GATE 2017 CS Set 2

33) A system shares 9 tape drives. The current allocation and maximum requirement of tape drives for 3 processes are shown below:

Process	Current Allocation	Maximum Requirement
P1	3	7
P2	1	6
P3	3	5

Which of the following best describes the current state of the system?

Safe, Deadlocked
Safe, Not Deadlocked
Not Safe, Deadlocked
Not Safe, Not Deadlocked

Answer: B

Explanation:

Given, Total Tape Drive = 9

Process	Current Allocation	Maximum Requirement	Need
P1	3	7	4
P2	1	6	5
P3	3	5	2

In the table, we can see that the currently allocated Tape Drive = 7

So, available Tape Drive is: 9 - 7 = 2

Now first we allocate the remaining Tape Drive to process P3 Which need is 2. After executing it releases 3 Tape Drive. So new available Tape Drive is: 3 + 2 = 5

Then Process P1 needs 4 Tape Drive. So it will be executed next and releases3 Tape Drive. So new available Tape Drive is: 3 + 5 = 8.

In last, Process P2 will be executed. Hence, all the Process in the Safe State with No Deadlock. Safe Thus, Safe Sequence will be P3→P2→P1 or P3→P1→P2. Therefore option (B) will be the right answer.

34) Consider a binary code that consists only four valid codewords as given below.

00000, 01011, 10101, 11110

Let minimum Hamming distance of code be p and maximum number of erroneous bits that can be corrected by the code be q. The value of p and q are:

p = 3 and q = 1
p = 3 and q = 2
p = 4 and q = 1
p = 4 and q = 2

Answer: A

Explanation:

Given,
Code1 = 00000, Code2 = 01011, Code3 = 10101, Code4 = 11110
Hamming distance -> The number of places between two vectors will differ.

Minimum of all hamming distances is = 3 ( Code1 = 00000, and Code2 = 01011 )
So, p = 3
Now, to find the maximum number of erroneous bits that can be corrected by the code we need Hamming Distance = 2d + 1
So, 2d + 1 = 3
d = 1

Therefore option (A) is the right answer.

35) Consider two hosts X and Y, connected by a single direct link of rate 10⁶ bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2 x 10⁸ m/s. Hosts X send a file of 50,000 bytes as one large message to hosts Y continuously. Let the transmission and propagation delays be p milliseconds and q milliseconds, respectively. Then the vales of p and q are:

p = 50 and q = 100
p = 50 and q = 400
p = 100 and q = 50
p = 400 and q = 50

Answer: D

Explanation:

Given,
Bandwidth(B) = 10⁶ bits/sec
Distance(D) = 10000 km = 10,000,000 m = 10⁷ m
Propogation Speed(V) = 2 * 10⁸ m/s
Data Length(L) = 50, 000 Bytes = 50,000 * 8 bits
∴ Transmission Time (p) = L / B
= (50, 000 * 8 ) / 10⁶
= 40 / 100 = 0.4 = 400 msec
∴ Propogation Time (q) = D / V
= 10⁷ / (2 * 10⁸)
= 1 / 20 = 0.05 = 50 msec
Therefore option (D) is the right answer.

36) The pre-order traversal of a binary search tree is given by 12, 8, 6, 2, 7, 9, 10, 16, 15, 19, 17, 20. Then the post-order traversal of this tree is:

2, 6, 7, 8, 9, 10, 12, 15, 16, 17, 19, 20
2, 7, 6, 10, 9, 8, 15, 17, 20, 19, 16, 12
7, 2, 6, 8, 9, 10, 20, 17, 19, 15, 16, 12
7, 6, 2, 10, 9, 8, 15, 16, 17, 20, 19, 12

Answer: B

Explanation:

Given,

Pre-order: 12, 8, 6, 2, 7, 9, 10, 16, 15, 19, 17, 20

In BST the first node of a pre-order traversal is always the root node. The BST will always be given in ascending order. So the corresponding Binary tree is given below-

∴ Post-order Traversal: 2, 7, 6, 10, 9, 8, 15, 17, 20, 19, 16, 12

Therefore option (B) is the right answer.

