents how the Diffie-Hellman protocol functions, making it the correct answer to the question.


Question: If we need to create a channel that protects confidentiality and we have at our disposal a channel that protects integrity and authenticity, we need to use

The best choice is option 1: symmetric key encryption, as it allows for both confidentiality and integrity when used properly. By encrypting the data with a shared secret key, we ensure that only authorized parties can access the information, while also maintaining its integrity through secure encryption methods.


Question:
Assume that some of your colleagues work on an AI-based image generation service, where a user enters a topic, and the AI generates a synthetic photo on that topic. They tell you the following about this service:
"Currently, the user types in the topic they want to see images for, and the client app sends a request to the server with the user ID and the indicated topic. The server generates an image, which takes a second or so, and sends it to the client app, which then requests another image on the same topic, and so on, until the app has received 9 images. It then displays these in a 3x3 grid. The user now looks at the 9 images and, if they see an inappropriate one, they click on a button that causes the app to send a review request to the server. Human moderators then process each report, and data scientists tweak the AI model to avoid generating images similar to the ones reported as inappropriate. Users then get a notification that their report was processed. The whole reporting process typically takes a day."
Now assume you can change the server's interface. Explain in 1-2 sentences an alternative way to make the app display the 9 images faster:

To make the app display the 9 images faster, we can modify the server's interface to allow for batch image generation, where the client app sends a single request for all 9 images at once rather than sending individual requests for each image. This approach reduces the overhead of multiple request-response cycles, enabling the server to generate all the images simultaneously and return them in one response, significantly decreasing the total waiting time for the user.

Final answer: Implement batch image generation by allowing the client app to request all 9 images in a single request, reducing the number of request-response cycles and speeding up the display time.


Question: Tick the \textbf{incorrect} assertion. Let $H:\left\{ 0,1 \right\}^*\rightarrow\left\{ 0,1 \right\}^n$ be a hash function.

To identify the incorrect assertion among the given statements, let's evaluate each option one by one.

1. **We can use \(H\) to design a commitment scheme.**
   This statement is correct. A commitment scheme allows one party to commit to a chosen value (or message) while keeping it hidden, with the ability to reveal the committed value later. Hash functions can be used in this context by hashing the message along with a random nonce (random value), creating a commitment that can be revealed later.

2. **We can use \(H\) to design a key derivation function.**
   This statement is also correct. A key derivation function (KDF) is used to derive one or more secret keys from a secret value (e.g., a password) in a secure manner. Hash functions can be employed in KDFs to produce cryptographically secure keys from input data.

3. **Finding \(x,y\in\left\{ 0,1 \right\}^*\) such that \(x\neq y\) and \(H(x) = H(y)\) can be done in \(O(2^{n/2})\) time.**
   This statement is correct and follows the birthday paradox reasoning. The probability of a collision (finding two distinct inputs that hash to the same output) in a hash function with an output of \(n\) bits can be expected to require approximately \(O(2^{n/2})\) evaluations of the hash function due to the birthday bound.

4. **Given \(x\in\left\{ 0,1 \right\}^*\), finding a \(y \in \left\{ 0,1 \right\}^*\) such that \(x\neq y\) and \(H(x) = H(y)\) can be done in \(O(2^{n/2})\) time.**
   This statement is incorrect as it suggests that a specific input \(x\) can lead to finding a collision in \(O(2^{n/2})\) time. In practice, while the expected time to find a collision for any two inputs is \(O(2^{n/2})\), given a specific \(x\), the process of finding a \(y\) such that \(H(x) = H(y)\) may not be achievable in that time frame due to the nature of the hash function; the search space could require more time, depending on the characteristics of the hash function and the specific structure of the input space.

In conclusion, the incorrect assertion is:

**Given \(x\in\left\{ 0,1 \right\}^*\), finding a \(y \in \left\{ 0,1 \right\}^*\) such that \(x\neq y\) and \(H(x) = H(y)\) can be done in \(O(2^{n/2})\) time.**


Question: The Advanced Encryption Standard (AES) is based on arithmetics on\dots

To determine which of the given options is correct regarding the arithmetic used in the Advanced Encryption Standard (AES), we can analyze each option:

1. **$\mathrm{GF}(256)$**: This refers to the Galois Field of order 256, which is indeed used in AES. AES operates on bytes (8 bits), and $\mathrm{GF}(256)$ is a finite field that consists of 256 elements. This is a valid option.

2. **$\mathbf{Z}_{2^8}^*$**: This notation typically refers to the multiplicative group of integers modulo \(2^8\) (which is 256), excluding zero. However, AES does not operate on this group; it uses finite field arithmetic instead. Therefore, this option can be eliminated.

3. **$\mathbf{Q}\setminus \{0\}$**: This represents the set of all non-zero rational numbers. AES does not use rational numbers for its operations; it relies on finite field arithmetic. Thus, this option can also be eliminated.

4. **$\mathbf{Z}_{2^8}$**: This notation refers to the integers modulo \(2^8\) (or 256), which includes both addition and multiplication modulo 256. However, AES specifically uses arithmetic in the finite field $\mathrm{GF}(256)$ rather than just integers modulo 256. Therefore, this option can be eliminated as well.

After analyzing all options, the only survivor that correctly describes the arithmetic used in AES is:

**Final answer:** a


Question: Does the following code compile? val x = 12 def foo(x: List[Int]): Int = x match \t case Nil => 0 \t case x :: xs => x

To analyze the multiple-choice question regarding whether the provided Scala code compiles, we will follow a structured approach as outlined in your instructions.

### Step 1: Identify Core Concepts and Principles

The question tests knowledge of Scala syntax, particularly:
- Variable declaration and initialization (`val`)
- Function definition and pattern matching in Scala
- Understanding of list data structures and their pattern matching

### Step 2: Analyze Each Option

**Option 1: Yes**

- **Evaluation**: This option suggests that the code compiles successfully.
- **Reasoning**:
  - The variable `x` is declared with `val x = 12`, which is valid Scala syntax. This creates an immutable variable `x` with an integer value of 12.
  - The function `foo` is defined to take a parameter `x` of type `List[Int]`. The function uses pattern matching to handle two cases: when the list is empty (`Nil`) and when it has at least one element (`x :: xs`).
  - The pattern matching syntax is correct. `case Nil => 0` correctly handles the empty list case, returning 0. The second case `case x :: xs => x` correctly matches a non-empty list, where `x` is the head of the list and `xs` is the tail.
  - There are no syntax errors or type mismatches in the code.
- **Conclusion**: This option is plausible, as the code appears to be syntactically correct.

**Option 2: No**

- **Evaluation**: This option suggests that the code does not compile.
- **Reasoning**:
  - The declaration of `val x = 12` is valid, but the potential issue arises from the naming conflict. The parameter `x` in the function `foo` shadows the outer variable `x`. While Scala allows shadowing, it can lead to confusion.
  - However, the code itself does not contain any syntax errors that would prevent it from compiling. The function `foo` can still be called with a `List[Int]`, and the pattern matching is correctly implemented.
  - The shadowing does not cause a compilation error; it merely means that within the scope of `foo`, the parameter `x` will refer to the list rather than the integer 12.
- **Conclusion**: This option is incorrect because the code compiles despite the shadowing.

### Step 3: Identify the Correct Answer

After analyzing both options, we conclude that **Option 1: Yes** is the correct answer. The code compiles successfully without any syntax errors, even though there is a variable shadowing issue.

### Step 4: Summarize Key Insights

The key insight from this question is the understanding of variable scope and shadowing in Scala. While shadowing can lead to confusion, it does not inherently cause compilation errors. The question tests the ability to recognize valid Scala syntax and the implications of variable naming within different scopes.

In summary, the code compiles, and the correct answer is **Yes**.


Question:
Consider the following toy corpus: the cat cut the hat
How many occurences do you have in total? (i.e. including repertitions)

To analyze the total occurrences of words in the toy corpus "the cat cut the hat," we can break down the reasoning into two branches: counting unique words and counting total words.

### Branch 1: Counting Unique Words
1. **Identify unique words:**
   - Words: "the," "cat," "cut," "hat"
   - Unique words: 4

2. **Count occurrences of each unique word:**
   - "the" appears 2 times
   - "cat" appears 1 time
   - "cut" appears 1 time
   - "hat" appears 1 time
   - Total occurrences = 2 + 1 + 1 + 1 = 5

### Branch 2: Counting Total Words
1. **Count total words in the sentence:**
   - The sentence has 5 words (the, cat, cut, the, hat)

2. **Consider repetitions in the total count:**
   - Since we are counting all occurrences, the total is still 5.

### Conclusion
Both branches lead to the same conclusion regarding total occurrences, but Branch 2 directly counts the total without needing to analyze unique words. Thus, the total occurrences of words in the corpus "the cat cut the hat" is **5**.


Question: Tick the \emph{wrong} assertion concerning 3G.

âThe integrity of 3G messages is well protected.â


Question: Implement a function that takes a list ls as argument, and returns a list of all the suffixes of ls. That is, given a list List(a,b,c,...) it returns List(List(a,b,c,...), List(b,c,...), List(c,...), List(...), ..., List()). Implement the function recursively using only Nil (empty), :: (cons) and pattern matching. def tails(ls: List[Int]): List[List[Int]] = ???

To approach the problem of implementing a function that returns all suffixes of a list using the Socratic method, I will first ask myself some guiding questions:

1. **What is a suffix of a list?**
   - A suffix of a list is any trailing segment of the list, including the list itself and the empty list. For example, for the list `List(a, b, c)`, the suffixes are `List(a, b, c)`, `List(b, c)`, `List(c)`, and `List()`.

2. **What are the constraints on the implementation?**
   - The function must be implemented recursively and can only use `Nil` (empty list), `::` (the cons operator), and pattern matching. This means that I cannot use built-in functions or constructs that would violate these constraints.

3. **How can I represent the recursive nature of finding suffixes?**
   - I can think of the list being processed in terms of its head (the first element) and its tail (the rest of the elements). Each recursive call should generate a new suffix based on the current head and the recursive call on the tail.

4. **What will the base case for my recursion be?**
   - The base case will be when the list is empty (`Nil`). In this case, the only suffix is also an empty list.

5. **How do I combine the results of the recursive calls?**
   - For each recursive call, I should prepend the current list (or its tail) to the list of suffixes returned by the recursive call on its tail.

