PREVIOUS YEAR SOLVED PAPERS - GATE 2018
- Option : B
- Explanation : I {ambncpdq | m + p = n + q} is clearly CFL since, we can rearrange the equation as m - n + p - q = 0 which can be done by push, pop, and pop and check if stack is empty at end. II {ambncpdq | m = n and p = q} is clearly CFL since, one comparison at a time can be done by pda III {ambncpdq | m = n = p and p ≠ q} is not CFL since m = n = p is a double comparison which cannot be done by PDA. IV {ambncpdq | mn = p + q} is not a CFL, since mn involves multiplying number of a’s and number b’s which cannot be done by a PDA. So, only I and II are CFL’s.
- Option : B
- Explanation :
// n takes 2^40
unsigned long int fun(unsigned long int n) {
// initialized sum = 0
unsigned long int i, j, sum = 0;
//First it takes i = n = 2^40,
//then it divides i by 2 and incremented once j
//each time, that's will make makes j = 40,
for( i=n; i>1; i=i/2) j++;
//Now the value of j = 40,
//it divides j by 2 and incremented once sum
//each time, that's will make makes sum = 5,
for( ; j>1; j=j/2) sum++;
//returns sum = 5
return sum;
}
So, answer is 5.
- Option : C
- Explanation :
Given query: r ⋈ ( σB<5(s))
A: σB<5(r ⋈ s)A B C a1 2 c1 a2 4 c1 a3 4 c1
B: σB<5(r ⋈ s)A B C a1 2 c1 a2 4 c1 a3 4 c1
C: r ⋈ (σB<5(s))A B C a1 2 c1 a2 4 c1 a3 4 c1
D: σ;B<5 (r) ⋈ sA B C a1 2 c1 a2 4 c1 a3 4 c1 a4 6 Null a5 6 Null
Option “C” query result not equal to given query.A B C a1 2 c1 a2 4 c1 a3 4 c1
| Allocation | Max | |||||||
| E | F | G | E | F | G | |||
| P0 | 1 | 0 | 1 | P0 | 4 | 3 | 1 | |
| P1 | 1 | 1 | 2 | P1 | 2 | 1 | 4 | |
| P2 | 1 | 0 | 3 | P2 | 1 | 3 | 3 | |
| P3 | 2 | 0 | 0 | P3 | 5 | 4 | 1 | |
From the perspective of deadlock avoidance, which one of the following is true?
- A
The system is in safe state
- B
The system is not in safe state, but would be safe if one more instance of E were available
- C
The system is not in safe state, but would be safe if one more instance of F were available
- D
The system is not in safe state, but would be safe if one more instance of G were available
- Option : A
- Explanation :
Safe sequence : P0, P2, P1, P3 Safe stateMax need Current allocation Current available Remaining need E F G E F G E(3) F(3) G(0) E F G P0 4 3 1 1 0 1 4 3 1 3 3 0 P1 2 1 4 1 1 2 5 3 4 1 0 2 P2 1 3 3 1 0 3 6 4 6 0 3 0 P3 5 4 1 2 0 0 8 4 6 3 4 1
- Option : C
- Explanation : Total number of scalar multiplications are 48 + 75 + 50 + 2000 = 2173 and optimal parenthesis is ((F1(F2(F3 F4))) F5). As concluded, F3, F4 are explicitly computed pairs.