37) Consider the C program fragment below which is meant to divide x by y using repeated subtractions. The variable x, y, q and r are all unsigned int.

while(r >= y)
{
   r = r - y;
   q = q + 1;
}

Which of the following conditions on the variables x, y, q and r before the execution of the fragment will ensure that the loop terminates in a state satisfying the condition x == (y*q + r)?

( q == r ) && ( r == 0)
( x > 0 ) && ( r == x ) && ( y > 0 )
( q == 0 ) && ( r == x ) && ( y > 0 )
( q == 0 ) && ( y > 0 )

Answer: C

Explanation:

Given, x = = (y * q + r)

According to question x= product, y= Multiplicand, q= Quotient and r= Remainder

Now, to divide a number with repeated subtraction, the quotient should be 0, and it should be incremented for each subtraction. Thus, if q = 0, r = x. Therefore option (C) will be the right answer.

38) Consider the following C function.

int fun(int n)
{
   int i, j;
   for (i = 1; i <= n ; i++)
     {
   for (j = 1;  j < n; j += i)
   {
       printf("%d %d", i, j);
   }
  }
}

Time complexity of fun in terms of θ notation is:

θ(n √n)
θ(n²)
θ(n log n)
θ(n² log n)

Answer: C

Explanation:

In the above question, we have to check how many times the innermost statement printf("%d %d," i, j); of inner for loop is executed. The inner for loop is dependent on i. So,

For i=1, the statement runs n times. Thus for ith iteration, statement runs θ(n/i) times.

Hence, Time complexity of fun in terms of θ notation is:

n- 1/1 + n-1/2 + n-1/3 +...........+ n-1/n-1 +1
n/1+n/2+n/3+......+n/n-1 -log(n-1)
n{1/1+1/2+1/3+.....+1/n-1}-log(n-1)
nlog(n-1)-log(n-1)
nlog(n-1)
nlogn
Therefore option (C) will be the right answer.

39) Let δ denote the transition function and α denoted the extended transition function of the ?-NFA whose transition table is given below:

δ	ε	a	b
?>q0	{q2}	{q1}	{q0}
q1	{q2}	{q2}	(q3}
q2	{q0)	ø	ø
q3	ø	ø	(q2}

Then, α (q2,aba) is

ø
{q1, q2, q3}
{q0, q1, q2}
{q0, q2, q3}

Answer: C

Explanation:

Given, α (q2,aba)
So, starting state is: q2
First find Epsilon closure of q2 = {q2,q0}
Then, Find transition on 'a': q0------------->q1
q2------------->'Ø ' (i.e nothing)
Find epsilon closure of 'q1'= {q1,q2,q0}
Then, Find transition on 'b': q1------------->q3
q0------------->q0
q2------------->'Ø '
Now, Find epsilon closure of 'q0': {q0,q2} and epsilon closure of 'q3': {q3}
Find epsilon closure of 'q0' Union 'q3': {q3}
Then, Find transition on 'a': q0------------->q1
q2------------->'Ø '
q3------------->'Ø '
Again, Find epsilon closure of 'q1': {q1,q0,q2}

Therefore option (C) is the right answer.

40) Consider the following languages.

L₁ = {a^p | p is a prime number}
L₂ = {aⁿb^mc^2m | n >= 0, m >= 0}
L₃ = {aⁿbⁿc²ⁿ | n >= 0}
L₄ = {aⁿbⁿ | n >= 1}

Which of the following are CORRECT ?

I. L₁ is context free but not regular.
II. L₂ is not context free.
III. L₃ is not context free but recursive
IV. L₄ is deterministic context free

I, II and IV only
II and III only
I and IV only
III and IV only

Answer: D

Explanation:

L₁ is Context Sensitive Language.

L₂ is Context Free Language

In L₃ it is not sure when to pop b and push a, because here the comparison is performed between three consecutive terminals. So it is not a Context Free Language. Since given language is CFL, so it is recursive also.

L₄ is Deterministic Context Free Language because we are sure to push a into stack first and on seeing b we are sure to pop a from the stack.

Therefore option (D) is the right answer.

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