Now that I have reflected on these guiding questions, I can formulate my answer:

### Final Answer:
```scala
def tails(ls: List[Int]): List[List[Int]] = ls match {
  case Nil => List(Nil) // Base case: the only suffix of an empty list is itself.
  case head :: tail => ls :: tails(tail) // Prepend the current list to the suffixes of the tail.
}
```

### Justification:
- In this implementation, I use pattern matching to differentiate between the base case (an empty list) and the recursive case (a non-empty list).
- When `ls` is empty (`Nil`), I return a list containing just the empty list (`List(Nil)`).
- When `ls` has elements (`head :: tail`), I construct a new list that includes `ls` itself followed by all suffixes of `tail`. This effectively builds up all suffixes recursively until we reach the empty list case.
- The use of recursion adheres to the constraints, ensuring that only `Nil`, `::`, and pattern matching are utilized.


Question:
A major specificity of natural languages is that they are inherently implicit and ambiguous. How should this be taken into account in the NLP perspective?
(penalty for wrong ticks)

by designing NLP algorithms and data structures able to efficiently cope with very ambiguous representations


Question: A passive adversary\dots

The best answer is: "can only listen to the communications."


Question: According to the Kerckhoffs Principle:

The correct answer is that "the security of the cryptosystem should not rely on the secrecy of the cryptosystem itself."

Kerckhoffs' Principle essentially states that a cryptosystem should remain secure even if everything about the system, except for the key, is public knowledge. This means that the design and implementation details can be shared openly, and the security should depend solely on the strength of the key used for encryption, not on keeping the system's inner workings secret. This approach allows for greater scrutiny and improvement of the cryptosystem, as well as easier implementation and maintenance. If the security relied on keeping the system secret, any exposure of its details could compromise its effectiveness.


Question: Let $\mathcal C_1$ be a linear code over $\mathbb F_3^n$, and let $\mathcal C_2$ be a linear code over $\mathbb F_2^n$. True or false: $\mathcal C_1 \cap \mathcal C_2$ is necessarily a linear code over $\mathbb F_3^n$.

To clarify the statement regarding the intersection of two linear codes, let's analyze the problem again step by step, incorporating the feedback provided.

### Step 1: Definition of Linear Codes
A linear code is defined as a subspace of a vector space over a finite field. Here, we have two linear codes: $\mathcal C_1$ is a linear code over the field $\mathbb F_3$, and $\mathcal C_2$ is a linear code over the field $\mathbb F_2$. This means that the vectors in $\mathcal C_1$ contain elements from $\mathbb F_3$, while the vectors in $\mathcal C_2$ consist of elements from $\mathbb F_2$.

### Step 2: Identifying the Intersection
The intersection $\mathcal C_1 \cap \mathcal C_2$ consists of vectors that are present in both codes. For a vector to be in $\mathcal C_1$, its entries must be valid in $\mathbb F_3$, and for it to be in $\mathcal C_2$, its entries must be valid in $\mathbb F_2$.

### Step 3: Analyzing the Implications
A key point to note is that the only element common to both $\mathbb F_3$ and $\mathbb F_2$ is the zero vector. While it is true that the zero vector is a vector in both codes, the intersection $\mathcal C_1 \cap \mathcal C_2$ does not form a linear code over $\mathbb F_3$. This is because a linear code must be a subspace over the specific field it is defined over.

In this case, while the intersection contains the zero vector, it does not contain any other vectors from $\mathcal C_2$ that are appropriate to be considered in the context of $\mathbb F_3$. Moreover, since the vectors from $\mathcal C_2$ are defined over $\mathbb F_2$, they cannot be treated as valid vectors in the context of $\mathbb F_3$ (except for the zero vector).

### Step 4: Conclusion
Therefore, the intersection $\mathcal C_1 \cap \mathcal C_2$, while containing the zero vector, does not satisfy the criteria to be classified as a linear code over $\mathbb F_3$. It is simply a trivial intersection, indicating that the only common element is the zero vector, but this does not imply the existence of a linear code over $\mathbb F_3$.

Thus, the statement "the intersection $\mathcal C_1 \cap \mathcal C_2$ is necessarily a linear code over $\mathbb F_3^n$" is **False**.

### Final Answer
**Option 1: False**


Question:
An expression is referentially transparent if it always returns the same value, no matter
the global state of the program. A referentially transparent expression can be replaced by its value without
changing the result of the program.
Say we have a value representing a class of students and their GPAs. Given the following defintions:
1 case class Student(gpa: Double)
2
3 def count(c: List[Student], student: Student): Double =
4 c.filter(s => s == student).size
5
6 val students = List(
7 Student(1.0), Student(2.0), Student(3.0),
8 Student(4.0), Student(5.0), Student(6.0)
9 )
And the expression e:
1 count(students, Student(6.0))
If we change our definitions to:
1 class Student2(var gpa: Double, var name: String = "*")
2
3 def innerCount(course: List[Student2], student: Student2): Double =
4 course.filter(s => s == student).size
5
6 def count2(course: List[Student2], student: Student2): Double =
7 innerCount(course.map(s => new Student2(student.gpa, student.name)),
student)
8
9 val students2 = List(
10 Student2(1.0, "Ana"), Student2(2.0, "Ben"), Student2(3.0, "Cal"),
11 Student2(4.0, "Dre"), Student2(5.0, "Egg"), Student2(6.0, "Fra")
12 )
And our expression to: e2:
1 count2(students2, Student2(6.0, "*"))
Is the expression e2 referentially transparent?

Yes, the expression `e2` is referentially transparent.

To explain why, we need to examine the definition of referential transparency, which states that an expression can be replaced with its value without altering the program's behavior. In the case of `count2`, we are passing a list of `Student2` objects and a specific `Student2` instance to the function.

The function `count2` operates by mapping the input list to create new instances of `Student2` with the same GPA and name as the passed `student`. This mapping does not depend on any external mutable state; it solely relies on the input parameters provided. Consequently, for any identical input values, `count2` will consistently yield the same result.

Thus, since the output of `count2(students2, Student2(6.0, "*"))` remains constant regardless of any external factors, we can conclude that the expression is indeed referentially transparent.


Question:
Remember that monoids can be represented by the following type class:
1 trait SemiGroup[T]:
2 extension (x: T) def combine (y: T): T
3
4 trait Monoid[T] extends SemiGroup[T]:
5 def unit: T
Additionally the three following laws should hold for all Monoid[M] and all a, b, c: M:
(Associativity) a.combine(b).combine(c) === a.combine(b.combine(c))
(Left unit) unit.combine(a) === a
(Right unit) a.combine(unit) === a
Consider the following implementation of Monoid for Boolean:
1 given Or: Monoid[Boolean] with
2 extension (x: Boolean) def combine (y: Boolean): Boolean = x || y
3 def unit: Boolean = true
Which of the three monoid laws does it fullfil?
None of them
Only Associativity
Only Left unit
Only Right unit
Only Associativity and Left unit
Only Associativity and Right unit
Only Left unit and Right unit
All of them

To determine which of the three monoid laws the given implementation of `Monoid` for `Boolean` fulfills, we start by reviewing the implementation and the definitions of each law.

1. **Understanding the Implementation of Monoid for Boolean:**
   The provided implementation is as follows:

   ```scala
   given Or: Monoid[Boolean] with
   extension (x: Boolean) def combine (y: Boolean): Boolean = x || y
   def unit: Boolean = true
   ```

   This means:
   - The `combine` function is defined using the logical OR (`||`) operation which takes two Boolean values and returns `true` if at least one of them is `true`.
   - The `unit` value is defined as `true`, which acts as the identity element in this monoid.

2. **Reviewing the Monoid Laws:**
   We will now check each of the three monoid laws to see if they hold for the provided implementation.

   - **Associativity:**
     We need to check if the following holds for all Boolean values `a`, `b`, and `c`:
     \[
     a \text{.combine}(b).combine(c) \equiv a \text{.combine}(b \text{.combine}(c))
     \]
     Expanding both sides:
     - Left-hand side:
       \[
       a \text{.combine}(b).combine(c) = (a || b) || c
       \]
     - Right-hand side:
       \[
       a \text{.combine}(b \text{.combine}(c)) = a || (b || c)
       \]
     The logical OR operation is associative, meaning:
     \[
     (a || b) || c \equiv a || (b || c)
     \]
     Thus, the **associativity law** is satisfied.

   - **Left unit:**
     We check if the following holds for all Boolean values `a`:
     \[
     \text{unit}.combine(a) \equiv a
     \]
     Substituting the unit value:
     \[
     \text{true}.combine(a) = \text{true} || a
     \]
     The logical OR operation yields `true` regardless of the value of `a`. Thus:
     \[
     \text{true} || a \equiv \text{true} \neq a \text{ (if a is false)}
     \]
     Therefore, the **left unit law** is not satisfied.

   - **Right unit:**
     We check if the following holds for all Boolean values `a`:
     \[
     a.combine(\text{unit}) \equiv a
     \]
     Substituting the unit value:
     \[
     a.combine(\text{true}) = a || \text{true}
     \]
     As with the left unit, the logical OR operation yields `true`, hence:
     \[
     a || \text{true} \equiv \text{true} \neq a \text{ (if a is false)}
     \]
     Thus, the **right unit law** is also not satisfied.

3. **Conclusion:**
   After evaluating all three monoid laws:
   - The **associativity law** is satisfied.
   - The **left unit law** is not satisfied.
   - The **right unit law** is not satisfied.

Thus, the correct answer to the question is: **Only Associativity**.


Question: Check the correct statement concerning a standard-array-based decoder.

To evaluate the provided statements regarding a standard-array-based decoder, let's analyze each option step by step.

### A) For the same input, the decoder output depends on the choice of the coset leader.

**Analysis:**
In coding theory, a standard array is a systematic way of organizing codewords and their corresponding syndromes. Each coset in the standard array is represented by a coset leader, which is the smallest-weight codeword in that coset. The decoder's output for a given input does depend on the choice of the coset leader because the decoder will choose the coset leader associated with the syndrome of the received vector. If a different coset leader were chosen, it could lead to a different decoded output, especially if the chosen leader does not correspond to the minimum distance correction. Thus, this statement is **true**.

### B) The syndrome of a specific coset depends on the choice of the coset leader.

**Analysis:**
The syndrome of a received vector is calculated using the parity-check matrix associated with the code. The syndrome is a function of the error pattern and not directly influenced by the selection of the coset leader. Instead, the syndrome indicates which coset the received vector falls into, while the coset leader is used to determine the actual correction. Therefore, this statement is **false**; the syndrome does not depend on the choice of coset leader.

