What is the count for the instruction CountMe as a function of n for the fragment below? (Assume that 2k-1 ≤ n < 2k).
Line 1: j = 1
Line 2: while (j<= n) {
Line 3: CountMe
Line 4: j = 2*j;
To determine the count for the instruction CountMe as a function of n, we need to consider how many times the instruction gets executed within the while loop.
The while loop condition is "j <= n", which means the loop will continue as long as j is less than or equal to n.
Line 1 initializes j to 1, and in each iteration of the while loop, j is multiplied by 2 (j = 2*j in Line 4).
Since we are assuming that 2k-1 ≤ n < 2k, where k is an integer, we can observe that j starts from 1 and doubles in each iteration.
The value of j in each iteration can be represented as 2^i, where i is the number of iterations.
So we need to find the number of iterations required for j to reach or exceed n. Let's solve this equation for i:
2^i > n
Taking the base 2 logarithm on both sides:
i > log₂(n)
Since i represents the number of iterations, it must be an integer, and we need to round up the result of log₂(n) to the nearest integer. Let's call this value i_max.
The count for the instruction CountMe is simply equal to the number of iterations, so the count can be represented as i_max.
Therefore, the count for the instruction CountMe as a function of n is equal to the smallest integer greater than or equal to log₂(n).
Mathematically:
CountMe(n) = ⌈log₂(n)⌉
So, to determine the count for a specific value of n, you can calculate the logarithm base 2 of n and round it up to the nearest integer.