Gate 2024 CSE Slot 2 Question Paper (Available)- Download Solution pdf with Answer Key

Solution: To correctly match Set X with Set Y, we analyze each component:
P. Lexical Analyzer: The lexical analyzer is responsible for generating tokens from the input stream. Hence, P – 2.
Q. Syntax Analyzer: The syntax analyzer generates a parse tree by analyzing the syntactical structure of tokens. Hence, Q – 3.
R. Intermediate Code Generator: The intermediate code generator creates an abstract syntax tree (AST) as part of its processing. Hence, R – 1.
S. Code Optimizer: The code optimizer applies techniques like constant folding to optimize the intermediate code. Hence, S – 4.

Correct match: P – 2, Q – 3, R – 1, S – 4.

Solution: The DFA has three states:
q₀: Initial state.
q₁: Intermediate state.
q₂: Accepting state.

Analysis:
From q₀, a 1 transitions to q₁.
From q₁, a 0 transitions to q₂, and another 1 transitions back to q₀.
From q₂, 0 loops back to q₂.
This DFA accepts strings that:
Start with 0* (any number of 0s at the beginning).
Have at least one 1 to reach q₁.
Follow by 0 + 10*1 repeatedly (to loop between q₁ and q₀).
Optionally end in 0* (to stay in q₂).
The equivalent regular expression is: 0*1(0 + 10*1)*

Solution: In a TCP/IP packet, the destination IP address remains constant and always points to the final destination node, which in this case is Y. Hence, the destination IP address is the IP address of Y, making Option (B) correct.
The MAC address, however, is updated at every hop and points to the next device in the path. When the packet leaves X, the destination MAC address is set to the MAC address of R (the next-hop router), not S or Y. Therefore, Option (B) is correct.

Solution:
(A) Allocate a new page table for a newly created process: This task is typically handled by the operating system, not the MMU. Hence, Option (A) is incorrect.
(B) Translate a virtual address to a physical address using the page table: This is one of the core functions of the MMU. The MMU performs the translation of virtual addresses to physical addresses based on the page table. Hence, Option (B) is correct.
(C) Raise a trap when a virtual address is not found in the page table: The MMU raises a page fault trap when a virtual address is not found in the page table. This prompts the operating system to handle the fault. Hence, Option (C) is correct.
(D) Raise a trap when a process tries to write to a page marked with read-only permission in the page table: The MMU enforces memory protection by checking permissions in the page table. If a process violates these permissions, such as writing to a read-only page, the MMU raises a trap. Hence, Option (D) is correct.

Solution: When process P makes a blocking system call (e.g., to read a block of data from the disk), it cannot proceed until the I/O operation completes. The OS places P in the blocked state, triggering a context switch. Therefore, Option (A) is correct.
If P tries to access a page that is not in memory but in swap space, a page fault occurs. The OS retrieves the page from disk while P is placed in the blocked state. This also triggers a context switch. Hence, Option (B) is correct.
An interrupt raised by the disk to deliver data requested by another process does not directly involve P. The OS handles the interrupt but does not always switch context for P. Thus, Option (C) is incorrect.
A timer interrupt raised by hardware does not always trigger a context switch. The OS scheduler may decide to continue running P or preempt it. Hence, Option (D) is incorrect.

Solution: In a scan operation, sequential access to data minimizes disk I/O. Sorted file organizations provide sequential access and are efficient for scan operations. Thus, Option (A) is correct.
Heap files store data without any particular order, but for a full scan, every block is accessed sequentially. This makes heap files I/O efficient for scan operations. Hence, Option (B) is correct.
Unclustered tree indexes, however, are not I/O efficient for scans because data blocks may not be accessed sequentially, leading to multiple random I/O operations. Hence, Option (C) is incorrect.
Unclustered hash indexes are designed for equality searches and not optimized for scan operations. Random I/O makes them inefficient for scans. Thus, Option (D) is incorrect.

