GATE IT 2006

Question 1
In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25°C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25°C, or at/below 25°C. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25°C?
Cross
0.4
Cross
0.6
Tick
0.8
Cross
0.9


Question 1-Explanation: 
 
Question 2
For the set N of natural numbers and a binary operation f : N x N → N, an element z ∊ N is called an identity for f, if f (a, z) = a = f(z, a), for all a ∊ N. Which of the following binary operations have an identity?
  1. f (x, y) = x + y - 3
  2. f (x, y) = max(x, y)
  3. f (x, y) = xy
 
Tick
I and II only
Cross
II and III only
Cross
I and III only
Cross
None of these


Question 2-Explanation: 
I f(x,y) = x+y-3 = x= y+x-3  =>  y=3 Here identity elements is 3 II f(x,y) = max(x,y)=x=max(y,x)  => y=1 Here identity elements is 1 (III f(x,y) =x^y is not same as f(y,x) = y^x. So no identity element.
Question 3
In the automaton below, s is the start state and t is the only final state.download1               Consider the strings u = abbaba, v = bab, and w = aabb. Which of the following statements is true?  
Cross
The automaton accepts u and v but not w
Cross
The automaton accepts each of u, v, and w
Cross
The automaton rejects each of u, v, and w
Tick
The automaton accepts u but rejects v and w


Question 3-Explanation: 
  For the acceptance and rejection of any string we can simply check for the movement on each input alphabet between states. A string is accepted if we stop at any final state of the DFA. For string u=abbaba the string ends at t (final state) hence it is accepted by the DFA. For string v=bab the string ends at s (non-final state) and hence rejected by the DFA. For string w=aabb the string ends at s (non-final state) and hence rejected by the DFA.   This solution is contributed by Yashika Arora.
Question 4
In the context-free grammar below, S is the start symbol, a and b are terminals, and ϵ denotes the empty string S → aSa | bSb | a | b | ϵ Which of the following strings is NOT generated by the grammar?  
Cross
aaaa
Tick
baba
Cross
abba
Cross
babaaabab


Question 4-Explanation: 
  The given Language is  Palindrome. A,C and D follow the language but  baba  is not a palindrome  , so B is the answer
Question 5
Which regular expression best describes the language accepted by the non-deterministic automaton below? download (7)  
Tick
(a + b)* a(a + b)b
Cross
(abb)*
Cross
(a + b)* a(a + b)* b(a + b)*
Cross
(a + b)*


Question 6
Given a boolean function f (x1, x2, ..., xn), which of the following equations is NOT true  
Cross
f (x1, x2, ..., xn) = x1'f(x1, x2, ..., xn) + x1f(x1, x2, ..., xn)
Cross
f (x1, x2, ..., xn) = x2f(x1, x2, …, xn) + x2'f(x1, x2, …,xn)
Cross
f (x1, x2, ..., xn) = xn'f(x1, x2, …, 0) + xnf(x1, x2, …,1)
Tick
f (x1, x2, ..., xn) = f(0, x2, …, xn) + f(1, x2, .., xn)


Question 6-Explanation: 
Option A: f (x1, x2, …, xn) = x1’f(x1, x2, …, xn) + x1f(x1, x2, …, xn) Case 1: taking x1=0 RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn) RHS =f(x1, x2, …, xn). Case 2: taking x1=1 RHS = 0.f(x1, x2, …, xn) + 1.f(x1, x2, …, xn) RHS =f(x1, x2, …, xn). In both cases RHS=LHS, so, (A) is true Option B: f (x1, x2, …, xn) = x2f(x1, x2, …, xn) + x2’f(x1, x2, …, xn) Case 1: taking x2=0 RHS= 0.f(x1, x2, …, xn) + 1.f(x1, x2…,xn) RHS =f(x1, x2, …, xn). Case 2: taking x2=1 RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn) RHS =f(x1, x2, …, xn). In both cases RHS=LHS, so, (B) is true. Option C: f (x1, x2, …, xn) = xn’f(x1, x2, …, 0) + xnf(x1, x2, …,1) Case 1: taking xn=0 RHS= 1.f(x1, x2, …, 0) + 0.f(x1, x2, …, 1) RHS =f(x1, x2, …, 0) Case 2: taking xn=1 RHS = 0.f(x1, x2, …, 0) + 1.f(x1, x2, …, 1) RHS =f(x1, x2, …, 1)In both cases RHS=LHS, so, (C) is true. Option D: f (x1, x2, …, xn) = f(0, x2, …, xn) + f(1, x2, .., xn) Here, no way to equate LHS and RHS so ‘NOT true’. NO term depends on value of ‘x1’.   This solution is contributed by Sandeep pandey.
Question 7
The addition of 4-bit, two's complement, binary numbers 1101 and 0100 results in  
Cross
0001 and an overflow
Cross
1001 and no overflow
Tick
0001 and no overflow
Cross
1001 and an overflow


