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

1. The name âstack canariesâ does indeed stem from the historical practice of using canaries in coal mines to detect toxic gases, as they would signal danger before humans were affected.
2. Stack canaries are designed specifically to protect against buffer overflow vulnerabilities, but they do not protect against format string vulnerabilities.
3. Stack canaries on Linux are generated per process, not per thread, as each process has its own stack.
4. Stack canary instrumentation is not inserted into every function by default; it is applied selectively based on compiler flags and function attributes.
5. As mentioned earlier, stack canaries on Linux are generated per process.

Short answer: The true statements are: the name comes from gas-detecting canaries, and they do not protect against format string vulnerabilities; stack canaries on Linux are generated per process.


Question: For which kind of bugs does default LLVM provide sanitizers?

Let’s analyze the types of bugs mentioned and the sanitizers provided by LLVM step by step:

1. **Memory leaks**
   - **Analysis:** LLVM provides a sanitizer called **AddressSanitizer (ASan)**, which can help detect memory leaks when combined with additional tools like **LeakSanitizer**. The primary focus of ASan is on memory access errors, but it can also help identify memory leaks.
   - **Verdict:** **Supported.**

2. **Buffer overflows**
   - **Analysis:** AddressSanitizer (ASan) is specifically designed to detect buffer overflows, as it checks for out-of-bounds accesses and use-after-free errors. It is one of the most effective tools for identifying these types of bugs.
   - **Verdict:** **Supported.**

3. **Race conditions between threads**
   - **Analysis:** LLVM provides **ThreadSanitizer (TSan)**, which is designed to detect data races in multithreaded programs. It identifies situations where two threads access the same variable concurrently, and at least one of them is a write.
   - **Verdict:** **Supported.**

4. **Logic bugs**
   - **Analysis:** LLVM does not provide specific sanitizers for logic bugs. Logic bugs typically involve incorrect program behavior that does not necessarily manifest as memory or threading issues and are generally harder to detect with sanitization tools. They require different debugging techniques such as code reviews, testing, or formal verification.
   - **Verdict:** **Not supported.**

**Conclusion:** The LLVM default sanitizers support detection for memory leaks, buffer overflows, and race conditions between threads. Logic bugs are not specifically addressed by LLVM sanitizers. Therefore, the supported bug types are:
- Memory leaks
- Buffer overflows
- Race conditions between threads


Question: Tick the \emph{incorrect} assertion. For a cipher $C$, decorrelation theory says that \ldots

'A decorrelation $0$ of order $1$ always protects against linear cryptanalysis.'


Question: Let P be the statement ∀x(x>-3 -> x>3). Determine for which domain P evaluates to true:

To evaluate the statement \( P \) defined as \( \forall x (x > -3 \to x > 3) \), we first need to unpack what it means in the context of different domains.

The expression \( x > -3 \to x > 3 \) is a conditional statement, which can be understood in terms of truth values. The implication \( A \to B \) is false only if \( A \) is true and \( B \) is false; in all other cases, it is true.

1. **Understanding the implication**:
   - \( A \) is \( x > -3 \)
   - \( B \) is \( x > 3 \)
   - Therefore, the implication \( x > -3 \to x > 3 \) is false if \( x > -3 \) but \( x \leq 3 \).

2. **Evaluating the statement \( P \)**:
   - The statement \( P \) asserts that for **all** \( x \) in the domain, if \( x > -3 \) is true, then \( x > 3 \) must also be true.
   - To find a domain where this statement holds true, we need to analyze when \( x > -3 \) is true without \( x > 3 \) being false.

3. **Testing the options**:
   - **Option 1: \( -3 < x < 3 \)**:
     - Here, \( x \) can take values like \( -2, 0, 2 \), etc. For these values, \( x > -3 \) is true, but \( x > 3 \) is false. Therefore, the implication fails, and \( P \) is false in this domain.
   - **Option 2: \( x > -3 \)**:
     - This includes all values greater than \( -3 \) (e.g., \( -2, 0, 2, 4 \), etc.). For \( x \) values between \( -3 \) and \( 3 \), the implication fails again, as shown above. Thus, \( P \) is false in this domain as well.
   - **Option 3: \( x > 3 \)**:
     - Here, every \( x \) is greater than \( 3 \). In this case, if \( x > -3 \) is true (which it is since all numbers greater than 3 are also greater than -3), then \( x > 3 \) is also true. Therefore, the implication holds for all \( x \) in this domain, making \( P \) true here.
   - **Option 4: None of the other options**:
     - Since we found that \( x > 3 \) makes \( P \) true, this option is not correct.

4. **Conclusion**:
The only domain for which \( P \) evaluates to true is when \( x > 3 \). Thus, the answer is:

**Answer: x > 3.**


Question: Given graph 1→2, 1→3, 2→3, 3→2, switching from Page Rank to Teleporting PageRank will have an influence on the value(s) of:

To analyze the influence of switching from PageRank to Teleporting PageRank on the values of the nodes in the given directed graph, we need to understand the mechanics of both algorithms.

First, let's define the graph based on the provided edges:
- Node 1 has outgoing edges to Nodes 2 and 3.
- Node 2 has an outgoing edge to Node 3.
- Node 3 has an outgoing edge back to Node 2.

The edges can be summarized as:
- \(1 \rightarrow 2\)
- \(1 \rightarrow 3\)
- \(2 \rightarrow 3\)
- \(3 \rightarrow 2\)

**PageRank Overview:**
PageRank is an algorithm that ranks nodes in a directed graph based on the number and quality of links. It assigns a score to each node based on the principle that "important" nodes are likely to receive more links from other nodes. The standard PageRank formula can be expressed recursively as follows:

\[
PR(v) = (1 - d) + d \sum_{u \in B(v)} \frac{PR(u)}{L(u)}
\]

where:
- \(PR(v)\) is the PageRank of node \(v\),
- \(d\) is the damping factor (typically set around 0.85),
- \(B(v)\) is the set of nodes that link to \(v\),
- \(L(u)\) is the number of outbound links from node \(u\).

**Teleporting PageRank Overview:**
Teleporting PageRank incorporates a teleportation mechanism, allowing a user to jump to any node with a certain probability. The modified formula becomes:

\[
PR(v) = (1 - d) + d \sum_{u \in B(v)} \frac{PR(u)}{L(u)} + \frac{d}{N}
\]

where \(N\) is the total number of nodes in the graph. The term \(\frac{d}{N}\) represents the probability of teleporting to node \(v\).

**Analyzing the Influence of Switching:**
1. **Effect on Node 1:**
   In both PageRank and Teleporting PageRank, Node 1 is an out-degree source with links to Nodes 2 and 3. However, because Node 1 has no incoming links, its PageRank would typically be lower. In Teleporting PageRank, Node 1 will still receive a teleportation boost of \(\frac{d}{N}\), which allows it to have a non-zero PageRank. Thus, Node 1's value will change when switching to Teleporting PageRank.

2. **Effect on Node 2 and Node 3:**
   For Nodes 2 and 3, which have mutual links, the situation is similar. In regular PageRank, the scores depend heavily on the incoming links. When we switch to Teleporting PageRank, both nodes will also benefit from the teleportation mechanism, which means they'll have their scores adjusted upward. The teleportation term will impact their values as it enables a direct influx of rank from the teleportation.

3. **Overall Network Effect:**
Since the teleportation mechanism influences the rank of every node, all nodes will show a change in their PageRank scores once we switch to Teleporting PageRank. The teleportation provides a uniform distribution contribution across all nodes, ensuring that no node is left unchanged.

Thus, the conclusion is that switching from standard PageRank to Teleporting PageRank will influence the values of all nodes in the graph.

The correct answer is: **All the nodes.**


Question:
How does a Prime+Probe cache attack works? What information does
                    it typically reveal to the attacker about the victim code?

A Prime+Probe cache attack is a side-channel attack that exploits the way modern processors manage cache memory to infer sensitive information from a victimâs execution. This type of attack is particularly effective in scenarios where an attacker can execute code on the same physical machine as the victim, such as in multi-tenant cloud environments or shared computing resources. Hereâs a step-by-step explanation of how this attack works and the information it typically reveals.

### Step-by-Step Explanation of Prime+Probe Attack

1. **Understanding Cache Architecture**: Modern CPUs use hierarchical memory systems, including caches (L1, L2, L3) that store frequently accessed data to speed up processing. The cache is organized in lines, typically 64 bytes each, and has a limited size.

2. **Attack Setup**:
   - The attacker identifies a specific cache line that may be used by the victim application. This could be based on known data structures or memory addresses utilized by the victim.
   - The attacker then primes the cache by loading data into the targeted cache line, filling it with their own data, thereby evicting the existing data (if any).

3. **Victim Execution**: The victim process is executed, and it may access certain data that resides in the cache. The victimâs code is expected to access memory addresses that may overlap with the attackerâs pre-primed cache lines.

4. **Probe Phase**: After the victim's execution, the attacker probes the cache. This involves checking whether the targeted cache line is still present in the cache:
   - If the cache line has been accessed by the victim, it will likely still be in the cache (cache hit).
   - If not, it has likely been evicted, indicating that the victim accessed other data (cache miss).

5. **Timing Measurement**: The attacker measures the time it takes to access the cache line. Cache hits are faster than misses; therefore, the attacker can deduce which cache lines were accessed by the victim based on the timing differences.

### Information Revealed to the Attacker

The Prime+Probe attack can reveal several types of sensitive information about the victim code:

1. **Memory Access Patterns**: By observing which cache lines were accessed, the attacker can infer which data structures or variables were used by the victim's code. This can help in understanding the execution flow of the victim program.

2. **Secret Data**: If the victim process is using cryptographic keys or sensitive data in its computations, the attacker might directly infer this information based on the cache line accesses.

3. **Control Flow Information**: The attack can also provide insights into the control flow of the victimâs execution, as certain branches in the code may lead to different memory accesses.

4. **Timing Information**: By analyzing the timing of accesses, the attacker can infer not only which data was accessed but also the frequency and order of those accesses, which can be critical in understanding the algorithmic complexity of the victimâs operations.

### Real-World Example

In practical terms, an attacker might use Prime+Probe to exploit vulnerabilities in cryptographic implementations, such as those found in SSL/TLS libraries. For instance, if a victim process performs operations based on a secret key (like during encryption/decryption), the attacker can learn about the key by monitoring which parts of memory are accessed in the cache during these operations.

### Conclusion

The Prime+Probe cache attack is a powerful technique that highlights the importance of considering side-channel attacks in system design. It underscores the need for secure programming practices, cache isolation, and considerations for shared resources in multi-tenant environments to mitigate potential vulnerabilities.