Solution: The Two Phase Locking (2PL) protocol ensures two distinct phases: the growing phase (acquiring locks) and the shrinking phase (releasing locks).
Option (A): 2PL permits only serializable schedules by ensuring conflict-serializability. Hence, (A) is correct.
Option (B): This statement is incorrect because locks are not acquired and released immediately for every operation. Locks are held until the shrinking phase.
Option (C): In 2PL, once a lock is released during the shrinking phase, no new locks can be acquired. This is a key property of 2PL. Hence, (C) is correct.
Option (D): Deadlocks can occur in 2PL due to circular wait conditions. Hence, (D) is correct.

Solution: IPv4 fragmentation occurs when the size of a datagram exceeds the Maximum Transmission Unit (MTU) of a network link.
Option (A): Incorrect. Fragmentation can occur at any intermediate router along the path, not just at the source.
Option (B): Correct. Any IP router that encounters an MTU smaller than the datagram size performs fragmentation.
Option (C): Correct. Reassembly of fragments is performed only at the destination to reconstruct the original datagram.
Option (D): Incorrect. Intermediate routers do not reassemble fragments; they forward them as they are.

Solution:
Option (A): This is true. In attribute grammars, functions in semantic rules are required to have no side effects, as they must not interfere with other computations.
Option (B): This is false. L-attributed definitions allow evaluation in a depth-first order since inherited attributes are passed only along the parse tree edges and are computable in one depth-first pass.
Option (C): This is true. Synthesized attributes are computed from the values of the children nodes and can be evaluated in a bottom-up parsing strategy.
Option (D): This is false. While L-attributed definitions can be evaluated using top-down parsing, they are not inherently suitable for bottom-up parsing strategies.

Solution: Option (A): True. Instruction I₁ writes to R3, and instruction I₂ reads from R3, creating a Read After Write (RAW) dependency between I₁ and I₂.
Option (B): False. Instruction I₃ writes to R3, but I₁ also writes to R3. This would be a Write After Write (WAW) dependency, not a WAR dependency.
Option (C): False. Instruction I₃ writes to R3, but I₂ does not read R3. Hence, there is no RAW dependency between I₂ and I₃.
Option (D): False. Both I₃ and I₄ write to R3, creating a Write After Write (WAW) dependency, but this dependency does not exist in the given instruction sequence.

Solution:
Option (A): The source IP address is not modified by routers. It remains the same throughout the packet's journey. Hence, Option (A) is incorrect.
Option (B): The protocol field indicates the higher-level protocol (e.g., TCP, UDP) and is not modified by routers. Hence, Option (B) is incorrect.
Option (C): The Time to Live (TTL) field is decremented by 1 by each router to prevent infinite loops in the network. This field is always modified by routers. Hence, Option (C) is correct.
Option (D): Since the TTL field is modified, the Header Checksum is recomputed by routers to ensure integrity. Hence, Option (D) is correct.

Solution:
Option (A): The string literal "abcd" is valid, and the function calculates the length of the string by iterating until the null character. Hence, Option (A) is correct.
Option (B): The character array c is a valid input for the function, and the function correctly calculates the length of the string. Hence, Option (B) is correct.
Option (C): This is false. The code is valid and will compile without any issues. Hence, Option (C) is incorrect.
Option (D): The character pointer c pointing to the string literal "abcd" is valid, and the function calculates the length of the string correctly. Hence, Option (D) is correct.

Solution: A total order is an extension of a partial order in which every pair of elements is comparable. Based on the given partial order P, we know the following relations: 1 ≤ 2, 3 ≤ 2, 3 ≤ 4. Using these relations, the valid total orders containing P are:
1. 1 < 3 < 4 < 2
2. 1 < 3 < 2 < 4
3. 1 < 4 < 3 < 2
4. 3 < 1 < 4 < 2
5. 3 < 1 < 2 < 4
Thus, there are 5 total orders that extend P.