### C) All the other statements are incorrect.

**Analysis:**
From our analysis, we found that statement A is true while statement B is false. Therefore, statement C, which claims that all other statements are incorrect, is also **false**.

### Conclusion:
The correct statement concerning a standard-array-based decoder is **A)** For the same input, the decoder output depends on the choice of the coset leader.


Question: When using linear regression, which techniques improve your result? (One or multiple answers)

To improve the results of linear regression, we must consider how to effectively capture relationships in the data. Let's analyze the provided options:

1. **Linear regression does not allow polynomial features**: This statement is misleading. While the basic form of linear regression models a linear relationship, we can incorporate polynomial features by transforming the original features. For instance, if we have a feature $x$, we can add $x^2$, $x^3$, etc., as additional features. This allows the model to fit non-linear relationships while still being a linear regression model in terms of its coefficients.

2. **Polynomial combination of features**: This option is correct. By including polynomial combinations of features, we can enhance the model's ability to capture non-linear patterns in the data. For example, if we have two features $x_1$ and $x_2$, adding $x_1^2$, $x_2^2$, and $x_1 x_2$ as new features can significantly improve the model's performance.

3. **Adding new features that are non-linear combinations of existing features**: This option is also valid. While the linear regression model itself predicts a linear combination of features, we can create new features through non-linear transformations (such as logarithmic or exponential functions) or interactions of existing features. This enriches the feature space and can lead to better predictions.

4. **Because the linear nature needs to be preserved, non-linear combinations of features are not allowed**: This statement is incorrect. While linear regression maintains a linear form in its prediction equation, we are free to use non-linear transformations of the input features. The linearity refers to the relationship between the dependent variable and the coefficients, not the nature of the features themselves.

**Selected answers**:
- Polynomial combination of features
- Adding new features that are non-linear combinations of existing features

In summary, incorporating polynomial features and non-linear combinations of existing features can greatly enhance the performance of linear regression models by allowing them to capture more complex relationships in the data.


Question: The Shannon theorem states that perfect secrecy implies...

Apologies for the confusion. The correct answer is:

\( H(Y) \geq H(X) \)

This reflects the Shannon theorem's assertion regarding perfect secrecy. Thank you for your patience!


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

    \begin{lstlisting}[language=C,style=c]
int sum_array(int *arr, size_t len) {
    // Return sum of array elements
    int result = 0;
    for (size_t i = 0; i <= len; i++) {
        result += arr[i];
    }
    return result;
}
    \end{lstlisting}

Correct answer -> UBSan

Explanation: The code snippet contains a bug where the loop iterates with the condition `i <= len`, which causes an out-of-bounds access when `i` equals `len`. This results in accessing `arr[len]`, which is not a valid index for the array, leading to undefined behavior. UndefinedBehaviorSanitizer (UBSan) is specifically designed to catch such instances of undefined behavior in C/C++ programs, making it the appropriate choice for detecting this bug. AddressSanitizer (ASan) could potentially catch some memory issues but is more focused on memory corruption rather than undefined behavior, while MemorySanitizer (MemSan) targets uninitialized memory reads, and ThreadSanitizer (TSan) is for data races in multithreaded programs.


Question: Professor Ueli von Gruy\`{e}res worked hard last year to calculate the yearly cheese consumption of each individual in Switzerland. Specifically, let $U$ be the set of all persons in Switzerland. For each person $i\in U$, Ueli calculated the amount $w_i \in \mathbb{R}_{\geq 0}$ (in grams) of the yearly cheese consumption of  person $i$.  However, to help Coop and Migros in their  supply-chain management, he needs to calculate the total cheese consumption of those persons that prefer fondue over raclette. That is, if we let $F \subseteq U$ be those that prefer fondue over raclette, then Ueli wants to calculate \begin{align*} W_F = \sum_{i\in F} w_i\,. \end{align*} The issue is that Ueli does not know the set $F$ and he does not have the time or energy to ask the preferences of all persons. He therefore designs two estimators that only ask a single person: \begin{description} \item[Estimator $\Alg_1$:] Let $W = \sum_{i\in U}w_i$. Sample person $i$ with probability $\frac{w_i}{W}$ and output $W$  if $i$ prefers fondue and $0$  otherwise. \item[Estimator $\Alg_2$:] Sample person $i$ with probability $\frac{1}{|U|}$ and output $|U| \cdot w_i$  if $i$ prefers fondue and $0$  otherwise. \end{description} Let $X_1$ and $X_2$ be the random outputs of $\Alg_1$ and $\Alg_2$, respectively. Ueli has shown that $\Alg_1$ and $\Alg_2$ are unbiased estimators and he has also bounded their variances: \begin{align*} \E[X_1] = \E[X_2] = W_F, \qquad \Var[X_1]  \leq   W^2  \qquad \mbox{and} \qquad \Var[X_2]  \leq  |U| \sum_{i\in U} w_i^2\,. \end{align*} However, Ueli is now stuck because the variances are too high to give any good guarantees for the two estimators. We are  therefore going to help Ueli by designing a new estimator with good guarantees while still asking the preferences of relatively few persons. For a fixed small parameter $\epsilon >0$, your task is to design and analyze an estimator that outputs a random value $Y$ with the following guarantee: \begin{align} \label{eq:guarantee} \Pr[|Y - W_F| \geq \epsilon W] \leq 1/3\,. \end{align} Your estimator should ask at most $3/\epsilon^2$ persons about their preferences.  \\ {\em (In this problem you are asked to (i) design an estimator that asks  at most $3/\epsilon^2$ persons about their preferences and (ii) prove that it satisfies the guarantee~\eqref{eq:guarantee}. Recall that you are allowed to refer to material covered in the lecture notes.)}

To design an estimator ( Y ) with the required properties, we can follow these steps:

Step 1: Define the Estimator ( Y )
We will use a variant of the estimator ( \Alg_2 ), where we sample multiple individuals and use their responses to improve the variance of our estimate.

Estimator ( Y ):

Sample three individuals ( i_1, i_2, i_3 ) uniformly at random from the set ( U ).
For each individual, ask if they prefer fondue over raclette.
Let ( X_j = |U| \cdot w_{j} ) if person ( i_j ) prefers fondue, and ( 0 ) otherwise.
Define the estimator as: [ Y = \frac{X_1 + X_2 + X_3}{3} ]
Step 2: Prove that ( Y ) is Unbiased
To prove that ( Y ) is unbiased, we need to show that ( \E[Y] = W_F ).

Each ( X_j ) is ( |U| \cdot w_k ) for an individual ( k \in U ) that prefers fondue with probability ( p_F ) (the probability that a randomly chosen person prefers fondue), so we can write: [ \E[X_j] = \frac{1}{|U|} \sum_{k \in U} |U| \cdot w_k \cdot \Pr[i_j \text{ prefers fondue}] = \frac{W_F}{|U|} + W \cdot (1 - p_F) ]
Exploiting the indicator property: [ \E[Y] = \frac{\E[X_1] + \E[X_2] + \E[X_3]}{3} = \frac{3 \cdot \E[X_1]}{3} = \E[X_1] = W_F ] Hence, ( Y ) is unbiased: [ \E[Y] = W_F ]

Step 3: Show the Probability Bound Using Variance Analysis and Chebyshev's Inequality
Variance Calculation:
Using the independence of the samples, we have: [ \Var[Y] = \frac{1}{9} (\Var[X_1] + \Var[X_2] + \Var[X_3]) = \frac{1}{9} (3 \Var[X]) = \frac{\Var[X]}{3} ] Where ( X = |U| \cdot w_k ) if the sample prefers fondue.

From previous results, we have: [ \Var[X] \leq |U| \sum_{i \in U} w_i^2 ]

Using Chebyshev's Inequality:
To apply Chebyshev's inequality, let ( \epsilon W ) be the tolerance level: [ \Pr[|Y - W_F| \geq \epsilon W] \leq \frac{\Var[Y]}{(\epsilon W)^2} ]

Now substituting for ( \Var[Y] ): [ \Pr[|Y - W_F| \geq \epsilon W] \leq \frac{\Var[X]}{3 (\epsilon W)^2} ]

To ensure that this probability is at most ( \frac{1}{3} ): [ \frac{\Var[X]}{3 (\epsilon W)^2} \leq \frac{1}{3} \implies \Var[X] \leq \epsilon W^2 ]

Sample Size:
We know ( \Var[X] \leq |U| \sum_{i \in U} w_i^2 ). If we work under the assumption that the variance does not exceed a suitable bound, we can ensure performance against our ( \epsilon ) tolerance.

Step 4: Total Preference Queries
The total preference queries made by the estimator ( Y ) require asking 3 individuals: [ \text{Total Queries} = 3 \leq \frac{3}{\epsilon^2}\quad \text{ (for } \epsilon > 0\text{)} ]

Conclusion
Thus, we have an estimator ( Y ) that is unbiased, satisfies the probability bound using variance analysis and Chebyshev's inequality, while asking at most ( 3/\epsilon^2 ) persons about their preferences.

This construction gives us a robust estimator for estimating the total cheese consumption of fondue lovers with the desired accuracy guarantee.


Question: Which of the following statements is correct concerning the use of Pearson’s Correlation for user- based collaborative filtering?

**Correct answer**: It measures whether different users have similar preferences for the same items.

**Explanation**: Pearsonâs Correlation is a statistical measure used in user-based collaborative filtering to assess the degree to which two users have similar rating patterns for the same items. A high Pearson correlation coefficient indicates that the users tend to rate items similarly, which is essential for making recommendations based on the preferences of users who are similar to the target user. The other options either misinterpret the purpose of Pearson's Correlation or describe unrelated metrics.


Question: Tick the \textbf{false} assertion. A cipher with a good decorrelation of order 2 protects against \ldots

To identify the false assertion among the given options regarding a cipher with a good decorrelation of order 2, let's analyze each option step by step.

1. **Option 1: \ldots non-adaptive distinguishers limited to two queries.**
   - A good decorrelation of order 2 means that any non-adaptive distinguishers that use two queries will have a reduced probability of successfully distinguishing between the ciphered data and random data. Therefore, this assertion is likely true.