Question: Implement a function that computes the support for each provided itemset by counting the number of its occurences in the original dataset of transactions.  You can use the following formula:  $$\mathrm{supp}(X) = \frac{|\{t \in T; X \subseteq t\}|}{|T|}$$

Answer: def compute_support(transactions, itemsets):
    support = {}
    total_transactions = len(transactions)
    for itemset in itemsets:
        count = sum(1 for transaction in transactions if set(itemset).issubset(transaction))
        support[itemset] = count / total_transactions if total_transactions > 0 else 0.0
    return support

 Explanation: This function computes the support for each itemset by iterating through each itemset and counting how many transactions contain the itemset as a subset. It uses a generator expression to count occurrences efficiently. The support is calculated as the ratio of the count of transactions containing the itemset to the total number of transactions. If there are no transactions, it returns 0.0 to prevent division by zero.

 Code snippets:
 1. total_transactions = len(transactions) - This line calculates the total number of transactions in the dataset, which is necessary for computing the support.
 2. count = sum(1 for transaction in transactions if set(itemset).issubset(transaction)) - This line counts how many transactions contain the itemset by checking if the itemset is a subset of each transaction, using a generator expression for efficiency.
 3. support[itemset] = count / total_transactions if total_transactions > 0 else 0.0 - This line calculates the support for the itemset and handles the case where there are no transactions to avoid division by zero, ensuring robustness of the function.


Question:
What is predication and why is it (almost) universal in VLIW
  processors? Could it make sense also in a RISC processor? Why?

Predication is a technique used in computer architecture that allows for the conditional execution of instructions without the need for branches. In VLIW (Very Long Instruction Word) processors, predication is particularly significant because it enables multiple instructions to be issued in parallel while avoiding stalls caused by branch mispredictions.

Predication works by associating a condition (or predicate) with an instruction, indicating whether that instruction should be executed based on the evaluation of the predicate. If the predicate evaluates to true, the instruction is executed; if false, it is effectively a no-op. This reduces the need for control flow changes and can increase instruction-level parallelism (ILP) since the processor can issue more instructions simultaneously without waiting for branches to resolve.

### Why is Predication Almost Universal in VLIW Processors?

1. **Instruction-Level Parallelism (ILP)**: VLIW architectures exploit ILP by allowing multiple instructions to be executed in parallel. Predication helps to maximize this parallelism by minimizing control hazards caused by branches.

2. **Reduced Branch Penalties**: By eliminating branches or minimizing their impact, predication reduces the performance penalties associated with mispredicted branches, which can stall the pipeline.

3. **Compiler Control**: VLIW architectures often rely on the compiler to schedule instructions. Predication allows compilers to generate more efficient instruction schedules without worrying about the overhead of branches.

4. **Simplicity in Execution**: VLIW processors typically feature a simpler execution model, where instructions are fetched and executed based on the instruction word rather than complex control logic, making predication a natural fit.

### Can Predication Make Sense in RISC Processors?

Yes, predication can also make sense in RISC (Reduced Instruction Set Computer) processors, although it is less common. Here are some reasons:

1. **Branch Prediction Improvements**: In modern RISC architectures, effective branch prediction techniques can minimize the overhead associated with branches, which may reduce the benefit of predication compared to VLIW.

2. **Simplified Control Flow**: RISC architectures are designed for simplicity and efficiency. Incorporating predication could potentially simplify control flow in certain applications, particularly in cases where branches are heavily mispredicted.

3. **Compiler Optimization**: Advanced compilers could benefit from predication to optimize loops and conditional statements by reducing branch instructions, thus improving performance in specific scenarios.

4. **Specific Use Cases**: Predication might be particularly advantageous in embedded systems or real-time applications where predictable performance is crucial, and avoiding branching latency is beneficial.

In summary, while predication is almost universal in VLIW processors due to their architecture and emphasis on ILP, it could also provide benefits in RISC architectures under certain conditions, particularly where branch prediction is less effective or predictable execution is required. However, the traditional RISC philosophy leans towards simplicity and efficiency, often relying on branch instructions rather than adopting predication extensively.


Question: What is the content of the inode?

The content of an inode is a critical aspect of file systems, particularly in Unix and Linux environments. An inode (index node) is a data structure that stores metadata about a file or directory. Here is a synthesis of the key information regarding what an inode contains:

1. **File Mode**: This information indicates the type of file (e.g., regular file, directory, symbolic link) and the permissions associated with it (read, write, execute) for the owner, group, and others.

2. **Hard Links Counter**: The inode keeps track of how many hard links point to the file. This counter is important for managing the file’s existence; when it reaches zero, the inode and its associated data blocks can be safely freed.

3. **File Size**: The inode contains the size of the file in bytes, which is essential for determining how much disk space the file occupies.

4. **Index Structure for Data Blocks**: The inode includes pointers (or an index structure) that indicate where the actual file data is stored on the disk. These pointers enable the file system to efficiently access the file's contents.

It is important to note that the inode does **not** store the filename itself; instead, the filename is stored in directory entries that map to the inode. Additionally, while the inode contains a user ID (UID) for the owner, it does not store the owner's name as a string. Information about the overall filesystem capacity is also not contained within the inode; that data is typically managed in the superblock.

In summary, the key elements contained in an inode are:
- **File mode**
- **Hard links counter**
- **File size**
- **Index structure for data blocks**

Understanding the function and content of inodes is essential for grasping how files are managed and accessed in a file system.


Question: What are the differences between statically scheduled HLS and dynamically scheduled HLS?

High-Level Synthesis (HLS) is a process that converts high-level programming languages (like C/C++) into hardware description languages (like Verilog or VHDL) for the design of digital circuits. HLS can be categorized into two main types based on scheduling: statically scheduled HLS and dynamically scheduled HLS. Here’s a step-by-step breakdown of the differences between these two approaches:

### 1. **Definition**
   - **Statically Scheduled HLS**: In this approach, the scheduling of operations is determined at compile time. The HLS tool analyzes the program and generates a fixed schedule for executing operations based on a predefined set of rules and constraints.
   - **Dynamically Scheduled HLS**: This approach allows scheduling decisions to be made at runtime. Operations can be reordered, and resources can be allocated as needed during execution, which provides greater flexibility.

### 2. **Scheduling Time**
   - **Static Scheduling**: The schedule is fixed and determined before runtime. Once the synthesis is complete, the schedule cannot change, which can lead to inefficiencies if the actual workload varies.
   - **Dynamic Scheduling**: The schedule can change during execution based on current conditions. This means that the system can adapt to different workloads or data dependencies as they arise.

### 3. **Complexity**
   - **Static Scheduling**: Generally simpler to implement and understand, as the scheduling algorithms do not need to account for runtime variations. However, this can lead to suboptimal resource utilization and throughput if the workload is not uniform.
   - **Dynamic Scheduling**: More complex, as it requires mechanisms to manage resource allocation and scheduling decisions at runtime. This can involve additional overhead in terms of control logic and can lead to increased latency.

### 4. **Resource Utilization**
   - **Static Scheduling**: Resource utilization is predetermined based on the static schedule. If there are idle resources due to fixed scheduling, this can result in wasted hardware.
   - **Dynamic Scheduling**: Can potentially improve resource utilization, as it allows for better adaptation to the actual data flow and resource availability. It can dynamically allocate resources to operations based on their readiness.

### 5. **Performance**
   - **Static Scheduling**: Performance can be predictable and consistent, but may not fully exploit parallelism if the schedule does not account for varying execution times or data dependencies.
   - **Dynamic Scheduling**: Typically offers better performance in scenarios with varying workloads or unpredictable data dependencies, as it can exploit parallelism more effectively.

### 6. **Use Cases**
   - **Static Scheduling**: Often used in applications with predictable workloads or where resource constraints are well understood, such as embedded systems with fixed functionality.
   - **Dynamic Scheduling**: More suited for applications that require adaptability and optimization based on varying workloads, like high-performance computing or real-time data processing.

### 7. **Example**
   - **Static Scheduling Example**: Consider a simple loop that processes an array. The HLS tool may statically determine the order of operations based on the loop structure and data dependencies, leading to a fixed schedule.
   - **Dynamic Scheduling Example**: In a scenario where the data being processed can change at runtime (e.g., streaming data), a dynamically scheduled HLS might reorder operations based on the current data availability, optimizing throughput on-the-fly.

### Conclusion
In summary, the choice between statically scheduled and dynamically scheduled HLS depends on the specific requirements of the application, such as the predictability of workloads, performance needs, and resource constraints. Statically scheduled HLS provides simplicity and predictability, while dynamically scheduled HLS offers flexibility and potentially better performance in diverse conditions.


Question: Explain why any fail-noisy consensus algorithm (one that uses an eventually perfect failure detector ◇P) requires a majority of the processes to be correct. More precisely, provide a “bad run” in the case where the majority of processes is faulty.

To understand why a fail-noisy consensus algorithm requires a majority of processes to be correct, let's break down the reasoning:

1. **Setup**: We have N processes, where N is even. Two groups, A and B, propose different values.

2. **Contradiction Assumption**: We assume that consensus can be achieved even if half (N/2) of the processes fail.

3. **Execution Scenarios**:
   - In Execution 1, all processes in A crash. Processes in B achieve consensus on value B.
   - In Execution 2, all processes in B crash. Processes in A achieve consensus on value A.

4. **Bad Execution**: In Execution 3, each group suspects the other, leading to confusion. Processes can’t distinguish between this scenario and the previous executions.

5. **Conclusion**: Since A decides A and B decides B, agreement is violated. This shows that if the majority (N/2) are faulty, consensus is impossible, necessitating a majority of correct processes for consensus to function reliably.

This approach effectively illustrates the fundamental requirement for achieving consensus amidst failures.


Question: What is a good distance metric to be used when you want to compute the similarity between documents independent of their length?A penalty will be applied for any incorrect answers.

When considering the problem of measuring similarity between documents while taking their length variability into account, one must delve into the properties of various distance metrics. The fundamental principle here revolves around how each metric treats the magnitude of vectors representing the documents.

Cosine similarity emerges as a strong candidate due to its unique ability to focus solely on the orientation of the document vectors rather than their magnitude. This is particularly beneficial because documents of varying lengths can skew metrics that are sensitive to absolute distances. Cosine similarity is computed as follows:

$$
\text{Cosine Similarity}(A, B) = \frac{A \cdot B}{\|A\| \|B\|}
$$

where $A \cdot B$ is the dot product of the vectors and $\|A\|$ and $\|B\|$ are their magnitudes. By normalizing the vectors, the cosine similarity effectively neutralizes length disparities, thereby allowing for a more accurate representation of the true similarity between documents.

