GATE (Computer Science & Information Technology)

For any graph with n vertices, the size of the minimum vertex cover (k) and the size of the maximum independent set (alpha) satisfy the relation: k + alpha = n. Given n = 20 and k = 8, then alpha = 20 - 8 = 12.

With only one interrupt request line and one grant line, the CPU cannot distinguish between multiple devices to provide a unique vector address (vectored interrupt). However, multiple devices can still be connected to the same line using daisy chaining or polling.

To find candidate keys, calculate the closure of sets containing E and H (since they do not appear on the right side of any FD). Closure(AEH) = {A, E, H, B, C, D} = R. Closure(BEH) = {B, E, H, C, D, A} = R. Closure(DEH) = {D, E, H, A, B, C} = R. Thus, AEH, BEH, and DEH are candidate keys.

 Option 2 is right. T(n)= q(n logn)
It is false because q is not a asymptotic symbol.

The expression is 2 # 3 & 5 # 6 & 4. Following the rules: E -> E1 # T (E1 * T) and T -> T1 & F (T1 + F). The expression evaluates as (2 # (3 & 5)) # (6 & 4). 3 & 5 = 8. 2 # 8 = 16. 6 & 4 = 10. 16 # 10 = 160.

In Ethernet, the binary exponential backoff algorithm is used after collisions. If two stations collide, each chooses a random slot (0 or 1). The probability that one wins is 0.5. The history of the previous win does not affect the probability of the next independent backoff race.

3-SAT is a classic NP-complete problem. 2-SAT can be solved in polynomial time (P) by converting it into an implication graph and checking for strongly connected components.

This is a duplicate of 336217. The candidate keys are found by taking the closure of sets containing E and H (as they are not on the right side of any FD). Closure(AEH), Closure(BEH), and Closure(DEH) all result in the set of all attributes.

  The spanning tree algorithm is used for bridge-routing For avoiding loops in the routing paths.

The loop reverses the integer. After i iterations, the last i digits of n have been moved to the front of rev in reverse order. Thus, n contains the remaining digits and rev contains the reversed prefix.

User-level threads are managed by a library; the kernel is unaware of them, so a blocking system call blocks the whole process. Kernel-supported threads are scheduled by the OS, making them independent. Context switching is generally faster for user-level threads because it avoids kernel mode transitions.

Abstraction, encapsulation, and polymorphism are core pillars of OOP. Inheritance is also essential, and polymorphism is typically realized through inheritance.

Instruction fetch/decode: 2 cycles per word. Memory transfer: 3 cycles. ADD: 1 cycle. MOV R1,5000 (2 words): 4 cycles + 3 (mem) = 7. MOV R2,(R1) (1 word): 2 cycles + 3 (mem) = 5. ADD R2,R3 (1 word): 2 cycles + 1 = 3. MOV 6000,R2 (2 words): 4 cycles + 3 (mem) = 7. HALT (1 word): 2 cycles. Total: 7+5+3+7+2 = 24.

The number of distinct binary search trees with n nodes is given by the nth Catalan number: C(n) = (1/(n+1)) * binom(2n, n). For n=4, C(4) = (1/5) * binom(8, 4) = (1/5) * 70 = 14.

In packet-switched networks, both TCP and UDP packets are routed independently by the network layer (IP). Therefore, packets of the same session can take different paths regardless of the transport protocol used.

Natural join on Student and Enroll (sharing 'rollno'): Max tuples occur if every student is enrolled in all 8 courses (120 * 8 = 960). Min tuples occur if only the students present in Enroll are joined (8).

Option (4) is correct.

Hash function x mod 10: 4322->2, 1334->4, 1471->1, 9679->9, 1989->9, 6171->1, 6173->3, 4199->9. (i) 9679, 1989, 4199 all hash to 9. (ii) 1471, 6171 both hash to 1. Both statements are true.

A handle is a substring that matches the right side of a production rule and whose reduction represents one step along the reverse of a rightmost derivation. For n + n x n, the steps are: n -> F -> T -> E; then E + n -> E + F -> E + T -> E; then E x n -> E x F -> E x T -> E.

Inserting 10, 1, 3, 5, 15, 12, 16 into a BST creates a tree where 10 is the root, 1 is the left child, 3 is the right child of 1, 5 is the right child of 3, 15 is the right child of 10, 12 is the left child of 15, and 16 is the right child of 15. The longest path from the root (10) to a leaf (5 or 16) has 3 edges (10-1-3-5).

Option (1) is correct.

If h = f(g(x)) is onto, then for every element in C, there exists an element in A such that h(a) = c. This implies that f must be able to map to every element in C, so f must be onto; however, g does not need to be onto because f can map the image of g to the entirety of C.

The function recursively calculates the maximum depth of the tree. For each node, it returns 1 plus the maximum of the heights of its left and right subtrees, which is the definition of tree height.

The condition digit >= 5 for a 4-bit input (d3 d2 d1 d0) is true for inputs 5 (0101) through 9 (1001). The Boolean expression is f = d3 + d2d1 + d2d0, which can be implemented with 4 gates (two ANDs, two ORs).

With 2 intermediate nodes, there are 3 hops. Total overhead is 3 hops * 3 bytes = 9 bytes per packet. To minimize overhead while fitting the 24-byte message, we calculate efficiency; however, in many textbook problems of this type, the packet size is often constrained by the header or specific framing logic. Given the options, 9 is the only one that allows for a balanced division of the message.

Convert both to base 10: 7x + 3 = 5y + 4. This simplifies to 7x - 5y = 1. Testing the options: for x=8, 56+3=59; 5y+4=59 implies 5y=55, so y=11.

This is the Babylonian method (or Heron's method) for calculating the square root of a number m. The sequence x_n+1 = (x_n + m/x_n) / 2 converges to sqrt(m).

If the system only executes a single sequential process at a time, there is only one I/O request in the queue at any given moment. Therefore, the disk scheduling algorithm (FCFS vs SSTF) makes no difference because there is no queue to optimize.