Solution: To calculate the distance, we need to determine the Longest Increasing Subsequence (LIS) of the array. The length of the LIS represents the largest subset of elements already in sorted order.
The given array is [2, 5, 3, 1, 4, 2, 6]. The LIS for this array is:
[2, 3, 4, 6] (length = 4).
The minimum number of elements to replace is given by:
Distance = Length of array – Length of LIS = 7 – 4 = 3.

Solution: The program initializes an array a[2] = {20.0, 25.0}, two pointers p and q, and performs the following steps:
1. p = a: Pointer p now points to the first element of the array, i.e., a[0] = 20.0.
2. q = p + 1: Pointer q now points to the second element of the array, i.e., a[1] = 25.0.
3. The statement q - p calculates the difference in pointer positions. Since q points to the second element and p points to the first element, q - p = 1.
4. The statement *q - *p calculates the difference between the values pointed to by q and p. This gives: *q - *p = a[1] - a[0] = 25.0 - 20.0 = 5.0.
5. The printf statement prints these values as integers: Output: (int)(q - p) = 1, (int)(*q - *p) = 5.

Solution:
Non-Preemptive SJF Scheduling:

Process | Arrival Time | Burst Time | Completion Time | Turnaround Time | Waiting Time
--------|--------------|------------|-----------------|-----------------|--------------
A       | 0            | 10         | 10              | 10              | 0
B       | 2            | 6          | 19              | 17              | 11
C       | 4            | 3          | 13              | 9               | 6
D       | 6            | 7          | 26              | 20              | 13
        

Process | Arrival Time | Burst Time | Completion Time | Turnaround Time | Waiting Time
--------|--------------|------------|-----------------|-----------------|--------------
A       | 0            | 10         | 26              | 26              | 16
B       | 2            | 6          | 11              | 9               | 3
C       | 4            | 3          | 7               | 3               | 0
D       | 6            | 7          | 18              | 12              | 5
       

Solution: The given range is 10.12.2.0 to 10.12.3.255.
The CIDR notation 10.12.2.0/23 has a subnet mask of 255.255.254.0, which covers the following range of IPs:
10.12.2.0 (start) to 10.12.3.255 (end).
Other options do not correctly represent this range:
(B) 10.12.2.0/24: Only covers 10.12.2.0 to 10.12.2.255.
(C) 10.12.0.0/22: Covers a larger range from 10.12.0.0 to 10.12.3.255.
(D) 10.12.2.0/22: Invalid as it overlaps incorrectly with the range 10.12.4.x.

Solution: In a Binary Search Tree (BST), the inorder traversal of the tree produces the elements in sorted order, regardless of the sequence in which they were inserted. Since V contains all the elements of T as a set, sorting V will give the inorder traversal of T. Hence, Option (A) is correct.
Option (B): The root node of T depends on the sequence of insertion and cannot be determined solely from V. Hence, Option (B) is incorrect.
Option (C): The preorder traversal depends on the structure of T, which in turn depends on the insertion order. Therefore, it cannot be determined from V. Hence, Option (C) is incorrect.
Option (D): The postorder traversal also depends on the structure of T and the insertion sequence, and cannot be determined from V. Hence, Option (D) is incorrect.

Solution: The given NFA M has five states (1, 2, 3, 4, 5) and includes ε-transitions. To determine the language accepted by M, consider the paths and transitions:
1. From state 1 to 2: The ε-transition allows state 2 to be reached directly without consuming any input.
2. From state 2 to 3: Input 0 can be consumed repeatedly (loop on 2) or proceed to 3 with a single 0. This corresponds to (00)*.
3. From state 1 to 4: The ε-transition allows state 4 to be reached directly without consuming any input.
4. From state 4 to 5: Input 1 can be consumed repeatedly (loop on 5) after consuming 1. This corresponds to (11)*.
5. Combining paths: The overall language includes all possible combinations:
- 0*, representing sequences of 0's from state 2.
- (1 + 0(00)*)(11)*, combining sequences starting with 1 or sequences starting with 0 followed by (00)*, and ending with (11)*.
The regular expression representing the language is: 0* + (1 + 0(00)*)(11)*.