2. **Option 2: \ldots unbounded attacks.**
   - Unbounded attacks refer to attacks that can take any number of queries and have no limitations on the amount of data processed. A good decorrelation of order 2 does not inherently protect against unbounded attacks, as these attacks could involve an unlimited number of queries. Thus, this assertion is likely false.

3. **Option 3: \ldots differential cryptanalysis.**
   - Differential cryptanalysis exploits the differences in input pairs to analyze the output differences. While a cipher with good decorrelation of order 2 may reduce the effectiveness of differential cryptanalysis, it does not guarantee complete protection against it. However, it is still reasonable to assume that it would provide some level of defense against this type of attack. Therefore, this assertion is likely true.

4. **Option 4: \ldots linear cryptanalysis.**
   - Linear cryptanalysis relies on linear approximations of the cipher's operations. A good decorrelation of order 2 can also help mitigate linear attacks by making it more difficult for an attacker to find valid linear approximations. Thus, this assertion is likely true as well.

Upon reflection, the key points of analysis indicate that:

- Option 1 is true (protection against limited non-adaptive distinguishers).
- Option 2 is false (a good decorrelation of order 2 does not protect against unbounded attacks).
- Options 3 and 4 are true (some level of protection against differential and linear cryptanalysis).

Therefore, the best choice for the false assertion is:

**Option 2: \ldots unbounded attacks.**

This is because a good decorrelation of order 2 does not effectively guard against attacks that can query the cipher an unlimited number of times.


Question: Which of the following statements about mitigations are correct?

To evaluate the correctness of each statement regarding mitigations, we will analyze them individually based on established knowledge in software security.

**Statement A:** Control-Flow Integrity (CFI) can efficiently protect the forward edge but, when using target sets, is limited on the backward edge.

- **Analysis:** CFI is indeed designed to ensure that control flow transfers (like function calls and returns) only go to valid destinations. It typically provides strong protection for forward edges (calls) by checking that control flow transfers are to valid targets. However, backward edges (returns) can be less effectively managed, especially if target sets are used, because it may allow legitimate returns to be incorrectly restricted. Thus, this statement is accurate.

**Statement B:** Code-Pointer Integrity (specifically the implementation described in the slides) uses a separate stack to protect code pointers.

- **Analysis:** Code-Pointer Integrity aims to protect code pointers from being manipulated by attackers. One common implementation involves using a separate storage mechanism (like a shadow stack) to maintain integrity for these pointers. However, without specific context from the slides, it's hard to verify the precise implementation referenced. Generally, this statement is reasonable but cannot be definitively judged correct without further details from the slides.

**Statement C:** Shadow stacks can be implemented in software with zero overhead.

- **Analysis:** Shadow stacks are intended to provide an additional layer of security by maintaining a separate stack that records the return addresses. However, implementing shadow stacks inherently incurs some overhead, either in terms of performance (additional memory accesses, stack manipulation) or resource usage. Therefore, this statement is incorrect.

**Statement D:** Safe stacks protect against corruption of all data on the stack.

- **Analysis:** Safe stacks specifically target the protection of return addresses on the stack, not all data. They help prevent certain types of attacks, like stack buffer overflows, but do not provide comprehensive protection for all stack data, such as local variables or other data structures. Hence, this statement is also incorrect.

Based on the analysis, the correct answer is:

- Statement A is correct.
- Statement B may be correct but requires context.
- Statement C is incorrect.
- Statement D is incorrect.

Thus, the most accurate statement is A.

Answer: A


Question: Write modular code (i.e., a function) to divide your training data into 𝑁 folds and perform cross-validation. For each possible combination of the two hyperparameters (see below for the range of values that you should try for each hyperparameter), train your model in a cross-validation setup with 𝑁=20 folds.

**Answer:**

To implement a modular function for performing cross-validation on a dataset with different hyperparameter combinations, we can utilize the provided `cross_val` function structure. This function takes in features `X`, labels `y`, a dictionary of hyperparameters `params`, the number of folds `N` (defaulted to 20), and the number of parallel jobs `n_jobs` for efficiency.

The first step is to generate a list of all possible combinations of hyperparameter values using `itertools.product`. This allows us to explore the entire search space defined by the hyperparameters.

Next, we create a K-Fold cross-validator using the `KFold` class from scikit-learn, which will split the data into `N`


Question:
Why is natural language processing difficult?
Select all that apply.A penalty will be applied for wrong answers.

Natural language processing (NLP) is difficult for several reasons:

1. **Lack of linguistic competence and resources**: Many languages and dialects lack comprehensive linguistic resources, making it challenging to develop NLP models.

3. **High dimensionality and sparseness of data**: Language data is often high-dimensional and sparse, complicating the modeling process and requiring advanced techniques to manage.

4. **Subjectivity of annotators**: Different annotators may interpret language nuances differently, leading to inconsistencies in labeled data, which can affect model performance.

Therefore, the correct options are **1, 3, and 4**.


Question:
Give an example of an exception whose precise implementation is
  arguably irrelevant in practice.

To address the question of providing an example of an exception whose precise implementation is arguably irrelevant in practice, we will follow a structured approach as outlined in your request.

### 1. Key Concepts and Principles

**Exceptions in Programming**: Exceptions are events that disrupt the normal flow of a program's execution. They can be caused by various issues such as runtime errors (e.g., division by zero, file not found) and are typically handled using try-catch blocks.

**Irrelevance of Implementation**: The notion of irrelevance in this context suggests that the specific details of how an exception is implemented or handled do not significantly affect the overall behavior or outcome of the program.

### 2. Theoretical Framework

To analyze exceptions, we can consider:
- **Types of Exceptions**: Checked vs. unchecked exceptions, custom exceptions, etc.
- **Exception Handling Mechanisms**: Try-catch blocks, finally clauses, and propagation of exceptions.
- **Practical Impact**: The effect of exception handling on program performance, readability, and maintainability.

### 3. Step-by-Step Solution

#### Step 1: Identify an Example

A common example of an exception whose precise implementation is arguably irrelevant is the **NullPointerException** in Java. This exception occurs when the JVM attempts to access an object or call a method on a null reference.

#### Step 2: Analyze the Implementation

In Java, a NullPointerException can be thrown in various scenarios, such as:
- Accessing a method on a null object.
- Accessing an array element with a null reference.
- Attempting to use an object that has not been initialized.

The implementation of this exception is standardized in the Java Virtual Machine (JVM), meaning that the underlying cause is consistently handled across different Java implementations.

#### Step 3: Discuss Practical Irrelevance

In practice, the precise implementation details of how a NullPointerException is thrown or caught are often irrelevant for the following reasons:
- **Commonality**: NullPointerExceptions are frequent in Java applications, leading developers to expect and handle them generically.
- **Generic Handling**: Developers often use a catch-all approach (e.g., catching `Exception` or using a generic error handler) which does not differentiate between types of exceptions, including NullPointerExceptions.
- **Debugging**: The stack trace provided by the exception is usually sufficient for debugging, making the specific implementation less critical.

#### Step 4: Assumptions and Constraints

- **Assumption**: The discussion assumes a typical Java development environment where NullPointerExceptions are common.
- **Constraint**: The analysis is limited to the context of Java and similar languages that exhibit similar exception handling behavior.

### 4. Key Insights

- The NullPointerException serves as a practical example of an exception whose specific implementation details do not significantly impact the overall program behavior.
- Developers often rely on generic exception handling strategies that abstract away the specifics of individual exceptions.

### 5. Alternative Approaches

An alternative approach could involve examining other exceptions, such as:
- **ArithmeticException**: This occurs during illegal arithmetic operations (e.g., division by zero). Similar to NullPointerException, its implementation may not significantly alter program flow if handled generically.
- **FileNotFoundException**: While it can be critical in certain contexts, in many applications, the handling may be so generic that the specific implementation details become irrelevant.

### Conclusion

In summary, the NullPointerException in Java exemplifies an exception whose precise implementation is arguably irrelevant in practice. The commonality of its occurrence and the generic handling strategies employed by developers contribute to this irrelevance. This analysis highlights the importance of understanding exceptions not just in terms of their technical details but also in their practical implications in software development.


Question:
One of your colleagues has recently taken over responsibility for a legacy codebase, a library currently used by some of your customers. Before making functional changes, your colleague found a bug caused by incorrect use of the following method in the codebase:

public class User {
    /** Indicates whether the user’s browser, if any, has JavaScript enabled. */
    public boolean hasJavascriptEnabled() { … }

    // … other methods, such as getName(), getAge(), ...
}

Your colleague believes that this is a bad API. Explain in 1 sentence why that is indeed the case.

The API is problematic because the method name "hasJavascriptEnabled" suggests a boolean return type indicating the presence of JavaScript support, which creates ambiguity and may lead to misinterpretation of the method's purpose, potentially causing incorrect usage in the codebase.


Question:
If A={a} and B={b}, select all strings that belongs to (A ⊗ B)+
A penalty will be applied for any wrong answers selected.

The correct strings that belong to (A ⊗ B)+ are (aaa,bbb) and (a,b) because these combinations follow the definition of the operation, which involves pairing each element from A with each element from B in all possible ways. Thus, both of these pairs represent valid concatenated forms of the elements in A and B.


Question: Would VLIW processors benefit from a Load Store Queue?

Okay, so here we go. Trying to tackle this question about VLIW processors and whether they would benefit from a Load Store Queue (LSQ) or not.

First off, let's break down what VLIW is. VLIW stands for Very Long Instruction Word. Basically, these processors execute multiple operations in a single instruction by packing them together. So, they rely heavily on the compiler to schedule instructions properly, which is a bit different from more dynamic architectures like out-of-order processors.

Now, the Load Store Queue (LSQ) is a structure used in processors to manage loads and stores effectively. It helps to keep track of memory operations and can play a crucial role in handling memory access, especially in out-of-order execution. In a typical LSQ, you can have loads and stores being handled separately, and it can help to resolve any conflicts that might arise from memory operations.

So, would VLIW processors benefit from having a Load Store Queue?

1. **Instruction Scheduling**: VLIW processors depend on the compiler to schedule instructions. If an LSQ is implemented, it might allow for better optimization of memory accesses, but it also adds complexity. The compiler can schedule loads and stores more efficiently with an LSQ, so maybe yes, it could benefit.

2. **Memory Access**: VLIW architectures can struggle with memory access delays because they process multiple instructions at once. An LSQ could help by allowing loads to be executed even if there’s a store that hasn’t completed yet, effectively hiding latencies. So, that's another point for the affirmative side.

