Question

Question 1. Below is the pseudo code for a divide and conquer algorithm that finds the minimum value in an array. Suppose that the input array, A, is of size n, analyze the computational cost of this algorithm in the form of Tin) and prove your conclusion findMin (A, left, right) if (left == right) return Alleft) center - (left + right) / 2 return min (findMin (A, left, center), findMin (A, center + 1, right)

Answer 1

Solution:

Let n = is the size of items in an array

Let T (n) = time required to apply the algorithm on an array of size n. Here we divide the terms as T(n/2).

1 here tends to the comparison of the minimum with minimum as in above given algorithm.

Thus, T(n) can be written as

T(n) = T(n/2) + T(n/2) + 1

or

T(n) = 2T(n/2) + 1

We can solve this using master's method as below:

a = 2, b = 2, f(n) = 1, c = 0 as n⁰ = 1, log_ba = log₂2 = 1

We observe that c < log_ba; case 1 applies. So Solution would be:

T(n) = theeta(n^log_ba) = theeta(n¹) = theeta(n).

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