Solution: The goal is to maintain the length of the longest subarray ending at the current index i with at most two distinct integers. The variables have the following roles:
1. len1: Length of the subarray ending at X[i] where all elements are equal to X[i].
2. len2: Length of the subarray ending at X[i] with at most two distinct integers.

Analysis of (P): When X[i] == second, len1 must be reset to 1 because this marks the start of a new sequence of identical integers. Thus, (P) = 1.
Analysis of (Q): When X[i] is neither first nor second, len2 must be updated to include the length of the subarray containing the new two distinct integers. Since the previous len1 represents the length of the contiguous subarray of first, the new len2 = len1 + 1 (the current element X[i] is added). Thus, (Q) = len1 + 1.
The correct replacements for (P) and (Q) are: (P) = 1, (Q) = len1 + 1.

Solution: To break down the expression x[i] = (p + r) * -s[i] + u/w into intermediate triples:
1. Triple (0): + operation for p and r. Already given as: (0) : + p r.
2. Triple (1): Assignment of -s[i]. Using the unary minus (uminus) operator: (1) := [] s[i].
3. Triple (2): Unary minus applied to (1). Already given as: (2) : uminus (1).
4. Triple (3): Multiplication of (0) and (2): (3) : * (0) (2).
5. Triple (4): Division of u and w. Already given as: (4) : / u w.
6. Triple (5): Addition of (3) and (4): (5) : + (3) (4).
7. Triple (6): Assignment of (5) to x[i]: (6) := [] x[i].
The correct missing entries are: (1) := [] s[i], (3) : * (0) (2), (6) := [] x[i].

Solution: The random variables x and y are in the interval [0, 1], and z = xy. For any pair of real values x,y ∈ [0,1], the value of xy is always less than or equal to x (since y ≤ 1). Therefore, the mean of z, z̄, satisfies: z̄ ≤ x̄.
This is always true, regardless of the dependence or independence of x and y.

Analysis of the other options:
(A) z̄ = x̄ȳ is true only when x and y are independent. This is not guaranteed here. Hence, (A) is incorrect.
(B) z̄ ≤ x̄ȳ is true due to the general inequality E[xy] ≤ E[x]E[y] (Cauchy-Schwarz inequality), but this is weaker than the stricter constraint z̄ ≤ x̄. Hence, (B) is not the strongest statement.
(C) z̄ ≥ x̄ȳ is false because the inequality E[xy] ≤ E[x]E[y] holds.
(D) z̄ ≤ x̄ is always true and the strongest statement in this case. Hence, (D) is correct.

Solution: The problem requires finding cities where at least 3 people reside. This can be accomplished as follows:
1. Use selection and projection to create a list of city values.
2. Perform cross product operations to create all combinations of rows for a given city, ensuring at least 3 unique entries for each city.
The minimum number of cross product operations required to verify at least 3 people in each city is 2.

Solution:
Outcome (A): This is not possible because T2 cannot proceed after T1 without acquiring s1, but s1 is released only when T1 completes. Thus, (A) is incorrect.
Outcome (B): T2 can run first, acquire s1, increment x, and print 1. Then T1 acquires s1, increments x, and prints 2. This outcome is possible.
Outcome (C): T1 acquires s1, increments x, and prints 1. However, it waits on s2, which is never signaled by T2, causing a deadlock. This outcome is possible.
Outcome (D): This is not possible because T2 signals s2 after printing, allowing T1 to proceed. Thus, (D) is incorrect.

Solution:
Option (A): When two rows of a matrix are swapped, the determinant of the new matrix is the negative of the determinant of the original matrix. Hence, (A) is correct.
Option (B): Swapping two rows of an invertible matrix does not affect its invertibility because the determinant remains non-zero. Hence, (B) is correct.
Option (C): Swapping rows of a symmetric matrix does not guarantee symmetry, as the symmetry condition A[i][j] = A[j][i] may be violated. Hence, (C) is incorrect.
Option (D): The trace of a matrix is the sum of its diagonal elements, which is independent of row swaps. Hence, if the trace of A is zero, the trace of B remains zero. This makes (D) incorrect.