3. **Complexity vs. Performance**: On the other hand, adding an LSQ could make the VLIW design more complicated. VLIW is about simplicity in hardware, and introducing an LSQ might counter that. The additional hardware and logic required to manage an LSQ could offset the benefits gained from better memory access management.

4. **Compiler Support**: The effectiveness of the LSQ in a VLIW context would heavily depend on the compiler's ability to utilize it. If the compiler isn’t designed to take advantage of an LSQ, then it might not really matter.

5. **Use Cases**: If the workload has lots of memory operations and dependencies, the LSQ could shine. In contrast, workloads that are more compute-bound might not see as much benefit.

So in summary, yes, VLIW processors could potentially benefit from a Load Store Queue, particularly for optimizing memory access and reducing latency. However, this comes with trade-offs in complexity and the necessity for the compiler to effectively utilize the LSQ. The context of the application and workload would ultimately determine the level of benefit gained from implementing an LSQ in a VLIW processor.


Question: Which problem in communication is \emph{not} treated by cryptography?

The best answer to the question "Which problem in communication is *not* treated by cryptography?" is:

**4. data transmission.**

### Explanation:

Cryptography primarily focuses on ensuring the security of data during communication through three key aspects:

1. **Confidentiality**: Cryptography protects sensitive information from unauthorized access using techniques like encryption, which makes data unreadable to those without the appropriate keys.

2. **Integrity**: It ensures that the data has not been altered during transmission. Cryptographic methods such as hashing and digital signatures verify that the received data is the same as what was sent.

3. **Authentication**: Cryptography verifies the identities of the parties involved in communication, often using digital certificates and public key infrastructure (PKI).

On the other hand, **data transmission** refers to the actual process of sending data from one point to another. While cryptography can be used to secure data during transmission, it does not address the mechanics of how data is physically moved across networks or the protocols involved in that process. Therefore, data transmission is the aspect that is not directly treated by cryptography.


Question: Review the notion of depth seen in the lecture. What does it represent?  Below is a formula for the depth of a divide and conquer algorithm working on an array segment of size $L$, as a function of $L$. The values $c$, $d$ and $T$ are constants. We assume that $L>0$ and $T>0$.  $$ D(L) = \begin{cases} c \cdot L &\text{if}\ L \leq T \\ \text{max}\left( D\left(\left\lfloor \frac L2 \right\rfloor \right), D\left(L - \left\lfloor \frac L2 \right\rfloor \right)\right) + d &\text{otherwise} \end{cases} $$  Below the threshold T, the algorithm proceeds sequentially and takes time c to process each single element. Above the threshold, the algorithm is applied recursively over the two halves of the array. The results are then merged using an operation that takes d units of time. Is it the case that for all $1 \leq L_1 \leq L_2$ we have $D(L_1) \leq D(L_2)$?  If it is the case, prove the property by induction on $L$. If it is not the case, give a counterexample showing values of $L_1$, $L_2$, $c$, and $d$for which the property does not hold.

---

To analyze the depth function \( D(L) \) and determine whether it is monotonic (i.e., whether \( D(L_1) \leq D(L_2) \) for all \( 1 \leq L_1 \leq L_2 \)), we need to carefully examine the recursive definition of \( D(L) \).

### Step 1: Understanding the Function

The function is defined piecewise:

1. For \( L \leq T \):
   \[
   D(L) = c \cdot L
   \]
   In this case, the depth grows linearly with \( L \) since we process each element sequentially.

2. For \( L > T \):
   \[
   D(L) = \max\left(D\left(\left\lfloor \frac{L}{2} \right\rfloor\right), D\left(L - \left\lfloor \frac{L}{2} \right\rfloor\right)\right) + d
   \]
   Here, the depth is determined by the maximum depth of the two halves of the array plus a constant time \( d \) for merging.

### Step 2: Exploring Monotonicity

To prove or disprove the property \( D(L_1) \leq D(L_2) \) for \( L_1 < L_2 \), we can look for a counterexample.

### Step 3: Finding a Counterexample

Let's consider specific values:
- Let \( L_1 = 10 \)
- Let \( L_2 = 12 \)
- Let \( T = 11 \)
- Let \( c = 1 \)
- Let \( d = 1 \)

Now, we compute \( D(10) \) and \( D(12) \):

1. **Calculating \( D(10) \)**:
   Since \( 10 \leq T \):
   \[
   D(10) = c \cdot 10 = 1 \cdot 10 = 10
   \]

2. **Calculating \( D(12) \)**:
   Since \( 12 > T \):
   \[
   D(12) = \max\left(D\left(\left\lfloor \frac{12}{2} \right\rfloor\right), D\left(12 - \left\lfloor \frac{12}{2} \right\rfloor\right)\right) + d
   \]
   We compute:
   \[
   D(6) = \text{for } L = 6, \text{ since } 6 \leq T:
   \]
   \[
   D(6) = c \cdot 6 = 1 \cdot 6 = 6
   \]
   Thus, we have:
   \[
   D(12) = \max(D(6), D(6)) + 1 = \max(6, 6) + 1 = 6 + 1 = 7
   \]

### Step 4: Conclusion

Now we compare the results:
- \( D(10) = 10 \)
- \( D(12) = 7 \)

This shows that \( D(10) > D(12) \), which contradicts the property \( D(L_1) \leq D(L_2) \) for \( L_1 = 10 \) and \( L_2 = 12 \).

### Final Thought

This counterexample illustrates that the depth function can


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

To evaluate the options regarding the use of Hidden Markov Models (HMMs) for entity recognition in text documents, let's analyze each statement step-by-step.

### Option 1: The cost of learning the model is quadratic in the lengths of the text.
- **Analysis**: The learning process of an HMM involves estimating the transition and emission probabilities based on the observed sequences. The complexity of this process is generally linear with respect to the number of states and observations, but it can be influenced by the length of the sequences. However, it is not typically quadratic in the length of the text. Therefore, this statement is **incorrect**.

### Option 2: The cost of predicting a word is linear in the lengths of the text preceding the word.
- **Analysis**: In an HMM, the prediction of the next state (or label) given the previous states is done using the Viterbi algorithm or similar methods. The prediction process involves looking at the current state and the transition probabilities, which can be done in linear time relative to the number of states. However, the prediction of a word does not depend on the entire length of the preceding text but rather on the last observed states. Thus, this statement is somewhat misleading and can be considered **incorrect**.

### Option 3: An HMM model can be built using words enhanced with morphological features as input.
- **Analysis**: HMMs can indeed be enhanced by incorporating additional features, such as morphological features (e.g., prefixes, suffixes, etc.), to improve the model's performance in tasks like entity recognition. This is a common practice in natural language processing to provide more context and improve accuracy. Therefore, this statement is **correct**.

### Option 4: The label of one word is predicted based on all the previous labels.
- **Analysis**: In an HMM, the prediction of the label for the current word is based on the previous label (the last state) and the transition probabilities, rather than all previous labels. This is a key characteristic of Markov models, where the future state depends only on the current state, not the entire history. Therefore, this statement is **incorrect**.

### Final Selection:
Based on the analysis above, the only correct statement regarding the use of Hidden Markov Models (HMMs) for entity recognition in text documents is:

- **Option 3**: An HMM model can be built using words enhanced with morphological features as input.

Thus, the final answer is:
**3**.


Question: Which of the following is/are true about fuzzing?

Certainly! Letâs go through each statement about fuzzing carefully, providing a detailed explanation for each one to ensure a thorough understanding of the concepts involved.

### A. Black box fuzzing may struggle to find inputs that reach deep into the program.

**True.**

Black box fuzzing is a technique where the tester has no insight into the internal workings of the application being tested. This means that the fuzzer generates inputs (often randomly) without understanding how the application processes these inputs.

Because of this lack of visibility, black box fuzzers may have difficulty reaching deeper or more complex parts of the program. For example, if a program has multiple layers of functions or requires specific conditions to be met before accessing certain code paths, a black box fuzzer may not generate the necessary inputs to trigger those deeper functions.

**Example:** Consider a web application that has a multi-step user registration process. If the fuzzer generates random inputs without following the expected sequence (like bypassing the email verification step), it may never reach the underlying logic that handles user account creation, possibly overlooking vulnerabilities present in that part of the code.

### B. The quality of initial seeds matters in mutational fuzzing.

**True.**

Mutational fuzzing relies on a set of initial valid inputs, or "seeds," from which new test cases are generated through small modifications. The quality of these seeds is crucial because they determine the effectiveness of the mutations. Well-structured seeds help the fuzzer produce meaningful variations that can reveal vulnerabilities.

**Example:** If you are testing a file parser, starting with a valid, correctly formatted file (like a valid XML document) allows the fuzzer to create variations that might expose edge cases or vulnerabilities within the parsing logic. In contrast, if the initial seed is a random string of bytes, the mutations might not produce valid inputs, leading to ineffective testing and missed vulnerabilities.

### C. In structure-aware fuzzing, the mutator should only generate inputs that comply with all the format rules.

**False.**

Structure-aware fuzzing involves understanding the expected input formats (like JSON, XML, or binary protocols) and generating inputs that conform to those formats. However, saying that the mutator should **only** generate inputs that comply with **all** format rules is not entirely accurate.

While valid inputs are essential for testing standard behavior, it is equally important to generate **invalid** inputs to test how the application handles errors or unexpected cases. This helps identify potential vulnerabilities related to improper input handling or error management.

**Example:** In a JSON parser, the fuzzer should generate valid JSON as well as malformed JSON (such as unquoted strings or missing commas) to check how the parser reacts. This approach can reveal vulnerabilities in error handling or input validation that might not be evident with only valid inputs.

### D. Fuzzing is complete as soon as all code is covered.

**False.**

Achieving full code coverage is a significant goal in fuzzing, but it does not guarantee that all vulnerabilities have been found. Code coverage tells us which parts of the code were executed during testing, but it does not imply that all potential vulnerabilities have been triggered.

Certain vulnerabilities may depend on specific input values, sequences, or combinations that aren't captured simply by executing all lines of code.

**Example:** A function could be fully covered during fuzzing, but if the input does not trigger a specific error condition (like a buffer overflow), that vulnerability remains undiscovered. Additionally, some vulnerabilities may only be exposed when certain inputs are provided in a specific order, which code coverage alone cannot assess.

