Question

T(n)=3T(n√2)+n
$t(n) = 3t(n \sqrt{2} ) + n {^{2}$

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Answer 1

Answer:

These types of recurrences are most easily solved by Master Theorem for analysis of algorithms which is explained as follows:

Let a be an integer greater than or equal to 1, b be a real number greater than 1, and c be a positive real number. Given a recurrence of the form -

T (n) = a * T(n/b) + nc where n > 1, then for n a power of b, if

Logba < c, T (n) = Θ(nc);

Logba = c, T (n) = Θ(nc * Log n);

Logba > c, T (n) = Θ(nlogba).

English translation of your recurrence

The most critical thing to understand in Master Theorem is the constants a, b, and c mentioned in the recurrence. Let's take your own recurrence - T(n) = 3T(n/2) + n - for example.

This recurrence is actually saying that the algorithm represented by it is such that,

(Time to solve a problem of size n) = (Time taken to solve 3 problems of size n/2) + n

The n at the end is the cost of merging the results of those 3 n/2 sized problems.

Now, intuitively you can understand that:

if the cost of "solving 3 problems of size n/2" has more weight than "n" then the first item will determine the overall complexity;

if the cost "n" has more weight than "solving 3 problems of size n/2" then the second item will determine the overall complexity; and,

if both parts are of same weight then solving the sub-problems and merging their results will have an overall compounded weight.

From the above three intuitive understanding, only the three cases of Master Theorem arise.

In your example, a = 3, b = 2 and c = 1. So it falls in case-3 as Logba = Log23 which is greater than 1 (the value of c).