Solution: The available operations allow for flexible reordering of elements in S1 and S2. By analyzing the given sequences:
1. Sequence (A): The sequence 100, 200, 400, 300 cannot be generated because 400 is above 300 in S1, and moving 400 before 300 without outputting 300 is not possible. Hence, (A) is not feasible.
2. Sequence (B): The sequence 200, 300, 400, 100 can be generated as follows: - Use GenerateOutput to output 200 and 300.
- Push 400 to S2 and output 100.
- Retrieve 400 from S2 and output it. Hence, (B) is feasible.
3. Sequence (C): The sequence 400, 200, 100, 300 can be generated by:
- Use GenerateOutput to output 400.
- Push 300 and 200 to S2.
- Output 100 from S1.
- Retrieve 300 from S2 and output it. Hence, (C) is feasible.
4. Sequence (D): The sequence 300, 200, 400, 100 can be generated by:
- Push 400 to S2.
- Use GenerateOutput to output 300 and 200.
- Retrieve 400 from S2 and output it.
- Finally, output 100. Hence, (D) is feasible.

Solution: The given number 224 in radix-5 can be converted to decimal (radix-10) as follows:
224₅ = 2 * 5² + 2 * 5¹ + 4 * 5⁰ = 50 + 10 + 4 = 64.
Now, check each option:
Option (A): 64 in radix-10 matches the decimal value of 224₅. Hence, (A) is correct.
Option (B): Convert 100₈ (radix-8) to decimal:
100₈ = 1 * 8² + 0 * 8¹ + 0 * 8⁰ = 64.
This matches 224₅. Hence, (B) is correct.
Option (C): Convert 50₁₆ (radix-16) to decimal:
50₁₆ = 5 * 16¹ + 0 * 16⁰ = 80.
This does not match 224₅. Hence, (C) is incorrect.
Option (D): Convert 121₇ (radix-7) to decimal:
121₇ = 1 * 7² + 2 * 7¹ + 1 * 7⁰ = 49 + 14 + 1 = 64.
This matches 224₅. Hence, (D) is correct.

Solution: If every spanning tree in G has an even weight, it implies a specific structure for the graph G.
Consider the properties:
Option (A): Incorrect. Not all edges in G need to have even weight; some edges can have odd weights as long as the total weight of spanning trees remains even.
Option (B): Incorrect. G may contain both even and odd weighted edges.
Option (C): Incorrect. It is not necessary for all edges in any cycle C to have even weight.
Option (D): Correct. For every cycle C in G, the parity of the edges must be consistent (all even or all odd) to ensure that every spanning tree has an even total weight.

Solution: The grammar G generates strings of the form: w = aⁿcb^n-1, where the derivation of c adds one b and one a. Hence, the relationships between n_a(w), n_b(w), n_c(w) can be analyzed:
Option (A): Incorrect. While n_a(w) is often greater than n_b(w), there is no guarantee, as n_a(w) and n_b(w) depend on the specific derivation.
Option (B): Correct. Since the derivation adds at least two a's for every c, the inequality n_a(w) > n_c(w) – 2 holds.
Option (C): Correct. Each derivation of S → aSbS adds one b for every c, making n_c(w) = n_b(w) + 1.
Option (D): Incorrect. The number of c's is not necessarily twice the number of b's.

Solution: The total time to read the file consists of:
Seek time: Average seek time is 5 ms.
Rotational latency: For a rotation speed of 6000 RPM, the time for one rotation is: Time per rotation = (60 sec) / 6000 = 0.01 sec. The average rotational latency is half the time per rotation: Average rotational latency = 0.01 / 2 = 0.005 sec = 5 ms.
Data transfer time per sector: Each track has 500 sectors. The time to transfer one sector is: Time per sector = (Time per rotation)/(Sectors per track) = 0.01 / 500 = 0.00002 sec = 20 μs.