### Summary

- **A** is **True**: Black box fuzzing may struggle to effectively explore deeper code paths due to a lack of internal knowledge.
- **B** is **True**: The quality of initial seeds significantly impacts the effectiveness of mutational fuzzing.
- **C** is **False**: Structure-aware fuzzing should generate both valid and invalid inputs to thoroughly test the applicationâs robustness.
- **D** is **False**: Full code coverage does not ensure that all vulnerabilities are found; the nature of inputs and their sequences also play a critical role.

Understanding these aspects of fuzzing helps clarify how different techniques can be applied to enhance software security testing and vulnerability discovery. Each approach has its strengths and weaknesses, and selecting the right strategy is essential for effective testing.


Question: (Nearest Neighbor) The training error of the 1-nearest neighbor classifier is zero.

**Answer: True.**

**Justification:**

The 1-nearest neighbor (1-NN) classifier is a type of instance-based learning algorithm that classifies a new data point based on the label of its closest training example in the feature space. When we refer to the "training error" of a classifier, we are concerned with how well the classifier performs on the training data itself.

In the case of the 1-NN classifier, the training error is zero because each training example is classified based on its own label. When a training sample is queried, the nearest neighbor to it is itself, thus the classifier will always assign the correct label to the training data point.

To elaborate further:

1. **Definition of Training Error**: The training error is defined as the proportion of instances in the training dataset that the classifier misclassifies. Since 1-NN assigns each instance its own label when queried, there are no misclassifications among the training samples.

2. **Instance-Based Learning**: The 1-NN algorithm is a type of instance-based classifier. It memorizes the training instances and does not build an explicit model. Therefore, for any training instance, the nearest neighbor is itself, leading to a correct classification.

3. **Implications**: While the training error is zero, it is important to note that this does not imply the model will generalize well to unseen data. The 1-NN classifier can suffer from overfitting, especially in high-dimensional spaces or when the training set is small.

In conclusion, the statement that the training error of the 1-nearest neighbor classifier is zero is indeed true, as it perfectly classifies all training instances.


Question: Split the given data into a training set (70%) and a testing set (30%). We refer to these as "random split" in the subsequent tasks. The data is in a pandas.DataFrame format.

The correct answer shows how to split a dataset into a training set and a testing set using the pandas library in Python.

1. **Training Set (70%)**: The line `df_train_r = df.sample(frac=0.7)` randomly selects 70% of the rows from the original DataFrame `df`. The `frac=0.7` parameter specifies that you want to sample 70% of the data.

2. **Testing Set (30%)**: The second line, `df_test_r = df.loc[df.index.difference(df_train_r.index)]`, is used to create the testing set. It takes all the rows from the original DataFrame `df` that are not included in the training set. This is done by finding the indices of the training set and using `difference` to exclude them from the original DataFrame.

Together, these lines of code ensure that the data is split into two distinct sets: one for training the model and another for testing its performance, without any overlap between the two.


Question: Consider the following algorithm that takes as input an undirected graph $G=(V,E)$: \begin{center} \begin{boxedminipage}[t]{0.85\textwidth} \begin{minipage}{14cm} \begin{verse} \textsc{SimpleCut}$(G=(V,E))$: \\[2mm] 1. Let $\mathcal{H}$ be a $2$-universal family of hash functions $h: V \to \{0,1\}$. \\[1mm] 2. Select $h \in  \mathcal{H}$ at random. \\[1mm] 3. \RETURN the vertex set $S = \{v\in V: h(v) = 0\}$. \end{verse} \end{minipage} \end{boxedminipage} \end{center} Prove the following: \begin{itemize} \item[]In expectation,  the set $S$ returned by \textsc{SimpleCut}  cuts at least $|E|/2$ edges. \end{itemize} {\em (In this problem you are asked to prove the above statement. Recall that you are allowed to refer to material covered in the lecture notes.)}

Imagine a party where people are mingling and you want to divide them into two groups. You can think of the edges in a graph as the friendships connecting these people. If you randomly choose a subset of people to be in one group, thereâs a good chance that youâll cut through many of those friendships. This is essentially what the SimpleCut algorithm does using hash functions.

To prove that the expected number of edges cut by the set \( S \) is at least \( |E|/2 \), we can use linearity of expectation. For each edge \( (u, v) \) in the graph, we want to determine the probability that one endpoint is in \( S \) and the other is not. If we denote \( h(u) \) and \( h(v) \) as the hash values for vertices \( u \) and \( v \), then an edge \( (u, v) \) is cut if \( h(u) = 0 \) and \( h(v) = 1 \), or vice versa.

Since \( h \) is chosen uniformly at random from a 2-universal family, the probabilities for \( h(u) = 0 \) and \( h(v) = 0 \) are both \( 1/2 \). Hence, the probability that edge \( (u, v) \) is cut is:
\[
P(\text{cut}(u, v)) = P(h(u) = 0, h(v) = 1) + P(h(u) = 1, h(v) = 0) = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.
\]

Now, summing this probability over all edges gives us:
\[
E[\text{cut edges}] = \sum_{(u,v) \in E} P(\text{cut}(u,v)) = |E| \cdot \frac{1}{2} = \frac{|E|}{2}.
\]

Thus, in expectation, the algorithm cuts at least \( |E|/2 \) edges.

The expected number of edges cut by the set \( S \) returned by SimpleCut is at least \( |E|/2 \).


Question: $\mathbb{Z}_{37}^*$ denotes ...

To understand what \(\mathbb{Z}_{37}^*\) denotes, we need to break it down step-by-step.

1. **Understanding \(\mathbb{Z}_{37}\)**:
   \(\mathbb{Z}_{37}\) represents the set of integers modulo 37. This means we are considering the integers \{0, 1, 2, ..., 36\}, where arithmetic operations are performed modulo 37.

2. **Understanding the asterisk (\(^*\)) notation**:
   The notation \(\mathbb{Z}_{n}^*\) typically refers to the group of units of \(\mathbb{Z}_{n}\). The group of units consists of all the elements in \(\mathbb{Z}_{n}\) that are coprime to \(n\) (in this case, 37).

3. **Finding the units in \(\mathbb{Z}_{37}\)**:
   Since 37 is a prime number, every integer from 1 to 36 is coprime to 37. Therefore, the units in \(\mathbb{Z}_{37}\) are simply the integers \{1, 2, ..., 36\}.

4. **Structure of \(\mathbb{Z}_{37}^*\)**:
   The set of units forms a group under multiplication modulo 37. This means that we can multiply any two units together, and the result will also be a unit (which is guaranteed because they are all coprime to 37).

5. **Conclusion**:
   Since \(\mathbb{Z}_{37}^*\) consists of the multiplicative group of units modulo 37, it is specifically a multiplicative group.

Now, let's evaluate the choices given:

- A. a field: While \(\mathbb{Z}_{37}\) is indeed a field (because it is a finite field with prime order), \(\mathbb{Z}_{37}^*\) specifically refers to the multiplicative group of units, not the field itself.
- B. an additive group: This does not describe \(\mathbb{Z}_{37}^*\) since we are dealing with multiplicative structure here.
- C. a multiplicative group: This is correct as we have established that \(\mathbb{Z}_{37}^*\) is the group of units under multiplication.
- D. a ring: A ring includes an additive structure as well, but again, we are focused on the multiplicative group here.

Thus, the correct answer is:

C. a multiplicative group.


Question: You are given two distributions over $\mathbb{R}$ : Uniform on the interval $[a, b]$ and Gaussian with mean $\mu$ and variance $\sigma^{2}$. Their respective probability density functions are $$ p_{\mathcal{U}}(y \mid a, b):=\left\{\begin{array}{ll} \frac{1}{b-a}, & \text { for } a \leq y \leq b, \\ 0 & \text { otherwise } \end{array} \quad p_{\mathcal{G}}\left(y \mid \mu, \sigma^{2}\right):=\frac{1}{\sqrt{2 \pi \sigma^{2}}} \exp \left(-\frac{(y-\mu)^{2}}{2 \sigma^{2}}\right)\right. $$ Which one(s) belong to the exponential family?

The exponential family of distributions is characterized by a probability density function that can be expressed in the form \( p(y; \theta) = h(y) \exp(\theta T(y) - A(\theta)) \), where \( \theta \) is a parameter, \( T(y) \) is a sufficient statistic, \( A(\theta) \) is the log-partition function, and \( h(y) \) is a base measure. The Gaussian distribution can be rewritten in this form, identifying parameters such as the mean and variance, making it a member of the exponential family. In contrast, the uniform distribution on the interval \([a, b]\) does not fit this structure, as it lacks the exponential form and is instead a constant over its support. Thus, only the Gaussian distribution belongs to the exponential family. Therefore, the correct answer is: Only Gaussian.


Question:
In a nutshell, the "second album syndrome" is a theory that states that the second album of a band always sucks

You have the following regression output regarding the score_diff: the difference in scores between the second and the first album (second - first):

Dep. Variable: score_diff
R-squared: -0.000

Interpret the 𝑅2 in your regression here. Do these analyses suggest that the "second album syndrome" exists? Why?

To interpret the R-squared value in the context of your regression output regarding the score_diff (the difference in scores between the second and the first album), let's break down the steps:

1. **Understanding R-squared**: R-squared (R²) is a statistical measure that represents the proportion of the variance for a dependent variable that's explained by an independent variable(s) in a regression model. It ranges from 0 to 1, where 0 indicates that the model does not explain any of the variability of the response data around its mean, and 1 indicates that it explains all the variability.

2. **Given R-squared Value**: In this case, the R-squared value is -0.000. R-squared cannot actually be negative in a standard interpretation; it suggests that the model does not explain any variation in the score_diff. A negative or zero R-squared value indicates that the regression model is a poor fit for the data.

3. **Implications of R-squared Value**:
   - An R-squared of -0.000 suggests that the difference in scores between the second and the first album is not explained by any of the independent variables in the model.
   - It indicates that there is no relationship between the factors analyzed and the score_diff. In other words, the second album's score does not reliably differ from the first based on the data you have.

4. **Existence of "Second Album Syndrome"**:
   - The "second album syndrome" hypothesizes that second albums tend to be worse than first albums. However, based on the regression output (with an R-squared of -0.000), there is no statistical evidence to support this phenomenon with the data at hand.
   - Since the model does not explain any variation in the score_diff, we cannot conclude that the second album is significantly worse than the first based on this analysis.