On the other hand, both Euclidean and Manhattan distances are heavily influenced by the lengths of the document vectors. For instance, Euclidean distance computes the straight-line distance in multi-dimensional space, which means longer documents will inherently have larger distances, thus skewing results. Similarly, Manhattan distance sums the absolute differences across dimensions, which also varies with length. Chi-squared distance, while useful in certain statistical contexts, similarly does not effectively adjust for length variability in document comparison.

In light of this analysis, the most appropriate choice for measuring document similarity, independent of their lengths, is indeed cosine similarity.

To summarize, cosine similarity is the recommended distance metric for evaluating document similarity without being affected by document length, while the other options are less suitable due to their inherent sensitivity to vector magnitudes.


Question: Consider an undirected graph $G=(V,E)$ and let $s\neq t\in V$. In the minimum (unweighted) $s,t$-cut problem, we wish to find a set $S\subseteq V$ such that $s\in S$, $t\not \in S$ and the number of edges crossing the cut is minimized. We shall use a linear program to solve this problem. Let ${P}$ be the set of all paths between $s$ and $t$ in the graph $G$. The linear program has a variable $y_e$ for each edge $e\in E$ and is defined as follows: \begin{equation*} \begin{array}{ll@{}ll} \text{minimize}  & & \displaystyle\sum_{e \in E} y_e &\\ \text{subject to}& & \displaystyle\sum_{e \in p} y_e \ge 1  &\forall p \in P,\\ & & y_e \ge 0 & \forall e \in E. \end{array} \end{equation*} For example, consider the following graph where the numbers on the edges depict the $y_e$-values of a feasible solution to the linear program: \begin{center} \input{cutExample} \end{center} The values on the edges depict a feasible but not optimal solution to the linear program. That it is feasible follows because each $y_e$ is non-negative and $\sum_{e\in p} y_e \geq 1$ for all $p\in P$. Indeed, for the path $s, b, a, t$ we have $y_{\{s,b\}}+ y_{\{b,a\}} + y_{\{a,t\}} = 1/4 + 1/4 + 1/2 = 1$, and similar calculations for each path $p$ between $s$ and $t$ show that $\sum_{e\in p} y_e \geq 1$. That the solution is not optimal follows because its value is $2.5$ whereas an optimal solution has value $2$. Prove that $\opt\leq \optlp$, where $\opt$ and $\optlp$ are defined as in {\bf 6a}. \\ Hint: Round a feasible linear programming solution $y$. In the (randomized) rounding it may be helpful to consider, for each vertex $v\in V$, the length of the shortest path from $s$ to $v$ in the graph where edge $e\in E$ has length $y_e$. For example, in the graph and linear programming solution depicted in the problem statement, we have that the length of the shortest path from $s$ to $a$ equals $1/2$. \\ {\em (In this problem you are asked to prove $\opt \leq \optlp$. Recall that you are allowed to refer to material covered in the lecture notes.)}

To prove that \(\opt \leq \optlp\), we start by considering \(y\), an optimal solution to the linear program, with a value of \(\optlp\). We will implement a randomized rounding technique to produce a cut \(S\) that, in expectation, will cut at most \(\optlp\) edges. For each vertex \(v \in V\), define \(x_v\) as the length of the shortest path from \(s\) to \(v\) in the modified graph where the length of each edge \(e\) is given by \(y_e\). Note that \(x_s = 0\) and \(x_t \geq 1\), since for every path \(p\) connecting \(s\) to \(t\), it holds that \(\sum_{e \in p} y_e \geq 1\). The key property we utilize is that for any edge \(\{u,v\} \in E\), we have \(y_{\{u,v\}} \geq |x_v - x_u|\). This implies we can derive a feasible solution for the linear program with a cost of at least \(\opt\). The rounding algorithm is as follows: we select \(\theta\) uniformly at random from \([0,1]\) and define the set \(S = \{v \in V : x_v < \theta\}\). This ensures \(s \in S\) and \(t \notin S\). The expected number of edges cut can be calculated by analyzing the probability that each edge \(e = \{u,v\}\) is cut. Specifically, if \(x_u < x_v\), then edge \(e\) is cut if \(\theta\) falls within the interval \([x_u, x_v]\), which occurs with probability at most \(x_v - x_u\). By our earlier observation, \(x_v - x_u \leq y_e\). Thus, the expected number of edges cut satisfies \(\E[\text{\# edges cut}] \leq \sum_{e \in E} y_e = \optlp\). Since any valid cut must cut at least \(\opt\) edges, we conclude that \(\opt \leq \optlp\).


Question: Using a block cipher, we can build \ldots

To evaluate what can be built using a block cipher, let's briefly examine the capabilities of block ciphers and the cryptographic constructs listed in the options.

1. **Hash Functions**: While block ciphers are not inherently hash functions, they can be used to construct hash functions through various schemes, such as the Merkle-Damgård construction. However, block ciphers alone don’t directly produce a hash function.

2. **Message Authentication Codes (MACs)**: Block ciphers can be directly used to create MACs, specifically through techniques like Cipher Block Chaining (CBC) or the more straightforward HMAC (Hash-based Message Authentication Code) when combined with hash functions.

3. **Stream Ciphers**: Block ciphers can also be utilized to create stream ciphers using techniques like Counter (CTR) mode or Output Feedback (OFB) mode, which transform the block cipher into a stream cipher by generating a keystream.

### Conclusion:
The option that correctly identifies what can be built using a block cipher is **hash functions, MACs, and stream ciphers**.

### Intuitive Example:
Consider a block cipher like AES. You can use AES to encrypt data in blocks, which can serve as the foundation for several cryptographic applications:
- **Building a MAC**: You can encrypt a message with AES and append a nonce to create a unique tag that verifies message integrity.
- **Creating a hash function**: You can repeatedly apply AES in a specific manner to derive a fixed-size output from varying input sizes.
- **Constructing a stream cipher**: By using AES in CTR mode, you generate a continuous keystream that can be XORed with plaintext to produce ciphertext, effectively functioning like a stream cipher.

This versatility illustrates the power of block ciphers in the cryptographic landscape.


Question: Select the \textbf{most accurate} answer. For a hash function to be secure (\textbf{but still efficient}) against collision attacks in 2015, the output length should be\dots

When evaluating the security of a hash function against collision attacks, the output length is a crucial factor. The principle behind this is related to the number of possible outputs versus the number of inputs that can be hashed.

In 2015, cryptographic standards leaned towards ensuring that hash functions could withstand collision attacks efficiently. Generally, a longer output length increases security but can also impact performance.

While 80 bits might be too short for robust security, lengths of 512 bits or 1024 bits might be excessive for many practical applications, potentially leading to inefficiencies. Thus, a balance must be struck.

Considering these factors, an output length in the mid-range, such as 160 bits, is often cited as a well-accepted standard that provides adequate security without significant efficiency loss. However, one could also contemplate the implications of choosing shorter or longer lengths based on specific use cases and performance requirements.


Question: Select the \emph{incorrect} statement. The Bluetooth project aims for

To determine the incorrect statement about the Bluetooth project, we need to analyze each option based on the goals of Bluetooth technology:

1. **Low Complexity**: Bluetooth aims to provide a simple and easy-to-use wireless communication method, so this statement is correct.
2. **Low Power**: One of the main objectives of Bluetooth technology is to operate efficiently with low power consumption, making this statement correct.
3. **Low Cost**: Bluetooth technology is designed to be cost-effective for both manufacturers and consumers, so this statement is also correct.
4. **Low Security**: Bluetooth includes security features to protect data during transmission. Aiming for low security contradicts the project's goals, making this statement incorrect.

Thus, the incorrect statement is related to low security.

4


Question:
For this question, one or more assertions can be correct. Tick only the correct assertion(s). There
will be a penalty for wrong assertions ticked.Using a 3-gram character model, which of the following expressions are equal to \( P(\text{opossum}) \) ?

A


Question: Using the same set of transformations as in the previous question, what is the final value you get for the edit distance between execution and exceuton, i.e. D(execution, exceuton)?Give your answer as a numerical value.

To calculate the edit distance between the two strings \"execution\" and \"exceuton\", we will use dynamic programming as previously described. \n\n1. **Initialize the distance matrix**:\n\n   Let `m` be the length of \"execution\" (which is 10) and `n` be the length of \"exceuton\" (which is 8). We will create a (m+1) x (n+1) matrix.\n\n   ```\n   D[i][0] = i (deletion cost for all characters of \"execution\")\n   D[0][j] = j (insertion cost for all characters of \"exceuton\")\n   ```\n\n2. **Fill the matrix**:\n\n   We will fill the matrix according to the rules for edit distance.\n\n   The initial state of the matrix looks like this:\n\n   ```\n           \"\"  e  x  c  e  u  t  o  n\n       0  0  1  2  3  4  5  6  7  8\n   e   1  1\n   x   2  2\n   e   3  3\n   c   4  4\n   u   5  5\n   t   6  6\n   i   7  7\n   o   8  8\n   n   9  9\n   ```\n\n   Now, we will fill in the matrix using the defined operations:\n\n   - For each cell `D[i][j]`, we will calculate the minimum cost based on the operations.\n\n   Filling each cell step-by-step, we find:\n\n   ```\n   D[1][1] = 0 (e == e)\n   D[1][2] = 1 (insertion)\n   D[1][3] = 2 (insertion)\n   D[1][4] = 3 (insertion)\n   D[1][5] = 4 (insertion)\n   D[1][6] = 5 (insertion)\n   D[1][7] = 6 (insertion)\n   D[1][8] = 7 (insertion)\n\n   D[2][1] = 1 (deletion)\n   D[2][2] = 0 (x == x)\n   D[2][3] = 1 (substitution)\n   D[2][4] = 2 (insertion)\n   D[2][5] = 3 (insertion)\n   D[2][6] = 4 (insertion)\n   D[2][7] = 5 (insertion)\n   D[2][8] = 6 (insertion)\n\n   D[3][1] = 2 (deletion)\n   D[3][2] = 1 (deletion)\n   D[3][3] = 0 (c == c)\n   D[3][4] = 1 (substitution)\n   D[3][5] = 2 (insertion)\n   D[3][6] = 3 (insertion)\n   D[3][7] = 4 (insertion)\n   D[3][8] = 5 (insertion)\n\n   D[4][1] = 3 (deletion)\n   D[4][2] = 2 (deletion)\n   D[4][3] = 1 (deletion)\n   D[4][4] = 0 (e == e)\n   D[4][5] = 1 (insertion)\n   D[4][6] = 2 (insertion)\n   D[4][7] = 3 (insertion)\n   D[4][8] = 4 (insertion)\n\n   D[5][1] = 4 (deletion)\n   D[5][2] = 3 (deletion)\n   D[5][3] = 2 (deletion)\n   D[5][4] = 1 (deletion)\n   D[5][5] = 0 (u == u)\n   D[5][6] = 1 (insertion)\n   D[5][7] = 2 (insertion)\n   D[5][8] = 3 (insertion)\n\n   D[6][1] = 5 (deletion)\n   D[6][2] = 4 (deletion)\n   D[6][3] = 3 (deletion)\n   D[6][4] = 2 (deletion)\n   D[6][5] = 1 (deletion)\n   D[6][6] = 0 (t == t)\n   D[6][7] = 1 (insertion)\n   D[6][8] = 2 (insertion)\n\n   D[7][1] = 6 (deletion)\n   D[7][2] = 5 (deletion)\n   D[7][3] = 4 (deletion)\n   D[7][4] = 3 (deletion)\n   D[7][5] = 2 (deletion)\n   D[7][6] = 1 (deletion)\n   D[7][7] = 0 (o == o)\n   D[7][8] = 1 (insertion)\n\n   D[8][1] = 7 (deletion)\n   D[8][2] = 6 (deletion)\n   D[8][3] = 5 (deletion)\n   D[8][4] = 4 (deletion)\n   D[8][5] = 3 (deletion)\n   D[8][6] = 2 (deletion)\n   D[8][7] = 1 (deletion)\n   D[8][8] = 0 (n == n)\n   ```\n\n3. **Final result**:\n\n   The value in the bottom-right cell `D[10][8]` gives us the edit distance, which is calculated to be **2**.\n\nThus, the final value for the edit distance \( D(\text{\"execution\"}, \text{\"exceuton\"}) \) is **2**.


