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I'm having trouble getting the classification of this recurrence relation using a recursion tree.

$$T(n) = 3T(n/2) + n^2$$

I have the tree written out correctly (I hope):

                                         n^2
                                      /   |   \
                                     /    |    \
                              (n/2)^2  (n/2)^2  (n/2)^2
                        /   |   \      /  |  \     /   |    \
                       /    |    \    /   |   \   /    |     \
               (n/4)^2   (n/4)^2 ...      ...         ...    (n/4)^2
                                         ...
                                         ...
              T(1) T(1) ... ... ... ... ... ... ... ... ... ... T(1) T(1)

And for each "level" I have the running times:

                                Level         Time
                                  0            n^2
                                  1            3(n/2)^2
                                  2            9(n/4)^2
                                 log_2{n}      ???

So I'm having trouble getting the time for the base case at level $log_2{n}$, and after that I'm unsure of what I'm supposed to do with all the different times to get the overall running time. Do I add them all up? Do I just take the running time of the $log_2{n}$ level?

Thank you for any help you can give.

UnworthyToast
  • 259
  • The following MSE link may be useful reading and can easily be adapted to your question, keeping in mind that the relation between work term and recursive term is different, but the method goes through the same. – Marko Riedel Oct 22 '14 at 19:17

1 Answers1

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At each level, the number of nodes triples and $n$ is cut in half. Extending the pattern in your table, we see that at level $k$, the time is $3^k(\tfrac{n}{2^k})^2 = (\tfrac{3}{4})^kn^2$. At the second-to-last level, we have $k = \log_2 n - 1$, which corresponds to the term $(\tfrac{3}{4})^{\log_2 n - 1}n^2$. At the last level when $k = \log_2 n \iff n = 2^k$, the base case kicks in. Assuming that $T(1) = c$ for some constant $c$, this level has running time $3^kT(1) = 3^{\log_2 n}c = cn^{\log_2 3}$.

Adding up all the levels, we observe that (other than the last level) we have a finite geometric series with initial term $a = n^2$, common ratio $r = \frac{3}{4}$, and a total of $N = \log_2 n$ terms. By applying the usual formula, we obtain: \begin{align*} T(n) &= [n^2 + (\tfrac{3}{4})n^2 + (\tfrac{3}{4})^2n^2 + \cdots + (\tfrac{3}{4})^{\log_2 n - 1}n^2] + cn^{\log_2 3} \\ &= \frac{n^2(1 - (\frac{3}{4})^{\log_2 n})}{1 - \frac{3}{4}} + cn^{\log_2 3} \\ &= 4n^2\left(1 - \frac{3^{\log_2 n}}{4^{\log_2 n}}\right) + cn^{\log_2 3} \\ &= 4n^2\left(1 - \frac{n^{\log_2 3}}{n^{\log_2 4}}\right) + cn^{\log_2 3} \\ &= \left(4n^2 - 4n^2 \cdot \frac{n^{\log_2 3}}{n^2}\right) + cn^{\log_2 3} \\ &= 4n^2 + (c - 4)n^{\log_2 3} \\ \end{align*} Notice that since $2 = \log_2 4 > \log_2 3$, it follows that $T(n) = \Theta(n^2)$.

Adriano
  • 41,576