A divide-and-conquer algorithm solves a problem by dividing its given instance i

Question

A divide-and-conquer algorithm solves a problem by dividing its given instance into several smaller instances, solving each of them recursively, and then, if necessary, combining the solutions to the smaller instances into a solution to the given instance. Assuming that all smaller instances have the same size n/b, with a of them being actually solved, we get the following recurrence valid for n = b^k, k = 0, 1,2,... T(n) = aT(n/b) + f(n), where a > 1, b > 2, and f(n) is a function that accounts for the time spent on dividing the problem into smaller ones and combining their solutions. Show that the following formula is the solution to this recurrence for n = b^k, The order of growth of solution T(n) depends on the values of the constants a and b and the order of growth of the function f(n).

Explanation / Answer

The recurrence for divide and conquer is mentioned as follows:

For and is the time spent in dividing the problem.

Consider

        

Expand the function T(n) after a

Expand the function T(n) after a

Proceed upto k and Take common

…(1)

Assume n=bk .

Since

Now equation (1) will be

The equation can be written as a summation.

Hence it is proved that

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