Question: A false negative in sampling can only occur for itemsets with support smaller than

In the context of sampling for itemset mining, a false negative occurs when an itemset that actually meets the support threshold is incorrectly identified as not meeting it due to the randomness of the sampling process.

To understand the options:

1. **the threshold s**: This is the direct support threshold that is used to determine whether an itemset is considered frequent. If an itemset has support smaller than this threshold, it is not frequent, so a false negative cannot occur for itemsets whose support is greater than or equal to the threshold.

2. **p*s**: Here, \( p \) typically represents the sampling fraction (the proportion of the dataset that is sampled). If an itemset has a true support that is less than \( p*s \), it may not be detected in the sampled data, resulting in a false negative. This option suggests that false negatives can occur for itemsets that are below a certain scaled threshold based on the sampling rate.

3. **p*m**: This option appears to represent a different concept. If \( m \) is the total number of transactions, \( p*m \) would represent the number of transactions sampled, but it does not directly relate to the support threshold for false negatives in a straightforward way.

4. **None of the above**: This option suggests that none of the previous options correctly identifies the condition under which a false negative can occur.

Given this analysis, the correct answer is:

**2. p*s**.

This is because a false negative can occur for itemsets with support smaller than the product of the sampling fraction and the support threshold.


Question: Two excellent students, Alice from EPFL and Bob from MIT, have both built their own spam filters. A spam filter is an algorithm that takes as input an email and outputs $1$ if the email is spam and $0$ otherwise.   Alice and Bob now want to compare their two spam filters. To perform the comparison, they both download the same huge data set consisting of $n$ emails out of which some are spam.  Alice then runs her spam filter on the data set  to obtain $a_1, a_2, \ldots, a_n$ where $a_i \in \{0,1\}$ is the output of her spam filter on the $i$:th email in the data set. Similarly, Bob runs his spam filter on the data set to obtain $b_1, b_2, \ldots, b_n$ where $b_i \in \{0,1\}$ is the output of his spam filter on the $i$:th email in the data set. Their goal is then to  determine whether their outputs are the same. An issue that they face is that $a_1, a_2,\ldots, a_n$ are stored on Alice's computer  and $b_1, b_2, \ldots, b_n$ are stored on Bob's computer. They thus need to transfer (or communicate) information to solve the problem. A trivial solution is for Alice to transfer all her outputs  $a_1, a_2,\ldots, a_n$ to Bob who then performs the comparison. However, this requires Alice to send $n$ bits of information to Bob; an operation that is very costly for a huge data set. In the following, we use randomization to achieve a huge improvement on the number of bits transfered between Alice and Bob. \\[0mm] Specifically, motivated  by something called pseudo-random generators, we  assume that Alice and Bob have access to the same randomness (called shared randomness). That is, Alice and Bob have access to the same infinite stream of random bits $r_1, r_2, \ldots$. Your task is now to  use this shared randomness to devise a randomized protocol of the following type: \begin{itemize} \item As a function of $a_1, a_2, \ldots, a_n$ and the random bits $r_1, r_2, \ldots$, Alice computes a message $m$ that consists of only $2$ bits. She then transmits this $2$-bit message $m$   to Bob. \item Bob then, as a function of $b_1, b_2, \ldots, b_n$, the message $m$, and the random bits $r_1, r_2, \ldots$, outputs \textsc{Equal} or \textsc{Not Equal}. \end{itemize} Bob's output is correct if he outputs $\textsc{Equal}$ when $a_i = b_i$ for all $i\in \{1,\ldots, n\}$ and $\textsc{Not Equal}$ otherwise. Your protocol should ensure that Bob outputs the correct answer with probability at least $2/3$, where the probability is over the random bits $r_1, r_2, \ldots $.\\ {\em (In this problem you are asked to (i) explain how Alice computes the message $m$ of $2$ bits (ii) explain how Bob calculates his output, and (iii) prove that Bob's output is correct  with probability at least $2/3$. A correct solution where Alice sends a message $m$ of $O(\log n)$ bits is rewarded $12$ points.  Recall that you are allowed to refer to material covered in the lecture notes.) }\\ \noindent {\small An interesting fact (but unrelated to the exam) is that any correct deterministic strategy would require Alice and Bob  to send  $n$ bits of information.}

Alice computes a 2-bit message \( m \) consisting of the parities of her outputs, while Bob computes his parities and compares them with \( m \); this ensures Bob's output is correct with a probability of at least \( 2/3 \).


Question:
Explain how precise exceptions are implemented in
  dynamically-scheduled out-of-order processors.

Precise exceptions are critical for ensuring that when an exception occurs in a dynamically scheduled out-of-order processor, the system can maintain a consistent and predictable state. This is essential for debugging, system calls, and handling various error conditions. Implementing precise exceptions in such processors involves several key components and techniques:

### Key Concepts for Precise Exceptions:

1. **In-Order Commit**: Even though instructions may be executed out of order, they should commit (i.e., complete their effects on the architectural state) in the original program order. This ensures that the state of the program remains consistent, allowing exceptions to be handled as if the program were executing in a strictly sequential manner.

2. **Retiring Instructions**: An instruction is considered to be "retired" or "committed" only when it has completed its execution and all prior instructions have also been retired. This is crucial for maintaining a precise exception model, as it guarantees that the state reflects the program's logical progression.

### Implementation Mechanisms:

1. **Reorder Buffer (ROB)**:
   - The reorder buffer is a key data structure used to track the status of instructions that have been issued for execution but have not yet been committed.
   - Each entry in the ROB corresponds to an instruction, storing its result, status, and destination register.
   - When an instruction is ready to commit, the ROB ensures that it is committed in the correct program order. This means that even if an instruction was executed out of order, it is not allowed to affect the architectural state until all earlier instructions have also been committed.

2. **Exception Handling Logic**:
   - When an exception occurs (e.g., due to a division by zero, invalid memory access, etc.), the processor must determine the precise point of failure.
   - The ROB helps in keeping track of which instructions have been completed and which have not, allowing the processor to identify the exact state of the program at the time the exception was raised.
   - If an exception is raised, instructions that have been committed (retired) are allowed to affect the state, while those that are still in the ROB can be effectively rolled back.

3. **State Recovery**:
   - Upon an exception, the processor will flush the ROB and any other buffers, removing instructions that have not yet committed.
   - The processor will then load the state information from the last committed instruction. This typically involves restoring the architectural registers and memory state to what it was at the last precise point before the exception occurred.
   - The processor can then invoke the appropriate exception handler.

4. **Instruction Status Tracking**:
   - Each instruction is tracked through various stages of execution (e.g., issue, execution, completion, commit) within the pipeline.
   - Status flags or bits can indicate whether an instruction has been issued, whether it has completed execution, and whether it is eligible for commit.

5. **Branch Prediction and Recovery**:
   - In the presence of branches, the processor must also ensure that any speculative execution does not lead to exceptions. If a branch prediction is incorrect, any speculative instructions must be discarded, and the state must revert to a point before the mispredicted branch.
   - This is managed in conjunction with the ROB, which ensures that only correctly predicted instructions can commit.

### Conclusion:

In summary, precise exceptions in dynamically scheduled out-of-order processors are implemented through a combination of in-order commit mechanisms, the use of reorder buffers to track instruction status and results, and careful management of exception handling and state recovery. By ensuring that all architectural changes occur in a well-defined order, the architecture maintains the illusion of sequential execution, allowing for robust exception handling and recovery. This design is crucial for supporting complex control flow and error conditions in modern processors.


Question: Let $G$ be a set and $*$ a commutative operation on pairs of elements from $G.$ Suppose there exists an element $e\in G$ such that $a*e=e*a=a$ for all $a \in G.$ Also, suppose there exist elements $b,c,d \in G$ such that $b*c=d*c.$. True or false: $(G,*)$ is a group if and only if $b=d.$

To determine whether the statement is true or false, we need to analyze the properties of the set \( G \) with the operation \( * \).

1. **Identity Element**: The statement specifies that there is an element \( e \) in \( G \) such that \( a * e = e * a = a \) for all \( a \in G \). This means that \( e \) serves as the identity element for the operation \( * \).

2. **Commutativity**: The operation \( * \) is commutative, meaning \( a * b = b * a \) for any \( a, b \in G \).

3. **Existence of Inverses**: For \( (G, *) \) to be a group, every element \( a \in G \) must have an inverse \( a^{-1} \) such that \( a * a^{-1} = e \).

4. **Given Condition**: We have \( b * c = d * c \). This implies that \( b * c \) and \( d * c \) are equal.

5. **Exploring Implications**: If we assume \( b \neq d \), we can manipulate the equation \( b * c = d * c \):
   - Rearranging gives \( b * c * c^{-1} = d \) (if \( c \) has an inverse, \( c^{-1} \)).
   - However, if \( c \) does not have an inverse or if the operation does not yield unique results for different inputs, this does not lead to a contradiction.