Question 7-Explanation: 
Its  -3+4=1, so no overflow So Answer is C.
Question 8

Which of the following DMA transfer modes and interrupt handling mechanisms will enable the highest I/O band-width?  

Cross

Transparent DMA and Polling interrupts

Cross

Cycle-stealing and Vectored interrupts

Tick

Block transfer and Vectored interrupts

Cross

Block transfer and Polling interrupts



Question 8-Explanation: 

Among the options provided, the DMA transfer mode and interrupt handling mechanism that enable the highest I/O bandwidth is option (C): Block transfer and Vectored interrupts.

Let's understand why this choice provides the highest I/O bandwidth:

  1. Transparent DMA: Transparent DMA is not the most efficient option for achieving high I/O bandwidth. It allows the DMA controller to directly access the memory and transfer data without CPU involvement. However, it doesn't utilize interrupts effectively for signaling completion or handling data transfer events.
  2. Cycle-stealing: Cycle-stealing DMA interrupts the CPU during its normal operation to transfer data. While it can improve I/O bandwidth, it interrupts the CPU frequently, affecting its performance.
  3. Block transfer: Block transfer DMA involves transferring a block of data in a single operation. It minimizes the number of interrupts needed for transferring large amounts of data, resulting in efficient data transfer.
  4. Polling interrupts: Polling interrupts require the CPU to check the status of the DMA controller repeatedly to determine if data transfer has completed. This approach involves continuous CPU involvement and wastes CPU cycles, making it less efficient for achieving high I/O bandwidth.
  5. Vectored interrupts: Vectored interrupts allow the DMA controller to directly inform the CPU about data transfer events, reducing the CPU overhead. The CPU can efficiently handle the interrupts and resume its normal operations.

Considering the characteristics described above, option (C) combines block transfer DMA, which minimizes interrupts for large data transfers, with vectored interrupts, which efficiently notify the CPU about data transfer events. This combination maximizes the I/O bandwidth by reducing CPU involvement and efficiently utilizing interrupts, making it the best choice among the provided options for achieving the highest I/O bandwidth.

Question 9
In a binary tree, the number of internal nodes of degree 1 is 5, and the number of internal nodes of degree 2 is 10. The number of leaf nodes in the binary tree is  
Cross
10
Tick
11
Cross
12
Cross
15


Question 9-Explanation: 
In a binary tree, the number of leaf nodes is always 1 more than number of internal nodes with 2 children, refer http://www.geeksforgeeks.org/handshaking-lemma-and-interesting-tree-properties/ So, Number of Leaf Nodes = Number of Internal nodes with 2 children + 1 Number of Leaf Nodes = 10 + 1 Number of Leaf Nodes = 11
Question 10

A problem in NP is NP-complete if  

Cross

It can be reduced to the 3-SAT problem in polynomial time

Tick

The 3-SAT problem can be reduced to it in polynomial time

Cross

It can be reduced to any other problem in NP in polynomial time

Cross

Some problem in NP can be reduced to it in polynomial time



Question 10-Explanation: 

A problem in NP becomes NPC if all NP problems can be reduced to it in polynomial time. This is the same as reducing any of the NPC problems to it. 3-SAT being an NPC problem, reducing it to an NP problem would mean that NP problem is NPC.

Please refer: http://www.geeksforgeeks.org/np-completeness-set-1/

There are 86 questions to complete.

  • Last Updated : 02 Dec, 2021

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