Consider the following code segment. int[][] multi = new int[4][4];

for (int rows = 0; rows < 4; rows++)

{

for (int cols = 0; cols < 4; cols++)

{

if (cols == 0)

{

multi[rows][cols] = 0;

}

else if (cols == 1)

{

multi[rows][cols] = 1;

}

else if (cols == 2)

{

multi[rows][cols] = 2;

}

if ((rows % 2 == 0) && (cols % 2 == 0))

{

if ((rows >= 2) && (cols <= 2))

{

multi[rows][cols] = 9;

}

}

}

}

As a result of executing the code segment, how many elements in the two-dimensional (2D) array multi will store the value 9 ?

A 0

B 1

C 2

D 4

E 6

In many applications, only a small percentage of the points in a given set p of n points are extreme. in such a case, the convex hull of p has less than n vertices. this can actually make our algorithm convexhull run faster than θ(nlogn). assume, for instance, that the expected number of extreme points in a random sample of p of size r is o(rα ), for some constant α < 1. (this is true when the set p has been created by picking points uniformly at random in a ball.) prove that under this condition, the running time of the algorithm is o(n).