Total time per sector: Total time to access one sector is: Time per sector = Seek time + Rotational latency + Transfer time. Time per sector = 5 ms + 5 ms + 20 μs = 10.02 ms.
Total time for 3000 sectors: Total time = 3000 × 10.02 ms = 30060 ms = 30.06 sec.

Solution: When a timeout occurs, TCP congestion control reduces the congestion window to 1 MSS and enters the slow-start phase. In slow start:
During the first RTT, the window size doubles: 1 → 2MSS.
During the second RTT, the window size doubles again: 2 → 4 MSS.
Thus, at the start of the third RTT, the congestion window size will be 4MSS.

Solution: The round-trip time for a signal to propagate across the segment is:
Time = 2 * (Distance / Propagation speed) = 2 * (500 / (2 * 10⁸)) = 5 μs.
To detect collisions, the frame transmission time must be at least equal to the round-trip time. The frame transmission time is: Frame transmission time = Frame size / Transmission speed.
Equating the two: Frame size / 10⁸ = 5 μs => Frame size = 10⁸ × 5 × 10^-6 = 500 bits.

Solution: For a relation R with 4 attributes {A, B, C, D}, we need to compute the total number of useful functional dependencies X → Y, where:
• X is a non-empty subset of the attributes.
• Y is a non-empty subset of the attributes.
• X and Y are disjoint (X ∩ Y = Ø).

Step 1: Total number of subsets of R
The total number of subsets of {A, B, C, D} is 2⁴ = 16. These subsets include:
{Ø, {A}, {B}, {C}, {D}, {A, B}, {A, C}, ..., {A, B, C, D}}.
Excluding the empty set, there are 15 non-empty subsets.

Step 2: Counting valid combinations of X and Y
Since X ∩ Y = Ø, Y must be chosen from the attributes that are not in X. For a fixed X:
• The size of X determines the attributes available for Y.
• If X contains k attributes, then 4 - k attributes are available for Y.
For every non-empty subset X (15 options), the number of possible non-empty subsets of the remaining attributes (Y) is: 2^4-|X| - 1.

Step 3: Summing over all valid combinations
To find the total number of useful functional dependencies, sum over all non-empty subsets X:
Total functional dependencies = ∑_k=1⁴ (⁴C_k) * (2^4-k - 1), where (⁴C_k) is the number of ways to choose k attributes for X.

Step 4: Calculate each term for k = 1, 2, 3, 4:
• For k = 1: (⁴C₁) * (2^4-1 - 1) = 4 * (8 - 1) = 4 * 7 = 28.
• For k = 2: (⁴C₂) * (2^4-2 - 1) = 6 * (4 - 1) = 6 * 3 = 18.
• For k = 3: (⁴C₃) * (2^4-3 - 1) = 4 * (2 - 1) = 4 * 1 = 4.
• For k = 4: (⁴C₄) * (2^4-4 - 1) = 1 * (1 - 1) = 0.

Step 5: Add the results
The total number of useful functional dependencies is: 28 + 18 + 4 + 0 = 50.

Solution: The 32-bit instruction format is divided as follows:
• 16 general-purpose registers: Each register requires 4 bits (2⁴ = 16).
• Two register operand fields: Each operand field is 4 bits, for a total of 8 bits.
• Addressing mode field: 3 bits to represent 8 addressing modes (2³ = 8).
• Scalar field: 16 bits.
The remaining bits are allocated to the opcode field: Opcode bits = 32 - (8 + 3 + 16) = 5 bits.
The total number of unique opcodes for every addressing mode is: 2^{Opcode bits} = 2⁵ = 32.