6. **Analyzing the Statement**: The statement claims that \( (G, *) \) is a group if and only if \( b = d \).
   - If \( b \neq d \), we can still have a set \( G \) where not all elements have an inverse, hence it may not form a group.
   - If \( b = d \), we do not directly conclude that \( G \) is a group unless we confirm that every element has an inverse.

7. **Conclusion**: The statement's condition \( b * c = d * c \) does not guarantee that \( b = d \) is necessary for \( (G, *) \) to be a group.

Therefore, the final answer is **False**.


Question: Select the \emph{incorrect} statement.

Let\'s analyze each statement one by one to identify the incorrect one:\n\n1. **"The order of an element is always a multiple of the order of its group."**\n   - This statement is incorrect. The order of an element in a group is defined as the smallest positive integer \\( k \\) such that \\( g^k = e \\) (the identity element), and this order can be less than the order of the group.\n\n2. **"An ideal \\( I \\) of a commutative ring \\( R \\) is a subgroup closed under multiplication by all elements of \\( R \\)."**\n   - This statement is correct. An ideal \\( I \\) of a commutative ring \\( R \\) is indeed a subgroup of \\( R \\) under addition and is closed under multiplication by any element of \\( R \\).\n\n3. **"Given a prime \\( p \\), we have \\( a^{p} = a \\) for every \\( a \\in \\mathbb{Z}_p \\)."**\n   - This statement is correct according to Fermat\'s Little Theorem, which states that for any integer \\( a \\) such that \\( a \\) is not divisible by the prime \\( p \\) (and a similar consideration can apply for \\( a = 0 \\)), \\( a^p \\equiv a \\mod p \\).\n\n4. **"Any element of order \\( \\varphi(n) \\) is a generator of \\( \\mathbb{Z}_n^*."**\n   - This statement is also incorrect. For an element to be a generator of the multiplicative group \\( \\mathbb{Z}_n^* \\), it must have order equal to the order of the group. While \\( \\varphi(n) \\) represents the number of elements in \\( \\mathbb{Z}_n^* \\), not every element of that order is guaranteed to be a generator.\n\nThus, the **incorrect** statements are:\n\n1. "The order of an element is always a multiple of the order of its group."\n4. "Any element of order \\( \\varphi(n) \\) is a generator of \\( \\mathbb{Z}_n^*."\n\nBased on this evaluation, the clear incorrect statement is: \n\n**"The order of an element is always a multiple of the order of its group."** \n\nHowever, if only one option should be chosen as \'incorrect,\' the first is typically more recognized. If forced to select the "most" incorrect as per conventional mathematics, it would be the first statement due to elemental group theory principles.


Question: Tick the \textbf{true} assertion. $x\in \mathbf{Z}_{n}$ is invertible iff \ldots

To determine the correct assertion regarding when an element \( x \) in \( \mathbf{Z}_n \) (the set of integers modulo \( n \)) is invertible, we need to understand the concept of invertibility in modular arithmetic.

An element \( x \in \mathbf{Z}_n \) is said to be **invertible** if there exists an integer \( y \) such that:

\[
x \cdot y \equiv 1 \ (\text{mod } n)
\]

This means that the product of \( x \) and \( y \) leaves a remainder of 1 when divided by \( n \).

### Key Concept: Invertibility Condition

The criterion for an element \( x \) to be invertible in \( \mathbf{Z}_n \) is that the **greatest common divisor (gcd)** of \( x \) and \( n \) must equal 1:

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

This condition indicates that \( x \) and \( n \) share no common factors other than 1, which allows for the existence of a multiplicative inverse.

### Evaluating the Choices

Now, letâs evaluate the provided options:

- **A. \( \varphi(n) = n - 1 \)**: This statement is not true in general. The function \( \varphi(n) \) (Euler's totient function) counts the number of integers up to \( n \) that are coprime to \( n \). While \( \varphi(n) \) does relate to invertibility, it does not define when a single \( x \) is invertible.

- **B. \( x \) is prime**: This is incorrect. While prime numbers greater than 1 are invertible modulo any number they do not divide, not all invertible elements must be prime. For example, \( x = 5 \) is invertible modulo \( 12 \) (since \( \text{gcd}(5, 12) = 1 \)), but \( 5 \) itself is prime.

- **C. \( x \) is not prime**: This is also incorrect. An invertible element can be prime, so this does not necessarily indicate invertibility.

- **D. \( \text{gcd}(x, n) = 1 \)**: This is the correct assertion. As stated earlier, for \( x \) to be invertible in \( \mathbf{Z}_n \), the gcd condition must hold.

### Conclusion

The correct answer is:

**D. \( \text{gcd}(x, n) = 1 \)**

This means that \( x \) is invertible in \( \mathbf{Z}_n \) if and only if \( x \) and \( n \) share no common factors other than 1.


Question:
Assume you're working for a startup that develops a university management app. You just received a description of what the app should do:

> This app will be the administrative backbone of the university.
> Almost all staff will use it.
> Human Resources will register each student, including their personal details, and use the system to ensure each student follows the rules concerning the duration of studies, the number of courses that must be taken, the payment of all applicable fees...
> Professors will use the app to input grades and to send informational messages to students in their courses.
> Students will be able to see the list of courses and register for a course.
> Staff members will also be able to update their personal details, including their banking coordinates for their salary.

Write a user story, in a single sentence using the below format, that summarizes this conversation:
> As a student, I want to ... so that ...

Your story must contain all necessary information and only that information.

As a student, I want to view the list of courses and register for a course so that I can manage my academic schedule effectively.


Question: Consider the following linear program for finding a  maximum-weight matching: \begin{align*} \text{Maximize}  \quad &\sum_{e\in E} x_e w_e\\ \text{Subject to} \quad &\sum_{e \in \delta(v)} x_e \leq 1 \quad \forall v \in V \\ &x_e \geq 0 \quad \forall e \in E \end{align*} (This is similar to the perfect matching problem seen in the lecture, except that we have inequality constraints instead of equality constraints.) Prove that, for bipartite graphs,  any extreme point is integral.

To prove that any extreme point of the linear program for finding a maximum-weight matching in bipartite graphs is integral, we will leverage key concepts from linear programming, properties of bipartite graphs, and the theory of matching.

### Key Concepts

1. **Linear Programming and Extreme Points**: In linear programming, an extreme point (or vertex) of the feasible region defined by the constraints is a solution that cannot be expressed as a convex combination of other feasible solutions. For a linear program with rational coefficients, if an extreme point exists, it can be shown that it can be represented with integer values under certain conditions.

2. **Bipartite Graphs**: A bipartite graph \( G = (U, V, E) \) consists of two disjoint sets of vertices \( U \) and \( V \) such that every edge \( e \in E \) connects a vertex in \( U \) to a vertex in \( V \).

3. **Integrality of Matchings**: For bipartite graphs, a fundamental property is that maximum matchings can be found efficiently using algorithms such as the Hungarian algorithm or the Hopcroft-Karp algorithm, which yield integral solutions.

### Mathematical Formulation

The linear program we are considering can be stated as follows:

\[
\text{Maximize} \quad \sum_{e \in E} x_e w_e
\]

\[
\text{Subject to} \quad \sum_{e \in \delta(v)} x_e \leq 1 \quad \forall v \in V
\]

\[
x_e \geq 0 \quad \forall e \in E
\]

Where \( \delta(v) \) is the set of edges incident to vertex \( v \).

### Proof Outline

1. **Feasibility and Boundedness**: The feasible region defined by the constraints is non-empty (since the zero vector is always a feasible solution) and bounded because the objective function is linear and the constraints form a polytope.

2. **Integral Solutions in Bipartite Graphs**: By the properties of bipartite graphs, any feasible solution to the maximum-weight matching problem can be transformed into an integer solution. This stems from the fact that the matching polytope of a bipartite graph has vertices corresponding to integral solutions.

3. **Extreme Points**: An extreme point of the feasible region corresponds to a solution where the constraints are tight (i.e., active). In the context of bipartite graphs, if we have a solution \( x \) that satisfies all constraints with strict inequalities, then it can be shown that such a solution can be adjusted slightly to yield another feasible solution, thereby contradicting the extremity.

4. **Rational Solutions**: Since the weights \( w_e \) can be assumed to be rational (as they typically are in practical applications), and since the linear program's constraints are also rational, by the integrality theorem for linear programming, any extreme point will be integral.

5. **Conclusion**: Hence, we conclude that any extreme point of the linear program for maximum-weight matching in bipartite graphs is integral.

### Real-World Applications

- **Network Design**: Maximum-weight matchings are applicable in resource allocation problems where resources (edges) need to be allocated to tasks (vertices) while maximizing the total benefit.
- **Supply Chain Management**: In logistics and supply chain, matching suppliers to retailers based on weights representing costs or benefits can be modeled using maximum-weight matchings.
- **Job Assignment Problems**: Assigning jobs to workers based on efficiency and skill levels can also be framed as a maximum-weight matching problem.

### Citations

- **Kuhn, H. W. (1955)**: "The Hungarian Method for the Assignment Problem," which describes an efficient algorithm for finding maximum matchings in bipartite graphs.
- **Edmonds, J., & Karp, R. (1972)**: "Theoretical Improvements in Algorithmic Efficiency for Network Flow Problems," which discusses matchings and provides foundational results in the context of bipartite graphs.

In summary, the linear program's structure, combined with properties specific to bipartite graphs, ensures that extreme points are integral, thus making this proof robust and applicable to various real-world scenarios.


Question:
The MIPS R10000 fetches four instructions at once and, therefore,
there are four such circuits working in parallel inside the processor. What is the function of the ``Old dest'' field in the ``Active
  List''? And what is the function of ``Log dest''? Why are they
  needed in the ``Active list''?

The "Old dest" field in the "Active List" of the MIPS R10000 processor is used to keep track of the destination register of the instruction that is currently being executed or has been issued but not yet committed. This helps in managing the state of the register file and ensuring that the correct data is used when the instruction is eventually committed.

The "Log dest" field, on the other hand, records the destination register of an instruction that is logged for future reference, typically during the process of handling exceptions or rollbacks. It ensures that the processor can track which registers have been modified and maintain the correct architectural state in case an exception occurs.

Both fields are necessary in the "Active List" to facilitate out-of-order execution while maintaining the correct program order for register writes and to ensure that the system can recover gracefully from exceptions or faults. They help manage dependencies and ensure that the correct data is available when needed, ultimately aiding in maintaining the integrity of the execution model.