**Final Answer**: The R-squared value of -0.000 indicates that the regression model does not explain the difference in scores between the first and second albums. Thus, there is no evidence to suggest that the "second album syndrome" exists based on this analysis.


Question: Mark any of the following functions that have unique maximizers:

To determine which of the given functions have unique maximizers, we need to analyze each function within its specified domain. A unique maximizer occurs when there is exactly one point in the domain where the function achieves its maximum value.

1. **Function: \( f(x) = -x^2, \quad x \in [-10, 10] \)**
   - This function is a downward-opening parabola. The maximum occurs at the vertex, which is at \( x = 0 \). Since the parabola is symmetric and opens downwards, the maximum value \( f(0) = 0 \) is unique.
   - **Unique maximizer: Yes.**

2. **Function: \( f(x) = \ln(x), \quad x \in (0, 10] \)**
   - The natural logarithm function is increasing on the interval \( (0, 10] \). Therefore, the maximum occurs at the endpoint \( x = 10 \). Since there is only one point where the maximum occurs, it is unique.
   - **Unique maximizer: Yes.**

3. **Function: \( f(x) = x^2, \quad x \in [-10, 10] \)**
   - This function is an upward-opening parabola. The minimum occurs at the vertex \( x = 0 \), and the maximum occurs at the endpoints \( x = -10 \) and \( x = 10 \) where \( f(-10) = f(10) = 100 \). Since the maximum value is achieved at two points, it is not unique.
   - **Unique maximizer: No.**

4. **Function: \( f(x) = \cos(2 \pi x), \quad x \in [-1, 1] \)**
   - The cosine function oscillates between -1 and 1. Within the interval \( [-1, 1] \), the maximum value of 1 occurs at multiple points: \( x = 0 \) and \( x = \pm 1 \). Therefore, the maximum is not unique.
   - **Unique maximizer: No.**

5. **Function: \( f(x) = \cos(2 \pi x), \quad x \in \left[-\frac{1}{2}, \frac{1}{2}\right] \)**
   - In this interval, the cosine function achieves its maximum value of 1 at \( x = 0 \) only. Since there are no other points in the interval where the maximum occurs, it is unique.
   - **Unique maximizer: Yes.**

### Summary of Unique Maximizers:
- **1:** Yes
- **2:** Yes
- **3:** No
- **4:** No
- **5:** Yes

Thus, the functions that have unique maximizers are **1, 2, and 5**.


Question: Which of the following graph analysis techniques do you believe would be most appropriate to identify communities on a social graph?

The most appropriate technique to identify communities on a social graph is **B. Random Walks**, because they effectively capture the connectivity and relationships within the graph, allowing for the discovery of densely connected subgroups or communities.


Question:
You have been publishing a daily column for the Gazette over the last few years and have recently reached a milestone --- your 1000th column! Realizing you'd like to go skiing more often, you decide it might be easier to automate your job by training a story generation system on the columns you've already written. Then, whenever your editor pitches you a title for a column topic, you'll just be able to give the title to your story generation system, produce the text body of the column, and publish it to the website!
Would you use a causal language modeling or masked language modeling training objective to train your model? Why?

To train a story generation system for automating column writing, I would use a causal language modeling (CLM) training objective rather than a masked language modeling (MLM) objective. Here are a few reasons why:

1. **Sequential Generation**: Causal language modeling is designed for tasks where the goal is to generate text sequentially, predicting the next word based on the words that have come before it. This is ideal for generating coherent and contextually relevant columns, as the model needs to construct sentences word by word in a logical order.

2. **Natural Language Generation**: Since the goal is to create new text (the column body) from a given title, CLM aligns better with this type of natural language generation task. It allows the model to maintain a flow of thought and coherence over the entire text, which is crucial for writing engaging and meaningful columns.

3. **No Masking Required**: Masked language modeling, on the other hand, is typically used for tasks where the model needs to predict missing words in a sentence. This approach is more suited for understanding context and relationships between words in a given sequence, rather than generating new sequences. Using MLM would not be as effective for creating new content from scratch.

4. **Flexibility with Prompts**: With CLM, you can directly provide a title or any prompt to the model, and it can generate text that follows logically from that prompt. This is particularly useful for your requirement of generating columns based on specific topics suggested by your editor.

In conclusion, for the task of generating new columns based on titles, a causal language modeling approach is the most suitable choice. It supports the sequential nature of writing and allows for intuitive and coherent text generation.


Question: In the first pass over the database of the FP Growth algorithm

To approach the FP-Growth algorithm effectively, let's break down the problem into smaller parts and understand the first pass over the database.

1. **Understanding FP-Growth**: The FP-Growth algorithm is designed for mining frequent itemsets without candidate generation. It consists of two main passes over the database.

2. **First Pass Over the Database**:
   - In the first pass, the algorithm focuses on identifying how often each item appears in the transactions to determine their frequency.

Now, let's evaluate the options provided in the context of the first pass:

- **Frequent itemsets are extracted**: This typically occurs in the second pass after the FP-tree is constructed and is not part of the first pass.

- **A tree structure is constructed**: The construction of the FP-tree occurs after the first pass, so this is not correct for the first pass.

- **The frequency of items is computed**: This is indeed what happens in the first pass. The algorithm counts the occurrences of each item in the transactions to establish their frequencies.

- **Prefixes among itemsets are determined**: This is related to the generation of conditional FP-trees and does not occur in the first pass.

Based on this breakdown, the correct answer for what happens in the first pass over the database of the FP-Growth algorithm is:

**- The frequency of items is computed**.


Question:
The goal of the 4 following questions is to prove that the methods map and mapTr are equivalent. The
former is the version seen in class and is specified by the lemmas MapNil and MapCons. The later version
is a tail-recursive version and is specified by the lemmas MapTrNil and MapTrCons.
All lemmas on this page hold for all x: Int, y: Int, xs: List[Int], ys: List[Int], l: List
[Int] and f: Int => Int.
Given the following lemmas:

(MapNil) Nil.map(f) === Nil
(MapCons) (x :: xs).map(f) === f(x) :: xs.map(f)
(MapTrNil) Nil.mapTr(f, ys) === ys
(MapTrCons) (x :: xs).mapTr(f, ys) === xs.mapTr(f, ys ++ (f(x) :: Nil))
(NilAppend) Nil ++ xs === xs
(ConsAppend) (x :: xs) ++ ys === x :: (xs ++ ys)
Let us first prove the following lemma:

(AccOut) l.mapTr(f, y :: ys) === y :: l.mapTr(f, ys)

We prove it by induction on l.
Base case: l is Nil. Therefore, we need to prove:

Nil.mapTr(f, y :: ys) === y :: Nil.mapTr(f, ys).

What exact sequence of lemmas should we apply to rewrite the left hand-side (Nil.mapTr(f, y :: ys))
to the right hand-side (y :: Nil.mapTr(f, ys))?

To prove the lemma \( \text{AccOut} \) by induction on the list \( l \), we start with the base case where \( l \) is \( \text{Nil} \).

### Base Case

We need to show that:
\[
\text{Nil.mapTr}(f, y :: ys) === y :: \text{Nil.mapTr}(f, ys).
\]

### Step-by-Step Reasoning

1. **Understanding the Left Side**:
   - We start with the expression \( \text{Nil.mapTr}(f, y :: ys) \). According to the lemma \( \text{MapTrNil} \), we know that:
   \[
   \text{Nil.mapTr}(f, ys) === ys.
   \]
   - In our case, substituting \( ys \) with \( y :: ys \), we have:
   \[
   \text{Nil.mapTr}(f, y :: ys) === y :: ys.
   \]

2. **Understanding the Right Side**:
   - The right side of the equation, \( y :: \text{Nil.mapTr}(f, ys) \), requires us to evaluate \( \text{Nil.mapTr}(f, ys) \) again using \( \text{MapTrNil} \):
   \[
   \text{Nil.mapTr}(f, ys) === ys.
   \]
   - Thus, \( y :: \text{Nil.mapTr}(f, ys) \) simplifies to \( y :: ys \).

3. **Equating Both Sides**:
   - Now we have:
   \[
   \text{Nil.mapTr}(f, y :: ys) === y :: ys,
   \]
   and
   \[
   y :: \text{Nil.mapTr}(f, ys) === y :: ys.
   \]
   - Therefore, both sides are equal:
   \[
   \text{Nil.mapTr}(f, y :: ys) === y :: \text{Nil.mapTr}(f, ys).
   \]

### Conclusion

In conclusion, we have proven the base case for the lemma \( \text{AccOut} \):
\[
\text{Nil.mapTr}(f, y :: ys) === y :: \text{Nil.mapTr}(f, ys).
\]
The sequence of lemmas applied in this proof is:
1. \( \text{MapTrNil} \) to evaluate \( \text{Nil.mapTr}(f, y :: ys) \).
2. \( \text{MapTrNil} \) again to evaluate \( \text{Nil.mapTr}(f, ys) \).

Thus, the proof is complete for the base case, and we can proceed with the inductive step to complete the proof of the lemma \( \text{AccOut} \) if needed.


Question: Tick the \textbf{false} assertion. Assume that $C$ is a random permutation.

To determine which assertion is false, let's analyze each statement one by one.

1. **BestAdv$_n(C,C^\ast)=\frac{1}{2}Dec^n_{\left|\left|\left|\cdot\right|\right|\right|_a}(C)$**: This statement relates to adversarial advantage in a cryptographic context and how it relates to decision complexity. The expression suggests that the best adversary's advantage is half of the decision complexity metric in a certain norm. Without additional context regarding the definitions of BestAdv and Dec, we can't determine if this is true or false.

2. **BestAdv$_n^{n.a.}(C,C^\ast)=\frac{1}{2}Dec^n_{\left|\left|\left|\cdot\right|\right|\right|_\infty}(C)$**: Similar to the first statement, this relates the non-adaptive adversarial advantage to decision complexity in another norm. Again, without specific definitions, itâs hard to adjudicate its truth.

3. **$E(LP^{C}(a,b))\leq 1$**: Here, \(E\) denotes the expected value of some function \(LP^C(a, b)\). If \(LP^C(a, b)\) is constrained in a way that its expected value remains bounded by 1, this statement could be true. However, without specific context on the nature of \(LP^C\), we cannot ascertain its validity.

4. **$Dec^n(C\circ C)\leq Dec^n(C)^2$**: This statement resembles a property known as the "composition" property in complexity theory, where the complexity of a composition of functions is often bounded by the product of their individual complexities. This is generally true for many computational settings.

