GATE 2017 CS Set 2

41) Let L(R) be the language represented by regular expression R. Let L(G) be the language generated by a context free grammar G. Let L(M) be the language accepted by a Turing machine M. Which of the following decision problems are undecidable?

I. Given a regular expression R and a string w, is w∈L(R)?
II. Given a context-free grammar G, is L(G)=Ø
III. Given a context-free grammar G, is L(G)=Σ∗ for some alphabet &Sigma?
IV. Given a Turing machine M and a string w, is w ∈ L(M)?

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

Answer: D

Explanation:

Given,
L ( R ) is the language represented by the regular expression
L (G) is the language generated by a context-free grammar
L (M) is the language accepted by Turing Machine
Option (I) is a membership problem which is decidable for Finite state machine and Regular expression.
Option (II) is an Emptiness problem which is decidable by checking the usefulness of start symbol.
Option (III) is a problem of CFL which is undecidable.
Option (IV) is a membership problem of regular expression language which is undecidable.

Therefore option (D) is the right answer.

42) The next state table of a 2-bit saturating up-counter is given below.

Q₁	Q₀	Q₁⁺	Q₀⁺
0	0	0	1
0	1	1	0
1	0	1	1
1	1	1	1

The counter is built as synchronous sequential circuit using T flip-flops. The value for T1 and T0 are

T₁ = Q₀Q₁ T₀ = Q'₀Q'₁
T₁ = Q'₁Q₀ T₀ = Q'₁ + Q'₀
T₁ = Q₁ + Q₀ T₀ = Q'₁ + Q'₀
T₁ = Q'₁Q₀ T₀ = Q₁ + Q₀

Answer: B

Explanation:

Q₁	Q₀	Q₁⁺	Q₀⁺	T₀	T₁
0	0	0	1	0	1
0	1	1	0	1	1
1	0	1	1	0	1
1	1	1	1	0	0

By using above Excitation table, we get

T₁ = Q'₁Q₀ T₀ = Q'₁ + Q'₀

Therefore option (B) is the right answer.

43) Consider the following snippet of a C program. Assume that swap(&x, &y) exchanges the contents of x and y.

int main()
{
   int array[] = {3, 5, 1, 4, 6, 2};
   int done = 0;
   int i;
   while (done == 0)
   {
       done  = 1;
       for (i = 0; i <= 4; i++)
       {
       if (array[i] < array[i+1])
       {
   swap(&array[i], &array[i+1]);
   done = 0;
       }
   }
   for (i = 5; i >= 1; i--)
   {
   if (array[i] > array[i-1])
   {
       swap(&array[i], &array[i-1]);
       done = 0;
   }
  }
 }
 printf("%d", array[3]);
}

The output of the program is _____.

Answer: C

Explanation:

Initially while loop starts execution with (while == 0)
After the execution of the for loop in first time, the content of array will be as follows: 5 3 4 6 2 1
After the execution of the for loop in Second time, the content of array will be as follows: 6 5 4 3 2 1
After the execution of the for loop in Third time, the content of array will be as follows: 6 5 4 3 2 1
Now, when while loop executes again, done = 1 and first and second for loop if condition will not be satisfied. So the final content of an array is 6, 5, 4, 3, 2, 1...The output of the program is = = 3.
Therefore option (C) is the right answer.

44) Two transactions T₁ and T₂ are given as:

T₁: r₁(X)w₁(X)r₁(Y)w₁(Y)
T₂: r₂(Y)w₂(Y)r₂(Z)w₂(Z)

where _ri(V) denotes a read operation by transaction T_i on a variable V and wi(V) denotes a write operation by transaction T_i on a variable V. The total number of conflict serializable schedules that can be formed by T₁ and T₂ is ______

Answer: A

Explanation:

Total number of schedules = Non conflict serializable schedules + Conflict serializable schedules

According to question, Total number of schedules: 8! / 4!*4! <=> 1*2*3*4*5*6*7*8 / 4*3*2*1* 4*3*2*1 = 70

Non conflict serializable schedules: A1_2_3(BCD4) + _1A2_3(BCD4) + _1_2A3(BCD4) + _1_2_3A(BCD4) = 4*(3+2?1) = 16

So total number of conflict serializable schedules are = 70 -16 = 54.

Therefore option (A) is the right answer.