Question: K-means can be equivalently written as the following Matrix Factorization $$ \begin{aligned} & \min _{\mathbf{z}, \boldsymbol{\mu}} \mathcal{L}(\mathbf{z}, \boldsymbol{\mu})=\left\|\mathbf{X}-\mathbf{M} \mathbf{Z}^{\top}\right\|_{\text {Frob }}^{2} \\ & \text { s.t. } \boldsymbol{\mu}_{k} \in \mathbb{R}^{D}, \\ & z_{n k} \in \mathbb{R}, \sum_{k=1}^{K} z_{n k}=1 . \end{aligned} $$

The given formulation represents K-means clustering as a matrix factorization problem, where \(\mathbf{X}\) is the data matrix, \(\mathbf{M}\) contains the cluster centroids \(\boldsymbol{\mu}\), and \(\mathbf{Z}\) is the assignment matrix. The Frobenius norm captures the distance between the data points and their corresponding cluster centroids.

However, there are key aspects to consider:

1. **Objective Function**: In K-means, the goal is to minimize the squared Euclidean distances between data points and their assigned cluster centers. This can indeed be expressed in terms of a loss function similar to what is presented but with certain constraints.

2. **Constraints**: The constraint \(z_{nk} \in \mathbb{R}\) is misleading because \(z_{nk}\) should actually be binary (0 or 1) indicating whether point \(n\) belongs to cluster \(k\). The condition \(\sum_{k=1}^{K} z_{nk}=1\) enforces that each point is assigned to exactly one cluster, which aligns with K-means behavior.

3. **Matrix Structure**: While the formulation suggests a continuous relaxation of assignments (using real numbers instead of binary), it does not accurately reflect standard K-means, which involves hard assignments.

Thus, while there is some validity in expressing K-means as a matrix factorization problem, the specific details regarding variable types and constraints make this formulation inaccurate for standard K-means clustering.

Therefore, the correct answer is:
(b) no


Question: You are doing your ML project. It is a regression task under a square loss. Your neighbor uses linear regression and least squares. You are smarter. You are using a neural net with 10 layers and activations functions $f(x)=3 x$. You have a powerful laptop but not a supercomputer. You are betting your neighbor a beer at Satellite who will have a substantially better scores. However, at the end it will essentially be a tie, so we decide to have two beers and both pay. What is the reason for the outcome of this bet?

To analyze the situation step by step, let's consider the components of your machine learning project:

1. **Task Type**: You are working on a regression task, which means you are trying to predict continuous values.

2. **Your Approach**: You are using a neural network with 10 layers and a linear activation function \( f(x) = 3x \). This means that despite having multiple layers, the output is still a linear transformation of the input, as any composition of linear functions is itself a linear function. Essentially, your 10-layer neural network is equivalent to a single linear regression model.

3. **Neighbor's Approach**: Your neighbor is using linear regression with least squares, which is a standard approach for regression tasks. This method efficiently finds the best-fitting line (or hyperplane) for the data.

4. **Computational Resources**: You have a powerful laptop, but it is not a supercomputer. Training deep networks can be computationally intensive, especially with many layers, but in this case, since your network is effectively linear, it doesn't require excessive resources.

5. **Outcome of the Bet**: Since your neural network with 10 layers and a linear activation function does not provide any additional expressive power over the linear regression model your neighbor is using, both models are likely to perform similarly on the regression task.

Given these considerations, the outcome of the bet being a tie can be attributed to the fact that both you and your neighbor are using essentially the same scheme for regression: a linear approach.

Therefore, the most appropriate answer to why the outcome of the bet is a tie is:

**Final Answer**: Because we use exactly the same scheme.


Question: Consider the following joint distribution that has the factorization $$ p\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=p\left(x_{1}\right) p\left(x_{2} \mid x_{1}\right) p\left(x_{3} \mid x_{2}\right) p\left(x_{4} \mid x_{1}, x_{3}\right) p\left(x_{5} \mid x_{4}\right) . $$ : (4 points.) Determine whether the following statement is correct. $$ X_{1} \perp X_{3} \mid X_{2}, X_{5} $$ Show your reasoning.

To determine whether the statement \( X_{1} \perp X_{3} \mid X_{2}, X_{5} \) is correct, we need to analyze the joint distribution given and understand the implications of conditional independence in the context of the factorization provided.

### Step 1: Understanding the Factorization

The joint distribution is given by:

\[
p(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = p(x_{1}) p(x_{2} \mid x_{1}) p(x_{3} \mid x_{2}) p(x_{4} \mid x_{1}, x_{3}) p(x_{5} \mid x_{4}).
\]

From this factorization, we can observe the following dependencies:

- \( X_2 \) depends on \( X_1 \).
- \( X_3 \) depends on \( X_2 \).
- \( X_4 \) depends on both \( X_1 \) and \( X_3 \).
- \( X_5 \) depends on \( X_4 \).

### Step 2: Analyzing the Conditional Independence

We want to check if \( X_1 \) is independent of \( X_3 \) given \( X_2 \) and \( X_5 \). In formal terms, we need to check if:

\[
p(X_1, X_3 \mid X_2, X_5) = p(X_1 \mid X_2, X_5) p(X_3 \mid X_2, X_5).
\]

### Step 3: Using the D-separation Criterion

To analyze the independence, we can use the concept of d-separation in a directed acyclic graph (DAG) representation of the joint distribution. The dependencies can be represented as follows:

- \( X_1 \rightarrow X_2 \rightarrow X_3 \)
- \( X_1 \rightarrow X_4 \)
- \( X_3 \) influences \( X_4 \)
- \( X_4 \rightarrow X_5 \)

In this graph:

- \( X_1 \) influences \( X_2 \), which in turn influences \( X_3 \).
- \( X_3 \) is influenced by \( X_2 \) and also influences \( X_4 \), which influences \( X_5 \).

### Step 4: Checking the Paths

To check for independence, we look at the paths between \( X_1 \) and \( X_3 \):

1. The path \( X_1 \rightarrow X_2 \rightarrow X_3 \) is active unless \( X_2 \) is conditioned on.
2. The other path \( X_1 \rightarrow X_4 \leftarrow X_3 \) is also active unless \( X_4 \) is conditioned on.

Since neither \( X_2 \) nor \( X_4 \) is conditioned on in the statement \( X_1 \perp X_3 \mid X_2, X_5 \), the presence of \( X_2 \) does not block the influence of \( X_1 \) on \( X_3 \) through \( X_2 \), and \( X_4 \) does not block the influence through \( X_3 \).

### Conclusion

Since there exists an active path between \( X_1 \) and \( X_3 \) that is not blocked by conditioning on \( X_2 \) and \( X_5 \), we conclude that:

\[
X_1 \not\perp X_3 \mid X_2, X_5.
\]

Thus, the statement \( X_{1} \perp X_{3} \mid X_{2}, X_{5} \) is **incorrect**.


Question: One can find a collision in a hash function $h\colon \{0,1\}^* \rightarrow \{0,1\}^n$ with expected time complexity\dots

The expected time complexity to find a collision in a hash function \( h\colon \{0,1\}^* \rightarrow \{0,1\}^n \) is \( \Theta(2^{n/2}) \).

This result is based on the birthday paradox, which states that the probability of a collision increases significantly as more values are hashed. Specifically, to find a collision with high probability, you need to hash about \( 2^{n/2} \) random inputs.

So, the concise answer is:
\(\Theta(2^{n/2})\).


Question: Which of the following properties is part of the RDF Schema Language?

To determine which properties are part of the RDF Schema (RDFS) language, let's analyze the options step by step.

### RDF Schema Overview
RDF Schema is a semantic extension of RDF (Resource Description Framework) that provides mechanisms to describe the relationships between resources. It allows for the definition of classes and properties, enabling a more structured representation of information.

### Evaluating the Options

1. **Type**:
   - In RDF, the term "type" is often associated with the property `rdf:type`, which is used to state that a resource is an instance of a class. In RDFS, `rdfs:Class` is used to define classes, and `rdf:type` is a fundamental part of how RDF describes the relationships between resources and classes. Thus, "Type" is indeed part of RDFS.

2. **Predicate**:
   - In RDF, a predicate is a property that describes a relationship between two resources (the subject and the object). While the term "predicate" itself is not explicitly defined in RDFS, properties in RDFS can be considered predicates. However, RDFS focuses more on defining properties and classes rather than explicitly labeling them as predicates. Therefore, while related, "Predicate" is not a specific property of RDFS.

3. **Description**:
   - The term "Description" is not a specific property or concept defined within RDFS. RDFS does not have a property called "Description." Instead, it focuses on classes, properties, and their relationships. Thus, this option is not correct.

4. **Domain**:
   - RDFS includes the concept of "domain" through the property `rdfs:domain`, which specifies the class of the subject of a property. This means that if a property has a domain defined, any subject using that property must be an instance of that domain class. Therefore, "Domain" is indeed part of RDFS.

### Conclusion
Based on the analysis:

- **Correct Options**: 1) Type, 4) Domain
- **Incorrect Options**: 2) Predicate, 3) Description

### Summary
In summary, "Type" and "Domain" are properties of the RDF Schema Language. "Type" relates to the classification of resources, while "Domain" specifies the class of subjects for properties. "Predicate" is a broader term related to RDF but not specifically a property of RDFS, and "Description" does not correspond to any defined property in RDFS.


Question: Freshly graduated from EPFL, you have been hired as contractors for a successful and rapidly growing bank. The bank has been experiencing problems with their money management system recently, which is written in Scala, and so they hired the best and brightest young engineer they could find: you! The system had been working perfectly fine so far, they tell you. In the past days, due to an increased number of customers, they had to switch from a single-threaded sequential execution environment to a multi-threaded concurrent one, in which the threads may perform transactions concurrently. That's when problems started, your manager says...  Here is the code responsible to withdraw money from the account from and transfer it to the account to, within the same bank:  def transfer(from: Account, to: Account, amount: BigInt): Unit = {   require(amount >= 0)    val balanceFrom = from.balance    if (balanceFrom >= amount) {     from.balance = balanceFrom - amount     val balanceTo = to.balance     to.balance = balanceTo + amount   } } For the bank, it is very important that the following two properties hold after any sequence of completed transfer transactions:  The balance of an account never goes below 0. The total sum of money held by the bank is constant. Does anything change in the setting where multiple threads can execute the transfer method concurrently? For each of the two desired properties of the system, check if it holds in this concurrent environment. If not, come up with an example execution which exhibits a violation of the property.

Certainly! In this analysis, we will examine the `transfer` method in a banking system, focusing on its implications in a multi-threaded environment with respect to two critical properties:

1. **The balance of an account never goes below 0.**
2. **The total sum of money held by the bank is constant.**

### Overview of the Code

Hereâs the code responsible for transferring money between accounts:

```scala
def transfer(from: Account, to: Account, amount: BigInt): Unit = {
  require(amount >= 0)
  val balanceFrom = from.balance
  if (balanceFrom >= amount) {
    from.balance = balanceFrom - amount
    val balanceTo = to.balance
    to.balance = balanceTo + amount
  }
}
```

### Property 1: The Balance of an Account Never Goes Below 0

This property is crucial for maintaining account integrity within the banking system. The implementation ensures that no account can go negative due to the following reasons:

- **Snapshot Mechanism**: Each thread takes a snapshot of the balance (`balanceFrom`) at the start of the transfer operation. This snapshot provides a consistent view of the balance at that moment.

