Suppose that, in a divide-and-conquer algorithm, we always divide an in stance of size \(n\) of a problem into 10 subinstances of size \(n / 3,\) and the dividing and combining steps take a time in \(\Theta\left(n^{2}\right)\). Write a recumence equation for the running time \(T(n),\) and solve the equation for \(T(n)\).

Short Answer

Expert verified
The running time T(n) of the algorithm is \(\Theta(n^{2}log(n))\).

Step by step solution

01

Identify the Dividing and Combining Operation

The problem states that the algorithm is a divide-and-conquer variety, with a problem of size n being divided into 10 subproblems of size n/3. This means we can say the division operation happens with 10 cases each of size n/3. Also, it further asserts that the divide and combine steps take \(\Theta\left(n^{2}\right)\) time.
02

Define the Recurrence Relation

Combining the above information, we formulate the recurrence relation for the run-time T(n) of the algorithm as \(T(n) = 10T(n/3) + \Theta\left(n^{2}\right)\). This equation is made up of two parts, the first term 10T(n/3) represents the time taken to solve the 10 subproblems of size n/3 and the second term \(\Theta\left(n^{2}\right)\) represents time taken to divide and later combine the solutions.
03

Solve the Recurrence Relation

The solution to the recurrence relation is obtained using the Master Theorem. The Master Theorem applies to recurrences of the form \(T(n) = aT(n/b) + f(n)\). In this case, a=10, b=3, and f(n) = \(\Theta\left(n^{2}\right)\). Comparing this with \(n^{log_{b}{a}} = n^{log_{3}{10}}\), it is seen that \(f(n)\) grows faster. Therefore, the time complexity T(n) belongs to \(\Theta(n^{2}log(n))\).

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