45) The read access times and the hit ratios for different caches in a memory hierarchy are as given below:

Cache	Read access time (in nanoseconds)	Hit ratio
I-cache	2	0.8
D-cache	2	0.9
L2-cache	8	0.9

The read access time of main memory in 90 nanoseconds. Assume that the caches use the referred-word-first read policy and the writeback policy. Assume that all the caches are direct mapped caches. Assume that the dirty bit is always 0 for all the blocks in the caches. In execution of a program, 60% of memory reads are for instruction fetch and 40% are for memory operand fetch. The average read access time in nanoseconds (up to 2 decimal places) is _________.

2.74
4.72
3.10
2.67

Answer: B

Explanation:

Given,
Instruction Fetch = 0.6
Operand fetch = 0.4
Average read time = ?

According to the question, L2 cache is shared between Instruction and Data. So,

Average read time = Fraction of Instruction Fetch * Average Instruction fetch time + Fraction of Data/Operand Fetch * Average Data/Operand Fetch Time

Now,

Average Instruction fetch Time = L1 access time + L1 miss rate * L2 access time + L1 miss rate * L2 miss rate * Memory access time = 2 + 0.2 * 8 + 0.2 * 0.1 * 90 = 5.4 ns

Average Data fetch Time = L1 access time + L1 miss rate * L2 access time + L1 miss rate * L2 miss rate * Memory access time = 2 + 0.1 * 8 + 0.1 * 0.1 * 90 = 3.7 ns

Thus, Average memory access time = 0.6 * 5.4 + 0.4 * 3.7 = 4.72 ns

Therefore option (B) is the right answer.

46) Consider the following database table named top_scorer.

Player	Country	Goals
Klose	Germany	16
Ronaldo	Brazil	15
G Miiller	Germany	14
Fontaine	France	13
Pele	Brazil	12
Klinsmann	Germany	11
Kocsis	Hungary	11
Batistuta	Argentina	10
Cubillas	Peru	10
Lato	Poland	10
Lineker	England	10
T Miiller	Germany	10
Rahn	Germany	10

Consider the following SQL query:

SELECT ta.player FROM top_scorer AS ta
WHERE ta.goals > ALL ( SELECT tb.goals
       FROM top_scorer AS tb
       WHERE tb.country = 'Spain' )
    AND ta.goals > ANY (SELECT tc.goals
       FROM top_scorer AS tc
      WHERE tc.country = 'Germany')

The number of tuples returned by the above SQL query is ____.

Answer: B

Explanation:

The first where condition select players which have goals greater than ALL players of Spain. Since Spain is not in the database, so this condition always returns TRUE.

The second where condition selects all those players who have higher than 10 Goals because we can see in the table that every German player have minimum 10 goals. Thus the first 7 rows of the table will be returned.

Player	Country	Goals
Klose	Germany	16
Ronaldo	Brazil	15
G Miiller	Germany	14
Fontaine	France	13
Pele	Brazil	12
Klinsmann	Germany	11
Kocsis	Hungary	11

Therefore option (B) will be the right answer.

47) If the ordinary generating function of a sequence is:

then a₃ - a₀ is equal to:

Answer: C

Explanation:

Given ordinary generating function = (1+z) (1-z)^-3
Now, (1-z)^-3 = 1.z⁰ + (³C₁).z¹ + (^{4^{C₂).z2 + (⁵C₃).z³ + .....+ ∞

So, (1+z) (1-z)^-3 = (1+z) . ( 1.z⁰ + (³C₁).z¹ + (⁴C₂).z² + (⁵C₃).z³ + .....+ ∞ )

Thus, Coefficient of a₀ = 1

Coefficient of a₃ = ⁴C₂ + ⁵C₃ = 4*3 / 2*1 + 5*4*3/ 3*2*1 = 6+10 = 16

then a3 - a0 is = 16 - 1 = 15

Therefore option (C) is the right answer.}}

48) If a random variable X has a Poisson distribution with mean 5, then the expression E[(X+2)²] equals _____.

Answer: B

Explanation:

As we know that in Poisson distribution, mean and variance are same. In the given question mean is 5. So variance should also be 5.
Also, In Poisson distribution: variance = E[X²] - (E[X])²
5 = E[X²] - 25
E[X²] = 25 + 5 = 30
By using Linearity of Expectation, we can write,
E[(X+2)²] = E[X² + 4X + 4]
= E(X²) + 4E(X) + 4
= 30 + 4*5 + 4 = 54
Therefore option (B) is the right answer.