- **Conditional Check**: The line `if (balanceFrom >= amount)` ensures that the transfer is only executed if there are sufficient funds in the `from` account. If the balance is insufficient, the transfer does not occur.

#### Example:

Assume `Account A` has $100:
- **Thread 1** attempts to withdraw $100 from `Account A` to `Account B`.
- **Thread 2** attempts to withdraw $50 from `Account A` to `Account C`.

1. **Thread 1** reads `Account A`: `balanceFrom = 100`.
2. **Thread 2** reads `Account A`: `balanceFrom = 100`.
3. **Thread 1** checks if `balanceFrom >= 100`: true.
4. **Thread 2** checks if `balanceFrom >= 50`: true.
5. **Thread 1** updates `Account A` to $0 (i.e., `100 - 100`) and `Account B` to $100 (i.e., `0 + 100`).
6. **Thread 2** tries to update `Account A`, but since it reads the original balance, it would attempt to set `Account A` to $50 (i.e., `100 - 50`).

Through this process, **Property 1 is upheld** since no account can go negative.

### Property 2: The Total Sum of Money Held by the Bank is Constant

This property can be violated in a concurrent environment due to race conditions that lead to double spending.

#### Example:

Using the same account setup:
1. **Initial Balances**:
   - `Account A`: $100
   - `Account B`: $0
   - `Account C`: $0
   - **Total = $100**

2. **Concurrent Transfers**:
   - **Thread 1** transfers $100 from `Account A` to `Account B`.
   - **Thread 2** transfers $50 from `Account A` to `Account C`.

### Execution Steps

1. **Thread 1** takes a snapshot of `Account A`'s balance:
   - `balanceFrom = 100`.

2. **Thread 2** takes a snapshot of `Account A`'s balance:
   - `balanceFrom = 100`.

Both threads check their conditions:
- **Thread 1** checks if `balanceFrom >= 100`: true.
- **Thread 2** checks if `balanceFrom >= 50`: true.

3. **Thread 1** executes its transfer:
   - It sets `Account A` to $0 (i.e., `100 - 100`) and `Account B` to $100 (i.e., `0 + 100`).

4. **Thread 2**, still holding its snapshot of `balanceFrom = 100`, proceeds to execute its transfer:
   - It sets `Account A` to $50 (i.e., `100 - 50`) and `Account C` to $50 (i.e., `0 + 50`).

### Final Balances

After both operations, the final balances are:
- `Account A`: $50 (after Thread 2's operation)
- `Account B`: $100 (after Thread 1's operation)
- `Account C`: $50 (after Thread 2's operation)

#### Total Calculation

- **Total = Account A + Account B + Account C = $50 + $100 + $50 = $200**.

### Conclusion

1. **Property 1 (No Negative Balances)**: This property is maintained because the snapshot mechanism and conditional checks ensure that no account can go below zero.

2. **Property 2 (Constant Total)**: This property is violated due to concurrent operations, where both threads utilize their snapshots of `Account A`'s balance. This leads to a situation where the total money in the bank increases from $100 to $200, demonstrating a double spending scenario.

### Recommendations

To prevent violations of these properties in a concurrent environment:
- **Use Locking Mechanisms**: Implement locks to ensure that only one thread can access and modify account balances at a time.
- **Atomic Operations**: Consider using atomic variables for account balances to ensure thread-safe updates.
- **Transactional Management**: Implement a transactional approach where operations are rolled back if inconsistencies occur.

By applying these strategies, you can maintain the integrity of both properties in a multi-threaded banking system.


Question: In Text Representation learning, which of the following statements are correct?

Here are my assessments of the provided statements:

1. **Correct**: "Learning GloVe word vectors can be done using the singular value decomposition, if the \( f_{dn} \) weights are set to 1 for all observed entries."
   - Explanation: GloVe (Global Vectors for Word Representation) can be formulated in a way that allows the use of Singular Value Decomposition (SVD) to derive word vectors, especially when simplifying the context matrix. Setting \( f_{dn} \) to 1 corresponds to treating all co-occurrences equally, which can facilitate this approach.

2. **Correct**: "The skip-gram model for learning original word2vec embeddings does learn a binary classifier for each word."
   - Explanation: In the skip-gram model, the objective is to predict context words given a target word. This can be framed as a binary classification problem where for each target word, the model learns to classify whether each word in the context is part of the context or not.

3. **Incorrect**: "FastText as discussed in the course learns word vectors and sentence representations which are specific to a supervised classification task."
   - Explanation: FastText primarily focuses on unsupervised learning of word embeddings by considering subword information (character n-grams) for better handling of out-of-vocabulary words. While it can be used in a supervised setting for classification tasks, its main purpose is not tied to supervised learning.

4. **Correct**: "Logistic regression used for text classification is faster at test time when using word vectors as opposed to bag-of-word representation of the input."
   - Explanation: Word vectors (like those from Word2Vec or GloVe) provide a dense representation of words in a continuous space, which typically results in fewer dimensions compared to the sparse representations of the bag-of-words model. This reduced dimensionality can lead to faster computation during inference.

In summary, the correct statements are: **1, 2, and 4**.


Question: A friend asked you to prove the following statement true or false: if a and b are rational numbers, a^b must be irrational. Examining the case where a is 1 and b is 2, what kind of proof are you using ?

To address the statement "if a and b are rational numbers, a^b must be irrational," we can examine the case where \( a = 1 \) and \( b = 2 \). Here, we find that:

\[
1^2 = 1
\]

Since 1 is a rational number, we have provided a counterexample that disproves the original statement. This means that the statement is false because we found a specific case where \( a \) and \( b \) are rational, but \( a^b \) is rational.

Now, let's analyze the options provided:

1. **Proof by contradiction**: This method involves assuming that the statement is true and then showing that this leads to a contradiction. In our case, we are not doing this; instead, we are simply showing a valid case that disproves the statement.

2. **Proof by counterexample**: This method involves providing a specific example that contradicts the statement. Since we provided \( 1^2 = 1 \) as a counterexample, this directly aligns with what we did.

3. **Exhaustive proof**: This would require checking every possible case to confirm or deny the statement. That is not what we've done here; we've only provided one specific example.

4. **Proof by cases**: This method involves breaking the problem into different scenarios and proving each one. We did not break down into different cases; we provided a single example.

5. **Existence proof**: This proves that at least one example exists that satisfies certain conditions. While we did show that an example exists, it is more specific to countering the statement than simply proving existence.

Upon evaluating the options, the most fitting method used in our approach is "Proof by counterexample," as we provided a specific example that disproves the general statement.

Final Answer: 2.


Question: An HMM model would not be an appropriate approach to identify

Let's evaluate each option to determine the scenarios where they might seem plausible but ultimately fail, and then clarify why the correct option holds true in all relevant scenarios.

### Option 1: Named Entities

**Scenario where it might seem plausible:** An HMM could appear to be effective in identifying named entities if one considers a simple case where the entities follow predictable patterns, such as being capitalized or positioned at the beginning of sentences. In a controlled dataset where named entities are clearly delineated, an HMM could perform adequately.

**Why it ultimately fails:** Named entity recognition (NER) often involves ambiguous contexts and requires understanding the relationships between words and their meanings, which HMMs, primarily relying on sequential probabilities, may struggle with. Furthermore, HMMs do not leverage rich features such as context and word embeddings, which are critical for accurately identifying entities in varied contexts.

### Option 2: Part-of-Speech Tags

**Scenario where it might seem plausible:** HMMs are quite useful for tasks like part-of-speech tagging due to their sequential nature. In a simplified text with clear syntactic structures, an HMM could seem effective in assigning parts of speech to words based on the observed sequences.

**Why it ultimately fails:** While HMMs can tag parts of speech reasonably well in some situations, they might struggle with more complex sentences or when words have multiple possible tags depending on context. Advanced models like conditional random fields (CRFs) or neural networks can incorporate more context and additional features to improve accuracy.

### Option 3: Concepts

**Scenario where it might seem plausible:** One might argue that HMMs can identify concepts within a text if those concepts consistently appear in specific sequences. For instance, if "climate change" frequently appears in similar contexts, an HMM could be seen as a potential candidate for recognizing that concept.

**Why it ultimately fails:** Concepts often require a deeper understanding of semantics and relationships between different words or phrases across sentences. HMMs, which are based solely on the probability of sequences, do not effectively capture the nuances needed to identify and differentiate complex concepts, especially when they involve multiple words or are context-dependent.

### Option 4: Word n-grams

**Correct Option Analysis:** Word n-grams are sequences of 'n' words used primarily in statistical language modeling and text processing. HMMs are not designed to work with fixed-length sequences like n-grams but rather with transitioning states based on previous observations. While it is possible to use an HMM to model sequences of words, it does not directly align with the concept of n-grams, which focuses on the immediate context of a fixed number of words without consideration of state transitions.

**Why it holds true in all relevant scenarios:** N-grams require a model that can handle fixed-length sequences of words effectively without relying on the transition probabilities of states (as in an HMM). HMMs are better suited for tasks where relationships evolve over time or across longer sequences, while n-grams are fundamentally about capturing local patterns of word co-occurrence without the state-based transitions that HMMs use.

### Conclusion
In summary, while HMMs may seem plausible for some tasks, they ultimately fail in handling complex dependencies, relationships, and contextual nuances required for tasks like identifying concepts. The correct choice, "Word n-grams," is fundamentally mismatched with HMM's state-transition model and retains its validity across all relevant scenarios.


Question: Tick the \emph{incorrect} assertion. In a multiplicative cyclic group $G$ of order $m > 1$ with neutral element $e_G$ \ldots

To determine the incorrect assertion among the given options concerning a multiplicative cyclic group \( G \) of order \( m > 1 \), we will analyze each option critically.

1. **Option 1: $\lambda = m$, where $\lambda$ is the exponent of $G$.**

   The exponent of a group \( G \), denoted \( \lambda \), is defined as the least common multiple of the orders of all elements in \( G \). In a cyclic group of order \( m \), the maximum order of any element is indeed \( m \). Therefore, the least common multiple of the orders of all elements, which includes elements of order \( 1, 2, \ldots, m \), will also be \( m \). Thus, this assertion is **correct**.

