- Option : B
- Explanation :

I {a^{m}b^{n}c^{p}d^{q}| 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 {a^{m}b^{n}c^{p}d^{q}| m = n and p = q} is clearly CFL since, one comparison at a time can be done by pda

III {a^{m}b^{n}c^{p}d^{q}| m = n = p and p ≠ q} is not CFL since m = n = p is a double comparison which cannot be done by PDA.

IV {a^{m}b^{n}c^{p}d^{q}| 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.

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- 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.

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Allocation | Max | |||||||

E | F | G | E | F | G | |||

P_{0} | 1 | 0 | 1 | P_{0} | 4 | 3 | 1 | |

P_{1} | 1 | 1 | 2 | P_{1} | 2 | 1 | 4 | |

P_{2} | 1 | 0 | 3 | P_{2} | 1 | 3 | 3 | |

P_{3} | 2 | 0 | 0 | P_{3} | 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 :
Max need Current allocation Current available Remaining need E F G E F G E(3) F(3) G(0) E F G P _{0}4 3 1 1 0 1 4 3 1 3 3 0 P _{1}2 1 4 1 1 2 5 3 4 1 0 2 P _{2}1 3 3 1 0 3 6 4 6 0 3 0 P _{3}5 4 1 2 0 0 8 4 6 3 4 1

Safe sequence : P_{0}, P_{2}, P_{1}, P_{3}

Safe state

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- Option : C
- Explanation :

Total number of scalar multiplications are 48 + 75 + 50 + 2000 = 2173 and optimal parenthesis is ((F_{1}(F_{2}(F_{3}F_{4}))) F_{5}). As concluded, F_{3}, F_{4}are explicitly computed pairs.

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