Let the selected chromosome X is 10110 represents array of five bits and the fitness function is f(x)= decimal (X) - (X[0] XOR X[2] XOR X[4] )*3. The cross over (must increase the fitness function) is one digit which is one of digits 0,2,4, then after cross over the chromosome become

a.10010

b.00110

c.10110

d.10111

sum = 0;

for (k=1; k<=n; k*=2)

for (j=n; j<=1; j/=2)

sum++;

The time complexity of the algorithm is, Select one:

a.O(log n²)

b.O((log n)²)

c.O(n²)

d.O(log² n)

Let the list: 7, 10, 15, 25, 22, 13, 11, 5, 9, 3, 6, 1 after applying the maxheap sort the list after the first heapify process becomes

a.25, 22, 15, 13, 11, 10, 9, 7, 6, 5, 3, 1

b.25, 22, 15, 10, 7, 13, 11, 5, 9, 3, 6, 1

c.25, 22, 15, 10, 13, 11, 9, 7, 6, 3, 5, 1

d.25, 22, 15, 7, 10, 11, 13, 9, 5, 6, 3, 1