How to analyze the time complexity $\Theta$ of this recurrence

Question

I am trying to understand how to show that

$$T(n) = T(n/2) + T(n/4) + n^2$$

is $\Theta(n^2)$ by using a recursion tree. I tried substitution at first but it got real messy real fast.

This is self-study. Problem is from this pdf, problem 1-12.

Answer 1

Suppose we start by solving the following recurrence: $$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + n^2$$ where $T(1) = 1$ and $T(0) = 0.$

Now let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$ be the binary representation of $n.$ We unroll the recursion to obtain an exact formula for $n\ge 1$ $$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor} [z^j] \frac{1}{1-z-z^2} \left(\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}\right)^2.$$

We recognize the generating function of the Fibonacci numbers, so the formula becomes $$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} \left( \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}\right)^2.$$

We now compute lower and upper bounds which are actually attained and cannot be improved upon. For the lower bound consider a one digit followed by a string of zeroes, to give

$$T(n) \ge \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} 4^{\lfloor \log_2 n \rfloor-j} = 4^{\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} 4^{-j}.$$

Now since $$\left|\frac{1+\sqrt{5}}{2}\right|<4$$ the sum term converges to a number, we have $$1 \le \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 4^{-j} \lt \sum_{j=0}^{\infty} F_{j+1} 4^{-j} = \frac{16}{11}$$

and we thus get for the asymptotics $$\frac{16}{11} 4^{\lfloor \log_2 n \rfloor}.$$

For an upper bound consider a string of one digits to get

$$T(n) \le \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} \left( \sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^{k-j}\right)^2 = \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} (2^{\lfloor \log_2 n \rfloor+1-j} - 1)^2 \\ = 4\times 4^{\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} 4^{-j} - 4\times 2^{\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} 2^{-j} + \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} \\ = 4\times 4^{\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} 4^{-j} - 4\times 2^{\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1} 2^{-j} + F_{\lfloor \log_2 n \rfloor + 3} - 1.$$

The same constant appears as in the lower bound as well as an additional constant. Computing these we get $$\frac{64}{11} \times 4^{\lfloor \log_2 n \rfloor} - 16\times 2^{\lfloor \log_2 n \rfloor} + F_{\lfloor \log_2 n \rfloor + 3} - 1.$$

Doing the asymptotics of these three terms we obtain a term in $n^2$, a term in $n$ and a term in $n^{\log_2 \varphi} \in o(n)$ where $\varphi$ is the golden ratio.

Joining the upper and the lower bound we get for the asymptotics of this recurrence that it is

$$T(n)\in\Theta\left(4^{\lfloor \log_2 n \rfloor}\right) = \Theta\left(2^{2\log_2 n}\right) = \Theta(n^2),$$ which, let it be said, could also have been obtained by inspection.

Remark. The evaluation of the constant is done by noting that the generating function of $$F_{j+1} 4^{-j}\quad \text{is}\quad\frac{1}{1-z/4-z^2/16}$$ which at $z=1$ evaluates to $\frac{1}{1-1/4-1/16} = \frac{16}{11}.$

Similarly we have the generating function of $$F_{j+1} 2^{-j}\quad \text{is}\quad\frac{1}{1-z/2-z^2/4}$$ which at $z=1$ evaluates to $\frac{1}{1-1/2-1/4} = 4.$

We have a certain flexibility as to what power of four to use in the constant but this does not affect the asymptotics.

A closely related recurrence that is somewhat simpler may be found at this MSE link.

Answer 2

First consider the homogeneous equation

$$T(n)=T(\frac n2)+T(\frac n4).$$

Letting $n=2^k$,

$$T(2^k)=T(2^{k-1})+T(2^{k-2)}$$ or $$S(k)=S(k-1)+S(k-2).$$

You should recognize the Fibonacci recurrence, with the asymptotic solution

$$T(n)=S(k)=O(\phi^k)=O(e^{log(\phi)log(n)/log(2)})=O(n^{0.6942\cdots})$$

Then you try a particular solution of the non-homogenous equation, such as $T(n)=an^2$. Plugging in the given equation,

$$an^2=\frac{an^2}4+\frac{an^2}{16}+n^2$$ and

$$T(n)=\frac{16}{11}n^2$$ works.

As this solution dominates the other, you can conclude $\Theta(n^2)$.