Ytukyg

I am trying to solve this recurrence $T(n) = 6 T(frac{n}{3}) + n$.

1st recurrence: $6^2T(frac{n}{3^2}) + frac{6n}{3} + n$

2nd: $6^3T(frac{n}{3^3}) + frac{6^3n}{3^2} + n$

3rd: $6^4T(frac{n}{3^4}) + frac{6^6n}{3^3} + n$

I am having trouble describing the general pattern after the k-th iteration.

I assume you are looking to find $T(n)$ after K iterations. First
$$ T(n) = 6 T(n/3) + n$$ then rewriting $T(n/3)$ in terms of $T(n/3^2)$ we conclude:
$$
T(n) = 6^2 T(n/3^2) + 6n/3 +n
$$
similarly $T(n)$ after K iterations becomes:
$$
T(n) = 6^K T(n/ 3^K) + sum_{i=1} ^K 6^{i-1}n/3^{i-1}
$$
or
$$
T(n) = 6^K T(n/ 3^K) + (2^{K}-1)n
$$

If you are trying to solve the recurrence, try finding the initial pattern and the number of terms - this will help solve the equation in most of the cases.

In the first step, we have $n$ work to do, we do $n$ work and have 6 problems of the same type but of size $frac{n}{3}$ this time.

In the second step, we have $frac{n}{3}$ work to do, we do $frac{n}{3}$ work and have 6 problems of the same type but of size $frac{n}{9}$ this time. We have 6 problems of this type.

In the third step, we have $frac{n}{9}$ work to do, we do $frac{n}{9}$ work and have 6 problems of the same type but of size $frac{n}{27}$ this time. We have 6 problems of this type.

This gives the equation,

$T(n) = n + 6(frac{n}{3}) + 36 (frac{n}{9}) + 216(frac{n}{27}) + ... + $

$T(n) = n + 2n + 4n + 8n + ... $

You will have $log_{3}n$ such terms, as you are cutting down the problem size by a factor of three all the time. This is a sum of a geometric series, given the number of terms and common ratio, it is easy to sum.

$T(n) = n + 2n + 4n + 8n + ... + 2^{log_{3}n}n $

Kth term will be of the form $2^{k}n$.

The solution should be $Theta(n^{log_{3}6})$.