Solution:
Step 1: Define throughput for a pipelined processor. The throughput of a pipelined processor is the number of instructions coming out from the last stage of the pipeline per unit time. In this case, the time unit is 1 clock cycle, which is 0.5 ns. This means that each stage in the pipelined processor takes only 1 clock cycle to operate (ignoring register delay and clock skew).
Step 2: Define CPI for pipelined and non-pipelined processors. The CPI (Clock Cycles Per Instruction) is defined as the average clock cycles per instruction. For a non-pipelined processor, it takes 6 clock cycles to complete an instruction, whereas for a pipelined processor, it takes only 1 clock cycle on average to complete an instruction. Thus: CPI (non-pipelined) = 6, CPI (pipelined) = 1.
Step 3: Define the speedup of the pipelined processor. The speedup of the pipelined processor compared to the non-pipelined processor is given by: Speedup = (CPI (non-pipelined)) / (Ideal CPI (pipelined) + Pipeline stall clock cycles). Here, the pipeline stall clock cycles are added because some clock cycles are wasted due to stalls in the pipeline.
Step 4: Calculate the speedup. For a program with a total instruction count of IC, the pipeline stall clock cycles are due to 20% of the instructions stalling for 2 cycles and another 20% of the instructions stalling for 3 cycles. The effective CPI for the pipelined processor becomes: Effective CPI (pipelined) = 1 + (20% * 2) + (20% * 3). Substitute the values: Effective CPI (pipelined) = 1 + 0.4 + 0.6 = 2. The speedup is then calculated as: Speedup = CPI (non-pipelined) / Effective CPI (pipelined) = 6 / 2 = 3.
Step 5: Conclude the result. The pipelined processor is 3.0 times faster than the non-pipelined processor.

Solution:
Step 1: Analyze the graph and find the minimum weight. The graph has 7 edges with weights: 2, 2, 2, 3, 3, 3, 1. To construct a minimum spanning tree (MST), the sum of edge weights must be minimized.
Step 2: Apply Kruskal's algorithm. Using Kruskal's algorithm:
- Select the edge with weight 1 (unique choice).
- Choose three edges of weight 2. These edges form a cycle, allowing multiple choices.
- Choose one edge of weight 3 to complete the MST.
Step 3: Count the distinct combinations.
- There are ³C₂ = 3 ways to choose two edges of weight 2.
- For each choice, there are 3 ways to select one edge of weight 3.
The total number of distinct MSTs is: 3 * 3 = 9.

Solution:
Step 1: Instruction format analysis. Each instruction is 32 bits in total. Based on the formats:
• R-type instruction fields:
OPCODE + UNUSED + DST Register + SRC Register1 + SRC Register2 = 32 bits.
• I-type instruction fields:
OPCODE + DST Register + SRC Register + Immediate Value/Address = 32 bits.

Step 2: Determine the number of bits for the registers. The processor has 50 architectural registers. The number of bits required to uniquely identify a register is: Register bits = [log₂(50)] = 6 bits. Thus, each register field (DST Register, SRC Register1, SRC Register2) uses 6 bits.

Step 3: Determine the number of bits for the OPCODE field. The ISA contains 150 distinct instructions, equally divided into R-type and I-type. Since 1 bit in the OPCODE field is used to distinguish between instruction types, the remaining bits encode the operations: OPCODE bits = [log₂(150)] = 8 bits.

Step 4: Calculate the number of bits for the UNUSED field (R-type) and Immediate Value/Address field (I-type).
• R-type: From the total 32 bits, the UNUSED field is: X = 32 - (OPCODE + 3 × Register bits) = 32 - (8 + 3 * 6) = 6 bits.
• I-type: The Immediate Value/Address field uses: Z = 32 - (OPCODE + 2 × Register bits) = 32 - (8 + 2 * 6) = 12 bits.

Step 5: Compute X + 2Y + Z. Substituting the values of X, Y, and Z: X + 2Y + Z = 6 + 2 * 8 + 12 = 34.

Solution:
Step 1: Define the set of language symbols. Let the set of language symbols be: Σ = {a, b} => |Σ| = 2.

Step 2: Define the regular language L₁. The regular language L₁ is defined as the set of strings where the number of a's is odd.

Step 3: Analyze the strings with an odd number of a's. For all strings with the alphabet set {a,b} of length n, half of them will have an odd number of a's. This can be expressed as: 2ⁿ / 2 , where the numerator 2ⁿ represents the total number of strings of length n (based on the number of input symbols).

