Time Complexity in $\theta$ Notation

Question

$$T(n) = 2T\left(\frac{n}{2}\right) + T\left(\frac{n}{4}\right) + 5$$

What is the time complexity of the given algorithm in $\theta$ notation.

Thanks in advance.

Answer 1

The shift $S(n)=T(n)+\frac52$ yields $$S(n)=\color{red}{\bf 2}\cdot S(n/\color{green}{\bf 2})+S(n/\color{green}{\bf 4}), $$ then a general result is that $$S(n)=\Theta(n^a), $$ where $a$ solves the equation $$1=\color{red}{\bf 2}\cdot\color{green}{\bf 2}^{-a}+\color{green}{\bf 4}^{-a},$$ that is, $$a=\log_2(\sqrt2+1)\approx1.2715533. $$ To prove this, proceed by induction, that is, assume that the property $$c\cdot n^a\leqslant S(n)\leqslant C\cdot n^a$$ holds for every $n\leqslant N-1$, for some suitable positive $c$ and $C$, and show that it holds for $n=N$ as well.

Answer 2

If $T(n)=2T\left(\frac n2\right)+T\left(\frac n4\right)+5$ then one expansion results in

$$T(n)=4T\left(\frac n4\right)+2T\left(\frac n8\right)+10+2T\left(\frac n8\right)+T\left(\frac n{16}\right)+5+5\\=4T\left(\frac n4\right)+4T\left(\frac n8\right)+T\left(\frac n{16}\right)+20$$

A second expansion brings

$$T(n)=8T\left(\frac n8\right)+4T\left(\frac n{16}\right)+20+8T\left(\frac n{16}\right)+4T\left(\frac n{32}\right)+20\\+2T\left(\frac n{32}\right)+T\left(\frac n{64}\right)+5+20\\ =8T\left(\frac n8\right)+12T\left(\frac n{16}\right)+6T\left(\frac n{32}\right)+T\left(\frac n{64}\right)+65$$

So we have an increasing number of terms per expansion, and apparent-quadratic behavior in the constant term $(5\cdot 1, 5\cdot 4, 5\cdot 13,\dots)$. One more expansion should help clarify these behaviors:

$$T(n)=16T\left(\frac n{16}\right)+8T\left(\frac n{32}\right)+40+24T\left(\frac n{32}\right)+12T\left(\frac n{64}\right)+60\\+12T\left(\frac n{64}\right)+6T\left(\frac n{128}\right)+30+2T\left(\frac n{128}\right)+T\left(\frac n{256}\right)+5+65\\= 16T\left(\frac n{16}\right)+32T\left(\frac n{32}\right)+24T\left(\frac n{64}\right)+8T\left(\frac n{128}\right)+T\left(\frac n{256}\right)+200$$

If we follow the constant term further, we now have $5\cdot 1, 5\cdot 4, 5\cdot 13, 5\cdot 40.$ The pattern of these constants is emerging, and appears to be $O(n^4)$. Continuing with the analysis in this direction should provide a satisfactory conclusion.

Answer 3

First of all note that this recurrence only depends on the number of bits of $n$ and not on their values, which simplifies things considerably.

Suppose we wish to solve $$T(n) = 2T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + 5$$ where we set $T(0)=0.$

We unroll the recursion to obtain an exact formula for all $n:$ $$T(n) = 5 \times \sum_{j=0}^{\lfloor \log_2 n \rfloor} [z^j] \frac{1}{1-2z-z^2}.$$

Now the roots of $1-2z-z^2 = 2-(z+1)^2$ are at $$-1 \pm \sqrt{2}$$ and their inverses are $$\rho_{1,2} = 1 \pm \sqrt{2}.$$ It follows that $$[z^j] \frac{1}{1-2z-z^2} = c_1 (1+\sqrt{2})^j + c_2 (1-\sqrt{2})^j.$$ Once we determine the coefficients we get $$[z^j] \frac{1}{1-2z-z^2} = \frac{2+\sqrt{2}}{4} (1+\sqrt{2})^j+\frac{2-\sqrt{2}}{4} (1-\sqrt{2})^j.$$ The conclusion is that the closed and exact formula for $T(n)$ is $$5\times \frac{2+\sqrt{2}}{\sqrt{2}\times 4} ((1+\sqrt{2})^{\lfloor \log_2 n \rfloor+1}-1)- 5\times \frac{2-\sqrt{2}}{\sqrt{2}\times 4} ((1-\sqrt{2})^{\lfloor \log_2 n \rfloor+1}-1).$$

If desired we may re-write this as $$-\frac{5}{2}+ 5\times \frac{2+\sqrt{2}}{\sqrt{2}\times 4} (1+\sqrt{2})^{\lfloor \log_2 n \rfloor+1}- 5\times \frac{2-\sqrt{2}}{\sqrt{2}\times 4} (1-\sqrt{2})^{\lfloor \log_2 n \rfloor+1}.$$

It follows that the complexity from the dominant asymptotic term is $$\Theta\left((1+\sqrt{2})^{\lfloor \log_2 n \rfloor}\right) = \Theta\left(2^{\log_2(1+\sqrt{2})\times \log_2 n}\right) = \Theta\left(n^{\log_2(1+\sqrt{2})}\right).$$

The exponent here on $n$ is $$ 1.271553303,$$confirming the analysis by @Did.

A similar Master Theorem calculation can be found at this MSE link.