2. **Option 2: The order of every element \( x \in G \) is \( m \).**

   In a cyclic group of order \( m \), there exists at least one element \( g \) (a generator) such that the order of \( g \) is \( m \). However, not all elements must have order \( m \). For example, in \( \mathbb{Z}/6\mathbb{Z} \), the element \( 2 \) has order \( 3 \) and \( 3 \) has order \( 2 \). In general, the order of an element \( x \) divides the order of the group \( m \). Therefore, this assertion is **incorrect**.

3. **Option 3: There exists \( g \in G \) that generates the whole group.**

   By definition, a cyclic group is generated by a single element \( g \). Therefore, there exists an element \( g \in G \) such that every element of \( G \) can be expressed as \( g^k \) for some integer \( k \). This assertion is **correct**.

4. **Option 4: For any \( x \in G \), we have that \( x^m = e_{G} \).**

   In any group, it holds that if \( x \) is an element of the group and the order of the group is \( m \), then \( x^m = e_G \). This is because the order of \( x \) must divide \( m \), so raising \( x \) to the power \( m \) will yield the identity element. Hence, this assertion is also **correct**.

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

The **incorrect assertion** is **Option 2**: "the order of every element \( x \in G \) is \( m \)."


Question: If we pick independent random numbers in $\{1, 2, \dots, N\}$ with uniform distribution, $\theta \sqrt{N}$ times, we get at least one number twice with probability\dots

To solve this problem, we can use the concept of the birthday problem, which deals with the probability of collisions (i.e., picking the same number more than once) when selecting random items from a finite set.

In this case, we are picking numbers uniformly from the set \(\{1, 2, \ldots, N\}\) a total of \(m = \theta \sqrt{N}\) times. We want to calculate the probability of getting at least one number twice.

### Step 1: Calculate the probability of no collisions

The probability of no collisions (i.e., all selected numbers are unique) can be calculated using the formula for the probability of unique selections in the birthday problem. The first number can be any of the \(N\) numbers. The second number can be any of the remaining \(N-1\) numbers, the third can be any of the remaining \(N-2\) numbers, and so forth.

Thus, the probability \(P(\text{no collisions})\) is given by:

\[
P(\text{no collisions}) = \frac{N}{N} \cdot \frac{N-1}{N} \cdot \frac{N-2}{N} \cdots \frac{N-m+1}{N} = \frac{N(N-1)(N-2)\cdots(N-m+1)}{N^m}
\]

### Step 2: Approximate the probability for large \(N\)

For large \(N\) and \(m = \theta \sqrt{N}\), we can approximate this product. The probability of no collisions can be rewritten using the exponential function:

\[
P(\text{no collisions}) \approx e^{-\frac{m(m-1)}{2N}} \approx e^{-\frac{(\theta \sqrt{N})(\theta \sqrt{N}-1)}{2N}} \approx e^{-\frac{\theta^2 N}{2N}} = e^{-\frac{\theta^2}{2}}
\]

### Step 3: Calculate the probability of at least one collision

Now, the probability of having at least one collision is the complement of the probability of no collisions:

\[
P(\text{at least one collision}) = 1 - P(\text{no collisions}) \approx 1 - e^{-\frac{\theta^2}{2}}
\]

### Conclusion

Thus, the probability of picking at least one number twice when selecting \(\theta \sqrt{N}\) numbers from the set \(\{1, 2, \ldots, N\}\) is:

\[
\boxed{1 - e^{-\frac{\theta^2}{2}}}
\]


Question: The SMART algorithm for query relevance feedback modifies? (Slide 11 Week 3)

**
To determine what the SMART algorithm for query relevance feedback modifies, we first need to understand its purpose and how it operates within the vector space model of information retrieval.

The SMART algorithm is designed to improve search result relevance by incorporating user feedback about which retrieved documents are relevant or irrelevant. This feedback is crucial because it allows the algorithm to adjust the search parameters to better meet user needs in future queries.

In the context of the vector space model, we have two key components that the SMART algorithm interacts with: document weight vectors and query weight vectors. Document weight vectors represent the importance of terms in the documents retrieved, while query weight vectors represent the importance of terms in the user's search query.

Now, let's analyze the answer choices:

1. **The original document weight vectors**: The SMART algorithm does not modify these vectors directly; instead, it focuses on improving the user's search query.

2. **The original query weight vectors**: This is a critical modification point for the SMART algorithm. It adjusts these vectors based on user feedback to enhance the relevance of future search results.

3. **The result document weight vectors**: These vectors reflect the outcome of the retrieval process and are not modified by the SMART algorithm, as it focuses on refining the query instead.

4. **The keywords of the original user query**: While the keywords are part of the process, the SMART algorithm modifies the weights associated with these keywords in the query vector rather than altering the keywords themselves.

After carefully considering each option, the correct answer is **the original query weight vectors**. This modification is central to how the SMART algorithm utilizes user feedback to improve search results.

---

**


Question:
In context of Meltdown Attack, what are the basic ideas of the attack and how they relate to
                    the snippet above? What is the microarchitectural mechanism targeted
                    by the attack? Is it likely to work on all processors with such
                    architectural mechanism or some processors may be intrinsically not
                    vulnerable?  Explain.

### Step-by-Step Breakdown of Meltdown Attack:

1. **Basic Idea of the Attack**:
   - The Meltdown attack exploits a vulnerability in modern processors that allows unauthorized access to protected memory. Specifically, it breaks the isolation between user applications and the operating system (OS).
   - It leverages speculative execution, a performance optimization where the CPU guesses which instructions will be needed and executes them ahead of time.

2. **Mechanism of the Attack**:
   - The attacker runs code that attempts to access memory that it should not have permissions for (e.g., kernel memory).
   - When the CPU speculatively executes this code, it processes the memory access, but the access is eventually rolled back if it’s determined to be unauthorized.
   - However, during this rollback, the CPU caches the results of the unauthorized access, which can be exploited by the attacker to infer sensitive information through timing variations.

3. **Microarchitectural Mechanism Targeted**:
   - The primary microarchitectural mechanism targeted by Meltdown is the **speculative execution** and the **cache architecture**. The attack relies on the side effects of speculative execution and how data is stored in the CPU cache.

4. **Vulnerability Across Processors**:
   - Not all processors are equally vulnerable. The Meltdown attack primarily affects Intel processors due to their specific implementation of speculative execution and memory protection.
   - Some other architectures (e.g., ARM, AMD) have mitigations or do not exhibit the same vulnerabilities, making them less susceptible or not vulnerable at all.

### Conclusion:
The Meltdown attack exploits speculative execution to access protected memory, targeting the CPU's cache behavior. While it affects many Intel processors, not all processors are vulnerable, as architectural differences can provide intrinsic protections against such attacks.


Question: Consider an RSA encryption where the $(p, q)$ are determined as $(53, 61)$. True or false: $(e,d) = (7, 223)$ are valid encoding/decoding exponents.

1.  The pair \( (e, d) = (7, 223) \) does not satisfy the RSA condition \( e \cdot d \equiv 1 \mod \phi(n) \).


Question: Let $\mathbf{A}, \mathbf{B} \in \mathbb{R}^{n \times n}$ be two symmetric matrices. Assume that $\mathbf{v} \in \mathbb{R}^{n}$ is an eigenvector for both matrices with associated eigenvalues $\lambda_{A}$ and $\lambda_{B}$ respectively. Show that $\mathbf{v}$ is an eigenvector of the matrix $\mathbf{A}+\mathbf{B}$. What is the corresponding eigenvalue?

To show that \(\mathbf{v}\) is an eigenvector of the matrix \(\mathbf{A} + \mathbf{B}\) with a corresponding eigenvalue, we will use the definition of eigenvectors and eigenvalues.

Recall that if \(\mathbf{A}\) is a matrix and \(\mathbf{v}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda_A\), then by definition we have:

\[
\mathbf{A} \mathbf{v} = \lambda_A \mathbf{v}
\]

Similarly, since \(\mathbf{v}\) is also an eigenvector of \(\mathbf{B}\) with eigenvalue \(\lambda_B\), we have:

\[
\mathbf{B} \mathbf{v} = \lambda_B \mathbf{v}
\]

Now, we want to show \(\mathbf{v}\) is an eigenvector of the matrix \(\mathbf{A} + \mathbf{B}\). We compute the action of \(\mathbf{A} + \mathbf{B}\) on \(\mathbf{v}\):

\[
(\mathbf{A} + \mathbf{B}) \mathbf{v} = \mathbf{A} \mathbf{v} + \mathbf{B} \mathbf{v}
\]

Now substituting the expressions we have for \(\mathbf{A} \mathbf{v}\) and \(\mathbf{B} \mathbf{v}\):

\[
(\mathbf{A} + \mathbf{B}) \mathbf{v} = \lambda_A \mathbf{v} + \lambda_B \mathbf{v}
\]

Factor out \(\mathbf{v}\):

\[
(\mathbf{A} + \mathbf{B}) \mathbf{v} = (\lambda_A + \lambda_B) \mathbf{v}
\]

This shows that \(\mathbf{v}\) is indeed an eigenvector of \(\mathbf{A} + \mathbf{B}\), and the corresponding eigenvalue is:

\[
\lambda_A + \lambda_B
\]

In conclusion, \(\mathbf{v}\) is an eigenvector of the matrix \(\mathbf{A} + \mathbf{B}\) with the eigenvalue \(\lambda_A + \lambda_B\).


Question: Suppose a file system used only for reading immutable files in random fashion. What is the best block allocation strategy?

In a file system designed specifically for reading immutable files in a random fashion, the best block allocation strategy would typically be:

**3. Index allocation with B-tree**

### Explanation:

1. **Linked-list allocation**: This method involves linking blocks together in a list. While it allows for efficient space utilization and is good for sequential access, it is not optimal for random access. Each block access requires following pointers, which can lead to poor performance when accessing blocks randomly.

2. **Continuous allocation**: This method allocates blocks in contiguous locations. While this can provide excellent performance for sequential access, it is not efficient for random access, especially if the files are of varying sizes or if free space is fragmented.

3. **Index allocation with B-tree**: This approach uses an index structure (like a B-tree) to keep track of the blocks of files. B-trees are well-suited for random accesses and allow for efficient searching, insertion, and deletion. Since the files are immutable and accessed randomly, this strategy would provide efficient block lookups and minimize the overhead associated with accessing non-contiguous blocks.

4. **Index allocation with Hash-table**: While a hash table can provide quick access, it might not be as efficient as a B-tree for range queries or when dealing with files that may have a large number of blocks, as it typically does not maintain order and can lead to clustering issues.

Given these considerations, **index allocation with B-tree** is the best choice for efficient random access in a file system dealing with immutable files.