Step 4: Calculate the total number of strings in L₁ for all lengths up to 4. We sum up half of the total strings of all lengths (up to 4) in L₁: 2⁰ / 2 + 2¹ / 2 + 2² / 2 + 2³ / 2 + 2⁴ / 2
Simplify the terms: 0 + 1 + 2 + 4 + 8 = 15.
The total number of strings in L₁ (up to length 4) is 15.

Solution:
Step 1: Define the property of an element being its own inverse. An element x in a group is its own inverse if: x + x = 0 (mod n). This simplifies to: 2x = 0 (mod n).

Step 2: Analyze each group.
• Z₂: The elements are {0,1}. 2x = 0 (mod 2) => x = 0,1.
• Z₃: The elements are {0, 1, 2}. 2x = 0 (mod 3) => x = 0.
• Z₄: The elements are {0, 1, 2, 3}. 2x = 0 (mod 4) => x = 0,2.

Step 3: Combine the results for Z₂ × Z₃ × Z₄. The total number of elements in Z₂ × Z₃ × Z₄ is: |Z₂ × Z₃ × Z₄| = 2 × 3 × 4 = 24. For an element (x₂, x₃, x₄) ∈ Z₂ × Z₃ × Z₄ to be its own inverse:
• x₂ ∈ {0,1} (2 choices).
• x₃ ∈ {0} (1 choice).
• x₄ ∈ {0,2} (2 choices).
The total number of such elements is: 2 * 1 * 2 = 4.

Solution:
Given:
• System address space (VA) = 32 bits.
• Page size (PS) = 4 KB.
• Total pages used by a process (N) = 2000 pages.
• Page table entry size (PTE) = 4 B.
It is given in the question that the OS allocates a page (means single page) for the outer page directory upon process creation.

Step 1: Calculate the total number of bits needed for the outer page table.
The total number of bits needed to represent entries in the outer page table is:
log₂(PS / PTE) = log₂((2¹²) / (2²)) = log₂(2¹⁰) = 10 bits.

Step 2: Calculate the total number of bits needed for the inner page table.
The total number of bits needed to represent entries in the inner page table is: 32 (total VA bits) - 10 (outer PT bits) - 12 (page offset bits) = 10 bits.

Step 3: Calculate X. In each inner page table, we can store 2¹⁰ entries. The minimum number of page tables (PT) needed to store 2000 pages will be: ⌈2000/2¹⁰⌉ = 2 PT. Thus: X = 2 + 1 (for the outer PT) = 3.

Step 4: Calculate Y. For occupying the maximum pages in a 2-level page table, we need to place at least one PTE from 2000 pages in every page of the inner page table. This equates to 2¹⁰ PT. Thus: Y = 2¹⁰ + 1 (for the outer PT) = 1025.

Step 5: Calculate X + Y. The value of X + Y is: X + Y = 3 + 1025 = 1028.

Solution:
Step 1: Analyze the given DFA diagram. The given DFA represents the items in the closure of I₀ and the GOTO function for the symbol S.

Step 2: Identify the items in the closure of I₀. The closure of I₀ contains the following items:
• S' → ·S
• S → ·SS
• S → ·Aa
• S → ·bAc
• S → ·BC
• S → ·bAb
• A → ·d#
• B → ·@

Step 3: Apply the GOTO function on the symbol S. When the GOTO function is applied to S, the following transitions occur:
• S' → S·
• S → S·S
• S → SS·
• S → ·Aa
• S → ·bAc
• S → ·BC
• S → ·bAb

Step 4: Count the total number of items. From the diagram, the GOTO function GOTO(closure(I₀), S) results in 9 items: S' → S·, S → S·S, S → SS·, S → Aa·, S → bAc·, S → BC·, S → bAb·, A → d#·, B → @·.
From the above DFA, we can conclude that: GOTO(closure(I₀), S) contains 9 items.