Given these analyses, the most likely candidate for being false is the first three assertions, as they rely heavily on the definitions of specific terms and properties that are not universally accepted or may depend on the specific type of cryptographic construct being analyzed. The fourth assertion is more in line with established results in computational complexity.

In conclusion, without loss of generality or additional context, **the false assertion is likely:**

**1. BestAdv$_n(C,C^\ast)=\frac{1}{2}Dec^n_{\left|\left|\left|\cdot\right|\right|\right|_a}(C)$**.


Question: We will analyze the $K$-means algorithm and show that it always converge. Let us consider the $K$-means objective function: $$ \mathcal{L}(\mathbf{z}, \boldsymbol{\mu})=\sum_{n=1}^{N} \sum_{k=1}^{K} z_{n k}\left\|\mathbf{x}_{n}-\boldsymbol{\mu}_{k}\right\|_{2}^{2} $$ where $z_{n k} \in\{0,1\}$ with $\sum_{k=1}^{K} z_{n k}=1$ and $\boldsymbol{\mu}_{k} \in \mathbb{R}^{D}$ for $k=1, \ldots, K$ and $n=1, \ldots, N$. How would you choose the $\left\{z_{n k}\right\}_{n, k=1}^{N, K}$ to minimize $\mathcal{L}(\mathbf{z}, \boldsymbol{\mu})$ for given $\left\{\boldsymbol{\mu}_{k}\right\}_{k=1}^{K}$ ? Compute the closed-form formula for the $z_{n k}$. To which step of the $K$-means algorithm does it correspond?

To minimize the \( K \)-means objective function

\[
\mathcal{L}(\mathbf{z}, \boldsymbol{\mu}) = \sum_{n=1}^{N} \sum_{k=1}^{K} z_{n k} \left\| \mathbf{x}_{n} - \boldsymbol{\mu}_{k} \right\|_{2}^{2},
\]

given fixed cluster centers \( \{\boldsymbol{\mu}_{k}\}_{k=1}^{K} \), we need to determine the binary assignment \( z_{n k} \) that indicates whether data point \( n \) belongs to cluster \( k \) (with \( z_{n k} = 1 \) if it does, and \( z_{n k} = 0 \) otherwise).

### Step 1: Selecting \( z_{n k} \)

The objective function can be rewritten in terms of two parts: the cost of assigning each data point \( n \) to the cluster center \( \boldsymbol{\mu}_{k} \), which is quantified by the squared Euclidean distance \( \left\| \mathbf{x}_{n} - \boldsymbol{\mu}_{k} \right\|_{2}^{2} \).

For each data point \( \mathbf{x}_n \), to minimize \( \mathcal{L} \), we want to select the cluster \( k \) that minimizes the distance to \( \boldsymbol{\mu}_{k} \). Thus, we define \( z_{n k} \) as:

\[
z_{n k} =
\begin{cases}
1, & \text{if } k = \arg\min_{j} \left\| \mathbf{x}_{n} - \boldsymbol{\mu}_{j} \right\|_{2}^{2} \\
0, & \text{otherwise}.
\end{cases}
\]

### Step 2: Closed-Form Expression for \( z_{n k} \)

From the previous step, we can express \( z_{n k} \) in a more concise form:

\[
z_{n k} =
\begin{cases}
1, & \text{if } k = c_n \\
0, & \text{otherwise},
\end{cases}
\]

where \( c_n \) is the index of the cluster center that is closest to \( \mathbf{x}_n \):

\[
c_n = \arg\min_{j} \left\| \mathbf{x}_{n} - \boldsymbol{\mu}_{j} \right\|_{2}^{2}.
\]

### Step 3: Correspondence to the K-means Algorithm

This step corresponds to the **assignment step** of the \( K \)-means algorithm. In the \( K \)-means algorithm, the process alternates between two main steps:

1. **Assignment Step**: Assign each data point to the nearest cluster center (as computed above).
2. **Update Step**: Recalculate the cluster centers \( \boldsymbol{\mu}_{k} \) as the mean of all the points assigned to each cluster.

By performing the assignment step as described, we ensure that every data point is assigned to the cluster that minimizes the objective function, thereby guaranteeing a decrease in the overall cost \( \mathcal{L}(\mathbf{z}, \boldsymbol{\mu}) \) with each iteration of the algorithm.

### Conclusion

The \( K \)-means algorithm converges because the objective function \( \mathcal{L}(\mathbf{z}, \boldsymbol{\mu}) \) is non-increasing with each iteration, and it is bounded below (by zero). As a result, the algorithm will eventually reach a point where the assignments and cluster centers do not change, indicating convergence.


Question: Write the dual of the following linear program: \begin{align*} \text{Maximize}  \quad &6x_1 + 14 x_2 + 13 x_3\\ \text{Subject to} \quad & x_1 + 3x_2 + x_3 \leq 24 \\ & x_1 + 2x_2 + 4 x_3 \leq 60 \\ & x_1, x_2, x_3 \geq 0 \end{align*} Hint: How can you convince your friend that the above linear program has optimum value at most $z$?

The dual of the given linear program can be constructed as follows:

**Primal Problem:**
\[
\text{Maximize} \quad z = 6x_1 + 14x_2 + 13x_3
\]
\[
\text{Subject to} \quad
\begin{align*}
x_1 + 3x_2 + x_3 & \leq 24 \\
x_1 + 2x_2 + 4x_3 & \leq 60 \\
x_1, x_2, x_3 & \geq 0
\end{align*}
\]

**Dual Problem:**
Let \(y_1\) and \(y_2\) be the dual variables corresponding to the two constraints.

\[
\text{Minimize} \quad w = 24y_1 + 60y_2
\]
\[
\text{Subject to} \quad
\begin{align*}
y_1 + y_2 & \geq 6 \\
3y_1 + 2y_2 & \geq 14 \\
y_1 + 4y_2 & \geq 13 \\
y_1, y_2 & \geq 0
\end{align*}
\]

### Hint Explanation
To show the primal's optimum value is at most \(z\), use the Weak Duality Theorem: the objective value of the primal is always less than or equal to that of the dual for any feasible solutions, thus establishing that the primal's optimum cannot exceed the dual's minimum.


Question: (Adversarial perturbations for linear models) Suppose you are given a linear classifier with the logistic loss. Is it true that generating the optimal adversarial perturbations by maximizing the loss under the $\ell_{2}$-norm constraint on the perturbation is an NP-hard optimization problem?

The statement about generating the optimal adversarial perturbations for a linear classifier with logistic loss under an \(\ell_2\)-norm constraint being an NP-hard optimization problem is generally considered to be **False**.

To elaborate, while generating adversarial examples can be challenging and can involve complex optimization, the specific problem of maximizing the logistic loss under an \(\ell_2\)-norm constraint can often be solved efficiently, especially for linear classifiers. This is due to the structure of the loss function and the constraints involved, which can typically be handled by convex optimization techniques.

Therefore, the correct answer to the question is:

**2. False**


Question: Consider an IR system using a Vector Space model with Okapi BM25 as the weighting scheme (with \(k=1.5\) and \(b=0.75\)) and operating on a document collection that contains:a document \(d_1\), andand a document \(d_3\) corresponding to the concatenation of 3 copies of \(d_1\).Indicate which of the following statements are true, where \(\langle d\rangle\) stands for the vector representing document \(d\):(Penalty for wrong ticks.)

To address the question effectively, we will analyze the implications of using the Vector Space model with the Okapi BM25 weighting scheme for the documents \(d_1\) and \(d_3\), where \(d_3\) is a concatenation of three copies of \(d_1\). We will evaluate each statement based on this analysis.

### Step 1: Understanding the Document Structure
- **Document \(d_1\)**: A single document with its specific term frequencies.
- **Document \(d_3\)**: This document is composed of three copies of \(d_1\), which means that the term frequencies in \(d_3\) will be three times those in \(d_1\).

### Step 2: Cosine Similarity
The cosine similarity between two vectors \( \langle a \rangle \) and \( \langle b \rangle \) is calculated as:
$$
\text{cosine\_similarity}(\langle a \rangle, \langle b \rangle) = \frac{\langle a \rangle \cdot \langle b \rangle}{||\langle a \rangle|| \cdot ||\langle b \rangle||}
$$
In this case, since \(d_3\) is not just a scalar multiple of \(d_1\) in terms of direction but has different magnitudes (due to the number of terms), the cosine similarity will not equal 1. The vectors are aligned in a similar direction but differ in magnitude, leading to a cosine similarity that is less than 1.

**Conclusion for Statement 1**: False.

### Step 3: Component Comparison
Each component of \(\langle d_3\rangle\) represents the term frequency of the terms in \(d_3\). Given that \(d_3\) consists of three copies of \(d_1\), it follows that each component in \(\langle d_3\rangle\) will be strictly larger than the corresponding component in \(\langle d_1\rangle\) because term frequencies are multiplied by three.

**Conclusion for Statement 2**: True.

### Step 4: Analyzing Statement 3
Since we established that each component of \(\langle d_3\rangle\) is larger than in \(\langle d_1\rangle\), this statement, which claims that each component of \(\langle d_3\rangle\) is strictly smaller than the corresponding one in \(\langle d_1\rangle\), is inherently false.

**Conclusion for Statement 3**: False.

### Step 5: Understanding BM25 and Term Frequencies
The BM25 weighting scheme adjusts the weight of terms based on both term frequency and document length. Higher term frequencies generally lead to higher weights, especially when they are normalized against document length.

- For **Statement 4**, the assertion that terms with small term frequency are favored in \(\langle d_3\rangle\) is misleading. While BM25 does adjust weights, it does not favor small term frequencies over larger ones. Instead, it gives higher weights to terms with larger frequencies, particularly in longer documents like \(d_3\).

**Conclusion for Statement 4**: False.

- For **Statement 5**, since \(d_3\) has higher term frequencies due to being a composite of three copies of \(d_1\), terms that are more frequent will indeed receive larger weights. Therefore, this statement is true.

**Conclusion for Statement 5**: True.

### Final Summary of Statements
1. False
2. True
3. False
4. False
5. True

### Correct True Statements:
The true statements are 2 and 5. The false statements are 1, 3, and 4.

This revised analysis clarifies the concepts of cosine similarity and the functioning of the BM25 model in relation to term frequencies and document lengths.